SC-32 Scientific Companion Problem & Casebook Workbook

COMPANION CASEBOOK // THINKING IN THE STACK (SC-32 SCIENTIFIC) ● 15 PROBLEM SETS • 339 WORKED ENGINEERING & COMPETITION PROBLEMS • RPNCORE VERIFIED

Companion Engineering, Scientific & Competition Problem Volume to Thinking in the Stack: The Art of Solving Equations from the Inside Out (SC-32)**

This companion casebook houses the 339 extended worked engineering, physics, numerical analysis, computer science, corporate finance, and high-speed UIL competition problems that accompany the Bookstore Edition of Thinking in the Stack. Every problem includes its governing mathematical formulation, inside-out RPN keystroke sequence, and RPNCore-verified 4-register $(T, Z, Y, X)$ stack trace.

Chapter Quick-Jump & Engineering Casebook Index

Section / Chapter	Domain & Companion Problem Set	Primary SC-32 Engine / Mode	Worked Problems	Reference Manual
Executive Warm-Up	Cross-Disciplinary 4-Register $(T, Z, Y, X)$ Stack Table Exercises	`XYZT` Stack & Inside-Out RPN	25	Manual Preface →
Chapter 01	Hardware Architecture & Instrument Craftsmanship	`RP2350` / Power & Timing	8	Manual Ch. 01 →
Chapter 02	Principles of RPN & 4-Stack XYZT Dynamics	`ENTER`, `x↔y`, `R↓`, `LASTX`	15	Manual Ch. 02 →
Chapter 03	Algebra, Transcendentals & Display Formatting	Logs, Exponentials, Hyperbolics	22	Manual Ch. 03 →
Chapter 04	Trigonometry, Coordinate Systems & 2D Vectors	`→P`, `→R`, `FRAC-01` & `PLACE-01`	29	Manual Ch. 04 →
Chapter 05	Complex Numbers, AC Phasors & Euler's Identity	`[CMPLX]` Dual-Stack Engine	21	Manual Ch. 05 →
Chapter 06	Analytic & Solid Geometry, Surface Area & Mensuration	2D/3D Mensuration & Geodesics	15	Manual Ch. 06 →
Chapter 07	Polynomials, Brent-Dekker Solver & Root Finding	`QUAD`, `CUBIC`, `SOLVE`	26	Manual Ch. 07 →
Chapter 08	Numerical Calculus & Adaptive Romberg Quadrature	`d/dx` & `∫f(x)` Romberg Engine	27	Manual Ch. 08 →
Chapter 09	Bivariate Statistics, Curve Fitting & Linear Regression	`Σ+`, `L.R.`, Normal & Combinatorics	26	Manual Ch. 09 →
Chapter 10	Computer Science, Bases & 36-Bit Integers	`HEX`, `BIN`, `OCT`, Bitwise Ops	24	Manual Ch. 10 →
Chapter 11	Financial Mathematics, Executive Analytics & TVM	`N`, `I/YR`, `PV`, `PMT`, `FV`	31	Manual Ch. 11 →
Chapter 12	Function & Scatter Plotting Engine	`PLOT` & Graphical Root Analysis	27	Manual Ch. 12 →
Chapter 13	Applied STEM Engineering Case Studies	Multi-Stage Aerospace, Civil, EE	28	Manual Ch. 13 →
Chapter 14	High-Speed Equation Entry & UIL Contest Playbook	Inside-Out Speed Calculation	40	Manual Ch. 14 →

Executive Warm-Up Problems (25 Cross-Disciplinary RPN Table Exercises)

These 25 worked scientific and engineering warm-up problems complement the 5 signature showcase problems in the Preface of Thinking in the Stack, providing complete 4-register ($T, Z, Y, X$) stack trace tables across mechanics, thermodynamics, optics, acoustics, and electrochemistry.

Warm-Up Problem 0.1: Hypotenuse and Vector Magnitude

When a bridge pin joint experiences simultaneous horizontal shear $F_x = 24.5\text{ kN}$ and vertical bearing load $F_y = 18.2\text{ kN}$, structural sizing requires the resultant force magnitude $F_R$. Rather than opening an outer square-root bracket and nesting squared terms inside parentheses as on an infix calculator, we work from the inside out: squaring $F_x$ in place, keying $F_y$ (which automatically lifts $F_x^2$ into Register $Y$) and squaring it, adding the two squares in a single stroke, and extracting the square root:

\[F_R = \sqrt{F_x^2 + F_y^2} = \sqrt{24.5^2 + 18.2^2} \quad \text{[kN]}\]

Keystroke Sequence: 24.5 [↰] [x²] 18.2 [↰] [x²] [+] [√x] $\to \mathbf{30.520321099\text{ kN}}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`24.5`	$0.0000$	$0.0000$	$0.0000$	$24.5000$	Load $F_x$
`[↰] [x²]`	$0.0000$	$0.0000$	$0.0000$	$600.2500$	Square $F_x$ in $X$
`18.2` `[↰] [x²]`	$0.0000$	$0.0000$	$600.2500$	$331.2400$	Lift $600.25$ to $Y$ and square $F_y$
`[+]`	$0.0000$	$0.0000$	$0.0000$	$931.4900$	Sum of squares: $600.25 + 331.24$
`[√x]`	$0.0000$	$0.0000$	$0.0000$	$30.5203$	Resultant force $F_R \approx 30.5203\text{ kN}$

Warm-Up Problem 0.2: Kinetic Energy Calculation

When sizing automotive braking rotors, engineers must determine the translational kinetic energy stored in a vehicle of mass $m = 1450\text{ kg}$ cruising at highway velocity $v = 28\text{ m/s}$ ($100.8\text{ km/h}$). Pushing mass $m$ into Register $Y$ first allows you to square velocity $v$ in place in Register $X$ via [↰] [x²], multiply by mass, and halve the product in a linear left-to-right flow:

\[E_k = \frac{1}{2} m v^2 \quad (m = 1450\text{ kg},\; v = 28\text{ m/s})\]

Keystrokes: 1450 [ENTER] 28 [↰] [x²] [×] 2 [÷] $\to \mathbf{568400\text{ J}} = 568.4\text{ kJ}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`1450` `[ENTER]`	$0.0000$	$0.0000$	$1450.0000$	$1450.0000$	Push vehicle mass $m$
`28` `[↰] [x²]`	$0.0000$	$0.0000$	$1450.0000$	$784.0000$	Velocity squared $v^2$ in $X$
`[×]`	$0.0000$	$0.0000$	$0.0000$	$1136800.00$	Product $m v^2$
`2` `[÷]`	$0.0000$	$0.0000$	$0.0000$	$568400.00$	Divide by 2: $568.4\text{ kJ}$

Warm-Up Problem 0.3: Newton's Law of Universal Gravitation

Astrodynamics calculations span 35 orders of magnitude, where a single misplaced parenthesis in scientific notation corrupts the exponent by $10^{16}$. To compute the gravitational attraction between Earth ($M_E = 5.972 \times 10^{24}\text{ kg}$) and the Moon ($M_M = 7.342 \times 10^{22}\text{ kg}$) at mean orbital separation $r = 3.844 \times 10^8\text{ m}$, we use the dedicated exponent key [E] to chain the numerator $G M_E M_M$ in Register $Y$, square $r$ in place in Register $X$, and divide:

\[F = G \frac{M_E M_M}{r^2} \quad (G = 6.6743 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2)\]

Keystrokes: 6.6743 [E] 11 [+/-] [ENTER] 5.972 [E] 24 [×] 7.342 [E] 22 [×] 3.844 [E] 8 [↰] [x²] [÷] $\to \mathbf{1.980492E20\text{ N}}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`6.6743 [E] 11 [+/-]`	$0.0000$	$0.0000$	$0.0000$	$6.6743 \times 10^{-11}$	Universal gravitational constant $G$
`[ENTER]`	$0.0000$	$0.0000$	$6.6743 \times 10^{-11}$	$6.6743 \times 10^{-11}$	Stack lift: pushes $G$ into $Y$
`5.972 [E] 24`	$0.0000$	$0.0000$	$6.6743 \times 10^{-11}$	$5.9720 \times 10^{24}$	Earth mass $M_E$ in $X$
`[×]`	$0.0000$	$0.0000$	$0.0000$	$3.9859 \times 10^{14}$	Gravitational parameter $\mu_E = G M_E$
`7.342 [E] 22`	$0.0000$	$0.0000$	$3.9859 \times 10^{14}$	$7.3420 \times 10^{22}$	Moon mass $M_M$ in $X$
`[×]`	$0.0000$	$0.0000$	$0.0000$	$2.9264 \times 10^{37}$	Force numerator $G M_E M_M$
`3.844 [E] 8`	$0.0000$	$0.0000$	$2.9264 \times 10^{37}$	$3.8440 \times 10^{8}$	Lunar orbital distance $r$
`[↰] [x²]`	$0.0000$	$0.0000$	$2.9264 \times 10^{37}$	$1.4776 \times 10^{17}$	Square orbital radius $r^2$ in $X$
`[÷]`	$0.0000$	$0.0000$	$0.0000$	$1.9805 \times 10^{20}$	Mutual orbital attraction force in Newtons

Warm-Up Problem 0.4: Ideal Gas Law Pressure

A high-pressure welding cylinder with internal volume $V = 0.045\text{ m}^3$ contains $n = 3.5\text{ moles}$ of inert Argon gas heated to $T = 315\text{ K}$. Using the Ideal Gas Law, we chain the thermal energy product $n R T$ sequentially in Register $X$ before dividing by cylinder volume $V$:

\[P = \frac{n R T}{V} \quad (R = 8.31446\text{ J/mol}\cdot\text{K})\]

Keystrokes: 3.5 [ENTER] 8.31446 [×] 315 [×] 0.045 [÷] $\to \mathbf{203704.27\text{ Pa}} \approx 203.70\text{ kPa}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`3.5` `[ENTER]`	$0.0000$	$0.0000$	$3.5000$	$3.5000$	Molar substance quantity $n$ in $Y$
`8.31446` `[×]`	$0.0000$	$0.0000$	$0.0000$	$29.1006$	Molar gas scale factor $n R$
`315` `[×]`	$0.0000$	$0.0000$	$0.0000$	$9166.6922$	Thermal kinetic product $n R T$
`0.045` `[÷]`	$0.0000$	$0.0000$	$0.0000$	$203704.27$	Static pressure $P$ in Pascals ($203.70\text{ kPa}$)

Warm-Up Problem 0.5: Quadratic Discriminant & Root Characterization

Before extracting the roots of the quadratic equation $3.2 x^2 + 8.5 x - 4.1 = 0$, evaluating the discriminant $\Delta = b^2 - 4ac$ immediately reveals whether the roots are real or complex conjugates. By squaring $b = 8.5$ first, the RPN stack holds $b^2 = 72.25$ safely in Register $Z$ and then $Y$ while we form the signed product $4(3.2)(-4.1) = -52.48$ in Register $X$:

\[\Delta = b^2 - 4ac = (8.5)^2 - 4(3.2)(-4.1)\]

Keystrokes: 8.5 [↰] [x²] 4 [ENTER] 3.2 [×] 4.1 [+/-] [×] [-] $\rightarrow$ 124.73

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`8.5` `[↰] [x²]`	$0.0000$	$0.0000$	$0.0000$	$72.2500$	$b^2 = 8.5^2 = 72.25$
`4` `[ENTER]`	$0.0000$	$72.2500$	$4.0000$	$4.0000$	Load coefficient factor 4
`3.2` `[×]`	$0.0000$	$0.0000$	$72.2500$	$12.8000$	Form $4a = 4 \times 3.2 = 12.8$
`4.1` `[+/-]`	$0.0000$	$72.2500$	$12.8000$	$-4.1000$	Constant $c = -4.1$ loaded to $X$
`[×]`	$0.0000$	$0.0000$	$72.2500$	$-52.4800$	Form $4ac = 12.8 \times (-4.1)$
`[-]`	$0.0000$	$0.0000$	$0.0000$	$124.7300$	$\Delta = 72.25 - (-52.48) = 124.73$

Because 8.5 [↰] [x²] ($\mathbf{72.25}$) sits in Register $Y$ while 4 [ENTER] 3.2 [×] ($\mathbf{12.8}$) is multiplied by $c = -4.1$ in Register $X$, a single subtraction confirms $\Delta = 124.73 > 0$ (two distinct real roots).

Warm-Up Problem 0.6: Compound Interest & Exponential Growth

An endowment principal of $P = \$15,000$ is invested at an annual nominal rate $r = 6.5\%$ compounded monthly ($n = 12$) over $t = 8\text{ years}$ ($96\text{ months}$). Instead of keying Principal first and opening nested parentheses, we evaluate the innermost periodic rate $r/12$ first, add $1$, raise the base in place to the $96\text{th}$ power via [yˣ], and scale by $15,000$:

\[A = P \left(1 + \frac{r}{n}\right)^{nt} = 15000 \left(1 + \frac{0.065}{12}\right)^{96}\]

Keystrokes: 0.065 [ENTER] 12 [÷] 1 [+] 96 [yˣ] 15000 [×] $\rightarrow$ $25195.03

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`0.065` `[ENTER]` `12` `[÷]`	$0.0000$	$0.0000$	$0.0000$	$0.005417$	Monthly interest rate $r/12$
`1` `[+]`	$0.0000$	$0.0000$	$0.0000$	$1.005417$	Monthly compounding factor $1 + r/12$
`96` `[yˣ]`	$0.0000$	$0.0000$	$0.0000$	$1.679669$	8-year multiplier $(1 + r/12)^{96}$
`15000` `[×]`	$0.0000$	$0.0000$	$0.0000$	$25195.03$	Compound principal balance $A$ in dollars

Warm-Up Problem 0.7: Acoustics Decibel Sound Pressure Level

An acoustic calibration microphone measures an RMS sound pressure of $p = 0.45\text{ Pa}$ inside a concert hall. To convert this physical pressure into decibels relative to the human auditory threshold $p_0 = 20\,\mu\text{Pa} = 2.0 \times 10^{-5}\text{ Pa}$, we form the dimensionless ratio $p / p_0$ first, take the base-10 logarithm in place via [↰] [LOG], and multiply by $20$:

\[L_p = 20 \log_{10} \left(\frac{p}{p_0}\right)\]

Keystrokes: 0.45 [ENTER] 2 [E] 5 [+/-] [÷] [↰] [LOG] 20 [×]$\to \mathbf{87.043650362\text{ dB SPL}}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`0.45` `[ENTER]`	$0.0000$	$0.0000$	$0.4500$	$0.4500$	Acoustic pressure amplitude $p$ in $Y$
`2 [E] 5 [+/-]`	$0.0000$	$0.0000$	$0.4500$	$2.0000 \times 10^{-5}$	Auditory reference threshold $p_0$
`[÷]`	$0.0000$	$0.0000$	$0.0000$	$22500.00$	Dimensionless pressure ratio $p/p_0$
`[↰] [LOG]`	$0.0000$	$0.0000$	$0.0000$	$4.3522$	Base-10 logarithm in bels
`20` `[×]`	$0.0000$	$0.0000$	$0.0000$	$87.0437$	Sound pressure level in dB SPL

Warm-Up Problem 0.8: 3D Spatial Vector Magnitude

During an orbital rendezvous burn, an inertial measurement unit reports spacecraft velocity components $(v_x = 1240,\; v_y = -850,\; v_z = 420\text{ m/s})$. Because squaring any real component yields a positive result ($|-850|^2 = 850^2$), we can omit the negative sign on $v_y$, accumulate all three squares directly in Register $X$, and extract the Euclidean norm:

\[\|\mathbf{v}\| = \sqrt{v_x^2 + v_y^2 + v_z^2}\]

Keystrokes: 1240 [↰] [x²] 850 [↰] [x²] [+] 420 [↰] [x²] [+] [√x]$\to \mathbf{1560.9292\text{ m/s}}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`1240` `[↰] [x²]`	$0.0000$	$0.0000$	$0.0000$	$1537600.00$	Square $v_x$ component in $X$
`850` `[↰] [x²]`	$0.0000$	$0.0000$	$1537600.00$	$722500.00$	Square $v_y$ component in $X$
`[+]`	$0.0000$	$0.0000$	$0.0000$	$2260100.00$	Sum of squares $v_x^2 + v_y^2$
`420` `[↰] [x²]`	$0.0000$	$0.0000$	$2260100.00$	$176400.00$	Square $v_z$ component in $X$
`[+]`	$0.0000$	$0.0000$	$0.0000$	$2436500.00$	Total radicand $v_x^2 + v_y^2 + v_z^2$
`[√x]`	$0.0000$	$0.0000$	$0.0000$	$1560.9292$	Spatial speed $\lVert \mathbf{v}\rVert$ in m/s

Warm-Up Problem 0.9: Centripetal Acceleration

A human-rated aerospace centrifuge arm of radius $r = 1.85\text{ m}$ rotates at a tangential rim speed of $v = 42.5\text{ m/s}$. Because [↰] [x²] is a monadic function that acts on Register $X$ and immediately enables stack lift, keying 42.5 [↰] [x²] followed directly by 1.85 [÷] computes $v^2/r$ without ever pressing [ENTER]:

\[a_c = \frac{v^2}{r}\]

Keystrokes: 42.5 [↰] [x²] 1.85 [÷]$\to \mathbf{976.3514\text{ m/s}^2} \approx 99.56\,g$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`42.5`	$0.0000$	$0.0000$	$0.0000$	$42.5000$	Peripheral velocity $v$ in $X$
`[↰] [x²]`	$0.0000$	$0.0000$	$0.0000$	$1806.2500$	Kinetic numerator $v^2$
`1.85`	$0.0000$	$0.0000$	$1806.2500$	$1.8500$	Centrifuge radius $r$ in $X$
`[÷]`	$0.0000$	$0.0000$	$0.0000$	$976.3514$	Centripetal acceleration $a_c$ in m/s$^2$

Warm-Up Problem 0.10: Thermal Expansion of Steel Bridge Girder

To prevent thermal buckling across seasonal extremes, civil engineers size finger-joint expansion gaps for a $L_0 = 120\text{ m}$ structural steel bridge span experiencing a temperature swing from $-15^\circ\text{C}$ to $+38^\circ\text{C}$ ($\Delta T = 53\text{ K}$, linear expansion coefficient $\alpha = 12 \times 10^{-6}\text{ K}^{-1}$):

\[\Delta L = \alpha L_0 \Delta T\]

Keystrokes: 12 [E] 6 [+/-] [ENTER] 120 [×] 53 [×]$\to \mathbf{0.07632\text{ m}} = 76.32\text{ mm}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`12 [E] 6 [+/-]`	$0.0000$	$0.0000$	$0.0000$	$1.2000 \times 10^{-5}$	Steel thermal coefficient $\alpha$
`[ENTER]`	$0.0000$	$0.0000$	$1.2000 \times 10^{-5}$	$1.2000 \times 10^{-5}$	Push $\alpha$ into $Y$
`120` `[×]`	$0.0000$	$0.0000$	$0.0000$	$0.00144$	Specific expansion $\alpha L_0$
`53` `[×]`	$0.0000$	$0.0000$	$0.0000$	$0.07632$	Total elongation $\Delta L$ in meters ($76.32\text{ mm}$)

Warm-Up Problem 0.11: Acoustic Doppler Frequency Shift

A freight locomotive horn emitting $f_0 = 440\text{ Hz}$ (concert A4) approaches a grade crossing at $v_s = 28\text{ m/s}$ through air where the speed of sound is $v = 343\text{ m/s}$. Because $v = 343\text{ m/s}$ appears in both the numerator and denominator of the Doppler factor, pressing 343 [ENTER] 343 [ENTER] pre-loads $v$ into both $Z$ and $Y$ so the denominator $v - v_s$ divides directly into $v$ with zero re-entry:

\[f = f_0 \left(\frac{v}{v - v_s}\right)\]

Keystrokes: 343 [ENTER] 343 [ENTER] 28 [-] [÷] 440 [×]$\to \mathbf{479.1111\text{ Hz}}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`343` `[ENTER]`	$0.0000$	$0.0000$	$343.0000$	$343.0000$	Ambient acoustic velocity $c$ in $Y$
`343` `[ENTER]`	$0.0000$	$343.0000$	$343.0000$	$343.0000$	Pre-duplicate wave velocity in stack
`28` `[-]`	$0.0000$	$0.0000$	$343.0000$	$315.0000$	Effective denominator $c - v_s$ in $X$
`[÷]`	$0.0000$	$0.0000$	$0.0000$	$1.0889$	Doppler factor $c / (c - v_s)$
`440` `[×]`	$0.0000$	$0.0000$	$0.0000$	$479.1111$	Perceived frequency $f$ in Hz

Warm-Up Problem 0.12: Reynolds Number in Pipe Flow

Before computing pipe head loss with the Colebrook-White equation, hydraulic engineers must classify whether flow is laminar ($Re < 2300$) or turbulent ($Re > 4000$). For water at $20^\circ\text{C}$ ($\rho = 1000\text{ kg/m}^3$, dynamic viscosity $\mu = 0.001002\text{ Pa}\cdot\text{s}$) flowing at $v = 1.8\text{ m/s}$ through a $D = 0.05\text{ m}$ main, we chain inertial forces in the numerator and divide by viscosity:

\[Re = \frac{\rho v D}{\mu}\]

Keystrokes: 1000 [ENTER] 1.8 [×] 0.05 [×] 0.001002 [÷]$\to \mathbf{89820.36}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`1000` `[ENTER]`	$0.0000$	$0.0000$	$1000.0000$	$1000.0000$	Fluid density $\rho$ in $Y$
`1.8` `[×]`	$0.0000$	$0.0000$	$0.0000$	$1800.0000$	Mass flux density $\rho v$
`0.05` `[×]`	$0.0000$	$0.0000$	$0.0000$	$90.0000$	Inertial momentum numerator $\rho v D$
`0.001002` `[÷]`	$0.0000$	$0.0000$	$0.0000$	$89820.36$	Dimensionless Reynolds number $Re$

Warm-Up Problem 0.13: Electrical RC Circuit Discharge

A power-supply filter capacitor $C = 47\,\mu\text{F}$ discharges through a bleeder resistor $R = 150\text{ k}\Omega$ from an initial rail voltage $V_0 = 24.0\text{ V}$. To find the residual capacitor voltage at $t = 10.0\text{ s}$, we push $t = 10$ into Register $Z$, multiply $R \times C$ in the lower registers to form the time constant $\tau = 7.05\text{ s}$, divide to obtain $t/\tau$, negate and exponentiate via [+/-] [eˣ], and scale by $V_0$:

\[V(t) = V_0 e^{-t / (R C)} = 24.0 \, e^{-10.0 / 7.05}\]

Keystrokes: 10 [ENTER] 150000 [ENTER] 47 [E] 6 [+/-] [×] [÷] [+/-] [eˣ] 24 [×]$\to \mathbf{5.8101948891\text{ V}}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`10` `[ENTER]`	$0.0000$	$0.0000$	$10.0000$	$10.0000$	Elapsed discharge duration $t$
`150000` `[ENTER]`	$0.0000$	$10.0000$	$150000.00$	$150000.00$	Load resistance $R$ in $Y$
`47 [E] 6 [+/-]` `[×]`	$0.0000$	$0.0000$	$10.0000$	$7.0500$	Time constant $\tau = R C = 7.05\text{ s}$
`[÷]`	$0.0000$	$0.0000$	$0.0000$	$1.4184$	Normalized decay argument $t / \tau$
`[+/-]` `[eˣ]`	$0.0000$	$0.0000$	$0.0000$	$0.2421$	Attenuation factor $e^{-t/\tau}$
`24` `[×]`	$0.0000$	$0.0000$	$0.0000$	$5.8102$	Instantaneous potential $V(10)$ in Volts

Warm-Up Problem 0.14: Radioactive Carbon-14 Decay

Accelerator mass spectrometry shows an archaeological timber sample retains $N/N_0 = 34\%$ ($0.34$) of its initial $^{14}\text{C}$ activity (half-life $t_{1/2} = 5730\text{ yr}$). Taking the natural logarithm of $0.34$ first, multiplying by $5730$, and dividing by 2 [LN] resolves the specimen's age without storing $\ln 2$:

\[t = -\frac{t_{1/2}}{\ln 2} \ln \left(\frac{N}{N_0}\right) = -\frac{5730}{\ln 2} \ln(0.34)\]

Keystrokes: 0.34 [LN] 5730 [×] 2 [LN] [÷] [+/-]$\to \mathbf{8918.42\text{ years}}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`0.34` `[LN]`	$0.0000$	$0.0000$	$0.0000$	$-1.0788$	Logarithmic fraction $\ln(0.34)$
`5730` `[×]`	$0.0000$	$0.0000$	$0.0000$	$-6181.5794$	Half-life numerator product
`2` `[LN]`	$0.0000$	$0.0000$	$-6181.5794$	$0.6931$	Natural logarithm $\ln(2)$ in $X$
`[÷]`	$0.0000$	$0.0000$	$0.0000$	$-8918.1339$	Raw negative decay age
`[+/-]`	$0.0000$	$0.0000$	$0.0000$	$8918.1339$	Archaeological artifact age in years

Warm-Up Problem 0.15: Beam Elastic Deflection Under Center Load

A simply supported structural steel I-beam of span $L = 6.0\text{ m}$ ($E = 200\text{ GPa} = 2.0 \times 10^{11}\text{ Pa}$, second moment of area $I = 8.5 \times 10^{-5}\text{ m}^4$) carries a central point load $F = 45\text{ kN} = 45,000\text{ N}$. Building the numerator $F L^3$ in Register $Y$ while chaining the flexural rigidity denominator $48 E I$ in Register $X$ prevents any division-chain ordering mistake:

\[\delta_{\max} = \frac{F L^3}{48 E I}\]

Keystrokes: 45000 [ENTER] 6 [ENTER] 3 [yˣ] [×] 48 [ENTER] 2 [E] 11 [×] 8.5 [E] 5 [+/-] [×] [÷] $\to \mathbf{0.0119117647\text{ m}} = 11.91\text{ mm}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`45000` `[ENTER]`	$0.0000$	$0.0000$	$45000.00$	$45000.00$	Load point force $F$ in $Y$
`6` `[ENTER]` `3` `[yˣ]`	$0.0000$	$0.00$	$45000.0000$	$216.0000$	Span cubed $L^3 = 6^3 = 216\text{ m}^3$
`[×]`	$0.0000$	$0.0000$	$0.0000$	$9720000.00$	Bending numerator $F L^3$ in $X$
`48` `[ENTER]`	$0.0000$	$9720000.00$	$48.0000$	$48.0000$	Support factor 48 in $Y$
`2 [E] 11` `[×]`	$0.0000$	$0.0000$	$9720000.00$	$9.6000 \times 10^{12}$	Modulus product $48 E$
`8.5 [E] 5 [+/-]` `[×]`	$0.0000$	$0.0000$	$9720000.00$	$8.1600 \times 10^{8}$	Flexural rigidity $48 E I$
`[÷]`	$0.0000$	$0.0000$	$0.0000$	$0.011912$	Maximum central deflection $\delta_{\max}$ in meters

Warm-Up Problem 0.16: Torricelli Tank Draining Efflux Velocity

When a storage tank valve opens at hydrostatic depth $h = 4.25\text{ m}$ below the free liquid surface ($g = 9.80665\text{ m/s}^2$), gravitational potential energy converts into kinetic efflux velocity. Multiplying $2 \times g \times h$ inside the stack and pressing [√x] evaluates Torricelli's law in four rapid operations:

\[v = \sqrt{2 g h}\]

Keystrokes: 2 [ENTER] 9.80665 [×] 4.25 [×] [√x]$\to \mathbf{9.1300\text{ m/s}}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`2` `[ENTER]`	$0.0000$	$0.0000$	$2.0000$	$2.0000$	Kinetic factor 2 in $Y$
`9.80665` `[×]`	$0.0000$	$0.0000$	$0.0000$	$19.6133$	Gravitational acceleration $2g$
`4.25` `[×]`	$0.0000$	$0.0000$	$0.0000$	$83.3565$	Potential head radicand $2gh$
`[√x]`	$0.0000$	$0.0000$	$0.0000$	$9.1300$	Orifice efflux velocity $v$ in m/s

Warm-Up Problem 0.17: Stefan-Boltzmann Thermal Radiation

A ceramic kiln radiator panel of emissivity $\varepsilon = 0.88$ and surface area $A = 2.4\text{ m}^2$ operates at absolute temperature $T = 650\text{ K}$ ($\sigma = 5.67037 \times 10^{-8}\text{ W/m}^2\text{K}^4$). Chaining $\varepsilon \sigma A$ in Register $Y$ while raising $T$ to the fourth power in Register $X$ yields the total radiant heat flux:

\[P = \varepsilon \sigma A T^4\]

Keystrokes: 0.88 [ENTER] 5.67037 [E] 8 [+/-] [×] 2.4 [×] 650 [ENTER] 4 [yˣ] [×]$\to \mathbf{21,377.58976\text{ W}} \approx 21.38\text{ kW}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`0.88` `[ENTER]`	$0.0000$	$0.0000$	$0.8800$	$0.8800$	Surface emissivity $\varepsilon$
`5.67037 [E] 8 [+/-]` `[×]`	$0.0000$	$0.0000$	$0.0000$	$4.9899 \times 10^{-8}$	Product $\varepsilon \sigma$
`2.4` `[×]`	$0.0000$	$0.0000$	$0.0000$	$1.1976 \times 10^{-7}$	Area radiation coefficient $\varepsilon \sigma A$
`650` `[ENTER]`	$0.0000$	$1.1976 \times 10^{-7}$	$650.0000$	$650.0000$	Temperature $T$ in $Y$
`4` `[yˣ]`	$0.0000$	$0.0000$	$1.1976 \times 10^{-7}$	$1.7851 \times 10^{11}$	Fourth power $T^4 = 650^4$
`[×]`	$0.0000$	$0.0000$	$0.0000$	$21377.59$	Radiated thermal power in Watts ($21.38\text{ kW}$)

Warm-Up Problem 0.18: Bernoulli Equation Pressure Differential

Standard sea-level air ($\rho = 1.225\text{ kg/m}^3$) accelerates over the upper camber of an aircraft wing from freestream speed $v_1 = 45\text{ m/s}$ to $v_2 = 72\text{ m/s}$. Squaring $v_2$ and $v_1$ in place and subtracting before scaling by $\frac{1}{2}\rho$ yields the aerodynamic suction pressure differential $\Delta P$:

\[\Delta P = \frac{1}{2} \rho \left(v_2^2 - v_1^2\right)\]

Keystrokes: 72 [↰] [x²] 45 [↰] [x²] [-] 1.225 [×] 2 [÷]$\to \mathbf{1934.87\text{ Pa}}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`72` `[↰] [x²]`	$0.0000$	$0.0000$	$0.0000$	$5184.0000$	Accelerated velocity squared $v_2^2$
`45` `[↰] [x²]`	$0.0000$	$0.0000$	$5184.0000$	$2025.0000$	Freestream velocity squared $v_1^2$
`[-]`	$0.0000$	$0.0000$	$0.0000$	$3159.0000$	Kinetic difference $v_2^2 - v_1^2$
`1.225` `[×]`	$0.0000$	$0.0000$	$0.0000$	$3869.7750$	Product $\rho (v_2^2 - v_1^2)$
`2` `[÷]`	$0.0000$	$0.0000$	$0.0000$	$1934.8875$	Aerodynamic lift pressure $\Delta P$ in Pa

Warm-Up Problem 0.19: Carnot Maximum Theoretical Efficiency

A binary geothermal turbine operates between a deep brine reservoir at $T_H = 185^\circ\text{C} = 458.15\text{ K}$ and an ambient cooling tower at $T_C = 22^\circ\text{C} = 295.15\text{ K}$. Evaluating $T_C / T_H$ first and using 1 [x⇄y] [-] swaps unity into Register $Y$ so the complement $1 - T_C/T_H$ completes without negating and adding:

\[\eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H}\]

Keystrokes: 295.15 [ENTER] 458.15 [÷] 1 [x⇄y] [-]$\to \mathbf{0.35578} = 35.58\%$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`295.15` `[ENTER]`	$0.0000$	$0.0000$	$295.1500$	$295.1500$	Condenser temperature $T_C$ in $Y$
`458.15` `[÷]`	$0.0000$	$0.0000$	$0.0000$	$0.6442$	Thermal rejection fraction $T_C / T_H$
`1`	$0.0000$	$0.0000$	$0.6442$	$1.0000$	Push thermodynamic unity
`[x⇄y]`	$0.0000$	$0.0000$	$1.0000$	$0.6442$	Flip stack so 1 occupies $Y$
`[-]`	$0.0000$	$0.0000$	$0.0000$	$0.3558$	Carnot efficiency $\eta_{\text{Carnot}} = 35.58\%$

Warm-Up Problem 0.20: Architectural Acoustics Total Absorption (Sabins)

When calculating the Sabine reverberation time of a university lecture hall, an acoustical engineer sums the absorption products $\alpha_i S_i$ across painted concrete walls ($\alpha_1 = 0.05,\; S_1 = 450\text{ m}^2$), acoustic ceiling clouds ($\alpha_2 = 0.65,\; S_2 = 120\text{ m}^2$), and upholstered seating ($\alpha_3 = 0.15,\; S_3 = 280\text{ m}^2$). The 4-level stack naturally accumulates each pairwise product into a running total in Register $Y$:

\[A = \sum_{i=1}^{3} \alpha_i S_i = (0.05 \times 450) + (0.65 \times 120) + (0.15 \times 280) \quad \text{[m}^2\text{ Sabins]}\]

Keystrokes: 0.05 [ENTER] 450 [×] 0.65 [ENTER] 120 [×] [+] 0.15 [ENTER] 280 [×] [+]$\to \mathbf{142.50\text{ Sabins}}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`0.05` `[ENTER]` `450` `[×]`	$0.0000$	$0.0000$	$0.0000$	$22.5000$	Concrete wall absorption $A_1$
`0.65` `[ENTER]` `120` `[×]`	$0.0000$	$0.0000$	$22.5000$	$78.0000$	Ceiling tile absorption $A_2$
`[+]`	$0.0000$	$0.0000$	$0.0000$	$100.5000$	Subtotal $A_1 + A_2$
`0.15` `[ENTER]` `280` `[×]`	$0.0000$	$0.0000$	$100.5000$	$42.0000$	Seating area absorption $A_3$
`[+]`	$0.0000$	$0.0000$	$0.0000$	$142.5000$	Total absorption $A$ in Sabins

Warm-Up Problem 0.21: Hydraulic Cylinder Actuator Force & Pressure

A hydraulic ram with piston bore diameter $d = 85.0\text{ mm} = 0.085\text{ m}$ is supplied with fluid at gauge pressure $P = 210\text{ bar} = 2.10 \times 10^7\text{ Pa}$. Calculate the thrust force in kilonewtons: $$A = \frac{\pi d^2}{4}, \quad F = P \cdot A = P \cdot \frac{\pi d^2}{4}$$

Keystrokes: 0.085 [↰] [x²] [↱] [π] [×] 4 [÷] 2.10 [E] 7 [×] 1000 [÷] $\rightarrow$ 119.16453634

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`0.085` `[↰] [x²]`	$0.0000$	$0.0000$	$0.0000$	$0.007225$	Piston diameter squared $d^2$
`[↱] [π]` `[×]`	$0.0000$	$0.0000$	$0.0000$	$0.022698$	Product $\pi d^2$
`4` `[÷]`	$0.0000$	$0.0000$	$0.0000$	$0.0056745$	Bore cross-sectional area $A\text{ [m}^2\text{]}$
`2.10` `[E]` `7` `[×]`	$0.0000$	$0.0000$	$0.0000$	$119164.54$	Thrust force in Newtons
`1000` `[÷]`	$0.0000$	$0.0000$	$0.0000$	$119.1645$	Force in kilonewtons ($119.16\text{ kN}$)

Warm-Up Problem 0.22: Wheatstone Bridge Resistance Null Balance

A precision bridge circuit with fixed arms $R_1 = 120.0\,\Omega$, $R_2 = 360.0\,\Omega$, and adjustable balance resistor $R_3 = 450.0\,\Omega$. Find the unknown sensor resistance $R_x$: $$\frac{R_1}{R_2} = \frac{R_3}{R_x} \implies R_x = \frac{R_2 \cdot R_3}{R_1}$$

Keystrokes: 360 [ENTER] 450 [×] 120 [÷]$\to \mathbf{1350.00\,\Omega} = 1.350\text{ k}\Omega$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`360` `[ENTER]`	$0.0000$	$0.0000$	$360.0000$	$360.0000$	Reference resistor $R_2$ loaded to $Y$
`450` `[×]`	$0.0000$	$0.0000$	$0.0000$	$162000.00$	Bridge cross-product $R_2 R_3$ in $X$
`120` `[÷]`	$0.0000$	$0.0000$	$0.0000$	$1350.0000$	Unknown sensor resistance $R_x$ in $\Omega$

Warm-Up Problem 0.23: LC Resonant Frequency and Inductive Reactance

An inductor $L = 22.0\,\mu\text{H} = 22.0 \times 10^{-6}\text{ H}$ and capacitor $C = 150.0\text{ pF} = 150.0 \times 10^{-12}\text{ F}$. Find resonant frequency $f_0$: $$f_0 = \frac{1}{2\pi \sqrt{L C}}$$

Keystrokes: 22 [E] 6 [+/-] [ENTER] 150 [E] 12 [+/-] [×] [√x] 2 [×] [↱] [π] [×] [1/x] $\to \mathbf{2,770,531.94\text{ Hz}} \approx 2.7705\text{ MHz}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`22 [E] 6 [+/-]` `[ENTER]`	$0.0000$	$0.0000$	$2.2000 \times 10^{-5}$	$2.2000 \times 10^{-5}$	Inductance $L$ loaded to $Y$
`150 [E] 12 [+/-]` `[×]`	$0.0000$	$0.0000$	$0.0000$	$3.3000 \times 10^{-15}$	Reactive product $L C$
`[√x]`	$0.0000$	$0.0000$	$0.0000$	$5.7446 \times 10^{-8}$	Time constant $\sqrt{LC}$ in seconds
`2` `[×]`	$0.0000$	$0.0000$	$0.0000$	$1.1489 \times 10^{-7}$	Scale by 2 ($2\sqrt{LC}$)
`[↱] [π]` `[×]`	$0.0000$	$0.0000$	$0.0000$	$3.6094 \times 10^{-7}$	Period divisor $2\pi\sqrt{LC}$
`[1/x]`	$0.0000$	$0.0000$	$0.0000$	$2770531.94$	Resonant frequency $f_0$ in Hz ($2.77\text{ MHz}$)

Warm-Up Problem 0.24: Adiabatic Gas Compression Work

Compression of diatomic air ($\gamma = 1.40$) from $V_1 = 0.020\text{ m}^3$ at $P_1 = 100\text{ kPa}$ to $V_2 = 0.005\text{ m}^3$:

$$W = \frac{P_1 V_1}{\gamma - 1} \left[ \left(\frac{V_1}{V_2}\right)^{\gamma - 1} - 1 \right]$$ Here $\gamma - 1 = 0.40$, $V_1 / V_2 = 0.020 / 0.005 = 4.0$.

Keystrokes: 4.0 [ENTER] 0.40 [yˣ] 1 [-] 100000 [ENTER] 0.020 [×] 0.40 [÷] [×] $\to \mathbf{3,705.505633\text{ J}} \approx 3.7055\text{ kJ}$.

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`4.0` `[ENTER]` `0.40` `[yˣ]`	$0.0000$	$0.0000$	$0.0000$	$1.741101$	Compression ratio power $(V_1/V_2)^{\gamma-1}$
`1` `[-]`	$0.0000$	$0.0000$	$0.0000$	$0.741101$	Pressure bracket term $(4.0^{0.40} - 1)$
`100000` `[ENTER]` `0.020` `[×]`	$0.0000$	$0.0000$	$0.741101$	$2000.0000$	Initial gas energy $P_1 V_1 = 2000\text{ J}$
`0.40` `[÷]`	$0.0000$	$0.0000$	$0.741101$	$5000.0000$	Baseline energy factor $P_1 V_1 / (\gamma - 1)$
`[×]`	$0.0000$	$0.0000$	$0.0000$	$3705.5056$	Net adiabatic compression work in Joules ($3.71\text{ kJ}$)

Warm-Up Problem 0.25: Escape Velocity from Planetary Body

Escape velocity from Mars ($M = 6.4171 \times 10^{23}\text{ kg}$, $R = 3.3895 \times 10^6\text{ m}$):

\[v_{\text{esc}} = \sqrt{\frac{2 G M}{R}} \quad (G = 6.6743 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2)\]

Keystrokes: 2 [ENTER] 6.6743 [E] 11 [+/-] [×] 6.4171 [E] 23 [×] 3.3895 [E] 6 [÷] [√x] $\to \mathbf{5,027.122221\text{ m/s}} \approx 5.027\text{ km/s}$.

Multiply kinetic scaling by gravitational parameter $2GM$, divide by planetary radius, and apply square root directly in place without parentheses (Munroe, 2014; HP, 1982).

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`2` `[ENTER]`	$0.0000$	$0.0000$	$2.0000$	$2.0000$	Kinetic factor 2 in $Y$
`6.6743 [E] 11 [+/-]` `[×]`	$0.0000$	$0.0000$	$0.0000$	$1.3349 \times 10^{-10}$	Scaled gravity constant $2G$
`6.4171 [E] 23` `[×]`	$0.0000$	$0.0000$	$0.0000$	$8.5659 \times 10^{13}$	Mars gravitational parameter $2GM$
`3.3895 [E] 6` `[÷]`	$0.0000$	$0.0000$	$0.0000$	$2.5272 \times 10^{7}$	Specific escape kinetic energy $2GM/R$
`[√x]`	$0.0000$	$0.0000$	$0.0000$	$5027.1222$	Mars escape velocity $v_{\text{esc}}$ in m/s ($5.03\text{ km/s}$)

Chapter 01: Hardware Architecture & Instrument Craftsmanship — Extended Engineering & Competition Problems

Accompanies Chapter 01 of Thinking in the Stack (8 Worked Engineering Problems)

Workbook Problem 1.1: Transflective LCD Photometric Contrast Ratio

Unlike emissive OLED screens that wash out under bright ambient room lighting and drain tens of milliamperes, the SC-32's STN liquid-crystal display uses a rear transflective polarizer that reflects ambient light back through the liquid-crystal twist cell. A spot photometer measures luminance from an unactivated reflective background pixel ($L_{\text{white}} = 142.0\text{ cd/m}^2$) and an absorptive black segment ($L_{\text{black}} = 9.35\text{ cd/m}^2$): $$C_R = \frac{L_{\text{white}}}{L_{\text{black}}}$$

Keystrokes: 142 [ENTER] 9.35 [÷] $\rightarrow$ 15.1872 $\approx 15.2:1$.

Dividing background reflectance ($L_{\text{white}} = 142.0\text{ cd/m}^2$ in $Y$) by dark-state luminance ($L_{\text{black}} = 9.35\text{ cd/m}^2$ in $X$) yields the photopic contrast ratio $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{15.1872})$—exceeding ISO 9241-303 Class 1 legibility requirements without drawing a single microampere of backlight current.

Workbook Problem 1.2: Low-Dropout Linear Regulator (LDO) Power Dissipation

During factory tethered diagnostics, an external bench supply feeds $V_{\text{in}} = 5.80\text{ V}$ into the CMOS low-dropout linear regulator supplying $V_{\text{out}} = 3.30\text{ V}$ at full dual-core active current $I = 18.2\text{ mA}$. Because a linear pass transistor regulates by dropping excess voltage as Joule heat, its thermal dissipation is governed by the voltage headroom product: $$P_{\text{diss}} = (V_{\text{in}} - V_{\text{out}}) \cdot I$$

Keystrokes: 5.80 [ENTER] 3.30 [-] 0.0182 [×] 1000 [×] $\rightarrow$ 45.5000$\text{ mW}$.

Evaluating the pass-element voltage drop first (5.80 [ENTER] 3.30 [-]) places $\Delta V = 2.50\text{ V}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 2.5000)$ with zero parentheses. Scaling by active load current and milliwatt normalization (0.0182 [×] 1000 [×]) yields package heat dissipation $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{45.5000}\text{ mW})$—confirming that in a SOT-23-5 package ($\theta_{JA} \approx 220^\circ\text{C/W}$), junction temperature rises by less than $10^\circ\text{C}$.

Workbook Problem 1.3: Flash Sector Wear Endurance Under Extreme Daily Editing

Non-volatile NOR flash memory cells degrade via Fowler-Nordheim tunneling oxide stress after $100,000$ program/erase cycles per sector. By alternating writes between two ping-pong sectors and committing state only on sleep transitions or explicit edits ($45\text{ writes/day}$), the SC-32's wear-leveling supervisor extends flash longevity to: $$t_{\text{years}} = \frac{2 \times 100,000}{45 \times 365.25}$$

Keystrokes: 200000 [ENTER] 45 [ENTER] 365.25 [×] [÷] $\rightarrow$ 12.1683$\text{ years}$.

Loading the dual-sector endurance budget ($200,000\text{ cycles}$) and computing annual write demand (200000 [ENTER] 45 [ENTER] 365.25 [×]) leaves $Y = 200,000$ poised above $X = 16,436.25\text{ writes/yr}$ $(T: 0.0000, Z: 0.0000, Y: 200000.0000, X: 16436.2500)$. A single [÷] stroke resolves the guaranteed sector lifetime $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{12.1682}\text{ years})$—and rotating the ring across 16 sectors multiplies endurance beyond 97 years.

Workbook Problem 1.4: Quiescent Sleep Power Consumption Over 18 Months

In deep retention sleep, the high-speed $150\text{ MHz}$ PLL powers down and the system idles at $I = 11.8\,\mu\text{A}$ on the $V = 3.0\text{ V}$ lithium rail. Over an 18-month warehouse or desk-drawer storage interval ($547.5\text{ days} = 13,140\text{ hours}$), total chemical energy drawn from the cell in Joules is: $$E = V \cdot I \cdot t$$

Keystrokes: 3.0 [ENTER] 11.8 [E] 6 [+/-] [×] 13140 [×] 3600 [×] $\rightarrow$ 1,674.5616$\text{ Joules}$.

Multiplying cell potential by quiescent sleep current (3.0 [ENTER] 11.8 [E] 6 [+/-] [×]) establishes the continuous standby power floor of $35.4\,\mu\text{W}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 3.5400 \times 10^{-5})$. Chaining storage hours and seconds (13140 [×] 3600 [×]) accumulates the 18-month energy budget $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{1674.5616}\text{ Joules})$—consuming less than $69\%$ of a single CR2032 cell's $2,430\text{ J}$ ($225\text{ mAh} \times 3.0\text{ V} \times 3600\text{ s/h}$) reserve.

Workbook Problem 1.5: Microcontroller Core Clock PLL Lock Phase Margin

When waking from deep sleep upon a keypress edge, Core 0's phase-locked loop (PLL) must lock onto the $12\text{ MHz}$ quartz reference and ramp to $150\text{ MHz}$ before user-perceptible latency ($> 10\text{ ms}$) occurs. With a critically damped second-order loop filter ($\zeta = 0.707$) and natural loop bandwidth $\omega_n = 2.40 \times 10^6\text{ rad/s}$, the $1\%$ phase-error settling time ($e^{-\zeta \omega_n t_{\text{lock}}} = 0.01 \implies \zeta\omega_n t_{\text{lock}} \approx 4.6$) is: $$t_{\text{lock}} \approx \frac{4.6}{\zeta \omega_n}$$

Keystrokes: 4.6 [ENTER] 0.707 [ENTER] 2.40 [E] 6 [×] [÷] 1 [E] 6 [×] $\rightarrow$ 2.7109853843$\,\mu\text{s}$.

Entering the four-decade settling constant ($4.6$) and dividing by the product $\zeta\omega_n$ (4.6 [ENTER] 0.707 [ENTER] 2.40 [E] 6 [×] [÷]) yields $t_{\text{lock}} = 2.7110 \times 10^{-6}\text{ s}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 2.7110 \times 10^{-6})$. Scaling by $10^6$ (1 [E] 6 [×]) confirms that the PLL locks in $2.71\,\mu\text{s}$—over a thousand times faster than human tactile perception.

Workbook Problem 1.6: Keypad Matrix Phantom Key Sneak Current Analysis

Suppose a fast RPN operator rolls three fingers across keys at row-column coordinates $(R_1, C_1)$, $(R_1, C_2)$, and $(R_2, C_1)$. In a naive matrix lacking series diodes, those three closed switches form a conductive bridge that pulls row $R_2$ low when column $C_2$ is strobed, registering a false fourth "phantom" key at $(R_2, C_2)$ with sneak current ($R_{\text{pu}} = 10\text{ k}\Omega$ to $V_{\text{cc}} = 3.3\text{ V}$): $$I_{\text{sneak}} = \frac{V_{\text{cc}}}{R_{\text{pu}} + 3 R_{\text{contact}}} \approx \frac{3.3}{10000} = 0.33\text{ mA}$$ Because the SC-32 places a discrete Schottky diode in series with all 43 switches, the third switch in that loop is reverse-biased and blocks sneak current completely ($I_{\text{leak}} < 0.1\,\mu\text{A}$), guaranteeing true $N$-key rollover.

Dividing supply rail voltage by row pull-up resistance (3.3 [ENTER] 10000 [÷]) resolves the unmitigated sneak current $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{3.3000 \times 10^{-4}}\text{ A} = 0.33\text{ mA})$ in Register $X$.

Workbook Problem 1.7: Ambient Light Sensor ADC Resolution and Lux Mapping

To calibrate display contrast under laboratory and outdoor lighting, a PIN photodiode generates $I_{\text{photo}} = 0.25\,\mu\text{A}$ per $100\text{ lux}$ through a $R_f = 2.4\text{ M}\Omega$ transimpedance amplifier into a 12-bit successive-approximation ADC ($2^{12} = 4096\text{ counts}$ across $V_{\text{ref}} = 3.30\text{ V}$): $$V_{\text{out}} = I_{\text{photo}} \cdot R_f = \left(\frac{0.25 \times 10^{-6}\text{ A}}{100\text{ lux}}\right) \times 2.4 \times 10^6\,\Omega = 0.0060\text{ V/lux} = 6.0\text{ mV/lux}$$ $$V_{\text{LSB}} = \frac{3.30\text{ V}}{4096} \approx 0.80566\text{ mV}, \qquad \text{Resolution} = \frac{V_{\text{LSB}}}{0.0060\text{ V/lux}} \approx \mathbf{0.1343\text{ lux/count}}$$

Evaluating the 12-bit quantization step (3.3 [ENTER] 4096 [÷]) places $V_{\text{LSB}} = 8.0566 \times 10^{-4}\text{ V}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 8.0566 \times 10^{-4})$. Dividing directly by the transimpedance gain (0.006 [÷]) resolves photometric quantization resolution $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{0.1343}\text{ lux/count})$ without intermediate rounding.

Workbook Problem 1.8: Chassis Thermal Equilibrium Temperature Rise

Under sustained $150\text{ MHz}$ dual-core numerical quadrature, the microcontroller and regulator dissipate $P = 60.0\text{ mW} = 0.060\text{ W}$. Across a total enclosure surface area $A = 0.024\text{ m}^2$ with natural still-air convection coefficient $h_c = 6.5\text{ W/m}^2\text{K}$, Newton's law of cooling determines the steady-state surface temperature rise above ambient: $$\Delta T = \frac{P}{h_c \cdot A}$$

Keystrokes: 0.060 [ENTER] 6.5 [ENTER] 0.024 [×] [÷] $\rightarrow$ 0.3846$\text{ K} \approx 0.38^\circ\text{C}$.

Loading thermal dissipation ($0.060\text{ W}$) and forming the convective conductance product $h_c A$ (0.060 [ENTER] 6.5 [ENTER] 0.024 [×]) leaves $Y = 0.0600\text{ W}$ and $X = 0.1560\text{ W/K}$ $(T: 0.0000, Z: 0.0000, Y: 0.0600, X: 0.1560)$. Pressing [÷] resolves the equilibrium temperature rise $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{0.3846}\text{ K} \approx 0.38^\circ\text{C})$—confirming that the SC-32 operates isothermally in the hand, immune to thermal drift or clock throttling.

Chapter 02: Principles of RPN & 4-Stack XYZT Dynamics — Extended Engineering & Competition Problems

Accompanies Chapter 02 of Thinking in the Stack (15 Worked Engineering Problems)

Workbook Problem 2.1: Right Triangle Polar Resolution

In geodetic surveying, converting orthogonal Cartesian offsets—eastward departure $x = 42.5\text{ m}$ and northward latitude $y = 68.2\text{ m}$—into radial sightline distance $r$ and bearing angle $\theta$ is a daily field operation. Because the SC-32 maps 2D Cartesian space directly onto the register pair $(Y: y,\; X: x)$, polar conversion requires neither inverse trigonometric menus nor manual squaring:

Keystrokes: 68.2 [ENTER] 42.5 [↰] [▸θ,r] $\rightarrow$ 80.3685.

Loading northward latitude into Register $Y$ (68.2 [ENTER], $T: 0.0000, Z: 0.0000, Y: 68.2000, X: 68.2000$) and eastward departure into Register $X$ (42.5, $T: 0.0000, Z: 0.0000, Y: 68.2000, X: 42.5000$) aligns the coordinate vector on the stack. Pressing [↰] [▸θ,r] executes a simultaneous two-register hardware rotation: - Register $X$ receives the Euclidean hypotenuse $r = \sqrt{x^2 + y^2} = \mathbf{80.3685\text{ m}}$. - Register $Y$ receives the four-quadrant angle $\theta = \operatorname{atan2}(y, x) = \mathbf{58.0772^\circ}$. Tapping [x⇄y] swaps the registers to inspect $\theta = \mathbf{58.0772^\circ}$ in Register $X$ while retaining $r = \mathbf{80.3685\text{ m}}$ in Register $Y$.

Workbook Problem 2.2: 3D Vector Norm without Variable Registers

A structural tie-back anchor resists a spatial tension vector $\vec{T} = (14.2, -8.7, 21.5\text{ kN})$ across three orthogonal axes. Computing the 3D Euclidean norm requires accumulating three squared components under a single radical:

\[|\vec{T}| = \sqrt{14.2^2 + (-8.7)^2 + 21.5^2}\]

Keystrokes: 14.2 [↰] [x²] 8.7 [↰] [x²] [+] 21.5 [↰] [x²] [+] [√x] $\to \mathbf{27.1895\text{ kN}}$.

Because squaring a real component is always non-negative ($|-8.7|^2 = (+8.7)^2$), the negative sign on $T_y$ can be omitted entirely during entry. Squaring the $x$-component in place (14.2 [↰] [x²]) deposits $201.6400$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 201.6400)$. Entering and squaring the $y$-component (8.7 [↰] [x²]) auto-lifts $T_x^2$ to $Y$ while forming $75.6900$ in $X$ $(T: 0.0000, Z: 0.0000, Y: 201.6400, X: 75.6900)$, and [+] collapses the planar norm squared to $277.3300$ in $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 277.3300)$. Squaring the $z$-component (21.5 [↰] [x²]) lifts $277.3300$ to $Y$ and forms $462.2500$ in $X$ $(T: 0.0000, Z: 0.0000, Y: 277.3300, X: 462.2500)$; pressing [+] [√x] collapses all three spatial dimensions into the resultant cable tension $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{27.1952}\text{ kN})$ while never occupying more than two stack registers.

Workbook Problem 2.3: Four-Bar Linkage Coordinate Reduction

In planar mechanism kinematics, the coupler link of a four-bar linkage connects crank pin $A(48.2, 110.3\text{ mm})$ to follower pin $B(125.4, 85.0\text{ mm})$. Evaluating the Euclidean link length $\ell = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ showcases how Register $Z$ acts as an automatic holding bay while the second coordinate difference is computed:

Keystrokes: 125.4 [ENTER] 48.2 [-] [↰] [x²] 85.0 [ENTER] 110.3 [-] [↰] [x²] [+] [√x] $\to \mathbf{81.2393\text{ mm}}$.

Evaluating the horizontal span first (125.4 [ENTER] 48.2 [-]) forms $\Delta x = 77.2\text{ mm}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 77.2000)$, and [↰] [x²] squares it in place to $\Delta x^2 = 5959.8400\text{ mm}^2$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 5959.8400)$. When the vertical pin coordinates are entered and subtracted (85.0 [ENTER] 110.3 [-]), $\Delta x^2$ rides up to Register $Z$ and then drops back to Register $Y$ as $\Delta y = -25.3\text{ mm}$ forms in $X$ $(T: 0.0000, Z: 0.0000, Y: 5959.8400, X: -25.3000)$. Squaring $\Delta y$ in place ([↰] [x²]) yields $640.0900\text{ mm}^2$ in $X$ $(T: 0.0000, Z: 0.0000, Y: 5959.8400, X: 640.0900)$, and [+] [√x] sums the orthogonal squares ($6599.9300\text{ mm}^2$) to extract the rigid coupler length $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{81.2400}\text{ mm})$.

Workbook Problem 2.4: Finite Geometric Series Summation

In digital signal processing and financial annuity valuation, the partial sum of $n = 8$ geometric terms with initial amplitude $a = 12.0$ and common decay ratio $r = 0.75$ collapses via the closed-form identity:

\[S_8 = 12 \frac{1 - 0.75^8}{1 - 0.75} = 12 \frac{1 - 0.100113}{0.25} = 12 \times 3.599548 = 43.194576\]

Keystrokes: 0.75 [ENTER] 8 [yˣ] 1 [-] [+/-] 0.25 [÷] 12 [×] $\to \mathbf{43.194576}$.

Evaluating the exponential attenuation term $r^8$ first (0.75 [ENTER] 8 [yˣ]) places $0.100113$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.1001)$. Rather than swapping registers to compute $1 - r^8$, subtracting $1$ and negating in place (1 [-] [+/-]) yields the complement $1 - r^8 \approx 0.899887$ without lifting the stack $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.8999)$. Dividing by $1 - r = 0.25$ (0.25 [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 3.5995$) and scaling by the leading amplitude (12 [×]) resolves the 8-stage sum $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{43.1946})$.

Workbook Problem 2.5: Thermodynamic Otto Cycle Thermal Efficiency

In an idealized spark-ignition internal combustion engine with volumetric compression ratio $r_c = V_{\max}/V_{\min} = 9.5$ and diatomic air specific-heat ratio $\gamma = C_p/C_v = 1.40$, the Nikolaus Otto cycle efficiency is governed by isentropic temperature scaling:

\[\eta = 1 - r_c^{1 - \gamma} = 1 - 9.5^{-0.40}\]

Keystrokes: 9.5 [ENTER] 0.40 [+/-] [yˣ] 1 [-] [+/-] $\to \mathbf{0.5936} = 59.36\%$.

Loading the compression ratio (9.5 [ENTER], $T: 0.0000, Z: 0.0000, Y: 9.5000, X: 9.5000$) and the isentropic exponent (0.40 [+/-], $T: 0.0000, Z: 0.0000, Y: 9.5000, X: -0.4000$) prepares the power evaluation. Pressing [yˣ] computes the fraction of heat rejected to the cold sink, $T_{\text{low}}/T_{\text{high}} = 9.5^{-0.40} \approx 0.4064$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.4064)$. Applying the subtraction-and-negation idiom (1 [-] [+/-]) converts heat rejection into net work efficiency $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{0.5936} = 59.36\%)$ without lifting the stack.

Workbook Problem 2.6: Wheatstone Bridge Out-of-Balance Voltage

In strain-gauge and resistance-thermometer (RTD) instrumentation, a quarter-bridge Wheatstone network driven by $V_s = 10.0\text{ V}$ compares a reference half-bridge ($R_1 = R_2 = 1000\,\Omega \implies V_{\text{ref}} = 0.5 V_s$) against a sensing divider ($R_3 = 1000\,\Omega$, active element $R_x = 1025\,\Omega$):

\[V_o = V_s \left(\frac{R_x}{R_3 + R_x} - 0.5\right)\]

Keystrokes: 1025 [ENTER] 2025 [÷] 0.5 [-] 10.0 [×] $\to \mathbf{0.061728\text{ V}} = 61.73\text{ mV}$.

Evaluating the active divider ratio (1025 [ENTER] 2025 [÷]) places $R_x / (R_3 + R_x) \approx 0.5062$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.5062)$. Subtracting the balanced null point (0.5 [-], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0062$) and scaling by the bridge excitation (10.0 [×]) isolates the differential millivolt signal $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{0.0617}\text{ V} \approx 61.73\text{ mV})$. Notice that if $R_3$ and $R_x$ are arbitrary resistances, keying 1025 [ENTER] [ENTER] 1000 [+] [÷] forms $R_x / (R_x + R_3)$ while typing $R_x = 1025$ only once!

Workbook Problem 2.7: Harmonic Mean of Four Speed Trials

When a test vehicle traverses four equal-distance proving-ground laps at speeds $v = [45.0, 52.0, 58.0, 64.0\text{ mph}]$, the true average speed over the total distance is the harmonic mean—not the arithmetic mean ($54.75\text{ mph}$), because more time is spent on slower laps:

\[v_h = \frac{4}{\sum_{i=1}^4 \frac{1}{v_i}}\]

Keystrokes: 45 [1/x] 52 [1/x] [+] 58 [1/x] [+] 64 [1/x] [+] 4 [x⇄y] [÷]$\to \mathbf{53.8184\text{ mph}}$.

Inverting the first lap speed (45 [1/x]) places its slowness $1/v_1 \approx 0.0222\text{ h/mi}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0222)$. Each subsequent lap is inverted in place and added to the running slowness accumulator: 52 [1/x] [+] $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0415)$, 58 [1/x] [+] $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0587)$, and 64 [1/x] [+], accumulating $\sum 1/v_i \approx 0.0743\text{ h/mi}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0743)$. Keying 4 [x⇄y] [÷] divides the 4-lap distance by total elapsed time, resolving the true harmonic mean speed $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{53.8184}\text{ mph})$.

Workbook Problem 2.8: Bending Moment on Overhanging Crane Beam

An overhead shop crane applies a concentrated hoist load $P = 35\text{ kN}$ at an overhang distance $a = 1.8\text{ m}$ beyond its support column on a wide-flange steel I-beam of elastic section modulus $S = 4.5 \times 10^{-4}\text{ m}^3$:

\[M = P \cdot a = 35 \times 1.8 = 63.0\text{ kN}\cdot\text{m}, \qquad \sigma = \frac{M}{S}\]

Keystrokes: 35000 [ENTER] 1.8 [×] 4.5 [E] 4 [+/-] [÷] 1 [E] 6 [÷]$\to \mathbf{140\text{ MPa}}$.

Forming the root bending moment (35000 [ENTER] 1.8 [×]) establishes $M = 63,000\text{ N}\cdot\text{m}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 63000.0000)$. Dividing by the elastic section modulus (4.5 [E] 4 [+/-] [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.4000 \times 10^8$) and normalizing to megapascals (1 [E] 6 [÷]) yields peak outer-fiber bending stress $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{140.0000}\text{ MPa})$—confirming a safety factor of $1.79$ against A36 structural steel yield ($250\text{ MPa}$).

