Aeronautical Engineering Math, Step by Step

The standard atmosphere

• Teens
• Community college
• Undergraduate (freshman–senior)

Track: Aeronautical Engineering · About 20 min · Numbers verified in build: 2026-10-16

Step by step

1. $$dp=-\rho g_0\,dh,\qquad p=\rho RT$$
   Hydrostatic balance plus the ideal gas law.
2. $$\frac{dp}{p}=-\frac{g_0}{RT}\,dh$$
   Divide to separate variables.
3. $$T=T_0-Lh\;\Rightarrow\;\frac{p}{p_0}=\left(\frac{T}{T_0}\right)^{g_0/(LR)},\quad L=0.0065\ \text{K/m}$$
   Troposphere: temperature falls linearly.
4. $$T=\text{const}\;\Rightarrow\;p=p_{11}\,e^{-g_0(h-h_{11})/(RT)}$$
   Above 11 km (to 20 km): isothermal, exponential decay.

import math
g0, R, L, T0, p0 = 9.80665, 287.05287, 0.0065, 288.15, 101325.0
h = 11000
T = T0 - L*h
p = p0 * (T/T0)**(g0/(L*R))
rho = p / (R*T)
print(T, p, rho, math.sqrt(1.4*R*T))

Auto-checked practice

1. ISA density at 5000 m, in kg/m³?

   kg/m³ Check Show a hint
   Full solution

   p = 54020 Pa; ρ = 0.7361 kg/m³.

Practice

1. Why does the formula change at 11 km?

   Show solution

   The model’s temperature stops falling (tropopause); with constant T the integral of dp/p gives an exponential instead of a power law.

References

• NOAA, NASA, USAF, U.S. Standard Atmosphere, 1976, U.S. Government Printing Office, 1976.
• J. D. Anderson Jr., Introduction to Flight, 8th ed., McGraw-Hill, 2016.
• NASA Glenn Research Center, Beginner’s Guide to Aeronautics.

Lift, the airfoil and thin-airfoil theory

• Community college
• Undergraduate (freshman–senior)

Track: Aeronautical Engineering · About 25 min · Numbers verified in build: 2026-10-16

Step by step

1. $$L=\tfrac12\rho V^2 S\,C_L$$
   Lift equation (definition of C_L).
2. $$\frac{1}{2\pi}\int_0^\pi\frac{\gamma(\theta)\sin\theta}{\cos\theta-\cos\theta_0}\,d\theta=V_\infty\left(\alpha-\frac{dz}{dx}\right)$$
   Thin-airfoil integral equation for the vortex sheet strength γ.
3. $$\gamma(\theta)=2\alpha V_\infty\frac{1+\cos\theta}{\sin\theta}\;\Rightarrow\;c_l=\frac{2\Gamma}{V_\infty c}=2\pi\alpha$$
   Symmetric airfoil solution and Kutta–Joukowski.
4. $$V_{s}=\sqrt{\frac{2W}{\rho S C_{L,\max}}}$$
   Stall speed: lift equals weight at the maximum C_L.

import math
m, S, clmax, rho, g0 = 1100, 16.2, 1.6, 1.225, 9.80665
Vs = math.sqrt(2*m*g0 / (rho*S*clmax))
print(Vs, Vs/0.514444, 2*math.pi*math.radians(5))

Auto-checked practice

1. Same aircraft at 1500 m ISA altitude: stall speed (true airspeed) in m/s?

   m/s Check Show a hint
   Full solution

   ρ(1500 m) = 1.0581 kg/m³, so V_s = 28.05 m/s (the indicated stall speed stays the same).

Practice

1. Why does thin-airfoil theory fail above about 12–15°?

   Show solution

   It assumes attached, inviscid flow and small angles; at high α the boundary layer separates (stall).

References

• J. D. Anderson Jr., Fundamentals of Aerodynamics, 6th ed., McGraw-Hill, 2017.
• J. D. Anderson Jr., Introduction to Flight, 8th ed., McGraw-Hill, 2016.
• NASA Glenn Research Center, Beginner’s Guide to Aeronautics.

