Exercises — Lift — Kutta-Joukowski theorem L = ρV∞Γ

What: direct plug-in of the K–J theorem. Why: we are given exactly the three quantities $\rho ,V_{\infty },\Gamma$ — nothing else is needed. $L^{\prime }=\rho V_{\infty }\Gamma =1.23\times 40\times 20=984\text{N/m}.$ Units check: $\frac{\text{kg}}{{\text{m}}^3}\cdot \frac{\text{m}}{\text{s}}\cdot \frac{{\text{m}}^2}{\text{s}}=\frac{\text{kg⋅m}}{{\text{s}}^2}\cdot \frac{1}{\text{m}}=\text{N/m}$ ✓

What: rearrange K–J to solve for the unknown. Why: $L^{\prime }$ depends linearly on $\Gamma$, so we just divide. $\Gamma =\frac{L^{\prime }}{\rho V_{\infty }}=\frac{1500}{1.23\times 50}=\frac{1500}{61.5}=24.39{\text{m}}^2/\text{s}.$

• $L^{\prime }$: $\text{N/m}$ (force per metre of span)
• $\rho$: ${\text{kg/m}}^3$
• $V_{\infty }$: $\text{m/s}$
• $\Gamma$: ${\text{m}}^2/\text{s}$

The prime (${}^{\prime }$) marks a per-unit-span quantity. A real wing's total lift is $L=L^{\prime }\times b$ where $b$ is the span.

Step 1 — convert to radians. Why: the thin-airfoil formula $\Gamma =\pi V_{\infty }c\alpha$ was derived with $\alpha$ measured in radians (it comes from a small-angle sine). $\alpha =4^{\circ }\times \frac{\pi }{180}=0.06981\text{rad}.$ Step 2 — circulation. $\Gamma =\pi V_{\infty }c\alpha =\pi \times 60\times 1.5\times 0.06981=19.74{\text{m}}^2/\text{s}.$ Step 3 — lift. $L^{\prime }=\rho V_{\infty }\Gamma =1.23\times 60\times 19.74=1457\text{N/m}.$

(a) Thin-airfoil prediction: $c_{\ell }=2\pi \alpha =2\pi (0.06981)=\mathrm{0.4386.}$ (b) From the computed lift: $\frac{1}{2}\rho V_{\infty }^{2}c=\frac{1}{2}(1.23)(60^2)(1.5)=\frac{1}{2}(1.23)(3600)(1.5)=3321\text{N/m}.$ $c_{\ell }=\frac{L^{\prime }}{3321}=\frac{1457}{3321}=\mathrm{0.4387.}$ They match ✓ — the tiny difference is rounding. This confirms $L^{\prime }=\rho V_{\infty }\Gamma$ and $L^{\prime }=\frac{1}{2}\rho V_{\infty }^{2}c{c}_{\ell }$ are the same statement seen two ways.

What: the Magnus Effect uses the same theorem — K–J does not care whether $\Gamma$ came from wing shape or from spin. $F^{\prime }=\rho V_{\infty }\Gamma =1.23\times 30\times 6=221.4\text{N/m}.$ The force is perpendicular to the ball's velocity (that's why the ball curves).

Key insight: at fixed $\alpha$ and $c$, the circulation itself scales with speed: $\Gamma =\pi V_{\infty }c\alpha \propto V_{\infty }$. So there are two hidden speed factors. $L^{\prime }=\rho V_{\infty }\Gamma =\rho V_{\infty }(\pi V_{\infty }c\alpha )=\pi \rho c\alpha V_{\infty }^2\propto V_{\infty }^2.$ Doubling $V_{\infty }$ multiplies $L^{\prime }$ by $2^2=\boxed{4}$. Why it's not just $\times 2$: the theorem is linear in $V_{\infty }$ and linear in $\Gamma$ — but $\Gamma$ also grew with speed, so the two multiply.

Wing A (clockwise, $\Gamma >0$): top flow speeds up, bottom slows down $\Rightarrow$ low pressure on top $\Rightarrow$ lift points upward. This is the normal lifting case. Wing B (counter-clockwise, $\Gamma <0$): everything mirrors — top slows, bottom speeds up $\Rightarrow$ high pressure on top $\Rightarrow$ force points downward (negative lift / down-force, like an upside-down wing on a race car). Why the sign flips cleanly: $L^{\prime }=\rho V_{\infty }\Gamma$ is odd in $\Gamma$ — reverse the swirl, reverse the force. The magnitude $|L^{\prime }|=\rho V_{\infty }|\Gamma |$ is identical.

Step 1 — how each factor changes.

• $\Gamma =\pi V_{\infty }c\alpha$: halving $\alpha$ halves $\Gamma$, factor $\times \frac{1}{2}$.
• $L^{\prime }=\rho V_{\infty }\Gamma$: density is halved, factor $\times \frac{1}{2}$.
• $V_{\infty }$, $c$ unchanged. Step 2 — combine (both act on $L^{\prime }$): $\frac{L_{\text{new}}^{\prime }}{L_{\text{old}}^{\prime }}=\underset{\rho }{\underbrace{\frac{1}{2}}}\times \underset{\alpha \Rightarrow \Gamma }{\underbrace{\frac{1}{2}}}=\frac{1}{4}.$ New lift is one-quarter of the original.