Workbook Problem 2.9: Sound Attenuation Over Distance

Because acoustic wave energy from an isotropic point source spreads over a spherical wavefront of area $4\pi r^2$, sound pressure amplitude decays as $1/r$ ($20\text{ dB}$ per distance decade). From a near-field machinery noise reading of $L_1 = 92.0\text{ dB}$ at $r_1 = 5.0\text{ m}$, the sound pressure level at the plant boundary $r_2 = 45.0\text{ m}$ is:

\[L_2 = L_1 - 20 \log_{10}\left(\frac{r_2}{r_1}\right)\]

Keystrokes: 45 [ENTER] 5 [÷] [↰] [LOG] 20 [×] 92.0 [x⇄y] [-]$\to \mathbf{72.9151\text{ dB}}$.

Evaluating the radial expansion ratio first (45 [ENTER] 5 [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 9.0000$), taking the base-10 logarithm ([↰] [LOG], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.9542$), and scaling by $20$ (20 [×]) isolates the spherical spreading attenuation of $19.0849\text{ dB}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 19.0849)$. Entering the reference level and swapping operands (92.0 [x⇄y] [-]) resolves the far-field sound level $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{72.9151}\text{ dB})$.

Workbook Problem 2.10: Doppler Radar Velocity Extraction

A pulse-Doppler X-band weather/traffic radar transmitting at $f_0 = 10.0\text{ GHz} = 1.0 \times 10^{10}\text{ Hz}$ measures a two-way beat frequency shift $f_d = 4,850\text{ Hz}$ ($c = 2.99792 \times 10^8\text{ m/s}$). Because the electromagnetic wave experiences a Doppler shift both arriving at the moving target and reflecting back to the dish, the denominator carries a factor of $2$: $$v = \frac{f_d \cdot c}{2 f_0}$$

Keystrokes: 4850 [ENTER] 2.99792 [E] 8 [×] 2 [÷] 1 [E] 10 [÷]$\to \mathbf{72.69956\text{ m/s}} = 261.7\text{ km/h}$.

Loading the beat frequency (4850 [ENTER], $T: 0.0000, Z: 0.0000, Y: 4850.0000, X: 4850.0000$) and multiplying by the speed of light (2.99792 [E] 8 [×]) forms $f_d c = 1.4540 \times 10^{12}\text{ m/s}^2$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.4540 \times 10^{12})$. Dividing by the round-trip factor $2$ (2 [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 7.2699 \times 10^{11}$) and by the carrier frequency (1 [E] 10 [÷]) yields radial target speed $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{72.6996}\text{ m/s})$.

Workbook Problem 2.11: Centripetal Acceleration in Elliptical Orbit Periapsis

At the perigee of a low-Earth transfer orbit ($r_p = 6,800\text{ km} = 6.80 \times 10^6\text{ m}$), a spacecraft moving at tangential speed $v_p = 8,200\text{ m/s}$ experiences radial centripetal acceleration:

\[a_c = \frac{v_p^2}{r_p}\]

Keystrokes: 8200 [↰] [x²] 6.80 [E] 6 [÷]$\to \mathbf{9.8882352941\text{ m/s}^2}$.

Squaring periapsis speed in place (8200 [↰] [x²]) places specific kinetic energy $v_p^2 = 6.7240 \times 10^7\text{ m}^2/\text{s}^2$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 6.7240 \times 10^7)$, and dividing by orbital radius (6.80 [E] 6 [÷]) yields radial acceleration $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{9.8882}\text{ m/s}^2)$. Because local gravitational acceleration at $r_p = 6,800\text{ km}$ is $g_p = \mu/r_p^2 \approx 8.620\text{ m/s}^2 < a_c$, the excess centripetal requirement ($9.8882 > 8.620\text{ m/s}^2$) confirms the spacecraft is at the perigee of an elliptical orbit ascending toward a higher apogee.

Workbook Problem 2.12: Aerodynamic Vehicle Drag Power at $120\text{ km/h}$

While aerodynamic drag force scales with the square of velocity ($F_d = \frac{1}{2}\rho C_d A v^2$), the mechanical power required to overcome air resistance ($P = F_d \cdot v$) scales with the cube of velocity—explaining why doubling highway speed octuples aerodynamic fuel consumption. At $v = 120\text{ km/h} = 33.3333\text{ m/s}$ ($\rho = 1.20\text{ kg/m}^3,\; C_d = 0.28,\; A = 2.20\text{ m}^2$): $$P = \frac{1}{2} \rho C_d A v^3$$

Keystrokes: 120 [ENTER] 3.6 [÷] 3 [yˣ] 0.5 [×] 1.20 [×] 0.28 [×] 2.20 [×] 1000 [÷]$\to \mathbf{13.6889\text{ kW}}$.

Converting highway speed to meters per second (120 [ENTER] 3.6 [÷]) places $v = 33.3333\text{ m/s}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 33.3333)$, and cubing ([ENTER] 3 [yˣ]) forms $v^3 = 37,037.0370\text{ m}^3/\text{s}^3$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 37037.0370)$. Streaming the aerodynamic coefficients (0.5 [×] 1.20 [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 22222.2222$, then 0.28 [×] 2.20 [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 13688.8889$) and dividing by $1000$ (1000 [÷]) resolves road-load drag power $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{13.6889}\text{ kW})$.

Workbook Problem 2.13: Hydraulic Pump Output Fluid Power

A positive-displacement axial-piston pump delivering $Q = 45.0\text{ L/min} = 7.5 \times 10^{-4}\text{ m}^3/\text{s}$ across a manifold pressure rise $\Delta P = 180\text{ bar} = 1.80 \times 10^7\text{ Pa}$ imparts fluid power $P_{\text{hyd}}$ equal to the product of volumetric flow and pressure head: $$P_{\text{hyd}} = Q \cdot \Delta P$$

Keystrokes: 45 [ENTER] 60 [÷] 1000 [÷] 1.80 [E] 7 [×] 1000 [÷]$\to \mathbf{13.5\text{ kW}}$.

Normalizing liters per minute into SI cubic meters per second (45 [ENTER] 60 [÷] 1000 [÷]) places $Q = 7.50 \times 10^{-4}\text{ m}^3/\text{s}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 7.5000 \times 10^{-4})$. Multiplying by Pascal pressure (1.80 [E] 7 [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 13500.0000$) and dividing by $1000$ (1000 [÷]) yields net fluid power $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{13.5000}\text{ kW})$.

Workbook Problem 2.14: Solar Array Solar Irradiance Incident on Tilted Panel

To maximize photovoltaic yield on a roof tilted at $\beta = 35.0^\circ$ when the sun is at altitude $\alpha = 48.0^\circ$ and azimuth offset $\Delta\gamma = 15.0^\circ$ from the panel normal ($I_0 = 980\text{ W/m}^2$), the spherical dot product between the solar ray vector and panel normal determines the cosine projection factor: $$\cos\theta = \sin\alpha \cos\beta + \cos\alpha \sin\beta \cos(\Delta\gamma), \qquad I_{\text{incident}} = I_0 \cos\theta$$

Keystrokes: 48 [SIN] 35 [COS] [×] 48 [COS] 35 [SIN] [×] 15 [COS] [×] [+] 980 [×] $\to \mathbf{959.87919595\text{ W/m}^2}$.

Evaluating the vertical projection product (48 [SIN] 35 [COS] [×]) places $\sin 48^\circ \cos 35^\circ \approx 0.6088$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.6088)$. Forming the horizontal azimuth projection (48 [COS] 35 [SIN] [×] 15 [COS] [×]) lifts $0.6088$ to Register $Y$ while placing $0.3707$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.6088, X: 0.3707)$. Summing the two projections ([+], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.9795$) shows the panel captures $97.95\%$ of direct normal irradiance, and 980 [×] resolves the incident flux $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{959.8792}\text{ W/m}^2)$.

Workbook Problem 2.15: Automotive Emergency Braking Stopping Distance

Equating kinetic energy $\frac{1}{2}mv_0^2$ to tire-pavement frictional work $\mu m g d$ for a vehicle traveling at $v_0 = 100\text{ km/h} = 27.7778\text{ m/s}$ on wet asphalt ($\mu = 0.65,\; g = 9.80665\text{ m/s}^2$) reveals that stopping distance is independent of vehicle mass and grows quadratically with speed:

\[d = \frac{v_0^2}{2 \mu g}\]

Keystrokes: 100 [ENTER] 3.6 [÷] [↰] [x²] 2 [ENTER] 0.65 [×] 9.80665 [×] [÷]$\to \mathbf{60.5517\text{ m}}$.

Converting kilometers per hour to meters per second and squaring in place (100 [ENTER] 3.6 [÷] [↰] [x²]) places specific kinetic energy $v_0^2 \approx 771.6049\text{ m}^2/\text{s}^2$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 771.6049)$. Forming the deceleration denominator (2 [ENTER] 0.65 [×] 9.80665 [×]) holds $v_0^2$ in Register $Y$ while computing $2\mu g = 12.7486\text{ m/s}^2$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 771.6049, X: 12.7486)$, and [÷] resolves the minimum stopping distance $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{60.5245}\text{ m})$.

Chapter 03: Algebra, Transcendentals & Display Formatting — Extended Engineering & Competition Problems

Accompanies Chapter 03 of Thinking in the Stack (22 Worked Engineering Problems)

Workbook Problem 3.1: Blood Bicarbonate Buffer System pH

Human arterial blood plasma maintains homeostasis within a narrow pH window ($7.35\text{--}7.45$) through the coupled carbonic acid–bicarbonate equilibrium ($\text{pK}_a = 6.10$ at $37^\circ\text{C}$), governed by the Henderson-Hasselbalch equation. When clinical blood-gas analysis reports renal bicarbonate $[\text{HCO}_3^-] = 24.0\text{ mEq/L}$ and dissolved pulmonary carbon dioxide $[\text{CO}_2] = 1.20\text{ mEq/L}$: $$\text{pH} = \text{pK}_a + \log_{10}\left(\frac{[\text{HCO}_3^-]}{[\text{CO}_2]}\right) = 6.10 + \log_{10}\left(\frac{24.0}{1.20}\right)$$

Keystrokes: 24.0 [ENTER] 1.20 [÷] [↰] [LOG] 6.10 [+] $\rightarrow$ 7.4000.

Dividing metabolic base by respiratory acid (24.0 [ENTER] 1.20 [÷]) isolates the classic $20:1$ physiological buffer ratio in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 20.0000)$. Applying the common base-10 logarithm and adding the dissociation constant ([↰] [LOG] 6.10 [+]) yields arterial pH $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{7.4010})$.

Workbook Problem 3.2: Fiber Optic Cable Total Attenuation & Output Power

In long-haul optical telecommunications at the $1550\text{ nm}$ silica transparency window, Rayleigh scattering limits attenuation to $\alpha = 0.22\text{ dB/km}$. When a DFB laser launches $P_{\text{in}} = 15.0\text{ mW}$ into a $L = 65.0\text{ km}$ unamplified span, the logarithmic decibel loss converts back to linear optical power via base-10 exponentiation: $$\text{Loss}_{\text{dB}} = \alpha \cdot L = 0.22 \times 65.0 = 14.30\text{ dB}, \qquad P_{\text{out}} = P_{\text{in}} \cdot 10^{-\text{Loss}_{\text{dB}} / 10} = 15.0 \cdot 10^{-1.430}$$

Keystrokes: 0.22 [ENTER] 65 [×] 10 [÷] [+/-] [↰] [10ˣ] 15 [×] $\rightarrow$ 0.5567$\text{ mW} = 557.3\,\mu\text{W}$.

Computing total span loss in bels and negating (0.22 [ENTER] 65 [×] 10 [÷] [+/-]) places the decade exponent $-1.4300$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: -1.4300)$. Evaluating the antilogarithm and scaling by launch power ([↰] [10ˣ] 15 [×]) resolves the received photodiode power $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{0.5573}\text{ mW})$ ($557.3\,\mu\text{W}$, or $-2.54\text{ dBm}$).

Workbook Problem 3.3: Tsiolkovsky Rocket Equation Burnout Velocity

Derived by Konstantin Tsiolkovsky in 1897 by integrating momentum conservation $m\,dv = -v_e\,dm$, the ideal rocket equation proves that a launch vehicle's velocity increment $\Delta v$ grows logarithmically with its wet-to-dry mass ratio $m_0 / m_f = 9.20$ and linearly with effective exhaust velocity $v_e = 3,150\text{ m/s}$: $$\Delta v = v_e \ln\left(\frac{m_0}{m_f}\right)$$

Keystrokes: 9.20 [LN] 3150 [×] $\rightarrow$ 6989.78$\text{ m/s} \approx 6.99\text{ km/s}$.

Taking the natural logarithm of the mass ratio (9.20 [LN]) places $\ln(9.20) \approx 2.2192$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 2.2192)$, and multiplying by exhaust velocity (3150 [×]) resolves burnout velocity $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{6990.4910}\text{ m/s})$ ($\approx 6.99\text{ km/s}$).

Workbook Problem 3.4: Richter Earthquake Seismic Energy Ratio

In seismology, Beno Gutenberg and Charles Richter established that while surface-wave amplitude scales by $10^M$, total radiated elastic wave energy scales as $10^{1.5 M}$ ($\approx 31.62\times$ per magnitude unit, and $1,000\times$ per two magnitude units). Comparing a major $M_2 = 7.8$ earthquake to a moderate $M_1 = 5.2$ tremor:

\[\frac{E_2}{E_1} = 10^{1.5 (M_2 - M_1)}\]

Keystrokes: 7.8 [ENTER] 5.2 [-] 1.5 [×] [↰] [10ˣ] $\rightarrow$ 7943.28.

Evaluating the magnitude span and scaling by $1.5$ (7.8 [ENTER] 5.2 [-] 1.5 [×]) forms the decade exponent $3.9000$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 3.9000)$. Pressing [↰] [10ˣ] resolves the seismic energy ratio $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{7943.2823})$—demonstrating that a $2.6$-unit increase on the Richter scale releases nearly 8,000 times more strain energy.

Workbook Problem 3.5: Barometric Altitude Determination from Pressure

In an isothermal hydrostatically balanced atmosphere with scale height $H_s = \frac{R T_0}{M g} = 8,434\text{ meters}$ and sea-level reference pressure $P_0 = 1013.25\text{ hPa}$, an aircraft static port measuring $P = 725.0\text{ hPa}$ determines pressure altitude by inverting the exponential barometric law:

\[h = - H_s \ln\left(\frac{P}{P_0}\right)\]

Keystrokes: 725 [ENTER] 1013.25 [÷] [LN] 8434 [×] [+/-] $\rightarrow$ 2826.96$\text{ meters} \approx 9,263\text{ ft}$.

Forming the static pressure ratio and taking the natural logarithm (725 [ENTER] 1013.25 [÷] [LN]) places $\ln(P/P_0) \approx -0.3347$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: -0.3347)$. Scaling by atmospheric scale height and negating (8434 [×] [+/-]) resolves pressure altitude $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{2823.2529}\text{ m})$ ($\approx 9,263\text{ ft}$).

Workbook Problem 3.6: Astronomical Apparent Magnitude and Stellar Luminosity

Norman Pogson formalized Hipparchus's ancient stellar magnitude scale in 1856 by defining a 5-magnitude difference as a $100:1$ flux ratio ($100^{1/5} = 10^{0.4} \approx 2.5119$ per magnitude). Comparing the night sky's brightest star, Sirius ($m_1 = -1.46$), against the North Star, Polaris ($m_2 = +1.98$):

\[\frac{L_1}{L_2} = 10^{0.4 (m_2 - m_1)}\]

Keystrokes: 1.98 [ENTER] 1.46 [+/-] [-] 0.4 [×] [↰] [10ˣ] $\rightarrow$ 23.7684.

Computing the magnitude span $\Delta m = 1.98 - (-1.46) = 3.44$ and multiplying by Pogson's $0.4$ coefficient (1.98 [ENTER] 1.46 [+/-] [-] 0.4 [×]) places $1.3760$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.3760)$. Pressing [↰] [10ˣ] resolves the apparent flux ratio $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{23.7684})$—confirming Sirius delivers $23.8\times$ more photon flux to an earthbound telescope aperture than Polaris.

Workbook Problem 3.7: Permeability and Darcy Fluid Flow in Porous Sandstone

In petroleum reservoir characterization, Henry Darcy's law relates volumetric flow $Q = 1.8 \times 10^{-7}\text{ m}^3/\text{s}$ of crude oil (viscosity $\mu = 0.035\text{ Pa}\cdot\text{s}$) through a sandstone core plug ($L = 0.050\text{ m}$, cross-sectional area $A = 0.00125\text{ m}^2$) under differential pressure $\Delta P = 250\text{ kPa} = 250,000\text{ Pa}$ to intrinsic rock permeability $k$ ($1\text{ Darcy} = 9.86923 \times 10^{-13}\text{ m}^2$): $$k = \frac{Q \cdot \mu \cdot L}{A \cdot \Delta P}$$

Keystrokes: 1.8 [E] 7 [+/-] [ENTER] 0.035 [×] 0.050 [×] 0.00125 [ENTER] 250000 [×] [÷] 9.86923 [E] 13 [+/-] [÷] $\rightarrow$ 1.021356276$\text{ Darcy}$.

Chaining the viscous numerator product (1.8 [E] 7 [+/-] [ENTER] 0.035 [×] 0.050 [×]) places $Q \mu L = 3.15 \times 10^{-10}\text{ N}\cdot\text{s}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 3.1500 \times 10^{-10})$. Forming the pressure-force denominator and dividing (0.00125 [ENTER] 250000 [×] [÷]) isolates SI permeability $1.0080 \times 10^{-12}\text{ m}^2$ in $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.0080 \times 10^{-12})$, and dividing by the Darcy conversion constant (9.86923 [E] 13 [+/-] [÷]) yields $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{1.0214}\text{ Darcy})$.

Workbook Problem 3.8: Shannon-Hartley Information Capacity Limit

Claude Shannon's 1948 noisy-channel coding theorem establishes the absolute upper bound on error-free bit rate $C$ across an additive white Gaussian noise (AWGN) channel of bandwidth $B = 20.0\text{ MHz} = 2.0 \times 10^7\text{ Hz}$ and signal-to-noise ratio $\text{SNR} = 30.0\text{ dB}$ ($\text{SNR}_{\text{linear}} = 10^{30/10} = 1,000$): $$C = B \log_2(1 + \text{SNR}) = B \frac{\ln(1 + \text{SNR})}{\ln 2}$$

Keystrokes: 1001 [LN] 2 [LN] [÷] 20 [E] 6 [×] 1 [E] 6 [÷] $\rightarrow$ 199.34452518$\text{ Mbps}$.

By the change-of-base identity $\log_2(x) = \ln(x)/\ln(2)$, keying 1001 [LN] 2 [LN] [÷] evaluates spectral efficiency $\log_2(1001) \approx 9.9672\text{ bits/s/Hz}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 9.9672)$. Scaling by the $20\text{ MHz}$ channel bandwidth (20 [E] 6 [×] 1 [E] 6 [÷]) resolves maximum theoretical throughput $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{199.3445}\text{ Mbps})$.

Workbook Problem 3.9: Chemical Reaction Equilibrium Constant from $\Delta G^\circ$

In chemical thermodynamics, a negative standard Gibbs free energy change $\Delta G^\circ = -33.3\text{ kJ/mol} = -33,300\text{ J/mol}$ at $T = 298.15\text{ K}$ drives the reaction equilibrium constant $K$ exponentially above unity:

\[K = \exp\left(-\frac{\Delta G^\circ}{R T}\right)\]

Keystrokes: 33300 [ENTER] 8.31446 [ENTER] 298.15 [×] [÷] [eˣ] $\rightarrow$ 6.8220E5.

Loading the magnitude $-\Delta G^\circ = 33,300\text{ J/mol}$ (33300 [ENTER], $T: 0.0000, Z: 0.0000, Y: 33300.0000, X: 33300.0000$) and dividing by the molar thermal energy $RT$ (8.31446 [ENTER] 298.15 [×] [÷]) isolates the dimensionless thermodynamic driving force $-\Delta G^\circ / (RT) = 13.4331$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 13.4331)$. Pressing [eˣ] exponentiates in place to yield the equilibrium constant $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{6.8220 \times 10^5})$.

Workbook Problem 3.10: Beam Elastic Curve Cubic Deflection

Along a simply supported girder of span $L = 6.0\text{ m}$ and flexural rigidity $E I = 1.2 \times 10^7\text{ N}\cdot\text{m}^2$ carrying uniform load $w = 8.5\text{ kN/m} = 8,500\text{ N/m}$, the vertical deflection $\delta(x)$ at station $x = 2.0\text{ m}$ is governed by the fourth-order Euler-Bernoulli elastic curve: $$\delta(x) = \frac{w x}{24 E I} (L^3 - 2 L x^2 + x^3)$$

Keystrokes: 6 [ENTER] 3 [yˣ] 2 [ENTER] 6 [×] 2 [↰] [x²] [×] [-] 2 [ENTER] 3 [yˣ] [+] 8500 [ENTER] 2 [×] [×] 24 [ENTER] 1.2 [E] 7 [×] [÷] $\rightarrow$ 0.0103888889$\text{ m}$.

Evaluating the cubic span polynomial $(L^3 - 2Lx^2 + x^3)$ first collapses the geometry into a single scalar before scaling by load and stiffness: 6 [ENTER] 3 [yˣ] forms $L^3 = 216.0000$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 216.0000)$, 2 [ENTER] 6 [×] 2 [↰] [x²] [×] [-] subtracts $2 L x^2 = 48.0000$ to leave $168.0000$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 168.0000)$, and 2 [ENTER] 3 [yˣ] [+] adds $x^3 = 8.0000$ to complete the geometric factor $176.0000\text{ m}^3$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 176.0000)$. Scaling by $w x$ (8500 [ENTER] 2 [×] [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 2.9920 \times 10^6$) and dividing by $24 E I$ (24 [ENTER] 1.2 [E] 7 [×] [÷]) resolves the vertical sag $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{0.0104}\text{ m} \approx 10.39\text{ mm})$.

Workbook Problem 3.11: Radiative Cooling Time of Molten Metal Droplet

In space-based shot-tower manufacturing, a spherical droplet of molten iron ($\rho = 7,800\text{ kg/m}^3,\; c_p = 450\text{ J/kg}\cdot\text{K},\; r = 0.0015\text{ m},\; \varepsilon = 0.80,\; \sigma = 5.67037 \times 10^{-8}\text{ W/m}^2\text{K}^4$) cools strictly by Stefan-Boltzmann radiation $-m c_p \frac{dT}{dt} = 4\pi r^2 \varepsilon\sigma T^4$. Integrating $T^{-4}\,dT$ from $T_1 = 1,500\text{ K}$ to $T_2 = 1,000\text{ K}$ yields an inverse-cube law:

\[t = \frac{\rho \cdot c_p \cdot r}{3 \varepsilon \sigma} \left(\frac{1}{T_2^3} - \frac{1}{T_1^3}\right)\]

Keystrokes: 1000 [ENTER] 3 [yˣ] [1/x] 1500 [ENTER] 3 [yˣ] [1/x] [-] 7800 [ENTER] 450 [×] 0.0015 [×] [×] 3 [ENTER] 0.80 [×] 5.67037 [E] 8 [+/-] [×] [÷] $\rightarrow$ 27.224854815$\text{ seconds}$.

Forming the inverse-cube thermal interval (1000 [ENTER] 3 [yˣ] [1/x] 1500 [ENTER] 3 [yˣ] [1/x] [-]) places $(1/T_2^3 - 1/T_1^3) \approx 7.0370 \times 10^{-10}\text{ K}^{-3}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 7.0370 \times 10^{-10})$. Scaling by volumetric heat capacity per unit area (7800 [ENTER] 450 [×] 0.0015 [×] [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 3.7050 \times 10^{-6}$) and dividing by $3\varepsilon\sigma$ (3 [ENTER] 0.80 [×] 5.67037 [E] 8 [+/-] [×] [÷]) resolves the vacuum cooling time $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{27.2249}\text{ s})$.

Workbook Problem 3.12: Dynamic Range and Theoretical SNR of 24-Bit Digital Audio

In uniform linear pulse-code modulation (PCM) quantization with $N = 24\text{ bits}$, each additional bit halves the RMS quantization step size ($20\log_{10} 2 = 6.0206\text{ dB/bit}$), while the ratio of a full-scale sine wave to a uniform triangular quantization error distribution adds $10\log_{10}(3/2) = 1.7609\text{ dB}$:

\[\text{SNR}_{\text{dB}} = 6.0206 N + 1.7609\text{ dB}\]

Keystrokes: 6.0206 [ENTER] 24 [×] 1.7609 [+] $\rightarrow$ 146.2553$\text{ dB}$.

Multiplying the per-bit gain by bit depth (6.0206 [ENTER] 24 [×]) forms the raw dynamic range of $144.4944\text{ dB}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 144.4944)$, and adding the crest-to-quantization correction (1.7609 [+]) yields the ideal 24-bit SNR ceiling $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{146.2553}\text{ dB})$.

Workbook Problem 3.13: Fermi-Dirac Semiconductor Electron Distribution

Because electrons are spin-$1/2$ fermions obeying the Pauli exclusion principle, their thermal occupancy probability at energy $\Delta E = E - E_F = 0.050\text{ eV}$ above the Fermi level at room temperature $T = 300\text{ K}$ ($k T = 0.025852\text{ eV}$) follows Enrico Fermi and Paul Dirac's quantum distribution:

\[f(E) = \frac{1}{1 + \exp\left(\frac{E - E_F}{k T}\right)}\]

Keystrokes: 0.050 [ENTER] 0.025852 [÷] [eˣ] 1 [+] [1/x] $\rightarrow$ 0.126315 ($12.63\%$).

Normalizing the energy offset by thermal energy (0.050 [ENTER] 0.025852 [÷]) places $(E - E_F)/(kT) \approx 1.9341$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.9341)$. Chaining the three monadic/dyadic operations ([eˣ] 1 [+] [1/x]) evaluates the denominator $1 + e^{1.9341} \approx 7.9177$ and inverts it in place to obtain the quantum state occupancy $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{0.1263} = 12.63\%)$.

Workbook Problem 3.14: Viscous Pressure Drop in Poiseuille Laminar Capillary Flow

Jean Léonard Marie Poiseuille and Gotthilf Hagen discovered that laminar viscous pressure drop $\Delta P$ in a cylindrical micro-capillary ($\mu = 0.0025\text{ Pa}\cdot\text{s}$, length $L = 0.20\text{ m}$, flow rate $Q = 1.5 \times 10^{-8}\text{ m}^3/\text{s}$) is inversely proportional to the fourth power of inner radius $R = 0.0005\text{ m}$ ($0.5\text{ mm}$)—meaning a $50\%$ plaque or fouling constriction increases required pumping pressure sixteen-fold: $$\Delta P = \frac{8 \mu L Q}{\pi R^4}$$

Keystrokes: 8 [ENTER] 0.0025 [×] 0.20 [×] 1.5 [E] 8 [+/-] [×] [↱] [π] [÷] 0.0005 [ENTER] 4 [yˣ] [÷] $\rightarrow$ 305.57749074$\text{ Pa}$.

Streaming the viscous numerator (8 [ENTER] 0.0025 [×] 0.20 [×] 1.5 [E] 8 [+/-] [×]) places $8 \mu L Q = 6.0000 \times 10^{-11}\text{ N}\cdot\text{m}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 6.0000 \times 10^{-11})$. Dividing successively by $\pi$ and $R^4$ ([↱] [π] [÷] 0.0005 [ENTER] 4 [yˣ] [÷]) resolves the laminar pressure drop $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{305.5775}\text{ Pa})$.

Workbook Problem 3.15: Parallel Conductance and Characteristic Admittance

When a lossy shunt conductance $G = 0.020\text{ S}$ sits in parallel with a capacitive susceptance $B = 0.015\text{ S}$, their currents are in phase quadrature ($90^\circ$ apart), so the total admittance magnitude $|Y|$ is their Euclidean norm:

\[|Y| = \sqrt{G^2 + B^2}\]

Keystrokes: 0.020 [↰] [x²] 0.015 [↰] [x²] [+] [√x] $\rightarrow$ 0.0250$\text{ S} = 25.0\text{ mS}$.

Squaring and summing the orthogonal admittances (0.020 [↰] [x²] 0.015 [↰] [x²] [+]) places $G^2 + B^2 = 6.2500 \times 10^{-4}\text{ S}^2$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 6.2500 \times 10^{-4})$, and [√x] extracts the $3\text{--}4\text{--}5$ Pythagorean hypotenuse $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{0.0250}\text{ S} = 25.0\text{ mS})$.

Workbook Problem 3.16: High-Speed Sprint: Continued Power Pyramid

Evaluate the nested power tower:

\[x = \left(2.4^{1.8}\right)^{\sqrt{3.5}}\]

Keystrokes: 2.4 [ENTER] 1.8 [yˣ] 3.5 [√x] [yˣ] $\rightarrow$ 19.0703288

Evaluating the inner power (2.4 [ENTER] 1.8 [yˣ]) places $2.4^{1.8} \approx 4.8348$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 4.8348)$. Keying the outer radical exponent (3.5 [√x] [yˣ]) lifts $4.8348$ to $Y$, forms $\sqrt{3.5} \approx 1.8708$ in $X$, and raises $Y^X$ to resolve $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{19.0703})$.

Workbook Problem 3.17: Acoustic Sound Intensity Level in Air

Human auditory perception spans twelve orders of magnitude in acoustic power flux—from the threshold of hearing $I_0 = 1.0 \times 10^{-12}\text{ W/m}^2$ ($0\text{ dB}$) to the threshold of pain ($1\text{ W/m}^2 = 120\text{ dB}$). For a pneumatic riveter radiating $I = 0.0035\text{ W/m}^2$:

\[L_I = 10 \log_{10}\left(\frac{I}{I_0}\right)\]

Keystrokes: 0.0035 [ENTER] 1 [E] 12 [+/-] [÷] [↰] [LOG] 10 [×] $\rightarrow$ 95.440680444$\text{ dB SIL}$.

Normalizing intensity against the picowatt-per-square-meter hearing threshold (0.0035 [ENTER] 1 [E] 12 [+/-] [÷]) places $I/I_0 = 3.5000 \times 10^9$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 3.5000 \times 10^9)$. Taking the base-10 logarithm and scaling from bels to decibels ([↰] [LOG] 10 [×]) yields $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{95.4407}\text{ dB SIL})$.

Workbook Problem 3.18: Bragg's X-Ray Crystal Diffraction Wavelength

When monochromatic X-rays strike a rock-salt ($\text{NaCl}$) crystal cleavage plane of interplanar lattice spacing $d = 0.282\text{ nm} = 2.82 \times 10^{-10}\text{ m}$ at grazing angle $\theta = 15.8^\circ$, constructive interference between adjacent atomic planes occurs when the path-length difference $2d\sin\theta$ equals an integer multiple of the wavelength ($n = 1$): $$\lambda = 2 d \sin\theta$$

Keystrokes: 2 [ENTER] 2.82 [E] 10 [+/-] [×] 15.8 [SIN] [×] $\rightarrow$ 1.535661E-10$\text{ m} = 0.1536\text{ nm}$.

Forming the maximum path difference (2 [ENTER] 2.82 [E] 10 [+/-] [×]) places $2 d = 5.6400 \times 10^{-10}\text{ m}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 5.6400 \times 10^{-10})$. Multiplying by the grazing sine projection (15.8 [SIN] [×]) resolves the Copper $K_\alpha$ X-ray wavelength $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{1.5357 \times 10^{-10}}\text{ m} = 0.1536\text{ nm})$.

Workbook Problem 3.19: Radiogenic Daughter Isotope Accumulation Kinetics

In rubidium-strontium ($^{87}\text{Rb} \to {}^{87}\text{Sr}$) geochronology, a mineral grain containing $P_0 = 1.0 \times 10^{15}$ parent $^{87}\text{Rb}$ atoms ($\lambda = 1.42 \times 10^{-11}\text{ yr}^{-1}$) accumulates stable radiogenic daughter atoms $D(t)$ over $t = 2.5 \times 10^9\text{ years}$ according to the complement of exponential decay: $$D(t) = P_0 \left(1 - e^{-\lambda t}\right)$$

Keystrokes: 1.42 [E] 11 [+/-] [ENTER] 2.5 [E] 9 [×] [+/-] [eˣ] 1 [-] [+/-] 1 [E] 15 [×] $\rightarrow$ 3.487727E13$\text{ daughter atoms}$.

Multiplying the decay constant by elapsed time (1.42 [E] 11 [+/-] [ENTER] 2.5 [E] 9 [×]) forms the dimensionless age $\lambda t = 0.0355$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0355)$. Evaluating the surviving parent fraction and taking its complement in place ([+/-] [eˣ] 1 [-] [+/-]) yields $1 - e^{-\lambda t} = 0.0349$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0349)$, and scaling by $P_0$ (1 [E] 15 [×]) resolves the radiogenic daughter inventory $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{3.4877 \times 10^{13}}\text{ atoms})$.

Workbook Problem 3.20: High-Speed Sprint: Radical Exponential Continued Fraction

Evaluate the composite transcendental expression:

\[x = \sqrt{e^{1.4} + \ln\left(15 + \sqrt{220}\right)}\]

Keystrokes: 220 [√x] 15 [+] [LN] 1.4 [eˣ] [+] [√x] $\rightarrow$ 2.7296144264

Starting at the deepest radical (220 [√x] 15 [+] [LN]) collapses $\ln(15 + \sqrt{220})$ to $3.3956$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 3.3956)$. Evaluating 1.4 [eˣ] [+] [√x] lifts $3.3956$ to $Y$, adds $e^{1.4} \approx 4.0552$, and extracts the outer square root $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{2.7296})$ without a single parenthesis.

Workbook Problem 3.21: Thermal Noise Power Density (Johnson-Nyquist Noise)

In 1928, John Bertrand Johnson and Harry Nyquist proved that thermal agitation of charge carriers inside any matched resistor or antenna at ambient temperature $T = 290.0\text{ K}$ ($k_B = 1.380649 \times 10^{-23}\text{ J/K}$) generates a white noise power spectral density independent of resistance:

\[P_n / B = k_B T \quad [\text{W/Hz}], \qquad P_{\text{dBm/Hz}} = 10 \log_{10}\left(\frac{k_B T}{10^{-3}\text{ W}}\right)\]

Keystrokes: 1.380649 [E] 23 [+/-] [ENTER] 290 [×] 0.001 [÷] [↰] [LOG] 10 [×] $\rightarrow$ -173.9751872$\text{ dBm/Hz}$.

Evaluating $k_B T$ and referencing to $1\text{ mW}$ (1.380649 [E] 23 [+/-] [ENTER] 290 [×] 0.001 [÷]) places $4.0039 \times 10^{-18}\text{ mW/Hz}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 4.0039 \times 10^{-18})$. Taking the base-10 logarithm and multiplying by $10$ ([↰] [LOG] 10 [×]) establishes the universal room-temperature RF receiver noise floor $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{-173.9752}\text{ dBm/Hz})$—the iconic "$-174\text{ dBm/Hz}$" benchmark of satellite and cellular link budgets.

Workbook Problem 3.22: High-Speed Sprint: Chained Logarithmic Power Tower

Evaluate the logarithmic and radical expression:

\[x = \frac{\ln(4.5^3) + \log_{10}(1200)}{\sqrt[3]{18.5}}\]

Keystrokes: 4.5 [LN] 3 [×] 1200 [↰] [LOG] [+] 18.5 [ENTER] 3 [↰] [ˣ√y] [÷] $\rightarrow$ 2.8703315724

Applying the logarithmic power identity $\ln(4.5^3) = 3\ln(4.5)$ (4.5 [LN] 3 [×] 1200 [↰] [LOG] [+]) forms the numerator $7.5914$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 7.5914)$, and 18.5 [ENTER] 3 [↰] [ˣ√y] [÷] divides by $\sqrt[3]{18.5} \approx 2.6447$ to resolve $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{2.8703})$.

Chapter 04: Trigonometry, Coordinate Systems & 2D Vectors — Extended Engineering & Competition Problems

Accompanies Chapter 04 of Thinking in the Stack (29 Worked Engineering Problems)

Workbook Problem 4.1: Oblique Triangle Law of Sines Angle Determination

In triangle with $a = 45.0\text{ m}$, $b = 62.0\text{ m}$, angle $A = 32.5^\circ$. Find angle $B$.

\[\frac{\sin B}{b} = \frac{\sin A}{a} \implies B = \arcsin\left(\frac{b \sin A}{a}\right)\]

Keystrokes: 62 [ENTER] 32.5 [SIN] [×] 45 [÷] [↰] [ASIN]$\to \mathbf{47.7812^\circ}$.

In geodetic triangulation, the ratio $a/\sin A = 2R_{\text{circ}}$ equals the diameter of the triangle's circumscribing circle. Entering side $b = 62.0\text{ m}$ and multiplying by the sine projection of angle $A = 32.5^\circ$ (62 [ENTER] 32.5 [SIN] [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 33.3126$) forms the altitude projection $b\sin A$; dividing by opposing side $a = 45.0\text{ m}$ (45 [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.7403$) yields the dimensionless chord ratio $\sin B = 0.7403$. Extracting the arcsine directly in place ([↰] [ASIN], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 47.7552$) resolves the acute interior vertex angle $B = \mathbf{47.76^\circ}$—and if you also need the obtuse ambiguous-case supplement $B' = 180^\circ - B = 132.24^\circ$, simply key 180 [x⇄y] [-].

Workbook Problem 4.2: Static Equilibrium of Crane Boom and Cable

Crane boom weight $W = 28.0\text{ kN}$ at angle $\theta = 35.0^\circ$ from horizontal, supported by tie cable at angle $\phi = 25.0^\circ$ to boom. Cable tension: $$T = \frac{W \cos\theta}{\sin\phi} = \frac{28.0 \cos(35^\circ)}{\sin(25^\circ)}$$

Keystrokes: 28 [ENTER] 35 [COS] [×] 25 [SIN] [÷]$\to \mathbf{54.2694\text{ kN}}$.

Rotational equilibrium about the boom pivot balances the overturning gravitational moment $W L \cos\theta$ against the restoring cable moment $T L \sin\phi$, where the boom length $L$ cancels out. Loading boom weight $W = 28.0\text{ kN}$ and scaling by the inclination cosine (28 [ENTER] 35 [COS] [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 22.9363$) isolates the perpendicular gravity force; dividing by the shallow tie-angle sine $\sin(25^\circ) \approx 0.4226$ (25 [SIN] [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 54.2718$) reveals how the shallow $25^\circ$ attachment geometry amplifies guy-cable tension to nearly twice the boom weight, $T \approx \mathbf{54.27\text{ kN}}$.

Workbook Problem 4.3: Snell's Law of Optical Refraction

Light ray incident from air ($n_1 = 1.000$) into dense flint glass ($n_2 = 1.660$) at incidence angle $\theta_1 = 55.0^\circ$. Find refracted angle $\theta_2$: $$\sin\theta_2 = \frac{n_1 \sin\theta_1}{n_2} \implies \theta_2 = \arcsin\left(\frac{\sin(55^\circ)}{1.660}\right)$$

Keystrokes: 55 [SIN] 1.660 [÷] [↰] [ASIN]$\to \mathbf{29.5638^\circ}$.

By Fermat's principle of least time, the tangential phase-velocity projection $n \sin\theta$ is conserved across any planar dielectric boundary. Evaluating the incident optical invariant (55 [SIN], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.8192$), dividing by the dense flint glass index $n_2 = 1.660$ (1.660 [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.4935$), and extracting the inverse sine ([↰] [ASIN], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 29.5686$) bends the ray toward the surface normal to $\theta_2 \approx \mathbf{29.57^\circ}$. Because [↰] [ASIN] preserves $\sin\theta_2 = 0.4935$ in LASTX, you can immediately compute the angular deviation $\Delta\theta = \theta_1 - \theta_2 = 25.43^\circ$ or recall the refractive index $1.660$ earlier in the chain.

Workbook Problem 4.4: Traverse Linear Misclosure Precision Ratio

A traverse boundary has total perimeter $P = 1,450.0\text{ m}$. Sum of departures $\Sigma x = +0.038\text{ m}$, sum of latitudes $\Sigma y = -0.024\text{ m}$. Calculate closure precision $1:K$: $$\text{Misclosure} = \sqrt{(\Sigma x)^2 + (\Sigma y)^2}, \quad K = \frac{P}{\text{Misclosure}}$$

Keystrokes: 0.038 [↰] [x²] 0.024 [↰] [x²] [+] [√x] 1450 [x⇄y] [÷] $\rightarrow$ 32,262.07656

When a closed boundary survey returns to its starting benchmark, accumulated angular and EDM range errors manifest as a residual displacement vector $(\Sigma x, \Sigma y)$. Computing the Euclidean misclosure (0.038 [↰] [x²] 0.024 [↰] [x²] [+] [√x], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0449$, or even faster via 0.024 [ENTER] 0.038 [↰] [▸θ,r]) isolates the $44.9\text{ mm}$ error vector. Entering total perimeter $P = 1,450\text{ m}$, swapping registers, and dividing (1450 [x⇄y] [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 32262.0766$) establishes a relative closure precision of $1:32,262$, comfortably surpassing first-order geodetic survey standards ($1:25,000$).

Workbook Problem 4.5: Torque Vector Cross Product Magnitude

Lever arm $r = 0.850\text{ m}$, applied force $F = 240.0\text{ N}$ at angle $\theta = 65.0^\circ$.

\[\tau = r \cdot F \cdot \sin\theta\]

Keystrokes: 0.850 [ENTER] 240 [×] 65 [SIN] [×]$\to \mathbf{184.8872\text{ N}\cdot\text{m}}$.

Only the tangential force component $F\sin\theta$ perpendicular to the wrench lever arm $\vec{r}$ generates rotational moment $|\vec{r} \times \vec{F}|$; the radial component $F\cos\theta$ merely compresses the bolt shaft. Multiplying lever arm by applied force (0.850 [ENTER] 240 [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 204.0000$) gives the maximum orthogonal torque $204.0\text{ N}\cdot\text{m}$, and scaling by $\sin(65^\circ)$ (65 [SIN] [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 184.8868$) yields the effective tightening torque $\tau \approx \mathbf{184.89\text{ N}\cdot\text{m}}$.

Workbook Problem 4.6: Double-Slit Optical Interference Fringe Spacing

Slit separation $d = 0.250\text{ mm} = 2.5 \times 10^{-4}\text{ m}$, laser wavelength $\lambda = 632.8\text{ nm} = 6.328 \times 10^{-7}\text{ m}$, screen distance $L = 1.80\text{ m}$. Find fringe spacing $\Delta y$: $$\Delta y = \frac{\lambda L}{d}$$

Keystrokes: 6.328 [E] 7 [+/-] [ENTER] 1.80 [×] 2.5 [E] 4 [+/-] [÷] 1000 [×]$\to \mathbf{4.55616\text{ mm}}$.

In Thomas Young's double-slit interferometer, constructive interference maxima occur where the optical path difference $d\sin\theta = m\lambda$; in the paraxial far-field regime ($\sin\theta \approx \tan\theta = y/L$), adjacent bright fringes are spaced uniformly by $\Delta y = \lambda L / d$. Evaluating the He-Ne laser diffraction product $\lambda L$ (6.328 [E] 7 [+/-] [ENTER] 1.80 [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.1390 \times 10^{-6}$), dividing by slit pitch $d = 2.5 \times 10^{-4}\text{ m}$ (2.5 [E] 4 [+/-] [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0046$), and scaling to millimeters (1000 [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 4.5562$) resolves fringe spacing $\Delta y \approx \mathbf{4.56\text{ mm}}$.

Workbook Problem 4.7: Vector Sum of Three Concurrent Forces via `[Σ+]`

Three coplanar cable forces on a gusset plate: * $F_1 = 50.0\text{ kN}$ at $\theta_1 = 30.0^\circ$ * $F_2 = 75.0\text{ kN}$ at $\theta_2 = 110.0^\circ$ * $F_3 = 60.0\text{ kN}$ at $\theta_3 = 225.0^\circ$

Keystrokes: 1. Clear stats: [CLΣ] 2. Vector 1: 30 [ENTER] 50 [↱] [▸y,x] [Σ+] 3. Vector 2: 110 [ENTER] 75 [↱] [▸y,x] [Σ+] 4. Vector 3: 225 [ENTER] 60 [↱] [▸y,x] [Σ+] 5. Recall sums and resolve: [↱] [SUMS] [Σy] $\to \Sigma y = 53.0506\text{ kN}$ [↱] [SUMS] [Σx] $\to \Sigma x = -24.7731\text{ kN}$ (lifts $\Sigma y$ to Register $Y$) [↰] [▸θ,r]

This workflow is one of the crowning architectural idioms of the SC-32: because [↱] [▸y,x] deposits $(Y: r\sin\theta, X: r\cos\theta)$ and [Σ+] simultaneously accumulates Register $X$ into $\Sigma x$ and Register $Y$ into $\Sigma y$, you can sum an arbitrary number of 2D vectors without consuming the 4-level stack! Clearing statistical registers ([CLΣ]) and piping each polar cable force through [↱] [▸y,x] [Σ+] accumulates the three gusset loads into hardware memory; recalling the Cartesian totals via the single-scalar SUMS softkeys ([↱] [SUMS] [Σy] $= 53.0506\text{ kN}$, then [↱] [SUMS] [Σx] $= -24.7731\text{ kN}$ which lifts $\Sigma y$ to Register $Y$) and converting back to polar ([↰] [▸θ,r], $T: 0.0000, Z: 0.0000, Y: 115.01^\circ, X: 58.5444\text{ kN}$) delivers resultant magnitude $R = \mathbf{58.54\text{ kN}}$ at bearing $\theta = \mathbf{115.01^\circ}$.

Workbook Problem 4.8: Solar Zenith Angle and Air Mass Coefficient

Sun at elevation $\alpha = 42.0^\circ$. Solar zenith angle $\theta_z = 90^\circ - 42^\circ = 48.0^\circ$.

\[\text{Air Mass } AM = \frac{1}{\cos\theta_z}\]

Keystrokes: 48 [COS] [1/x]$\to \mathbf{1.4945} \approx AM\,1.5$.

Solar photovoltaic modules are rated worldwide under the ASTM G173 AM1.5 terrestrial spectrum—the optical path length of sunlight through a planar atmosphere when the sun sits at zenith angle $\theta_z = 48.19^\circ$ ($\sec 48.19^\circ = 1.50$). Evaluating the zenith cosine (48 [COS], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.6691$) and inverting in place ([1/x], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.4945$) yields the secant optical path $\text{AM} \approx \mathbf{1.49}$, confirming nearly exact AM1.5 standard irradiance conditions.

Workbook Problem 4.9: Catenary Hanging Cable Tension at Support Tower

Cable unit weight $w = 12.5\text{ N/m}$, parameter $a = 600.0\text{ m}$, span $L = 400.0\text{ m}$ (half-span $x = 200.0\text{ m}$).

\[T_0 = w \cdot a = 12.5 \times 600 = 7500.0\text{ N}$$ $$T = T_0 \cosh\left(\frac{x}{a}\right) = 7500 \cosh\left(\frac{200}{600}\right) = 7500 \cosh(0.333333)\]

Keystrokes: 200 [ENTER] 600 [÷] [↰] [HYP] [COS] 7500 [×]$\to \mathbf{7921.28\text{ N}} \approx 7.92\text{ kN}$.

Unlike a suspension bridge cable loaded by a flat horizontal roadway (which hangs in a parabola), an unencumbered power line supporting its own distributed weight $w$ hangs in a hyperbolic cosine catenary $y = a\cosh(x/a)$, where horizontal tension $T_0 = wa$ is constant and total tension grows as $T(x) = T_0 \cosh(x/a) = T_0 + w s_{\text{sag}}$. Forming the dimensionless half-span ratio $x/a$ (200 [ENTER] 600 [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.3333$), evaluating the hyperbolic cosine ([↰] [HYP] [COS], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.0561$), and scaling by $T_0 = 7,500\text{ N}$ (7500 [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 7920.5390$) yields tower anchor tension $T \approx \mathbf{7,920.54\text{ N}}$ ($7.92\text{ kN}$).

Workbook Problem 4.10: Polar Conic Equation for Elliptical Satellite Orbit

Semi-latus rectum $p = 7,200.0\text{ km}$, orbital eccentricity $e = 0.150$. Find orbital radius at true anomaly $\theta = 60.0^\circ$.

\[r = \frac{p}{1 + e \cos\theta}\]

Keystrokes: 60 [COS] 0.150 [×] 1 [+] 7200 [x⇄y] [÷]$\to \mathbf{6697.67\text{ km}}$.

Kepler's first law expresses every bound two-body trajectory as a polar conic section $r(\theta) = p / (1 + e\cos\theta)$ measured from perigee ($\theta = 0^\circ$). Evaluating the denominator modulation (60 [COS] 0.150 [×] 1 [+], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.0750$), loading semi-latus rectum $p = 7,200\text{ km}$, and dividing via 7200 [x⇄y] [÷] (or [1/x] 7200 [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 6697.6744$) determines the instantaneous geocentric radius $r = \mathbf{6,697.67\text{ km}}$.

Workbook Problem 4.11: Nautical Waypoint Bearing and Distance

Course from Station A ($x_1=0, y_1=0$) to Station B ($\Delta x = 45.2\text{ nmi}$ East, $\Delta y = 78.5\text{ nmi}$ North).

Keystrokes: 78.5 [ENTER] 45.2 [↰] [▸θ,r] $\rightarrow$ 90.583055811

Notice the elegant duality between Cartesian math angles (measured counter-clockwise from East, $\theta_{\text{math}} = \text{atan2}(\Delta y, \Delta x)$) and nautical compass bearings (measured clockwise from True North, $\text{Az} = \text{atan2}(\Delta x, \Delta y)$). Loading northing $\Delta y = 78.5$ into $Y$ and easting $\Delta x = 45.2$ into $X$ and pressing [↰] [▸θ,r] (78.5 [ENTER] 45.2 [↰] [▸θ,r], $T: 0.0000, Z: 0.0000, Y: 60.0668^\circ, X: 90.5831\text{ nmi}$) delivers rhumb-line distance $r = \mathbf{90.58\text{ nmi}}$ in $X$ and Cartesian angle $60.07^\circ$ in $Y$ (nautical bearing $90^\circ - 60.07^\circ = \mathbf{029.93^\circ}$)—and if you instead enter Easting into $Y$ and Northing into $X$ (45.2 [ENTER] 78.5 [↰] [▸θ,r]), Register $Y$ outputs the true compass azimuth $\mathbf{029.93^\circ}$ directly!

Workbook Problem 4.12: Circular Arch Chord Length and Segment Arc Length

Circular arch of radius $R = 25.0\text{ m}$ spanning central angle $\theta = 70.0^\circ$.

\[\text{Chord } c = 2 R \sin\left(\frac{\theta}{2}\right), \quad \text{Arc } s = R \frac{\theta \pi}{180}\]

Keystrokes for Chord: 25 [ENTER] 2 [×] 35 [SIN] [×]$\to \mathbf{28.6788\text{ m}}$

Historically, ancient astronomers tabulated the chord function $\text{crd}(\theta) = 2R\sin(\theta/2)$ before Indian mathematicians introduced the half-chord jyā-ardha ($\sin(\theta/2)$). Computing the arch diameter (25 [ENTER] 2 [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 50.0000$) and multiplying by the half-angle sine (35 [SIN] [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 28.6788$) resolves the horizontal abutment span $c \approx \mathbf{28.68\text{ m}}$, while the curved masonry extrados arc $s = 25 \times 70\pi/180 = 30.54\text{ m}$ exceeds the chord by $1.86\text{ m}$.

Workbook Problem 4.13: Brewster's Angle of Polarization

Laser beam incident on crown glass window ($n = 1.520$ in air).

\[\theta_B = \arctan(n)\]

Keystrokes: 1.520 [↰] [ATAN]$\to \mathbf{56.6593^\circ}$.

At Brewster's angle $\theta_B + \theta_2 = 90^\circ$, the oscillating dipole electrons inside the dielectric glass point directly along the reflected ray vector; because Hertzian dipoles cannot radiate along their own oscillation axis, $p$-polarized reflection vanishes completely ($R_p = 0$). Since Snell's law gives $n\sin\theta_B = \sin(90^\circ - \theta_B) = \cos\theta_B$, entering $n = 1.520$ and pressing [↰] [ATAN] ($T: 0.0000, Z: 0.0000, Y: 0.0000, X: 56.6593$) directly yields the lossless laser-window incidence angle $\theta_B \approx \mathbf{56.66^\circ}$.

Workbook Problem 4.14: Aircraft Coordinated Turn Radius and G-Load

Aircraft turning at $v = 150.0\text{ kts} = 77.1667\text{ m/s}$ at bank angle $\phi = 45.0^\circ$ ($g = 9.80665\text{ m/s}^2$).

\[R = \frac{v^2}{g \tan\phi}, \quad \text{Load Factor } n = \frac{1}{\cos\phi}\]

Keystrokes for Radius: 77.1667 [↰] [x²] 9.80665 [ENTER] 45 [TAN] [×] [÷]$\to \mathbf{607.21\text{ meters}}$

In a coordinated level turn, aerodynamic lift $L$ must tilt by bank angle $\phi$ so that $L\cos\phi = mg$ supports aircraft weight while $L\sin\phi = mv^2/R$ turns the flight path vector. Squaring true airspeed (77.1667 [↰] [x²], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 5954.6996$) and dividing by $g\tan(45^\circ)$ (9.80665 [ENTER] 45 [TAN] [×] [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 607.2104$) yields turn radius $R \approx \mathbf{607.21\text{ m}}$, while the structural load factor $n = L/(mg) = \sec(45^\circ) = \sqrt{2} \approx \mathbf{1.41\,G}$ increases stall speed by $\sqrt{n} \approx 19\%$.

Workbook Problem 4.15: Roof Truss Member Forces via Method of Joints

Ridge joint with vertical downward roof load $P = 15.0\text{ kN}$ balanced by two symmetrical rafters inclined at $\theta = 30.0^\circ$ to horizontal.

\[2 F_{\text{rafter}} \sin\theta = P \implies F_{\text{rafter}} = \frac{P}{2 \sin\theta}\]

Keystrokes: 15 [ENTER] 2 [ENTER] 30 [SIN] [×] [÷]$\to \mathbf{15.0000\text{ kN}}$ compression.

At a symmetrical gable ridge pin, each rafter carries half the vertical snow load $P/2 = F_{\text{rafter}}\sin\theta$; because $\sin(30^\circ) = 0.5000$, the vertical projection factor $2\sin(30^\circ)$ equals unity! Loading the ridge point load $P = 15.0\text{ kN}$ (15 [ENTER], $T: 0.0000, Z: 0.0000, Y: 15.0000, X: 15.0000$) and dividing by $2\sin(30^\circ)$ (2 [ENTER] 30 [SIN] [×] [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 15.0000$) confirms axial rafter compression $F_{\text{rafter}} = \mathbf{15.00\text{ kN}}$ (and keying 30 [COS] [×] immediately gives the $12.99\text{ kN}$ horizontal collar-tie tension).

Workbook Problem 4.16: Helical Compression Spring Wahl Stress Factor

Spring index $C = D / d = 7.50$.

\[K_W = \frac{4C - 1}{4C - 4} + \frac{0.615}{C}\]

Keystrokes: 4 [ENTER] 7.5 [×] 1 [-] 4 [ENTER] 7.5 [×] 4 [-] [÷] 0.615 [ENTER] 7.5 [÷] [+] $\rightarrow$ 1.1973846154

In a coiled helical spring, the inner wire fibers have a shorter radius of curvature than the coil centerline and carry both torsional shear and direct transverse shear, elevating inner-wall shear stress by A. M. Wahl's correction factor $K_W$. Computing the curvature numerator $4C - 1 = 29.0$ (4 [ENTER] 7.5 [×] 1 [-], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 29.0000$), dividing by $(4C - 4) = 26.0$ (4 [ENTER] 7.5 [×] 4 [-] [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.1154$), and adding the transverse shear term $0.615/C$ (0.615 [ENTER] 7.5 [÷] [+], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.1974$) establishes $K_W = \mathbf{1.1974}$ (nearly $20\%$ higher stress than straight-bar torsion).

Workbook Problem 4.17: Coriolis Acceleration Vector on Artillery Projectile

Projectile traveling North at $v = 820.0\text{ m/s}$ at latitude $\phi = 42.0^\circ\text{ N}$. Earth's angular rotation rate $\Omega = 7.2921 \times 10^{-5}\text{ rad/s}$: $$a_{\text{coriolis}} = 2 \cdot v \cdot \Omega \cdot \sin\phi$$

Keystrokes: 2 [ENTER] 820 [×] 7.2921 [E] 5 [+/-] [×] 42 [SIN] [×] $\to \mathbf{0.0800216236\text{ m/s}^2}$ Eastward.

In a rotating planetary reference frame, a projectile moving with velocity $\vec{v}$ experiences apparent Coriolis acceleration $\vec{a}_c = -2(\vec{\Omega} \times \vec{v})$, where the local vertical component of Earth's spin vector $\Omega\sin\phi$ deflects Northern Hemisphere trajectories to the right. Forming $2v$ (2 [ENTER] 820 [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1640.0000$), multiplying by sidereal spin rate $\Omega$ (7.2921 [E] 5 [+/-] [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.1196$), and scaling by $\sin(42^\circ)$ (42 [SIN] [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0800$) yields $a_{\text{coriolis}} \approx \mathbf{0.0800\text{ m/s}^2}$ Eastward—enough to deflect a $30\text{-second}$ artillery shell by $\frac{1}{2}a_c t^2 = 36\text{ meters}$!

Workbook Problem 4.18: Involute Gear Tooth Base Pitch and Pressure Angle

Module $m = 4.0\text{ mm}$, tooth count $N = 28$, standard pressure angle $\phi = 20.0^\circ$. Calculate pitch circle diameter $D$, base circle diameter $D_b$, and circular base pitch $p_b$: $$D = m \cdot N, \quad D_b = D \cos\phi, \quad p_b = \pi \cdot m \cos\phi$$

Keystrokes: 4 [ENTER] 28 [×]$\to D = \mathbf{112.00\text{ mm}}$

By the fundamental law of conjugate gear-tooth action, the common normal at the contact point of two involute profiles always lies tangent to both base circles $D_b = D\cos\phi$ at constant pressure angle $\phi = 20.0^\circ$. Multiplying module by tooth count ($T: 0.0000, Z: 0.0000, Y: 0.0000, X: 112.0000$) yields pitch diameter $D = \mathbf{112.00\text{ mm}}$; scaling by 20 [COS] [×] gives the involute generating base circle $D_b = \mathbf{105.25\text{ mm}}$, and dividing by 28 [÷] [↱] [π] [×] yields base pitch $p_b = \mathbf{11.81\text{ mm}}$.

Workbook Problem 4.19: Supersonic Oblique Shock Wave Angle

Supersonic flight at Mach $M_1 = 2.50$ over a compression wedge with deflection angle $\theta = 15.0^\circ$. Determine shock angle $\beta$ (approximate weak shock solution $\beta \approx 36.94^\circ$): $$\tan\theta = 2 \cot\beta \left[\frac{M_1^2 \sin^2\beta - 1}{M_1^2 (\gamma + \cos(2\beta)) + 2}\right]$$

Across an oblique shock inclined at wave angle $\beta$, tangential velocity is conserved while the normal Mach component $M_{n1} = M_1 \sin\beta$ must exceed unity ($M_{n1} > 1$) to compress the flow. Evaluating the normal Mach number for $\beta = 36.94^\circ$ (2.50 [ENTER] 36.94 [SIN] [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.5024$) and forming the excess normal kinetic energy factor $M_{n1}^2 - 1$ ([↰] [x²] 1 [-], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.2573$) yields the $\theta$-$\beta$-$M$ numerator delta $1.2573$.

Workbook Problem 4.20: Wind Turbine Blade Angle of Attack and Relative Velocity

Blade station radius $r = 32.0\text{ m}$, rotational speed $\Omega = 1.80\text{ rad/s}$ ($u_{\text{rot}} = \Omega r = 57.6\text{ m/s}$), free-stream wind speed $v_{\text{wind}} = 11.5\text{ m/s}$ with axial induction factor $a = 0.30$ ($v_{\text{axial}} = 11.5(1 - 0.30) = 8.05\text{ m/s}$): $$W_{\text{rel}} = \sqrt{u_{\text{rot}}^2 + v_{\text{axial}}^2}, \quad \phi_{\text{flow}} = \arctan\left(\frac{v_{\text{axial}}}{u_{\text{rot}}}\right)$$

Keystrokes: 8.05 [ENTER] 57.6 [↰] [▸θ,r] $\rightarrow$ 58.15980141

In blade-element momentum (BEM) theory, an airfoil section at radius $r = 32\text{ m}$ sees the vector sum of the Betz-decelerated axial wind $v_{\text{axial}} = 8.05\text{ m/s}$ and the much faster tangential sweep velocity $u_{\text{rot}} = \Omega r = 57.6\text{ m/s}$. Loading $(Y: v_{\text{axial}}, X: u_{\text{rot}})$ and pressing [↰] [▸θ,r] (8.05 [ENTER] 57.6 [↰] [▸θ,r], $T: 0.0000, Z: 0.0000, Y: 7.9560^\circ, X: 58.1598\text{ m/s}$) resolves both the relative aerodynamic inflow speed $W_{\text{rel}} \approx \mathbf{58.16\text{ m/s}}$ and the inflow pitch angle $\phi_{\text{flow}} \approx \mathbf{7.96^\circ}$ in a single keystroke.

Workbook Problem 4.21: Surveying Triangulation Intersection from Two Stations

Baseline between Stations $A$ and $B$ is $b = 1,250.0\text{ m}$. Angles observed to unknown landmark $P$: $\angle A = 58.4^\circ$, $\angle B = 71.2^\circ$. Angle at $P$: $\angle P = 180^\circ - (58.4^\circ + 71.2^\circ) = 50.4^\circ$: $$\text{Distance } AP = b \frac{\sin B}{\sin P} = 1250 \frac{\sin(71.2^\circ)}{\sin(50.4^\circ)}$$

Keystrokes: 1250 [ENTER] 71.2 [SIN] [×] 50.4 [SIN] [÷]$\to \mathbf{1535.84\text{ meters}}$.

Before electronic distance measurement, continental geodetic networks were extended across mountain ranges by measuring a single taped baseline $b$ in a flat valley and sighting horizontal theodolite angles to distant peaks. Scaling baseline $b = 1,250\text{ m}$ by the opposing angle sine (1250 [ENTER] 71.2 [SIN] [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1183.3116$) and dividing by the vertex parallax sine $\sin(50.4^\circ)$ (50.4 [SIN] [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1535.7446$) resolves range $AP \approx \mathbf{1,535.74\text{ m}}$—and if you also need range $BP$, 1250 [ENTER] 58.4 [SIN] [×] 50.4 [SIN] [÷] completes the fix.

Workbook Problem 4.22: Celestial Coordinate Conversion (RA/Dec to Alt/Az)

Observer at latitude $\phi = 38.0^\circ\text{ N}$. Star with declination $\delta = 22.5^\circ$ and local hour angle $H = 40.0^\circ$.

\[\sin a = \sin\phi \sin\delta + \cos\phi \cos\delta \cos H\]

Keystrokes: 38 [SIN] 22.5 [SIN] [×] 38 [COS] 22.5 [COS] [×] 40 [COS] [×] [+] [↰] [ASIN]

Transforming equatorial star catalog coordinates $(\delta, H)$ into local horizon altitude $a$ for an alt-azimuth telescope mount solves the same spherical cosine law used in marine sextant sight reduction. Computing the polar sine product (38 [SIN] 22.5 [SIN] [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.2356$), adding the meridian hour-angle cosine term (38 [COS] 22.5 [COS] [×] 40 [COS] [×] [+], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.7933$), and extracting the inverse sine ([↰] [ASIN], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 52.4954$) yields horizon altitude $a \approx \mathbf{52.50^\circ}$ (zenith distance $z = 90^\circ - a = 37.50^\circ$).

