§AERE 310 Homework 5

Sep 25, 2025

§1.

yu=16xc(1−xc) mm y_u = 16 \frac{x}{c} (1 - \frac{x}{c}) ~ \text{mm}
yl=8xc(1−xc) mm y_l = 8 \frac{x}{c} (1 - \frac{x}{c}) ~ \text{mm}
c=4 cm=40 mm c = 4 ~ \text{cm} = 40 ~ \text{mm}
The average of both surfaces should be the camber line:
ηc=yu+yl2=12[16xc(1−xc)+8xc(1−xc)]=12xc(1−xc) \eta_c = \frac{y_u + y_l}{2} = \frac{1}{2} \left [ 16 \frac{x}{c} (1 - \frac{x}{c}) + 8 \frac{x}{c} (1 - \frac{x}{c}) \right] = 12 \frac{x}{c} (1 - \frac{x}{c})
For ease of differentiation, I am declaring:
x′=xc x' = \frac{x}{c}
dx′dx=1c \frac{d x'}{dx} = \frac{1}{c}
ηc=12x′(1−x′) \eta_c = 12 x' (1 - x')
dηcdx′=12(1−x′)+12x′(−1)=12−12x′−12x′=12−24x′ \frac{d \eta_c}{dx'} = 12 (1 - x') + 12x'(-1) = 12 - 12x' - 12x' = 12 - 24x'
dηcdx=dηcdx′dx′dx=(12−24x′)1c=(12−24xc)1c=12c−24c2x \frac{d \eta_c}{dx} = \frac{d \eta_c}{dx'} \frac{d x'}{dx} = (12 - 24x') \frac{1}{c} = (12 - 24 \frac{x}{c}) \frac{1}{c} = \frac{12}{c} - \frac{24}{c^2} x
x=c2(1−cos⁡θ) x = \frac{c}{2} (1 - \cos \theta)
dηcdx=12c−24c2c2(1−cos⁡θ)=12c−12c(1−cos⁡θ) \frac{d \eta_c}{dx} = \frac{12}{c} - \frac{24}{c^2} \frac{c}{2} (1 - \cos \theta) = \frac{12}{c} - \frac{12}{c} (1 - \cos \theta)
dηcdx=12c(1−1+cos⁡θ)=12ccos⁡θ \frac{d \eta_c}{dx} = \frac{12}{c} (1 - 1 + \cos \theta) = \frac{12}{c} \cos \theta
dηcdx=A0+A1cos⁡θ+A2cos⁡2θ \frac{d \eta_c}{dx} = A_0 + A_1 \cos \theta + A_2 \cos 2 \theta
  ⟹  A0=0, A1=12c, A2=0 \implies A_0 = 0, ~ A_1 = \frac{12}{c}, ~ A_2 = 0
αL=0=A0−A12=1212c=6c \alpha_{L=0} = \cancel{A_0} - \frac{A_1}{2} = \frac{1}{2} \frac{12}{c} = \frac{6}{c}
cl=2π(α−αL=0)=2π(α−6c)=2π(α−640)=2π(α−320) c_l = 2 \pi (\alpha - \alpha_{L=0}) = 2 \pi (\alpha - \frac{6}{c}) = 2 \pi (\alpha - \frac{6}{40}) = \boxed{2 \pi (\alpha - \frac{3}{20})}
cmc/4=π4(A2−A1)=−π4A1=−π412c=−π41240=−3π40 c_{m_{c/4}} = \frac{\pi}{4} (\cancel{A_2} - A_1) = -\frac{\pi}{4} A_1 = -\frac{\pi}{4} \frac{12}{c} = -\frac{\pi}{4} \frac{12}{40} = \boxed{-\frac{3 \pi}{40}}

§2.

αL=0=−3°=−0.05236 \alpha_{L=0} = -3\degree = -0.05236
a=0.11°=5.730 a = \frac{0.1}{1\degree} = 5.730
cL=a(α−αL=0) c_L = a (\alpha - \alpha_{L=0})

§2. (a)

α=5°=0.08727 \alpha = 5\degree = 0.08727
cL=5.730∗(0.08727−(−0.05236))=0.8001 c_L = 5.730 * (0.08727 - (-0.05236)) = \boxed{0.8001}

§2. (b)