Finite wings: lifting-line theory and induced drag

• Undergraduate (freshman–senior)

Track: Aeronautical Engineering · About 30 min · Numbers verified in build: 2026-10-16

Step by step

1. $$\Gamma(\theta)=2bV_\infty\sum_{n}A_n\sin n\theta,\qquad y=-\tfrac b2\cos\theta$$
   Fourier series for the circulation along the span.
2. $$C_L=\pi AR\,A_1,\qquad C_{D_i}=\pi AR\sum_n nA_n^2=\frac{C_L^2}{\pi e AR}$$
   Lift needs only A₁; every other term adds drag (e ≤ 1).
3. $$a=\frac{a_0}{1+a_0/(\pi e AR)}$$
   Finite-wing lift slope (elliptic-type correction).

import math
AR, e, CL, a0 = 9, 0.8, 0.5, 2*math.pi
print(CL**2/(math.pi*e*AR), a0/(1 + a0/(math.pi*e*AR)))

Auto-checked practice

1. Same wing, AR raised to 12 (e = 0.8, C_L = 0.5): C_Di?

   Check Show a hint
   Full solution

   C_Di = 0.25/(π · 0.8 · 12) = 0.00829, 25% less than at AR = 9.

Practice

1. Which spanwise lift distribution gives e = 1?

   Show solution

   The elliptic distribution (only A₁ ≠ 0), which also gives uniform downwash.

References

• J. D. Anderson Jr., Fundamentals of Aerodynamics, 6th ed., McGraw-Hill, 2017.
• MIT OpenCourseWare, 16.100 Aerodynamics (CC BY-NC-SA 4.0).

Aircraft performance: best glide and the Breguet range equation

• Undergraduate (freshman–senior)

Track: Aeronautical Engineering · About 30 min · Numbers verified in build: 2026-10-16

Step by step

1. $$C_D=C_{D_0}+kC_L^2,\qquad k=\frac{1}{\pi eAR}$$
   Parabolic drag polar.
2. $$\frac{d}{dC_L}\!\left(\frac{C_L}{C_D}\right)=0\;\Rightarrow\;C_{D_0}=kC_L^2,\qquad \left(\frac{L}{D}\right)_{\max}=\frac{1}{2\sqrt{kC_{D_0}}}$$
   Maximise L/D: friction drag equals induced drag.
3. $$-\frac{dW}{dt}=c\,T=c\,\frac{W}{L/D}\;\Rightarrow\;R=\int V\,dt=\frac{V}{c}\frac{L}{D}\ln\frac{W_i}{W_f}$$
   Fuel burn rate c (TSFC) at constant V and L/D; integrate.

import math
AR, e, CD0 = 9, 0.8, 0.02
k = 1/(math.pi*e*AR)
print(1/(2*math.sqrt(k*CD0)), math.sqrt(CD0/k))
V, c, LD, wr = 230, 0.6/3600, 17, 1.25
print(V/c*LD*math.log(wr)/1000, "km")

Auto-checked practice

1. With C_D0 = 0.015 (cleaner airframe), same AR and e: (L/D)max?

   Check Show a hint
   Full solution

   (L/D)max = 19.42, about 15% better.

Practice

1. Why is c divided by 3600 in the worked example?

   Show solution

   TSFC is quoted per hour; V is in m/s, so c must be per second for R to come out in metres.

References

• J. D. Anderson Jr., Introduction to Flight, 8th ed., McGraw-Hill, 2016.
• NASA Glenn Research Center, Beginner’s Guide to Aeronautics.