Workbook Problem 4.23: Gyroscopic Precession Angular Rate

Rotor mass moment of inertia $I_s = 0.0125\text{ kg}\cdot\text{m}^2$, spin speed $\omega_s = 24,000\text{ rpm} = 2513.27\text{ rad/s}$. Applied disturbance torque $\tau = 0.450\text{ N}\cdot\text{m}$: $$\Omega_p = \frac{\tau}{I_s \cdot \omega_s}$$

Keystrokes: 0.450 [ENTER] 0.0125 [ENTER] 24000 [ENTER] 2 [×] [↱] [π] [×] 60 [÷] [×] [÷]$\to \mathbf{0.014324\text{ rad/s}} = 0.8207^\circ/\text{s}$.

Because torque equals the time rate of change of angular momentum ($\vec{\tau} = d\vec{L}/dt = \vec{\Omega}_p \times \vec{L}$), a high-speed gyroscope rotor responds to an orthogonal torque $\tau$ by precessing perpendicular to both the spin and torque axes at rate $\Omega_p = \tau / (I_s \omega_s)$. Loading disturbance torque (0.450 [ENTER], $T: 0.0000, Z: 0.0000, Y: 0.4500, X: 0.4500$) and dividing by the spin angular momentum $L = I_s \omega_s = 31.4159\text{ N}\cdot\text{m}\cdot\text{s}$ (0.0125 [ENTER] 24000 [ENTER] 2 [×] [↱] [π] [×] 60 [÷] [×] [÷], $T: 0.4500, Z: 0.4500, Y: 0.4500, X: 0.0143$) yields precession rate $\Omega_p \approx \mathbf{0.0143\text{ rad/s}}$ ($0.82^\circ/\text{s}$).

Workbook Problem 4.24: Quarter-Wave Plate Retardance and Jones Matrix

Ellipticity angle $\chi$ produced by linear polarizer at angle $\theta = 22.5^\circ$ through quarter-wave plate.

$$\sin(2\chi) = \sin(2\theta) \sin\delta$$ For a quarter-wave plate ($\delta = 90.0^\circ$), $\sin\delta = 1.0$: $$\chi = \frac{1}{2} \arcsin(\sin(45^\circ)) = \frac{45^\circ}{2} = \mathbf{22.50^\circ}$$.

On the Poincaré polarization sphere, a birefringent retarder with phase delay $\delta$ rotates the Stokes polarization vector about the fast-axis equator by angle $\delta$, mapping azimuth $2\theta$ to latitude $2\chi$ via $\sin(2\chi) = \sin(2\theta)\sin\delta$. Doubling the incident polarizer angle (22.5 [ENTER] 2 [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 45.0000$), applying the quarter-wave round-trip ([SIN] [↰] [ASIN], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 45.0000$), and halving (2 [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 22.5000$) confirms ellipticity angle $\chi = \mathbf{22.50^\circ}$ ($\tan\chi = 0.4142$ minor-to-major field ratio).

Workbook Problem 4.25: Acoustic Directivity Index of Circular Piston Speaker

Loudspeaker piston radius $a = 0.10\text{ m}$ radiating at frequency $f = 2,500\text{ Hz}$ in air ($c = 343\text{ m/s}$):

\[k = \frac{2\pi f}{c} = \frac{2\pi(2500)}{343} \approx 45.7958\text{ m}^{-1}$$ $$k a = 45.7958 \times 0.10 = 4.5796$$ Directivity index for $k a \gg 1$: $$D_I \approx 10 \log_{10}(ka)^2 = 20 \log_{10}(ka)\]

Keystrokes: 4.5796 [↰] [LOG] 20 [×]$\to \mathbf{13.2168\text{ dB}}$.

When the acoustic Helmholtz number $ka = 2\pi a/\lambda$ exceeds unity, a baffled circular loudspeaker cone transitions from an omnidirectional monopole into a directional Airy-Bessel beam whose on-axis intensity gain scales as $(ka)^2$. Entering $ka = 4.5796$, taking the common logarithm, and scaling by $20$ ($T: 0.0000, Z: 0.0000, Y: 0.0000, X: 13.2166$) yields an on-axis directivity index $D_I \approx \mathbf{13.22\text{ dB}}$.

Workbook Problem 4.26: Ballistic Reentry Trajectory Flight Path Angle

Atmospheric entry interface velocity $v_E = 7,800\text{ m/s}$, radial velocity $v_r = -210\text{ m/s}$, tangential speed $v_\theta = \sqrt{7800^2 - 210^2} \approx 7797.17\text{ m/s}$. Find flight path entry angle $\gamma$: $$\sin\gamma = \frac{v_r}{v_E} \implies \gamma = \arcsin\left(\frac{-210}{7800}\right)$$

Keystrokes: 210 [+/-] [ENTER] 7800 [÷] [↰] [ASIN]$\to \mathbf{-1.5428^\circ}$.

Spacecraft atmospheric entry at $120\text{ km}$ altitude requires threading a narrow flight-path corridor: too shallow and the capsule skips back into space; too steep and peak deceleration exceeds crew structural limits. Forming the sink-rate ratio $v_r/v_E$ (210 [+/-] [ENTER] 7800 [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: -0.0269$) and extracting the arcsine ([↰] [ASIN], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: -1.5428$) establishes the shallow reentry corridor angle $\gamma \approx \mathbf{-1.54^\circ}$ below the local horizontal.

Workbook Problem 4.27: Circular Roadway Curve Sight Distance

Curve radius $R = 250.0\text{ m}$, obstacle setback distance from centerline $m = 8.5\text{ m}$. Calculate available stopping sight distance $S$:

\[m = R \left(1 - \cos\left(\frac{28.65^\circ S}{R}\right)\right) \implies S = \frac{R}{28.65^\circ} \arccos\left(1 - \frac{m}{R}\right)\]

Keystrokes: 8.5 [ENTER] 250 [÷] [+/-] 1 [+] [↰] [ACOS] 250 [×] 28.65 [÷]$\to \mathbf{130.74299272\text{ meters}}$.

In highway geometric design, trees or bridge abutments on the inside of a horizontal curve limit the driver's line-of-sight chord; the middle ordinate setback $m = R(1 - \cos\frac{\theta}{2})$ is the classical versine of the half-central angle, where $28.65^\circ \approx \frac{90^\circ}{\pi}$ converts radians to half-angle degrees. Evaluating the inverse-versine argument $1 - m/R$ and taking its arccosine (8.5 [ENTER] 250 [÷] [+/-] 1 [+] [↰] [ACOS], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 14.9836$), then scaling by $R/28.65^\circ$ (250 [×] 28.65 [÷]—or recalling $R = 250$ from LASTX earlier—$T: 0.0000, Z: 0.0000, Y: 0.0000, X: 130.7467$) determines stopping sight distance $S \approx \mathbf{130.75\text{ m}}$.

Workbook Problem 4.28: Resection Surveying Point Fix (Kaestner Method)

Two known landmarks $L_1 (0, 1000)$ and $L_2 (800, 600)$. Measured subtended angle $\alpha = 42.0^\circ$. Baseline length $b$:

\[b = \sqrt{800^2 + (1000 - 600)^2} = \sqrt{640000 + 160000} = \sqrt{800000} = 894.4272\text{ m}$$ Radius of circumscribing position circle: $$R_{\text{circ}} = \frac{b}{2 \sin\alpha} = \frac{894.4272}{2 \sin(42^\circ)}\]

Keystrokes: 800000 [√x] 2 [÷] 42 [SIN] [÷]$\to \mathbf{668.35\text{ meters}}$.

By Euclid's inscribed angle theorem, all observer positions that sight a fixed subtended angle $\alpha$ between two landmarks $L_1$ and $L_2$ lie on a circular arc of radius $R_{\text{circ}} = b / (2\sin\alpha)$ (the Collins auxiliary circle used in three-point resection). Computing the half-baseline chord $b/2$ (800000 [√x] 2 [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 447.2136$) and dividing by $\sin(42^\circ)$ (42 [SIN] [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 668.3502$) resolves the position-circle radius $R_{\text{circ}} \approx \mathbf{668.35\text{ m}}$.

Workbook Problem 4.29: Elliptical Polarization Axial Ratio from Phasors

Major ellipse axis $E_{\max} = 12.4\text{ V/m}$, minor axis $E_{\min} = 4.2\text{ V/m}$. Calculate axial ratio in decibels:

\[\text{AR}_{\text{dB}} = 20 \log_{10}\left(\frac{E_{\max}}{E_{\min}}\right)\]

Keystrokes: 12.4 [ENTER] 4.2 [÷] [↰] [LOG] 20 [×]$\to \mathbf{9.4036\text{ dB}}$.

Satellite communication antennas rely on circular polarization ($\text{AR} = 0\text{ dB}$) to prevent Faraday rotation fading through the ionosphere; any cross-polar mismatch elongates the polarization locus into an ellipse quantified by the voltage axial ratio $E_{\max}/E_{\min}$. Forming the semi-axis ratio (12.4 [ENTER] 4.2 [÷], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 2.9524$), taking the base-10 logarithm, and scaling by $20$ ([↰] [LOG] 20 [×], $T: 0.0000, Z: 0.0000, Y: 0.0000, X: 9.4034$) yields $\text{AR}_{\text{dB}} \approx \mathbf{9.40\text{ dB}}$.

Chapter 05: Complex Numbers, AC Phasors & Euler's Identity — Extended Engineering & Competition Problems

Accompanies Chapter 05 of Thinking in the Stack (21 Worked Engineering Problems)

Workbook Problem 5.1: Principal Value of Natural Logarithm of Negative Real: $\ln(-5)$

In complex analysis, $\ln(z) = \ln|z| + i \operatorname{Arg}(z)$. For $z = -5 + 0i$, $|z| = 5$ and $\operatorname{Arg}(z) = \pi$:

\[\ln(-5) = \ln(5) + i\pi \approx 1.609438 + 3.141593i\]

While the real-valued logarithm [LN] halts on negative inputs, extending $\ln(z)$ to the complex Riemann surface reveals that a negative real number $-5 = 5 e^{i\pi}$ has log-modulus $\ln(5)$ as its real part and the half-turn phase angle $\pi$ radians as its imaginary part. Load $z = -5 + 0i$ into $(Y, X)$ and evaluate the principal complex logarithm via 5 [+/-] [ENTER] 0 [↰] [CMPLX] [LN] $\rightarrow$ 3.1415926536, transitioning the stack to $(T: 0, Z: 0, Y: 1.609438, X: 3.141593)$.

Workbook Problem 5.2: Complex Reciprocal of High-Q Inductor Admittance

An RF inductor has impedance $\mathbf{Z}_L = 0.5 + 120.0i\,\Omega$. Calculate its complex admittance $\mathbf{Y} = 1/\mathbf{Z}$:

\[\mathbf{Y} = \frac{1}{0.5 + 120i} = \frac{0.5 - 120i}{0.5^2 + 120^2} = \frac{0.5 - 120i}{0.25 + 14400} = 3.4722 \times 10^{-5} - 8.3333 \times 10^{-3}i\,\text{S}\]

Keystrokes: 0.5 [ENTER] 120 [↰] [CMPLX] [1/x] $\rightarrow$ -0.008333189

In parallel RF tank circuits, series resistance $R_s = 0.5\,\Omega$ and reactance $X_L = +120\,\Omega$ ($Q = 240$) cannot be inverted term-by-term; complex inversion [↰] [CMPLX] [1/x] automatically divides the conjugate $R - jX$ by $|Z|^2 = R^2 + X^2$ using overflow-safe Smith scaling. Load the inductor impedance via 0.5 [ENTER] 120, establishing $(T: 0, Z: 0, Y: 0.5, X: 120)$. Press [↰] [CMPLX] [1/x] to invert the complex phasor into admittance $\mathbf{Y} = G + jB$ in Siemens, arriving at $(T: 0, Z: 0, Y: 3.4722 \times 10^{-5}, X: -8.3333 \times 10^{-3})$.

Workbook Problem 5.3: Series RLC Circuit Total Impedance

At angular frequency $\omega = 1,000\text{ rad/s}$, a series circuit has $R = 50\,\Omega$, $L = 80\text{ mH}$ ($X_L = \omega L = 80\,\Omega$), and $C = 20\,\mu\text{F}$ ($X_C = 1/(\omega C) = 50\,\Omega$): $$\mathbf{Z} = R + j(X_L - X_C) = 50 + j(80 - 50) = 50 + 30i\,\Omega$$

Keystrokes: 50 [ENTER] 80 [ENTER] 0 [ENTER] 50 [+/-] [↰] [CMPLX] [+] $\rightarrow$ 30

In a series RLC branch, inductive reactance $+j\omega L$ (where voltage leads current by $90^\circ$) and capacitive reactance $-j/(\omega C)$ (where voltage lags current by $90^\circ$) directly oppose one another along the imaginary axis. Load the RL branch $\mathbf{Z}_1 = 50 + 80i$ with 50 [ENTER] 80 [ENTER], then key the capacitor $\mathbf{Z}_2 = 0 - 50i$ with 0 [ENTER] 50 [+/-], priming the 4-level stack to $(T: 50, Z: 80, Y: 0, X: -50)$. Tap [↰] [CMPLX] [+] to execute dyadic phasor addition, leaving $(T: 50, Z: 50, Y: 50.0000, X: 30.0000)$, confirming net inductive impedance $50.0 + 30.0i\,\Omega$.

Workbook Problem 5.4: RLC Circuit Current Phasor via Ohm's Law

The circuit from Workbook Problem 5.3 ($\mathbf{Z} = 50 + 30i\,\Omega$) is driven by voltage $\mathbf{V} = 120 + 0i\text{ V}$. Find current phasor $\mathbf{I} = \mathbf{V} / \mathbf{Z}$: $$\mathbf{I} = \frac{120}{50 + 30i} = \frac{120(50 - 30i)}{2500 + 900} = \frac{6000 - 3600i}{3400} = 1.7647 - 1.0588i\text{ A}$$

Keystrokes: 120 [ENTER] 0 [ENTER] 50 [ENTER] 30 [↰] [CMPLX] [÷] $\rightarrow$ -1.0588235294

If you are continuing directly from Workbook Problem 5.3 with $\mathbf{Z} = 50 + 30i$ already sitting in $(Y, X)$, you can invert and scale without re-entering $\mathbf{Z}$: key [↰] [CMPLX] [1/x] 120 [ENTER] 0 [↰] [CMPLX] [×]! Or starting fresh, enter numerator $\mathbf{V}$ via 120 [ENTER] 0 [ENTER] followed by denominator $\mathbf{Z}$ via 50 [ENTER] 30, giving $(T: 120, Z: 0, Y: 50, X: 30)$. Press [↰] [CMPLX] [÷] to evaluate complex Ohm's law, yielding lagging current phasor $(T: 120, Z: 120, Y: 1.764706, X: -1.058824)$, confirming $1.7647 - 1.0588i\text{ Amperes}$ ($|\mathbf{I}| = 2.058\text{ A}\angle -30.96^\circ$).

Workbook Problem 5.5: AC Apparent, Real, and Reactive Power Complex Vector

Complex apparent power is $\mathbf{S} = \mathbf{V} \cdot \mathbf{I}^*$. For $\mathbf{V} = 120 + 0i\text{ V}$ and $\mathbf{I} = 1.7647 - 1.0588i\text{ A} \implies \mathbf{I}^* = 1.7647 + 1.0588i\text{ A}$: $$\mathbf{S} = 120 \times (1.7647 + 1.0588i) = 211.7647 + 127.0560i\text{ VA}$$ Real power $P = 211.76\text{ Watts}$, Reactive power $Q = 127.06\text{ VAR}$.

Keystrokes: 120 [ENTER] 0 [ENTER] 1.764706 [ENTER] 1.058824 [↰] [CMPLX] [×] $\rightarrow$ 127.05888

Why does AC power engineering define complex power as $\mathbf{S} = \mathbf{V}\mathbf{I}^*$ rather than $\mathbf{V}\mathbf{I}$? Conjugating the current phasor ensures that the phase difference $\theta_v - \theta_i$ is time-invariant, making $+Q = |\mathbf{V}||\mathbf{I}|\sin(\theta_v - \theta_i) > 0$ positive for lagging inductive loads. Continuing directly from Workbook Problem 5.4 where $\mathbf{I}$ sits in $(Y, X)$, simply press [+/-] to conjugate $\mathbf{I} \to \mathbf{I}^*$ in place and multiply by $\mathbf{V}$! From scratch, enter voltage $\mathbf{V}$ via 120 [ENTER] 0 [ENTER] and conjugate current $\mathbf{I}^*$ via 1.764706 [ENTER] 1.058824, loading $(T: 120, Z: 0, Y: 1.764706, X: 1.058824)$. Press [↰] [CMPLX] [×] to evaluate $\mathbf{S} = P + jQ$, resulting in $(T: 120, Z: 120, Y: 211.7647, X: 127.0589)$, confirming $P = 211.76\text{ W}, Q = 127.06\text{ VAR}$.

Workbook Problem 5.6: Power Factor from Complex Apparent Power

For complex power $\mathbf{S} = 211.7647 + 127.0589i\text{ VA}$ (in radian mode), compute magnitude $|\mathbf{S}|$ and power factor $\cos\theta$:

\[|\mathbf{S}| = \sqrt{P^2 + Q^2} \approx 247.0\text{ VA}, \quad \text{PF} = \frac{P}{|\mathbf{S}|} = \frac{211.7647}{247.0588} \approx 0.8572\]

Keystrokes: 211.7647 [ENTER] 127.0589 [x⇄y] [↰] [▸θ,r] [x⇄y] [COS] $\rightarrow$ 0.8571673007

Bridging [↰] [CMPLX] rectangular registers $(Y: \text{Real}, X: \text{Imag})$ to the 2D polar converter [↰] [▸θ,r] $(Y: y, X: x)$ requires a single [x⇄y] swap before pressing [↰] [▸θ,r]. Key 211.7647 [ENTER] 127.0589 into $(Y, X)$ (or continue directly from Workbook Problem 5.5's stack!), swap to Euclidean $(Y=Q, X=P)$, and convert to polar form via [x⇄y] [↰] [▸θ,r] $(T: 0, Z: 0, Y: 0.540420, X: 247.0588)$, revealing phase angle $\theta = 0.5404\text{ rad}$ in $Y$ and apparent power $|\mathbf{S}| = 247.06\text{ VA}$ in $X$. Tap [x⇄y] [COS] to extract the lagging power factor, ending with $(T: 0, Z: 0, Y: 247.0588, X: 0.857167)$, confirming $\text{PF} = 0.8572\text{ (lagging)}$ with $|\mathbf{S}|$ preserved in Register $Y$.

Workbook Problem 5.7: Second Root of Unity from de Moivre

From Textbook Showcase Problem 5.5, the real root is $z_1 = -2.0 + 0.0i$, and the third root is $z_2 = 1.0 - 1.732051i$. Confirm $z_1^3$: Keystrokes: 2 [+/-] [ENTER] 0 [ENTER] 3 [ENTER] 0 [↰] [CMPLX] [yˣ]$\implies$ $-8.0000 + 0.0000i$.

By de Moivre's theorem, the three cube roots of $-8$ sit at the vertices of an equilateral triangle inscribed in the circle $|z| = 2$ at angles $\theta_k = \frac{\pi + 2\pi k}{3} \in \{60^\circ, 180^\circ, 300^\circ\}$. Enter the real root $z_1 = 2 e^{i\pi} = -2.0 + 0.0i$ into $(T, Z)$ and the power $3.0 + 0.0i$ into $(Y, X)$ via 2 [+/-] [ENTER] 0 [ENTER] 3 [ENTER] 0, creating $(T: -2, Z: 0, Y: 3, X: 0)$. Executing [↰] [CMPLX] [yˣ] confirms $(-2+0i)^3 = -8.0 + 0.0i$, leaving $(T: -2, Z: -2, Y: -8.0000, X: 0.0000)$.

Workbook Problem 5.8: Smith Chart Bilinear Transformation

Compute normalized load $z = r + jx$ from reflection coefficient $\Gamma = 0.2 + 0.4i$:

\[z = \frac{1 + \Gamma}{1 - \Gamma} = \frac{1 + 0.2 + 0.4i}{1 - 0.2 - 0.4i} = \frac{1.2 + 0.4i}{0.8 - 0.4i}\]

Keystrokes: 1.2 [ENTER] 0.4 [ENTER] 0.8 [ENTER] 0.4 [+/-] [↰] [CMPLX] [÷] $\rightarrow$ 1

Phillip H. Smith's 1939 transmission-line chart is a conformal Möbius transformation $z = (1+\Gamma)/(1-\Gamma)$ that maps the right-half impedance plane $\text{Re}(z) \ge 0$ onto the interior of the unit reflection disk $|\Gamma| \le 1$, turning circles of constant resistance and reactance into orthogonal families of circles. Load numerator $1 + \Gamma = 1.2 + 0.4i$ via 1.2 [ENTER] 0.4 [ENTER] and denominator $1 - \Gamma = 0.8 - 0.4i$ via 0.8 [ENTER] 0.4 [+/-], loading $(T: 1.2, Z: 0.4, Y: 0.8, X: -0.4)$. Striking [↰] [CMPLX] [÷] evaluates the inverse bilinear mapping to $z = 1.0 + 1.0i$, with the stack resting at $(T: 1.2, Z: 1.2, Y: 1.000000, X: 1.000000)$ (a matched real resistance $R = Z_0$ in series with inductive reactance $X = +Z_0$).

Workbook Problem 5.9: Acoustic Wave Impedance at Boundary

Characteristic impedance of air $Z_1 = 415\text{ Pa}\cdot\text{s/m}$, porous fiberglass absorber $Z_2 = 600 - 450i\text{ Pa}\cdot\text{s/m}$. Find acoustic reflection coefficient: $$\Gamma_a = \frac{Z_2 - Z_1}{Z_2 + Z_1} = \frac{(600 - 415) - 450i}{(600 + 415) - 450i} = \frac{185 - 450i}{1015 - 450i}$$

Keystrokes: 185 [ENTER] 450 [+/-] [ENTER] 1015 [ENTER] 450 [+/-] [↰] [CMPLX] [÷] $\rightarrow$ -0.302987

Plane sound waves striking an architectural fiberglass treatment obey the exact same Fresnel reflection equation $\Gamma_a = (Z_2 - Z_1)/(Z_2 + Z_1)$ as electromagnetic waves on a transmission line, where the porous absorber's compliance introduces a negative (capacitive) acoustic reactance $-450i\text{ Rayls}$. Key numerator differential impedance $(Z_2 - Z_1) = 185 - 450i$ with 185 [ENTER] 450 [+/-] [ENTER] and denominator $(Z_2 + Z_1) = 1015 - 450i$ with 1015 [ENTER] 450 [+/-], yielding $(T: 185, Z: -450, Y: 1015, X: -450)$. Press [↰] [CMPLX] [÷] to resolve the complex pressure reflection coefficient $(T: 185, Z: 185, Y: 0.316595, X: -0.302987)$—and keying [x⇄y] [↰] [▸θ,r] [↰] [x²] 1 [x⇄y] [-] immediately computes the Sabine normal-incidence sound absorption coefficient $\alpha = 1 - |\Gamma_a|^2 = \mathbf{0.8080}$ ($80.8\%$ of acoustic energy absorbed!).

Workbook Problem 5.10: Mutual Inductive Coupling in Transformer

Primary inductance $L_1 = 50\,\mu\text{H}$, secondary $L_2 = 20\,\mu\text{H}$, coupling factor $k = 0.8$. Mutual inductance $M = k\sqrt{L_1 L_2}$:

$$M = 0.8 \sqrt{50 \times 20} = 0.8 \sqrt{1000} \approx 25.2982\,\mu\text{H}$$ Keystrokes: 50 [ENTER] 20 [×] [√x] 0.8 [×]$\implies$ $25.2982\,\mu\text{H}$.

In coupled-coil RF transformers, magnetic flux leakage limits the mutual inductance $M$ to the fraction $k \in [0, 1]$ of the geometric mean $\sqrt{L_1 L_2}$, introducing a mutual transfer impedance $Z_m = j\omega M$ between primary and secondary mesh equations. Enter 50 [ENTER] 20 [×] to form the inductive product $1000\,\mu\text{H}^2$ in $X$ $(T: 0, Z: 0, Y: 0, X: 1000)$, take the geometric mean with [√x] $(T: 0, Z: 0, Y: 0, X: 31.6228)$, and scale by coupling coefficient with 0.8 [×], resolving $(T: 0, Z: 0, Y: 0, X: 25.2982)$.

Workbook Problem 5.11: Waveguide Attenuation and Propagation Constant ($\gamma$)

Propagation constant $\gamma = \alpha + j\beta$. In cutoff band: $\alpha = 14.5\text{ Np/m}, \beta = 0\text{ rad/m}$. Attenuation in decibels per meter:

\[\text{Atten} = 20 \log_{10}(e) \times \alpha = 8.68589 \times 14.5 = \mathbf{125.945\text{ dB/m}}\]

Below a rectangular waveguide's cutoff frequency $f_c$, the propagation constant $\gamma = \sqrt{k_c^2 - k^2} = \alpha + j0$ becomes purely real: rather than propagating as a traveling wave, the field decays evanescently as $e^{-\alpha z}$ at $20\log_{10}(e) \approx 8.68589\text{ dB}$ per Neper (the basis of waveguide-below-cutoff precision attenuators). Key 1 [eˣ] [↰] [LOG] 20 [×] to generate the exact Nepers-to-dB constant ($8.685890\text{ dB/Np}$) on the stack $(T: 0, Z: 0, Y: 0, X: 8.685890)$, then multiply by $\alpha = 14.5\text{ Np/m}$ with 14.5 [×], resolving total attenuation $(T: 0, Z: 0, Y: 0, X: 125.945405)$, confirming $125.945\text{ dB/m}$.

Workbook Problem 5.12: Complex Impedance of Parallel RLC Tank at Resonance

Tank circuit with $R = 10\text{ k}\Omega, L = 10\,\mu\text{H}, C = 100\text{ pF}$. Resonant frequency $\omega_0 = 1/\sqrt{LC} = 31.6228\text{ Mrad/s}$. At resonance, reactive admittances cancel ($j\omega C + 1/(j\omega L) = 0$), leaving pure resistance: $$\mathbf{Z}_{\text{tank}} = R = 10,000.0 + 0.0i\,\Omega$$

In a parallel RLC bandpass resonator at $\omega_0 = 1/\sqrt{LC}$, energy sloshes losslessly between the magnetic field of the inductor and the electric field of the capacitor with equal and opposite susceptances ($\pm j\sqrt{C/L} = \pm j\,3.1623\text{ mS}$), so the parallel tank presents a maximum, purely resistive impedance $\mathbf{Z}_{\text{tank}} = R$ to the driving RF stage. Evaluate the reactive product with 10 [E] 6 [+/-] [ENTER] 100 [E] 12 [+/-] [×] $(T: 0, Z: 0, Y: 0, X: 1.0 \times 10^{-15})$, extract the resonant angular frequency with [√x] [1/x] $(T: 0, Z: 0, Y: 0, X: 3.1623 \times 10^7)$, then load the resonant tank impedance 10000 [ENTER] 0 across $(Y, X)$, confirming purely resistive impedance $(T: 0, Z: 3.1623 \times 10^7, Y: 10000.0000, X: 0.0000)$ of $10,000.0 + 0.0i\,\Omega$.

Workbook Problem 5.13: Complex Root Extraction via de Moivre ($z^4 = -16$)

Roots of $z^4 = -16 = 16 e^{i\pi}$: $r = 16^{1/4} = 2.0$. First root $\theta = \pi/4 = 45^\circ$:

$$z_0 = 2(\cos 45^\circ + i\sin 45^\circ) = \sqrt{2} + i\sqrt{2} \approx 1.414214 + 1.414214i$$ Keystrokes: 45 [ENTER] 2 [↱] [▸y,x]$\implies$ $1.414214$ in $X$ and $Y$.

The four fourth roots of $-16$ form the $45^\circ$-rotated square vertices $2 e^{i(\pi/4 + k\pi/2)}$—the exact pole pattern of a 2nd-order Butterworth filter! Key the polar angle and radial modulus 45 [ENTER] 2 into $(Y, X)$ $(T: 0, Z: 0, Y: 45, X: 2)$, then convert from polar to rectangular via [↱] [▸y,x], extracting principal root $z_0 = 1.414214 + 1.414214i$ across $(T: 0, Z: 0, Y: 1.414214, X: 1.414214)$.

Workbook Problem 5.14: Skin Effect in Solid Copper Conductor

Skin depth $\delta = \sqrt{2/(\omega \mu \sigma)}$. For copper ($\sigma = 5.8 \times 10^7\text{ S/m}$) at $f = 10\text{ MHz}$ ($\omega = 6.283185 \times 10^7\text{ rad/s}$): $$\delta = \sqrt{\frac{2}{6.283185\cdot 10^7 \times 4\pi\cdot 10^{-7} \times 5.8\cdot 10^7}} \approx 2.089 \times 10^{-5}\text{ m} = \mathbf{20.89\,\mu\text{m}}$$

At radio frequencies, alternating magnetic flux inside a conductor induces eddy currents that oppose current flow in the core and crowd the current density $J(r) = J_0 e^{-(1+j)d/\delta}$ into a thin surface annulus of thickness $\delta$. Form free-space permeability $\mu_0$ with 4 [↱] [π] [×] 1 [E] 7 [+/-] [×] $(T: 0, Z: 0, Y: 0, X: 1.2566 \times 10^{-6})$, multiply by angular frequency and conductivity 6.283185 [E] 7 [×] 5.8 [E] 7 [×] $(T: 0, Z: 0, Y: 0, X: 4.5795 \times 10^9)$, invert into ratio $2/(\omega\mu\sigma)$ with 2 [x⇄y] [÷] $(T: 0, Z: 0, Y: 0, X: 4.3639 \times 10^{-10})$, and take the square root [√x], arriving at skin depth $(T: 0, Z: 0, Y: 0, X: 2.0890 \times 10^{-5})$, confirming $\delta = 20.89\,\mu\text{m}$.

Workbook Problem 5.15: Dispersion Relation in Fiber Optics (Group Velocity Dispersion)

Complex propagation constant $\beta(\omega) = \beta_0 + \beta_1(\omega - \omega_0) + \frac{1}{2}\beta_2(\omega - \omega_0)^2$. Pulse dispersion parameter $D = -2\pi c \beta_2 / \lambda^2$. For $\beta_2 = -20\text{ ps}^2/\text{km}$ at $\lambda = 1.55\,\mu\text{m}$: $$D = \frac{-2\pi(3\cdot 10^8)(-20\cdot 10^{-27})}{(1.55\cdot 10^{-6})^2} \approx \mathbf{15.68\text{ ps/(nm}\cdot\text{km)}}$$

In single-mode silica telecom fiber operating in the $1550\text{ nm}$ C-band, negative group-velocity dispersion ($\beta_2 = -20\text{ ps}^2/\text{km}$, the anomalous dispersion regime where optical solitons form) causes shorter-wavelength blue spectral components to travel faster than red components by $D \approx 15.68\text{ ps}$ per nanometer of spectral width per kilometer. Compute numerator $2\pi c |\beta_2|$ with 2 [↱] [π] [×] 3 [E] 8 [×] 20 [E] 27 [+/-] [×] $(T: 0, Z: 0, Y: 0, X: 3.7699 \times 10^{-17})$, square wavelength with 1.55 [E] 6 [+/-] [↰] [x²] $(T: 0, Z: 0, Y: 3.7699 \times 10^{-17}, X: 2.4025 \times 10^{-12})$, and divide and scale via [÷] 1 [E] 6 [×], arriving at dispersion parameter $(T: 0, Z: 0, Y: 0, X: 15.6841)$, confirming $15.68\text{ ps/(nm}\cdot\text{km)}$.

Workbook Problem 5.16: Fractional Pole Filter: $s^{0.5}$ Fractional Calculus Impedance

Warburg impedance in electrochemistry: $\mathbf{Z}_W = \frac{A_W}{\sqrt{\omega}} (1 - j)$. For $A_W = 200\,\Omega\cdot\text{s}^{-1/2}$ at $\omega = 100\text{ rad/s}$: $$\mathbf{Z}_W = \frac{200}{10} (1 - j) = \mathbf{20.0000 - 20.0000i\,\Omega}$$

In electrochemical impedance spectroscopy (EIS) of lithium-ion battery electrodes, semi-infinite Fickian ion diffusion obeys a half-order fractional differential equation whose Laplace transform $(j\omega)^{-1/2} = \frac{1-j}{\sqrt{2\omega}}$ produces a constant $-45^\circ$ phase line on a Nyquist plot (the Warburg element). Evaluate $\sqrt{\omega}$ with 100 [√x] $(T: 0, Z: 0, Y: 0, X: 10)$, compute scale factor $A_W/\sqrt{\omega} = 20\,\Omega$ with 200 [x⇄y] [÷] $(T: 0, Z: 0, Y: 0, X: 20)$, and duplicate and negate ([ENTER] [+/-] or 20 [ENTER] 20 [+/-]), confirming Warburg impedance $(T: 20.0000, Z: -20.0000, Y: 20.0000, X: -20.0000)$.

Workbook Problem 5.17: Complex Poynting Vector & Electromagnetic Radiation Density

$\mathbf{S}_{\text{cpx}} = \frac{1}{2} \mathbf{E} \times \mathbf{H}^*$. For $\mathbf{E} = 10.0 + 5.0i\text{ V/m}$ and $\mathbf{H} = 0.02 + 0.01i\text{ A/m}$: $$\mathbf{H}^* = 0.02 - 0.01i \implies \mathbf{E} \cdot \mathbf{H}^* = (10+5i)(0.02-0.01i) = (0.2 + 0.05) + (-0.1 + 0.1)i = 0.25 + 0i$$ Average radiated power: $\text{Re}(\mathbf{S}_{\text{cpx}}) = 0.5 \times 0.25 = \mathbf{0.125\text{ W/m}^2}$.

The complex Poynting theorem separates electromagnetic energy flux into time-averaged radiated power density $\text{Re}(\frac{1}{2}\mathbf{E} \times \mathbf{H}^*)$ and reactive near-field stored energy density $\text{Im}(\frac{1}{2}\mathbf{E} \times \mathbf{H}^*)$. Because $\mathbf{E}$ and $\mathbf{H}$ here share the exact same phase angle $\arctan(0.5) = 26.57^\circ$ (with wave impedance $E/H = 500\,\Omega$), their cross-product $\mathbf{E}\mathbf{H}^*$ is purely real! Load electric field $\mathbf{E}$ via 10 [ENTER] 5 [ENTER] and conjugate magnetic field $\mathbf{H}^*$ via 0.02 [ENTER] 0.01 [+/-], placing $(T: 10, Z: 5, Y: 0.02, X: -0.01)$. Execute [↰] [CMPLX] [×] to obtain $0.25 + 0i$ $(T: 10, Z: 10, Y: 0.25, X: 0)$, swap the real part into Register $X$, and divide by 2 with [x⇄y] 2 [÷], giving average radiation density $(T: 10, Z: 10, Y: 0, X: 0.1250)$, confirming $0.125\text{ W/m}^2$.

Workbook Problem 5.18: Sagnac Ring Interferometer Phase Shift

Fiber optic gyroscope: $\Delta\Phi = \frac{8\pi A N}{c \lambda} \Omega$. For coil area $A = 0.01\text{ m}^2$, $N = 1000\text{ turns}$, $\lambda = 1.31\,\mu\text{m}$, rotation rate $\Omega = 1.0\text{ rad/s}$: $$\Delta\Phi = \frac{8\pi(0.01)(1000)}{(3\cdot 10^8)(1.31\cdot 10^{-6})} \approx \mathbf{0.6395\text{ radians}}$$

In Georges Sagnac's relativistic ring interferometer, counter-propagating laser beams traversing a rotating fiber coil of total enclosed area $N A = 10\text{ m}^2$ experience a non-reciprocal transit-time difference $\Delta t = 4AN\Omega/c^2$, shifting their interference phasor $e^{i\Delta\Phi}$ by $\Delta\Phi = \frac{8\pi AN}{c\lambda}\Omega$. Form the optical area-turns product with 8 [↱] [π] [×] 0.01 [×] 1000 [×] $(T: 0, Z: 0, Y: 0, X: 251.3274)$, compute $c\lambda$ with 3 [E] 8 [ENTER] 1.31 [E] 6 [+/-] [×] $(T: 0, Z: 0, Y: 251.3274, X: 393.0000)$, and divide with [÷], resolving Sagnac phase shift $(T: 0, Z: 0, Y: 0, X: 0.6395)$.

Workbook Problem 5.19: Transformer T-Model Core Loss Branch

Parallel magnetizing branch: $R_c = 50\text{ k}\Omega \parallel X_m = 15\text{ k}\Omega$. Total complex excitation admittance:

$$\mathbf{Y}_m = \frac{1}{50000} - j\frac{1}{15000} = 2.0 \times 10^{-5} - 6.6667 \times 10^{-5}i\,\text{S}$$ Impedance: $\mathbf{Z}_m = 1/\mathbf{Y}_m = \mathbf{4,128.44 + 13,761.47i\,\Omega}$.

In a power transformer's Steinmetz equivalent circuit, eddy-current/hysteresis core loss is modeled by shunt resistance $R_c = 50\text{ k}\Omega$ in parallel with magnetizing reactance $+j X_m = +j\,15\text{ k}\Omega$. Converting this parallel excitation branch into its equivalent series impedance $\mathbf{Z}_m = R_s + j X_s$ is effortless on the stack: key core conductance 50000 [1/x] $(T: 0, Z: 0, Y: 0, X: 2.0 \times 10^{-5})$ and inductive susceptance 15000 [1/x] [+/-] $(T: 0, Z: 0, Y: 2.0 \times 10^{-5}, X: -6.6667 \times 10^{-5})$, then take the complex reciprocal with [↰] [CMPLX] [1/x], arriving at equivalent branch impedance $(T: 0, Z: 0, Y: 4128.4404, X: 13761.4679)$, confirming $4,128.44 + 13,761.47i\,\Omega$ (or use 50000 [ENTER] 0 [ENTER] 0 [ENTER] 15000 [↰] [CMPLX] [x∥y] directly!).

Workbook Problem 5.20: Elliptical Polarization Axial Ratio from Orthogonal E-Phasors

$\mathbf{E}_x = 10\angle 0^\circ = 10 + 0i$, $\mathbf{E}_y = 6\angle 90^\circ = 0 + 6i$. Major axis $a = 10.0$, minor axis $b = 6.0$. Axial ratio:

\[\text{AR} = \frac{10.0}{6.0} = 1.6667 \implies \text{AR}_{\text{dB}} = 20 \log_{10}(1.6667) = \mathbf{4.4370\text{ dB}}\]

When two orthogonal dipole elements are driven in quadrature ($90^\circ$ phase offset) with unequal field amplitudes $E_x = 10\text{ V/m}$ and $E_y = 6\text{ V/m}$, the tip of the electric field vector traces an upright polarization ellipse of axial ratio $a/b = 10/6$. Key 10 [ENTER] 6 [÷] $(T: 0, Z: 0, Y: 0, X: 1.666667)$ to divide the semi-axes, take the base-10 logarithm with [↰] [LOG] $(T: 0, Z: 0, Y: 0, X: 0.221849)$, and scale with 20 [×] to determine polarization axial ratio in decibels $(T: 0, Z: 0, Y: 0, X: 4.436975)$, confirming $4.4370\text{ dB}$.

Workbook Problem 5.21: High-Speed Sprint: Radical Exponential Nested Complex Identity

Evaluate $\sqrt{e^{i\pi/2}}$:

$$e^{i\pi/2} = i \implies \sqrt{i} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \approx 0.707107 + 0.707107i$$ Keystrokes: 0 [↱] [π] 2 [÷] [↰] [CMPLX] [eˣ] [↰] [CMPLX] [√x] $\rightarrow$ 0.7071067812

This closing sprint links Euler's exponential map $e^{i\theta} = \cos\theta + i\sin\theta$ directly to de Moivre root extraction on the live stack. Enter the pure imaginary exponent $0 + i(\pi/2)$ via 0 [↱] [π] 2 [÷] $(T: 0, Z: 0, Y: 0, X: 1.570796)$, evaluate the complex exponential with [↰] [CMPLX] [eˣ] $(T: 0, Z: 0, Y: 0.000000, X: 1.000000)$, and extract the principal complex square root with [↰] [CMPLX] [√x], producing $(T: 0, Z: 0, Y: 0.707107, X: 0.707107)$ in 4 seconds flat!

Stack outputs: $Y = 0.707107$, $X = 0.707107$ in 4 seconds flat!

Chapter 06: Analytic & Solid Geometry, Surface Area & Mensuration — Extended Engineering & Competition Problems

Accompanies Chapter 06 of Thinking in the Stack (15 Worked Engineering Problems)

Workbook Problem 6.1: Elliptical Pressure Vessel Manway Area & Perimeter

A vessel inspection hatch is elliptical with semimajor axis $a = 250.0\text{ mm}$ and semiminor axis $b = 150.0\text{ mm}$. 1. Area $A = \pi a b = \pi \times 250 \times 150 = 37,500\pi \approx \mathbf{117,809.72\text{ mm}^2}$ 2. Ramanujan's approximate perimeter $P \approx \pi [3(a+b) - \sqrt{(3a+b)(a+3b)}]$: $h = \frac{(a-b)^2}{(a+b)^2} = \frac{100^2}{400^2} = 0.0625 \implies P \approx \pi(a+b)(1 + \frac{3h}{10 + \sqrt{4 - 3h}}) \approx \mathbf{1,276.26\text{ mm}}$

Keystrokes (Area): 250 [ENTER] 150 [×] [↱] [π] [×] $\rightarrow$ 117,809.7245

Boiler inspection manways are made elliptical so the cover plate can be rotated $90^\circ$, slipped edgewise through its own opening, and seated against the inside rim so internal boiler pressure seals the gasket automatically! While the area $A = \pi ab$ is an exact affine scaling of a circle, the perimeter requires a complete elliptic integral of the second kind (approximated to $10^{-9}$ accuracy by Srinivasa Ramanujan's rational $h$-formula). Push semimajor axis $a = 250$ via 250 [ENTER], multiply by semiminor axis $b = 150$ via 150 [×] to obtain planar product $37,500.0$, and apply [↱] [π] [×] to resolve exact manway opening area at $(T: 0, Z: 0, Y: 0, X: 117809.7245)$, confirming $117,809.72\text{ mm}^2$ ($1,178.10\text{ cm}^2$).

Workbook Problem 6.2: Torus Volume and Surface Area (Hydraulic O-Ring)

An elastomer O-ring has major mean radius $R = 45.0\text{ mm}$ and cross-sectional circular radius $r = 3.5\text{ mm}$. 1. Volume $V = 2\pi^2 R r^2 = 2 \pi^2 (45) (3.5^2) = 2 \pi^2 (45) (12.25) \approx \mathbf{10,880.89\text{ mm}^3}$ 2. Surface Area $A = 4\pi^2 R r = 4 \pi^2 (45) (3.5) \approx \mathbf{6,217.85\text{ mm}^2}$

Keystrokes (Volume): [↱] [π] [↰] [x²] 2 [×] 45 [×] 3.5 [↰] [x²] [×] $\rightarrow$ 10,881.23885

By Pappus's centroid theorems, sweeping a circle of area $\pi r^2$ and circumference $2\pi r$ along a circular centerline of length $2\pi R$ generates volume $V = (\pi r^2)(2\pi R) = 2\pi^2 R r^2$ and surface area $A = (2\pi r)(2\pi R) = 4\pi^2 R r$—exactly the same as an unwrapped cylinder of length $2\pi R$! Evaluate $2\pi^2$ via [↱] [π] [↰] [x²] 2 [×], scale by major centerline radius $R = 45\text{ mm}$ via 45 [×], and multiply by cross-sectional radius squared $r^2 = 3.5^2$ via 3.5 [↰] [x²] [×], establishing torus body volume at $(T: 0, Z: 0, Y: 0, X: 10880.8872)$.

Keystrokes (Surface Area): [↱] [π] [↰] [x²] 4 [×] 45 [×] 3.5 [×] $\rightarrow$ 6217.6500

Notice that $A / V = 2 / r$: if you already have $V = 10,880.89\text{ mm}^3$ in Register $X$, keying 2 [×] 3.5 [÷] yields $A = 6,217.85\text{ mm}^2$ in three keystrokes! Or from scratch, compute the $4\pi^2$ envelope constant via [↱] [π] [↰] [x²] 4 [×], then chain major and minor radii via 45 [×] 3.5 [×] to yield total contact surface area at $(T: 0, Z: 0, Y: 0, X: 6217.8508)$, confirming $V = 10,880.89\text{ mm}^3$, $A = 6,217.85\text{ mm}^2$.

Workbook Problem 6.3: Truncated Pyramidal Monument Volume

A stone pedestal is a truncated square pyramid: lower base side $a = 3.20\text{ m}$, upper base side $b = 1.80\text{ m}$, height $h = 4.50\text{ m}$. Base areas: $A_1 = 3.2^2 = 10.24\text{ m}^2$, $A_2 = 1.8^2 = 3.24\text{ m}^2$: $$V = \frac{h}{3} (A_1 + A_2 + \sqrt{A_1 A_2}) = \frac{4.5}{3} (10.24 + 3.24 + \sqrt{33.1776}) = 1.5 \times (13.48 + 5.76) = 1.5 \times 19.24 = \mathbf{28.86\text{ m}^3}$$

Keystrokes: 3.2 [↰] [x²] [STO] 1 1.8 [↰] [x²] [STO] 2 [+] [RCL] 1 [RCL] 2 [×] [√x] [+] 4.5 [×] 3 [÷] $\rightarrow$ 28.86

Recorded in the 1850 BCE Egyptian Moscow Mathematical Papyrus (Problem 14), the truncated square pyramid formula $V = \frac{h}{3}(a^2 + ab + b^2) = \frac{h}{3}(A_1 + A_2 + \sqrt{A_1 A_2})$ is one of the oldest jewels of solid geometry. Store lower base $A_1 = 10.24$ into Register 1 via 3.2 [↰] [x²] [STO] 1, and upper base $A_2 = 3.24$ into Register 2 via 1.8 [↰] [x²] [STO] 2. Add them via [+] for sum $13.48$. Compute geometric mean $\sqrt{A_1 A_2} = 5.76$ via [RCL] 1 [RCL] 2 [×] [√x], sum with base areas via [+], and multiply by $h/3$ via 4.5 [×] 3 [÷], yielding pedestal volume at $(T: 0, Z: 0, Y: 0, X: 28.8600)$.

Workbook Problem 6.4: Regular Icosahedron Dice Volume and Area

A 20-sided gaming die has edge length $a = 18.0\text{ mm}$. 1. Surface Area $A = 5\sqrt{3} a^2 = 5 \sqrt{3} (18^2) = 5\sqrt{3}(324) \approx \mathbf{2,805.9223\text{ mm}^2}$ 2. Volume $V = \frac{5(3 + \sqrt{5})}{12} a^3 = \frac{5(5.236068)}{12} (5832) \approx \mathbf{12,723.65\text{ mm}^3}$

Keystrokes (Volume): 5 [√x] 3 [+] 5 [×] 12 [÷] 18 [ENTER] 3 [yˣ] [×] $\rightarrow$ 12,723.64519

The vertices of a regular icosahedron lie on three mutually orthogonal golden rectangles of aspect ratio $\varphi = (1+\sqrt{5})/2$, so its volume constant $\frac{5(3+\sqrt{5})}{12} = \frac{5\varphi^2}{6}$ is governed directly by the square of the golden ratio! Evaluate icosahedral volume factor $\frac{5(3+\sqrt{5})}{12} \approx 2.181695$ via 5 [√x] 3 [+] 5 [×] 12 [÷]. Cube die edge $a = 18\text{ mm}$ via 18 [ENTER] 3 [yˣ] and multiply via [×], arriving at total polyhedron volume at $(T: 0, Z: 0, Y: 0, X: 12723.6452)$, confirming $V = 12,723.65\text{ mm}^3$, $A = 2,805.92\text{ mm}^2$.

Workbook Problem 6.5: Regular Dodecahedron Volume

12-sided solid with edge $a = 12.0\text{ mm}$:

\[V = \frac{15 + 7\sqrt{5}}{4} a^3 \approx \frac{15 + 15.652476}{4} (1728) = 7.663119 \times 1728 = \mathbf{13,241.87\text{ mm}^3}\]

Keystrokes: 5 [√x] 7 [×] 15 [+] 4 [÷] 12 [ENTER] 3 [yˣ] [×] $\rightarrow$ 13,241.86956

Dual to the icosahedron, the regular dodecahedron can be constructed by erecting roof-like "tents" on the six faces of an inscribed cube of edge $\varphi a$, giving volume constant $\frac{15+7\sqrt{5}}{4} \approx 7.663119$. Compute this constant via 5 [√x] 7 [×] 15 [+] 4 [÷], cube facet edge $a = 12\text{ mm}$ via 12 [ENTER] 3 [yˣ], and multiply with [×], yielding $(T: 0, Z: 0, Y: 0, X: 13241.8694)$, confirming $13,241.87\text{ mm}^3$.

Workbook Problem 6.6: 3D Euclidean Distance Between Aerospace Telemetry Nodes

Station $A$ at coordinates $(1250, -420, 850)\text{ m}$, Station $B$ at $(2100, 680, 1400)\text{ m}$:

\[\Delta x = 2100 - 1250 = 850\text{ m}, \quad \Delta y = 680 - (-420) = 1100\text{ m}, \quad \Delta z = 1400 - 850 = 550\text{ m}$$ $$d = \sqrt{850^2 + 1100^2 + 550^2} = \sqrt{722500 + 1210000 + 302500} = \sqrt{2235000} \approx \mathbf{1,494.9916\text{ meters}}\]

Keystrokes: 2100 [ENTER] 1250 [-] [↰] [x²] 680 [ENTER] 420 [+/-] [-] [↰] [x²] [+] 1400 [ENTER] 850 [-] [↰] [x²] [+] [√x] $\rightarrow$ 1,494.991639

Computing a 3D Euclidean norm $\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}$ on the stack never requires more than two active registers when each coordinate difference is squared and added immediately to the running sum of squares in Register $Y$. Evaluate $\Delta x^2 = 850^2 = 722,500$ via 2100 [ENTER] 1250 [-] [↰] [x²], accumulate $\Delta y^2 = 1100^2 = 1,210,000$ with 680 [ENTER] 420 [+/-] [-] [↰] [x²] [+], and add $\Delta z^2 = 550^2 = 302,500$ via 1400 [ENTER] 850 [-] [↰] [x²] [+]. Extract the square root via [√x] to resolve 3D spatial separation at $(T: 0, Z: 0, Y: 0, X: 1494.9916)$, confirming $1,494.99\text{ meters}$.

Workbook Problem 6.7: Vector Cross Product in 3-Space ($\vec{A} \times \vec{B}$)

$\vec{A} = (3, -2, 4)$, $\vec{B} = (1, 5, -2)$:

\[\vec{C} = \vec{A} \times \vec{B} = (A_y B_z - A_z B_y)\hat{i} + (A_z B_x - A_x B_z)\hat{j} + (A_x B_y - A_y B_x)\hat{k}$$ * $C_x = (-2)(-2) - (4)(5) = 4 - 20 = \mathbf{-16}$ * $C_y = (4)(1) - (3)(-2) = 4 - (-6) = \mathbf{10}$ * $C_z = (3)(5) - (-2)(1) = 15 - (-2) = \mathbf{17}$ $$\vec{C} = (-16, 10, 17) \implies |\vec{C}| = \sqrt{256 + 100 + 289} = \sqrt{645} \approx \mathbf{25.3969}\]

Keystrokes (Magnitude): 16 [↰] [x²] 10 [↰] [x²] [+] 17 [↰] [x²] [+] [√x] $\rightarrow$ 25.396850198

Geometrically, the cross product $\vec{C} = \vec{A} \times \vec{B}$ is normal to the plane spanned by $\vec{A}$ and $\vec{B}$, and its Euclidean magnitude $|\vec{C}| = |\vec{A}||\vec{B}|\sin\theta = \sqrt{645}$ equals the area of the 3D parallelogram formed by the two vectors (and half of $|\vec{C}|$ is the area of the spatial triangle). Square each Cartesian cross-product component sequentially: 16 [↰] [x²] 10 [↰] [x²] [+] 17 [↰] [x²] [+] produces sum of squares $645.0$ in Register $X$. Taking the square root via [√x] delivers vector magnitude $|\vec{C}|$ at $(T: 0, Z: 0, Y: 0, X: 25.3969)$, confirming $\vec{C} = (-16, 10, 17), |\vec{C}| = 25.40$.

Workbook Problem 6.8: Dihedral Angle Between Two Intersecting Planes

Plane 1: $2x + y - 2z = 5 \implies \vec{n}_1 = (2, 1, -2)$. Plane 2: $x - 2y + 2z = 3 \implies \vec{n}_2 = (1, -2, 2)$:

\[\cos\theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|} = \frac{|(2)(1) + (1)(-2) + (-2)(2)|}{\sqrt{4+1+4} \sqrt{1+4+4}} = \frac{|2 - 2 - 4|}{3 \times 3} = \frac{4}{9} \approx 0.444444$$ $$\theta = \arccos(4/9) \approx \mathbf{63.6122^\circ}\]

Keystrokes: 4 [ENTER] 9 [÷] [↰] [ACOS] $\rightarrow$ 63.612200039

The acute dihedral fold angle $\theta$ between two intersecting sheet-metal or roof planes equals the angle between their outward surface normals $\vec{n}_1$ and $\vec{n}_2$ (which here are both Pythagorean triples of length $\sqrt{2^2 + 1^2 + 2^2} = 3$). Evaluate cosine quotient $4/9 \approx 0.444444$ via 4 [ENTER] 9 [÷]. Taking inverse cosine via [↰] [ACOS] extracts the dihedral intersection angle in degrees, settling the stack at $(T: 0, Z: 0, Y: 0, X: 63.6122)$.

Workbook Problem 6.9: Radius of Gyration ($k$) of Rotating Turbine Disk

Rotor disk with moment of inertia $I = 14.8\text{ kg}\cdot\text{m}^2$ and mass $M = 42.0\text{ kg}$:

\[k = \sqrt{\frac{I}{M}} = \sqrt{\frac{14.8}{42.0}} = \sqrt{0.352381} = \mathbf{0.593617\text{ meters}} = 593.6\text{ mm}\]

Keystrokes: 14.8 [ENTER] 42 [÷] [√x] $\rightarrow$ 0.5936168397

The radius of gyration $k = \sqrt{I/M}$ is the radial distance at which the entire mass $M$ of a complex profiled turbine rotor could be concentrated into an infinitesimally thin hoop ($I = M k^2$) without altering its rotational kinetic energy $\frac{1}{2}I\omega^2$. Divide moment of inertia $I = 14.8$ by mass $M = 42.0$ via 14.8 [ENTER] 42 [÷] to form $k^2 = 0.352381\text{ m}^2$. Extracting the square root via [√x] gives radius of gyration $k$ in meters at $(T: 0, Z: 0, Y: 0, X: 0.593617)$, confirming $k = 0.5936\text{ m}$ ($593.6\text{ mm}$).

Workbook Problem 6.10: Parabolic Solar Reflector Surface Area

Deep parabolic dish $z = \frac{x^2 + y^2}{4f}$ with focal length $f = 1.20\text{ m}$ and rim radius $R = 2.0\text{ m}$. Surface area of revolution:

\[A = \frac{8\pi f^2}{3} \left[ \left(1 + \frac{R^2}{4f^2}\right)^{3/2} - 1 \right]$$ For $f = 1.2, R = 2.0$: $\frac{R^2}{4f^2} = \frac{4}{4(1.44)} = \frac{1}{1.44} \approx 0.694444$. $1 + 0.694444 = 1.694444 \implies (1.694444)^{1.5} \approx 2.205739$: $$A = \frac{8\pi(1.44)}{3} [2.205739 - 1] = 3.84\pi \times 1.205739 \approx \mathbf{14.5458\text{ m}^2}\]

Integrating the surface-of-revolution element $2\pi r\sqrt{1 + (dz/dr)^2}\,dr$ for a paraboloid $z = r^2/(4f)$ yields an exact $u$-substitution in $u = 1 + r^2/(4f^2)$ that requires no elliptic integrals. Evaluate base factor $1 + R^2/(4f^2) \approx 1.694444$ via 2 [↰] [x²] 4 [÷] 1.2 [↰] [x²] [÷] 1 [+]. Raise to power $1.5$ and subtract unity via 1.5 [yˣ] 1 [-] to obtain $1.205739$. Form leading coefficient $\frac{8\pi f^2}{3} \approx 12.063716$ via 8 [↱] [π] [×] 1.2 [↰] [x²] [×] 3 [÷] and multiply via [×] to resolve reflector mirror area at $(T: 0, Z: 0, Y: 0, X: 14.5458)$ (compared to flat aperture $\pi R^2 = 12.5664\text{ m}^2$).

Workbook Problem 6.11: Hydraulic Mean Depth & Wetted Perimeter

Trapezoidal irrigation canal: bottom width $b = 4.0\text{ m}$, water depth $y = 1.8\text{ m}$, side slope $z = 1.5$ ($1.5\text{ H} : 1\text{ V}$): * Flow area $A = (b + z y) y = (4.0 + 1.5 \times 1.8) \times 1.8 = (4.0 + 2.7) \times 1.8 = 6.7 \times 1.8 = \mathbf{12.06\text{ m}^2}$ * Wetted perimeter $P = b + 2 y \sqrt{1 + z^2} = 4.0 + 2(1.8)\sqrt{1 + 2.25} = 4.0 + 3.6\sqrt{3.25} \approx 4.0 + 6.49 = \mathbf{10.4892\text{ m}}$ * Hydraulic radius $R_h = A / P = 12.06 / 10.4892 = \mathbf{1.1498\text{ meters}}$

In open-channel hydraulics, the hydraulic radius $R_h = A/P$ measures the ratio of gravity-driven cross-sectional flow area $A$ to boundary-shear wetted perimeter $P$, driving Manning's velocity $V = \frac{1}{n} R_h^{2/3} S^{1/2}$. Evaluate flow area $A = (b+zy)y = 12.06\text{ m}^2$ with 1.5 [ENTER] 1.8 [×] 4 [+] 1.8 [×], then compute wetted perimeter $P = b + 2y\sqrt{1+z^2} \approx 10.4892\text{ m}$ inside-out via 1.5 [↰] [x²] 1 [+] [√x] 2 [×] 1.8 [×] 4 [+], using only two active stack registers so $A = 12.06$ stays safely in Register $Y$ while $P$ forms in Register $X$. Pressing [÷] divides the two directly to establish hydraulic radius at $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.1498)$.

Workbook Problem 6.12: Circular Segment Area (Partially Filled Pipe)

Storm sewer diameter $D = 1.20\text{ m}$ ($r = 0.60\text{ m}$). Water depth $d = 0.40\text{ m}$. Subtended angle:

\[\theta = 2 \arccos\left(\frac{r - d}{r}\right) = 2 \arccos\left(\frac{0.2}{0.6}\right) = 2 \arccos(0.333333) = 2(70.5288^\circ) = 141.0575^\circ = 2.461914\text{ rad}$$ $$\text{Area } A = \frac{1}{2} r^2 (\theta - \sin\theta) = 0.5(0.36)(2.461914 - \sin(141.0575^\circ)) = 0.18(2.461914 - 0.628539) = 0.18(1.833375) = \mathbf{0.330008\text{ m}^2}\]

Notice how stack duplication eliminates re-typing the pipe radius $r = 0.60\text{ m}$ and subtended angle $\theta$: duplicating $r = 0.60$ via 0.6 [ENTER] [ENTER] 0.4 [-] [x⇄y] [÷] [↰] [ACOS] 2 [×] determines subtended angle $\theta = 141.0576^\circ$ while typing $0.6$ once (or work directly in [RAD] mode where $\theta - \sin\theta$ requires no $\pi/180$ conversion!). In [DEG] mode, duplicate $\theta$ into Register $Y$ and convert Register $X$ to radians via [ENTER] [↱] [π] [×] 180 [÷], subtract $\sin\theta$ via [x⇄y] [SIN] [-] to obtain the chord-segment factor $1.833375$, and scale by $r^2/2$ via 0.6 [↰] [x²] [×] 2 [÷] to deliver the wetted flow area at $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.3300)$.

Workbook Problem 6.13: Slosh Baffle Circular Perforation Net Area

Circular baffle plate $R = 400.0\text{ mm}$ perforated with 48 circular holes of diameter $d = 35.0\text{ mm}$ ($r_h = 17.5\text{ mm}$): * Solid plate area: $\pi(400^2) = 160,000\pi \approx 502,654.82\text{ mm}^2$ * Perforations total: $48 \times \pi(17.5^2) = 48 \times 306.25\pi = 14,700\pi \approx 46,181.41\text{ mm}^2$ * Net remaining solid area: $(160,000 - 14,700)\pi = 145,300\pi \approx \mathbf{456,473.41\text{ mm}^2}$

In liquid-propellant rocket tanks, perforated anti-slosh ring baffles damp propellant sloshing modes via orifice vortex shedding while minimizing structural mass. Evaluating gross solid plate area $\pi R^2 \approx 502,654.82\text{ mm}^2$ with 400 [↰] [x²] [↱] [π] [×] and subtracting the 48 perforation holes $48 \times \pi r_h^2 \approx 46,181.41\text{ mm}^2$ via 17.5 [↰] [x²] [↱] [π] [×] 48 [×] [-] leaves net solid baffle area at $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 456473.4126)$, confirming $456,473.41\text{ mm}^2$ (a $9.19\%$ open-area porosity ratio)—or factor $\pi$ out to the end via 400 [↰] [x²] 17.5 [↰] [x²] 48 [×] [-] [↱] [π] [×] to save two keystrokes!

Workbook Problem 6.14: Inscribed and Circumscribed Spheres of Regular Octahedron

Octahedron with edge length $a = 40.0\text{ mm}$: * Circumscribed radius: $R = \frac{a}{\sqrt{2}} = \frac{40}{\sqrt{2}} \approx \mathbf{28.2843\text{ mm}}$ * Inscribed radius: $r = \frac{a}{\sqrt{6}} = \frac{40}{\sqrt{6}} \approx \mathbf{16.3299\text{ mm}}$ Ratio $R/r = \sqrt{3} \approx 1.732051$.

Keystrokes: 40 [ENTER] 2 [√x] [÷] 40 [ENTER] 6 [√x] [÷] [÷] $\rightarrow$ 1.7321

In any regular octahedron (and in its dual, the cube!), the ratio of the circumscribed sphere radius $R$ (passing through the 6 vertices) to the inscribed sphere radius $r$ (tangent to the 8 equilateral-triangle faces) is the scale-invariant constant $R/r = \sqrt{6}/\sqrt{2} = \sqrt{3}$. Evaluate circumradius $R = a/\sqrt{2} \approx 28.2843\text{ mm}$ via 40 [ENTER] 2 [√x] [÷], then inradius $r = a/\sqrt{6} \approx 16.3299\text{ mm}$ via 40 [ENTER] 6 [√x] [÷] (leaving both radii visible in $Y$ and $X$). Dividing the two via [÷] yields exact sphere ratio $R/r = \sqrt{3} \approx 1.7321$ at $(T: 0, Z: 0, Y: 0, X: 1.7321)$, and squaring via [↰] [x²] confirms the dimension-free surface-area ratio $(R/r)^2 = 3.0000$.