The slope's the same but now the intercept is opposite:
αL=0=3°=0.05236 \alpha_{L=0} = 3\degree = 0.05236
α=−5°=−0.08727 \alpha = -5\degree = -0.08727
cL=5.730∗(−0.08727−0.05236)=−0.8001 c_L = 5.730 * (-0.08727 - 0.05236) = \boxed{-0.8001}

§2. (c)

Still, the intercept is different here.
cL(5°)=0.8001=a(α−αL=0) c_L(5\degree) = 0.8001 = a (\alpha - \alpha_{L=0})
α=cL(5°)a+αL=0=0.80015.730+0.05236=0.1920=11.00° \alpha = \frac{c_L(5\degree)}{a} + \alpha_{L=0} = \frac{0.8001}{5.730} + 0.05236 = 0.1920 = \boxed{11.00\degree}

§3.

ηcc={0.25[0.8xc−(xc)2]for 0≤xc≤0.40.111[0.2+0.8xc−(xc)2]for 0.4≤xc≤1 \frac{\eta_c}{c} = \begin{cases} 0.25 \left[ 0.8 \frac{x}{c} - (\frac{x}{c})^2 \right] & \text{for } 0 \leq \frac{x}{c} \leq 0.4 \\ 0.111 \left[ 0.2 + 0.8 \frac{x}{c} - (\frac{x}{c})^2 \right] & \text{for } 0.4 \leq \frac{x}{c} \leq 1 \end{cases}

§3. (a)