Compressible flow and normal shocks

• Undergraduate (freshman–senior)
• Graduate

Track: Aeronautical Engineering · About 35 min · Numbers verified in build: 2026-10-16

Step by step

1. $$c_pT+\tfrac12V^2=c_pT_0\;\Rightarrow\;\frac{T_0}{T}=1+\frac{\gamma-1}{2}M^2$$
   Steady energy equation with a = √(γRT).
2. $$\frac{p_0}{p}=\left(\frac{T_0}{T}\right)^{\gamma/(\gamma-1)}$$
   Isentropic deceleration to rest.
3. $$\rho_1u_1=\rho_2u_2,\quad p_1+\rho_1u_1^2=p_2+\rho_2u_2^2,\quad h_1+\tfrac12u_1^2=h_2+\tfrac12u_2^2$$
   Conservation across the shock.
4. $$M_2^2=\frac{1+\frac{\gamma-1}{2}M_1^2}{\gamma M_1^2-\frac{\gamma-1}{2}},\qquad \frac{p_2}{p_1}=1+\frac{2\gamma}{\gamma+1}(M_1^2-1)$$
   Normal-shock relations (solve the three equations).

import math
g, M1 = 1.4, 2.0
t = lambda M: 1 + (g-1)/2*M*M
M2 = math.sqrt(t(M1) / (g*M1**2 - (g-1)/2))
p21 = 1 + 2*g/(g+1)*(M1**2 - 1)
p0r = p21 * t(M2)**(g/(g-1)) / t(M1)**(g/(g-1))
print(t(M1), t(M1)**3.5, M2, p21, p0r)

Auto-checked practice

1. Static-to-total pressure: p₀/p at M = 1 (γ = 1.4)?

   Check Show a hint
   Full solution

   p₀/p = 1.2^3.5 = 1.8929 (the "critical" ratio, about 1.89).

Practice

1. Can a normal shock occur in subsonic flow?

   Show solution

   No: for M₁ < 1 the relations would require entropy to decrease, violating the second law.

References

• J. D. Anderson Jr., Fundamentals of Aerodynamics, 6th ed., McGraw-Hill, 2017.
• NASA Glenn Research Center, Beginner’s Guide to Aeronautics.

Jet propulsion: thrust and the Brayton cycle

• Undergraduate (freshman–senior)

Track: Aeronautical Engineering · About 30 min · Numbers verified in build: 2026-10-16

Step by step

1. $$F=\dot m\,(V_e-V_0)+(p_e-p_0)A_e$$
   Momentum balance on the engine (fuel mass neglected).
2. $$\eta_{th}=1-\frac{q_{out}}{q_{in}}=1-\frac{T_4-T_1}{T_3-T_2}$$
   Heat in at constant pressure (2→3), out (4→1).
3. $$\frac{T_2}{T_1}=\frac{T_3}{T_4}=\pi_c^{(\gamma-1)/\gamma}\;\Rightarrow\;\eta_{th}=1-\pi_c^{-(\gamma-1)/\gamma}$$
   Isentropic compression and expansion share the same pressure ratio.

g, pic = 1.4, 30
print(1 - pic**(-(g-1)/g))
mdot, V0, Ve = 100, 250, 600
print(mdot*(Ve - V0)/1000, "kN")

Auto-checked practice

1. Ideal Brayton efficiency at π_c = 40?

   Check Show a hint
   Full solution

   η = 1 − 40^(−0.2857) = 0.6514.

Practice

1. Why not raise the pressure ratio without limit?

   Show solution

   Compressor exit temperature rises too, leaving less room for heat addition before the turbine’s material temperature limit.

References

• J. D. Mattingly, Elements of Propulsion: Gas Turbines and Rockets, AIAA, 2006.
• NASA Glenn Research Center, Beginner’s Guide to Aeronautics.

Stability and control: the short-period mode

• Graduate

Track: Aeronautical Engineering · About 45 min · Numbers verified in build: 2026-10-16

Step by step

1. $$\begin{bmatrix}\dot\alpha\\ \dot q\end{bmatrix}=\begin{bmatrix}Z_\alpha/V & 1\\ M_\alpha+M_{\dot\alpha}Z_\alpha/V & M_q+M_{\dot\alpha}\end{bmatrix}\begin{bmatrix}\alpha\\ q\end{bmatrix}$$
   Short-period approximation (speed held constant).
2. $$\det(s\mathbf I-\mathbf A)=s^2-\left(\tfrac{Z_\alpha}{V}+M_q+M_{\dot\alpha}\right)s+\left(\tfrac{Z_\alpha}{V}M_q-M_\alpha\right)=0$$
   Characteristic equation.
3. $$s^2+2\zeta\omega_ns+\omega_n^2=0\;\Rightarrow\;\omega_n=\sqrt{\tfrac{Z_\alpha}{V}M_q-M_\alpha},\quad \zeta=\frac{-\left(\tfrac{Z_\alpha}{V}+M_q+M_{\dot\alpha}\right)}{2\omega_n}$$
   Match to the standard second-order form.