Workbook Problem 6.15: Prolate Spheroid Volume and Eccentricity

Streamlined airship envelope with semimajor axis $a = 60.0\text{ m}$, equatorial radius $b = 15.0\text{ m}$:

\[V = \frac{4}{3} \pi a b^2 = \frac{4}{3} \pi (60)(15^2) = 80 \pi (225) = 18,000\pi \approx \mathbf{56,548.67\text{ m}^3}$$ $$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{225}{3600}} = \sqrt{1 - 0.0625} = \sqrt{0.9375} \approx \mathbf{0.968246}\]

Keystrokes: 4 [ENTER] 3 [÷] [↱] [π] [×] 60 [×] 15 [↰] [x²] [×] $\rightarrow$ 56548.6678; 15 [↰] [x²] 60 [↰] [x²] [÷] 1 [x⇄y] [-] [√x] $\rightarrow$ 0.9682

A prolate spheroid of fineness ratio $a/b = 60/15 = 4:1$ approaches the ideal low-drag Prandtl-Glauert hull shape for subsonic airships and submarine hulls. Evaluate volume constant $\frac{4\pi}{3} \approx 4.188790$ via 4 [ENTER] 3 [÷] [↱] [π] [×] and multiply by $a \cdot b^2 = 60 \times 15^2$ via 60 [×] 15 [↰] [x²] [×], establishing envelope volume $56,548.67\text{ m}^3$ in Register $X$. For eccentricity, form squared ratio $b^2/a^2 = 0.0625$ with 15 [↰] [x²] 60 [↰] [x²] [÷], subtract from unity, and extract the square root via 1 [x⇄y] [-] [√x] to settle prolate eccentricity at $(T: 0.0000, Z: 0.0000, Y: 56548.6678, X: 0.9682)$.

Chapter 07: Polynomials, Brent-Dekker Solver & Root Finding — Extended Engineering & Competition Problems

Accompanies Chapter 07 of Thinking in the Stack (26 Worked Engineering Problems)

Workbook Problem 7.1: Structural Column Buckling Transcendental Mode

For a slender elastic column of length $L$ and flexural rigidity $EI$ clamped at its base and pinned against lateral sway at its top, enforcing zero bending moment at the pinned support on the fourth-order Euler-Bernoulli beam-column differential equation $EI w'''' + P w'' = 0$ (with wavenumber $k = \sqrt{P/EI}$) yields the transcendental buckling eigenvalue equation:

\[\tan(x) - x = 0 \quad (\text{with } x = k L)\]

Set the angular baseline mode to [RAD] and program the characteristic equation into [↱] [EQN] under label BUCK: X [TAN] X [-].

Keying 4.5 [STO] X loads the initial trial eigenvalue estimate $x_0 = 4.5000\text{ rad}$ (just below the vertical asymptote at $3\pi/2 \approx 4.7124$) into variable $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 4.5000)$. Striking [↱] [SOLVE] X executes Brent's hybrid root solver, evaluating $\tan(x) - x$ and rapidly contracting the interval via inverse quadratic interpolation to isolate the first non-trivial buckling root $(T: 0.0000, Z: 0.0000, Y: 4.5000, X: \mathbf{4.493409})$ in Register $X$ ($\approx 1.4303\pi$, corresponding to an effective column length factor $K = \pi / x = 0.69916$).

Workbook Problem 7.2: Blackbody Planck Peak Wavelength (Wien's Law Derivation)

Differentiating Max Planck's (1900) quantum blackbody spectral radiance $B_\lambda(\lambda, T) = \frac{2hc^2}{\lambda^5 (e^{hc/\lambda k_B T} - 1)}$ with respect to wavelength $\lambda$ and setting $\partial B_\lambda / \partial \lambda = 0$ under the dimensionless photon-energy substitution $x = \frac{hc}{\lambda k_B T}$ yields the universal transcendental root governing Wien's displacement law:

\[5(1 - e^{-x}) - x = 0 \quad (\text{where } x = h c / (\lambda k_B T))\]

Program the equation into [↱] [EQN] under label WIEN: 1 X [+/-] [eˣ] [-] 5 [×] X [-].

Entering 5.0 [STO] X stores initial estimate $x_0 = 5.0$ (the high-energy Wien asymptotic limit where $e^{-x} \ll 1$) into variable $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 5.0000)$. Striking [↱] [SOLVE] X executes Brent's solver; within 6 iterations of guarded inverse quadratic interpolation, the root converges to $10^{-12}$ precision $(T: 0.0000, Z: 0.0000, Y: 5.0000, X: \mathbf{4.965114})$ in Register $X$, establishing Wien's displacement constant $b = \frac{hc}{x k_B} = 2.897772 \times 10^{-3}\text{ m}\cdot\text{K}$.

Workbook Problem 7.3: Adiabatic Flame Temperature Equilibrium

During constant-pressure adiabatic combustion of methane in air, integrating the temperature-dependent molar heat capacity $C_p(T) = c_0 + c_1 T$ of the flue gas products ($\text{CO}_2, \text{H}_2\text{O}, \text{N}_2$) against the lower heating value enthalpy release yields a quadratic first-law energy balance:

\[H_{\text{prod}}(T) - H_{\text{react}} = 0 \implies 0.05 T^2 + 120 T - 280,000 = 0\]

Keystrokes: [↱] [POLY] 0.05 [A] 120 [B] 280000 [+/-] [C] 0 [D] [XEQ] [RCL] 3 $\rightarrow$ 1,453.299832

In [↱] [POLY], keying 0.05 [A] 120 [B] registers enthalpy temperature coefficients $A$ and $B$ $(T: 0.0000, Z: 0.0000, Y: 0.0500, X: 120.0000)$. Entering 280000 [+/-] [C] 0 [D] loads enthalpy release $C = -280,000$ and depressed constant $D = 0.0$ $(T: 0.0500, Z: 120.0000, Y: -280000.0000, X: 0.0000)$. Pressing [XEQ] computes Cardano roots; keying [RCL] 5 extracts the physical positive adiabatic flame temperature $(T: 120.0000, Z: -3853.3000, Y: 0.0000, X: \mathbf{1453.299832}\text{ K})$ in Register $X$ (with unphysical negative rebound root $-3,853.30\text{ K}$ stored in variable 1), confirming $T_{\text{flame}} = 1,453.30\text{ Kelvin}$.

Workbook Problem 7.4: Bungee Jumper Maximum Cord Stretch

A jumper of mass $m = 75\text{ kg}$ steps from a bridge tethered to an elastic bungee cord of unstretched length $L = 30\text{ m}$ and Hookean spring stiffness $k = 40\text{ N/m}$. At the instant of maximum dynamic arrest, kinetic energy vanishes and total lost gravitational potential energy $mg(L + x)$ equals stored elastic strain energy $\frac{1}{2}kx^2$:

\[20 x^2 - 735.75 x - 22,072.5 = 0\]

Keystrokes: [↱] [POLY] 20 [A] 735.75 [+/-] [B] 22072.5 [+/-] [C] 0 [D] [XEQ] [RCL] 5 $\rightarrow$ 56.366833086

In [↱] [POLY], keying 20 [A] 735.75 [+/-] [B] loads quadratic coefficient $A = 20.0$ and linear coefficient $B = -735.75$ $(T: 0.0000, Z: 0.0000, Y: 20.0000, X: -735.7500)$. Entering 22072.5 [+/-] [C] stores constant term $C = -22,072.5$ (with cubic constant $D$ defaulting to $0.0$ automatically for depressed quadratic solving, or explicitly keyed as 0 [D]) $(T: 20.0000, Z: -735.7500, Y: -22072.5000, X: 0.0000)$. Striking [XEQ] computes Cardano roots; keying [RCL] 5 extracts the physical positive maximum dynamic cord elongation $(T: -735.7500, Z: -19.579333, Y: 0.0000, X: \mathbf{56.366833}\text{ m})$ in Register $X$ (with unphysical negative rebound root $-19.5793\text{ m}$ stored in variable 1), confirming $x = 56.3668\text{ meters}$ (total fall distance $L + x = 86.37\text{ m}$).

Workbook Problem 7.5: Chemical Equilibrium Dissociation Constant ($K_c$)

Dinitrogen tetroxide dissociates endothermically into nitrogen dioxide gas ($N_2 O_4 \rightleftharpoons 2 NO_2$) in a sealed reactor with molar equilibrium constant $K_c = 0.040$ and initial concentration $[N_2 O_4]_0 = 0.50\text{ M}$. By the Guldberg-Waage Law of Mass Action, an ICE (Initial-Change-Equilibrium) reaction extent $x$ yields $[NO_2] = 2x$ and $[N_2 O_4] = 0.50 - x$, satisfying the quadratic equilibrium relation $\frac{(2x)^2}{0.50 - x} = K_c$:

\[4 x^2 + 0.040 x - 0.020 = 0\]

Keystrokes: [↱] [POLY] 4 [A] 0.04 [B] 0.02 [+/-] [C] 0 [D] [XEQ] [RCL] 5 $\rightarrow$ 0.0658872344

In [↱] [POLY], entering 4 [A] 0.040 [B] loads stoichiometry coefficients $A$ and $B$ $(T: 0.0000, Z: 0.0000, Y: 4.0000, X: 0.0400)$. Entering 0.020 [+/-] [C] loads constant $C = -0.020$ (with $D$ defaulting to $0.0$ automatically for depressed quadratic embedding, or explicitly keyed as 0 [D]) $(T: 4.0000, Z: 0.0400, Y: -0.0200, X: 0.0000)$. Striking [XEQ] evaluates quadratic roots; keying [RCL] 5 isolates the positive dissociated equilibrium extent $(T: 0.0400, Z: -0.075887, Y: 0.0000, X: \mathbf{0.065887}\text{ M})$ in Register $X$ ($65.89\text{ mM}$, corresponding to $[NO_2] = 2x = 0.1318\text{ M}$ and $13.18\%$ dissociation).

Workbook Problem 7.6: Cantilever Beam Deflection Under Distributed Load

For a statically indeterminate propped cantilever beam of span $L = 10\text{ m}$ carrying a uniform distributed load, integrating the Euler-Bernoulli elastic curve $EI v''(x) = M(x)$ yields a cubic compatibility polynomial whose root confirms zero deflection and slope at the clamped support:

\[2 x^3 - 3 L x^2 + L^3 = 0 \implies \text{for } L = 10\text{ m}: \quad 2 x^3 - 30 x^2 + 1000 = 0\]

Keystrokes: [↱] [POLY] 2 [A] 30 [+/-] [B] 0 [C] 1000 [D] [XEQ] [RCL] 3 $\rightarrow$ 10

In [↱] [POLY], keying 2 [A] 30 [+/-] [B] registers cubic and quadratic flexural coefficients $(T: 0.0000, Z: 0.0000, Y: 2.0000, X: -30.0000)$. Entering 0 [C] 1000 [D] sets linear coefficient $C = 0.0$ and boundary constant $D = 1000$ $(T: 2.0000, Z: -30.0000, Y: 0.0000, X: 1000.0000)$. Striking [XEQ] runs Cardano's closed-form cubic resolvent, factoring $2(x+5)(x-10)^2 = 0$; keying [RCL] 3 extracts the physical clamped-support double root $(T: -30.0000, Z: -5.0000, Y: 0.0000, X: \mathbf{10.000000}\text{ m})$ in Register $X$, where the repeated multiplicity confirms both $v(L) = 0$ and $v'(L) = 0$.

Workbook Problem 7.7: Rocket Staging Mass Ratio Optimization

A two-stage orbital launch vehicle with payload $m_p = 500\text{ kg}$ and stage structural mass fractions $\epsilon = 0.10$ must maximize total Tsiolkovsky burnout velocity $\Delta v = c_1 \ln R_1 + c_2 \ln R_2$ under a fixed gross liftoff mass constraint. Applying Lagrange multipliers to the coupled stage mass ratios yields a symmetric quadratic optimality condition:

\[\lambda^2 - 2 \mu \lambda + \mu = 0 \implies \lambda^2 - 4.828427 \lambda + 2.414214 = 0\]

Keystrokes: [↱] [POLY] 1 [A] 4.828427 [+/-] [B] 2.414214 [C] 0 [D] [XEQ] [RCL] 5 $\rightarrow$ 4.2619723651

In [↱] [POLY], entering 1 [A] 4.828427 [+/-] [B] enters unit quadratic and negative coupling term $(T: 0.0000, Z: 0.0000, Y: 1.0000, X: -4.828427)$. Entering 2.414214 [C] sets structural constant $C = 2.414214$ (with $D$ defaulting to $0.0$ automatically for quadratic solving, or keyed as 0 [D]) $(T: 1.0000, Z: -4.828427, Y: 2.414214, X: 0.0000)$. Striking [XEQ] evaluates Cardano roots; keying [RCL] 5 yields the optimal stage mass ratio $(T: -4.828427, Z: 0.0000, Y: 0.0000, X: \mathbf{4.261972})$ in Register $X$ (with secondary reciprocal root $\lambda_2 = 0.566455$ in variable 3), confirming $\lambda = 4.261972$.

Workbook Problem 7.8: Economic Production Lot Size with Quadratic Storage Congestion

In a high-density distribution warehouse where floor congestion imposes a quadratic marginal holding cost ($h_2 = \$0.05/\text{unit}^2\cdot\text{yr}$) atop the classical Ford W. Harris (1913) linear inventory holding rate ($h_1 = \$7.35/\text{unit}\cdot\text{yr}$) and annualized setup-demand product ($5,000\text{ units}^2/\text{yr}$), differentiating total annual cost $\text{TC}(Q) = \frac{K D}{Q} + \frac{h_1 Q}{2} + \frac{h_2 Q^2}{4}$ with respect to lot size $Q$ yields the quadratic equilibrium condition: $$0.05 Q^2 + 7.35 Q - 5,000 = 0$$

In [↱] [POLY], keying 0.05 [A] 7.35 [B] loads quadratic congestion coefficient $A = 0.05$ and linear holding coefficient $B = 7.35$ $(T: 0.0000, Z: 0.0000, Y: 0.0500, X: 7.3500)$. Entering 5000 [+/-] [C] [XEQ] loads setup constant $C = -5,000.0$ (with $D = 0.0$ by default) and computes the Cardano roots $(T: 0.0000, Z: 0.0500, Y: -398.1571, X: 0.0000)$. Keying [RCL] 5 recalls the positive optimal production lot size $(T: 0.0500, Z: -398.1571, Y: 0.0000, X: \mathbf{251.157127}\text{ units})$ into Register $X$, yielding $Q^* = 251.16\text{ units}$ (smaller than the congestion-free Wilson EOQ of $\sqrt{2(5000)/7.35} = 368.86\text{ units}$).

Workbook Problem 7.9: Laser Resonator Cavity Gaussian Beam Waist Quadratic

A thermal lens with focal parameter $f = 350\text{ mm}$ inside a flat-flat laser resonator cavity of length $L = 600\text{ mm}$ establishes a confocal Rayleigh range squared $z_0^2 = L(2f - L) = 600(700 - 600) = 60,000\text{ mm}^2$ ($z_0 = 244.9490\text{ mm}$). By Kogelnik's $ABCD$ complex beam parameter transformation, the intracavity TEM$_{00}$ beam waist positions $z$ satisfy the quadratic resonator stability equation: $$z^2 - L z + z_0^2 = 0 \implies z^2 - 600 z + 60,000 = 0$$

In [↱] [POLY], keying 1 [A] 600 [+/-] [B] loads quadratic coefficient $A = 1.0$ and cavity span $B = -600.0$ $(T: 0.0000, Z: 0.0000, Y: 1.0000, X: -600.0000)$. Entering 60000 [C] [XEQ] sets $C = 60,000.0$ and solves the quadratic $(T: 0.0000, Z: 1.0000, Y: 0.0000, X: 0.0000)$. Keying [RCL] 3 extracts the primary intracavity waist position $(T: 1.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{126.794919}\text{ mm})$, and keying [RCL] 5 recalls the conjugate symmetric waist location $(T: 0.0000, Z: 0.0000, Y: 126.7949, X: \mathbf{473.205081}\text{ mm})$ (with $z_1 + z_2 = L = 600.0\text{ mm}$ by Viète's sum of roots).

Workbook Problem 7.10: Centrifugal Pump and Piping System Operating Point Quadratic

A centrifugal pump with backward-curved impeller blades exhibits manometric head curve $H_p(Q) = 60 + 0.15 Q - 0.002 Q^2$, while the pipeline system resistance (static lift plus Darcy-Weisbach friction loss) is $H_s(Q) = 15 + 0.005 Q^2$. Equating pump supply head to pipeline demand head determines the steady-state hydraulic operating point: $$60 + 0.15 Q - 0.002 Q^2 = 15 + 0.005 Q^2 \implies 0.007 Q^2 - 0.15 Q - 45 = 0$$

In [↱] [POLY], entering 0.007 [A] 0.15 [+/-] [B] loads combined hydraulic resistance $A = 0.0070$ and linear Euler swirl coefficient $B = -0.1500$ $(T: 0.0000, Z: 0.0000, Y: 0.0070, X: -0.1500)$. Keying 45 [+/-] [C] [XEQ] sets static head differential $C = -45.0$ and solves for the roots $(T: 0.0000, Z: 0.0070, Y: -70.1768, X: 0.0000)$. Pressing [RCL] 5 isolates the physical positive operating discharge $(T: 0.0070, Z: -70.1768, Y: 0.0000, X: \mathbf{91.605369}\text{ m}^3/\text{hr})$, yielding $Q = 91.61\text{ m}^3/\text{hr}$ at operating head $H = 56.97\text{ m}$.

Workbook Problem 7.11: Combined Stokes-Newton Aerodynamic Drag Equilibrium

A spherical meteorological drop-sonde descending through the atmosphere experiences both linear Stokes viscous shear drag ($c_1 v = 6\pi\mu r v$, with $c_1 = 0.27\text{ N}\cdot\text{s/m}$) and quadratic Newton pressure-wake drag ($c_2 v^2 = \frac{1}{2}\rho C_D A v^2$, with $c_2 = 0.018\text{ N}\cdot\text{s}^2/\text{m}^2$) balancing its net buoyant weight $W = 31.5\text{ N}$: $$c_2 v^2 + c_1 v - W = 0 \implies 0.018 v^2 + 0.27 v - 31.5 = 0$$

In [↱] [POLY], loading 0.018 [A] 0.27 [B] enters quadratic pressure-drag coefficient $A = 0.0180$ and linear viscous coefficient $B = 0.2700$ $(T: 0.0000, Z: 0.0000, Y: 0.0180, X: 0.2700)$. Keying 31.5 [+/-] [C] [XEQ] loads net weight $C = -31.5$ and solves the quadratic $(T: 0.0000, Z: 0.0180, Y: -50.0000, X: 0.0000)$. Pressing [RCL] 5 recalls the positive terminal equilibrium velocity $(T: 0.0180, Z: -50.0000, Y: 0.0000, X: \mathbf{35.000000}\text{ m/s})$, yielding $v = 35.00\text{ m/s}$ ($126.0\text{ km/h}$).

Workbook Problem 7.12: Non-Linear Torsional Pendulum Oscillation Frequency

A hardening Duffing-type torsional suspension exhibits cubic nonlinear restoring torque $\tau(\theta) = -k \theta - \alpha \theta^3$ under static external torque $\tau_{\text{ext}} = 5.0\text{ N}\cdot\text{m}$. For linear stiffness $k = 10\text{ N}\cdot\text{m/rad}$ and strain-hardening coefficient $\alpha = 2\text{ N}\cdot\text{m/rad}^3$, static angular deflection $\theta$ satisfies a naturally depressed cubic polynomial ($B = 0$):

\[2 \theta^3 + 10 \theta - 5 = 0\]

Keystrokes: [↱] [POLY] 2 [A] 0 [B] 10 [C] 5 [+/-] [D] [XEQ] [x⇄y] $\rightarrow$ 0.478138005

In [↱] [POLY], entering 2 [A] loads non-linear cubic coefficient $A = 2.0$ (with quadratic coefficient $B$ defaulting to $0.0$ automatically, or explicitly keyed as 0 [B]) $(T: 0.0000, Z: 0.0000, Y: 2.0000, X: 0.0000)$. Entering 10 [C] 5 [+/-] [D] loads linear restoring torque $C = 10.0$ and external load $D = -5.0$ $(T: 2.0000, Z: 0.0000, Y: 10.0000, X: -5.0000)$. Striking [XEQ] computes Cardano roots; because $p = C/A = 5 > 0$, the cubic is strictly monotonic and the unique real equilibrium deflection angle emerges in Register $Y$ $(T: 2.0000, Z: 0.0000, Y: \mathbf{0.478138}, X: 0.000000)$ with the complex conjugate pair stored in variables 3–6, or recalled directly via [RCL] 1, confirming $\theta = 0.478138\text{ radians}$ ($27.40^\circ$).

Workbook Problem 7.13: Annular Clutch Cubic Torque Allocation Root

For a uniform-wear annular friction clutch of outer radius $R = 150.0\text{ mm}$ where axial contact pressure varies inversely with radius ($p(r) \propto 1/r$), normalizing the inner radius $x = r_i / R \in (0, 1)$ produces normalized frictional torque capacity $T(x) \propto x(1 - x^2)$, which peaks at $x^* = 1/\sqrt{3} \approx 0.57735$. Sizing an annular lining to deliver a specified fraction of theoretical capacity requires solving the depressed cubic polynomial ($\Delta < 0$, casus irreducibilis with three distinct real roots): $$x^3 - 3 x + 1.5 = 0$$

In [↱] [POLY], keying 1 [A] 0 [B] sets leading cubic coefficient $A = 1.0$ and depressed quadratic coefficient $B = 0.0$ $(T: 0.0000, Z: 0.0000, Y: 1.0000, X: 0.0000)$. Entering 3 [+/-] [C] 1.5 [D] [XEQ] loads $C = -3.0, D = 1.5$ and executes François Viète's trigonometric trisection $(T: 1.0000, Z: 0.0000, Y: -1.9422, X: 0.0000)$. Keying [RCL] 3 150 [×] recalls the unique physical inner-to-outer radius ratio $x = 0.557875 \in (0, 1)$ and scales by $R = 150.0\text{ mm}$ $(T: -1.9422, Z: -1.9422, Y: 0.0000, X: \mathbf{83.681205}\text{ mm})$, yielding $r_i = 83.68\text{ mm}$.

Workbook Problem 7.14: Hydraulic Jump Conjugate Depth Quadratic (Bélanger Equation)

A supercritical open-channel spillway flow with upstream depth $y_1 = 0.40\text{ m}$ and Froude number $\text{Fr}_1 = 4.5$ ($\text{Fr}_1^2 = 20.25$) undergoes a turbulent stationary shock—a hydraulic jump—into tranquil subcritical flow. Although mechanical energy is dissipated across the turbulent roller, Jean-Baptiste Ch. Bélanger (1838) showed that specific momentum flux $\frac{1}{2}y^2 + \frac{q^2}{gy}$ is conserved across the jump, yielding the Bélanger quadratic equation for downstream conjugate (sequent) depth $y_2$: $$y_2^2 + y_1 y_2 - 2 y_1^2 \text{Fr}_1^2 = 0 \implies y_2^2 + 0.40 y_2 - 6.48 = 0 \implies y_2 = \mathbf{2.353429\text{ meters}}$$

In [↱] [POLY], entering 1 [A] 0.40 [B] loads unit quadratic coefficient $A = 1.0$ and upstream depth $B = 0.40\text{ m}$ $(T: 0.0000, Z: 0.0000, Y: 1.0000, X: 0.4000)$. Keying 6.48 [+/-] [C] [XEQ] loads momentum constant $C = -2(0.40)^2(20.25) = -6.48$ and extracts the roots $(T: 0.0000, Z: 1.0000, Y: -2.7534, X: 0.0000)$. Pressing [RCL] 5 recalls the positive subcritical conjugate depth $(T: 1.0000, Z: -2.7534, Y: 0.0000, X: \mathbf{2.353429}\text{ m})$, yielding $y_2 = 2.3534\text{ meters}$ (a $5.88\times$ depth jump dissipating $\Delta E = \frac{(y_2 - y_1)^3}{4 y_1 y_2} = 1.98\text{ m}$ of hydraulic head).

Workbook Problem 7.15: Acoustical Helmholtz Resonator Neck-Radius Quadratic

An acoustic Helmholtz absorber chamber of volume $V = 1.93793 \times 10^{-3}\text{ m}^3$ and physical neck length $L = 50.0\text{ mm}$ ($c = 343\text{ m/s}$) is tuned to attenuate a $f_0 = 120\text{ Hz}$ tonal resonance. Because the oscillating air plug in the neck radiates piston-like acoustic inertia into both the cavity and free space, Lord Rayleigh's flanged end-correction extends the effective inertial length to $L' = L + 1.7a$, coupling neck radius $a$ (in $\text{mm}$) into both numerator area $A = \pi a^2$ and denominator mass length inside $f_0 = \frac{c}{2\pi}\sqrt{\frac{\pi a^2}{V(L + 1.7a)}}$: $$a^2 - \left(\frac{6.8\pi f_0^2 V}{c^2}\right) a - \left(\frac{4\pi f_0^2 V L}{c^2}\right) = 0 \implies a^2 - 5.066225 a - 149.006625 = 0$$

In [↱] [POLY], keying 1 [A] 5.066225 [+/-] [B] loads $A = 1.0$ and Rayleigh end-correction coefficient $B = -5.066225\text{ mm}$ $(T: 0.0000, Z: 0.0000, Y: 1.0000, X: -5.066225)$. Entering 149.006625 [+/-] [C] [XEQ] sets cavity constant $C = -149.006625\text{ mm}^2$ and solves the quadratic $(T: 0.0000, Z: 1.0000, Y: -9.9338, X: 0.0000)$. Striking [RCL] 5 isolates the physical positive neck radius $(T: 1.0000, Z: -9.9338, Y: 0.0000, X: \mathbf{15.000000}\text{ mm})$, giving $a = 15.00\text{ mm}$ (neck diameter $d = 2a = 30.00\text{ mm}$ and effective acoustic neck length $L' = 50.0 + 1.7(15.0) = 75.50\text{ mm}$).

Workbook Problem 7.16: Relativistic Proton Kinetic Energy Quadratic

A synchrotron proton ($m_0 c^2 = 938.272\text{ MeV}$) circulates with relativistic momentum $p c = 2,500.0\text{ MeV}$. Substituting total relativistic energy $E = K + m_0 c^2$ into Einstein's Lorentz-invariant energy-momentum four-vector norm $E^2 - (pc)^2 - (m_0 c^2)^2 = 0$ eliminates the rest-mass square and yields a quadratic equation directly in kinetic energy $K$: $$K^2 + 2(m_0 c^2) K - (pc)^2 = 0 \implies K^2 + 1876.544 K - 6,250,000 = 0 \implies K = \mathbf{1,732.0003\text{ MeV}}$$

In [↱] [POLY], keying 1 [A] 1876.544 [B] enters $A = 1.0$ and rest-energy coupling $B = 2 m_0 c^2 = 1876.544\text{ MeV}$ $(T: 0.0000, Z: 0.0000, Y: 1.0000, X: 1876.5440)$. Entering 6250000 [+/-] [C] [XEQ] loads $C = -(pc)^2 = -6,250,000\text{ MeV}^2$ and solves for the energy branches $(T: 0.0000, Z: 1.0000, Y: -3608.5443, X: 0.0000)$. Pressing [RCL] 5 recalls the positive-energy particle kinetic energy $(T: 1.0000, Z: -3608.5443, Y: 0.0000, X: 0.000000)$, yielding $K = 1,732.000\text{ MeV}$ (total energy $E = K + m_0 c^2 = 2,670.272\text{ MeV}$, Lorentz factor $\gamma = E / m_0 c^2 = 2.8459$, while the lower root $r_1 = -3,608.544\text{ MeV} = -E - m_0 c^2$ in Register 3 reflects Paul Dirac's negative-energy sea).

Workbook Problem 7.17: Heat Exchanger Implicit Terminal Temperature Root (LMTD)

A counter-flow shell-and-tube heat exchanger required to maintain Log Mean Temperature Difference $\text{LMTD} = 41.8624^\circ\text{C}$ operates with hot-end temperature approach $\Delta T_1 = 65.0^\circ\text{C}$. Because $\Delta T_2 = X$ appears simultaneously in the linear numerator difference and the logarithmic denominator ratio of $\text{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}$, isolating the cold-end terminal temperature approach requires solving the implicit transcendental LMTD equation: $$f(X) = \frac{65 - X}{\ln(65 / X)} - 41.8624 = 0 \implies X = \Delta T_2 = \mathbf{25.0000^\circ\text{C}}$$

Program 65 X [-] 65 X [÷] [LN] [÷] 41.8624 [-] into [↱] [EQN].

Keying 30 [STO] X seeds initial terminal approach guess $X_0 = 30.0^\circ\text{C}$ (safely below the removable $0/0$ singularity at $X = 65.0^\circ\text{C}$ where $\lim_{X\to 65} \text{LMTD} = 65.0^\circ\text{C}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 30.0000)$. Striking [↱] [SOLVE] X executes Brent's hybrid solver to isolate the cold-end terminal temperature difference $(T: 0.0000, Z: 0.0000, Y: 30.0000, X: \mathbf{25.000000}^\circ\text{C})$, yielding $\Delta T_2 = 25.00^\circ\text{C}$.

Workbook Problem 7.18: Aircraft Cambered Drag-Polar Quadratic Root

An aircraft wing of aspect ratio $\text{AR} = 9.5$ and Oswald span efficiency $e = 0.85$ has Prandtl lifting-line induced-drag factor $k = \frac{1}{\pi e \text{AR}} = 0.039419$, cambered airfoil minimum-drag shift $b = -0.0050$, and parasitic zero-lift drag $C_{D0} = 0.0200$. At a cruise thrust setting corresponding to total drag coefficient $C_D = 0.033405$, the operating lift coefficient $C_L$ satisfies the cambered parabolic drag polar: $$k C_L^2 + b C_L + (C_{D0} - C_D) = 0 \implies 0.039419 C_L^2 - 0.0050 C_L - 0.013405 = 0$$

In [↱] [POLY], entering 0.039419 [A] 0.0050 [+/-] [B] loads induced-drag factor $A = 0.039419$ and camber offset $B = -0.0050$ $(T: 0.0000, Z: 0.0000, Y: 0.0394, X: -0.0050)$. Keying 0.013405 [+/-] [C] [XEQ] sets net drag margin $C = -0.013405$ and extracts the roots $(T: 0.0000, Z: 0.0394, Y: -0.5232, X: 0.0000)$. Pressing [RCL] 5 recalls the positive cruise lift coefficient $(T: 0.0394, Z: -0.5232, Y: 0.0000, X: \mathbf{0.650010})$, yielding $C_L = 0.6500$ (where vortex-induced drag is $C_{Di} = k C_L^2 = 0.016655$ and lift-to-drag ratio is $L/D = C_L / C_D = 19.46$).

Workbook Problem 7.19: Planetary Escaping Orbit Hyperbolic Excess Velocity Quadratic

During a Trans-Mars injection burn at low-Earth parking radius $r = 6,671\text{ km}$ ($\mu_E = 398,600\text{ km}^3/\text{s}^2$, local parabolic escape energy $\frac{2\mu_E}{r} = 119.5023\text{ km}^2/\text{s}^2$), a spacecraft reaches geocentric burnout speed $v = 11.45\text{ km/s}$. Conservation of specific orbital mechanical energy $\mathcal{E} = \frac{v^2}{2} - \frac{\mu_E}{r} = \frac{v_\infty^2}{2}$ yields the vis-viva quadratic for the asymptotic hyperbolic excess velocity $v_\infty$ (and characteristic injection energy $C_3 = v_\infty^2$): $$v_\infty^2 - v^2 + \frac{2\mu_E}{r} = 0 \implies v_\infty = \sqrt{11.45^2 - \frac{2(398600)}{6671}} = \sqrt{131.1025 - 119.5023} = \sqrt{11.6002} = \mathbf{3.4059\text{ km/s}}$$

Pushing 11.45 [↰] [x²] squares injection burn velocity to obtain $v^2 = 131.1025\text{ km}^2/\text{s}^2$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 131.1025)$. Keying 2 [ENTER] 398600 [×] 6671 [÷] [-] evaluates local gravitational escape energy $2\mu_E / r = 119.5023\text{ km}^2/\text{s}^2$ and subtracts to isolate $C_3 = v_\infty^2 = 11.6002\text{ km}^2/\text{s}^2$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 11.6002)$. Striking [√x] extracts the positive quadratic root in place, resolving asymptotic interplanetary departure velocity $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{3.405903}\text{ km/s})$, yielding $v_\infty = 3.4059\text{ km/s}$.

Workbook Problem 7.20: Planar Couette-Poiseuille Channel Velocity Profile Root

Incompressible Newtonian fluid ($\mu = 0.10\text{ Pa}\cdot\text{s}$, $\nu = 10^{-4}\text{ m}^2/\text{s}$) flows between a stationary lower wall ($y = 0$) and a parallel upper wall at $h = 0.050\text{ m}$ translating at $U = 2.0\text{ m/s}$ ($\text{Re} = U h / \nu = 1,000.0$) against an adverse stream-wise pressure gradient $dp/dx = +120\text{ Pa/m}$. Integrating the Navier-Stokes momentum balance $\mu \frac{d^2 u}{dy^2} = \frac{dp}{dx}$ superimposes linear Couette shear and parabolic Poiseuille backflow into $u(y) = 600 y^2 + 10 y$. Find the channel elevation $y$ where local velocity reaches $u(y) = 1.00\text{ m/s}$: $$600 y^2 + 10 y - 1.00 = 0 \implies y = \mathbf{0.033333\text{ meters}}$$

In [↱] [POLY], loading 600 [A] 10 [B] enters pressure-gradient curvature $A = \frac{1}{2\mu}\frac{dp}{dx} = 600.0$ and wall-shear gradient $B = 10.0$ $(T: 0.0000, Z: 0.0000, Y: 600.0000, X: 10.0000)$. Keying 1 [+/-] [C] [XEQ] loads target velocity offset $C = -1.00$ and solves the quadratic $(T: 0.0000, Z: 600.0000, Y: -0.0500, X: 0.0000)$. Pressing [RCL] 5 isolates the physical channel elevation $(T: 600.0000, Z: -0.0500, Y: 0.0000, X: \mathbf{0.033333}\text{ m})$, yielding $y = 33.33\text{ mm}$ ($2/3$ of the channel gap $h$).

Workbook Problem 7.21: Electromagnetic Skin-Depth Attenuation Wavenumber Root

At industrial plasma RF excitation frequency $f = 13.56\text{ MHz}$ in an aluminum conductor ($\sigma = 3.5 \times 10^7\text{ S/m}$, $\mu_0 = 4\pi \times 10^{-7}\text{ H/m}$), Maxwell's curl equations in a good conductor ($\sigma \gg \omega\varepsilon$) reduce to the magnetic diffusion wave equation $\nabla^2 \mathbf{E} = j\omega\mu_0\sigma \mathbf{E}$, whose attenuation wavenumber $k = 1/\delta$ satisfies $k^2 - \pi f \mu_0 \sigma = 0$: $$\delta = \frac{1}{k} = \frac{1}{\sqrt{\pi f \mu_0 \sigma}} = \mathbf{2.311 \times 10^{-5}\text{ m}} = 23.11\,\mu\text{m}$$

Keying [↱] [π] 13.56 [E] 6 [×] computes angular factor $\pi f \approx 4.2600 \times 10^7\text{ rad/s}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 4.2600 \times 10^7)$. Keying 4 [↱] [π] [×] 1 [E] 7 [+/-] [×] [×] multiplies by permeability $\mu_0 = 4\pi \times 10^{-7}\text{ H/m}$, yielding $53.5327$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 53.5327)$. Keying 3.5 [E] 7 [×] multiplies by conductivity $\sigma = 3.5 \times 10^7\text{ S/m}$ to form $k^2 = 1.8736 \times 10^9\text{ m}^{-2}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.8736 \times 10^9)$. Striking [√x] [1/x] extracts the wavenumber root $k$ and inverts in place to obtain the electromagnetic skin depth $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{2.311 \times 10^{-5}}\text{ m})$ ($23.11\,\mu\text{m}$).

Workbook Problem 7.22: Standard-Atmosphere Cubic Barometric Polynomial via Horner's Method

Across the troposphere, static pressure $P(h)$ (in $\text{kPa}$) as a function of geopotential altitude $h$ (in $\text{km}$) is modeled by the cubic barometric calibration polynomial: $$P(h) = -0.025 h^3 + 0.95 h^2 - 12.0 h + 101.325 = ((-0.025 h + 0.95) h - 12.0) h + 101.325$$ Evaluate $P(h)$ at the isothermal atmospheric scale height $h = \frac{R T_0}{M g} = 8.435\text{ km}$ by flooding the T-register so 8.435 is typed only once:

Keying 8.435 [ENTER] [ENTER] [ENTER] fills all four stack registers with $h = 8.435\text{ km}$ $(T: 8.4350, Z: 8.4350, Y: 8.4350, X: 8.4350)$, turning Register $T$ into an inexhaustible constant source for Horner multiplication. Keying 0.025 [+/-] [×] 0.95 [+] evaluates the innermost Horner binomial $-0.025(8.435) + 0.95 = 0.739125$ while $T$ replicates $8.4350$ downward into $Z$ and $Y$ $(T: 8.4350, Z: 8.4350, Y: 8.4350, X: 0.7391)$. Keying [×] 12.0 [-] multiplies directly by the copy of $h = 8.4350$ waiting in Register $Y$ and subtracts $12.0$ to form the quadratic accumulator $-5.765481$ $(T: 8.4350, Z: 8.4350, Y: 8.4350, X: -5.7655)$. Striking [×] 101.325 [+] multiplies by the fresh copy of $8.4350$ in Register $Y$ and adds sea-level standard pressure $101.325\text{ kPa}$, resolving static pressure at $h = 8.435\text{ km}$ $(T: 8.4350, Z: 8.4350, Y: 8.4350, X: \mathbf{52.693171}\text{ kPa})$ without ever re-keying $h$.

Workbook Problem 7.23: Thick-Walled Pressure Vessel Lamé Wall-Thickness Quadratic

A high-pressure cylindrical autoclave of outer radius $R_o = 200\text{ mm}$ ($D = 400\text{ mm}$) containing internal pressure $P = 45.0\text{ MPa}$ must limit maximum inner-bore circumferential (hoop) stress—governed by Gabriel Lamé's (1852) thick-shell elasticity solution $\sigma_\theta(R_i) = P \frac{R_o^2 + (R_o - t)^2}{R_o^2 - (R_o - t)^2}$—to allowable working stress $\sigma_{\text{allow}} = 125.0\text{ MPa}$, yielding the wall-thickness quadratic: $$17 t^2 - 6,800 t + 360,000 = 0 \implies t = \mathbf{62.801132\text{ mm}}$$

In [↱] [POLY], entering 17 [A] 6800 [+/-] [B] loads Lamé stress sum $A = (\sigma_{\text{allow}} + P)/10 = 17.0$ and linear bore coefficient $B = -2 R_o A = -6,800.0$ $(T: 0.0000, Z: 0.0000, Y: 17.0000, X: -6800.0000)$. Keying 360000 [C] [XEQ] loads pressure-radius constant $C = 360,000.0$ and extracts the roots $(T: 0.0000, Z: 17.0000, Y: 0.0000, X: 0.0000)$. Pressing [RCL] 3 recalls the physical wall thickness $t < R_o$ $(T: 17.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{62.801132}\text{ mm})$, yielding $t = 62.80\text{ mm}$ (inner bore radius $R_i = R_o - t = 137.20\text{ mm}$, compared to the unconservative thin-wall Barlow approximation $t_{\text{thin}} = P R_o / \sigma_{\text{allow}} = 72.0\text{ mm}$).

Workbook Problem 7.24: Gyroscopic Rotor Nutation & Steady Precession Quadratic

A rapidly spinning axisymmetric gyroscopic rotor with transverse moment of inertia $I_t = 0.030\text{ kg}\cdot\text{m}^2$, polar spin angular momentum $H = I_p \omega_s = 18.0\text{ N}\cdot\text{m}\cdot\text{s}$, and gravitational overturning torque $M g L = 2.40\text{ N}\cdot\text{m}$ executes steady conical motion at nutation angle $\theta \approx 90^\circ$ when its precession angular velocity $\Omega$ satisfies Euler's gyroscopic quadratic: $$I_t \Omega^2 - H \Omega + M g L = 0 \implies 0.030 \Omega^2 - 18.0 \Omega + 2.40 = 0$$

In [↱] [POLY], keying 0.030 [A] 18.0 [+/-] [B] loads transverse inertia $A = 0.0300$ and negative spin angular momentum $B = -18.0000$ $(T: 0.0000, Z: 0.0000, Y: 0.0300, X: -18.0000)$. Entering 2.40 [C] [XEQ] loads gravitational torque $C = 2.40$ and solves for both modal frequencies $(T: 0.0000, Z: 0.0300, Y: 0.0000, X: 0.0000)$. Keying [RCL] 3 extracts the slow steady precession root $(T: 0.0300, Z: 0.0000, Y: 0.0000, X: \mathbf{0.133363}\text{ rad/s})$ (closely matching the elementary asymptotic approximation $\Omega_1 \approx M g L / H = 0.133333\text{ rad/s}$), while [RCL] 5 recalls the fast retrograde nutational mode $(T: 0.0000, Z: 0.0000, Y: 0.1334, X: \mathbf{599.866637}\text{ rad/s} \approx H / I_t)$.

Workbook Problem 7.25: Geostationary Orbital Altitude Pure Cubic Root

For Earth's sidereal rotation period $T = 86,164.1\text{ s}$ ($23\text{ h } 56\text{ m } 4.1\text{ s}$) and geocentric gravitational constant $\mu = 3.9860044 \times 10^{14}\text{ m}^3/\text{s}^2$ ($R_E = 6,378.14\text{ km}$), Kepler's Third Law ($r^3 \omega^2 = \mu$) dictates that the Clarke geostationary orbital radius $r$ (expressed in megameters, $1\text{ Mm} = 1,000\text{ km}$) satisfies the pure cubic equation: $$r^3 - \frac{\mu T^2}{4\pi^2} = 0 \implies r^3 - 74,960.202 = 0 \implies r = \mathbf{42.164173\text{ Mm}} = 42,164.17\text{ km}$$

In [↱] [POLY], keying 1 [A] 74960.202 [+/-] [D] [XEQ] sets $A = 1.0$ and $D = -74,960.202$ (leaving $B = C = 0$ by default) and extracts the unique real cube root in Register $Y$ $(T: 0.0000, Z: 0.0000, Y: 42.1642, X: 0.0000)$ alongside the two non-physical complex conjugate roots $r e^{\pm j 2\pi/3}$ in variables 3–6. Keying [x⇄y] 1000 [×] 6378.14 [-] brings $r$ to Register $X$, converts to kilometers, and subtracts Earth's equatorial radius $R_E$, resolving equatorial geostationary altitude $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{35786.0326}\text{ km})$.

Workbook Problem 7.26: High-Speed Sprint: Cardano Cubic Instant Root

Solve the depressed cubic $x^3 - 7x - 6 = 0$. Factorization: $(x+1)(x+2)(x-3) = 0 \implies x_1 = -1, x_2 = -2, x_3 = 3$. The Cardano/Viète hardware solver outputs $x = 3.0$ in under 2 seconds! To verify root $x = 3$ in a single pass, stream the depressed cubic $x^3 - 7x - 6 = (x^2 - 7)x - 6$ through the stack via Horner's method: Executing 3 [ENTER] [ENTER] [×] loads candidate root $x = 3$ into $Y$ and squares to obtain $x^2 = 9.0000$ in $X$ $(T: 0.0000, Z: 0.0000, Y: 3.0000, X: 9.0000)$. Keying 7 [-] subtracts 7 to establish the inner Horner factor $x^2 - 7 = 2.0000$ while $x = 3.0000$ waits untouched in Register $Y$ $(T: 0.0000, Z: 0.0000, Y: 3.0000, X: 2.0000)$. Striking [×] 6 [-] multiplies directly by the candidate root waiting in $Y$ ($3 \times 2 = 6$) and subtracts 6, yielding an exact zero residual $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{0.0000})$ and confirming $x = 3.0$ ($f(3) = 0$).

Chapter 08: Numerical Calculus & Adaptive Romberg Quadrature — Extended Engineering & Competition Problems

Accompanies Chapter 08 of Thinking in the Stack (27 Worked Engineering Problems)

Workbook Problem 8.1: Exponential Distribution CDF Integral

In reliability engineering, the memoryless exponential distribution models the survival of electronic components subjected to random Poisson failure shocks. The cumulative probability that a component with unit hazard rate $\lambda = 1$ fails before normalized time $t = 1$ is $F(1) = \int_0^1 e^{-x} dx$:

\[\int_0^1 e^{-x} \, dx = 1 - e^{-1} \approx 1 - 0.3678794 = \mathbf{0.632120559}\]

Program the hazard density into [↱] [EQN] as X [+/-] [eˣ]. Staging the unit lifetime interval 0 [ENTER] 1 $(T: 0.0000, Z: 1.0000, Y: 0.0000, X: 1.0000)$ and invoking [↱] [∫] X constructs the Romberg tableau across the concave decay curve, depositing the certified Richardson truncation bound $4.82 \times 10^{-12}$ into Register $Y$ and the canonical $1 - e^{-1} \approx \mathbf{0.632120559}$ ($63.21\%$ one-time-constant failure probability) into Register $X$ $(T: 0.0000, Z: 1.0000, Y: 4.82\times 10^{-12}, X: 0.632120559)$.

Analytical Stack Verification: 1 [ENTER] 1 [+/-] [eˣ] [-] $\implies \mathbf{0.632121}$.

Workbook Problem 8.2: Uniform Distribution Probability Integral

A quantizer or phase-locked loop whose phase error is uniformly distributed across $[0, 10]$ has constant density $f(x) = 0.10$. Integrating over the half-span $[0, 5]$ provides an immediate diagnostic baseline for zero-curvature quadrature:

\[\int_0^5 0.10 \, dx = 0.10 \times 5 = \mathbf{0.500000}\]

Program the constant density 0.1 into [↱] [EQN], stage the interval 0 [ENTER] 5 $(T: 0.0000, Z: 0.1000, Y: 0.0000, X: 5.0000)$, and execute [↱] [∫] X. Because all higher derivatives $f^{(2m)}(x)$ vanish identically in the Euler-Maclaurin expansion, the very first trapezoidal panel $R(0,0)$ is exact, returning an identically zero truncation error bound $0.0000$ in Register $Y$ and $\mathbf{0.500000000}$ in Register $X$ $(T: 0.0000, Z: 0.1000, Y: 0.0000, X: 0.500000000)$.

Analytical Stack Verification: 0.1 [ENTER] 5 [×] $\implies \mathbf{0.500000}$.

Workbook Problem 8.3: Gaussian Normal 1-Sigma Confidence Interval ($68.27\%$)

The Gaussian density $\phi(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ possesses no elementary antiderivative—a theorem proved by Joseph Liouville in 1835. Integrating across the central inflection points from $z = -1.0$ to $z = +1.0$ establishes the $1\sigma$ metrological confidence band:

\[P(-1 \le Z \le 1) = \frac{1}{\sqrt{2\pi}} \int_{-1}^1 e^{-x^2 / 2} \, dx \approx \mathbf{0.682689492}\]

With the standard normal curve NORM active, staging the symmetric $1\sigma$ bounds via 1 [+/-] [ENTER] 1 (or 1 [+/-] [ENTER] [+/-]) $(T: -1.0000, Z: 1.0000, Y: -1.0000, X: 1.0000)$ and pressing [↱] [∫] X eliminates the inflection curvature through Richardson extrapolation, returning error bound $1.85 \times 10^{-11}$ in Register $Y$ and the $68.27\%$ confidence probability $\mathbf{0.682689492}$ in Register $X$ $(T: -1.0000, Z: 1.0000, Y: 1.85\times 10^{-11}, X: 0.682689492)$.

Workbook Problem 8.4: Gaussian Normal 2-Sigma Confidence Interval ($95.45\%$)

Extending the integration domain to $z \in [-2.0, +2.0]$ spans both the concave core ($|z| < 1$) and the convex shoulders ($|z| > 1$) of the Gaussian bell curve:

\[P(-2 \le Z \le 2) = \frac{1}{\sqrt{2\pi}} \int_{-2}^2 e^{-x^2 / 2} \, dx \approx \mathbf{0.954499736}\]

Using the same NORM equation, stage the $2\sigma$ limits 2 [+/-] [ENTER] 2 $(T: 0.0000, Z: 0.0000, Y: -2.0000, X: 2.0000)$ and invoke [↱] [∫] X. The Romberg tableau resolves the inflection transition with error bound $3.14 \times 10^{-11}$ in $Y$ and deposits the two-sided $95.45\%$ confidence integral $\mathbf{0.954499736}$ into Register $X$ $(T: 0.0000, Z: 0.0000, Y: 3.14\times 10^{-11}, X: 0.954499736)$.

Workbook Problem 8.5: Gaussian Normal 3-Sigma Confidence Interval ($99.73\%$)

In statistical process control (Walter Shewhart's $\pm 3\sigma$ control limits), integrating from $z = -3.0$ to $z = +3.0$ captures all but $0.27\%$ ($2,700\text{ ppm}$) of random variation:

\[P(-3 \le Z \le 3) = \frac{1}{\sqrt{2\pi}} \int_{-3}^3 e^{-x^2 / 2} \, dx \approx \mathbf{0.997300204}\]

Stage the $3\sigma$ bounds 3 [+/-] [ENTER] 3 $(T: 0.0000, Z: 0.0000, Y: -3.0000, X: 3.0000)$ and execute [↱] [∫] X. Adaptive Richardson extrapolation integrates out into the asymptotic tails to error bound $5.22 \times 10^{-11}$ in $Y$, placing the canonical $99.73\%$ confidence area $\mathbf{0.997300204}$ into Register $X$ $(T: 0.0000, Z: 0.0000, Y: 5.22\times 10^{-11}, X: 0.997300204)$.

Workbook Problem 8.6: Polytropic Gas Expansion Work: $P V^{1.4} = C$

An ideal diatomic gas expands reversibly in an insulated cylinder from $V_1 = 0.02\text{ m}^3$ to $V_2 = 0.08\text{ m}^3$. Initial chamber pressure $P_1 = 600.0\text{ kPa} = 600,000\text{ Pa}$. Polytropic index $n = 1.4$. State constant $C = P_1 V_1^{1.4} = 600,000 \times (0.02)^{1.4} \approx 2,492.357\text{ Pa}\cdot\text{m}^{4.2}$. Boundary expansion work $W = \int_{V_1}^{V_2} P(V) \, dV$: $$W = C \int_{0.02}^{0.08} V^{-1.4} \, dV = \frac{P_1 V_1 \left[ 1 - (V_1/V_2)^{0.4} \right]}{0.4} = \frac{12,000 \left[ 1 - (0.25)^{0.4} \right]}{0.4}$$ $$(0.25)^{0.4} \approx 0.57434918 \implies 1 - 0.57434918 = 0.42565082$$ $$W = \frac{12,000 \times 0.42565082}{0.4} = 30,000 \times 0.42565082 = \mathbf{12,769.52\text{ Joules}} = \mathbf{12.7695\text{ kJ}}$$

Program the adiabatic pressure law V 1.4 [+/-] [yˣ] 2492.357 [×] into [↱] [EQN]. Stage the expansion volume bounds 0.02 [ENTER] 0.08 $(T: 0.0000, Z: 0.0000, Y: 0.0200, X: 0.0800)$ and press [↱] [∫] V $(T: 0.0000, Z: 0.0000, Y: 2.18\times 10^{-10}, X: 12769.5248)$. Romberg quadrature integrates the steep inverse-power pressure decay across the $4:1$ expansion ratio, placing error bound $2.18 \times 10^{-10}$ in Register $Y$ and mechanical boundary work $12,769.5248\text{ J}$ in Register $X$, which 1000 [÷] scales to kilojoules $(T: 0.0000, Z: 0.0000, Y: 2.18\times 10^{-10}, X: 12.7695)$.

Workbook Problem 8.7: Surface Area of Paraboloid Radar Dish of Revolution

A parabolic satellite reflector dish is generated by revolving $y = 0.25 x^2$ about the vertical $y$-axis from center $x = 0$ to outer rim radius $R = 2.0\text{ m}$. Surface area of revolution formula: $$A = 2\pi \int_0^R x \sqrt{1 + [y'(x)]^2} \, dx = 2\pi \int_0^2 x \sqrt{1 + (0.5 x)^2} \, dx = 2\pi \int_0^2 x \sqrt{1 + 0.25 x^2} \, dx$$ Analytical evaluation via $u = 1 + 0.25 x^2, du = 0.5 x dx \implies x dx = 2 du$: $$A = 4\pi \int_1^2 u^{1/2} \, du = \frac{8\pi}{3} \left[ u^{3/2} \right]_1^2 = \frac{8\pi}{3} (2\sqrt{2} - 1) = \frac{8\pi}{3}(1.8284271) \approx \mathbf{15.317764\text{ m}^2}$$

Program the surface-element integrand X 0.5 [×] [↰] [x²] 1 [+] [√x] X [×] 2 [×] [↱] [π] [×] into [↱] [EQN]. Staging the aperture radial bounds 0 [ENTER] 2 $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 2.0000)$ and pressing [↱] [∫] X integrates the curved paraboloid shell—which exceeds the flat aperture disc area $\pi R^2 = 12.5664\text{ m}^2$ by $21.9\%$—returning error bound $2.14 \times 10^{-10}$ in Register $Y$ and surface area $\mathbf{15.317764\text{ m}^2}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 2.14\times 10^{-10}, X: 15.317764)$ ($A = 15.3178\text{ m}^2$).

Workbook Problem 8.8: Thermal Heat Conduction Across Temperature-Dependent Wall

High-temperature furnace wall of thickness $L = 0.20\text{ m}$ and cross-sectional area $A = 10.0\text{ m}^2$. Material thermal conductivity is temperature-dependent: $k(T) = 0.50 + 0.002 T\text{ W/(m}\cdot\text{K)}$. Internal hot face $T_1 = 800.0\text{ K}$, exterior cold face $T_2 = 300.0\text{ K}$. One-dimensional steady heat flow rate $q$ via Fourier's Law: $$q = \frac{A}{L} \int_{300}^{800} k(T) \, dT = \frac{10.0}{0.20} \int_{300}^{800} (0.50 + 0.002 T) \, dT = 50.0 \left[ 0.50 T + 0.001 T^2 \right]_{300}^{800}$$ $$q = 50.0 \left[ 0.50(500) + 0.001(800^2 - 300^2) \right] = 50.0 [250 + 0.001(640,000 - 90,000)] = 50.0 [250 + 550]$$ $$q = 50.0 \times 800 = \mathbf{40,000.0\text{ Watts}} = \mathbf{40.000\text{ kW}}$$

Program Kirchhoff's integral transform T 0.002 [×] 0.5 [+] 50 [×] into [↱] [EQN]. Staging the cold-to-hot wall boundary temperatures 300 [ENTER] 800 $(T: 0.0000, Z: 25.0000, Y: 300.0000, X: 800.0000)$ and pressing [↱] [∫] T $(T: 0.0000, Z: 25.0000, Y: 0.0000, X: 40000.0000)$ integrates the temperature-dependent refractory conductivity with zero truncation error ($0.0000$ in $Y$, since Trapezoidal/Simpson quadrature is exact on linear and cubic polynomials), yielding $40,000.0\text{ W}$ in $X$ before 1000 [÷] scales the flux to kilowatts $(T: 25.0000, Z: 25.0000, Y: 0.0000, X: 40.0000)$ ($q = 40.000\text{ kW}$).

Workbook Problem 8.9: Fresnel Sine Integral in Diffraction Optics ($S(x)$)

When coherent light passes across a sharp razor edge, Augustin-Jean Fresnel (1818) showed that the near-field diffraction fringes are governed by the non-elementary Fresnel sine integral:

$$S(x) = \int_0^x \sin\left(\frac{\pi}{2} t^2\right) dt$$ Evaluate $S(2.0)$ at normalized aperture coordinate $x = 2.0$.

With radian mode engaged ([RAD]), program the quadratic-phase chirp integrand T [↰] [x²] [↱] [π] [×] 2 [÷] [SIN] into [↱] [EQN]. Staging 0 [ENTER] 2 $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 2.0000)$ and pressing [↱] [∫] T resolves the accelerating phase oscillations across two full half-cycles, returning certified error bound $1.48 \times 10^{-10}$ in Register $Y$ and the Fresnel sine coordinate $S(2.0) = \mathbf{0.343416}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 1.48\times 10^{-10}, X: 0.343416)$.

Workbook Problem 8.10: Fresnel Cosine Integral ($C(x)$)

The companion Fresnel cosine integral defines the horizontal coordinate of Marie Alfred Cornu's clothoid spiral (1874), whose arc length equals $x$ and whose curvature grows linearly with arc length ($\kappa = \pi x$):

\[C(x) = \int_0^x \cos\left(\frac{\pi}{2} t^2\right) dt$$ Evaluate $C(2.0)$ at $x = 2.0$. Together with $S(2.0) = 0.343416$ stored in Register `A` from Workbook Problem 8.9, this defines the Cornu spiral phasor position: $$r = \sqrt{C^2(2) + S^2(2)} = \sqrt{0.488253^2 + 0.343416^2} = \sqrt{0.238391 + 0.117935} = \sqrt{0.356326} \approx \mathbf{0.596930}\]

In radian mode ([RAD]), with the companion cosine integrand T [↰] [x²] [↱] [π] [×] 2 [÷] [COS] in [↱] [EQN], first store $S(2.0)$ in Register A, then integrate $C(2.0)$ and combine both components on the stack without re-typing intermediate floats:

Store the Fresnel sine component 0 [ENTER] 2 [↱] [∫] T [STO] A $(T: 0.0000, Z: 0.0000, Y: 1.48\times 10^{-10}, X: 0.343416)$, then set integration limits 0 [ENTER] 2 $(T: 0.0000, Z: 0.3434, Y: 0.0000, X: 2.0000)$ and press [↱] [∫] T $(T: 0.0000, Z: 0.3434, Y: 2.22\times 10^{-10}, X: 0.488253)$ to evaluate the Fresnel cosine integral $C(2.0) = 0.488253$ with error bound $2.22 \times 10^{-10}$ in Register $Y$. Squaring, recalling $S(2.0)$ from Register A to square and add, and taking the square root ([↰] [x²] [RCL] A [↰] [x²] [+] [√x]) yields the Cornu spiral phasor amplitude $\mathbf{0.596930}$ $(T: 0.3434, Z: 0.3434, Y: 2.22\times 10^{-10}, X: 0.596930)$.

Workbook Problem 8.11: Gravitational Attraction of Thin Rod on End Particle

A slender spacecraft boom of length $L = 5.0\text{ m}$ and uniform mass $M = 100.0\text{ kg}$ is positioned along the $x$-axis from $x = 1.0\text{ m}$ to $x = 6.0\text{ m}$. A sensor payload of mass $m = 2.0\text{ kg}$ is positioned at the origin $x = 0$. Linear mass density $\lambda = M/L = 100.0 / 5.0 = 20.0\text{ kg/m}$. Gravitational constant $G = 6.6743 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$. Newtonian attraction force $F$: $$F = G m \lambda \int_1^6 \frac{dx}{x^2} = (6.6743 \times 10^{-11}) \times 2.0 \times 20.0 \left[ -\frac{1}{x} \right]_1^6 = 2.66972 \times 10^{-9} \left[ 1 - \frac{1}{6} \right]$$ $$F = 2.66972 \times 10^{-9} \times \frac{5}{6} = \mathbf{2.224767 \times 10^{-9}\text{ Newtons}} = \mathbf{2.224767\text{ nN}}$$

Program the Newtonian differential attraction X [↰] [x²] [1/x] 2.66972 [E] 9 [+/-] [×] into [↱] [EQN]. Staging the boom span limits 1 [ENTER] 6 $(T: 0.0000, Z: 0.0000, Y: 1.0000, X: 6.0000)$ and pressing [↱] [∫] X $(T: 0.0000, Z: 0.0000, Y: 1.15\times 10^{-16}, X: 2.224767\times 10^{-9})$ integrates the inverse-square field—demonstrating why distributed-mass gravity cannot be approximated at the geometric midpoint $x = 3.5\text{ m}$, but acts at the harmonic mean distance $\sqrt{x_1 x_2} = \sqrt{6} \approx 2.449\text{ m}$. Multiplying by $10^9$ (1 [E] 9 [×]) scales the self-gravity perturbation to nanonewtons $(T: 0.0000, Z: 0.0000, Y: 1.15\times 10^{-16}, X: 2.224767)$.

Workbook Problem 8.12: Fourier Series 1st Harmonic Sine Coefficient ($b_1$)

An odd square wave of fundamental period $2\pi$ has value $+1$ on $(0, \pi)$ and $-1$ on $(-\pi, 0)$. The fundamental Fourier harmonic sine coefficient $b_1$ is: $$b_1 = \frac{2}{\pi} \int_0^\pi \sin(x) \, dx = \frac{2}{\pi} \left[ -\cos(x) \right]_0^\pi = \frac{2}{\pi} (1 - (-1)) = \frac{4}{\pi} \approx \mathbf{1.273240}$$

With radian mode engaged ([RAD]), program the orthogonal harmonic projection X [SIN] into [↱] [EQN]. Staging the half-period interval 0 [↱] [π] $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 3.141593)$ and pressing [↱] [∫] X $(T: 0.0000, Z: 0.0000, Y: 1.82\times 10^{-11}, X: 2.000000)$ evaluates the sine arch area to $2.000000$ with error bound $1.82 \times 10^{-11}$ in $Y$. Normalizing by 2 [×] [↱] [π] [÷] resolves the fundamental Fourier amplitude $b_1 = 4/\pi \approx \mathbf{1.273240}$ $(T: 0.0000, Z: 0.0000, Y: 1.82\times 10^{-11}, X: 1.273240)$—explaining why the first harmonic of a unit square wave overshoots the flat $\pm 1$ plateau by $27.3\%$.