αL=0=3°=0.05236 \alpha_{L=0} = 3\degree = 0.05236
ηc={0.25[0.8x−1cx2]for 0≤x≤0.4c0.111[0.2c+0.8x−1cx2]for 0.4c≤x≤c \eta_c = \begin{cases} 0.25 \left[ 0.8 x - \frac{1}{c}x^2 \right] & \text{for } 0 \leq x \leq 0.4c \\ 0.111 \left[ 0.2c + 0.8x - \frac{1}{c}x^2 \right] & \text{for } 0.4c \leq x \leq c \end{cases}
dηcdx={0.25[0.8−2cx]for 0≤x≤0.4c0.111[0.8−2cx]for 0.4c≤x≤c \frac{d \eta_c}{dx} = \begin{cases} 0.25 \left[ 0.8 - \frac{2}{c}x \right] & \text{for } 0 \leq x \leq 0.4c \\ 0.111 \left[ 0.8 - \frac{2}{c}x \right] & \text{for } 0.4c \leq x \leq c \end{cases}
dηcdx={0.2−0.5cxfor 0≤x≤0.4c0.0888−0.222cxfor 0.4c≤x≤c \frac{d \eta_c}{dx} = \begin{cases} 0.2 - \frac{0.5}{c}x & \text{for } 0 \leq x \leq 0.4c \\ 0.0888 - \frac{0.222}{c}x & \text{for } 0.4c \leq x \leq c \end{cases}
The first bound:
x=c2(1−cos⁡θ) x = \frac{c}{2} (1 - \cos \theta)
x=0  ⟹  0=c2(1−cos⁡θ) x = 0 \implies 0 = \frac{c}{2} (1 - \cos \theta)
0=1−cos⁡θ 0 = 1 - \cos \theta
cos⁡θ=1  ⟹  θ=0 \cos \theta = 1 \implies \theta = 0
The second bound:
x=0.4c  ⟹  0.4c=c2(1−cos⁡θ) x = 0.4c \implies 0.4\cancel{c} = \frac{\cancel{c}}{2} (1 - \cos \theta)
0.8=1−cos⁡θ  ⟹  θ=1.369 0.8 = 1 - \cos \theta \implies \theta = 1.369
The third bound:
x=c  ⟹  c=c2(1−cos⁡θ) x = c \implies \cancel{c} = \frac{\cancel{c}}{2} (1 - \cos \theta)
2=1−cos⁡θ  ⟹  θ=π 2 = 1 - \cos \theta \implies \theta = \pi
dηcdx={0.2−0.5cc2(1−cos⁡θ)for 0≤θ≤1.3690.0888−0.222cc2(1−cos⁡θ)for 1.369≤x≤π \frac{d \eta_c}{dx} = \begin{cases} 0.2 - \frac{0.5}{c} \frac{c}{2} (1 - \cos \theta) & \text{for } 0 \leq \theta \leq 1.369 \\ 0.0888 - \frac{0.222}{c} \frac{c}{2} (1 - \cos \theta) & \text{for } 1.369 \leq x \leq \pi \end{cases}
dηcdx={0.25cos⁡θ−0.05for 0≤θ≤1.3690.111cos⁡θ−0.0222for 1.369≤x≤π \frac{d \eta_c}{dx} = \begin{cases} 0.25 \cos \theta - 0.05 & \text{for } 0 \leq \theta \leq 1.369 \\ 0.111 \cos \theta - 0.0222 & \text{for } 1.369 \leq x \leq \pi \end{cases}
To reduce the syntax insanity, I am declaring:
dηcdx={HAfor 0≤θ≤1.369HBfor 1.369≤x≤π \frac{d \eta_c}{dx} = \begin{cases} H_A & \text{for } 0 \leq \theta \leq 1.369 \\ H_B & \text{for } 1.369 \leq x \leq \pi \end{cases}
Where:
HA=0.25cos⁡θ−0.05 H_A = 0.25 \cos \theta - 0.05
HB=0.111cos⁡θ−0.0222 H_B = 0.111 \cos \theta - 0.0222
Thus:
A0=1π∫0πdηcdxdθ=1π[∫01.369HA dθ+∫1.369πHB dθ] A_0 = \frac{1}{\pi} \int_0^\pi \frac{d \eta_c}{dx} d\theta = \frac{1}{\pi} \left [ \int_0^{1.369} H_A ~ d\theta + \int_{1.369}^\pi H_B ~ d\theta \right]
∫01.369HA dθ=∫01.3690.25cos⁡θ−0.05 dθ=0.1765 \int_0^{1.369} H_A ~ d\theta = \int_0^{1.369} 0.25 \cos \theta - 0.05 ~ d\theta = 0.1765
∫1.369πHB dθ=∫1.369π0.111cos⁡θ−0.0222 dθ=−0.1481 \int_{1.369}^\pi H_B ~ d\theta = \int_{1.369}^\pi 0.111 \cos \theta - 0.0222 ~ d\theta = -0.1481
A0=1π[0.1765−0.1481]=9.040×10−3 A_0 = \frac{1}{\pi} [ 0.1765 - 0.1481] = 9.040 \times 10^{-3}
Moreover:
A1=2π∫0πdηcdxcos⁡θ dθ=2π[∫01.369HAcos⁡θ dθ+∫1.369πHBcos⁡θ dθ] A_1 = \frac{2}{\pi} \int_0^\pi \frac{d \eta_c}{dx} \cos \theta ~ d\theta = \frac{2}{\pi} \left[ \int_0^{1.369} H_A \cos \theta ~ d\theta + \int_{1.369}^\pi H_B \cos \theta ~ d\theta \right]
∫01.369HAcos⁡θ dθ=∫01.369(0.111cos⁡θ−0.0222)cos⁡θ dθ=0.06513 \int_0^{1.369} H_A \cos \theta ~ d\theta = \int_0^{1.369} (0.111 \cos \theta - 0.0222) \cos \theta ~ d\theta = 0.06513
∫1.369πHBcos⁡θ dθ=∫1.369π(0.111cos⁡θ−0.0222)cos⁡θ dθ=0.1092 \int_{1.369}^\pi H_B \cos \theta ~ d\theta = \int_{1.369}^\pi (0.111 \cos \theta - 0.0222) \cos \theta ~ d\theta = 0.1092
A1=2π[0.06513+0.1092]=0.1110 A_1 = \frac{2}{\pi} \left[ 0.06513 + 0.1092 \right] = 0.1110
Finally:
αL=0=A0−A12=9.040×10−3−0.11102=−0.04646 \alpha_{L=0} = A_0 - \frac{A_1}{2} = 9.040 \times 10^{-3} - \frac{0.1110}{2} = -0.04646
cl=2π(α−αL=0)=2π(0.05236−(−0.04646))=0.6209 c_l = 2\pi (\alpha - \alpha_{L=0}) = 2\pi (0.05236 - (-0.04646)) = \boxed{0.6209}

§3. (b)