import numpy as np
ZaV, Mq, Mad, Ma = -1.0, -1.5, -0.3, -8.0
A = np.array([[ZaV, 1.0], [Ma + Mad*ZaV, Mq + Mad]])
lam = np.linalg.eigvals(A)
wn = np.sqrt(np.prod(lam).real); zeta = -lam.sum().real / (2*wn)
print(lam, wn, zeta)

Auto-checked practice

1. Same aircraft with M_q = −0.5 s⁻¹ (weaker pitch damping): damping ratio ζ?

   Check Show a hint
   Full solution

   ζ = 0.309: below 0.35, so it falls out of Level 1 (pilots would feel the nose bobbing).

Practice

1. What does a positive real part of an eigenvalue mean physically?

   Show solution

   The disturbance grows exponentially: the aircraft is unstable in that mode.

References

• B. Etkin, L. D. Reid, Dynamics of Flight: Stability and Control, 3rd ed., Wiley, 1996.
• U.S. Department of Defense, MIL-STD-1797A, Flying Qualities of Piloted Aircraft, 1990 (approved for public release).
• MIT OpenCourseWare, 16.100 Aerodynamics (CC BY-NC-SA 4.0).

Rotorcraft: momentum and blade-element theory

• Undergraduate (freshman–senior)
• Graduate
• PhD / research

Track: Aeronautical Engineering · About 45 min · Numbers verified in build: 2026-10-16

Step by step

1. $$T=\dot m\,w,\qquad \dot m=\rho A v_i$$
   Momentum: thrust equals mass flow times the far-wake velocity w.
2. $$Tv_i=\tfrac12\dot m\,w^2\;\Rightarrow\;w=2v_i$$
   Energy: power at the disk equals kinetic energy flux in the wake.
3. $$v_i=\sqrt{\frac{T}{2\rho A}},\qquad P_{ideal}=Tv_i=\frac{T^{3/2}}{\sqrt{2\rho A}},\qquad FM=\frac{P_{ideal}}{P_{actual}}$$
   Induced velocity, ideal power, figure of merit.
4. $$dT=\tfrac12\rho\,(\Omega r)^2\,c\,a\,(\theta-\phi)\,N_b\,dr,\qquad \phi\approx\frac{v_i}{\Omega r}$$
   Blade-element theory: each strip of blade is a small airfoil at angle θ − φ, where the inflow angle φ comes from the induced velocity.
5. $$C_T=\frac{\sigma a}{2}\left(\frac{\theta_{0.75}}{3}-\frac{\lambda}{2}\right),\qquad \sigma=\frac{N_b c}{\pi R},\quad \lambda=\frac{v_i}{\Omega R}$$
   Integrate from root to tip with uniform inflow: thrust coefficient in terms of solidity σ, lift slope a and pitch at 75 % radius.
6. $$\lambda=\sqrt{C_T/2}\;\Rightarrow\;\lambda=\frac{\sigma a}{16}\left(\sqrt{1+\frac{64\,\theta_{0.75}}{3\sigma a}}-1\right)$$
   Combined blade-element momentum theory (BEMT): set the blade-element thrust equal to the momentum-theory thrust and solve the quadratic for λ.
7. $$C_P=\frac{C_T^{3/2}}{\sqrt2}+\frac{\sigma C_{d0}}{8},\qquad FM=\frac{C_T^{3/2}/\sqrt2}{C_P}$$
   Power = induced part + profile part from blade drag. This is why FM stays below 1.

import math
m, R, rho = 2700, 5.1, 1.225
T = m*9.80665; A = math.pi*R*R
vi = math.sqrt(T/(2*rho*A)); P = T*vi
print(T/A, vi, P/1000, P/1000/400)

Auto-checked practice

1. Same helicopter with R = 6.0 m: ideal hover power in kW?

   kW Check Show a hint
   Full solution

   A = 113.10 m²; P_ideal = 258.8 kW (vs 304.5 kW at R = 5.1 m).