Workbook Problem 8.13: Sound Exposure Level (SEL) from Transient Flyover Noise

An aircraft flyover noise pulse at an airfield boundary is modeled by acoustic pressure $p(t) = p_0 e^{-t^2 / 20}\text{ Pa}$ with peak pressure $p_0 = 2.00\text{ Pa}$ over $t \in [-15, +15]\text{ s}$. Reference acoustic pressure $p_{\text{ref}} = 20\,\mu\text{Pa} = 2.0 \times 10^{-5}\text{ Pa}$, reference time $t_0 = 1.0\text{ s}$. Sound exposure integral $E = \int_{-15}^{15} p^2(t) \, dt = \int_{-15}^{15} 4.0 e^{-t^2 / 10} \, dt$: $$E \approx 4.0 \sqrt{10\pi} = 4.0 \times 5.604991 = \mathbf{22.41996\text{ Pa}^2\cdot\text{s}}$$ $$\text{SEL} = 10 \log_{10}\left( \frac{E}{p_{\text{ref}}^2 \cdot t_0} \right) = 10 \log_{10}\left( \frac{22.41996}{4.0 \times 10^{-10}} \right) = 10 \log_{10}(5.604991 \times 10^{10}) = \mathbf{107.4857\text{ dBA}}$$

Because certification standards compress an entire transient acoustic event into a 1-second equivalent energy dose, entering the squared-pressure envelope T [↰] [x²] 10 [÷] [+/-] [eˣ] 4 [×] into [↱] [EQN] and sweeping across the 30-second flyover window 15 [+/-] [ENTER] 15 $(T: 0.0000, Z: 4.0000, Y: -15.0000, X: 15.0000)$ via [↱] [∫] T $(T: 0.0000, Z: 4.0000, Y: 4.88\times 10^{-11}, X: 22.419964)$ integrates total sound exposure $E = 22.419964\text{ Pa}^2\cdot\text{s}$ in Register $X$ (with quadrature uncertainty $4.88 \times 10^{-11}$ in $Y$). Normalizing against the human auditory threshold $p_{\text{ref}}^2 t_0 = 4.0 \times 10^{-10}\text{ Pa}^2\cdot\text{s}$ and mapping to the decibel scale (4 [E] 10 [+/-] [÷] [↰] [LOG] 10 [×]) establishes the certified Sound Exposure Level $\mathbf{107.4857\text{ dBA}}$ $(T: 4.0000, Z: 4.0000, Y: 4.88\times 10^{-11}, X: 107.485743)$ ($\text{SEL} = 107.49\text{ dBA}$).

Workbook Problem 8.14: Aerodynamic Circulation Line Integral & Prandtl Elliptic Spanwise Lift

An aircraft wing of span $b = 12.0\text{ m}$ ($y \in [-6, +6]\text{ m}$) flying at freestream airspeed $V_\infty = 60.0\text{ m/s}$ in air of density $\rho_\infty = 1.225\text{ kg/m}^3$ exhibits Prandtl's minimum-induced-drag elliptic bound circulation distribution with root vortex circulation $\Gamma_0 = 45.0\text{ m}^2/\text{s}$:

$$\Gamma(y) = \Gamma_0 \sqrt{1 - \left(\frac{2y}{b}\right)^2} = 45.0 \sqrt{1 - \left(\frac{y}{6}\right)^2}\text{ m}^2/\text{s}$$ By the Kutta-Joukowski theorem, sectional lift is $L'(y) = \rho_\infty V_\infty \Gamma(y)$ (peaking at $L'_0 = 1.225 \times 60.0 \times 45.0 = 3,307.50\text{ N/m}$ at the wing root), and total wing lift is the spanwise integral: $$L = \rho_\infty V_\infty \int_{-6}^{6} 45.0 \sqrt{1 - (y/6)^2} \, dy = (1.225)(60.0)(45.0)\left(6\cdot\frac{\pi}{2}\right) = 73.50 \times 424.1150 = \mathbf{31,172.45\text{ N}} = \mathbf{31.17\text{ kN}}$$ Equivalent sustained aircraft mass: $M = L / g = 31,172.45 / 9.80665 = \mathbf{3,178.71\text{ kg}} \approx \mathbf{3.18\text{ metric tonnes}}$.

In Ludwig Prandtl's lifting-line theory, an elliptical spanwise circulation profile produces uniform downwash across the trailing vortex sheet and minimizes induced drag. Defining Y 6 [÷] [↰] [x²] 1 [x⇄y] [-] [√x] 45 [×] in [↱] [EQN] and integrating from wingtip to wingtip 6 [+/-] [ENTER] 6 $(T: 0.0000, Z: 45.0000, Y: -6.0000, X: 6.0000)$ via [↱] [∫] Y $(T: 0.0000, Z: 45.0000, Y: 3.84\times 10^{-10}, X: 424.1150)$ yields the integrated vortex volume $\int_{-6}^6 \Gamma(y)\,dy = 135\pi \approx 424.1150\text{ m}^3/\text{s}$. Scaling by the Kutta-Joukowski momentum flux $\rho_\infty V_\infty$ (1.225 [×] 60 [×], $(T: 45.0000, Z: 45.0000, Y: 3.84\times 10^{-10}, X: 31172.4525)$) gives total aerodynamic lift $L = \mathbf{31,172.45\text{ N}}$ ($\mathbf{31.17\text{ kN}}$), which supports a steady-flight aircraft gross mass of $\mathbf{3.18\text{ tonnes}}$ (9.80665 [÷] 1000 [÷], $(T: 45.0000, Z: 45.0000, Y: 3.84\times 10^{-10}, X: 3.1787)$).

Workbook Problem 8.15: Damped Harmonic Oscillator Energy Dissipated Per Cycle

Viscous dashpot damper force $F_d = c \dot{x}$ with damping coefficient $c = 15.0\text{ N}\cdot\text{s/m}$. Natural frequency $\omega = 20.0\text{ rad/s}$, harmonic displacement amplitude $X_0 = 0.020\text{ m}$ ($20\text{ mm}$). Velocity $v(t) = -\omega X_0 \sin(\omega t) = -0.40 \sin(20 t)\text{ m/s}$. Oscillation period $T = \frac{2\pi}{\omega} = \frac{\pi}{10} \approx 0.314159\text{ s}$. Energy dissipated per steady-state cycle: $$\Delta E = \int_0^T c v^2(t) \, dt = 15.0 \int_0^{\pi/10} [ -0.40 \sin(20 t) ]^2 dt = 2.40 \int_0^{\pi/10} \sin^2(20 t) \, dt$$ $$\Delta E = 2.40 \times \left( \frac{T}{2} \right) = 2.40 \times \frac{\pi}{20} = 0.12 \pi \approx \mathbf{0.376991\text{ Joules}}$$

Under steady-state harmonic excitation, the area enclosed by the elliptical force-displacement hysteresis loop equals the viscous thermal dissipation over one full period $T = \pi/10\text{ s}$. With radian mode ([RAD]) active and the instantaneous power dissipation T 20 [×] [SIN] [↰] [x²] 2.4 [×] entered into [↱] [EQN], staging the single-cycle bounds 0 [↱] [π] 10 [÷] $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.314159)$ and pressing [↱] [∫] T integrates the $\sin^2(\omega t)$ waveform to an error bound of $1.12 \times 10^{-11}$ in $Y$ and per-cycle thermal loss $\mathbf{0.376991\text{ Joules}}$ in $X$ $(T: 0.0000, Z: 0.0000, Y: 1.12\times 10^{-11}, X: 0.376991)$ ($\Delta E = 0.3770\text{ Joules}$).

Workbook Problem 8.16: Continuous Present Value of Perpetual Cash Flow Stream

Continuous discount rate $r = 0.06$, cash flow intensity $C(t) = 50,000 e^{-0.02 t}$:

\[PV = \int_0^{30} 50000 e^{-0.08 t} \, dt = \frac{50000}{0.08} [1 - e^{-2.4}] = 625,000 [1 - 0.090718] = \mathbf{\$568,301.25}\]

For a depleting petroleum reservoir or royalty trust whose production decays exponentially at $2\%/\text{yr}$ while capital is discounted continuously at $6\%/\text{yr}$, the net present value is governed by the combined hazard rate $\lambda + r = 0.08\text{ yr}^{-1}$. Programming T 0.08 [×] [+/-] [eˣ] 50000 [×] into [↱] [EQN] and integrating over the 30-year concession horizon 0 [ENTER] 30 $(T: 0.0000, Z: 50000.0000, Y: 0.0000, X: 30.0000)$ via [↱] [∫] T resolves the continuous valuation $\mathbf{\$568,301.25}$ in Register $X$ with quadrature error $4.12 \times 10^{-11}$ in Register $Y$ $(T: 0.0000, Z: 50000.0000, Y: 4.12\times 10^{-11}, X: 568301.25)$.

Workbook Problem 8.17: Second Moment of Area of Structural I-Beam Web ($I_{xx}$)

For a vertical steel web of height $h = 300\text{ mm}$ ($y \in [-150, +150]\text{ mm}$) and thickness $t = 10\text{ mm}$, the flexural bending stiffness depends on the second moment of area $I_{xx}$: $$I_{xx} = \int_{-150}^{150} t y^2 \, dy = 10 \left[ \frac{y^3}{3} \right]_{-150}^{150} = \frac{20}{3}(3,375,000) = \mathbf{22,500,000\text{ mm}^4} = 2.25 \times 10^7\text{ mm}^4$$

In Euler-Bernoulli beam bending, fibers farthest from the neutral axis carry linearly increasing normal stress $\sigma_x = -My/I_{xx}$, weighting the flexural rigidity by $y^2$. Entering the differential moment integrand Y [↰] [x²] 10 [×] into [↱] [EQN] and integrating across the web depth 150 [+/-] [ENTER] 150 $(T: 0.0000, Z: 0.0000, Y: -150.0000, X: 150.0000)$ via [↱] [∫] Y yields exact parabolic convergence on the first Richardson column, returning error bound $1.85 \times 10^{-10}$ in $Y$ and second moment of area $I_{xx} = \mathbf{22,500,000.0\text{ mm}^4}$ ($2.25 \times 10^7\text{ mm}^4$) in $X$ $(T: 0.0000, Z: 0.0000, Y: 1.85\times 10^{-10}, X: 22500000.0)$.

Workbook Problem 8.18: Physical-Optics Aperture Integral & Radar Cross Section (RCS)

Physical-optics surface phase integration over a square flat conducting plate of side $a = 0.50\text{ m}$ ($x \in [-0.25, +0.25]\text{ m}$) at X-band radar wavelength $\lambda = 0.03\text{ m}$ ($10\text{ GHz}$) integrates the illuminated surface area $A = \int_{-0.25}^{0.25} a\,dx = 0.25\text{ m}^2$ to determine the peak specular Radar Cross Section: $$\sigma_{\max} = \frac{4\pi}{\lambda^2}\left[\int_{-a/2}^{a/2} a\,dx\right]^2 = \frac{4\pi A^2}{\lambda^2} = \frac{4\pi (0.25^2)}{0.0009} \approx \mathbf{872.6646\text{ m}^2} \quad (\mathbf{29.41\text{ dBsm}})$$

At normal incidence (broadside flash), every surface patch across the conducting plate scatters in coherent phase, so the Kirchhoff physical-optics integral reduces to integrating the constant transverse width 0.5 in [↱] [EQN] over 0.25 [+/-] [ENTER] 0.25 $(T: 0.0000, Z: 0.5000, Y: -0.2500, X: 0.2500)$. Executing [↱] [∫] X $(T: 0.0000, Z: 0.5000, Y: 0.0000, X: 0.2500)$ yields the illuminated aperture area $A = 0.2500\text{ m}^2$, and applying the Fraunhofer diffraction gain factor $4\pi A^2/\lambda^2$ ([↰] [x²] 4 [×] [↱] [π] [×] 0.03 [↰] [x²] [÷], $(T: 0.5000, Z: 0.5000, Y: 0.0000, X: 872.6646)$) reveals why a half-meter flat plate appears to an X-band radar as an $\mathbf{872.6646\text{ m}^2}$ target ($\mathbf{29.41\text{ dBsm}}$ via [↰] [LOG] 10 [×], $(T: 0.5000, Z: 0.5000, Y: 0.0000, X: 29.4085)$).

Workbook Problem 8.19: Thermal Expansion Clearance of Cryogenic Propellant Line

When a 304 stainless-steel LOX/LN2 transfer line of length $L = 12.0\text{ m}$ is chilled from ambient $T_1 = 293\text{ K}$ down to liquid nitrogen boiling temperature ($T_2 = 77\text{ K}$), the instantaneous thermal expansion coefficient follows $\alpha(T) = (5.5022 + 0.04158\,T)\times 10^{-6}\text{ K}^{-1}$: $$\Delta L = L \int_{293}^{77} \alpha(T) \, dT = 12.0 \int_{293}^{77} (5.5022 + 0.04158\,T)\times 10^{-6} \, dT = 12.0 \times (-0.002850) = \mathbf{-34.20\text{ mm}}\text{ (axial contraction)}$$ Engineering Insight: Staging initial ambient temperature ($a = 293\text{ K}$) in Register $Y$ and cryogenic terminal temperature ($b = 77\text{ K}$) in Register $X$ before pressing [↱] [∫] T automatically produces the negative sign indicating contraction, reminding bellows designers to provide at least $35\text{ mm}$ of axial expansion-joint travel.

Because lattice specific heat and anharmonic phonon vibrations freeze out near liquid-nitrogen temperatures (Debye $T^3$ regime), $\alpha(T)$ varies strongly across the cooldown range. Defining T 0.04158 [×] 5.5022 [+] 1 [E] 6 [+/-] [×] in [↱] [EQN] and integrating in the physical cooling direction 293 [ENTER] 77 $(T: 0.0000, Z: 0.0000, Y: 293.0000, X: 77.0000)$ via [↱] [∫] T $(T: 0.0000, Z: 0.0000, Y: 1.05\times 10^{-13}, X: -0.002850)$ yields the integrated thermal strain $\varepsilon_{\text{th}} = -0.002850$, which scales over the $12\text{ m}$ manifold (12 [×] 1000 [×], $(T: 0.0000, Z: 0.0000, Y: 1.05\times 10^{-13}, X: -34.2000)$) to an axial shrinkage of $\Delta L = \mathbf{-34.20\text{ mm}}$.

Workbook Problem 8.20: Planetary Atmospheric Jeans Escape Flux

Integrating the Maxwell-Boltzmann molecular velocity distribution above a planet's escape velocity $v_{\text{esc}}$ (with exobase density $n_0 = 1.0 \times 10^{13}\text{ m}^{-3}$, thermal speed $v_{\text{th}} = 6,356.7\text{ m/s}$, and Jeans parameter $\lambda = (v_{\text{esc}}/v_{\text{th}})^2 = 9.50$) yields the outward molecular escape flux at the exobase: $$F_J = \frac{n_0 v_{\text{th}}}{2\sqrt{\pi}} \int_{\lambda}^{\infty} u e^{-u} \, du = \frac{n_0 v_{\text{th}}}{2\sqrt{\pi}} (1 + \lambda) e^{-\lambda} = \left(1.7932 \times 10^{16}\right)(10.50) e^{-9.50} \approx \mathbf{1.4102 \times 10^{13}\text{ molecules/(m}^2\cdot\text{s)}}$$

In planetary aeronomy, only molecules in the high-energy Maxwell-Boltzmann tail ($u = v^2/v_{\text{th}}^2 \ge \lambda$) located above the collisionless exobase can escape into interplanetary space. Entering the dimensionless energy-tail integrand U [+/-] [eˣ] U [×] into [↱] [EQN] and integrating from the escape threshold $\lambda = 9.5$ to the negligible upper tail $u = 35$ (9.5 [ENTER] 35, $(T: 0.0000, Z: 0.0000, Y: 9.5000, X: 35.0000)$) via [↱] [∫] U $(T: 0.0000, Z: 0.0000, Y: 4.15\times 10^{-12}, X: 0.000786)$ evaluates $(1 + \lambda)e^{-\lambda} = 0.0007863$; scaling by the upward effusive flux $n_0 v_{\text{th}}/(2\sqrt{\pi})$ (6.3567 [E] 16 [×] [↱] [π] [√x] 2 [×] [÷], $(T: 0.0000, Z: 0.0000, Y: 4.15\times 10^{-12}, X: 1.4102\times 10^{13})$) determines the Jeans escape rate $F_J = \mathbf{1.4102 \times 10^{13}\text{ molecules/(m}^2\cdot\text{s)}}$.

Workbook Problem 8.21: Torsional Rigidity of an Elliptical Shaft (Saint-Venant Warping Integral)

Integrating Saint-Venant warping shear stress over an elliptical shaft cross-section with semiaxes $a = 40\text{ mm}$ and $b = 25\text{ mm}$ reduces to the normalized cross-sectional integral $\int_{-1}^{1}(1 - u^2)^{3/2}\,du = \frac{3\pi}{8} \approx 1.178097$, scaled by $\frac{8 a^3 b^3}{3(a^2 + b^2)}$: $$J = \frac{8 a^3 b^3}{3(a^2 + b^2)} \int_{-1}^{1} (1 - u^2)^{3/2} \, du = \frac{\pi a^3 b^3}{a^2 + b^2} = \frac{\pi (64,000)(15,625)}{1,600 + 625} \approx \mathbf{1,411,951.76\text{ mm}^4}$$

Unlike a circular shaft, an elliptical bar under torsion undergoes out-of-plane axial warping $\psi(x,y) \propto xy$, reducing its torsional stiffness below the polar moment of inertia $I_p$. Entering the normalized Prandtl stress-function integrand U [↰] [x²] 1 [x⇄y] [-] 1.5 [yˣ] into [↱] [EQN] and integrating over 1 [+/-] [ENTER] 1 $(T: 0.0000, Z: 1.0000, Y: -1.0000, X: 1.0000)$ via [↱] [∫] U $(T: 0.0000, Z: 1.0000, Y: 2.85\times 10^{-10}, X: 1.178097)$ evaluates the dimensionless geometric shape factor $3\pi/8 = 1.178097$, which scales by the semiaxis prefactor $\frac{8 a^3 b^3}{3(a^2 + b^2)}$ (8 [×] 3 [÷] 40 [ENTER] 3 [yˣ] [×] 25 [ENTER] 3 [yˣ] [×] 40 [↰] [x²] 25 [↰] [x²] [+] [÷], $(T: 0.0000, Z: 0.0000, Y: 2.85\times 10^{-10}, X: 1411951.76)$) to the Saint-Venant torsion constant $J = \mathbf{1,411,951.76\text{ mm}^4}$.

Workbook Problem 8.22: Second-Order Chemical Batch Reactor Residence Time

For an isothermal second-order reaction ($-r_A = k C_{A0}^2 (1-x)^2$) with characteristic rate product $k C_{A0} = \frac{1}{15}\text{ min}^{-1}$ ($0.06667\text{ min}^{-1}$), integrating the design equation from conversion $x = 0$ to $X = 0.75$ gives the required batch residence time: $$t = \frac{1}{k C_{A0}} \int_0^{0.75} \frac{dx}{(1-x)^2} = 15.0 \left[ \frac{X}{1-X} \right]_{X=0.75} = 15.0 \times \frac{0.75}{0.25} = \mathbf{45.0\text{ minutes}}$$

Because bimolecular collision frequency drops quadratically as reactant $A$ is consumed ($-r_A \propto (1-x)^2$), each additional percentage point of conversion requires disproportionately longer reactor holding time. Entering the Levenspiel batch-design integrand 1 X [-] [↰] [x²] [1/x] 15 [×] into [↱] [EQN] and integrating over 0 [ENTER] 0.75 $(T: 0.0000, Z: 15.0000, Y: 0.0000, X: 0.7500)$ via [↱] [∫] X $(T: 0.0000, Z: 15.0000, Y: 3.12\times 10^{-11}, X: 45.0000)$ yields the exact batch holding time $t = \mathbf{45.0000\text{ minutes}}$, matching the direct stack evaluation of $15 X/(1-X)$ via 0.75 [ENTER] [ENTER] 1 [x⇄y] [-] [÷] 15 [×] $(T: 45.0000, Z: 45.0000, Y: 45.0000, X: 45.0000)$.

Workbook Problem 8.23: Terzaghi Soil Consolidation Degree Integral

Integrating excess pore-water pressure dissipation $u(z, T_v)/u_0 = \frac{4}{\pi}\sin\!\left(\frac{\pi\zeta}{2}\right) e^{-\pi^2 T_v / 4}$ across normalized clay drainage depth $\zeta = z/H_d \in [0, 2]$ at dimensionless time factor $T_v = 0.50$ yields average settlement consolidation: $$U(T_v) = 1 - \frac{1}{2H_d}\int_0^{2H_d} \frac{u(z, T_v)}{u_0} \, dz \approx 1 - \frac{8}{\pi^2} e^{-\pi^2 (0.50) / 4} \approx \mathbf{76.40\%}$$

In Karl Terzaghi's one-dimensional consolidation theory, foundation settlement tracks the spatial average of pore-water dissipation across a doubly-drained clay stratum. With radian mode ([RAD]) active and the spatial eigenmode Z [↱] [π] [×] 2 [÷] [SIN] 2 [×] [↱] [π] [÷] loaded into [↱] [EQN], integrating across the normalized layer thickness 0 [ENTER] 2 $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 2.0000)$ via [↱] [∫] Z $(T: 0.0000, Z: 0.0000, Y: 1.94\times 10^{-11}, X: 0.810569)$ recovers the fundamental Fourier coefficient $8/\pi^2 = 0.810569$. Attenuating by the diffusive decay factor $e^{-\pi^2(0.50)/4}$ and taking the complement ([↱] [π] [↰] [x²] 0.5 [×] 4 [÷] [+/-] [eˣ] [×] 1 [x⇄y] [-], $(T: 0.0000, Z: 0.0000, Y: 1.94\times 10^{-11}, X: 0.7640)$) establishes that $U(0.50) = \mathbf{76.40\%}$ of ultimate primary settlement has completed.

Integrating gravitational restoring torque around Earth's mean radius $R_E = 6,371\text{ km}$ yields the classic Schuler pendulum period $T_S = 2\pi\sqrt{R_E / g_0} = \mathbf{84.4\text{ minutes}}$, while vertical rocket ascent velocity over burn time $t_b = 6.0\text{ s}$ integrates net acceleration $a_{\text{net}}(t) = 10.266667\,t - g_0$ (ideal Tsiolkovsky impulse $\int_0^6 10.266667\,t\,dt = 184.8000\text{ m/s}$ minus the gravity loss integral $\int_0^{t_b} g_0 \, dt = g_0 t_b$): $$v(t_b) = \int_0^6 \left(10.266667\,t - 9.80665\right) dt = 184.8000 - 58.8399 = \mathbf{125.9601\text{ m/s}}$$

During vertical boost, an inertial measurement unit (IMU) senses specific force $a_{\text{thrust}}(t)$ while navigation computer firmware subtracts gravitational acceleration $g_0$ before integrating velocity. Entering T 10.266667 [×] 9.80665 [-] into [↱] [EQN] and integrating over 0 [ENTER] 6 $(T: 0.0000, Z: -9.8066, Y: 0.0000, X: 6.0000)$ via [↱] [∫] T $(T: 0.0000, Z: -9.8066, Y: 0.0000, X: 125.9601)$ gives net vertical burnout speed $v = \mathbf{125.9601\text{ m/s}}$, which decomposes directly on the stack into ideal thrust impulse minus integrated gravity loss (9.80665 [ENTER] 6 [×] 184.8 [x⇄y] [-] $\implies \mathbf{125.9601\text{ m/s}}$).

Workbook Problem 8.25: Supersonic Sears-Haack Minimum Wave-Drag Body

A minimum-drag Sears-Haack fuselage of length $L = 30.0\text{ m}$ ($u = 2x/L \in [-1, +1]$) and maximum mid-body cross-sectional area $S_{\max} = \frac{16 V}{3\pi L} = 2.546479\text{ m}^2$ has cross-sectional area profile $S(u) = S_{\max}(1 - u^2)^{3/2}$. Integrating the cross-sectional area over the fuselage length recovers total volume $V = \frac{L}{2}\int_{-1}^{1} S_{\max}(1 - u^2)^{3/2}\,du = 45.0\text{ m}^3$, and evaluating the von Kármán slender-body wave-drag integral at flight dynamic pressure $q_\infty = 25,000\text{ Pa}$ yields: $$D_{\text{wave}} = q_\infty \frac{128 V^2}{\pi L^4} = 25,000 \times \frac{128(2025)}{\pi(810,000)} \approx \mathbf{2,546.48\text{ N}}$$ Engineering Insight: Because wave drag scales with $L^{-4}$, lengthening the nose-to-tail fineness ratio by just $10\%$ reduces supersonic wave drag by $1 - 1.10^{-4} = 31.7\%$.

By shaping the longitudinal cross-sectional area distribution as $S(u) = S_{\max}(1 - u^2)^{3/2}$, the Sears-Haack body minimizes the von Kármán supersonic acoustic source integral for a fixed enclosed volume $V$. Programming the differential volume element U [↰] [x²] 1 [x⇄y] [-] 1.5 [yˣ] 38.197186 [×] into [↱] [EQN] and integrating from nose to tail 1 [+/-] [ENTER] 1 $(T: 0.0000, Z: 38.1972, Y: -1.0000, X: 1.0000)$ via [↱] [∫] U $(T: 0.0000, Z: 38.1972, Y: 4.12\times 10^{-10}, X: 45.0000)$ confirms enclosed volume $V = 45.0000\text{ m}^3$; applying the slender-body wave-drag scaling $128 q_\infty V^2 / (\pi L^4)$ ([↰] [x²] 128 [×] 25000 [×] 30 [ENTER] 4 [yˣ] [↱] [π] [×] [÷], $(T: 0.0000, Z: 0.0000, Y: 4.12\times 10^{-10}, X: 2546.4791)$) determines the supersonic wave drag $D_{\text{wave}} = \mathbf{2,546.48\text{ N}}$.

Workbook Problem 8.26: Stratospheric Ozone Column Integration (Dobson Units)

Satellite UV backscatter sounding across the stratosphere ($z \in [5, 35]\text{ km}$ with step $h = 15\text{ km}$) measures ozone concentration $D(5) = 3.0\text{ DU/km}$ near the tropopause, $D(20) = 13.0\text{ DU/km}$ at the stratospheric peak, and $D(35) = 5.0\text{ DU/km}$ in the upper stratosphere, corresponding to the quadratic vertical profile $D(z) = -0.04\,z^2 + 1.6667\,z - 4.3333\text{ DU/km}$. Integrating $D(z)$ via [↱] [∫] Z or Romberg's parabolic column $R(1,1) = \frac{h}{3}[D(5) + 4 D(20) + D(35)]$ yields the exact total column abundance: $$\Omega = \int_{5}^{35} D(z) \, dz = \frac{15}{3}\left[ 3.0 + 4(13.0) + 5.0 \right] = \mathbf{300.0\text{ Dobson Units}} \equiv \mathbf{8.0601 \times 10^{22}\text{ molecules/m}^2}$$ (where $1\text{ DU} = 2.6867 \times 10^{20}\text{ molecules/m}^2$, equivalent to a $0.01\text{ mm}$ layer of pure ozone at STP).

Because the first Richardson extrapolation column $R(1,1)$ is algebraically identical to Simpson's $1/3$ rule, Romberg quadrature integrates quadratic atmospheric profiles with zero truncation error at level $n = 1$. Defining Z 0.04 [+/-] [×] 1.666667 [+] Z [×] 4.333333 [-] in [↱] [EQN] and integrating across 5 [ENTER] 35 $(T: 0.0000, Z: -4.3333, Y: 5.0000, X: 35.0000)$ via [↱] [∫] Z $(T: 0.0000, Z: -4.3333, Y: 0.0000, X: 300.0000)$ yields total ozone abundance $\Omega = \mathbf{300.0\text{ DU}}$, which matches the 3-node Simpson combination 3 [ENTER] 4 [ENTER] 13 [×] [+] 5 [+] 15 [×] 3 [÷] $(T: 300.0000, Z: 300.0000, Y: 300.0000, X: 300.0000)$ and converts via 2.6867 [E] 20 [×] $(T: 300.0000, Z: 300.0000, Y: 300.0000, X: 8.0601\times 10^{22})$ to an overhead molecular column of $N_{\text{O3}} = \mathbf{8.0601 \times 10^{22}\text{ molecules/m}^2}$.

Workbook Problem 8.27: High-Speed Sprint: Romberg Richardson Extrapolation on $\int_0^4 3x^2\,dx$

Evaluate $\int_0^4 3x^2 \, dx = [x^3]_0^4 = 4^3 = \mathbf{64.0}$ via [↱] [∫] X and verify how the first Richardson extrapolation column $R(1,1) = \frac{4 R(1,0) - R(0,0)}{3}$ cancels the trapezoidal truncation error on $f(x) = 3x^2$ ($R(0,0) = 96.0$ at $h_0 = 4$, $R(1,0) = 72.0$ at $h_1 = 2$):

Whereas a single trapezoid $R(0,0) = 96.0$ overestimates the concave-up parabola $3x^2$ by $32.0$ and two trapezoids $R(1,0) = 72.0$ overestimate it by $8.0$ (exactly one-fourth the error), Richardson extrapolation weights $R(1,0)$ by $4/3$ and subtracts $R(0,0)/3$ to annihilate the $O(h^2)$ error term completely. Entering X [↰] [x²] 3 [×] into [↱] [EQN] and integrating over 0 [ENTER] 4 $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 4.0000)$ via [↱] [∫] X $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 64.0000)$ returns $\mathbf{64.000000}$, identical to the manual Richardson combination 72 [ENTER] 4 [×] 96 [-] 3 [÷] $(T: 0.0000, Z: 0.0000, Y: 64.0000, X: 64.0000)$. Analytical Stack Verification: 4 [ENTER] 3 [yˣ] $\implies \mathbf{64.0000}$.

Chapter 09: Bivariate Statistics, Curve Fitting & Linear Regression — Extended Engineering & Competition Problems

Accompanies Chapter 09 of Thinking in the Stack (26 Worked Engineering Problems)

Workbook Problem 9.1: Gaussian Normal PDF Peak at the Mean ($Z = 0$)

Calculate the maximum height (mode) of the normal distribution for $\mu = 100, \sigma = 15$ at $x = 100$:

\[f(100) = \frac{1}{15\sqrt{2\pi}} e^0 = \frac{1}{15\sqrt{2\pi}}\]

Keystrokes: 15 [ENTER] 2 [↱] [π] [×] [√x] [×] [1/x] $\rightarrow$ 0.026596152

At the distribution centroid $x = \mu$, the quadratic Mahalanobis distance $(x - \mu)^2/\sigma^2$ vanishes ($e^0 = 1$), so the peak probability density is governed solely by the normalization factor $(\sigma\sqrt{2\pi})^{-1}$. Staging $\sigma = 15.0$ via 15 [ENTER] $(T: 0.0000, Z: 0.0000, Y: 15.0000, X: 15.0000)$, forming the Gaussian radical 2 [↱] [π] [×] [√x] ($\sqrt{2\pi} \approx 2.5066$, $(T: 0.0000, Z: 0.0000, Y: 15.0000, X: 2.5066)$), multiplying by $\sigma$ via [×] ($\sigma\sqrt{2\pi} \approx 37.5994$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 37.5994)$), and inverting with [1/x] resolves the modal density $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{0.02659615})$ in 7 keystrokes without auxiliary storage.

Workbook Problem 9.2: Gaussian Normal PDF at $Z = +2.0$ ($x = 130$)

Calculate the probability density for a score two standard deviations above the mean ($x = 130$):

\[f(130) = \frac{1}{15\sqrt{2\pi}} e^{-2^2 / 2} = \frac{e^{-2}}{15\sqrt{2\pi}}\]

Keystrokes: 2 [+/-] [eˣ] 15 [ENTER] 2 [↱] [π] [×] [√x] [×] [÷] $\rightarrow$ 0.0035993978

At two standard deviations above the mean ($Z = +2.0$), the Gaussian ordinate attenuates by $e^{-Z^2/2} = e^{-2} \approx 0.135335$ relative to the mode. Evaluating the exponential attenuation factor 2 [+/-] [eˣ] $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.1353)$, constructing the dispersion normalizer $15\sqrt{2\pi} \approx 37.5994$ in $X$ while $e^{-2}$ waits in $Y$ (15 [ENTER] 2 [↱] [π] [×] [√x] [×], $(T: 0.1353, Z: 0.1353, Y: 0.1353, X: 37.5994)$), and dividing via [÷] determines the $2\sigma$ tail ordinate $(T: 0.1353, Z: 0.1353, Y: 0.1353, X: \mathbf{0.00359940})$.

Workbook Problem 9.3: Cumulative Normal Probability Critical Z-Score ($Z = 1.96$)

Evaluate $\Phi(1.96)$ for the two-tailed 95% confidence interval ($\alpha = 0.05$):

$$\int_0^{1.96} e^{-t^2/2} \, dt = 1.1906427$$ Scaling by $1/\sqrt{2\pi}$: $1.1906427 \times 0.39894228 = 0.475000$. Cumulative probability: $0.500000 + 0.475000 = \mathbf{0.975000}$ ($97.50\%$).

To verify Ronald Fisher's classic $1.96\sigma$ benchmark for a two-sided $95\%$ confidence interval ($\alpha = 0.05$, leaving $2.5\%$ in each tail), keep the unnormalized Gaussian kernel $e^{-x^2/2}$ (X [↰] [x²] 2 [÷] [+/-] [eˣ]) in [↱] [EQN] mode and integrate over 0 [ENTER] 1.96 $(T: 0.0000, Z: 1.0000, Y: 0.0000, X: 1.9600)$ via [↱] [∫] $(T: 0.0000, Z: 1.0000, Y: 0.0000, X: 1.19064270)$, which places the converged integral $1.19064270$ in Register $X$ and the Romberg truncation estimate ($2.04 \times 10^{-12}$) in Register $Y$. Normalizing by $\sqrt{2\pi}$ (2 [↱] [π] [×] [√x] [÷]) isolates the half-interval probability $0.47500000$, and adding the lower half-plane area (0.5 [+]) confirms $\Phi(1.96) = \mathbf{0.97500000}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{0.97500000})$ ($97.50\%$).

Workbook Problem 9.4: Cumulative Normal Probability at $Z = 2.576$ (99% Two-Tailed)

Evaluate $\Phi(2.576)$ ($\alpha = 0.01$):

$$\int_0^{2.576} e^{-t^2/2} \, dt = 1.240767$$ Scaling: $1.240767 \times 0.39894228 = 0.495000$. Cumulative probability: $0.500000 + 0.495000 = \mathbf{0.995000}$ ($99.50\%$).

Extending the same [↱] [EQN] kernel $e^{-x^2/2}$ to the $99\%$ two-tailed threshold ($\alpha = 0.01$, leaving $0.5\%$ in the upper tail) requires only entering the new limits 0 [ENTER] 2.576 and pressing [↱] [∫] $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.24076700)$. Dividing the unnormalized integral $1.24076700$ by $\sqrt{2\pi}$ (2 [↱] [π] [×] [√x] [÷]) yields $0.49500000$ above the mean, and adding 0.5 [+] establishes the cumulative quantile $\Phi(2.576) = \mathbf{0.99500000}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{0.99500000})$ ($99.50\%$).

Workbook Problem 9.5: Semiconductor Diode Shockley I-V Curve (Exponential Fit)

A silicon diode satisfies $I = I_s e^{V / V_T}$. * $(V_1, I_1) = (0.55\text{ V}, 0.24\text{ mA})$ * $(V_2, I_2) = (0.60\text{ V}, 1.65\text{ mA})$ * $(V_3, I_3) = (0.65\text{ V}, 11.40\text{ mA})$ * $(V_4, I_4) = (0.70\text{ V}, 78.50\text{ mA})$

Keystrokes: 1. [CLΣ] 2. .24 [LN] .55 [Σ+] 3. 1.65 [LN] .60 [Σ+] 4. 11.40 [LN] .65 [Σ+] 5. 78.50 [LN] .70 [Σ+] 6. Slope: [↱] [L.R.] [m] $\rightarrow$ 38.6015 ($\text{V}^{-1}$) 7. Thermal voltage: [1/x] $\rightarrow$ 0.025905 V ($25.91\text{ mV}$ at $300\text{ K}$) 8. Intercept: [b] $\rightarrow$ -22.6583 9. Reverse saturation current $I_s$: [eˣ] $\rightarrow$ $\mathbf{1.444 \times 10^{-10}\text{ mA}}$ ($1.444 \times 10^{-13}\text{ A}$).

In William Shockley's p-n junction theory, forward current grows exponentially with bias voltage according to $\ln I = \ln I_s + V/(n V_T)$. Applying [LN] to each current reading on the stack prior to [Σ+] maps the semi-logarithmic diode curve directly into the linear regression registers: [↱] [L.R.] [m] yields the reciprocal slope $1/(n V_T) = 38.6015\text{ V}^{-1}$, whose inverse [1/x] isolates the thermal voltage $X = \mathbf{0.025906\text{ V}}$ ($25.91\text{ mV}$ at $300\text{ K}$), while [b] [eˣ] exponentiates the intercept $\ln I_s = -22.6583$ to recover the reverse saturation current $I_s = 1.444 \times 10^{-10}\text{ mA}$.

Workbook Problem 9.6: Arrhenius Chemical Reaction Activation Energy

Reaction rate constant $k$ measured at different Kelvin temperatures: $\ln k = \ln A - \frac{E_a}{R}\frac{1}{T}$ * $(T_1, k_1) = (300\text{ K}, 0.0014\text{ s}^{-1})$ * $(T_2, k_2) = (320\text{ K}, 0.0082\text{ s}^{-1})$ * $(T_3, k_3) = (340\text{ K}, 0.0385\text{ s}^{-1})$

Stack entry: $y' = \ln k, x' = 1/T$. 1. .0014 [LN] 300 [1/x] [Σ+] 2. .0082 [LN] 320 [1/x] [Σ+] 3. .0385 [LN] 340 [1/x] [Σ+] 4. Slope: [m] $\rightarrow$ -8451.92 K (equals $-E_a / R$) 5. Activation energy $E_a$: 8.314462618 [×] [+/-] $\rightarrow$ 70,273 J/mol ($70.27\text{ kJ/mol}$).

In Svante Arrhenius's transition-state rate law, plotting $\ln k$ against reciprocal absolute temperature $1/T$ yields a straight line of slope $m = -E_a / R$. Transforming each $(k_i, T_i)$ pair on the stack via [LN] in $Y$ and [1/x] in $X$ before [Σ+] extracts the Arrhenius slope $m = -8451.92\text{ K}$ via [m], and multiplying by the universal gas constant $R = 8.314462618\text{ J/(mol}\cdot\text{K)}$ (8.314462618 [×] [+/-]) determines the molar activation barrier $X = \mathbf{70,273.18\text{ J/mol}}$ ($70.27\text{ kJ/mol}$).

Workbook Problem 9.7: Hooke's Law Spring Constant Calibration

Force $F$ vs. elongation $\Delta x$: * $(0.02\text{ m}, 5.0\text{ N})$ * $(0.04\text{ m}, 9.9\text{ N})$ * $(0.06\text{ m}, 15.1\text{ N})$ * $(0.08\text{ m}, 20.0\text{ N})$

5.0 [ENTER] .02 [Σ+]
9.9 [ENTER] .04 [Σ+]
15.1 [ENTER] .06 [Σ+]
20.0 [ENTER] .08 [Σ+]
Spring constant $k = m$: [↱] [L.R.] [m] $\rightarrow$ 250.5 N/m
Intercept: [b] $\rightarrow$ -0.03 N
Correlation: [r] $\rightarrow$ 0.99998.

Fitting load-extension pairs $(F_i, \Delta x_i)$ by ordinary least squares filters out random extensometer reading jitter across the four calibration stations: [↱] [L.R.] [m] identifies the effective Hookean stiffness $k = 250.5000\text{ N/m}$, [b] reveals a negligible fixture zero-offset $b = -0.0300\text{ N}$, and [r] confirms elastic proportionality ($r = 0.99998$) across the test span.

Workbook Problem 9.8: Precision Resistor Ohm's Law V-I Sweep

Voltage vs. Current across precision shunt: * $(I = 0.10\text{ A}, V = 1.002\text{ V})$ * $(I = 0.20\text{ A}, V = 2.005\text{ V})$ * $(I = 0.50\text{ A}, V = 5.012\text{ V})$

1.002 [ENTER] .10 [Σ+]
2.005 [ENTER] .20 [Σ+]
5.012 [ENTER] .50 [Σ+]
Resistance $R = m$: [m] $\rightarrow$ 10.0250 Ω.

In four-wire Kelvin resistance metrology, regressing sense voltage $V$ against excitation current $I$ separates differential resistance $R = dV/dI$ from parasitic thermoelectric EMF offsets ($b$). Accumulating the three $(V_i, I_i)$ points and querying [m] isolates the true shunt resistance $X = \mathbf{10.0250\ \Omega}$ free of lead thermoelectric bias.

Workbook Problem 9.9: NTC Thermistor Beta Parameter Extraction

Thermistor resistance $R = R_0 e^{\beta (1/T - 1/T_0)}$ linearized as $\ln R = C + \beta (1/T)$. * $(T_1 = 273.15\text{ K}, R_1 = 32,650\text{ }\Omega)$ * $(T_2 = 298.15\text{ K}, R_2 = 10,000\text{ }\Omega)$ * $(T_3 = 323.15\text{ K}, R_3 = 3,530\text{ }\Omega)$

32650 [LN] 273.15 [1/x] [Σ+]
10000 [LN] 298.15 [1/x] [Σ+]
3530 [LN] 323.15 [1/x] [Σ+]
Slope $\beta = m$: [m] $\rightarrow$ 3924.95 K.

Semiconductor negative-temperature-coefficient (NTC) thermistors exhibit thermally activated carrier hopping governed by the two-parameter Steinhart-Hart relation $\ln R = \ln R_\infty + \beta/T$. Transforming $(R_i, T_i)$ into $(\ln R_i, 1/T_i)$ on the stack before pressing [Σ+] extracts the characteristic material constant $\beta = \mathbf{3924.95\text{ K}}$ directly in Register $X$ via [m].

Workbook Problem 9.10: Boyle's Law Isothermal Gas Compression

Ideal gas isothermal pressure-volume relation: $\ln P = \ln C - \ln V$. * $(V = 100\text{ mL}, P = 1.00\text{ atm})$ * $(V = 50\text{ mL}, P = 2.01\text{ atm})$ * $(V = 25\text{ mL}, P = 3.98\text{ atm})$

1.00 [LN] 100 [LN] [Σ+]
2.01 [LN] 50 [LN] [Σ+]
3.98 [LN] 25 [LN] [Σ+]
Slope: [m] $\rightarrow$ -0.9962 (Confirms $P \propto V^{-1}$)
Intercept: [b] [eˣ] $\rightarrow$ 98.5946 atm·mL ($C = e^b$).

Transforming both manometer pressure and syringe volume with [LN] tests for polytropic behavior $P V^\gamma = C$: querying [m] yields $-\gamma = -0.9962 \approx -1.0$ (confirming isothermal Boyle's law compression), while [b] [eˣ] recovers the molar gas product $P V = e^b$ $(T: 0.0000, Z: 0.0000, Y: -0.9964, X: \mathbf{98.5946}\text{ atm}\cdot\text{mL})$.

Workbook Problem 9.11: Acoustic Inverse Square Law

Sound intensity $I = \frac{P}{4\pi r^2} \implies \ln I = \ln K - 2 \ln r$. * $(r = 1\text{ m}, I = 100\text{ W/m}^2)$ * $(r = 2\text{ m}, I = 25\text{ W/m}^2)$ * $(r = 4\text{ m}, I = 6.25\text{ W/m}^2)$

100 [LN] 1 [LN] [Σ+]
25 [LN] 2 [LN] [Σ+]
6.25 [LN] 4 [LN] [Σ+]
Slope: [m] $\rightarrow$ -2.000000 (Exact inverse-square exponent).

In anechoic free-field acoustics, spherical wavefront expansion distributes source power over $4\pi r^2$, producing a $-6.02\text{ dB}$ drop per distance doubling. Accumulating $(\ln I_i, \ln r_i)$ and pressing [m] verifies the free-field spreading exponent $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{-2.000000})$ with zero reflection distortion.

Workbook Problem 9.12: Traffic Flow Greenshields Model

Speed $v$ vs. Density $k$: $v = v_f (1 - k / k_j) = v_f - (v_f / k_j) k$. * $(k = 20\text{ veh/mi}, v = 55\text{ mph})$ * $(k = 40\text{ veh/mi}, v = 42\text{ mph})$ * $(k = 60\text{ veh/mi}, v = 28\text{ mph})$ * $(k = 80\text{ veh/mi}, v = 15\text{ mph})$

55 [ENTER] 20 [Σ+]
42 [ENTER] 40 [Σ+]
28 [ENTER] 60 [Σ+]
15 [ENTER] 80 [Σ+]
Free-flow speed $v_f = b$: [b] $\rightarrow$ 68.3000 mph
Jam density $k_j = -b / m$: [b] [m] [÷] [+/-] $\rightarrow$ 102.2388 veh/mi.

In Bruce Greenshields' foundational traffic-stream model, average space-mean speed decreases linearly with vehicle density from the unobstructed free-flow speed $v_f$ ($y$-intercept [b] $= 68.3000\text{ mph}$) down to complete standstill at the bumper-to-bumper jam density $k_j = -b/m$ ($x$-intercept), which [b] [m] [÷] [+/-] resolves to $(T: 0.0000, Z: 0.0000, Y: 68.5000, X: \mathbf{102.2388}\text{ veh/mi})$.

Workbook Problem 9.13: Weighted University Course GPA

Calculate semester GPA given courses with credits $c_i$ and grades $g_i$: * Calculus: 4 credits, Grade 3.7 * Physics: 4 credits, Grade 3.3 * Literature: 3 credits, Grade 4.0 * Chemistry Lab: 1 credit, Grade 3.0

Accumulate $(y = \text{credits}, x = \text{grade})$: 1. 4 [ENTER] 3.7 [Σ+] 2. 4 [ENTER] 3.3 [Σ+] 3. 3 [ENTER] 4.0 [Σ+] 4. 1 [ENTER] 3.0 [Σ+] 5. Weighted GPA: [x̄w] $\rightarrow$ 3.5833.

Because [Σ+] accumulates the cross-product $\Sigma(x_i y_i)$ in Register $\Sigma xy$ and the weight sum $\Sigma y_i$ in Register $\Sigma y$, staging credit hours $c_i$ in $Y$ and grade values $g_i$ in $X$ enables [x̄w] to evaluate the credit-weighted GPA $\Sigma(c_i g_i)/\Sigma c_i$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{3.5833})$ in a single keystroke.

Workbook Problem 9.14: Concrete Compressive Strength Aging

Compressive strength follows logarithmic curing: $\sigma(t) = a + b \ln(t)$. * $(t = 3\text{ days}, \sigma = 18.0\text{ MPa})$ * $(t = 7\text{ days}, \sigma = 26.5\text{ MPa})$ * $(t = 14\text{ days}, \sigma = 33.2\text{ MPa})$ * $(t = 28\text{ days}, \sigma = 40.0\text{ MPa})$

Stack entry: $(y = \sigma, x' = \ln t)$. 1. 18.0 [ENTER] 3 [LN] [Σ+] 2. 26.5 [ENTER] 7 [LN] [Σ+] 3. 33.2 [ENTER] 14 [LN] [Σ+] 4. 40.0 [ENTER] 28 [LN] [Σ+] 5. Slope: [m] $\rightarrow$ 9.836 MPa 6. Intercept: [b] $\rightarrow$ 7.261 MPa 7. Predict 56-day strength: 56 [LN] [ŷ] $\rightarrow$ 46.8487.

Portland cement hydration slows logarithmically as calcium-silicate-hydrate gel diffuses around unhydrated clinker grains ($\sigma(t) = b + m\ln t$). Transforming curing age $t$ via [LN] prior to [Σ+] yields the semi-log hydration rate [m] $= 9.8361\text{ MPa}$, and feeding $\ln(56)$ into the linear predictor (56 [LN] [ŷ]) extrapolates the 56-day structural cylinder strength $(T: 0.0000, Z: 11.7585, Y: 0.9911, X: 49.6697)$.

Workbook Problem 9.15: Photovoltaic Solar Cell Irradiance vs. Current

Solar short-circuit current $I_{sc}$ vs. solar irradiance $G$ ($\text{W/m}^2$): * $(G = 200, I_{sc} = 1.82\text{ A})$ * $(G = 500, I_{sc} = 4.54\text{ A})$ * $(G = 800, I_{sc} = 7.28\text{ A})$ * $(G = 1000, I_{sc} = 9.10\text{ A})$

1.82 [ENTER] 200 [Σ+]
4.54 [ENTER] 500 [Σ+]
7.28 [ENTER] 800 [Σ+]
9.10 [ENTER] 1000 [Σ+]
Responsivity: [m] $\rightarrow$ 0.00910 A/(W/m²) ($9.10\text{ mA/(W/m²)}$)
Intercept: [b] $\rightarrow$ 0.000 A.

Because photogenerated carrier collection in a crystalline silicon solar cell is directly proportional to incident photon flux across 1-sun operating conditions, regressing $I_{\text{sc}}$ against solar irradiance $G$ yields the spectral responsivity slope $X = \mathbf{0.00910\text{ A/(W/m}^2)}$ ($9.10\text{ mA/(W/m}^2)$) via [m] with zero dark-current offset ($b = 0.0000\text{ A}$).

Workbook Problem 9.16: Centrifugal Pump Head Loss Curve

Pump head follows $H = H_0 - A Q^2$. Linearize via $(y = H, x' = Q^2)$: * $(Q = 10\text{ gpm}, H = 98.0\text{ ft})$ * $(Q = 20\text{ gpm}, H = 92.0\text{ ft})$ * $(Q = 30\text{ gpm}, H = 82.0\text{ ft})$

98.0 [ENTER] 10 [↰] [x²] [Σ+]
92.0 [ENTER] 20 [↰] [x²] [Σ+]
82.0 [ENTER] 30 [↰] [x²] [Σ+]
Shutoff head $H_0 = b$: [b] $\rightarrow$ 100.0 ft
Loss coefficient $A = -m$: [m] $\rightarrow$ -0.020 ft/gpm².

Squaring volumetric discharge $Q$ via [↰] [x²] before pressing [Σ+] maps the parabolic centrifugal pump characteristic $H(Q) = H_0 - A Q^2$ into a linear regression against $Q^2$: [b] recovers the zero-flow shutoff head $H_0 = 100.0000\text{ ft}$, while [m] isolates the internal impeller hydraulic resistance slope $m = \mathbf{-0.0200\text{ ft/gpm}^2}$ (with $H_0 = 100.0000\text{ ft}$ lifted into Register $Y$).

Workbook Problem 9.17: Laminar Pipe Friction vs. Reynolds Number

Laminar Darcy friction $f = 64 / Re \implies \ln f = \ln 64 - \ln Re$. * $(Re = 400, f = 0.160)$ * $(Re = 800, f = 0.080)$ * $(Re = 1600, f = 0.040)$

.160 [LN] 400 [LN] [Σ+]
.080 [LN] 800 [LN] [Σ+]
.040 [LN] 1600 [LN] [Σ+]
Slope: [m] $\rightarrow$ -1.000000
Intercept: [b] [eˣ] $\rightarrow$ 64.0000 ($e^b$).

On a Moody diagram below the critical transition $Re < 2100$, Hagen-Poiseuille viscous shear produces a straight log-log line of slope $-1$. Accumulating $(\ln f_i, \ln Re_i)$ confirms the exact reciprocal exponent [m] $= -1.000000$ and recovers the Poiseuille laminar constant $e^b = \mathbf{64.0000}$ in Register $X$ (with $m = -1.0000$ in Register $Y$) via [b] [eˣ].

Workbook Problem 9.18: Aerodynamic Drag vs. Airspeed

$F_d = C v^2 \implies \ln F_d = \ln C + 2 \ln v$. * $(v = 10\text{ m/s}, F_d = 25\text{ N})$ * $(v = 20\text{ m/s}, F_d = 100\text{ N})$ * $(v = 30\text{ m/s}, F_d = 225\text{ N})$

25 [LN] 10 [LN] [Σ+]
100 [LN] 20 [LN] [Σ+]
225 [LN] 30 [LN] [Σ+]
Slope: [m] $\rightarrow$ 2.000000 (Exact quadratic velocity dependence)
Coefficient: [b] [eˣ] $\rightarrow$ 0.2500 N/(m/s)².

At high Reynolds numbers where pressure form drag dominates skin friction, aerodynamic drag scales quadratically with freestream velocity ($F_d = \frac{1}{2}\rho C_D A v^2 = C v^2$). Regressing $\ln F_d$ against $\ln v$ confirms the velocity exponent [m] $= 2.000000$ and extracts the dimensional drag factor $C = e^b = \mathbf{0.2500\text{ N/(m/s)}^2}$ in Register $X$ (with $m = 2.0000$ in Register $Y$) via [b] [eˣ].

Workbook Problem 9.19: Cantilever Beam Deflection Elasticity

Point load deflection $\delta = \frac{P L^3}{3 E I}$. Given $L = 2.0\text{ m}, I = 5 \times 10^{-6}\text{ m}^4$: * $(P = 500\text{ N}, \delta = 0.00127\text{ m})$ * $(P = 1000\text{ N}, \delta = 0.00254\text{ m})$ * $(P = 1500\text{ N}, \delta = 0.00381\text{ m})$

.00127 [ENTER] 500 [Σ+]
.00254 [ENTER] 1000 [Σ+]
.00381 [ENTER] 1500 [Σ+]
Compliance $m = \frac{\delta}{P}$: [m] $\rightarrow$ $\mathbf{2.54 \times 10^{-6}\text{ m/N}}$
Young's modulus $E = \frac{L^3}{3 I m}$: 2 [ENTER] 3 [yˣ] 3 [÷] 5 [E] 6 [+/-] [÷] 2.54 [E] 6 [+/-] [÷] $\rightarrow$ 2.099738E11 ($210\text{ GPa}$, Structural Steel).

Regressing tip deflection $\delta$ against applied point load $P$ determines the flexural compliance $m = d\delta/dP = L^3/(3EI) = 2.54 \times 10^{-6}\text{ m/N}$ in Register $X$, which inverts via $E = L^3/(3Im)$ directly on the stack to identify structural steel with Young's modulus $(T: 0.0000, Z: 0.0000, Y: 2.54 \times 10^{-6}, X: \mathbf{2.0997 \times 10^{11}}\text{ Pa})$ ($210\text{ GPa}$).

Workbook Problem 9.20: Pharmacokinetic Drug Washout Clearance

Blood concentration $C(t) = C_0 e^{-k_e t}$. * $(t = 1\text{ hr}, C = 45.2\text{ mg/L})$ * $(t = 3\text{ hr}, C = 24.8\text{ mg/L})$ * $(t = 6\text{ hr}, C = 10.1\text{ mg/L})$

45.2 [LN] 1 [Σ+]
24.8 [LN] 3 [Σ+]
10.1 [LN] 6 [Σ+]
Elimination rate $k_e = -m$: [m] $\rightarrow$ -0.2997 hr^{-1}
Elimination half-life: 2 [LN] [x⇄y] [+/-] [÷] $\rightarrow$ 2.3129 hours.

In a one-compartment intravenous bolus pharmacokinetic model, plasma concentration decays exponentially as renal and hepatic clearance eliminate the drug ($\ln C(t) = \ln C_0 - k_e t$). Accumulating $(\ln C_i, t_i)$ and pressing [m] extracts the first-order elimination rate constant $-k_e = -0.2997\text{ hr}^{-1}$, which converts via 2 [LN] [x⇄y] [+/-] [÷] to the biological elimination half-life $t_{1/2} = (\ln 2)/k_e$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{2.3129}\text{ hours})$.

Workbook Problem 9.21: Financial Stock Beta in Capital Asset Pricing Model (CAPM)

Calculate stock volatility Beta ($\beta$) against market benchmark: * Month 1: Market excess return $x = 2.0\%$, Stock excess return $y = 3.1\%$ * Month 2: $x = -1.5\%, y = -2.2\%$ * Month 3: $x = 4.0\%, y = 5.8\%$ * Month 4: $x = -3.0\%, y = -4.4\%$

3.1 [ENTER] 2.0 [Σ+]
-2.2 [ENTER] -1.5 [Σ+]
5.8 [ENTER] 4.0 [Σ+]
-4.4 [ENTER] -3.0 [Σ+]
Beta $\beta = m$: [m] $\rightarrow$ 1.468 (High-beta growth asset)
Alpha $\alpha = b$: [b] $\rightarrow$ 0.045%.

In William Sharpe's Capital Asset Pricing Model, regressing an equity's excess return $R_i - R_f$ ($Y$) against the market portfolio's excess return $R_m - R_f$ ($X$) partitions total return into systematic market sensitivity $\beta = \text{Cov}(R_i, R_m)/\text{Var}(R_m)$ ([m], $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{1.4676})$) and idiosyncratic risk-adjusted outperformance, or Jensen's Alpha $\alpha$ ([b] $= 0.045\%$).

Workbook Problem 9.22: Vector Graphic Scatter Plotting & Trendline Overlay

Accessible in App Mode via Right Shift [↱] then [PLOT] (excluded from physical firmware for standardized-exam compliance)

When bivariate Σ+ data is accumulated and no algebraic equations are stored in EQN, pressing [↱] [PLOT] in App Mode transitions the high-contrast dot-matrix display to the bivariate scatter & regression plotter: * The independent variable $x$ scales along the horizontal axis ($X_{\min} \to X_{\max}$). * The dependent variable $y$ scales along the vertical axis ($Y_{\min} \to Y_{\max}$). * Each accumulated point appears as a $2\times 2$ pixel reticle. * The linear regression line $y = m x + b$ is rendered as an uninterrupted 1-pixel vector trendline across the display field. * Navigation: * Press [4] to pan left, [6] to pan right. * Press [8] to pan up, [2] to pan down. * Press [+] to zoom in, [-] to zoom out. * Press [C] to exit plot mode and return to the RPN stack.

As Francis Anscombe famously demonstrated with Anscombe's Quartet, summary regression statistics ($m, b, r$) alone cannot distinguish a true linear relationship from high-leverage outliers or unmodeled curvature. Launching [↱] [PLOT] in App Mode overlays the least-squares vector trendline $y = mx + b$ directly across the $2\times 2$ pixel data reticles while preserving the operational stack registers $(T, Z, Y, X)$ untouched in background memory; panning ([4]/[6], [8]/[2]) and zooming ([+]/[-]) inspect residual structure visually before [C] returns to the four-register stack $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0000)$.

Workbook Problem 9.23: Residual Analysis and Standard Error of the Estimate ($s_e$)

In linear regression $y = m x + b$, the standard error of the estimate measures the dispersion of observed values around the regression line:

\[s_e = \sqrt{\frac{\sum_{i=1}^n (y_i - \hat{y}_i)^2}{n - 2}} = \sqrt{\frac{\Sigma y^2 - b \Sigma y - m \Sigma xy}{n - 2}}\]

For the pressure sensor dataset from Textbook Showcase Problem 9.1: * $n = 4$, degrees of freedom $n - 2 = 2$ * $\Sigma y = 12.45$, $\Sigma y^2 = 46.5075$, $\Sigma xy = 373.5$ * $m = 0.1245$, $b = 0.0000$

Compute residual sum of squares: $$\text{SS}_{\text{res}} = 46.5075 - 0.0000(12.45) - 0.1245(373.5) = 46.5075 - 46.50075 = 0.00675\text{ V}^2$$ $$s_e = \sqrt{\frac{0.00675}{2}} = \sqrt{0.003375} \approx 0.05809\text{ V}$$

Keystrokes: 46.5075 [ENTER] .1245 [ENTER] 373.5 [×] [-] 2 [÷] [√x] $\rightarrow$ 0.0580947502

Because the six statistical registers already hold the exact cross-product moments, the residual sum of squares $\text{SS}_{\text{res}} = \Sigma y^2 - b\Sigma y - m\Sigma xy$ can be evaluated without re-visiting individual observations ($b = 0$ here). Staging $\Sigma y^2$ via 46.5075 [ENTER] $(T: 0.0000, Z: 0.0000, Y: 46.5075, X: 46.5075)$ and subtracting the regression sum of squares $m \Sigma xy$ via 0.1245 [ENTER] 373.5 [×] [-] isolates $\text{SS}_{\text{res}} = 0.00675\text{ V}^2$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.00675)$; normalizing by $n - 2 = 2$ degrees of freedom (2 [÷], $s_e^2 = 0.003375$) and taking the square root ([√x]) determines the root-mean-square calibration scatter $s_e = \mathbf{0.0580947502\text{ V}}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0580947502)$.

Workbook Problem 9.24: Binomial Distribution Probability via Combinatorics `[Cn,r]`

In an automated manufacturing testing line, component defect probability is $p = 0.03$. In a batch of $n = 20$ components, calculate the probability of observing exactly $k = 2$ defective units: $$P(X = 2) = \binom{20}{2} (0.03)^2 (0.97)^{18}$$

Keystrokes: 20 [ENTER] 2 [↱] [PROB] [Cn,r] .03 [↰] [x²] [×] .97 [ENTER] 18 [yˣ] [×] $\rightarrow$ 0.0988; 0.1710 [ENTER] .97 [ENTER] 18 [yˣ] [×] $\rightarrow$ 0.0988304453

In Bernoulli lot acceptance sampling, the number of distinct defect pairings across $n = 20$ items is counted by the binomial coefficient 20 [ENTER] 2 [↱] [PROB] [Cn,r] ($\binom{20}{2} = \mathbf{190}$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 190.0000)$). Multiplying by the defective pair probability $p^2$ via .03 [↰] [x²] [×] ($190 \times 0.0009 = 0.1710$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.1710)$) and weighting by the survival probability of the remaining 18 good units (.97 [ENTER] 18 [yˣ] [×], $0.97^{18} \approx 0.5779$) resolves the exact binomial mass $P(X = 2) = \mathbf{0.0988304453}$ ($9.88\%$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0988304453)$.

Workbook Problem 9.25: Hypergeometric Sampling Without Replacement

In a batch of $N = 50$ aerospace fasteners, $K = 6$ are defective. A quality inspector selects $n = 5$ fasteners at random without replacement. Find the probability that exactly $k = 1$ selected fastener is defective: $$P(X = 1) = \frac{\binom{K}{k} \binom{N - K}{n - k}}{\binom{N}{n}} = \frac{\binom{6}{1} \binom{44}{4}}{\binom{50}{5}}$$

Keystrokes: 6 [ENTER] 1 [↱] [PROB] [Cn,r] 44 [ENTER] 4 [↱] [PROB] [Cn,r] [×] 50 [ENTER] 5 [↱] [PROB] [Cn,r] [÷]; 6 [ENTER] 44 [ENTER] 4 [↱] [PROB] [Cn,r] [×]; 50 [ENTER] 5 [↱] [PROB] [Cn,r] [÷] $\rightarrow$ 0.384426

When sampling a finite aerospace lot without replacement ($n/N = 10\%$), each draw alters the defect concentration of the remaining population, requiring the exact hypergeometric ratio of combinatorial counts. Evaluating 6 [ENTER] 1 [↱] [PROB] [Cn,r] ($\binom{6}{1} = 6.0000$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 6.0000)$), multiplying by the conforming combinations 44 [ENTER] 4 [↱] [PROB] [Cn,r] [×] ($\binom{44}{4} = 135,751 \implies 814,506$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 814506.0000)$), and dividing by the total sample space 50 [ENTER] 5 [↱] [PROB] [Cn,r] [÷] ($\binom{50}{5} = 2,118,760$) determines $P(X = 1) = \mathbf{0.384426}$ ($38.44\%$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.384426)$.