αL=0=3°=0.05236 \alpha_{L=0} = 3\degree = 0.05236
A2=2π∫0πdηcdxcos⁡2θ dθ=2π[∫01.369HAcos⁡2θ dθ+∫1.369πHBcos⁡2θ dθ] A_2 = \frac{2}{\pi} \int_0^\pi \frac{d \eta_c}{dx} \cos 2\theta ~ d\theta = \frac{2}{\pi} \left[ \int_0^{1.369} H_A \cos 2\theta ~ d\theta + \int_{1.369}^\pi H_B \cos 2\theta ~ d\theta \right]
∫01.369HAcos⁡2θ dθ=∫01.369(0.25cos⁡θ−0.05)cos⁡2θ dθ=0.0784 \int_0^{1.369} H_A \cos 2\theta ~ d\theta = \int_0^{1.369} (0.25 \cos \theta - 0.05) \cos 2\theta ~ d\theta = 0.0784
∫1.369πHBcos⁡2θ dθ=∫1.369π(0.111cos⁡θ−0.0222)cos⁡2θ dθ=−0.03480 \int_{1.369}^\pi H_B \cos 2\theta ~ d\theta = \int_{1.369}^\pi (0.111 \cos \theta - 0.0222) \cos 2\theta ~ d\theta = -0.03480
A2=2π[0.0784−0.03480]=0.02776 A_2 = \frac{2}{\pi} \left[ 0.0784 - 0.03480 \right] = 0.02776
cmc/4=π4(A2−A1)=π4(0.02776−0.1110)=−0.06538 c_{m_{c/4}} = \frac{\pi}{4} (A_2 - A_1) = \frac{\pi}{4} (0.02776 - 0.1110) = \boxed{-0.06538}
xcp=c4[1+π(A1−A2)cl] x_{cp} = \frac{c}{4} \left[ 1 + \frac{\pi (A_1 - A_2)}{c_l} \right]
xcp/c=14[1+π(A1−A2)cl] x_{cp}/c = \frac{1}{4} \left[ 1 + \frac{\pi (A_1 - A_2)}{c_l} \right]
xcp/c=14[1+π(0.1110−0.02776)0.6209]=0.3553 x_{cp}/c = \frac{1}{4} \left[ 1 + \frac{\pi (0.1110 - 0.02776)}{0.6209} \right] = \boxed{0.3553}

§3. (c)

Pretty darn close.

§3. (d)

h=3km h = 3km
V∞=60m/s V_\infty = 60m/s
c=2m c = 2m
ρ=9.0925×10−1kgm3 \rho = 9.0925 \times 10^{-1} \frac{kg}{m^3}
q=12ρV∞2=12∗9.0925×10−1kgm3∗(60m/s)2=1.637kPa q = \frac{1}{2} \rho V_\infty^2 = \frac{1}{2} * 9.0925 \times 10^{-1} \frac{kg}{m^3} * (60m/s)^2 = 1.637kPa
L′=clqc=0.6209∗1.637kPa∗2m=2.033kN/m L' = c_l q c = 0.6209 * 1.637kPa * 2m = \boxed{2.033 kN/m}

§4.

The piecewise function:
ηc={ηAfor 0≤x≤0.1cηBfor 0.1c≤x≤0.7cηCfor 0.7c≤x≤c \eta_c = \begin{cases} \eta_A & \text{for } 0 \leq x \leq 0.1c \\ \eta_B & \text{for } 0.1c \leq x \leq 0.7c \\ \eta_C & \text{for } 0.7c \leq x \leq c \\ \end{cases}
The pieces:
ηA=0.02c0.1cx−0.02c=0.2x−0.02c \eta_A = \frac{0.02c}{0.1c}x - 0.02c = 0.2x - 0.02c
ηB=0 \eta_B = 0
ηC=−0.05c0.3c(x−0.7c)=0.1167c−0.1667x \eta_C = \frac{-0.05c}{0.3c}(x - 0.7c) = 0.1167 c - 0.1667 x
Derivatives:
HA=dηadx=0.2 H_A = \frac{d \eta_a}{dx} = 0.2
HB=dηbdx=0 H_B = \frac{d \eta_b}{dx} = 0
HC=dηcdx=−0.1667 H_C = \frac{d \eta_c}{dx} = -0.1667
Bounds in polar:
x=0  ⟹  θ=0 x = 0 \implies \theta = 0
x=0.1c=c2(1−cos⁡θ)  ⟹  x=0.6435 x = 0.1c = \frac{c}{2} (1 - \cos \theta) \implies x = 0.6435
x=0.7c=c2(1−cos⁡θ)  ⟹  x=1.982 x = 0.7c = \frac{c}{2} (1 - \cos \theta) \implies x = 1.982
x=c  ⟹  θ=π x = c \implies \theta = \pi
dηcdx={0.2for 0≤θ≤0.64350for 0.6435≤θ≤1.982−0.1667for 1.982≤θ≤π \frac{d\eta_c}{dx} = \begin{cases} 0.2 & \text{for } 0 \leq \theta \leq 0.6435 \\ 0 & \text{for } 0.6435 \leq \theta \leq 1.982 \\ -0.1667 & \text{for } 1.982 \leq \theta \leq \pi \\ \end{cases}