2. Blade-element momentum theory, hover: solidity σ = 0.08, lift slope a = 5.73 /rad, blade pitch at 75 % radius 8°. What is the thrust coefficient C_T?

   Check Show a hint
   Full solution

   θ = 0.13963 rad, λ = 0.04980, C_T = 0.00496. With C_d0 = 0.011: C_P = 0.000357, FM = 0.692. At 10° pitch C_T = 0.00670 and FM = 0.779: a more heavily loaded rotor is more efficient.

Practice

1. Why can FM never reach 1 in reality?

   Show solution

   Real rotors also spend power on blade profile drag, non-uniform inflow, tip losses and swirl, which momentum theory ignores.

2. In BEMT, why do we quote blade pitch at 75 % radius?

   Show solution

   Thrust grows with r², so the outer blade dominates; for linearly twisted blades the pitch at 0.75R gives the same C_T as the full twist distribution under uniform inflow.

References

• J. G. Leishman, Principles of Helicopter Aerodynamics, 2nd ed., Cambridge University Press, 2006.
• NASA Glenn Research Center, Beginner’s Guide to Aeronautics.

CFD basics: finite differences, accuracy and stability

• Graduate
• PhD / research

Track: Aeronautical Engineering · About 45 min · Numbers verified in build: 2026-10-16

Step by step

1. $$f(x\pm h)=f\pm hf^{\prime}+\tfrac{h^2}{2}f^{\prime\prime}\pm\tfrac{h^3}{6}f^{\prime\prime\prime}+\dots$$
   Taylor expansions about x.
2. $$\frac{f(x+h)-f(x)}{h}=f^{\prime}+O(h),\qquad \frac{f(x+h)-f(x-h)}{2h}=f^{\prime}+O(h^2)$$
   Forward difference is first order; central is second order.
3. $$u_t=\nu u_{xx}:\quad u_j^{n+1}=u_j^n+r\left(u_{j+1}^n-2u_j^n+u_{j-1}^n\right),\quad r=\frac{\nu\Delta t}{\Delta x^2}\le\frac12$$
   Explicit scheme for diffusion; von Neumann analysis gives the stability limit.
4. $$\rho\left(\frac{\partial\mathbf u}{\partial t}+\mathbf u\cdot\nabla\mathbf u\right)=-\nabla p+\mu\nabla^2\mathbf u,\qquad \nabla\cdot\mathbf u=0$$
   Incompressible Navier–Stokes: the viscous term is this same diffusion operator.

import math
x, h = 1.0, 0.1
fwd = (math.sin(x+h) - math.sin(x)) / h
cen = (math.sin(x+h) - math.sin(x-h)) / (2*h)
print(fwd, cen, abs(fwd - math.cos(x)), abs(cen - math.cos(x)))

Auto-checked practice

1. Explicit diffusion scheme with ν = 1×10⁻⁵ m²/s and Δx = 1 mm: largest stable Δt, in seconds?

   s Check Show a hint
   Full solution

   Δt ≤ 0.5 · 10⁻⁶ / 10⁻⁵ = 0.05 s.

Practice

1. Halving Δx in the explicit diffusion scheme changes the maximum Δt by what factor?

   Show solution

   It is divided by 4, because Δt_max ∝ Δx². Implicit schemes avoid this limit.

References

• J. D. Anderson Jr., Computational Fluid Dynamics: The Basics with Applications, McGraw-Hill, 1995.
• J. D. Anderson Jr., Fundamentals of Aerodynamics, 6th ed., McGraw-Hill, 2017.
• MIT OpenCourseWare, 16.100 Aerodynamics (CC BY-NC-SA 4.0).

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