Workbook Problem 9.26: Log-Normal Aerosol Droplet Size Distribution

In medical nebulizer sprays, aerosol droplet diameters $D$ follow a log-normal distribution with geometric mean $\mu_{\ln} = \ln(4.5\text{ }\mu\text{m}) = 1.504077$ and geometric standard deviation $\sigma_{\ln} = 0.35$. Calculate the probability density for a droplet of diameter $D = 5.0\text{ }\mu\text{m}$: $$f(D) = \frac{1}{D \sigma_{\ln} \sqrt{2\pi}} \exp\left( -\frac{(\ln D - \mu_{\ln})^2}{2\sigma_{\ln}^2} \right)$$

Keystrokes: 5.0 [LN] 1.504077 [-] [↰] [x²] .35 [↰] [x²] 2 [×] [÷] [+/-] [eˣ] 5.0 [ENTER] .35 [×] 2 [↱] [π] [×] [√x] [×] [÷]; 0.0111009219 [ENTER] .35 [↰] [x²] 2 [×] [÷] [+/-] [eˣ]; 0.955701 [ENTER] 5.0 [ENTER] .35 [×] 2 [↱] [π] [×] [√x] [×] [÷] $\rightarrow$ 0.217865884

Because multiplicative breakup of liquid sheets in a pneumatic nebulizer produces normal scatter in $\ln D$ rather than $D$, the Jacobian transformation $d(\ln D)/dD = 1/D$ introduces an explicit diameter prefactor into the log-normal density. Evaluating the squared log-ratio 5.0 [LN] 4.5 [LN] [-] [↰] [x²] ($(\ln 5.0 - \ln 4.5)^2 \approx 0.011101$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.011101)$), dividing by $2\sigma_{\ln}^2$ and exponentiating via .35 [↰] [x²] 2 [×] [÷] [+/-] [eˣ] ($\exp(-0.045310) \approx 0.955701$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.955701)$), and normalizing by $D\sigma_{\ln}\sqrt{2\pi}$ (5.0 [ENTER] .35 [×] 2 [↱] [π] [×] [√x] [×] [÷]) resolves $f(5.0) = \mathbf{0.217865884\ \mu\text{m}^{-1}}$ $(T: 0.9557, Z: 0.9557, Y: 0.9557, X: 0.217865884)$.

Chapter 10: Computer Science, Bases & 36-Bit Integers — Extended Engineering & Competition Problems

Accompanies Chapter 10 of Thinking in the Stack (24 Worked Engineering Problems)

Workbook Problem 10.1: IPv4 CIDR /27 Subnet (30 Usable Hosts)

A server has IP 10.20.30.142 /27. * $10 = \text{0Ah}, 20 = \text{14h}, 30 = \text{1Eh}, 142 = \text{8Eh} \implies \text{0A141E8Eh}$. * Subnet Mask /27: $27$ ones, $5$ zeros $\implies$ block size $2^5 = 32_{10} = \text{20h}$ ($255.255.255.224$).

Octet 4 Representation	Upper 3 Subnet Bits	Lower 5 Host Bits	Hex / Decimal Value
Host Octet 4	`100`	`01110`	`8Eh` ($142_{10}$)
Subnet Mask (`/27`)	`111`	`00000`	`E0h` ($224_{10}$)
Network Base (`÷R 20h × 20h`)	`100`	`00000`	`80h` ($128_{10}$)
Broadcast (`+ 1Fh`)	`100`	`11111`	`9Fh` ($159_{10}$)

Keystrokes in [HEX] Mode: 1. 0A141E8E [ENTER] 20 [↱] [÷R] 20 [×] $\rightarrow$ A141E80h (10.20.30.128) 2. Wildcard mask: $32 - 27 = 5$ bits $\implies 2^5 - 1 = 31_{10} = \text{1Fh}$. 3. Broadcast: 0A141E80 [ENTER] 1F [+] $\rightarrow$ A141E9Fh (10.20.30.159) 4. Usable host range: 10.20.30.129 through 10.20.30.158 ($30$ usable hosts).

In CIDR block allocation, a /27 prefix reserves 27 upper network bits and leaves $32 - 27 = 5$ host bits, partitioning each /24 into eight 32-address (20h) blocks. Staging the packed address 0A141E8E [ENTER] (10.20.30.142) in [HEX] mode $(T: 0\text{h}, Z: 0\text{h}, Y: \text{A141E8Eh}, X: \text{A141E8Eh})$ and dividing by the block stride 20 [↱] [÷R] splits the 27-bit network prefix quotient $\text{50A0F4h}$ into Register $X$ and the 5-bit host index $\text{Eh}$ ($14_{10}$) into Register $Y$ $(T: 0\text{h}, Z: 0\text{h}, Y: \text{Eh}, X: \text{50A0F4h})$. Multiplying by the stride 20 [×] aligns the network base A141E80h (10.20.30.128), and adding the 5-bit wildcard mask 1F [+] ($2^5 - 1 = 31_{10}$) establishes the directed broadcast address $\mathbf{\text{A141E9Fh}}$ (10.20.30.159).

Workbook Problem 10.2: IPv4 Point-to-Point /30 Subnet (WAN Link)

A WAN router serial link connects two routers using 172.16.5.9 /30. * $172 = \text{ACh}, 16 = \text{10h}, 5 = \text{05h}, 9 = \text{09h} \implies \text{AC100509h}$. * Subnet Mask /30: $30$ ones, $2$ zeros $\implies$ block size $2^2 = 4_{10} = \text{4h}$ ($255.255.255.252$).

Keystrokes in [HEX] Mode: 1. AC100509 [ENTER] 4 [↱] [÷R] 4 [×] $\rightarrow$ AC100508h (172.16.5.8) 2. Wildcard: $2^2 - 1 = 3 = \text{03h}$. 3. Broadcast: AC100508 [ENTER] 3 [+] $\rightarrow$ AC10050Bh (172.16.5.11) 4. Router A interface: 172.16.5.9, Router B interface: 172.16.5.10 ($2$ usable hosts).

Point-to-point WAN links traditionally use /30 prefixes (stride $2^2 = 4\text{h}$) to conserve IPv4 address space while providing exactly two routable endpoints between network and broadcast boundaries. Entering AC100509 [ENTER] in [HEX] mode and partitioning with 4 [↱] [÷R] separates host offset $1\text{h}$ in Register $Y$ from the 30-bit prefix quotient $\text{2B040142h}$ in Register $X$ $(T: 0\text{h}, Z: 0\text{h}, Y: 1\text{h}, X: \text{2B040142h})$; scaling back by 4 [×] recovers the wire network ID $\mathbf{\text{AC100508h}}$ (172.16.5.8), and adding the 2-bit wildcard 3 [+] determines the link broadcast address $\mathbf{\text{AC10050Bh}}$ (172.16.5.11).

Workbook Problem 10.3: IPv4 Host Route (/32) and Loopback

A BGP loopback interface is assigned 198.51.100.1 /32. * In Hex: IP is C6336401h, Mask is FFFFFFFFh ($255.255.255.255$). * Network ID = Host IP = Broadcast = 198.51.100.1 (C6336401h). * Usable hosts: $2^{32-32} = 2^0 = \mathbf{1\text{ host}}$.

Keystrokes in [HEX] & [DEC] Modes: 1. Load Host IP: C6336401 [ENTER] $\rightarrow$ C6336401h 2. Wildcard mask: FFFFFFFF [ENTER] FFFFFFFF [-] $\rightarrow$ 0h (0.0.0.0) 3. Broadcast Address: [+] $\rightarrow$ C6336401h (198.51.100.1) 4. Host Capacity: Switch to [DEC], 2 [ENTER] 0 [yˣ] $\rightarrow$ 1.

Autonomous system border routers advertise /32 loopback interfaces into BGP and OSPF so routing tables track a virtual node that never goes down when a single physical port flaps. With C6336401 [ENTER] staged on the stack $(T: 0\text{h}, Z: 0\text{h}, Y: \text{C6336401h}, X: \text{C6336401h})$, complementing the 32-bit host mask (FFFFFFFF [ENTER] FFFFFFFF [-]) yields a zero wildcard offset 0h, confirming via [+] that network, host, and broadcast collapse into the single address $\mathbf{\text{C6336401h}}$ ($2^0 = \mathbf{1}$ via [↰] [BASE] [DEC] 2 [ENTER] 0 [yˣ]).

Workbook Problem 10.4: CIDR Route Aggregation (Supernetting)

An ISP wants to summarize four contiguous subnets into a single route: * Subnet 1: 192.168.16.0 /24 (C0A81000h) * Subnet 2: 192.168.17.0 /24 (C0A81100h) * Subnet 3: 192.168.18.0 /24 (C0A81200h) * Subnet 4: 192.168.19.0 /24 (C0A81300h)

Inspect the third octet: $16 = 00010000_2$, $17 = 00010001_2$, $18 = 00010010_2$, $19 = 00010011_2$. The upper 6 bits ($000100_2$) are identical across all four routes! New network mask length: $16 + 6 = \mathbf{22\text{ bits}}$ (/22). Supernet Mask: 255.255.252.0 (FFFFFC00h). Summarized Route: 192.168.16.0 /22.

Keystrokes in [HEX] Mode: 1. Base Network ID: C0A81000 [ENTER] $\rightarrow$ C0A81000h (192.168.16.0) 2. Supernet block span: 4 [ENTER] 100 [×] $\rightarrow$ 400h ($1024_{10}$ addresses) 3. Wildcard Mask: 1 [-] $\rightarrow$ 3FFh (0.0.3.255) 4. Broadcast Address: [+] $\rightarrow$ C0A813FFh (192.168.19.255) 5. Usable Hosts: 3FF [ENTER] 1 [-] $\rightarrow$ 3FEh ($1022_{10}$ hosts).

CIDR route aggregation (RFC 4632) prevents global routing-table explosion by folding contiguous power-of-two /24 blocks into a shorter prefix. Staging the aligned base network C0A81000 [ENTER] and evaluating the 4-block span 4 [ENTER] 100 [×] (400h $= 1024_{10}$ addresses, $(T: 0\text{h}, Z: \text{C0A81000h}, Y: 4\text{h}, X: 400\text{h})$) followed by 1 [-] derives the 10-bit supernet wildcard mask 3FFh (0.0.3.255). Adding the wildcard to the waiting base address via [+] resolves the summarized broadcast boundary $\mathbf{\text{C0A813FFh}}$ (192.168.19.255), while 3FF [ENTER] 1 [-] confirms $\mathbf{\text{3FEh}}$ ($1022_{10}$) usable host addresses.

Workbook Problem 10.5: Wildcard Mask Generation via 32-Bit Complement Subtraction

In Cisco ACLs, calculate the wildcard mask for /26 ($255.255.255.192$): 1. In [HEX]: Mask is FFFFFFC0h. 2. Subtract from the 32-bit all-ones mask FFFFFFFFh: key FFFFFFFF [ENTER] FFFFFFC0 [-] $\rightarrow$ 3Fh. 3. In [DEC]: 3Fh = 63. 4. Dotted decimal wildcard: 0.0.0.63.

Because a bitwise one's complement on an $N$-bit unsigned word is algebraically identical to subtracting from $2^N - 1$, subtracting the /26 netmask FFFFFFC0h from the 32-bit all-ones constant FFFFFFFF [ENTER] $(T: 0\text{h}, Z: 0\text{h}, Y: \text{FFFFFFFFh}, X: \text{FFFFFFFFh})$ via FFFFFFC0 [-] $(T: 0\text{h}, Z: 0\text{h}, Y: 0\text{h}, X: \text{3Fh})$ isolates the inverted access-control wildcard 3Fh, which [↰] [BASE] [DEC] displays as $\mathbf{63}$ (0.0.0.63).

Workbook Problem 10.6: Embedded GPIO Pin Bit-Clearing (`-`)

Clear Bit 11 (Disable motor enable line, currently $1$) in register holding 0x00002F80. Bit 11 weight: $2^{11} = 2048_{10} = \text{800h}$.

Keystrokes in [HEX] Mode: 2F80 [ENTER] 800 [-] $\rightarrow$ 2780h

When a target bit $k$ is known to be asserted ($1$), subtracting its positional weight $2^k$ executes a borrow-free bitwise clear (R & ~(1 << k)). Staging 2F80 [ENTER] in [HEX] mode and subtracting Bit 11's weight $2^{11} = \text{800h}$ via 800 [-] clears the motor-enable bit cleanly without disturbing adjacent flags, leaving $\mathbf{\text{2780h}}$ in Register $X$ $(T: 0\text{h}, Z: 0\text{h}, Y: 0\text{h}, X: \text{2780h})$.

Workbook Problem 10.7: Multi-Bit Field Insertion (Read-Modify-Write)

In a peripheral timer control register currently holding 0x4521, update the 4-bit prescaler field at bits [11:8] (currently 5h) with value 0xA ($10_{10}$), without altering any of the surrounding configuration bits.

Insertion Stage	Binary Nibbles (Bits 15–0)	Hexadecimal Value
Original Timer Register	`0100 0101 0010 0001`	`4521h`
Subtract Old Field (`- 500h`)	`0100 0000 0010 0001`	`4021h`
Shifted New Field (`A × 100h`)	`0000 1010 0000 0000`	`0A00h`
Updated Register (`+`)	`0100 1010 0010 0001`	`4A21h`

Keystrokes in [HEX] Mode: 1. Clear existing field: 4521 [ENTER] 500 [-] $\rightarrow$ 4021h 2. Prepare shifted new field: A [ENTER] 100 [×] $\rightarrow$ A00h 3. Merge fields: [+] $\rightarrow$ 4A21h

Firmware read-modify-write sequences (R = (R & ~MASK) | (VAL << SHIFT)) translate cleanly onto the hexadecimal stack: subtracting the old field 4521 [ENTER] 500 [-] zeroes bits [11:8] to leave the masked register 4021h in Register $X$ $(T: 0\text{h}, Z: 0\text{h}, Y: 0\text{h}, X: \text{4021h})$, left-shifting the new nibble via A [ENTER] 100 [×] forms A00h in $X$ while 4021h waits in $Y$ $(T: 0\text{h}, Z: 0\text{h}, Y: \text{4021h}, X: \text{A00h})$, and pressing [+] merges the updated control word $\mathbf{\text{4A21h}}$.

Workbook Problem 10.8: RGB565 to RGB888 Color Unpacking & Expansion

Given packed RGB565 word 0xE5B2, extract individual color channels using [÷] and [↱] [÷R] and expand them to standard 8-bit values ($0 \dots 255$): * Red channel (bits 15–11): divide by $2^{11} = \text{800h}$ * Green channel (bits 10–5): take remainder modulo $\text{800h}$ via [↱] [÷R] [x⇄y], then divide by $2^5 = \text{20h}$ * Blue channel (bits 4–0): take remainder modulo $\text{20h}$ via [↱] [÷R] [x⇄y]

Keystrokes in [HEX] & [DEC] Modes: 1. Extract Red: E5B2 [ENTER] 800 [÷] $\rightarrow$ 1Ch ($28_{10}$) * Expand in [DEC]: 28 [ENTER] 255 [×] 31 [÷] $\rightarrow$ 230.32258065 2. Extract Green: E5B2 [ENTER] 800 [↱] [÷R] [x⇄y] 20 [÷] $\rightarrow$ 2Dh ($45_{10}$) * Expand in [DEC]: 45 [ENTER] 255 [×] 63 [÷] $\rightarrow$ 182.14285714 3. Extract Blue: E5B2 [ENTER] 20 [↱] [÷R] [x⇄y] $\rightarrow$ 12h ($18_{10}$) * Expand in [DEC]: 18 [ENTER] 255 [×] 31 [÷] $\rightarrow$ 148.06451613 * Unpacked TrueColor: RGB(230, 182, 148) (#E6B694).

Because [↱] [÷R] simultaneously returns the upper bitfield quotient in Register $X$ and the lower bitfield remainder in Register $Y$, a 16-bit RGB565 pixel can be unpacked either channel-by-channel (E5B2 [ENTER] 800 [÷] $\rightarrow \mathbf{\text{1Ch}} = 28_{10} \implies R = \mathbf{230}$; E5B2 [ENTER] 800 [↱] [÷R] [x⇄y] 20 [÷] $\rightarrow \mathbf{\text{2Dh}} = 45_{10} \implies G = \mathbf{182}$; E5B2 [ENTER] 20 [↱] [÷R] [x⇄y] $\rightarrow \mathbf{\text{12h}} = 18_{10} \implies B = \mathbf{148}$) or in a single-entry cascade: keying E5B2 [ENTER] 800 [↱] [÷R] [x⇄y] 20 [↱] [÷R] unpacks all three channels onto the stack at once (Z: 1Ch Red, Y: 12h Blue, X: 2Dh Green) with E5B2 typed only once!

Workbook Problem 10.9: RGBA4444 16-Bit Texture Color Packing

In embedded OpenGL ES graphics engines, 16-bit RGBA4444 allocates 4 bits per channel: Pack channels $R=14$ (Eh), $G=9$ (9h), $B=3$ (3h), $A=15$ (Fh): $$(R \times \text{1000h}) + (G \times \text{100h}) + (B \times \text{10h}) + A$$

Keystrokes in [HEX] Mode: E [ENTER] 1000 [×] 9 [ENTER] 100 [×] [+] 3 [ENTER] 10 [×] [+] F [+] $\rightarrow$ E93Fh

In 4-bit-per-channel RGBA4444 sprite textures, each channel aligns with a single hexadecimal digit weight ($16^3, 16^2, 16^1, 16^0$). Shifting Red via E [ENTER] 1000 [×] ($\text{E000h}$, $(T: 0\text{h}, Z: 0\text{h}, Y: 0\text{h}, X: \text{E000h})$) and accumulating Green (9 [ENTER] 100 [×] [+] $\rightarrow \text{E900h}$), Blue (3 [ENTER] 10 [×] [+] $\rightarrow \text{E930h}$), and Alpha (F [+]) assembles the packed 16-bit texel $\mathbf{\text{E93Fh}}$ in Register $X$.

Workbook Problem 10.10: Two's Complement Negation Proof ($(2^{36} - 1 - X) + 1$)

Verify that subtracting $X$ from all ones (FFFFFFFFFh, 9 hex digits for 36 bits) and adding $1$ produces the exact mathematical two's complement negation [+/-]: * Key in $42$ in [DEC] mode and press [+/-]: produces -42. * Switch to [↰] [BASE] [HEX]: displays as FFFFFFFD6h ($36$-bit two's complement representation, $2^{36} - 42$). * Verify by one's complement subtraction: FFFFFFFFF [ENTER] 2A [-] 1 [+] $\rightarrow$ FFFFFFFD6h (-42 in [DEC]).

In an $N$-bit two's complement ALU ($N = 36$ on the SC-32), additive inverse is defined modulo $2^{36}$ via $-X \equiv (2^{36} - 1 - X) + 1 \pmod{2^{36}}$. Negating 42 [+/-] in [DEC] mode yields $\mathbf{-42}$, whose underlying 36-bit word appears in [↰] [BASE] [HEX] as $\mathbf{\text{FFFFFFFD6h}}$—identical to forming the 9-nibble one's complement FFFFFFFFF [ENTER] 2A [-] ($\text{FFFFFFFD5h}$) and incrementing by the carry-in 1 [+].

Workbook Problem 10.11: Arithmetic Right Shift via Signed Integer Division

For signed negative integer $-64$: * In [DEC]: 64 [+/-] [ENTER] 4 [÷] $\rightarrow$ -16 (Exact arithmetic right shift by 2 bits, sign-extended).

Whereas a logical right shift (>>) pads vacated upper bits with zeros, an arithmetic right shift (>>> or ASR) replicates the sign bit (Bit 35) to preserve negative two's complement magnitudes. Staging 64 [+/-] [ENTER] in [DEC] mode and dividing by $2^2 = 4$ via 4 [÷] yields $\mathbf{-16}$ $(T: 0, Z: 0, Y: 0, X: -16)$, which [↰] [BASE] [HEX] confirms as the sign-extended 36-bit word $\mathbf{\text{FFFFFFFFF0h}}$.

Workbook Problem 10.12: Parity Bit Generation

Data byte 0x3B ($00111011_2$, five ones $\implies$ odd parity). For even parity, the MSB parity bit (Bit 7, weight $2^7 = 128_{10} = \text{80h}$) must be set to $1$: 3B [ENTER] 80 [+] $\rightarrow$ BBh.

Because the 7-bit ASCII payload 3Bh ($00111011_2$) contains an odd population count of five 1 bits, asserting the MSB parity bit ($2^7 = \text{80h}$) via 3B [ENTER] 80 [+] brings total population weight to six 1 bits, forming the even-parity UART frame byte $\mathbf{\text{BBh}}$ ($10111011_2$).

Workbook Problem 10.13: Longitudinal Redundancy Check (LRC)

Calculate LRC across message bytes 0x02, 0x41, 0x42.

2 [ENTER] 41 [+] 42 [+] $\rightarrow$ 85h.

LRC = $(256 - \text{Sum}) \bmod 256$: 100 [ENTER] 85 [-] $\rightarrow$ 7Bh.

In Intel HEX and Modbus ASCII serial framing, the Longitudinal Redundancy Check (LRC) byte is the two's complement of the modulo-256 payload sum, ensuring that the sum of payload plus LRC equals 00h $(\bmod\text{ 100h})$ at the receiver. Summing the three frame bytes 2 [ENTER] 41 [+] 42 [+] yields 85h $(T: 0\text{h}, Z: 0\text{h}, Y: 0\text{h}, X: \text{85h})$, and subtracting from 100h (100 [ENTER] 85 [-], or 100 [x⇄y] [-] without re-typing 85) generates the transmitted LRC byte $\mathbf{\text{7Bh}}$.

Workbook Problem 10.14: Ethernet FCS 32-Bit Inversion & Frame Check Sequence Post-Conditioning

In IEEE 802.3 Ethernet frame transmission, the final 32-bit CRC remainder (Frame Check Sequence, FCS) is post-conditioned prior to transmission by inverting all 32 bits (one's complement). Given an Ethernet polynomial check state of 0xEDB88320: 1. Invert the 32-bit word by subtracting from the all-ones mask FFFFFFFFh: $$FFFFFFFF_{16} - EDB88320_{16} = \mathbf{12477CDF}_{16} \ (306,678,751_{10})$$

Keystrokes in [HEX] Mode: 1. Load 32-bit inversion mask: FFFFFFFF [ENTER] $\rightarrow$ FFFFFFFFh 2. Subtract Ethernet generator polynomial: EDB88320 [-] $\rightarrow$ 12477CDFh 3. Switch radix to inspect in decimal: [↰] [BASE] [DEC] $\rightarrow$ 306678751.

In IEEE 802.3 CRC-32 hardware, post-inverting the shift-register state against FFFFFFFFh ensures that appending trailing zero bytes to a valid frame alters its CRC. Subtracting the reversed generator polynomial EDB88320h from FFFFFFFF [ENTER] via EDB88320 [-] produces the one's-complement word $\mathbf{\text{12477CDFh}}$, which [↰] [BASE] [DEC] renders losslessly as $\mathbf{306,678,751}$.

Workbook Problem 10.15: SPI Clock Prescaler Selection

A SPI peripheral runs from a $75.0\text{ MHz}$ bus clock. The connected flash memory specifies a maximum operating frequency of $10.0\text{ MHz}$. The SPI prescaler offers powers of two: $2, 4, 8, 16, 32, 64, 128$. Determine the optimal prescaler divider and actual SPI operating frequency.

Keystrokes: 75 [ENTER] 10 [÷] $\rightarrow$ $7.5$.

Dividing the $75.0\text{ MHz}$ peripheral bus clock by the $10.0\text{ MHz}$ NOR-flash ceiling yields the minimum divider ratio $7.5000$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 7.5000)$. Rounding up to the next binary prescaler tap $2^{\lceil \log_2 7.5 \rceil} = \mathbf{8}$ (8 [ENTER]) and dividing the bus clock via 75 [ENTER] 8 [÷] establishes the safe operating SPI clock $f_{\text{SPI}} = \mathbf{9.375\text{ Hz}}$ ($9.375\text{ MHz}$).

Workbook Problem 10.16: I2C Fast-Mode Plus SCL Timing Register Calculations

Target I2C clock frequency $f_{\text{SCL}} = 1.0\text{ MHz}$ ($1000\text{ ns}$ cycle time) derived from a $150.0\text{ MHz}$ core clock period ($T_{\text{clk}} = \frac{1}{150\times 10^6} \approx 6.6667\text{ ns}$). Total clocks per SCL period: $150\text{ cycles}$. In Fast-Mode Plus, asymmetric duty cycle allocates $t_{\text{LOW}} \ge 500\text{ ns}$ and $t_{\text{HIGH}} \ge 260\text{ ns}$. Allocating $\frac{1}{3}$ high and $\frac{2}{3}$ low: * $t_{\text{HIGH}} = 50\text{ cycles} \times 6.6667\text{ ns} = \mathbf{333.3\text{ ns}}$ ($> 260\text{ ns}$) * $t_{\text{LOW}} = 100\text{ cycles} \times 6.6667\text{ ns} = \mathbf{666.7\text{ ns}}$ ($> 500\text{ ns}$).

Keystrokes: 1. Total cycles: 150 [ENTER] 1 [÷] $\rightarrow$ $150$ cycles 2. High phase clocks: 3 [÷] $\rightarrow$ $50$ cycles 3. Timing durations: 1000 [ENTER] 3 [÷] $\rightarrow$ 333.3 ns ($t_{\text{HIGH}}$); 2 [×] $\rightarrow$ 666.7 ns ($t_{\text{LOW}}$).

Open-drain I2C buses require an asymmetric 1:2 high-to-low clock ratio at $1\text{ MHz}$ (Fast-Mode Plus) so active pull-down transistors hold SCL low long enough ($t_{\text{LOW}} \ge 500\text{ ns}$) after passive pull-up RC rise times. Dividing the $150\text{-cycle}$ period (150 [ENTER] 1 [÷]) by 3 [÷] assigns $50\text{ cycles}$ to SCLL_H and $100\text{ cycles}$ to SCLL_L; partitioning the $1000\text{ ns}$ bit period (1000 [ENTER] 3 [÷]) confirms $t_{\text{HIGH}} = \mathbf{333.3\text{ ns}} > 260\text{ ns}$ and 2 [×] confirms $t_{\text{LOW}} = \mathbf{666.7\text{ ns}} > 500\text{ ns}$.

Workbook Problem 10.17: Fixed-Point Q16.16 Arithmetic Multiplication

In digital signal processing without FPU acceleration, numbers are formatted as signed Q16.16 (16 integer bits, 16 fractional bits, scaling factor $2^{16} = 65,536$). Multiply $A = 3.5$ and $B = 2.25$: * $A = 3.5 \times 65536 = 229,376 = \text{38000h}$ * $B = 2.25 \times 65536 = 147,456 = \text{24000h}$ * Product formula: $C = (A \times B) \gg 16 = (A \times B) / 65536$.

Keystrokes in [HEX] Mode: 38000 [ENTER] 24000 [×] 10000 [÷] $\rightarrow$ 7E000h

Because multiplying two Q16.16 fixed-point numbers $(A \cdot 2^{16})(B \cdot 2^{16}) = AB \cdot 2^{32}$ doubles the binary-point scaling to Q32.32, a 32-bit ALU would overflow on the intermediate product $\text{38000h} \times \text{24000h} = \text{7E0000000h}$. The SC-32's 36-bit integer datapath accommodates the full 35-bit intermediate product (38000 [ENTER] 24000 [×]) without truncation, and right-shifting by 16 bits (10000 [÷]) normalizes the result back to Q16.16 word $\mathbf{\text{7E000h}}$ ($516,096_{10} / 65,536 = \mathbf{7.875}$).

Workbook Problem 10.18: Permutations ($nPr$) in Cryptography S-Box Design

Calculate the total number of ways to construct an ordered 4-element substitution key from a pool of 16 distinct nibbles:

\[P(16, 4) = \frac{16!}{(16 - 4)!} = 16 \times 15 \times 14 \times 13 = \mathbf{43,680}\]

Keystrokes: 16 [ENTER] 4 [↱] [PROB] [Pn,r] $\rightarrow$ 43,680

In block-cipher substitution-box (S-Box) construction, selecting an ordered sequence of $r = 4$ distinct 4-bit nibbles without replacement from the 16-element hexadecimal alphabet spans the falling factorial $16^{\underline{4}} = P(16, 4)$. Keying 16 [ENTER] 4 [↱] [PROB] [Pn,r] evaluates the exact keyspace cardinality $P(16, 4) = \mathbf{43,680}$ ($43,680$).

Workbook Problem 10.19: Combinations ($nCr$) and Hamming Sphere

In a $[7, 4]$ Hamming code, calculate the number of possible 2-bit error patterns:

\[\binom{7}{2} = \frac{7!}{2! \times 5!} = \mathbf{21}\]

Keystrokes: 7 [ENTER] 2 [↱] [PROB] [Cn,r] $\rightarrow$ 21

In Richard Hamming's $[7, 4]$ single-error-correcting block code, the surface area of the Hamming sphere of radius $t = 2$ around any 7-bit codeword counts all unordered pairs of corrupted bit positions. Evaluating 7 [ENTER] 2 [↱] [PROB] [Cn,r] returns $\binom{7}{2} = \mathbf{21}$ ($21$ distinct double-bit error patterns).

Workbook Problem 10.20: Factorials and Stirling's Approximation Comparison

Calculate $15!$: Exact: 15 [↱] [PROB] [↰] [x!] $\rightarrow$ 1,307,674,368,000.

Stirling's approximation: $\sqrt{2\pi n} (n/e)^n \approx 1.3004 \times 10^{12}$ (Error $< 0.6\%$).

Evaluating 15 [↱] [PROB] [↰] [x!] computes the exact permutation count of a 15-element array, $\mathbf{1,307,674,368,000}$ ($1,307,674,368,000$), whereas James Stirling's asymptotic leading term $\sqrt{2\pi n}(n/e)^n \approx 1.30043 \times 10^{12}$ underestimates the exact factorial by the series correction factor $e^{-1/(12n)} \approx 1 - \frac{1}{180}$ ($0.55\%$).

Workbook Problem 10.21: Pseudo-Random Number Generation (`RAND`)

Generate 3 uniform random floats: [↱] [PROB] [RAND] $\rightarrow$ $0.724194$ [↱] [PROB] [RAND] $\rightarrow$ $0.183022$ [↱] [PROB] [RAND] $\rightarrow$ $0.941581$.

Because [↱] [PROB] [RAND] is a non-consuming generator that enables stack lift before pushing each uniform variate $U_k \in [0, 1)$ into Register $X$, three consecutive invocations populate a 3-dimensional Monte Carlo sample vector directly across Registers $Z, Y,$ and $X$: the first press generates $U_1 = \mathbf{0.724194}$ in $X$, the second lifts $U_1 \to Y$ while placing $U_2 = \mathbf{0.183022}$ in $X$, and the third lifts $U_1 \to Z, U_2 \to Y$ while placing $U_3 = \mathbf{0.941581}$ in $X$.

Workbook Problem 10.22: Pseudo-Random Die Roll Simulation

Simulate a standard 6-sided die roll: [↱] [PROB] [RAND] 6 [×] 1 [+] [↱] [PARTS] [IP] $\rightarrow$ Integer in $1 \dots 6$.

Mapping a continuous uniform variate $U \in [0, 1)$ ([↱] [PROB] [RAND], e.g., $0.724194$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.724194)$) onto a discrete equiprobable integer set $\{1, \dots, 6\}$ uses the affine floor transformation $\lfloor 6U + 1 \rfloor$: scaling by the cardinality 6 [×] ($4.345164 \in [0, 6)$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 4.345164)$), shifting the origin 1 [+] ($5.345164 \in [1, 7)$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 5.345164)$), and truncating via [↱] [PARTS] [IP] yields the unbiased die face $\mathbf{5}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 5.0000)$.

Workbook Problem 10.23: Decimal to Binary Conversion & Bit Inspection

Convert decimal 13 into its 4-bit binary representation and verify its bit weights ($8 + 4 + 0 + 1$): Keystrokes: 13 [ENTER] [↰] [BASE] [BIN] $\rightarrow$ 1101b ($1101_2$).

Entering 13 [ENTER] in [DEC] mode and switching radix via [↰] [BASE] [BIN] formats the underlying integer as $\mathbf{1101b}$ ($1101_2 = 2^3 + 2^2 + 2^0 = 8 + 4 + 1$), demonstrating that radix switching is a pure display projection that leaves all four stack registers unaltered.

Workbook Problem 10.24: Binary to Hexadecimal & Decimal Reconstruction

Convert binary 1101b back to decimal (13) and hexadecimal (Dh) via [↰] [BASE] [DEC] and [↰] [BASE] [HEX], confirming lossless radix conversion across all stack registers.

Staging the binary nibble 1101 [ENTER] in [BIN] mode $(T: 0\text{b}, Z: 0\text{b}, Y: 1101\text{b}, X: 1101\text{b})$ and cycling through [↰] [BASE] [HEX] $(T: 0\text{h}, Z: 0\text{h}, Y: \text{Dh}, X: \mathbf{Dh})$ and [↰] [BASE] [DEC] $(T: 0, Z: 0, Y: 13, X: \mathbf{13})$ verifies bijective, lossless radix projection across all four operational registers.

Chapter 11: Financial Mathematics, Executive Analytics & TVM — Extended Engineering & Competition Problems

Accompanies Chapter 11 of Thinking in the Stack (31 Worked Engineering Problems)

Workbook Problem 11.1: Annual Revenue Acceleration (%CHG)

Revenues grew from $\$42.50\text{M}$ in fiscal year 2024 to $\$51.85\text{M}$ in fiscal year 2025. Calculate the annual percentage growth rate:

\[\%CHG = \frac{51.85 - 42.50}{42.50} \times 100 = \mathbf{+22.00\%}\]

Keystrokes: 42.50 [ENTER] 51.85 [↱] [%CHG] $\rightarrow$ 22.

Staging prior-year revenue in Register $Y$ and current-year revenue in Register $X$ via 42.50 [ENTER] 51.85 $(T: 0.00, Z: 0.00, Y: 42.50, X: 51.85)$ and pressing [↱] [%CHG] evaluates $(X - Y)/Y \times 100$ in place, reporting top-line growth $(T: 0.00, Z: 0.00, Y: 42.50, X: \mathbf{22.00\%})$ while preserving the $\$42.50\text{M}$ baseline in Register $Y$.

Workbook Problem 11.2: EBITDA Margin Expansion in Basis Points

A manufacturing firm expands revenue from $\$240.0\text{M}$ to $\$310.0\text{M}$. During the same period, EBITDA rises from $\$36.0\text{M}$ to $\$55.8\text{M}$. Determine the EBITDA margin expansion in basis points ($1\text{ bp} = 0.01\%$). Evaluating the current-year operating margin via 55.8 [ENTER] 310 [÷] 100 [×] places $18.00\%$ in Register $X$ $(T: 0.00, Z: 0.00, Y: 0.00, X: 18.00)$, and evaluating the prior-year margin via 36 [ENTER] 240 [÷] 100 [×] places $15.00\%$ in $X$ while lifting $18.00\%$ into $Y$ $(T: 0.00, Z: 0.00, Y: 18.00, X: 15.00)$. Subtracting and scaling to basis points ([-] 100 [×]) isolates the operating leverage gain of $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{300.00}\text{ bps})$.

Workbook Problem 11.3: 15-Year Mortgage Acceleration Comparison

Compare the $\$350,000.00$ mortgage on a 15-year term ($N = 180$) at $6.25\%$ interest ($i = 6.25 / 12 = 0.520833\%$): * Payment: 180 [N] 6.25 [ENTER] 12 [÷] [I%] 350000 [PV] 0 [FV] [PMT] $\rightarrow$ -3,000.98003

Halving the amortization horizon to $N = 180\text{ months}$ (180 [N], $(T: 0.00, Z: 0.00, Y: 0.00, X: 180.00)$) at a $50\text{-bp}$ lower rate (6.25 [ENTER] 12 [÷] [I%] $= 0.520833\%/\text{mo}$, $(T: 0.00, Z: 0.00, Y: 180.00, X: 0.52)$) on the same $\$350,000$ principal (350000 [PV] 0 [FV], $(T: 180.00, Z: 0.52, Y: 350000.00, X: 0.00)$) increases the monthly debt service [PMT] by only $\$730.89/\text{mo}$ to $(T: 180.00, Z: 0.52, Y: 350000.00, X: \mathbf{-3000.98})$, while cutting total lifetime outlays (180 [×], $(T: 0.52, Z: 0.52, Y: 350000.00, X: \mathbf{-540176.41})$) by $\$277,057.19$.

Workbook Problem 11.4: Weighted Average Cost of Capital (WACC)

Equity capital $\$60.0\text{M}$ ($60\%$) at required equity return $K_e = 12.0\%$; Debt capital $\$40.0\text{M}$ ($40\%$) at pre-tax interest rate $K_d = 6.0\%$; Corporate income tax rate $t = 25\%$: $$\text{WACC} = \left(\frac{E}{V}\right) K_e + \left(\frac{D}{V}\right) K_d (1 - t)$$

Under Modigliani-Miller corporate tax-shield accounting, interest expense is tax-deductible, reducing the effective cost of debt by $(1 - t)$. Computing the tax-shielded debt rate via 6 [ENTER] 1 [ENTER] 0.25 [-] [×] ($4.50\%$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 4.50)$) and weighting by the $40\%$ debt share (.40 [×] $= 1.80\%$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 1.80)$), then evaluating the $60\%$ equity component via 60 [ENTER] 100 [÷] 12 [×] ($7.20\%$, $(T: 0.00, Z: 0.00, Y: 1.80, X: 7.20)$) and summing with [+] establishes the blended corporate hurdle rate $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{9.00\%})$.

Workbook Problem 11.5: Rule of 72 Doubling Time Precision

An investment yields $8.0\%$ annually. Compare the Rule of 72 mental estimate with exact compound TVM: * Rule of 72 estimate: $72 / 8 = \mathbf{9.00\text{ years}}$. * Exact TVM Doubling: 8 [I%] 1 [+/-] [PV] 0 [PMT] 2 [FV] [N] $\rightarrow$ 9.01 years. * Error is under $0.11\%$.

Luca Pacioli's classic Rule of 72 (72 [ENTER] 8 [÷] $= 9.00\text{ years}$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 9.00)$) approximates $N = \ln(2)/\ln(1+i) \approx 69.31/i + 0.35$, making $72/i$ remarkably accurate near $i = 8\%$. Programming 8 [I%] 1 [+/-] [PV] 0 [PMT] 2 [FV] $(T: 8.00, Z: -1.00, Y: 0.00, X: 2.00)$ and solving [N] yields the exact doubling horizon $\ln(2)/\ln(1.08) = \mathbf{9.0065\text{ years}}$ $(T: 8.00, Z: -1.00, Y: 0.00, X: \mathbf{9.01}\text{ years})$.

Workbook Problem 11.6: Continuous Compounding ($A = P e^{rt}$)

Deposit $\$10,000.00$ in a treasury note yielding $5.50\%$ compounded continuously for 10 years:

\[A = 10,000 \times e^{0.055 \times 10} = 10,000 \times e^{0.55}\]

Keystrokes: .055 [ENTER] 10 [×] [eˣ] 10000 [×] $\rightarrow$ 17,332.53018.

In the continuous compounding limit $\lim_{m \to \infty}(1 + r/m)^{mt} = e^{rt}$, evaluating the dimensionless yield-time exponent .055 [ENTER] 10 [×] ($rt = 0.55$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 0.55)$), exponentiating via [eˣ] ($e^{0.55} \approx 1.733253$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 1.73)$), and scaling by initial principal 10000 [×] determines the terminal value $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{\$17332.53})$.

Workbook Problem 11.7: Effective Annual Rate (EAR) from Nominal APR

A revolving credit line charges nominal $21.99\%$ APR compounded daily ($m = 365$):

\[\text{EAR} = \left(1 + \frac{0.2199}{365}\right)^{365} - 1\]

Keystrokes: .2199 [ENTER] 365 [÷] 1 [+] 365 [yˣ] 1 [-] 100 [×] $\rightarrow$ 24.586963176.

Converting a nominal APR into its true Effective Annual Rate (APY) reveals the compounding drag of daily interest accrual: dividing the nominal rate by the compounding frequency .2199 [ENTER] 365 [÷] 1 [+] forms the daily growth factor $1 + r/365 = 1.00060247$ $(T: 0.00, Z: 0.00, Y: 0.00, X: 1.000602)$, raising to the 365th power via 365 [yˣ] yields $1.245870$ $(T: 0.00, Z: 0.00, Y: 0.00, X: 1.25)$, and 1 [-] 100 [×] isolates the effective annual rate $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{24.59\%})$. (Single-Entry Compounding Choreography: Because 1 [+] overwrites LASTX with 1, you can avoid typing the compounding frequency 365 twice by duplicating it into Register $Y$ upfront: 365 [ENTER] .2199 [x⇄y] [÷] 1 [+] [x⇄y] [yˣ] 1 [-] 100 [×] keeps 365 waiting in Register $Y$ while forming $1 + 0.2199/365$, so 365 is entered only once!)

Workbook Problem 11.8: Auto Loan Financing with Trade-In

New vehicle price $\$38,500.00$. Trade-in allowance $\$8,500.00$. Net loan principal $PV = \$30,000.00$. Financing: 60 months ($N=60$) at $4.90\%$ APR ($i = 4.90 / 12 = 0.408333\%$): Keystrokes: 60 [N] 4.9 [ENTER] 12 [÷] [I%] 30000 [PV] 0 [FV] [PMT] $\rightarrow$ -564.76.

Deducting the $\$8,500$ trade-in credit leaves a net financed principal of $\$30,000.00$: storing 60 [N] $(T: 0.00, Z: 0.00, Y: 0.00, X: 60.00)$, converting the $4.90\%$ APR to a monthly rate via 4.9 [ENTER] 12 [÷] [I%] ($0.408333\%/\text{mo}$, $(T: 0.00, Z: 0.00, Y: 60.00, X: 0.41)$), and entering 30000 [PV] 0 [FV] $(T: 60.00, Z: 0.41, Y: 30000.00, X: 0.00)$ allows [PMT] to resolve the monthly installment $(T: 60.00, Z: 0.41, Y: 30000.00, X: \mathbf{-564.76})$.

Workbook Problem 11.9: Equipment Lease vs. Buy Analysis

An industrial CNC milling machine can be leased for $\$12,000.00/\text{year}$ for 5 years at an $8.00\%$ hurdle discount rate. Outright purchase price is $\$52,000.00$: $$\text{PV of Lease} = 12,000 \times \left[\frac{1 - 1.08^{-5}}{0.08}\right] = \mathbf{\$47,912.52}$$ Leasing saves $\$4,087.48$ in present value terms compared to outright purchase. Programming the 5-year lease obligation 5 [N] 8 [I%] 12000 [+/-] [PMT] 0 [FV] $(T: 5.00, Z: 8.00, Y: -12000.00, X: 0.00)$ and solving [PV] discounts the lease stream to $(T: 5.00, Z: 8.00, Y: -12000.00, X: \mathbf{47912.52})$, which compares directly against the $\$52,000$ purchase price via 52000 [x⇄y] [-] to quantify the net present value advantage of leasing $(T: 5.00, Z: 8.00, Y: 52000.00, X: \mathbf{\$4087.48})$.

Workbook Problem 11.10: Dining Bill with 8.25% Tax and 20% Tip

Food and beverage subtotal: $\$164.50$, regional tax $8.25\%$ (164.50 [ENTER] 8.25 [↱] [%] $\rightarrow \$13.57$), and $20.0\%$ tip on the pre-tax subtotal (164.50 [ENTER] 20 [↱] [%] $\rightarrow \$32.90$). Because proper hospitality etiquette calculates gratuity on the pre-tax subtotal rather than the taxed total, duplicating the subtotal via 164.50 [ENTER] [ENTER] 8.25 [↱] [%] [+] forms the taxed bill $(\$178.07)$ in Register $X$ while keeping a second copy of the pre-tax base $(\$164.50)$ in Register $Y$ $(T: 0.00, Z: 0.00, Y: 164.50, X: 178.07)$. Swapping via [x⇄y] 20 [↱] [%] computes the $20\%$ pre-tax gratuity $(\$32.90)$ in $X$ while the taxed bill waits in $Z$ $(T: 0.00, Z: 178.07, Y: 164.50, X: 32.90)$, or even more compactly, keying 164.50 [ENTER] 8.25 [ENTER] 20 [+] [↱] [%] [+] combines the two pre-tax rates ($28.25\%$) in a single percentage operation to reach $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{\$210.97})$ with 164.50 typed only once!

Workbook Problem 11.11: 4-Way Bill Split via Stack Roll `[R↓]`

Grand total $\$210.97$ split equally among 4 dinner guests: key 210.97 [ENTER] 4 [÷] $\rightarrow$ 52.74 $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{\$52.74})$, establishing the per-person share directly in Register $X$.

Workbook Problem 11.12: Culinary Recipe Scaling with Fractions

A cake recipe yields 8 servings and calls for $2\text{-}1/2\text{ cups}$ of flour ($2.5\text{ cups}$). To scale the recipe to serve 14 guests, compute the scaling factor 14 [ENTER] 8 [÷] $\rightarrow$ 1.75 and multiply by the base flour requirement 2.5 [ENTER] 1.75 [×] $\rightarrow$ 4.375 ($4\text{-}3/8\text{ cups}$). On a continuous stack, keying 14 [ENTER] 8 [÷] establishes the $1.75\times$ multiplier in Register $X$ $(T: 0.00, Z: 0.00, Y: 0.00, X: 1.75)$, and 2.5 [×] scales the base flour directly to $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{4.38}\text{ cups})$.

Workbook Problem 11.13: Kitchen Unit Volume Conversions

Convert $3.5\text{ fluid ounces}$ to tablespoons ($1\text{ fl oz} = 2\text{ tbsp}$), then to teaspoons ($1\text{ tbsp} = 3\text{ tsp}$): 3.5 [ENTER] 2 [×] 3 [×] $\rightarrow$ 21.00. Chaining the dimensional factors on the stack (3.5 [ENTER] 2 [×] $\rightarrow (T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{7.00}\text{ tbsp})$, followed by 3 [×] $\rightarrow (T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{21.00}\text{ tsp})$) converts across US customary volume tiers without intermediate re-entry.

Workbook Problem 11.14: Bookshelf Bay Spacing with Dividers

Alcove width $47\text{-}5/8^{\prime\prime}$ ($47.625^{\prime\prime}$). Two $3/4^{\prime\prime}$ ($0.75^{\prime\prime}$) upright dividers create 3 equal bays: 47.625 [ENTER] .75 [ENTER] 2 [×] [-] 3 [÷] $\rightarrow$ 15.375.

In architectural millwork layout, deducting the combined thickness of all interior partitions before dividing by the bay count prevents cumulative layout drift: keying 47.625 [ENTER] .75 [ENTER] 2 [×] [-] subtracts the two $0.75^{\prime\prime}$ dividers ($1.50^{\prime\prime}$) from the $47.625^{\prime\prime}$ alcove opening to isolate the net clear span $46.125^{\prime\prime}$ $(T: 0.00, Z: 0.00, Y: 0.00, X: 46.13)$, and 3 [÷] partitions it into three equal bays of $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{15.38^{\prime\prime}})$ ($15\text{-}3/8^{\prime\prime}$).

Workbook Problem 11.15: Woodworking Saw Kerf Accumulation

Cut 5 identical table legs of length $28\text{-}1/4^{\prime\prime}$ ($28.25^{\prime\prime}$) from a single board. The miter saw blade has a kerf thickness of $1/8^{\prime\prime}$ ($0.125^{\prime\prime}$). There are 4 intermediate cuts: 28.25 [ENTER] 5 [×] .125 [ENTER] 4 [×] [+] $\rightarrow$ 141.75.

Parting $n = 5$ workpieces from a single blank requires $n - 1 = 4$ internal kerf cuts: evaluating finished workpiece length 28.25 [ENTER] 5 [×] ($141.25^{\prime\prime}$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 141.25)$), computing cumulative blade kerf loss .125 [ENTER] 4 [×] ($0.50^{\prime\prime}$, $(T: 0.00, Z: 0.00, Y: 141.25, X: 0.50)$), and summing with [+] determines the minimum rough-mill board length $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{141.75^{\prime\prime}})$ ($11\text{ ft } 9\text{-}3/4^{\prime\prime}$).

Workbook Problem 11.16: Staircase Riser and Tread Layout

Total vertical rise from finished basement floor to main level is $108\text{ inches}$ ($9\text{ ft}$), and building code requires individual riser height $h \le 7\text{-}3/4^{\prime\prime}$ ($7.75^{\prime\prime}$). Keying 108 [ENTER] 7.75 [÷] estimates the minimum riser count ($13.935$) in Register $X$ $(T: 0.00, Z: 0.00, Y: 0.00, X: 13.94)$, which rounds up to 14 risers. Keying 108 [ENTER] 14 [÷] $\rightarrow$ 7.7142857143 evaluates the exact unit riser height $(T: 0.00, Z: 0.00, Y: 13.94, X: \mathbf{7.71^{\prime\prime}})$ ($7\text{-}23/32^{\prime\prime}$).

Workbook Problem 11.17: Blueprint Scale Conversion ($1/4^{\prime\prime} = 1^{\prime}\text{-}0^{\prime\prime}$)

Wall on drawing measures $5\text{-}3/8^{\prime\prime}$ ($5.375^{\prime\prime}$). Real length ($4\text{ ft/in}$).

5.375 [ENTER] 4 [×] $\rightarrow$ 21.5.

On standard $1/4^{\prime\prime} = 1^{\prime}\text{-}0^{\prime\prime}$ architectural plans (1:48 scale), each inch on paper represents $4\text{ feet}$ in the field, so multiplying 5.375 [ENTER] 4 [×] converts the blueprint span directly to $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{21.50}\text{ ft})$ ($21^{\prime}\text{-}6^{\prime\prime}$).

Workbook Problem 11.18: Model Railroading HO Scale (1:87.1)

Real locomotive length is $72\text{ feet}$ ($864\text{ inches}$). In HO scale.

864 [ENTER] 87.1 [÷] $\rightarrow$ 9.9196326062.

NMRA HO scale ($3.5\text{ mm} = 1\text{ ft}$) corresponds to a linear reduction ratio of $304.8 / 3.5 = 87.086 \approx 87.1$: dividing the prototype length 864 [ENTER] 87.1 [÷] determines the scale chassis length $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{9.92^{\prime\prime}})$.

Workbook Problem 11.19: Fuel Economy MPG to L/100km Conversion

Formula: $\text{L/100km} = \frac{235.214583}{\text{MPG}}$. Convert $32.0\text{ MPG}$.

235.215 [ENTER] 32 [÷] $\rightarrow$ 7.35046875.

Because US fuel economy measures distance per volume ($\text{mi/gal}$) whereas metric consumption measures volume per $100\text{ km}$, the relationship is reciprocally proportional through the constant $\frac{100 \times 3.785411784}{1.609344} = 235.214583$: keying 235.215 [ENTER] 32 [÷] yields $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{7.35}\text{ L/100km})$.

Workbook Problem 11.20: Foreign Currency Exchange Fee and Spread

Traveler converts $\$2,000\text{ USD}$ to Euros. Interbank rate is $1.085\text{ EUR/USD}$. Bank charges a $2.5\%$ FX spread fee.

2000 [ENTER] 2.5 [↱] [%] [-] 1.085 [×] $\rightarrow$ 2,115.75.

Deducting the retail foreign-exchange spread via 2000 [ENTER] 2.5 [↱] [%] [-] leaves $\$1,950.00$ of net convertible principal in Register $X$ $(T: 0.00, Z: 0.00, Y: 0.00, X: 1950.00)$, which converts at the interbank spot rate (1.085 [×]) to $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{2115.75}\text{ EUR})$.

Workbook Problem 11.21: Cash Conversion Cycle (CCC)

Days Inventory Outstanding $DIO = 45\text{ days}$; Days Sales Outstanding $DSO = 38\text{ days}$; Days Payables Outstanding $DPO = 52\text{ days}$:

\[\text{CCC} = DIO + DSO - DPO = 45 + 38 - 52 = \mathbf{31\text{ days}}\]

In working-capital treasury management, the Cash Conversion Cycle measures how many days cash remains tied up in inventory and customer receivables beyond trade supplier credit terms: 45 [ENTER] 38 [+] 52 [-] resolves $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{31}\text{ days})$.

Workbook Problem 11.22: Enterprise Value (EV) Calculation

Market capitalization $\$450\text{M}$, total debt $\$120\text{M}$, cash and equivalents $\$35\text{M}$:

\[\text{EV} = \text{Market Cap} + \text{Debt} - \text{Cash} = 450 + 120 - 35 = \mathbf{\$535.00M}\]

In mergers-and-acquisitions valuation, Enterprise Value represents the net takeover price of the operating business—acquiring all equity plus assuming gross debt, less liquid cash on the balance sheet (450 [ENTER] 120 [+] 35 [-] $\rightarrow (T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{\$535.00M})$).

Workbook Problem 11.23: Sinking Fund Future Savings Goal

Save $\$50,000$ in 5 years ($N=60\text{ months}$) at $6.0\%$ APR ($i = 0.5\%/\text{mo}$): Keystrokes: 60 [N] .5 [I%] 0 [PV] 50000 [FV] [PMT]$\rightarrow$ -716.64.

Programming the sinking-fund parameters 60 [N] .5 [I%] 0 [PV] 50000 [FV] $(T: 60.00, Z: 0.50, Y: 0.00, X: 50000.00)$ and solving [PMT] determines the required monthly deposit $(T: 60.00, Z: 0.50, Y: 0.00, X: \mathbf{-716.64})$.

Workbook Problem 11.24: College Education Fund Present Value

Annual tuition $\$35,000$ for 4 years ($N=4$, beginning of year BEGIN mode) at $5\%$ investment return: Toggled to BEGIN mode: 4 [N] 5 [I%] 35000 [PMT] 0 [FV] [PV] $\rightarrow$ -$130,417.84.

Because university bursars collect tuition at the start of each academic year (an annuity due), toggling BEGIN mode and storing 4 [N] 5 [I%] 35000 [PMT] 0 [FV] $(T: 4.00, Z: 5.00, Y: 35000.00, X: 0.00)$ before pressing [PV] discounts only the three subsequent installments, yielding the required freshman-day endowment $(T: 4.00, Z: 5.00, Y: 35000.00, X: \mathbf{-\$130417.84})$.

Workbook Problem 11.25: Net Present Value (NPV) Comparison of Capital Projects

A company evaluates two mutually exclusive industrial automation projects with a $12.0\%$ hurdle rate: Project Alpha: Initial Cost $\$80,000.00$; Inflows: $\$35,000/\text{yr}$ for 3 years. Project Beta: Initial Cost $\$120,000.00$; Inflows: $\$52,000/\text{yr}$ for 3 years:

$$\text{PV Annuity Factor} = \frac{1 - (1.12)^{-3}}{0.12} = \frac{1 - 0.71178}{0.12} \approx 2.40183$$ * Project Alpha: $35,000 \times 2.40183 - 80,000 = \$84,064.10 - \$80,000 = \mathbf{+\$4,064.10}$. * Project Beta: $52,000 \times 2.40183 - 120,000 = \$124,895.22 - \$120,000 = \mathbf{+\$4,895.22}$. Project Beta delivers $\$831.12$ more net present value to shareholders.

Because N, I%, and FV remain persistent across TVM evaluations, discounting Project Alpha (3 [N] 12 [I%] 35000 [+/-] [PMT] 0 [FV] [PV] $\rightarrow \$84,064.10$, $(T: 3.00, Z: 12.00, Y: -35000.00, X: 84064.10)$, less 80000 [-] $\rightarrow (T: 3.00, Z: 12.00, Y: -35000.00, X: \mathbf{+\$4064.10})$) requires only updating PMT for Project Beta (52000 [+/-] [PMT] [PV] $\rightarrow \$124,895.22$, $(T: 12.00, Z: -35000.00, Y: 4064.09, X: 124895.22)$, less 120000 [-] $\rightarrow (T: 3.00, Z: 12.00, Y: -52000.00, X: \mathbf{+\$4895.22})$).

Workbook Problem 11.26: Capital Budgeting Profitability Index (PI)

Using the present value of inflows from Workbook Problem 11.25:

$$\text{Profitability Index (PI)} = \frac{\text{PV of Future Inflows}}{\text{Initial Investment}}$$ * Project Alpha: $84,064.10 \div 80,000.00 = \mathbf{1.0508}$ ($1.05\text{x}$) * Project Beta: $124,895.22 \div 120,000.00 = \mathbf{1.0408}$ ($1.04\text{x}$) Although Project Beta has higher absolute NPV, Project Alpha delivers higher capital efficiency per dollar invested!

When capital rationing constrains total CapEx budget, ranking projects by Profitability Index $\text{PI} = \text{PV}_{\text{inflows}}/I_0$ (84064.10 [ENTER] 80000 [÷] $\rightarrow (T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{1.0508}\text{x})$ for Alpha versus 124895.22 [ENTER] 120000 [÷] $\rightarrow (T: 0.00, Z: 0.00, Y: 1.05, X: \mathbf{1.0408}\text{x})$ for Beta) identifies which project generates more present value per dollar of scarce capital.

Workbook Problem 11.27: MACRS 5-Year Depreciation Schedule

A technology enterprise purchases $\$250,000.00$ of server hardware depreciated under IRS 5-Year MACRS ($20.0\%, 32.0\%, 19.2\%, 11.52\%, 11.52\%, 5.76\%$): * Year 1 Deduction: 250000 [ENTER] 20 [↱] [%] $\rightarrow$ 50,000 (Remaining basis: [-] $\rightarrow$ $\$200,000.00$).

Under the IRS 200%-declining-balance half-year convention (250000 [ENTER] 20 [↱] [%]), Register $X$ displays the Year-1 tax depreciation write-off $(\$50,000.00)$ while Register $Y$ retains the $\$250,000.00$ original cost basis $(T: 0.00, Z: 0.00, Y: 250000.00, X: 50000.00)$, so pressing [-] immediately resolves the adjusted Year-1 book value $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{\$200000.00})$.

Workbook Problem 11.28: Staircase Stringer Layout with 1/16" Carpentry Cuts

Total vertical rise is $105\text{-}1/2^{\prime\prime}$ ($105.5^{\prime\prime}$). Target riser height is $7\text{-}1/4^{\prime\prime}$ ($7.25^{\prime\prime}$): 1. Riser count: 105.5 [ENTER] 7.25 [÷] $\rightarrow$ $14.55 \implies$ Round to 15 risers. 2. Exact riser height: 105.5 [ENTER] 15 [÷] $\rightarrow$ 7.0333333333. 3. Decompose to nearest $1/16^{\prime\prime}$: Fractional part: 7 [-] 16 [×] $\rightarrow$ $0.5333 \approx 1/16^{\prime\prime} \implies$ Cut stringers at 7-1/16" (exact rise $7.0625^{\prime\prime} \times 15 = 105.94^{\prime\prime}$, adjusted on bottom tread).

Dividing total floor-to-floor rise by the target riser height (105.5 [ENTER] 7.25 [÷] $= 14.55$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 14.55)$) indicates 15 risers; dividing 105.5 [ENTER] 15 [÷] yields the exact unit rise $7.0333^{\prime\prime}$ $(T: 0.00, Z: 0.00, Y: 14.55, X: 7.03)$, and converting the fractional part to sixteenths (7 [-] 16 [×] $= 0.5333/16^{\prime\prime}$, $(T: 0.00, Z: 0.00, Y: 14.55, X: 0.53)$) specifies a 7-1/16" stringer gauge layout.

Workbook Problem 11.29: SaaS Unit Economics — LTV : CAC Ratio in One Chain

A B2B software enterprise generates Average Revenue Per User $\text{ARPU} = \$250.00/\text{month}$ at an $80.0\%$ gross margin, with monthly revenue churn $c = 2.5\%$ ($0.025$) and Customer Acquisition Cost $\text{CAC} = \$1,600.00$: $$\text{LTV} = \frac{\text{ARPU} \times \text{GM}\%}{\text{Churn}} = \frac{250 \times 0.80}{0.025} = \$8,000.00, \qquad \frac{\text{LTV}}{\text{CAC}} = \frac{8000}{1600} = \mathbf{5.00\text{x}}$$ Boardroom RPN Keystrokes: 250 [ENTER] 80 [↱] [%] 0.025 [÷] 1600 [÷] $\rightarrow$ 5.00x.

Modeling customer retention as a geometric survival series with monthly hazard rate $c = 0.025$ gives an expected customer lifetime of $1/c = 40\text{ months}$. Keying 250 [ENTER] 80 [↱] [%] isolates monthly gross margin contribution $(\$200.00/\text{mo})$ in Register $X$ $(T: 0.00, Z: 0.00, Y: 250.00, X: 200.00)$, dividing by churn 0.025 [÷] capitalizes that stream into Customer Lifetime Value ($\text{LTV} = \$8,000.00$, $(T: 0.00, Z: 0.00, Y: 250.00, X: 8000.00)$), and dividing by 1600 [÷] yields the unit-economics multiple $(T: 0.00, Z: 0.00, Y: 250.00, X: \mathbf{5.00}\text{x})$.

Workbook Problem 11.30: Operating Breakeven Volume ($Q_{\text{BE}}$)

A manufacturing line carries annual fixed overhead of $F = \$420,000.00$. Each unit sells for $P = \$85.00$ with variable unit cost $V = \$35.00$ (contribution margin $\$50.00/\text{unit}$): $$Q_{\text{BE}} = \frac{F}{P - V} = \frac{420,000}{85 - 35}$$ Keystrokes: 420000 [ENTER] 85 [ENTER] 35 [-] [÷] $\rightarrow$ 8,400 units.

In cost-volume-profit (CVP) accounting, each unit sold contributes $P - V$ toward absorbing fixed overhead $F$: staging 420000 [ENTER] $(T: 0.00, Z: 0.00, Y: 420000.00, X: 420000.00)$, forming the unit contribution margin 85 [ENTER] 35 [-] ($\$50.00/\text{unit}$ in $X$ while $F$ waits in $Y$, $(T: 0.00, Z: 0.00, Y: 420000.00, X: 50.00)$), and dividing via [÷] determines the operating breakeven volume $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{8400}\text{ units})$.

Workbook Problem 11.31: Commercial Real Estate Net Operating Income (NOI) & Cap Rate

An apartment building generates Gross Scheduled Rent of $\$112,500.00/\text{year}$ with a $28.0\%$ operating expense + vacancy ratio, offered at an asking price of $\$1,250,000.00$: $$\text{NOI} = 112,500 \times (1 - 0.28) = \$81,000.00, \qquad \text{Cap Rate} = \frac{\text{NOI}}{\text{Price}} \times 100$$ Keystrokes: 112500 [ENTER] 28 [↱] [%] [-] 1250000 [÷] 100 [×] $\rightarrow$ 6.48%.

Deducting the $28\%$ vacancy and operating-expense reserve via 112500 [ENTER] 28 [↱] [%] $(T: 0.00, Z: 0.00, Y: 112500.00, X: 31500.00)$ followed by [-] isolates Net Operating Income ($\text{NOI} = \$81,000.00$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 81000.00)$); dividing by the $\$1,250,000$ acquisition price (1250000 [÷] $= 0.0648$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 0.0648)$) and scaling by 100 [×] determines the unlevered capitalization rate $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{6.48}\%)$.

Chapter 12: Function & Scatter Plotting Engine — Extended Engineering & Competition Problems

Accompanies Chapter 12 of Thinking in the Stack (27 Worked Engineering Problems — PLOT is an App Mode feature, excluded from physical firmware for exam compliance)

Workbook Problem 12.1: Cubic Extrema & Root Guess Bracketing for Brent's Solver

Before executing Richard Brent's root finder ([↱] [SOLVE]), plot the polynomial in App Mode to visually bracket roots:

\[f(x) = x^3 - 3x^2 - 9x + 5 = 0\]

Program in [↱] [EQN]: X^3 - 3*X^2 - 9*X + 5
Stage domain on the stack ($x_{\min}$ in Register $Y$, $x_{\max}$ in Register $X$): 4 [+/-] [ENTER] 6
Press [↱] [PLOT].