§4. (a)

A0=1π∫0πdηcdxdθ=1π[0.2(0.6435−0)+0(1.982−0.6435)−0.1667(π−1.982)]=−0.02056 A_0 = \frac{1}{\pi} \int_0^\pi \frac{d \eta_c}{dx} d\theta = \frac{1}{\pi} \left[ 0.2(0.6435 - 0) + 0(1.982 - 0.6435) - 0.1667(\pi - 1.982) \right] = -0.02056
A1=2π∫0πdηcdxcos⁡θ dθ=2π[∫00.64350.2cos⁡θ dθ+∫0.64351.9820cos⁡θ dθ−∫1.982π0.1667cos⁡θ dθ] A_1 = \frac{2}{\pi} \int_0^\pi \frac{d \eta_c}{dx} \cos \theta ~ d\theta = \frac{2}{\pi} \left[ \int_0^{0.6435} 0.2 \cos \theta ~ d\theta + \cancel{\int_{0.6435}^{1.982} 0 \cos \theta ~ d\theta} - \int_{1.982}^\pi 0.1667 \cos \theta ~ d\theta \right]
A1=2π[0.12−(−0.152804)]=0.1737 A_1 = \frac{2}{\pi}[0.12 - (-0.152804)] = 0.1737
αL=0=A0−A12=−0.02056−0.17372=−0.1074=−6.154° \alpha_{L=0} = A_0 - \frac{A_1}{2} = -0.02056 - \frac{0.1737}{2} = -0.1074 = \boxed{-6.154\degree}

§4. (b)

ρ=ρ∞=1.2kgm3 \rho = \rho_\infty = 1.2\frac{kg}{m^3}
V∞=50m/s V_\infty = 50m/s
α=2°=0.03491 \alpha = 2\degree = 0.03491
c=2m c = 2m
cl=2π(α−αL=0)=2π(0.03491−(−0.1074))=0.8942 c_l = 2\pi (\alpha - \alpha_{L=0}) = 2\pi (0.03491 - (-0.1074)) = 0.8942
q=12ρV∞2=12∗1.2kgm3∗(50m/s)2=1.5kPa q = \frac{1}{2} \rho V_\infty^2 = \frac{1}{2} * 1.2\frac{kg}{m^3} * (50m/s)^2 = 1.5kPa
L′=clqc=0.8942∗1.5kPa∗2m=2.683kN/m L' = c_l q c = 0.8942 * 1.5kPa * 2m = \boxed{2.683kN/m}
A2=2π∫0πdηcdxcos⁡2θ dθ=2π[∫00.64350.2cos⁡2θ dθ+∫0.64351.9820cos⁡2θ dθ−∫1.982π0.1667cos⁡2θ dθ] A_2 = \frac{2}{\pi} \int_0^\pi \frac{d \eta_c}{dx} \cos 2\theta ~ d\theta = \frac{2}{\pi} \left[ \int_0^{0.6435} 0.2 \cos 2\theta ~ d\theta + \cancel{\int_{0.6435}^{1.982} 0 \cos 2\theta ~ d\theta} - \int_{1.982}^\pi 0.1667 \cos 2\theta ~ d\theta \right]
A2=2π[0.09600−0.06108]=0.02223 A_2 = \frac{2}{\pi} \left[ 0.09600 - 0.06108 \right] = 0.02223
cmc/4=π4(A2−A1)=π4(0.02223−0.1737)=−0.1190 c_{m_{c/4}} = \frac{\pi}{4} (A_2 - A_1) = \frac{\pi}{4} (0.02223 - 0.1737) = -0.1190
cmLE=cmc/4−14cl=−0.1190−14∗0.8942=−0.3426 c_{m_{LE}} = c_{m_{c/4}} - \frac{1}{4}c_l = -0.1190 - \frac{1}{4} * 0.8942 = -0.3426
MLE′=qc2cmLE=1.5kPa∗(2m)2∗−0.3426=−2.06kN M_{LE}' = q c^2 c_{m_{LE}} = 1.5kPa * (2m)^2 * -0.3426 = \boxed{-2.06kN}