(a) Lift per unit span — total lift must equal weight, spread over span $b$: $L^{\prime }=\frac{W}{b}=\frac{8000}{8}=1000\text{N/m}.$ (b) Circulation — invert K–J: $\Gamma =\frac{L^{\prime }}{\rho V_{\infty }}=\frac{1000}{1.23\times 45}=\frac{1000}{55.35}=18.07{\text{m}}^2/\text{s}.$ (c) Angle of attack — invert the thin-airfoil relation $\Gamma =\pi V_{\infty }c\alpha$: $\alpha =\frac{\Gamma }{\pi V_{\infty }c}=\frac{18.07}{\pi \times 45\times 1.4}=\frac{18.07}{197.9}=0.09130\text{rad}.$ $\alpha =0.09130\times \frac{180}{\pi }=5.231^{\circ }.$ Chain of ideas used: weight → $L^{\prime }$ (per-span) → $\Gamma$ (K–J) → $\alpha$ (thin-airfoil). Each link is one formula.

Step 1: $c_{\ell }=2\pi \alpha =2\pi (0.09130)=0.5736$. Step 2: dynamic-pressure-times-chord: $\frac{1}{2}\rho V_{\infty }^{2}c=\frac{1}{2}(1.23)(45^2)(1.4)=\frac{1}{2}(1.23)(2025)(1.4)=1743.5\text{N/m}.$ Step 3: $L^{\prime }=1743.5\times 0.5736=1000\text{N/m}$ ✓ Why it must agree: $\frac{1}{2}\rho V_{\infty }^{2}c(2\pi \alpha )$ and $\rho V_{\infty }(\pi V_{\infty }c\alpha )$ are algebraically identical — both equal $\pi \rho V_{\infty }^{2}c\alpha$.

What: lift scales linearly with $\alpha$ (through $\Gamma$), so multiply the previous angle by the lift ratio. Step 1: new $L^{\prime }=11200/8=1400\text{N/m}$. Step 2: $L^{\prime }\propto \alpha$ at fixed $\rho ,V_{\infty },c$. So ${\alpha }_{\text{new}}={\alpha }_{\text{old}}\times \frac{1400}{1000}=5.231^{\circ }\times 1.4=7.323^{\circ }.$ Sanity: still a small angle (< $10^{\circ }$), so thin-airfoil theory remains valid.

The trap: the student treated $\Gamma$ as constant. But at fixed geometry $\Gamma =\pi V_{\infty }c\alpha \propto V_{\infty }$, so $\Gamma$ also doubles when speed doubles. $L^{\prime }=\pi \rho c\alpha V_{\infty }^2.$

• Density $\times \frac{1}{2}$: factor $\frac{1}{2}$.
• Speed $\times 2$ inside $V_{\infty }^2$: factor $2^2=4$. $\frac{L_{\text{new}}^{\prime }}{L_{\text{old}}^{\prime }}=\frac{1}{2}\times 4=2.$ The student is wrong — lift doubles, it does not stay the same.

Direction: a cross product $\vec{A}\times \vec{B}$ is always perpendicular to both $\vec{A}$ and $\vec{B}$. Since ${\vec{L}}^{\prime }=\rho {\vec{V}}_{\infty }\times \vec{\Gamma }$, the lift is perpendicular to ${\vec{V}}_{\infty }$ — i.e. at right angles to the wind, not to the ground. Magnitude (freestream and circulation-axis are perpendicular, so $\sin 90^{\circ }=1$): $|L^{\prime }|=\rho V_{\infty }\Gamma =1.20\times 70\times 22=1848\text{N/m}.$ d'Alembert: in ideal (inviscid, incompressible) 2-D flow the net force along ${\vec{V}}_{\infty }$ — the drag — is exactly zero. All of K–J lives in the perpendicular component. Real drag needs viscosity, which this theory omits.

(a) Myth's top speed — same travel time, longer path: $V_{\text{top}}=V_{\infty }\times \frac{1.05c}{c}=50\times 1.05=52.5\text{m/s}.$ (b) Bernoulli pressure difference (no height change): $\Delta p=\frac{1}{2}\rho \left(V_{\text{top}}^2-V_{\infty }^2\right)=\frac{1}{2}(1.23)(52.5^2-50^2).$ $52.5^2-50^2=2756.25-2500=\mathrm{256.25.}$ $\Delta p=\frac{1}{2}(1.23)(256.25)=157.6\text{Pa},L^{\prime }=\Delta p\times c=157.6\text{N/m}.$ (c) The lesson: real wings of this size lift ten-plus times more than $157.6\text{N/m}$. So the equal-transit assumption under-predicts lift badly — proving air does not rejoin (top air actually arrives ahead). Lift is set by the Kutta Condition-selected circulation, then delivered by $L^{\prime }=\rho V_{\infty }\Gamma$, not by any rejoining rule.

Did the number come out in $\text{N/m}$? ::: Yes if you used $\rho V_{\infty }\Gamma$ and never multiplied by span. Was every angle in radians before entering $\Gamma =\pi V_{\infty }c\alpha$? ::: Convert with $\times \pi /180$ first. If geometry was fixed, did $\Gamma$ ride along with $V_{\infty }$? ::: Then $L^{\prime }\propto V_{\infty }^2$, not $V_{\infty }$. Did you divide by span exactly once (only to get $L^{\prime }$ from total $L$)? ::: One division, no more, no less.