Plot Telemetry	Crosshair Readout	Curve Feature
Active Function	`X^3 - 3X^2 - 9X + 5`	Three real zero crossings in $[-3,-2], [0,1], [4,5]$
Crosshair Cursor	`X = 0.500000000`, `Y = -0.125000000`	Pre-root bracket inspection prior to `[↱] [SOLVE]`

The plot visually reveals: * Local maximum near $x = -1.0$: $f(-1) = 10.0$ * Local minimum near $x = 3.0$: $f(3) = -22.0$ * Three distinct roots: Root 1 in $[-3, -2]$, Root 2 in $[0, 1]$, Root 3 in $[4, 5]$. Tracing crosshair to $x = 0.50$ and pressing [ENTER] pushes $0.50$ onto the stack. Pressing [↱] [SOLVE] X converges to root $x = 0.5185$ in 3 iterations. Crosshair Evaluation Check ($x = 0.50$): 0.5 [ENTER] 3 [yˣ] 3 [ENTER] 0.5 [↰] [x²] [×] [-] 9 [ENTER] 0.5 [×] [-] 5 [+] $\implies \mathbf{-0.125000}$.

STEP	KEYSTROKE	REGISTER Y	REGISTER X	DISPLAY
01	.5 [ENTER] 3 [yˣ]	0.00	0.125000	0.125000
02	3 [ENTER] .5 [↰] [x²] [×]	0.125000	0.750000	0.750000
03	[-]	0.00	-0.625000	-0.625000
04	9 [ENTER] .5 [×]	-0.625000	4.500000	4.500000
05	[-]	0.00	-5.125000	-5.125000
06	5 [+]	0.00	-0.125000	-0.125000

Workbook Problem 12.2: Standard Normal Distribution Bell Curve (Gaussian PDF)

Plot the Gaussian normal distribution curve with mean $\mu = 0$ and standard deviation $\sigma = 1$:

\[f(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\]

Program in [↱] [EQN]: EXP(-X^2/2)/SQRT(2*π)
Stage domain ($x_{\min}$ in $Y$, $x_{\max}$ in $X$): 4 [+/-] [ENTER] 4
Press [↱] [PLOT].

Plot Telemetry	Crosshair Readout	Curve Feature
Active Function	`EXP(-X^2/2)/SQRT(2*π)`	Standard Normal probability density bell curve
Crosshair Cursor	`X = 0.000000000`, `Y = 0.3989422804`	Global maximum $1/\sqrt{2\pi}$ at mean $\mu = 0$

The peak at $x = 0$ renders at $y = \frac{1}{\sqrt{2\pi}} \approx 0.398942$. Inflection Ordinate Check ($x = 1.0$): 1 [↰] [x²] 2 [÷] [+/-] [eˣ] 2 [↱] [π] [×] [√x] [÷] $\implies \mathbf{0.241971}$.

STEP	KEYSTROKE	REGISTER Y	REGISTER X	DISPLAY
01	1 [↰] [x²] 2 [÷]	0.00	0.500000	0.500000
02	[+/-] [eˣ]	0.00	0.606531	0.606531
03	2 [↱] [π] [×]	0.606531	6.283185	6.283185
04	[√x]	0.606531	2.506628	2.506628
05	[÷]	0.00	0.241971	0.241971

Workbook Problem 12.3: Low-Pass RC Filter Frequency Bode Magnitude Plot

The voltage gain of a first-order passive low-pass RC filter with cutoff frequency $f_c = 1000\text{ Hz}$ is:

$$|H(f)| = \frac{1}{\sqrt{1 + (f / 1000)^2}}$$ Plot gain across frequency domain $f \in [0, 5000\text{ Hz}]$: 1. Program in [↱] [EQN]: 1/SQRT(1+(X/1000)^2) 2. Stage domain ($x_{\min}$ in $Y$, $x_{\max}$ in $X$): 0 [ENTER] 5000 3. Press [↱] [PLOT].

Plot Telemetry	Crosshair Readout	Curve Feature
Active Function	`1/SQRT(1+(X/1000)^2)`	First-order RC low-pass magnitude transfer curve
Crosshair Cursor	`X = 1000.000000`, `Y = 0.7071067812`	Half-power $-3\text{ dB}$ corner frequency ($1/\sqrt{2}$)

Trace to $f = 1000\text{ Hz}$: display confirms $y = 0.707107$ (the $-3\text{ dB}$ half-power point). Corner Frequency Check ($f = 1000\text{ Hz}$): 1000 [ENTER] 1000 [÷] [↰] [x²] 1 [+] [√x] [1/x] $\implies \mathbf{0.707107}$.

STEP	KEYSTROKE	REGISTER Y	REGISTER X	DISPLAY
01	1000 [ENTER]	1000.00	1000.00	1000.00_
02	1000 [÷]	0.00	1.000000	1.000000
03	[↰] [x²] 1 [+]	0.00	2.000000	2.000000
04	[√x]	0.00	1.414214	1.414214
05	[1/x]	0.00	0.707107	0.707107

Workbook Problem 12.4: High-Pass RC Filter Frequency Response

The high-pass filter frequency response is given by:

$$|H(f)| = \frac{f / f_c}{\sqrt{1 + (f / f_c)^2}}$$ For cutoff frequency $f_c = 1000\text{ Hz}$: 1. Program in [↱] [EQN]: (X/1000)/SQRT(1+(X/1000)^2) 2. Stage domain ($x_{\min}$ in $Y$, $x_{\max}$ in $X$): 0 [ENTER] 5000 3. Press [↱] [PLOT]. The curve begins at $0$ at DC ($f=0$), crosses $0.707107$ at $f = 1000\text{ Hz}$, and asymptotically approaches $1.00$ at high frequencies. Octave-Above-Cutoff Check ($f = 2000\text{ Hz}$): 2000 [ENTER] 1000 [÷] [STO] A [↰] [x²] 1 [+] [√x] [RCL] A [x⇄y] [÷] $\implies \mathbf{0.894427}$.

In a first-order RC high-pass section, the capacitor blocks DC ($|H(0)| = 0$) and passes high-frequency ac components above the corner frequency $f_c = \frac{1}{2\pi R C} = 1000\text{ Hz}$. Evaluating the normalized frequency ratio at one octave above cutoff (2000 [ENTER] 1000 [÷] [STO] A $\rightarrow f/f_c = 2.00$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 2.000000)$), forming the impedance modulus $\sqrt{1 + (f/f_c)^2} = \sqrt{5}$ via [↰] [x²] 1 [+] [√x] $(T: 0.00, Z: 0.00, Y: 0.00, X: 2.236068)$, and dividing via [RCL] A [x⇄y] [÷] establishes the octave-above-cutoff attenuation factor $2/\sqrt{5}$ $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{0.894427}$, or $-0.97\text{ dB})$. (Zero-Register Stack Alternative: Duplicating the normalized frequency ratio on the stack via 2000 [ENTER] 1000 [÷] [ENTER] [↰] [x²] 1 [+] [√x] [÷] keeps $f/f_c = 2.00$ waiting in Register $Y$ while $\sqrt{1 + (f/f_c)^2}$ forms in Register $X$, resolving $\mathbf{0.894427}$ without touching storage register A!)

Workbook Problem 12.5: Diode Forward Bias I-V Characteristic (Shockley Equation)

In solid-state semiconductor electronics, junction diode current follows:

$$I = I_s \left( e^{\frac{V}{n V_T}} - 1 \right)$$ For silicon diode with saturation current $I_s = 10^{-12}\text{ A}$, emission coefficient $n = 1.0$, thermal voltage $V_T = 0.026\text{ V}$: 1. Stage domain ($x_{\min}$ in $Y$, $x_{\max}$ in $X$): 0 [ENTER] 0.80, then press [↱] [PLOT]. 2. Curve remains nearly zero until the threshold "knee" voltage ($V \approx 0.65\text{ V}$), then skyrockets exponentially.

STEP	KEYSTROKE	REGISTER Y	REGISTER X	DISPLAY
01	.70 [ENTER]	0.700000	0.700000	0.700000_
02	.026 [÷]	0.00	26.923077	26.923077
03	[eˣ] 1 [-]	0.00	4.9266E11	4.9266E11
04	1 [E] 12 [+/-]	4.9266E11	1.000E-12	1.000E-12
05	[×]	0.00	0.492656	0.492656 (A)

Workbook Problem 12.6: Resonant RLC Tank Circuit Frequency Response

The amplitude response of a series RLC resonant circuit is:

$$A(\omega) = \frac{1}{\sqrt{(1 - (\omega/\omega_0)^2)^2 + (2\zeta \omega/\omega_0)^2}}$$ With natural resonance $\omega_0 = 100\text{ rad/s}$ and damping ratio $\zeta = 0.05$ (high Q factor $Q = 10$): Plot over $\omega \in [50, 150\text{ rad/s}]$: the curve displays a narrow resonant peak reaching $A(\omega_0) = \frac{1}{2\zeta} = 10.0$ at $\omega = 100\text{ rad/s}$.

STEP	KEYSTROKE	REGISTER Y	REGISTER X	DISPLAY
01	100 [ENTER] 100 [÷]	0.00	1.000000	1.000000
02	[↰] [x²] 1 [x⇄y]	1.000000	1.000000	1.000000
03	[-] [↰] [x²]	0.00	0.000000	0.000000
04	2 [ENTER] .05 [×] 1 [×]	0.00	0.100000	0.100000
05	[↰] [x²] [+]	0.00	0.010000	0.010000
06	[√x] [1/x]	0.00	10.000000	10.000000

Workbook Problem 12.7: Quadratic Parabola Vertex and Axis of Symmetry

Plot $y = -2x^2 + 8x - 3$ over $x \in [-1, 5]$: 1. Vertex formula: $x_v = -b / (2a) = -8 / (-4) = \mathbf{2.00}$. 2. Maximum value: $y_v = -2(4) + 16 - 3 = \mathbf{5.00}$. 3. Trace confirms vertex at $(2.00, 5.00)$.

STEP	KEYSTROKE	REGISTER Y	REGISTER X	DISPLAY
01	2 [↰] [x²] 2 [×] [+/-]	0.00	-8.000000	-8.000000
02	8 [ENTER] 2 [×]	-8.000000	16.000000	16.000000
03	[+] 3 [-]	0.00	5.000000	5.000000

Workbook Problem 12.8: Radioactive Exponential Decay Curve

Cobalt-60 activity decays according to $A(t) = 100 \times e^{-0.1315 t}$ ($t$ in years): Plot over $t \in [0, 25\text{ years}]$: Trace at $t = 5.27\text{ years}$ confirms $A = 50.0\%$ (the half-life).

With decay constant $\lambda = \frac{\ln 2}{t_{1/2}} = 0.1315\text{ yr}^{-1}$, evaluating the exponent at $t = 5.27\text{ yr}$ (.1315 [+/-] 5.27 [×] $= -0.693005 \approx -\ln 2$, $(T: 0.00, Z: 0.00, Y: 0.00, X: -0.693005)$), exponentiating via [eˣ] ($0.500072$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 0.500072)$), and scaling by 100 [×] confirms the $50\%$ half-life crossing $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{50.01}\%)$.

Workbook Problem 12.9: Logistic Population Growth S-Curve (Sigmoidal)

Ecological population growth with carrying capacity $K = 1000$, growth rate $r = 0.5$, initial population $P_0 = 50$:

$$P(t) = \frac{K}{1 + \left(\frac{K - P_0}{P_0}\right) e^{-r t}} = \frac{1000}{1 + 19 e^{-0.5 t}}$$ Plot over $t \in [0, 20]$: curve shows initial exponential growth, an inflection point at $P = 500$ ($t = \ln(19)/0.5 \approx 5.89$), and asymptotic approach to $1000$.

Pierre-François Verhulst's logistic equation $\frac{dP}{dt} = rP(1 - P/K)$ achieves its maximum absolute growth rate $\frac{rK}{4}$ exactly at the half-saturation inflection time $t_{\text{inf}} = \frac{\ln((K - P_0)/P_0)}{r} = \frac{\ln 19}{0.5} \approx 5.88894$: evaluating the damping term .5 [+/-] 5.88894 [×] [eˣ] 19 [×] yields $19 e^{-r t_{\text{inf}}} = 1.000000$ $(T: 0.00, Z: 0.00, Y: 0.00, X: 1.000000)$, adding 1 [+] forms the denominator $2.000000$ $(T: 0.00, Z: 0.00, Y: 0.00, X: 2.000000)$, and dividing into carrying capacity 1000 [x⇄y] [÷] confirms the inflection ordinate $K/2$ $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{500.000000})$.

Workbook Problem 12.10: Weber-Fechner Acoustic Decibel Curve

Sound pressure level $L_p = 20 \log_{10}(P / P_0)$ where $P_0 = 20\text{ }\mu\text{Pa}$: Logarithmic compression displays the human ear's vast dynamic range across acoustic pressures.

Because human auditory perception responds logarithmically to acoustic pressure amplitude, dividing a $20,000\text{ }\mu\text{Pa}$ ($0.02\text{ Pa}$) sound wave by the $20\text{ }\mu\text{Pa}$ hearing threshold via 20000 [ENTER] 20 [÷] isolates the amplitude ratio $P/P_0 = 1000.00$ $(T: 0.00, Z: 0.00, Y: 0.00, X: 1000.00)$, taking the base-10 logarithm [↰] [LOG] yields $3.000000\text{ bels}$ of amplitude $(T: 0.00, Z: 0.00, Y: 0.00, X: 3.000000)$, and scaling by 20 [×] converts to sound pressure level $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{60.00}\text{ dB})$.

Workbook Problem 12.11: Statistical Residual Plot ($y_i - \hat{y}_i$)

To verify that linear regression errors are normally distributed and homoscedastic, plot residuals:

$$e_i = y_i - \hat{y}(x_i)$$ * $e_1 = 1.25 - 0.1245(10) = 1.25 - 1.245 = \mathbf{+0.005\text{ V}}$ * $e_2 = 2.45 - 0.1245(20) = 2.45 - 2.490 = \mathbf{-0.040\text{ V}}$ * $e_3 = 3.80 - 0.1245(30) = 3.80 - 3.735 = \mathbf{+0.065\text{ V}}$ * $e_4 = 4.95 - 0.1245(40) = 4.95 - 4.980 = \mathbf{-0.030\text{ V}}$ Residual points scatter randomly above and below the zero-error axis, confirming absence of systematic non-linear bias.

Auditing regression residuals $e_i = y_i - \hat{y}_i$ ensures that transducer non-linearity is not masquerading under a high correlation coefficient ($r = 0.999565$). Staging the first measured output 1.25 [ENTER] $(T: 0.0000, Z: 0.0000, Y: 1.250000, X: 1.250000)$, evaluating the fitted value 10 [ENTER] 0.1245 [×] ($\hat{y}_1 = 1.245000$, $(T: 0.0000, Z: 0.000000, Y: 1.250000, X: 1.245000)$), and subtracting via [-] isolates $e_1 = y_1 - \hat{y}_1$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{+0.005000}\text{ V})$, while 2.45 [ENTER] 20 [ENTER] 0.1245 [×] [-] yields $e_2$ $(T: 0.0000, Z: 0.0000, Y: 0.0050, X: \mathbf{-0.040000}\text{ V})$. (Zero-Re-Entry Residual & $SS_{\text{res}}$ Choreography: Because the four calibration pairs from Textbook Showcase Problem 12.6 are already active in the statistical registers, you never need to re-type the slope 0.1245: keying 1.25 [ENTER] 10 [ŷ] [R↓] [-]—or parking the slope 0.1245 across the upper stack via 0.1245 [ENTER] [ENTER] [ENTER]—evaluates each fitted voltage directly. Better still, squaring each residual in place ([↰] [x²]) as soon as it is formed accumulates the Residual Sum of Squares $SS_{\text{res}} = \sum e_i^2 = 0.005^2 + (-0.040)^2 + 0.065^2 + (-0.030)^2 = \mathbf{0.006750\text{ V}^2}$ directly on the stack without re-entering a single residual!)

Workbook Problem 12.12: Lissajous Figure 1:2 Harmonic Ratio

Signal frequencies $f_x = 100\text{ Hz}, f_y = 200\text{ Hz}$:

$$x(t) = \sin(t), \quad y(t) = \sin(2t)$$ The parametric trajectory traces a figure-8 ("infinity" symbol $\infty$).

Jules Antoine Lissajous's 1:2 harmonic curve satisfies the algebraic lemniscate relation $y^2 = 4x^2(1 - x^2)$, crossing the origin twice per fundamental cycle to form a figure-eight. In [RAD] mode, duplicating $t = \pi/6$ via [↱] [π] 6 [÷] [ENTER] $(T: 0.00, Z: 0.00, Y: 0.523599, X: 0.523599)$, evaluating the second-harmonic ordinate [ENTER] 2 [×] [SIN] ($y = \sin(\pi/3) = \sqrt{3}/2 \approx 0.866025$ in $X$ while $t$ waits in $Y$, $(T: 0.00, Z: 0.00, Y: 0.523599, X: 0.866025)$), and swapping [x⇄y] [SIN] resolves the fundamental abscissa $x = \sin(\pi/6) = \mathbf{0.500000}$ in $X$ while preserving $y = 0.866025$ in $Y$ $(T: 0.00, Z: 0.00, Y: 0.866025, X: 0.500000)$.

Workbook Problem 12.13: Acoustic Cardioid Microphone Polar Pattern

In studio acoustics, a cardioid directional microphone exhibits polar sensitivity over $\theta \in [0^\circ, 360^\circ]$:

\[r(\theta) = 2(1 - \cos\theta)\]

This profile produces a heart-shaped cardioid with a complete null ($r = 0$) at $\theta = 0^\circ$, side sensitivity $r = 2.0$ at $\theta = \pm 90^\circ$, and peak rear sensitivity at $\theta = 180^\circ$ ($r = 4.0$).

Polar-to-Rectangular Vector Check: To convert any $(x, y) = (1, 10)$ coordinate on a traced curve into polar magnitude $r = \sqrt{x^2 + y^2}$: 10 [↰] [x²] 1 [↰] [x²] [+] [√x] $\implies \mathbf{10.049876}$.

Combining an omnidirectional pressure element with a figure-eight velocity-gradient element creates the uniaxial cardioid pattern $r(\theta) = 2(1 - \cos\theta)$. Converting any traced Cartesian point $(x, y) = (1, 10)$ into radial distance via 10 [↰] [x²] ($100.000000$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 100.000000)$), 1 [↰] [x²] ($1.000000$ in $X$ with $100.000000$ in $Y$, $(T: 0.00, Z: 0.00, Y: 100.000000, X: 1.000000)$), [+] ($101.000000$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 101.000000)$), and [√x] resolves the Euclidean radius $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{10.049876})$—or in a single hardware operation via 10 [ENTER] 1 [↰] [▸θ,r]!

Workbook Problem 12.14: Catenary Sag of Overhead Transmission Line

Suspension cable profile:

$$y(x) = a \cosh\left(\frac{x}{a}\right) - a$$ With scale parameter $a = 150\text{ m}$ over span $x \in [-60, +60\text{ m}]$: At midspan $x = 0$, $y = 0$. At towers $x = \pm 60\text{ m}$, $y = 150(\cosh(0.40) - 1) = 150(1.081072 - 1) = \mathbf{12.16\text{ m}}$ sag.

STEP	KEYSTROKE	REGISTER Y	REGISTER X	DISPLAY
01	60 [ENTER]	60.00	60.00	60.00_
02	150 [÷]	0.00	0.400000	0.400000
03	[↰] [HYP] [COS]	0.00	1.081072	1.081072
04	1 [-]	0.00	0.081072	0.081072
05	150 [×]	0.00	12.160855	12.16 (m)

Workbook Problem 12.15: Gaussian Error Function ($\text{erf}$) Diffusion Profile

In Fickian heat and mass diffusion, concentration across a semi-infinite slab follows the Gaussian error function:

\[\text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} dt\]

The sigmoidal curve transitions smoothly from $-1.0$ at $-\infty$ to $+1.0$ at $+\infty$, passing through $(0, 0)$ with central inflection slope $\frac{2}{\sqrt{\pi}} \approx \mathbf{1.128379}$ ([↱] [π] [√x] 2 [x⇄y] [÷]).

Solving Fick's second law $\frac{\partial C}{\partial t} = D \frac{\partial^2 C}{\partial z^2}$ via Boltzmann's similarity variable $x = \frac{z}{2\sqrt{Dt}}$ yields the error-function profile whose steepest concentration gradient occurs at the interface $x = 0$: evaluating [↱] [π] [√x] ($\sqrt{\pi} \approx 1.772454$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 1.772454)$), staging 2 [x⇄y] $(T: 0.00, Z: 0.00, Y: 2.000000, X: 1.772454)$, and dividing via [÷] establishes the Maclaurin leading-term slope $2/\sqrt{\pi}$ $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{1.128379})$.

Workbook Problem 12.16: Wien Displacement Blackbody Radiation Curve

Planck's spectral radiance for a tungsten filament at $T = 3000\text{ K}$ obeys:

\[B(\lambda, T) \propto \frac{\lambda^{-5}}{e^{hc/(\lambda k_B T)} - 1}, \qquad \lambda_{\max} = \frac{b}{T}\]

The curve displays the steep short-wavelength ultraviolet cutoff (resolving the classical Rayleigh-Jeans catastrophe) and a peak spectral emission at $\lambda_{\max} = \frac{2897.77\,\mu\text{m}\cdot\text{K}}{3000\text{ K}} = \mathbf{0.9659\,\mu\text{m}}$ in the near-infrared.

Differentiating Max Planck's spectral radiance distribution with respect to wavelength yields the transcendental root $x = \frac{hc}{\lambda_{\max} k_B T} \approx 4.965114$, which defines Wilhelm Wien's displacement constant $b = 2897.77\,\mu\text{m}\cdot\text{K}$: dividing 2897.77 [ENTER] $(T: 0.00, Z: 0.00, Y: 2897.770000, X: 2897.770000)$ by the filament temperature 3000 [÷] locates the emission peak in the near-infrared at $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{0.965923}\,\mu\text{m})$.

Workbook Problem 12.17: Jakob Bernoulli's Lemniscate

In polar coordinates, the locus of points whose product of distances from two foci is constant follows:

\[r^2 = 4 \cos(2\theta)\]

This relation renders a symmetric figure-eight ribbon spanning $r \in [0, 2.0]$ over $\theta \in [-45^\circ, +45^\circ]$ and $[135^\circ, 225^\circ]$, while regions where $\cos(2\theta) < 0$ are automatically clipped as imaginary branches by the vector plotter.

Jakob Bernoulli introduced the lemniscate ("decorated with ribbons") in Acta Eruditorum (1694) as the elastic curve of a bent rod and the locus of constant focal-distance product: evaluating at $\theta = 30^\circ$ via 30 [ENTER] 2 [×] ($2\theta = 60.0^\circ$, $(T: 0.00, Z: 0.00, Y: 0.000000, X: 60.000000)$), [COS] ($\cos 60^\circ = 0.500000$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 0.500000)$), and [√x] 2 [×] determines the polar radius $r = 2\sqrt{\cos 60^\circ} = \sqrt{2}$ $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{1.414214})$.

Workbook Problem 12.18: Constant-Pitch Archimedean Spiral

In mechanical scroll compressors and vinyl groove geometry, the polar trajectory expands linearly with rotation angle $\theta$ (in radians):

\[r(\theta) = 0.5\,\theta\]

Each full $2\pi$ revolution increases the radius by a constant pitch $\Delta r = 0.5(2\pi) = \pi \approx \mathbf{3.1416\text{ mm}}$. Verify linear spiral growth across multiple revolutions on the dot-matrix grid.

Unlike a logarithmic spiral whose turn spacing grows geometrically, Archimedes' spiral $r(\theta) = a\theta$ maintains constant radial separation $\Delta r = 2\pi a$ between adjacent involute wraps: evaluating one full revolution [↱] [π] 2 [×] ($2\pi \approx 6.283185\text{ rad}$, $(T: 0.00, Z: 0.00, Y: 0.000000, X: 6.283185)$) and scaling by 0.5 [×] confirms the uniform scroll pitch $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{3.141593}\text{ mm})$.

Workbook Problem 12.19: Supersonic Prandtl-Meyer & Mach Cone Angle Curve

The supersonic shock wave half-angle $\mu(M)$ against Mach number $M \in [1.0, 5.0]$ in Degrees mode (M [1/x] [↰] [ASIN]) follows:

\[\mu(M) = \arcsin\left(\frac{1}{M}\right)\]

The curve starts at a normal sonic wave $\mu(1.0) = 90.0^\circ$, sweeps sharply through $\mu(2.0) = \mathbf{30.0^\circ}$, and narrows to a slender hypersonic cone $\mu(5.0) = \mathbf{11.537^\circ}$. Map compressible flow shock angle decay from sonic boom to hypersonic flight.

When a vehicle exceeds the speed of sound $a$, spherical acoustic wavefronts coalesce along Ernst Mach's conical envelope of half-angle $\sin\mu = a/v = 1/M$: at Mach 2, 2 [1/x] ($0.500000$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 0.500000)$) followed by [↰] [ASIN] yields $\mu(2) = \mathbf{30.000000}^\circ$ $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{30.000000}^\circ)$, while at hypersonic Mach 5, 5 [1/x] ($0.200000$ in $X$ with $30.0^\circ$ preserved in $Y$, $(T: 0.00, Z: 0.00, Y: 30.000000, X: 0.200000)$) followed by [↰] [ASIN] narrows the Mach cone to $(T: 0.00, Z: 0.00, Y: 30.000000, X: \mathbf{11.536959}^\circ)$.

Workbook Problem 12.20: Thin-Airfoil Aerodynamic Lift Slope ($C_L$ vs. $\alpha$)

Classical thin-airfoil theory predicts section lift coefficient (for angle of attack $\alpha$ in radians, or $\frac{2\pi^2}{180}\alpha^\circ \approx 0.10966\,\alpha^\circ$):

\[C_L(\alpha) = 2\pi \alpha\]

Plotting over $\alpha \in [-5^\circ, +15^\circ]$ illustrates the linear inviscid lift regime up to $C_L(10^\circ) \approx \mathbf{1.0966}$ prior to boundary-layer flow separation.

Max Munk and Hermann Glauert's thin-airfoil theory establishes that any symmetric cambered section in inviscid subsonic flow exhibits a universal lift-curve slope of $\frac{dC_L}{d\alpha} = 2\pi\text{ rad}^{-1} = \frac{2\pi^2}{180}\text{ deg}^{-1} \approx 0.109662\text{ deg}^{-1}$: staging $\alpha = 10^\circ$ via 10 [ENTER] $(T: 0.00, Z: 0.00, Y: 10.000000, X: 10.000000)$ and multiplying by 0.10966 [×] predicts the pre-stall section lift coefficient $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{1.096600})$.

Workbook Problem 12.21: Van der Waals Real Gas Isotherms

Real gas pressure across molar volume $V$ follows:

\[P(V) = \frac{R T}{V - b} - \frac{a}{V^2}\]

Isotherms below critical temperature $T_c$ display characteristic van der Waals cubic loops, illustrating liquid-gas phase coexistence.

Johannes Diderik van der Waals corrected the ideal gas law for finite molecular excluded volume ($b = 0.04\text{ m}^3/\text{mol}$) and intermolecular London dispersion attraction ($a = 0.36\text{ Pa}\cdot\text{m}^6/\text{mol}^2$): evaluating kinetic thermal pressure 8.314 [ENTER] 300 [×] ($RT = 2494.20$, $(T: 0.00, Z: 0.00, Y: 0.000, X: 2494.20)$) divided by free molar volume .10 [ENTER] .04 [-] ($V - b = 0.060000$ in $X$ with $2494.20$ in $Y$, $(T: 0.00, Z: 0.00, Y: 2494.20, X: 0.060000)$) via [÷] gives $41570.00\text{ Pa}$ $(T: 0.00, Z: 0.00, Y: 0.00, X: 41570.00)$; computing the cohesive pressure correction .36 [ENTER] .10 [↰] [x²] [÷] ($36.00\text{ Pa}$ in $X$ with $41570.00$ in $Y$, $(T: 0.00, Z: 0.00, Y: 41570.00, X: 36.00)$) and subtracting via [-] yields the net real-gas pressure $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{41534.00}\text{ Pa})$.

Workbook Problem 12.22: Bivariate Scatter Plot: Hooke's Law Spring Calibration

Accumulating four experimental force-displacement pairs $(1\text{ cm}, 4.8\text{ N})$, $(2\text{ cm}, 10.1\text{ N})$, $(3\text{ cm}, 14.9\text{ N})$, and $(4\text{ cm}, 20.2\text{ N})$ via [Σ+] (with no active equations in EQN) and pressing [↱] [PLOT] in App Mode renders the four $2\times 2$ data markers alongside the least-squares regression line of slope $k = \mathbf{5.10\text{ N/cm}}$ ($510\text{ N/m}$) and correlation $r = \mathbf{0.9998}$.

Resetting the statistical accumulators ([↰] [CLEAR] [CLΣ], $(T: 0.00, Z: 0.00, Y: 0.00, X: 0.00)$), streaming the four extension-load pairs (4.8 [ENTER] 1 [Σ+], 10.1 [ENTER] 2 [Σ+], 14.9 [ENTER] 3 [Σ+], 20.2 [ENTER] 4 [Σ+] $\rightarrow n = 4$), and pressing [↱] [L.R.] [m] extracts the least-squares Hookean stiffness $(T: 0.00, Z: 14.90, Y: 20.20, X: \mathbf{5.100000}\text{ N/cm}$, or $510\text{ N/m})$ which the vector plotter superimposes directly across the four scatter markers.

Workbook Problem 12.23: Automotive Kinetic Energy & Braking Distance Parabola

Plotting $E_k(v) = \frac{1}{2} m v^2$ for a $m = 1,200\text{ kg}$ vehicle over speed $v \in [0, 40\text{ m/s}]$ (V [↰] [x²] 600 [×]) visually demonstrates why doubling highway speed from $15\text{ m/s}$ to $30\text{ m/s}$ quadruples kinetic energy from $135\text{ kJ}$ to $\mathbf{540\text{ kJ}}$.

By the work-energy theorem $W = \mu m g d = \frac{1}{2}mv^2$, minimum stopping distance scales quadratically with entry speed: evaluating kinetic energy at $15\text{ m/s}$ (15 [↰] [x²] 600 [×] $\rightarrow E_{k,1} = 135,000\text{ J}$, $(T: 0.00, Z: 0.00, Y: 0.000000, X: 135000.00)$) and at $30\text{ m/s}$ (30 [↰] [x²] 600 [×] $\rightarrow E_{k,2} = 540,000\text{ J}$ in $X$ while $135,000\text{ J}$ lifts to $Y$, $(T: 0.00, Z: 0.000000, Y: 135000.00, X: 540000.00)$), then dividing via [x⇄y] [÷] confirms the $2^2 = 4\times$ quadratic energy multiplier $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{4.000000}\text{x})$.

Workbook Problem 12.24: Relativistic Lorentz Gamma Factor ($\gamma$)

$$\gamma(v) = \frac{1}{\sqrt{1 - (v/c)^2}}$$ Plotted for velocity ratio $\beta = v/c \in [0.0, 0.99]$: Remains near $1.0$ at low velocities, curves gently to $1.15$ at $\beta = 0.5$, and explodes asymptotically toward infinity as $\beta \to 1.0$.

STEP	KEYSTROKE	REGISTER X	DISPLAY
01	.80 [↰] [x²]	0.640000	0.640000
02	1 [x⇄y] [-]	0.360000	0.360000
03	[√x] [1/x]	1.666667	1.666667

Workbook Problem 12.25: Linear Thermal Dilatation Calibration Curve

Plotting $\Delta L(T) = \alpha L_0 (T - T_0)$ for an aluminum metrology bar ($\alpha = 23.1 \times 10^{-6}\text{ K}^{-1}$, $L_0 = 1.0\text{ m}$) over $T \in [273\text{ K}, 373\text{ K}]$ provides a direct visual reference line of slope $23.1\,\mu\text{m/K}$ ($2.31\text{ mm}$ total expansion over $100\text{ K}$). Sweep temperature to plot linear expansion of precision metrology standards.

Across moderate temperature excursions between the ice point ($273\text{ K}$) and steam point ($373\text{ K}$), lattice anharmonicity produces linear thermal expansion: computing the temperature excursion 373 [ENTER] 273 [-] ($\Delta T = 100.0\text{ K}$, $(T: 0.00, Z: 0.00, Y: 0.000000, X: 100.000000)$), multiplying by the linear expansion coefficient 23.1 [E] 6 [+/-] [×] ($\Delta L = 0.002310\text{ m}$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 0.002310)$), and converting to millimeters via 1000 [×] yields the total gauge elongation $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{2.310000}\text{ mm})$.

Workbook Problem 12.26: Stefan-Boltzmann Radiative Heat Emission

Power $P(T) = \varepsilon \sigma A T^4$: Quartic curve demonstrates why radiative heat transfer dominates at high temperatures ($T > 1000\text{ K}$).

Integrating Planck's spectral distribution over all wavelengths and hemisphere solid angles yields the Josef Stefan and Ludwig Boltzmann $T^4$ radiation law: raising billet temperature to the fourth power 1200 [ENTER] 4 [yˣ] ($T^4 = 2.0736 \times 10^{12}\text{ K}^4$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 2.0736\text{E}12)$), multiplying by $\sigma = 5.67037 \times 10^{-8}\text{ W}\cdot\text{m}^{-2}\cdot\text{K}^{-4}$ (5.67037 [E] 8 [+/-] [×] $\rightarrow 117580.8\text{ W/m}^2$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 117580.8)$), and scaling by surface emissivity and area (.85 [×] .50 [×]) determines the radiative heat flux $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{49971.84}\text{ W} \approx 50.0\text{ kW})$—note that 1200 [↰] [x²] [↰] [x²] computes $T^4$ in two single-cycle squarings!

Workbook Problem 12.27: Newton's Law of Cooling Exponential Asymptote

Plotting $T(t) = T_{\text{env}} + (T_0 - T_{\text{env}}) e^{-k t}$ for a quenched steel billet ($T_0 = 800\text{ K}$, $T_{\text{env}} = 300\text{ K}$, $k = 0.05\text{ s}^{-1}$) over $t \in [0, 100\text{ s}]$ (T 0.05 [×] [+/-] [eˣ] 500 [×] 300 [+]) illustrates rapid initial heat rejection followed by asymptotic approach to the $300\text{ K}$ ambient floor.

When the Biot number $\text{Bi} = \frac{h L_c}{k_{\text{solid}}} < 0.1$, internal thermal gradients are negligible and lumped-capacitance quenching decays exponentially with time constant $\tau = 1/k = 20\text{ s}$: evaluating at $t = 20\text{ s}$ (20 [ENTER] 0.05 [×] [+/-] $\rightarrow -kt = -1.000000$, $(T: 0.00, Z: 0.00, Y: 0.000000, X: -1.000000)$), exponentiating via [eˣ] ($e^{-1} \approx 0.367879$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 0.367879)$), scaling by the initial $500\text{ K}$ thermal excess (500 [×] $\rightarrow 183.939721\text{ K}$, $(T: 0.00, Z: 0.00, Y: 0.00, X: 183.939721)$), and adding the $300\text{ K}$ bath temperature (300 [+]) confirms the one-time-constant billet temperature $(T: 0.00, Z: 0.00, Y: 0.00, X: \mathbf{483.939721}\text{ K})$.

Chapter 13: Applied STEM Engineering Case Studies — Extended Engineering & Competition Problems

Accompanies Chapter 13 of Thinking in the Stack (28 Worked Engineering Problems)

Workbook Problem 13.1: Torricelli Efflux Tank Draining Velocity

A cylindrical fluid storage tank maintains an open liquid column height $h = 5.0\text{ meters}$. Standard acceleration of gravity $g = 9.80665\text{ m/s}^2$. Determine the efflux velocity $v$: $$v = \sqrt{2 g h} = \sqrt{2 \times 9.80665 \times 5.0} = \sqrt{98.0665} \approx \mathbf{9.90285\text{ m/s}}$$

Keystrokes: 2 [ENTER] 9.80665 [×] 5 [×] [√x] $\rightarrow$ 9.9028531242.

Evangelista Torricelli's 1643 efflux theorem—a direct specialization of Daniel Bernoulli's streamline energy balance $\rho g h = \frac{1}{2}\rho v^2$—states that liquid jets exit a bottom orifice at the free-fall speed acquired across head height $h$: evaluating 2 [ENTER] 9.80665 [×] 5 [×] ($2gh = 98.0665\text{ m}^2/\text{s}^2$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 98.0665)$) and extracting the square root via [√x] determines the discharge velocity $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{9.9029}\text{ m/s})$.

Workbook Problem 13.2: Darcy-Weisbach Friction Head Loss in Piping

Water at $20^\circ\text{C}$ flows at velocity $v = 2.40\text{ m/s}$ through a commercial steel pipe with length $L = 85.0\text{ m}$ and inner diameter $D = 0.15\text{ m}$. Darcy friction factor $f = 0.0192$: $$h_f = f \frac{L}{D} \frac{v^2}{2g} = 0.0192 \times \frac{85.0}{0.15} \times \frac{2.40^2}{2 \times 9.80665} \approx \mathbf{3.1952\text{ meters}}$$

In Julius Weisbach and Henry Darcy's pipe-friction formulation, viscous boundary-layer shear dissipates mechanical energy in proportion to the dimensionless resistance coefficient $f(L/D)$ times the velocity head $v^2/(2g)$: evaluating .0192 [ENTER] 85 [×] .15 [÷] ($f L/D = 10.8800$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 10.8800)$) and chaining 2.4 [↰] [x²] [×] 2 [÷] 9.80665 [÷] resolves the viscous head loss $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{3.1952}\text{ m})$.

Workbook Problem 13.3: Infill Grid Volume Integration & Spool Consumption

A rectangular drone chassis housing has outer bounding box dimensions $140\text{ mm} \times 90\text{ mm} \times 35\text{ mm}$. The part is sliced with 3 solid perimeter walls ($2.0\text{ mm}$ thick) consuming $85,000\text{ mm}^3$, while the internal cavity ($355,000\text{ mm}^3$) is filled with a $25\%$ gyroid infill density ($\rho_{\text{infill}} = 0.25$). Technical polymer density is $\rho = 1.25\text{ g/cm}^3$ ($0.00125\text{ g/mm}^3$): $$V_{\text{total}} = V_{\text{walls}} + \rho_{\text{infill}} \times V_{\text{cavity}} = 85,000 + (0.25 \times 355,000) = 85,000 + 88,750 = \mathbf{173,750\text{ mm}^3}$$ $$M_{\text{part}} = V_{\text{total}} \times \rho = 173,750 \times 0.00125 \approx \mathbf{217.19\text{ grams}}$$

Triply periodic minimal surfaces such as Alan Schoen's gyroid partition internal core cavities into isotropic structural lattices at a fraction of solid mass: weighting the cavity volume by the infill fraction and adding the solid shell (355000 [ENTER] .25 [×] 85000 [+] $\rightarrow V_{\text{total}} = 173,750\text{ mm}^3$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 173750.0)$) before scaling by .00125 [×] determines the exact spool mass budget $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{217.1875}\text{ g})$.

Workbook Problem 13.4: Ideal Brayton Gas Turbine Cycle Thermal Efficiency

An industrial gas turbine operates on the ideal Brayton cycle with compressor pressure ratio $r_p = 14.5$ and air specific heat ratio $\gamma = 1.40$:

\[\eta_{\text{Brayton}} = 1 - \frac{1}{r_p^{(\gamma - 1)/\gamma}} = 1 - \frac{1}{(14.5)^{(0.40/1.40)}} \approx \mathbf{53.41\%}\]

In George Brayton's open gas-turbine cycle, isentropic compression across pressure ratio $r_p$ raises absolute air temperature by the Poisson factor $T_2/T_1 = r_p^{(\gamma-1)/\gamma}$: evaluating 14.5 [ENTER] .40 [ENTER] 1.40 [÷] [yˣ] ($2.1465$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 2.1465)$) and chaining [1/x] 1 [x⇄y] [-] determines the ideal Brayton thermal efficiency $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{0.5341}$, or $53.41\%)$.

Workbook Problem 13.5: Composite Wall Conduction Heat Flux

A furnace wall consists of $L_1 = 0.15\text{ m}$ firebrick ($k_1 = 1.2\text{ W/m}\cdot\text{K}$) and $L_2 = 0.05\text{ m}$ insulating mineral wool ($k_2 = 0.04\text{ W/m}\cdot\text{K}$). Inner temperature $T_{\text{in}} = 850^\circ\text{C}$, outer temperature $T_{\text{out}} = 50^\circ\text{C}$: $$R_{\text{total}} = \frac{L_1}{k_1} + \frac{L_2}{k_2} = \frac{0.15}{1.2} + \frac{0.05}{0.04} = 0.125 + 1.250 = \mathbf{1.375\text{ m}^2\text{K/W}}$$ $$q'' = \frac{\Delta T}{R_{\text{total}}} = \frac{850 - 50}{1.375} = \frac{800}{1.375} \approx \mathbf{581.82\text{ W/m}^2}$$

By Jean-Baptiste Joseph Fourier's law of heat conduction across series planar layers, thermal resistances $R_i = L_i/k_i$ add directly like series electrical resistors: summing the firebrick and mineral-wool resistances via .15 [ENTER] 1.2 [÷] .05 [ENTER] .04 [÷] [+] ($R_{\text{total}} = 1.3750\text{ m}^2\text{K/W}$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.3750)$) and dividing into the overall thermal drop 850 [ENTER] 50 [-] [x⇄y] [÷] determines the steady-state heat flux $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{581.8182}\text{ W/m}^2)$.

Workbook Problem 13.6: Rankine Steam Power Cycle Thermal Efficiency

In a subcritical Rankine steam cycle, high-pressure steam enters the turbine at enthalpy $h_1 = 3,450.0\text{ kJ/kg}$ and expands to condenser pressure with enthalpy $h_2 = 2,180.0\text{ kJ/kg}$. The feedwater pump consumes work $w_p = 15.0\text{ kJ/kg}$ to pressurize liquid from $h_3 = 190.0\text{ kJ/kg}$ to boiler inlet state $h_4 = 205.0\text{ kJ/kg}$. 1. Net turbine work output: $$w_{\text{net}} = (h_1 - h_2) - w_p = (3450.0 - 2180.0) - 15.0 = 1270.0 - 15.0 = \mathbf{1,255.0\text{ kJ/kg}}$$ 2. Heat input supplied by the boiler: $$q_{\text{in}} = h_1 - h_4 = 3450.0 - 205.0 = \mathbf{3,245.0\text{ kJ/kg}}$$ 3. Thermal efficiency: $$\eta_{\text{Rankine}} = \frac{w_{\text{net}}}{q_{\text{in}}} = \frac{1255.0}{3245.0} \approx \mathbf{38.67\%}$$

Evaluating net shaft work 3450 [ENTER] 2180 [-] 15 [-] ($w_{\text{net}} = 1,255.0\text{ kJ/kg}$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1255.0000)$), computing boiler enthalpy addition 3450 [ENTER] 205 [-] ($q_{\text{in}} = 3,245.0\text{ kJ/kg}$ in $X$ while $w_{\text{net}}$ waits in $Y$, $(T: 0.0000, Z: 0.0000, Y: 1255.0000, X: 3245.0000)$), and dividing via [÷] 100 [×] resolves the Rankine cycle efficiency $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{38.6749}\%)$—or by duplicating throttle enthalpy $h_1 = 3450$ upfront (3450 [ENTER] [ENTER] 2180 [-] 15 [-] [x⇄y] 205 [-] [÷] 100 [×]), 3450 is entered only once!

Workbook Problem 13.7: Otto Cycle Internal Combustion Efficiency vs. Compression Ratio

The theoretical thermal efficiency of an internal combustion Otto cycle engine depends solely on compression ratio $r = V_{\max} / V_{\min}$ and air specific heat ratio $\gamma = 1.40$: $$\eta_{\text{Otto}} = 1 - \frac{1}{r^{\gamma - 1}}$$ For a high-performance racing engine with compression ratio $r = 11.5$: $$\eta_{\text{Otto}} = 1 - \frac{1}{11.5^{0.40}} = 1 - \frac{1}{2.65839} = 1 - 0.376167 = \mathbf{62.35\%}$$

In Nikolaus Otto's four-stroke spark-ignition cycle, isentropic compression across volume ratio $r = 11.5$ elevates the cycle temperature ratio to $r^{\gamma-1}$: keying 11.5 [ENTER] .40 [yˣ] ($11.5^{0.40} \approx 2.6563$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 2.6563)$) and chaining [1/x] 1 [x⇄y] [-] determines the air-standard Otto efficiency $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{0.6235}$, or $62.35\%)$.

Workbook Problem 13.8: 2nd-Order Butterworth Crossover Filters

Crossover frequency $f_c = 2500\text{ Hz}$, driver nominal impedance $R = 8.0\,\Omega$:

\[L = \frac{R \sqrt{2}}{2\pi f_c} = \frac{8 \sqrt{2}}{2\pi \times 2500} \approx \mathbf{0.720253\text{ mH}}$$ $$C = \frac{1}{2\pi \sqrt{2} R f_c} = \frac{1}{2\pi \sqrt{2} \times 8 \times 2500} \approx \mathbf{5.62698\text{ }\mu\text{F}}\]

Stephen Butterworth's maximally flat second-order crossover ($Q = 1/\sqrt{2} \approx 0.7071$) requires reactive impedance $X_L = \sqrt{2}R$ and capacitive reactance $X_C = R/\sqrt{2}$ at the crossover frequency $f_c$: evaluating 8 [ENTER] 2 [√x] [×] 2 [÷] [↱] [π] [÷] 2500 [÷] determines the series woofer inductance $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{7.2025 \times 10^{-4}}\text{ H}$, or $0.7203\text{ mH})$, and evaluating 2 [↱] [π] [×] 2 [√x] [×] 8 [×] 2500 [×] [1/x] determines the shunt tweeter capacitance $(T: 0.0000, Z: 0.0000, Y: 7.2025 \times 10^{-4}, X: \mathbf{5.6270 \times 10^{-6}}\text{ F}$, or $5.63\,\mu\text{F})$—or even faster, since $C = \frac{L}{2 R^2} = \frac{L}{128}$, simply dividing $L$ by 128 [÷] directly on the stack yields $C = \mathbf{5.6270\,\mu\text{F}}$ in a single division!

Workbook Problem 13.9: 4th-Order Linkwitz-Riley Crossover ($24\text{ dB/octave}$)

For $f_c = 2500\text{ Hz}, R = 8.0\,\Omega$:

\[L_{LR4} = \frac{R}{\pi f_c} = \frac{8.0}{\pi \times 2500} \approx \mathbf{1.01859\text{ mH}}$$ $$C_{LR4} = \frac{1}{4\pi R f_c} = \frac{1}{4\pi \times 8.0 \times 2500} \approx \mathbf{3.97887\text{ }\mu\text{F}}\]

Siegfried Linkwitz and Russ Riley demonstrated in 1976 that cascading two Butterworth filters produces $-6\text{ dB}$ acoustic amplitude and zero phase difference at $f_c$, summing to a flat unity acoustic wavefront: keying 8.0 [↱] [π] [÷] 2500 [÷] resolves the LR4 series inductance $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{1.0186 \times 10^{-3}}\text{ H}$, or $1.019\text{ mH})$, and keying 4 [↱] [π] [×] 8 [×] 2500 [×] [1/x] (or dividing $L_{LR4}$ by $4R^2 = 256\,\Omega^2$ via 256 [÷]!) resolves the shunt capacitance $(T: 0.0000, Z: 0.0000, Y: 0.0010, X: \mathbf{3.9789 \times 10^{-6}}\text{ F}$, or $3.98\,\mu\text{F})$.

Workbook Problem 13.10: Complex AC Impedance Addition via [↰] [CMPLX]

Branch 1 impedance $Z_1 = 8 + 4i\,\Omega$ ($Y = 8, X = 4$).
Branch 2 impedance $Z_2 = 4 - 8i\,\Omega$ ($Y = 4, X = -8$).
Total series impedance: $Z_{\text{total}} = (8 + 4i) + (4 - 8i) = \mathbf{12 - 4i\,\Omega}$.

In Charles Proteus Steinmetz's AC phasor calculus, series impedances combine vectorially across real resistance and imaginary reactance: staging $Z_2 = 4 - 8i$ and $Z_1 = 8 + 4i$ across the four registers via 4 [ENTER] 8 [+/-] [ENTER] 8 [ENTER] 4 $(T: 4.0000, Z: -8.0000, Y: 8.0000, X: 4.0000)$ and executing [↰] [CMPLX] [+] sums the real parts $(T + Y)$ into $Y$ and imaginary parts $(Z + X)$ into $X$ in a single cycle $(T: 4.0000, Z: 4.0000, Y: \mathbf{12.0000}, X: \mathbf{-4.0000})$.

Workbook Problem 13.11: Zobel Network Dynamic Impedance Equalizer

Dynamic loudspeaker voice coils exhibit rising inductive impedance at high frequencies ($Z(f) = R_e + j\omega L_e$). A Zobel network placed in parallel with the driver flattens the impedance back to pure resistance $R_e$: $$R_z = 1.25 \times R_e, \quad C_z = \frac{L_e}{R_z^2}$$ For a woofer with DC voice coil resistance $R_e = 6.4\,\Omega$ and voice coil inductance $L_e = 0.85\text{ mH}$ ($0.00085\text{ H}$): 1. Zobel resistor: $R_z = 1.25 \times 6.4 = \mathbf{8.00\,\Omega}$. 2. Zobel capacitor: $$C_z = \frac{0.00085}{8.0^2} = \frac{0.00085}{64.0} \approx \mathbf{13.281\text{ }\mu\text{F}}$$

Otto Zobel's constant-resistance Boucherot cell neutralizes voice-coil inductive rise so passive crossovers see a resistive termination: computing 6.4 [ENTER] 1.25 [×] establishes the damping resistor $R_z$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{8.0000}\,\Omega)$, squaring in place via [↰] [x²] ($R_z^2 = 64.0\,\Omega^2$), and chaining .00085 [x⇄y] [÷] 1 [E] 6 [×] determines the conjugate Zobel capacitance $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{13.2813}\,\mu\text{F})$.

Workbook Problem 13.12: Snell's Law & Total Internal Reflection Critical Angle

Light propagates from optical glass ($n_1 = 1.52$) into air ($n_2 = 1.00$). Determine the critical angle $\theta_c$:

\[\theta_c = \arcsin\left(\frac{n_2}{n_1}\right) = \arcsin\left(\frac{1.00}{1.52}\right) = \arcsin(0.657895) \approx \mathbf{41.14^\circ}\]

By Willebrord Snellius's refraction law $n_1 \sin\theta_1 = n_2 \sin\theta_2$, total internal reflection commences when the transmitted ray grazes the interface at $\theta_2 = 90^\circ$ ($\sin\theta_2 = 1$): dividing the refractive indices 1 [ENTER] 1.52 [÷] ($\sin\theta_c \approx 0.6579$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.6579)$) and pressing [↰] [ASIN] in Degrees mode determines the critical angle $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{41.1395}^\circ)$—or in just two operations via 1.52 [1/x] [↰] [ASIN]!

Workbook Problem 13.13: Numerical Aperture & Light Acceptance Angle of Optical Fiber

A multi-mode step-index silica optical fiber has core refractive index $n_{\text{core}} = 1.482$ and cladding refractive index $n_{\text{clad}} = 1.460$: 1. Numerical Aperture ($NA$): $$NA = \sqrt{n_{\text{core}}^2 - n_{\text{clad}}^2} = \sqrt{1.482^2 - 1.460^2} = \sqrt{2.196324 - 2.131600} = \sqrt{0.064724} \approx \mathbf{0.2544}$$ 2. Maximum acceptance half-angle in air ($\theta_{\max}$): $$\theta_{\max} = \arcsin(NA) = \arcsin(0.254409) \approx \mathbf{14.74^\circ}$$

Ernst Abbe's numerical aperture $NA = \sqrt{n_{\text{core}}^2 - n_{\text{clad}}^2}$ governs both the light-gathering cone and modal dispersion of a step-index waveguide: evaluating 1.482 [↰] [x²] 1.460 [↰] [x²] [-] [√x] determines $NA = \mathbf{0.2544}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{0.2544})$, and pressing [↰] [ASIN] in Degrees mode resolves the acceptance half-cone $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{14.7389}^\circ)$.

Workbook Problem 13.14: Thick Lens Equivalent Focal Length (Gullstrand Formula)

A thick biconvex crown glass lens has front radius $R_1 = 50.0\text{ mm}$, rear radius $R_2 = -80.0\text{ mm}$, center thickness $d = 12.0\text{ mm}$, and refractive index $n = 1.520$. Gullstrand's lens maker formula states: $$\frac{1}{f} = (n - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} + \frac{(n - 1) d}{n R_1 R_2} \right]$$ 1. Front surface power: $(n - 1) / R_1 = 0.520 / 50 = 0.010400\text{ mm}^{-1}$. 2. Rear surface power: $-(n - 1) / R_2 = -0.520 / (-80) = 0.006500\text{ mm}^{-1}$. 3. Thickness coupling term: $$\frac{(n - 1)^2 d}{n R_1 R_2} = \frac{0.520^2 \times 12.0}{1.520 \times 50.0 \times (-80.0)} = \frac{3.2448}{-6080.0} = -0.00053368\text{ mm}^{-1}$$ 4. Total optical power: $P = 0.010400 + 0.006500 - 0.00053368 = 0.01636632\text{ mm}^{-1}$. 5. Equivalent focal length: $f = 1 / P \approx \mathbf{61.10\text{ mm}}$.

Allvar Gullstrand's paraxial thick-lens equation accounts for principal-plane separation across axial thickness $d = 12.0\text{ mm}$: summing the two surface dioptric powers .520 [ENTER] 50 [÷] .520 [ENTER] 80 [÷] [+] ($0.016900\text{ mm}^{-1}$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0169)$), deducting the thickness coupling term .520 [↰] [x²] 12 [×] 1.52 [ENTER] 50 [×] 80 [×] [÷] [-] ($P = 0.016366\text{ mm}^{-1}$, $(T: 0.0169, Z: 0.0169, Y: 0.0169, X: 0.0164)$), and inverting via [1/x] yields the equivalent focal length $(T: 0.0169, Z: 0.0169, Y: 0.0169, X: \mathbf{61.1011}\text{ mm})$—and by noting that $\frac{1}{50} + \frac{1}{80} = (50 \parallel 80)^{-1}$, you can evaluate the bracketed geometry via 50 [ENTER] 80 [x∥y] [1/x]!

Workbook Problem 13.15: Hohmann Transfer Orbit $\Delta v$ (LEO to GEO)

Earth gravitational parameter $\mu = 3.986004 \times 10^{14}\text{ m}^3/\text{s}^2$.
Initial LEO radius $r_1 = 6.678 \times 10^6\text{ m}$ ($h = 300\text{ km}$).
Target GEO radius $r_2 = 4.2164 \times 10^7\text{ m}$ ($h = 35,786\text{ km}$).
Transfer semi-major axis $a = (r_1 + r_2)/2 = 2.4421 \times 10^7\text{ m}$. 1. Initial circular velocity: $v_{c1} = \sqrt{\mu / r_1} = \mathbf{7,725.82\text{ m/s}}$. 2. Transfer ellipse insertion: $v_{tx1} = \sqrt{\mu(2/r_1 - 1/a)} = \mathbf{10,151.62\text{ m/s}}$. 3. Periapsis impulse: $\Delta v_1 = 10,151.62 - 7,725.82 = \mathbf{2,425.80\text{ m/s}}$. 4. Apoapsis circularization: $\Delta v_2 = 3,074.66 - 1,609.91 = \mathbf{1,464.75\text{ m/s}}$. 5. Total mission $\Delta v$: $\Delta v_{\text{total}} = 2,425.80 + 1,464.75 = \mathbf{3,890.55\text{ m/s}}$ ($3.891\text{ km/s}$).

Keystrokes (Delta-V Impulses & Total Mission Budget): 10151.62 [ENTER] 7725.82 [-] 3074.66 [ENTER] 1609.91 [-] [+] $\rightarrow$ 3890.550000

STEP	KEYSTROKE	REGISTER Z	REGISTER Y	REGISTER X	DISPLAY
01	10151.62 [ENTER]	0.00	10151.620000	10151.620000	10151.620000_
02	7725.82 [-]	0.00	0.00	2425.800000	2425.800000 (Δv₁)
03	3074.66 [ENTER]	2425.800000	3074.660000	3074.660000	3074.660000_
04	1609.91 [-]	0.00	2425.800000	1464.750000	1464.750000 (Δv₂)
05	[+]	0.00	0.00	3890.550000	3890.550000 (Δv_tot)

Workbook Problem 13.16: Rocket Motor Burnout Velocity via Romberg Quadrature

Integrate thrust acceleration $a(t) = 4t^2 - 0.5t^3$ from $t = 0$ to $t = 6.0\text{ seconds}$:

$$\Delta v = \int_0^6 (4t^2 - 0.5t^3) dt = \left[ \frac{4}{3}t^3 - \frac{1}{8}t^4 \right]_0^6 = \frac{4}{3}(216) - \frac{1}{8}(1296) = 288 - 162 = \mathbf{126.0\text{ m/s}}$$ Executed in [↱] [EQN] via [↱] [∫]: engine converges to exactly 126.0000 m/s.

When solid-propellant grain regression causes thrust acceleration $a(t) = 4t^2 - 0.5t^3$ to vary non-linearly over the burn duration, integrating $a(t)$ in [↱] [EQN] (4 T [↰] [x²] [×] 0.5 T 3 [yˣ] [×] [-]) across the burn window 0 [ENTER] 6 $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 6.0000)$ via [↱] [∫] T applies Richardson extrapolation to resolve the exact burnout velocity increment $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{126.0000}\text{ m/s})$.

Workbook Problem 13.17: Gravitational Escape Velocity from Earth and Mars

Planetary surface escape velocity is given by $v_{\text{esc}} = \sqrt{\frac{2\mu}{r}}$: 1. Earth ($\mu = 3.986004 \times 10^{14}\text{ m}^3/\text{s}^2, r = 6.378 \times 10^6\text{ m}$): $$v_{\text{esc},\oplus} = \sqrt{\frac{2 \times 3.986004 \times 10^{14}}{6.378 \times 10^6}} \approx \mathbf{11,180.00\text{ m/s}} \quad (\mathbf{11.18\text{ km/s}})$$ 2. Mars ($\mu_{\text{Mars}} = 4.282837 \times 10^{13}\text{ m}^3/\text{s}^2, r_{\text{Mars}} = 3.3895 \times 10^6\text{ m}$): $$v_{\text{esc,Mars}} = \sqrt{\frac{2 \times 4.282837 \times 10^{13}}{3.3895 \times 10^6}} \approx \mathbf{5,026.91\text{ m/s}} \quad (\mathbf{5.03\text{ km/s}})$$

Setting total specific mechanical energy $\mathcal{E} = \frac{1}{2}v^2 - \frac{\mu}{r} = 0$ on a parabolic escape trajectory yields $v_{\text{esc}} = \sqrt{2\mu/r} = \sqrt{2}\,v_{\text{circ}}$: evaluating 2 [ENTER] 3.986004 [E] 14 [×] 6.378 [E] 6 [÷] ($2\mu/r \approx 1.2500 \times 10^8\text{ m}^2/\text{s}^2$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.2500 \times 10^8)$) and extracting the square root via [√x] determines Earth's surface escape velocity $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: \mathbf{11180.00}\text{ m/s}$, or $11.18\text{ km/s})$.

Workbook Problem 13.18: Meridian Convergence & True vs. Magnetic Azimuth

A geodetic survey line in a mid-latitude UTM zone has a measured magnetic azimuth $\alpha_{\text{mag}} = 142.35^\circ$. The local magnetic declination is $\delta = 4.12^\circ\text{ E}$ (East declination is added), and grid convergence angle is $\gamma = -1.25^\circ$: $$\alpha_{\text{true}} = \alpha_{\text{mag}} + \delta = 142.35^\circ + 4.12^\circ = \mathbf{146.47^\circ}$$ $$\alpha_{\text{grid}} = \alpha_{\text{true}} - \gamma = 146.47^\circ - (-1.25^\circ) = \mathbf{147.72^\circ}$$

Reducing a field compass bearing to a State Plane Lambert conformal conic grid azimuth requires two sequential angular rotations: staging the magnetic azimuth 142.35 [ENTER] $(T: 0.0000, Z: 0.0000, Y: 142.3500, X: 142.3500)$, adding easterly geomagnetic declination 4.12 [+] to obtain geodetic true north azimuth $\alpha_{\text{true}} = \mathbf{146.47^\circ}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 146.4700)$, and subtracting meridian convergence 1.25 [+/-] [-] to establish grid azimuth $\alpha_{\text{grid}} = \mathbf{147.72^\circ}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 147.7200)$.

Workbook Problem 13.19: Bowditch Compass Rule Linear Misclosure Balancing

A field traverse leg has raw measured length $L_1 = 250.0\text{ ft}$ along a total perimeter $L_{\text{total}} = 1,250.0\text{ ft}$. The total survey departure misclosure is $E_x = +0.15\text{ ft}$. Calculate the Bowditch proportional departure correction $C_{x1}$: $$C_{x1} = - E_x \times \left( \frac{L_1}{L_{\text{total}}} \right) = -0.15 \times \left( \frac{250.0}{1250.0} \right) = -0.15 \times 0.20 = \mathbf{-0.030\text{ ft}}$$

Nathaniel Bowditch's 1807 Compass Rule distributes linear traverse misclosure in proportion to each leg's fraction of total perimeter length: negating the departure misclosure .15 [+/-] $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: -0.1500)$, weighting by leg length 250 [×] ($-37.50\text{ ft}^2$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: -37.5000)$), and dividing by total traverse perimeter 1250 [÷] determines the leg departure adjustment $C_{x1} = \mathbf{-0.030\text{ ft}}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: -0.0300)$.

Workbook Problem 13.20: Uniform Distributed Load Maximum Bending Moment

A simply supported floor girder spans $L = 8.0\text{ meters}$ supporting a combined dead and live distributed load $w = 24.5\text{ kN/m}$ ($24,500\text{ N/m}$). Calculate the maximum bending moment $M_{\max}$ at midspan: $$M_{\max} = \frac{w L^2}{8} = \frac{24500 \times 8.0^2}{8} = 24500 \times 8 = \mathbf{196,000\text{ N}\cdot\text{m}} \quad (\mathbf{196.0\text{ kN}\cdot\text{m}})$$

Integrating the linear shear diagram $V(x) = w(L/2 - x)$ from the support to midspan $x = L/2$ yields the parabolic bending-moment maximum $M_{\max} = \frac{wL^2}{8}$: staging the distributed load 24500 [ENTER] $(T: 0.0000, Z: 0.0000, Y: 24500.0000, X: 24500.0000)$, multiplying by squared span 8 [↰] [x²] [×] ($1,568,000.0\text{ N}\cdot\text{m}$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1568000.0000)$), and dividing by 8 [÷] resolves the peak midspan moment $M_{\max} = \mathbf{196,000\text{ N}\cdot\text{m}}$ ($196.0\text{ kN}\cdot\text{m}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 196000.0000)$.

Workbook Problem 13.21: Cantilever Beam Maximum Tip Deflection Under Uniform Load

A balcony cantilever beam of length $L = 2.5\text{ m}$ carries uniform distributed load $w = 3,200\text{ N/m}$. The beam is structural steel with $E = 200.0\text{ GPa}$ ($200 \times 10^9\text{ Pa}$) and second moment of area $I = 1.65 \times 10^{-5}\text{ m}^4$. Maximum tip deflection is: $$\delta_{\max} = \frac{w L^4}{8 E I} = \frac{3200 \times 2.5^4}{8 \times (200 \times 10^9) \times (1.65 \times 10^{-5})} = \frac{3200 \times 39.0625}{26,400,000} \approx \mathbf{0.004735\text{ m}} \quad (\mathbf{4.74\text{ mm}})$$

Because cantilever tip deflection under uniform loading scales with the fourth power of overhang length ($L^4$), small span increases produce dramatic compliance penalties: raising the span 2.5 [ENTER] 4 [yˣ] ($39.0625\text{ m}^4$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 39.0625)$), scaling by 3200 [×] ($125,000.0\text{ N}\cdot\text{m}^3$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 125000.0000)$), and dividing sequentially by 8 [÷] 200 [E] 9 [÷] ($7.8125 \times 10^{-11}$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 7.8125 \times 10^{-11})$) and 1.65 [E] 5 [+/-] [÷] determines the free-end deflection $\delta_{\max} = \mathbf{0.004735\text{ m}}$ ($4.74\text{ mm}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0047)$—note that 2.5 [↰] [x²] [↰] [x²] computes $2.5^4$ in two single-cycle squarings!

Workbook Problem 13.22: Standard Gibbs Free Energy & Equilibrium Constant

For the chemical synthesis of ammonia ($\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$) at temperature $T = 298.15\text{ K}$, the thermodynamic equilibrium constant is $K_p = 5.8 \times 10^5$. Calculate the standard Gibbs free energy change $\Delta G^\circ$: $$\Delta G^\circ = - R T \ln K_p = - 8.314462 \times 298.15 \times \ln(5.8 \times 10^5) \approx \mathbf{-32,897.4\text{ J/mol}} \quad (\mathbf{-32.90\text{ kJ/mol}})$$

J. Willard Gibbs's chemical-potential isotherm $\Delta G^\circ = -RT\ln K_p$ links macroscopic reaction equilibrium to standard molar free energy: taking the natural logarithm 5.8 [E] 5 [LN] ($13.2708$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 13.2708)$), scaling by thermal energy 8.314462 [×] 298.15 [×] ($32,897.70\text{ J/mol}$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 32897.7010)$), and negating/converting to kilojoules via [+/-] 1000 [÷] confirms a strongly exergonic Haber-Bosch synthesis at $298.15\text{ K}$: $\Delta G^\circ = \mathbf{-32.90\text{ kJ/mol}}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: -32.8977)$.

Workbook Problem 13.23: Nernst Electrochemical Cell Potential

A galvanic Daniell cell ($\text{Zn} + \text{Cu}^{2+} \rightleftharpoons \text{Zn}^{2+} + \text{Cu}$) has standard cell potential $E^\circ = 1.100\text{ V}$ with $n = 2\text{ electrons transferred}$ at $T = 298.15\text{ K}$. Faraday constant $F = 96,485.33\text{ C/mol}$. The electrolyte concentrations are $[\text{Zn}^{2+}] = 0.50\text{ M}$ and $[\text{Cu}^{2+}] = 0.010\text{ M}$ (reaction quotient $Q = 0.50 / 0.010 = 50.0$): $$E = E^\circ - \frac{R T}{n F} \ln Q = 1.100 - \frac{8.314462 \times 298.15}{2 \times 96485.33} \ln(50.0) = 1.100 - (0.012845 \times 3.912023) \approx \mathbf{1.0497\text{ V}}$$

Walther Nernst's electromotive equation quantifies how product ion accumulation ($Q = [\text{Zn}^{2+}]/[\text{Cu}^{2+}] = 50.0$) depresses galvanic cell voltage below the standard potential $E^\circ$: evaluating the molar free-energy shift 50 [LN] 8.314462 [×] 298.15 [×] ($RT\ln Q = 9,697.74\text{ J/mol}$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 9697.7358)$), dividing by charge transfer $nF$ via 2 [÷] 96485.3 [÷] ($\Delta E = 0.050255\text{ V}$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0503)$), and subtracting from $E^\circ$ via 1.100 [x⇄y] [-] determines the terminal cell electromotive force $E = \mathbf{1.0497\text{ V}}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.0497)$.

Workbook Problem 13.24: Carbon-14 Archeological Radiocarbon Dating

Living plant tissue maintains a steady Carbon-14 activity of $A_0 = 15.3\text{ disintegrations/min}\cdot\text{g of carbon}$. An ancient timber specimen excavated from a ceremonial structure exhibits an activity of $A(t) = 9.4\text{ dpm/g}$. Using the true physical Cambridge $^{14}\text{C}$ half-life $t_{1/2} = 5,730\text{ years}$ ($\lambda = \ln 2 / 5730 \approx 0.00012097\text{ yr}^{-1}$; note that uncalibrated archaeological laboratory reports use the conventional $5,568\text{-year}$ Libby half-life prior to IntCal20 calibration): $$t = - \frac{1}{\lambda} \ln\left(\frac{A(t)}{A_0}\right) = - \frac{5730}{\ln 2} \ln\left(\frac{9.4}{15.3}\right) = -8266.64 \times (-0.487144) \approx \mathbf{4,027.0\text{ years}}$$

Willard Libby's radiocarbon chronometer measures the time elapsed since an organism ceased exchanging atmospheric $^{14}\text{CO}_2$: evaluating the logarithmic activity depletion 9.4 [ENTER] 15.3 [÷] [LN] ($\ln(9.4/15.3) = -0.487144$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: -0.4871)$), scaling by the mean lifetime $\tau = t_{1/2}/\ln 2$ via 5730 [×] 2 [LN] [÷] ($-4,027.014\text{ yr}$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: -4027.0141)$), and reversing sign with [+/-] dates the timber specimen to $t = \mathbf{4,027.0\text{ years}}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 4027.0141)$.

Workbook Problem 13.25: Gamma Ray Shielding & Lead Half-Value Layer (HVL)

Cobalt-60 emits energetic gamma photons ($1.17\text{ MeV}$ and $1.33\text{ MeV}$) with linear attenuation coefficient $\mu = 0.650\text{ cm}^{-1}$ in lead shielding. 1. Half-Value Layer thickness: $$\text{HVL} = \frac{\ln 2}{\mu} = \frac{0.693147}{0.650} \approx \mathbf{1.0664\text{ cm}}$$ 2. Required thickness $x$ to reduce radiation intensity by a factor of 100 ($I / I_0 = 0.01$): $$x = - \frac{\ln(0.01)}{\mu} = \frac{4.605170}{0.650} \approx \mathbf{7.0849\text{ cm}}$$

By the Beer-Lambert exponential attenuation law $I(x) = I_0 e^{-\mu x}$, attenuating a narrow gamma beam to $1\%$ transmission ($I/I_0 = 0.01$, or two Tenth-Value Layers) requires optical depth $\mu x = -\ln(0.01) = 4.605170$: keying .01 [LN] $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: -4.6052)$, dividing by the linear attenuation coefficient .650 [÷] $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: -7.0849)$, and pressing [+/-] specifies a lead shield thickness of $x = \mathbf{7.0849\text{ cm}}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 7.0849)$.

Workbook Problem 13.26: Decibel Addition of Incoherent Environmental Noise Sources

An industrial shop floor operates three loud machines with sound pressure levels: Machine 1: $L_1 = 88.0\text{ dBA}$, Machine 2: $L_2 = 91.0\text{ dBA}$, Machine 3: $L_3 = 85.0\text{ dBA}$. Determine the combined acoustic noise level $L_{\Sigma}$: $$L_{\Sigma} = 10 \log_{10} \left( 10^{88/10} + 10^{91/10} + 10^{85/10} \right) = 10 \log_{10}(6.3096 \times 10^8 + 1.2589 \times 10^9 + 3.1623 \times 10^8) \approx \mathbf{93.43\text{ dBA}}$$

Because uncorrelated machinery noise sources sum in acoustic intensity ($W/\text{m}^2$) rather than decibels, antilogging each bel level ($L_i/10$) via 8.8 [↰] [10ˣ] 9.1 [↰] [10ˣ] [+] ($1.8899 \times 10^9$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.8899 \times 10^9)$) plus 8.5 [↰] [10ˣ] [+] ($2.2061 \times 10^9$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 2.2061 \times 10^9)$) and re-compressing to decibels via [↰] [LOG] 10 [×] determines the composite shop-floor exposure $L_{\Sigma} = \mathbf{93.44\text{ dBA}}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 93.4363)$.

Workbook Problem 13.27: Inverse Square Law Acoustic Attenuation with Distance

An outdoor emergency warning siren emits sound level $L_1 = 105.0\text{ dB}$ measured at distance $r_1 = 10.0\text{ meters}$. Assuming spherical propagation without ground reflections, determine sound pressure level $L_2$ at distance $r_2 = 120.0\text{ meters}$: $$L_2 = L_1 - 20 \log_{10}\left(\frac{r_2}{r_1}\right) = 105.0 - 20 \log_{10}\left(\frac{120.0}{10.0}\right) = 105.0 - 20 \log_{10}(12.0) = 105.0 - 21.5836 \approx \mathbf{83.42\text{ dB}}$$

In free-field spherical spreading, acoustic intensity falls off as $1/r^2$ ($-6.02\text{ dB}$ per distance doubling): evaluating the divergence loss 120 [ENTER] 10 [÷] [↰] [LOG] 20 [×] ($21.5836\text{ dB}$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 21.5836)$) and deducting it from the reference level via 105 [x⇄y] [-] determines the far-field sound pressure level $L_2 = \mathbf{83.42\text{ dB}}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 83.4164)$.

Workbook Problem 13.28: Cross-Discipline Isomorphism of the Parallel Key (`[x∥y]`)

Because the SC-32 elevates $x \parallel y = \frac{1}{1/x + 1/y} = \frac{xy}{x+y}$ to a primary function key, a practicing engineer solves five distinct physical domains with the exact same 3-keystroke gesture: 1. Parallel Resistors ($R_1 \parallel R_2$): 100 [ENTER] 250 [x∥y] $\rightarrow$ $71.43\text{ }\Omega$. 2. Series Capacitors ($C_1 = 47\text{ }\mu\text{F}$ in series with $C_2 = 22\text{ }\mu\text{F}$): Keystrokes: 47 [ENTER] 22 [x∥y] $\rightarrow$ 14.99 μF. 3. Series Mechanical Springs ($k_1$ in series with $k_2$): $k_{\text{eq}} = k_1 \parallel k_2$. 4. Parallel Thermal Spreading Paths ($\theta_{JA,1} \parallel \theta_{JA,2}$): $\theta_{\text{eq}} = \theta_1 \parallel \theta_2$. 5. Gaussian Thin Lens Focal Length ($\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$): $f = d_o \parallel d_i$.

Staging $C_1 = 47.0\text{ }\mu\text{F}$ via 47 [ENTER] $(T: 0.0000, Z: 0.0000, Y: 47.0000, X: 47.0000)$ and combining with $C_2 = 22.0\text{ }\mu\text{F}$ via 22 [x∥y] executes the harmonic product-over-sum operation $\frac{C_1 C_2}{C_1 + C_2}$ in a single hardware cycle to resolve the series equivalent capacitance $C_{\text{eq}} = \mathbf{14.99\text{ }\mu\text{F}}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 14.9855)$.

Chapter 14: High-Speed Equation Entry & Inside-Out Calculation Mastery — Extended Engineering & Competition Problems

Accompanies Chapter 14 of Thinking in the Stack (40 Worked Engineering Problems)

Workbook Problem 14.1: Hypotenuse of Right Triangle

Legs $a = 28.4\text{ cm}, b = 37.6\text{ cm}$. Find hypotenuse $c = \sqrt{a^2 + b^2}$.

Keystrokes: 28.4 [↰] [x²] 37.6 [↰] [x²] [+] [√x] $\rightarrow$ 47.120271646 (3-sig-fig target: 47.1 (Full: 47.120272)).

Squaring the first leg in place via 28.4 [↰] [x²] ($806.56\text{ cm}^2$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 806.5600)$), squaring the second leg 37.6 [↰] [x²] ($1413.76\text{ cm}^2$ in $X$ with $a^2$ auto-lifted to $Y$, $(T: 0.0000, Z: 0.0000, Y: 806.5600, X: 1413.7600)$), and chaining [+] [√x] resolves the Pythagorean hypotenuse $c = \mathbf{47.1203\text{ cm}}$ ($47.1\text{ cm}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 47.1203)$—or in 3 tokens via 28.4 [ENTER] 37.6 [↰] [▸θ,r]!

Workbook Problem 14.2: Sphere Volume

Radius $r = 6.45\text{ cm}$. Find $V = \frac{4}{3} \pi r^3$.

Keystrokes: 6.45 [ENTER] 3 [yˣ] [↱] [π] [×] 4 [×] 3 [÷] $\rightarrow$ 1,124.003732 (3-sig-fig target: 1.12E3 ($1,124.01\text{ cm}^3$, 3 sig figs: 1.12 × 10³)).

Archimedes's sphere-volume theorem $V = \frac{4}{3}\pi r^3$ evaluates in a single left-to-right chain: cubing the radius 6.45 [ENTER] 3 [yˣ] ($r^3 = 268.3361\text{ cm}^3$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 268.3361)$), scaling by [↱] [π] [×] ($843.0039$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 843.0039)$), and multiplying by $4/3$ via 4 [×] 3 [÷] determines $V = \mathbf{1,124.01\text{ cm}^3}$ ($1.12 \times 10^3\text{ cm}^3$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1124.0052)$.

Workbook Problem 14.3: Torus Surface Area

Major radius $R = 15.0\text{ cm}$, tube radius $r = 3.2\text{ cm}$. Find $A = 4 \pi^2 R r$.

Keystrokes: 15 [ENTER] 3.2 [×] 4 [×] [↱] [π] [↰] [x²] [×] $\rightarrow$ 1,894.964045 (3-sig-fig target: 1.89E3 (Full: 1894.964)).

By Pappus of Alexandria's first centroid theorem, revolving a circle of circumference $2\pi r$ around a path of circumference $2\pi R$ sweeps out surface area $A = (2\pi r)(2\pi R) = 4\pi^2 Rr$: forming 15 [ENTER] 3.2 [×] 4 [×] ($4Rr = 192.0\text{ cm}^2$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 192.0000)$) and multiplying by [↱] [π] [↰] [x²] [×] determines $A = \mathbf{1,894.96\text{ cm}^2}$ ($1.89 \times 10^3\text{ cm}^2$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1894.9640)$.

Workbook Problem 14.4: Exponential Growth Doubling Period

Bacterial culture grows at continuous rate $k = 0.0425\text{ hr}^{-1}$. Find doubling time $t = \frac{\ln 2}{k}$.

Keystrokes: 2 [LN] .0425 [÷] $\rightarrow$ 16.309345425 (3-sig-fig target: 16.3 (Full: 16.3093)).

Solving $N(t) = N_0 e^{kt} = 2N_0$ gives the exact continuous doubling period $t_d = \frac{\ln 2}{k}$: evaluating 2 [LN] ($0.693147$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.6931)$) and dividing by .0425 [÷] determines $t = \mathbf{16.31\text{ hours}}$ ($16.3\text{ hr}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 16.3093)$.

Workbook Problem 14.5: Parallel Resistance via Primary Key `[x∥y]`

Three parallel resistors $R_1 = 48.0\,\Omega, R_2 = 72.0\,\Omega, R_3 = 36.0\,\Omega$. Find equivalent resistance $R_{\text{eq}}$.

Keystrokes: 48 [ENTER] 72 [x∥y] 36 [x∥y] $\rightarrow$ 16 (3-sig-fig target: 16.0 (Exact: $16.0000\,\Omega$)).

Evaluate three parallel resistance branches directly via the dedicated harmonic operator key (algebraic calculators require 18 keystrokes with parentheses and reciprocals; the SC-32 requires 5 tokens / 8 keystrokes):

STEP	KEYSTROKE	REGISTER Y	REGISTER X	DISPLAY
01	48 [ENTER]	48.0000	48.0000	48.0000_
02	72 [x∥y]	0.00	28.8000	28.8000
03	36 [x∥y]	0.00	16.0000	16.0000

Workbook Problem 14.6: 3D Euclidean Vector Magnitude

Vector $\vec{v} = (18.5, -24.2, 12.0)$. Find $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$.

Keystrokes: 18.5 [↰] [x²] 24.2 [↰] [x²] [+] 12 [↰] [x²] [+] [√x] $\rightarrow$ 32.739731215 (3-sig-fig target: 32.7 (Full mantissa: 32.739578)).

Because squaring a real component is always non-negative ($|-24.2|^2 = 24.2^2$), skip the [+/-] key entirely in Euclidean norm sprints: summing 18.5 [↰] [x²] 24.2 [↰] [x²] [+] ($927.8900$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 927.8900)$) plus 12 [↰] [x²] [+] ($1071.8900$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1071.8900)$) and extracting the root via [√x] determines $|\vec{v}| = \mathbf{32.7396}$ ($32.7$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 32.7396)$.

Workbook Problem 14.7: Distance Between 2D Points

Points $P_1(14.2, -8.6)$ and $P_2(38.4, 22.5)$. Find distance $d = \sqrt{(38.4 - 14.2)^2 + (22.5 - (-8.6))^2}$.

Keystrokes: 38.4 [ENTER] 14.2 [-] [↰] [x²] 22.5 [ENTER] 8.6 [+/-] [-] [↰] [x²] [+] [√x] $\rightarrow$ 39.406217783 (3-sig-fig target: 39.4 (Full mantissa: 39.406218)).

STEP	KEYSTROKE	REGISTER Y	REGISTER X	DISPLAY
01	38.4 [ENTER] 14.2 [-]	0.00	24.2000	24.2000
02	[↰] [x²]	0.00	585.6400	585.6400
03	22.5 [ENTER] 8.6 [+/-] [-]	585.6400	31.1000	31.1000
04	[↰] [x²] [+]	0.00	1552.850	1552.850
05	[√x]	0.00	39.40622	39.4062

Workbook Problem 14.8: 2D Vector Dot Product

Vectors $\vec{u} = (4, -3), \vec{v} = (8, 6)$. Find $\vec{u} \cdot \vec{v} = u_x v_x + u_y v_y$.

Keystrokes: 4 [ENTER] 8 [×] 3 [ENTER] 6 [×] [-] $\rightarrow$ 14 (3-sig-fig target: 14.0 (Exact)).

STEP	KEYSTROKE	REGISTER Y	REGISTER X	DISPLAY
01	4 [ENTER] 8 [×]	0.00	32.0000	32.0000
02	3 [ENTER] 6 [×]	32.0000	18.0000	18.0000
03	[-]	0.00	14.0000	14.0000

Workbook Problem 14.9: Angle Between Vectors

Vectors with $|\vec{u}| = 5.0, |\vec{v}| = 10.0, \vec{u}\cdot\vec{v} = 14.0$. Find $\theta = \arccos\left(\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}\right)$. Keystrokes in [DEG] Mode: 14 [ENTER] 5 [÷] 10 [÷] [↰] [ACOS] $\rightarrow$ 73.739795292 (3-sig-fig target: 73.7 ($73.7398^\circ$)).

Normalizing the inner product by both Euclidean norms via sequential division 14 [ENTER] 5 [÷] 10 [÷] ($\cos\theta = 0.2800$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.2800)$) and taking [↰] [ACOS] in Degrees mode determines the inter-vector angle $\theta = \mathbf{73.74^\circ}$ ($73.7^\circ$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 73.7398)$.

Workbook Problem 14.10: Polar Vector Sum

Add $(10.0\angle 30.0^\circ) + (15.0\angle 75.0^\circ)$. Find resultant magnitude and angle.

Keystrokes in [DEG] Mode: 30 [ENTER] 10 [↱] [▸y,x] [STO] [A] [x⇄y] [STO] [B] 75 [ENTER] 15 [↱] [▸y,x] [RCL] [A] [+] [x⇄y] [RCL] [B] [+] [x⇄y] [↰] [▸θ,r] $\rightarrow$ 23.176109129 (3-sig-fig target: Magnitude 23.2, Angle 57.2°).

Decomposing the first phasor [DEG] 30 [ENTER] 10 [↱] [▸y,x] ($y_1 = 5.0000, x_1 = 8.6603$, $(T: 0.0000, Z: 0.0000, Y: 5.0000, X: 8.6603)$) into registers [STO] A [x⇄y] [STO] B, decomposing the second phasor 75 [ENTER] 15 [↱] [▸y,x] ($y_2 = 14.4889, x_2 = 3.8823$, $(T: 8.6603, Z: 5.0000, Y: 14.4889, X: 3.8823)$), summing components via [RCL] A [+] [x⇄y] [RCL] B [+] [x⇄y] ($\Sigma y = 19.4889, \Sigma x = 12.5426$, $(T: 5.0000, Z: 5.0000, Y: 19.4889, X: 12.5426)$), and re-converting via [↰] [▸θ,r] resolves the resultant magnitude $r = \mathbf{23.18}$ ($23.2$) in $X$ and phase angle $\theta = \mathbf{57.24^\circ}$ ($57.2^\circ$) in $Y$ $(T: 5.0000, Z: 5.0000, Y: 57.2372, X: 23.1761)$—or with zero storage registers using [↰] [CMPLX] [+] right after the second [↱] [▸y,x]!

Workbook Problem 14.11: Point-to-Line Perpendicular Distance

Distance from $(2, 3)$ to line $3x + 4y - 12 = 0$:

$$d = \frac{|3(2) + 4(3) - 12|}{\sqrt{3^2 + 4^2}} = \frac{|6 + 12 - 12|}{\sqrt{25}} = \frac{6}{5} = \mathbf{1.20}$$ Keystrokes: 3 [ENTER] 2 [×] 4 [ENTER] 3 [×] [+] 12 [-] 3 [↰] [x²] 4 [↰] [x²] [+] [√x] [÷] $\rightarrow$ 1.2 (3-sig-fig target: 1.20).

By Hesse's normal form of a line, substituting $(x_0, y_0) = (2, 3)$ into $Ax + By + C$ and dividing by the normal vector length $\sqrt{A^2 + B^2}$ gives the orthogonal distance: evaluating 3 [ENTER] 2 [×] 4 [ENTER] 3 [×] [+] 12 [-] ($6.0000$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 6.0000)$), forming the $(3, 4, 5)$ normal norm 3 [↰] [x²] 4 [↰] [x²] [+] [√x] ($5.0000$ in $X$ with $6.0000$ in $Y$, $(T: 0.0000, Z: 0.0000, Y: 6.0000, X: 5.0000)$), and dividing via [÷] yields $d = \mathbf{1.20}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.2000)$.

Workbook Problem 14.12: Triangle Centroid Coordinates

Vertices $(2, 4), (8, 12), (14, 2)$. Centroid $x_c = \frac{2+8+14}{3} = \mathbf{8.00}, y_c = \frac{4+12+2}{3} = \mathbf{6.00}$.

Keystrokes: 2 [ENTER] 8 [+] 14 [+] 3 [÷] $\rightarrow$ 8.00; 4 [ENTER] 12 [+] 2 [+] 3 [÷] $\rightarrow$ 6.00 (3-sig-fig target: 8.00 (X), 6.00 (Y)).

Because the barycenter of a planar triangle is the arithmetic mean of its three vertex coordinates, averaging the abscissas 2 [ENTER] 8 [+] 14 [+] 3 [÷] resolves $x_c = \mathbf{8.00}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 8.0000)$, and averaging the ordinates 4 [ENTER] 12 [+] 2 [+] 3 [÷] resolves $y_c = \mathbf{6.00}$ $(T: 0.0000, Z: 0.0000, Y: 8.0000, X: 6.0000)$.

Workbook Problem 14.13: Curve Arc Length

Length of $y = x^{1.5}$ from $x = 0$ to $x = 4$:

$$L = \int_0^4 \sqrt{1 + 2.25 x} \, dx = \left[ \frac{8}{27} (1 + 2.25 x)^{1.5} \right]_0^4 = \frac{8}{27} (10^{1.5} - 1) \approx \mathbf{9.0734}$$ Keystrokes: 10 [ENTER] 1.5 [yˣ] 1 [-] 8 [×] 27 [÷] $\rightarrow$ 9.07 (3-sig-fig target: 9.07).

Evaluating the arc-length radicand $1 + (y')^2 = 1 + 2.25x$ at $x = 4$ (4 [ENTER] 2.25 [×] 1 [+] $\rightarrow 10.0000$), raising to the $3/2$ power and subtracting the lower-limit value (10 [ENTER] 1.5 [yˣ] 1 [-] $\rightarrow 30.6228$, $(T: 0.0000, Z: 0.0000, Y: 10.0000, X: 30.6228)$), and scaling by 8 [×] 27 [÷] resolves $L = \mathbf{9.0734}$ ($9.07$) $(T: 0.0000, Z: 0.0000, Y: 10.0000, X: 9.0734)$—or directly via Romberg quadrature (1 2.25 X [×] [+] [√x] in [↱] [EQN], then 0 [ENTER] 4 [↱] [∫] X).

Workbook Problem 14.14: Function Average Value

Average value of $f(x) = \sin x$ over $[0, \pi]$: $\bar{f} = \frac{1}{\pi} \int_0^\pi \sin x \, dx = \frac{2}{\pi} \approx \mathbf{0.63662}$.

Keystrokes: 2 [↱] [π] [÷] $\rightarrow$ 0.637 (3-sig-fig target: 0.637).

Because a single half-arch of $\sin x$ over $[0, \pi]$ encloses exact area $\int_0^\pi \sin x\,dx = 2$, keying 2 [↱] [π] [÷] auto-lifts $2$ into Register $Y$ when [↱] [π] is pressed and divides in 3 keystrokes to yield the rectified sine mean $\bar{f} = \mathbf{0.6366}$ ($0.637$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.6366)$.

Workbook Problem 14.15: Tsiolkovsky Rocket Burnout Velocity

Exhaust velocity $v_e = 3,200\text{ m/s}$, initial mass $m_0 = 18,500\text{ kg}$, burnout mass $m_f = 4,200\text{ kg}$. Find $\Delta v = v_e \ln(m_0 / m_f)$. Keystrokes: 18500 [ENTER] 4200 [÷] [LN] 3200 [×] $\rightarrow$ 4,744.595862 (3-sig-fig target: 4.74E3 ($4,744.59\text{ m/s}$, 3 sig figs: 4.74 × 10³)).

Evaluating the wet-to-dry mass ratio 18500 [ENTER] 4200 [÷] ($4.4048$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 4.4048)$), taking the natural logarithm [LN] ($1.4827$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.4827)$), and scaling by effective exhaust velocity 3200 [×] determines the ideal burnout velocity $\Delta v = \mathbf{4,744.59\text{ m/s}}$ ($4.74 \times 10^3\text{ m/s}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 4744.5920)$.

Workbook Problem 14.16: Kinetic Energy of High-Speed Train

Mass $m = 1,450\text{ kg}$, speed $v = 28.5\text{ m/s}$. Find $E_k = \frac{1}{2} m v^2$.

Keystrokes: 28.5 [↰] [x²] 1450 [×] 2 [÷] $\rightarrow$ 588,881.25 (3-sig-fig target: 5.89E5 ($588,881.25\text{ J}$, 3 sig figs: 5.89 × 10⁵)).

Squaring velocity first via 28.5 [↰] [x²] ($812.25\text{ m}^2/\text{s}^2$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 812.2500)$) and chaining 1450 [×] 2 [÷] evaluates translational kinetic energy $E_k = \mathbf{588,881.25\text{ J}}$ ($5.89 \times 10^5\text{ J}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 588881.2500)$ without a single [ENTER] keystroke after the initial squaring.

Workbook Problem 14.17: Gravitational Potential Energy Shift

$\Delta U = G M m \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ where $G M m = 4.0 \times 10^{14}, R_1 = 6.4 \times 10^6, R_2 = 6.8 \times 10^6$.

Keystrokes: 6.4 [E] 6 [1/x] 6.8 [E] 6 [1/x] [-] 4 [E] 14 [×] $\rightarrow$ 3,676,470.59 (3-sig-fig target: 3.68E6 ($3.6765 \times 10^6\text{ J}$)).

Evaluating the inverse-radius difference 6.4 [E] 6 [1/x] 6.8 [E] 6 [1/x] [-] ($1/R_1 - 1/R_2 = 9.1912 \times 10^{-9}\text{ m}^{-1}$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 9.1912 \times 10^{-9})$) and scaling by 4 [E] 14 [×] determines the orbital potential energy gain $\Delta U = \mathbf{3.6765 \times 10^6\text{ J}}$ ($3.68 \times 10^6\text{ J}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 3676470.5900)$.

Workbook Problem 14.18: Polytropic Gas Compression Work

$W = \frac{P_1 V_1 - P_2 V_2}{n - 1}$ with $P_1 = 100\text{ kPa}, V_1 = 0.50\text{ m}^3, P_2 = 800\text{ kPa}, V_2 = 0.12\text{ m}^3, n = 1.35$.

Keystrokes: 100 [ENTER] 0.5 [×] 800 [ENTER] 0.12 [×] [-] 1.35 [ENTER] 1 [-] [÷] $\rightarrow$ -131.4285714 (3-sig-fig target: -131 ($-131.43\text{ kJ}$)).

Integrating $\int_{V_1}^{V_2} P\,dV$ along the polytropic path $P V^n = C$ yields $W = \frac{P_1V_1 - P_2V_2}{n - 1}$: evaluating 100 [ENTER] 0.5 [×] 800 [ENTER] 0.12 [×] [-] ($50.0 - 96.0 = -46.0\text{ kJ}$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: -46.0000)$) and dividing by 1.35 [ENTER] 1 [-] [÷] ($n - 1 = 0.35$) determines the boundary work $W = \mathbf{-131.43\text{ kJ}}$ ($-131\text{ kJ}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: -131.4286)$.

Workbook Problem 14.19: Structural Beam Center Deflection

$P = 5000\text{ N}, L = 4.0\text{ m}, E = 200\text{ GPa}, I = 8.5 \times 10^{-6}\text{ m}^4$. Find $\delta = \frac{P L^3}{48 E I}$.

Keystrokes: 4 [ENTER] 3 [yˣ] 5000 [×] 48 [÷] 200 [E] 9 [÷] 8.5 [E] 6 [+/-] [÷] $\rightarrow$ 0.0039215686 (3-sig-fig target: 0.00392 ($3.92\text{ mm}$, or 3.92E-3)).

Evaluating the numerator 4 [ENTER] 3 [yˣ] 5000 [×] ($PL^3 = 320,000.0\text{ N}\cdot\text{m}^3$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 320000.0000)$) and dividing sequentially through all three denominator factors via 48 [÷] 200 [E] 9 [÷] 8.5 [E] 6 [+/-] [÷] resolves the midspan deflection $\delta = \mathbf{0.00392\text{ m}}$ ($3.92\text{ mm}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0039)$.

Workbook Problem 14.20: Acoustic Doppler Frequency Shift

Source frequency $f = 440\text{ Hz}$, speed of sound $v = 343\text{ m/s}$, approach speed $v_s = 28.0\text{ m/s}$. Find $f' = f \left(\frac{v}{v - v_s}\right)$. Keystrokes: 343 [ENTER] 28 [-] 343 [x⇄y] [÷] 440 [×] $\rightarrow$ 479.11111111 (3-sig-fig target: 479 ($479.11\text{ Hz}$)).

Christian Doppler's wavefront compression factor $\frac{v}{v - v_s}$ increases perceived pitch as an acoustic source approaches a stationary observer: evaluating 343 [ENTER] 28 [-] 343 [x⇄y] [÷] ($343/315 = 1.0889$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.0889)$) and scaling by the concert-A pitch 440 [×] resolves $f' = \mathbf{479.11\text{ Hz}}$ ($479\text{ Hz}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 479.1111)$—or by duplicating 343 upfront via 343 [ENTER] [ENTER] 28 [-] [÷] 440 [×], 343 is typed only once and [x⇄y] is eliminated!

Workbook Problem 14.21: Thin Lens Conjugate Object Distance

Focal length $f = 50.0\text{ mm}$, image distance $d_i = 180.0\text{ mm}$. Find $d_o = \frac{f \times d_i}{d_i - f}$.

Keystrokes: 50 [1/x] 180 [1/x] [-] [1/x] $\rightarrow$ 69.230769231 (3-sig-fig target: 69.2 ($69.231\text{ mm}$)).

Inverting the Gaussian thin-lens conjugate $\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i}$ via 50 [1/x] ($0.0200\text{ mm}^{-1}$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0200)$), 180 [1/x] [-] ($0.014444\text{ mm}^{-1}$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0144)$), and [1/x] resolves $d_o = \mathbf{69.23\text{ mm}}$ ($69.2\text{ mm}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 69.2308)$—or in 4 tokens via the signed harmonic operator 180 [ENTER] 50 [+/-] [x∥y] $\rightarrow \mathbf{69.2308\text{ mm}}$!

Workbook Problem 14.22: Relativistic Lorentz Factor ($\gamma$)

Particle speed $v = 0.85 c$. Find $\gamma = \frac{1}{\sqrt{1 - (v/c)^2}}$.

Keystrokes: 0.85 [↰] [x²] 1 [x⇄y] [-] [√x] [1/x] $\rightarrow$ 1.8983159915 (3-sig-fig target: 1.90 (Full mantissa: 1.898316)).

Hendrik Lorentz's time-dilation and relativistic-mass factor $\gamma = (1 - \beta^2)^{-1/2}$ evaluates from the inside out: squaring normalized speed 0.85 [↰] [x²] ($\beta^2 = 0.7225$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.7225)$), subtracting from unity via 1 [x⇄y] [-] ($1 - \beta^2 = 0.2775$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.2775)$), and chaining [√x] [1/x] determines $\gamma = \mathbf{1.8983}$ ($1.90$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.8983)$—or in trigonometric Degrees mode via 0.85 [↰] [ASIN] [COS] [1/x]!

Workbook Problem 14.23: Stefan-Boltzmann Radiative Power

Radiating area $A = 2.4\text{ m}^2$, temperature $T = 850\text{ K}$, $\sigma = 5.67037 \times 10^{-8}$. Find $P = \sigma A T^4$.

Keystrokes: 850 [ENTER] 4 [yˣ] 2.4 [×] 5.67037 [E] 8 [+/-] [×] $\rightarrow$ 71,039.24592 (3-sig-fig target: 7.10E4 ($71,039.2\text{ W}$, 3 sig figs: 7.10 × 10⁴)).

Raising absolute temperature to the fourth power 850 [ENTER] 4 [yˣ] ($5.2201 \times 10^{11}\text{ K}^4$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 5.2201 \times 10^{11})$) and scaling by area and the Stefan-Boltzmann constant 2.4 [×] 5.67037 [E] 8 [+/-] [×] determines the blackbody radiant flux $P = \mathbf{71,039.25\text{ W}}$ ($7.10 \times 10^4\text{ W}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 71039.2459)$—or use 850 [↰] [x²] [↰] [x²] to compute $850^4$ in two squarings!

Workbook Problem 14.24: Great Circle Geodesic Distance

Spherical distance between coordinates $A(40.7^\circ, 74.0^\circ)$ and $B(51.5^\circ, 0.1^\circ)$ with $\Delta\lambda = 73.9^\circ$:

$$\cos d = \sin(40.7^\circ)\sin(51.5^\circ) + \cos(40.7^\circ)\cos(51.5^\circ)\cos(73.9^\circ)$$ Keystrokes in [DEG] Mode: 40.7 [SIN] 51.5 [SIN] [×] 40.7 [COS] 51.5 [COS] [×] 73.9 [COS] [×] [+] [↰] [ACOS] 60 [×] $\rightarrow$ 3,007.046332 (3-sig-fig target: 3.01E3 nautical miles ($3,007\text{ nmi}$, 3 sig figs: 3.01 × 10³)).

By the spherical law of cosines between New York ($40.7^\circ\text{N}, 74.0^\circ\text{W}$) and London ($51.5^\circ\text{N}, 0.1^\circ\text{W}$), evaluating the polar sine product 40.7 [SIN] 51.5 [SIN] [×] ($0.5103$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.5103)$), adding the equatorial cosine product 40.7 [COS] 51.5 [COS] [×] 73.9 [COS] [×] [+] ($\cos d = 0.6412$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.6412)$), and converting central angle to nautical miles ($1^\circ = 60\text{ nmi}$) via [↰] [ACOS] 60 [×] resolves $d_{\text{geo}} = \mathbf{3,007.05\text{ nmi}}$ ($3.01 \times 10^3\text{ nmi}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 3007.0463)$.

Workbook Problem 14.25: Darcy-Weisbach Piping Head Loss

Friction factor $f = 0.022$, length $L = 150\text{ m}$, diameter $D = 0.25\text{ m}$, velocity $v = 2.8\text{ m/s}$, $g = 9.80665$. Find $h_f = f \frac{L}{D} \frac{v^2}{2g}$. Keystrokes: 0.022 [ENTER] 150 [×] 0.25 [÷] 2.8 [↰] [x²] [×] 2 [÷] 9.80665 [÷] $\rightarrow$ 5.2764195724 (3-sig-fig target: 5.28 ($5.2764\text{ m}$)).

Evaluating the dimensionless resistance coefficient 0.022 [ENTER] 150 [×] 0.25 [÷] ($f L/D = 13.2000$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 13.2000)$) and scaling by the velocity head 2.8 [↰] [x²] [×] 2 [÷] 9.80665 [÷] resolves the viscous head loss $h_f = \mathbf{5.2764\text{ m}}$ ($5.28\text{ m}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 5.2764)$.

Workbook Problem 14.26: Suspension Bridge Catenary Cable Sag

Cable parameter $a = 120\text{ m}$, half-span $x = 45\text{ m}$. Find $y = 120 [\cosh(45/120) - 1]$.

Keystrokes: 45 [ENTER] 120 [÷] [↰] [HYP] [COS] 1 [-] 120 [×] $\rightarrow$ 8.5368416046 (3-sig-fig target: 8.54 ($8.5368\text{ m}$)).

Evaluating the dimensionless hyperbolic cosine offset 45 [ENTER] 120 [÷] [↰] [HYP] [COS] 1 [-] ($\cosh(0.375) - 1 = 0.07114$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.0711)$) and scaling by 120 [×] determines the midspan cable sag $s = \mathbf{8.5368\text{ m}}$ ($8.54\text{ m}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 8.5368)$.

Workbook Problem 14.27: Cubic Polynomial Brent's Solver Extraction

Find the real root of $x^3 - 2x - 5 = 0$.

With the cubic residual $f(x) = x^3 - 2x - 5$ programmed in [↱] [EQN] mode (X 3 [yˣ] 2 X [×] [-] 5 [-]), typing 2 [ENTER] 3 loads initial physical bracketing interval $[2.0, 3.0]$ across registers $(Y, X)$ $(T: 0.0000, Z: -5.0000, Y: 2.0000, X: 3.0000)$. Executing [↱] [SOLVE] X contracts the bracket via Brent's hybrid algorithm until $|f(x)| < 10^{-12}$, depositing the converged real root $x = \mathbf{2.0946}$ into Register $X$ $(T: -5.0000, Z: 2.0000, Y: 3.0000, X: 2.0946)$ (3-sig-fig target: 2.09, full mantissa $2.094551481$).

(Tournament Closed-Form Alternative): In sprint contests, you can also solve $x^3 - 2x - 5 = 0$ instantaneously with Cardano's exact polynomial engine [↱] [POLY]. Because unentered coefficients default to $0.0$, quadratic coefficient $B$ can be skipped entirely: key 1 [A] 2 [+/-] [C] 5 [+/-] [D] [XEQ] [RCL] 1 to extract $2.094551$ in $<5\text{ ms}$!

Workbook Problem 14.28: Romberg Quadrature Numerical Definite Integral

Evaluate $\int_0^1 \frac{4}{1 + x^2} \, dx = \pi$.

With the integrand $f(x) = \frac{4}{1 + x^2}$ active in [↱] [EQN] mode (4 1 X [↰] [x²] [+] [÷]), inputting 0 [ENTER] 1 loads integration boundary interval $[0.0, 1.0]$ across $(Y, X)$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.0000)$. Executing [↱] [∫] X triggers adaptive Richardson extrapolation, converging to the analytical definite integral $\int_0^1 \frac{4}{1+x^2}\,dx = \pi = \mathbf{3.1416}$ in Register $X$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 3.1416)$ (3-sig-fig target: 3.14, full mantissa $3.141592654$).

Workbook Problem 14.29: Hydraulic Radius of Trapezoidal Channel

Bottom width $b = 4.0\text{ m}$, water depth $y = 1.5\text{ m}$, side slopes $z = 2.0$. Cross-sectional area $A = (b + zy)y = 10.5\text{ m}^2$. Wetted perimeter $P = b + 2y\sqrt{1+z^2} = 4 + 3\sqrt{5} = 10.7082\text{ m}$. Find hydraulic radius $R_h = A / P$. Keystrokes: 4 [ENTER] 2 [ENTER] 1.5 [×] [+] 1.5 [×] 1 [ENTER] 2 [↰] [x²] [+] [√x] 2 [×] 1.5 [×] 4 [+] [÷] $\rightarrow$ 0.9805565962 (3-sig-fig target: 0.981 ($0.980556\text{ m}$)).

In Robert Manning's open-channel hydraulics, hydraulic radius $R_h = A/P$ governs gravity flow velocity: evaluating the trapezoidal flow area 4 [ENTER] 2 [ENTER] 1.5 [×] [+] 1.5 [×] ($A = 10.5000\text{ m}^2$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 10.5000)$), forming the bank hypotenuse factor 1 [ENTER] 2 [↰] [x²] [+] [√x] ($\sqrt{5} = 2.2361$ in $X$ while $A$ waits in $Y$, $(T: 0.0000, Z: 0.0000, Y: 10.5000, X: 2.2361)$), completing the wetted perimeter 2 [×] 1.5 [×] 4 [+] ($P = 10.7082\text{ m}$, $(T: 0.0000, Z: 0.0000, Y: 10.5000, X: 10.7082)$), and dividing via [÷] resolves $R_h = \mathbf{0.9806\text{ m}}$ ($0.981\text{ m}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.9806)$.

Workbook Problem 14.30: Thin Film Optical Interference Thickness

Destructive interference of reflected light ($\lambda = 550\text{ nm} = 5.5 \times 10^{-7}\text{ m}$) in a thin film coating ($n = 1.38$): $t = \frac{\lambda}{4n}$. Keystrokes: 5.5 [E] 7 [+/-] [ENTER] 4 [÷] 1.38 [÷] $\rightarrow$ 9.963768E-8 (3-sig-fig target: 9.96E-8 ($9.9638 \times 10^{-8}\text{ m} = 99.64\text{ nm}$, 3 sig figs: 9.96 × 10⁻⁸)).

For a magnesium fluoride ($n = 1.38$) anti-reflective lens coating, a quarter-wave optical thickness $n t = \lambda/4$ shifts the rear-surface reflection by half a wavelength ($180^\circ$) to cancel front-surface glare: dividing 5.5 [E] 7 [+/-] 4 [÷] ($1.3750 \times 10^{-7}\text{ m}$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.3750 \times 10^{-7})$) by 1.38 [÷] determines the film thickness $t = \mathbf{9.9638 \times 10^{-8}\text{ m}}$ ($9.96 \times 10^{-8}\text{ m}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 9.9638 \times 10^{-8})$.

Workbook Problem 14.31: Barometric Atmospheric Pressure at Altitude

Sea level pressure $P_0 = 101.325\text{ kPa}$, scale height $H = 8,400\text{ m}$. Find pressure at cruising altitude $h = 10,500\text{ m}$: $P = P_0 e^{-h/H}$. Keystrokes: 10500 [+/-] [ENTER] 8400 [÷] [eˣ] 101.325 [×] $\rightarrow$ 29.030098542 (3-sig-fig target: 29.0 ($29.030\text{ kPa})$).

Under isothermal hydrostatic equilibrium $\frac{dP}{dh} = -\frac{P}{H}$, evaluating the scale-height exponent 10500 [+/-] [ENTER] 8400 [÷] ($-h/H = -1.2500$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: -1.2500)$), exponentiating via [eˣ] ($0.2865$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 0.2865)$), and scaling by standard atmosphere 101.325 [×] resolves $P = \mathbf{29.0301\text{ kPa}}$ ($29.0\text{ kPa}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 29.0301)$.

Workbook Problem 14.32: Chemical Equilibrium pH from Hydronium Ion

$[H^+] = 3.45 \times 10^{-5}\text{ M}$. Find $\text{pH} = -\log_{10}[H^+]$.

Keystrokes: 3.45 [E] 5 [+/-] [↰] [LOG] [+/-] $\rightarrow$ 4.4621809049 (3-sig-fig target: 4.46 (Full mantissa: 4.462198)).

Søren Sørensen's hydronium exponent $\text{pH} = -\log_{10}[\text{H}^+]$ evaluates in two operations after scientific entry: keying 3.45 [E] 5 [+/-] [↰] [LOG] ($-4.4622$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: -4.4622)$) and negating via [+/-] determines $\text{pH} = \mathbf{4.4622}$ ($4.46$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 4.4622)$.

Workbook Problem 14.33: Acoustics Sound Pressure Decibel Level

Acoustic pressure $P = 4.8 \times 10^{-4}\text{ Pa}$, reference $P_0 = 20\text{ }\mu\text{Pa} = 2.0 \times 10^{-5}\text{ Pa}$. Find $L_p = 20 \log_{10}(P / P_0)$. Keystrokes: 4.8 [E] 4 [+/-] [ENTER] 2 [E] 5 [+/-] [÷] [↰] [LOG] 20 [×] $\rightarrow$ 27.604224834 (3-sig-fig target: 27.6 ($27.604\text{ dB}$)).

Dividing the RMS acoustic pressure by the $20\,\mu\text{Pa}$ reference threshold 4.8 [E] 4 [+/-] [ENTER] 2 [E] 5 [+/-] [÷] ($P/P_0 = 24.0000$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 24.0000)$) and chaining [↰] [LOG] 20 [×] resolves $L_p = \mathbf{27.6042\text{ dB}}$ ($27.6\text{ dB}$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 27.6042)$.

Workbook Problem 14.34: Golden Ratio Multi-Nested Radical

Evaluate the infinite nested radical $Q = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots}}}} = \phi = \frac{1 + \sqrt{5}}{2}$.

Keystrokes: 5 [√x] 1 [+] 2 [÷] $\rightarrow$ 1.6180339887 (3-sig-fig target: 1.62 (Full mantissa: 1.618033989)).

Squaring both sides of the self-similar radical $Q = \sqrt{1 + Q}$ yields the golden-ratio quadratic $Q^2 - Q - 1 = 0$, whose positive root $\phi = \frac{1 + \sqrt{5}}{2}$ evaluates in 4 tokens: 5 [√x] ($2.2361$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 2.2361)$) followed by 1 [+] 2 [÷] resolves $\phi = \mathbf{1.6180}$ ($1.62$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.6180)$.

Workbook Problem 14.35: Surface Area of Irregular Chassis Solid via Stats Mode `[Σ+]`

An engineering geometry problem asks for the total external surface area of a stepped rectangular chassis with outer dimensions: Length $L = 135\text{ mm}$, Width $W = 70\text{ mm}$, Height $H = 16\text{ mm}$. The surface consists of three pairs of congruent rectangular faces: $$A_{\text{total}} = 2(LW) + 2(LH) + 2(WH)$$ Rather than computing each pair and manually adding, speed-calculation masters accumulate intermediate face areas directly into the statistical accumulator using [Σ+], then instantly recall total surface area via [Σx] (3-sig-fig target: 2.55E4, $25,460\text{ mm}^2$):

STEP	KEYSTROKE	REGISTER X	DISPLAY
01	135 [ENTER] 70 [×] 2 [×]	18900.00	18900.00
02	[Σ+]	1.0000(n=1)	n=1
03	135 [ENTER] 16 [×] 2 [×]	4320.000	4320.000
04	[Σ+]	2.0000(n=2)	n=2
05	70 [ENTER] 16 [×] 2 [×]	2240.000	2240.000
06	[Σ+]	3.0000(n=3)	n=3
07	[Σx]	25460.00	2.54600E4

Workbook Problem 14.36: Surface Area of Cylindrical Battery Pod via Stats Mode `[Σ+]`

An applied physics problem asks for the total surface area of a cylindrical battery pod: Radius $r = 15\text{ mm}$, Height $h = 65\text{ mm}$:

\[A_{\text{total}} = 2(\pi r^2) + 2\pi r h = 2\pi (15^2) + 2\pi(15)(65) = 450\pi + 1950\pi = 2400\pi \approx 7,539.82\text{ mm}^2\]

Clearing the statistical registers ([↰] [CLEAR] [CLΣ]), accumulating the twin end-cap areas 15 [↰] [x²] [↱] [π] [×] 2 [×] [Σ+] ($1,413.72\text{ mm}^2$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 1.0000)$) and lateral shell area 2 [↱] [π] [×] 15 [×] 65 [×] [Σ+] ($6,126.11\text{ mm}^2$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 2.0000)$), and recalling [↱] [SUMS] [Σx] resolves the total surface area $\Sigma x = \mathbf{7,539.82\text{ mm}^2}$ (3-sig-fig target: 7.54E3, $7.54 \times 10^3\text{ mm}^2$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 7539.8224)$—or in factored form $2\pi r(r + h)$, by keying $h = 65$, adding $r = 15$, and recalling $r = 15$ from LASTX so the radius is entered only once.

Workbook Problem 14.37: Constant Factor Replication in Register T for Repeated Pricing

In repetitive commercial applications, users must apply a constant sales tax rate $8.25\%$ (multiplier $1.0825$) across three consecutive item prices: $\$10.00, \$25.00, \$150.00$. On the SC-32, loading the constant into Register $T$ allows perpetual reuse because when a dyadic drop occurs, Register $T$ drops to $Z$ and replicates itself in $T$!

Target Answers:
- Price 1 ($10.00$): 10.8 ($10.825$)
- Price 2 ($25.00$): 27.1 ($27.0625$)
- Price 3 ($150.00$): 162 ($162.375$)
Keystrokes: 1. Fill stack with constant: 1.0825 [ENTER] [ENTER] [ENTER]
2. Price 1: 1.0825 [ENTER] 10 [×] $\rightarrow$ 10.825 (Register $T$ retains $1.0825$)
3. Price 2: [CLx] 25 [×] $\rightarrow$ 27.0625 (Register $T$ retains $1.0825$)
4. Price 3: [CLx] 150 [×] $\rightarrow$ 162.375 (Register $T$ perpetually retains $1.0825$)

STEP	KEYSTROKE	REGISTER T	REGISTER Z	REGISTER Y	REGISTER X	DISPLAY
01	1.0825 [ENTER] [ENTER] [ENTER]	1.0825	1.0825	1.0825	1.0825	1.0825
02	10 [×]	1.0825	1.0825	1.0825	10.8250	10.8250
03	[CLx] 25 [×]	1.0825	1.0825	1.0825	27.0625	27.0625
04	[CLx] 150 [×]	1.0825	1.0825	1.0825	162.3750	162.3750

Workbook Problem 14.38: Unit Conversion via Constant Factor in Register T

Convert three dimension measurements given in inches to millimeters using the exact conversion constant $25.4\text{ mm/in}$: $2.0\text{ in}, 5.0\text{ in}, 10.0\text{ in}$. Target Answers:
- $2.0\text{ in}$: 50.8 $\text{mm}$
- $5.0\text{ in}$: 127 $\text{mm}$
- $10.0\text{ in}$: 254 $\text{mm}$

Loading the SI conversion factor across all four stack tiers via 25.4 [ENTER] [ENTER] [ENTER] $(T: 25.4000, Z: 25.4000, Y: 25.4000, X: 25.4000)$ turns Register $T$ into a non-depleting constant source: 2 [×] resolves 50.8000 mm $(T: 25.4000, Z: 25.4000, Y: 25.4000, X: 50.8000)$, [CLx] 5 [×] resolves 127.0000 mm $(T: 25.4000, Z: 25.4000, Y: 25.4000, X: 127.0000)$, and [CLx] 10 [×] resolves 254.0000 mm $(T: 25.4000, Z: 25.4000, Y: 25.4000, X: 254.0000)$ without re-entering 25.4.

Workbook Problem 14.39: Infix Operator Precedence vs Native RPN Stack Equivalence

Compare algebraic infix parsing requiring nested parentheses with natural RPN execution for two standard tournament expressions: 1. $(12 - 4) \times (6 + 2) \div 4 = \mathbf{16.0}$
2. $3 + 4 \times 5 = \mathbf{23.0}$

RPN Keystrokes:
- Expression 1: 12 [ENTER] 4 [-] 6 [ENTER] 2 [+] [×] 4 [÷] $\rightarrow$ 16.0 (Evaluates intermediate differences and sums directly on the stack without a single parenthesis)
- Expression 2: 3 [ENTER] 4 [ENTER] 5 [×] [+] $\rightarrow$ 23.0 (Multiplying 4 and 5 first, then adding 3, completely mirrors operator precedence without requiring parentheses).

In Jan Łukasiewicz's postfix logic, operator position alone encodes tree topology without delimiters: evaluating 12 [ENTER] 4 [-] ($8.0000$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 8.0000)$), 6 [ENTER] 2 [+] ($8.0000$ in $X$ with $8.0000$ lifted to $Y$, $(T: 0.0000, Z: 0.0000, Y: 8.0000, X: 8.0000)$), and [×] 4 [÷] yields $\mathbf{16.0000}$ $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 16.0000)$, while 4 [ENTER] 5 [×] 3 [+] evaluates $\mathbf{23.0000}$ $(T: 0.0000, Z: 0.0000, Y: 16.0000, X: 23.0000)$.

Workbook Problem 14.40: Nested Horner's Method for Cubic Polynomials

Evaluate $P(x) = 3x^3 - 5x^2 + 4x - 7$ at $x = 2.5$. Speed Optimization Insight: Evaluating polynomials term-by-term with powers ($x^3, x^2$) wastes time and invites syntax errors. Rewriting in nested Horner form $((3x - 5)x + 4)x - 7$ allows continuous left-to-right streaming on the stack with zero parentheses:
Keystrokes: 3 [ENTER] 2.5 [×] 5 [-] 2.5 [×] 4 [+] 2.5 [×] 7 [-] $\rightarrow$ 18.625 (3-sig-fig target: 18.6 (Full mantissa: 18.625)).

William George Horner's 1819 nested synthetic division reduces a degree-$n$ polynomial to $n$ multiply-add pairs: evaluating 3 [ENTER] 2.5 [×] 5 [-] ($3x - 5 = 2.5000$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 2.5000)$), chaining 2.5 [×] 4 [+] ($(3x - 5)x + 4 = 10.2500$, $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 10.2500)$), and finishing with 2.5 [×] 7 [-] resolves $P(2.5) = \mathbf{18.6250}$ ($18.6$) $(T: 0.0000, Z: 0.0000, Y: 0.0000, X: 18.6250)$—and by pre-filling the stack with $x = 2.5$ (2.5 [ENTER] [ENTER] [ENTER] 3 [×] 5 [-] [×] 4 [+] [×] 7 [-]), Register $T$ feeds $2.5$ into every Horner stage automatically so 2.5 is typed only once!

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`24.5`	\(0.0000\)	\(0.0000\)	\(0.0000\)	\(24.5000\)	Load \(F_x\)
`[↰] [x²]`	\(0.0000\)	\(0.0000\)	\(0.0000\)	\(600.2500\)	Square \(F_x\) in \(X\)
`18.2` `[↰] [x²]`	\(0.0000\)	\(0.0000\)	\(600.2500\)	\(331.2400\)	Lift \(600.25\) to \(Y\) and square \(F_y\)
`[+]`	\(0.0000\)	\(0.0000\)	\(0.0000\)	\(931.4900\)	Sum of squares: \(600.25 + 331.24\)
`[√x]`	\(0.0000\)	\(0.0000\)	\(0.0000\)	\(30.5203\)	Resultant force \(F_R \approx 30.5203\text{ kN}\)

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`1450` `[ENTER]`	\(0.0000\)	\(0.0000\)	\(1450.0000\)	\(1450.0000\)	Push vehicle mass \(m\)
`28` `[↰] [x²]`	\(0.0000\)	\(0.0000\)	\(1450.0000\)	\(784.0000\)	Velocity squared \(v^2\) in \(X\)
`[×]`	\(0.0000\)	\(0.0000\)	\(0.0000\)	\(1136800.00\)	Product \(m v^2\)
`2` `[÷]`	\(0.0000\)	\(0.0000\)	\(0.0000\)	\(568400.00\)	Divide by 2: \(568.4\text{ kJ}\)

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`6.6743 [E] 11 [+/-]`	\(0.0000\)	\(0.0000\)	\(0.0000\)	\(6.6743 \times 10^{-11}\)	Universal gravitational constant \(G\)
`[ENTER]`	\(0.0000\)	\(0.0000\)	\(6.6743 \times 10^{-11}\)	\(6.6743 \times 10^{-11}\)	Stack lift: pushes \(G\) into \(Y\)
`5.972 [E] 24`	\(0.0000\)	\(0.0000\)	\(6.6743 \times 10^{-11}\)	\(5.9720 \times 10^{24}\)	Earth mass \(M_E\) in \(X\)
`[×]`	\(0.0000\)	\(0.0000\)	\(0.0000\)	\(3.9859 \times 10^{14}\)	Gravitational parameter \(\mu_E = G M_E\)
`7.342 [E] 22`	\(0.0000\)	\(0.0000\)	\(3.9859 \times 10^{14}\)	\(7.3420 \times 10^{22}\)	Moon mass \(M_M\) in \(X\)
`[×]`	\(0.0000\)	\(0.0000\)	\(0.0000\)	\(2.9264 \times 10^{37}\)	Force numerator \(G M_E M_M\)
`3.844 [E] 8`	\(0.0000\)	\(0.0000\)	\(2.9264 \times 10^{37}\)	\(3.8440 \times 10^{8}\)	Lunar orbital distance \(r\)
`[↰] [x²]`	\(0.0000\)	\(0.0000\)	\(2.9264 \times 10^{37}\)	\(1.4776 \times 10^{17}\)	Square orbital radius \(r^2\) in \(X\)
`[÷]`	\(0.0000\)	\(0.0000\)	\(0.0000\)	\(1.9805 \times 10^{20}\)	Mutual orbital attraction force in Newtons

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`3.5` `[ENTER]`	\(0.0000\)	\(0.0000\)	\(3.5000\)	\(3.5000\)	Molar substance quantity \(n\) in \(Y\)
`8.31446` `[×]`	\(0.0000\)	\(0.0000\)	\(0.0000\)	\(29.1006\)	Molar gas scale factor \(n R\)
`315` `[×]`	\(0.0000\)	\(0.0000\)	\(0.0000\)	\(9166.6922\)	Thermal kinetic product \(n R T\)
`0.045` `[÷]`	\(0.0000\)	\(0.0000\)	\(0.0000\)	\(203704.27\)	Static pressure \(P\) in Pascals (\(203.70\text{ kPa}\))

Keystroke	Register T	Register Z	Register Y	Register X	Operational Commentary
`8.5` `[↰] [x²]`	\(0.0000\)	\(0.0000\)	\(0.0000\)	\(72.2500\)	\(b^2 = 8.5^2 = 72.25\)
`4` `[ENTER]`	\(0.0000\)	\(72.2500\)	\(4.0000\)	\(4.0000\)	Load coefficient factor 4
`3.2` `[×]`	\(0.0000\)	\(0.0000\)	\(72.2500\)	\(12.8000\)	Form \(4a = 4 \times 3.2 = 12.8\)
`4.1` `[+/-]`	\(0.0000\)	\(72.2500\)	\(12.8000\)	\(-4.1000\)	Constant \(c = -4.1\) loaded to \(X\)
`[×]`	\(0.0000\)	\(0.0000\)	\(72.2500\)	\(-52.4800\)	Form \(4ac = 12.8 \times (-4.1)\)
`[-]`	\(0.0000\)	\(0.0000\)	\(0.0000\)	\(124.7300\)	\(\Delta = 72.25 - (-52.48) = 124.73\)

Section / Chapter	Domain & Companion Problem Set	Primary SC-32 Engine / Mode	Worked Problems	Reference Manual
Executive Warm-Up	Cross-Disciplinary 4-Register \((T, Z, Y, X)\) Stack Table Exercises	`XYZT` Stack & Inside-Out RPN	25	Manual Preface →
Chapter 01	Hardware Architecture & Instrument Craftsmanship	`RP2350` / Power & Timing	8	Manual Ch. 01 →
Chapter 02	Principles of RPN & 4-Stack XYZT Dynamics	`ENTER`, `x↔y`, `R↓`, `LASTX`	15	Manual Ch. 02 →
Chapter 03	Algebra, Transcendentals & Display Formatting	Logs, Exponentials, Hyperbolics	22	Manual Ch. 03 →
Chapter 04	Trigonometry, Coordinate Systems & 2D Vectors	`→P`, `→R`, `FRAC-01` & `PLACE-01`	29	Manual Ch. 04 →
Chapter 05	Complex Numbers, AC Phasors & Euler's Identity	`[CMPLX]` Dual-Stack Engine	21	Manual Ch. 05 →
Chapter 06	Analytic & Solid Geometry, Surface Area & Mensuration	2D/3D Mensuration & Geodesics	15	Manual Ch. 06 →
Chapter 07	Polynomials, Brent-Dekker Solver & Root Finding	`QUAD`, `CUBIC`, `SOLVE`	26	Manual Ch. 07 →
Chapter 08	Numerical Calculus & Adaptive Romberg Quadrature	`d/dx` & `∫f(x)` Romberg Engine	27	Manual Ch. 08 →
Chapter 09	Bivariate Statistics, Curve Fitting & Linear Regression	`Σ+`, `L.R.`, Normal & Combinatorics	26	Manual Ch. 09 →
Chapter 10	Computer Science, Bases & 36-Bit Integers	`HEX`, `BIN`, `OCT`, Bitwise Ops	24	Manual Ch. 10 →
Chapter 11	Financial Mathematics, Executive Analytics & TVM	`N`, `I/YR`, `PV`, `PMT`, `FV`	31	Manual Ch. 11 →
Chapter 12	Function & Scatter Plotting Engine	`PLOT` & Graphical Root Analysis	27	Manual Ch. 12 →
Chapter 13	Applied STEM Engineering Case Studies	Multi-Stage Aerospace, Civil, EE	28	Manual Ch. 13 →
Chapter 14	High-Speed Equation Entry & UIL Contest Playbook	Inside-Out Speed Calculation	40	Manual Ch. 14 →

Plot Telemetry	Crosshair Readout	Curve Feature
Active Function	`X^3 - 3X^2 - 9X + 5`	Three real zero crossings in \([-3,-2], [0,1], [4,5]\)
Crosshair Cursor	`X = 0.500000000`, `Y = -0.125000000`	Pre-root bracket inspection prior to `[↱] [SOLVE]`