§AERE 355 Homework 1

Sep 25, 2025
Flying wing, sea level, steady flight.
W=3000lb W = 3000 lb
S=150ft2 S = 150ft^2
h=0.14 h = 0.14
hac=0.26 h_{ac} = 0.26
cˉ=5ft \bar{c} = 5ft
Cmac=0.015 C_{m_{ac}} = 0.015
CLα=0.08deg⁡−1 C_{L_\alpha} = 0.08 \deg^{-1}
Cmα=CLα(h−hac) C_{m_\alpha} = C_{L_\alpha} (h - h_{ac})
Cmα=0.08deg⁡−1(0.14−0.26)=−0.0096deg⁡−1 C_{m_\alpha} = 0.08 \deg^{-1} (0.14 - 0.26) = \boxed{-0.0096 \deg^{-1}}
Cm=Cmac+Cmαα C_m = C_{m_{ac}} + C_{m_\alpha} \alpha
α=Cm−CmacCmα \alpha = \frac{C_m - C_{m_{ac}}}{C_{m_\alpha}}
Trim flight:
α=Cm−CmacCmα=−CmacCmα=−0.015−0.0096deg⁡−1=1.56deg⁡ \alpha = \frac{\cancel{C_m} - C_{m_{ac}}}{C_{m_\alpha}} = -\frac{C_{m_{ac}}}{C_{m_\alpha}} = -\frac{0.015}{-0.0096 \deg^{-1}} = \boxed{1.56 \deg}

§b.

Cm=0=Cmac+CL(h−hac) C_m = 0 = C_{m_{ac}} + C_L (h - h_{ac})
CL=−Cmach−hac=−0.0150.14−0.26=0.125 C_L = - \frac{C_{m_{ac}}}{h - h_{ac}} = - \frac{0.015}{0.14 - 0.26} = 0.125
Sea level:
ρ=2.3769∗10−3slug/ft3 \rho = 2.3769 * 10^{-3} slug/ft^3
W=L=CLqS W = L = C_L q S
q=12ρV2 q = \frac{1}{2} \rho V^2
W=12CLSρV2 W = \frac{1}{2} C_L S \rho V^2
V=2WCLSρ=2∗3000lb0.125∗150ft2∗2.3769∗10−3slug/ft3 V = \sqrt{\frac{2 W}{C_L S \rho}} = \sqrt{\frac{2 * 3000 lb}{0.125 * 150ft^2 * 2.3769 * 10^{-3} slug/ft^3}}
V=366.92ft/s V = \boxed{366.92 ft/s}

§c.

Wp=500lb W_p = 500lb
xp=xcgold+Δ x_p = x_{cg}^{old} + \Delta
The new center of gravity will just be the weighted average:
xcgnew=Wxcgold+WpxpW+Wp x_{cg}^{new} = \frac{W x_{cg}^{old} + W_p x_p}{W + W_p}
xcgnew=Wxcgold+Wp(xcgold+Δ)W+Wp x_{cg}^{new} = \frac{W x_{cg}^{old} + W_p (x_{cg}^{old} + \Delta)}{W + W_p}
xcgnew=Wxcgold+Wpxcgold+WpΔW+Wp x_{cg}^{new} = \frac{W x_{cg}^{old} + W_p x_{cg}^{old} + W_p \Delta}{W + W_p}
xcgnew=(W+Wp)xcgold+WpΔW+Wp x_{cg}^{new} = \frac{(W + W_p) x_{cg}^{old} + W_p \Delta}{W + W_p}
xcgnew=(W+Wp)xcgoldW+Wp+WpΔW+Wp x_{cg}^{new} = \frac{\cancel{(W + W_p)} x_{cg}^{old}}{\cancel{W + W_p}} + \frac{W_p \Delta}{W + W_p}
xcgnew=xcgold+WpW+WpΔ \boxed{x_{cg}^{new} = x_{cg}^{old} + \frac{W_p}{W + W_p} \Delta}
Alternatively, with the numbers (units cancel out):
xcgnew=xcgold+5003000+500Δ \boxed{x_{cg}^{new} = x_{cg}^{old} + \frac{500}{3000 + 500} \Delta}

§d.

Neutral stability:
Cmac=Cm=0 C_{m_{ac}} = C_m = 0
Cm=Cmac+CL(hnew−hac) \cancel{C_m} = \cancel{C_{m_{ac}}} + C_L (h^{new} - h_{ac})
0=CL(hnew−hac) 0 = C_L (h^{new} - h_{ac})
hnew=hac=0.26 h^{new} = h_{ac} = 0.26
Recovering xx:
xcgold=cˉh=5ft∗0.14=0.7ft x_{cg}^{old} = \bar{c} h = 5ft * 0.14 = 0.7ft
xcgnew=cˉhnew=5ft∗0.26=1.3ft x_{cg}^{new} = \bar{c} h^{new} = 5ft * 0.26 = 1.3ft
Solved for Δ\Delta:
Δ=(xcgnew−xcgold)(W+Wp)Wp \Delta = \frac{(x_{cg}^{new} - x_{cg}^{old})(W + W_p)}{W_p}
Δ=(1.3ft−0.7ft)(3000lb+500lb)500lb=4.2ft \Delta = \frac{(1.3ft - 0.7ft)(3000lb + 500lb)}{500lb} = \boxed{4.2ft}

§2.

CLα=4.44 C_{L_\alpha} = 4.44
CL=0.41 C_L = 0.41
CL=CLαα+CLδeδe=CLαα C_L = C_{L_\alpha} \alpha + \cancel{C_{L_{\delta_e}} \delta_e} = C_{L_\alpha} \alpha
α=CLCLα=0.414.44=0.0923rad=5.291° \alpha = \frac{C_L}{C_{L_\alpha}} = \frac{0.41}{4.44} = 0.0923rad = \boxed{5.291\degree}
α0L=−5° \alpha_{0_L} = -5\degree
αgeo=α−α0L=5.291°−(−5°)=10.291° \alpha_{geo} = \alpha - \alpha_{0_L} = 5.291\degree - (-5\degree) = \boxed{10.291\degree}

§b.

V=176ft/s V = 176 ft/s
W=2750lb W = 2750lb
S=184ft2 S = 184 ft^2
ρ=2.3769∗10−3slug/ft3 \rho = 2.3769 * 10^{-3} slug/ft^3
q=12ρV2=12∗2.3769∗10−3slug/ft3∗(176ft/s)2=36.81lb/ft2 q = \frac{1}{2} \rho V^2 = \frac{1}{2} * 2.3769 * 10^{-3} slug/ft^3 * (176 ft/s)^2 = 36.81 lb/ft^2
L=W=CLqS L = W = C_L q S
CL=WqS=2750lb36.81lb/ft2∗184ft2=0.406 C_L = \frac{W}{q S} = \frac{2750lb}{36.81 lb/ft^2 * 184 ft^2} = \boxed{0.406}
That's pretty close to 0.410.41 from the table.

§c.

Target:
Cmδe=−ltcˉCLδe C_{m_{\delta_e}} = - \frac{l_t}{\bar{c}} C_{L_{\delta_e}}
Identities:
VH=ltStcˉS V_H = \frac{l_t S_t}{\bar{c} S}
CLδe=ητStSCLαt C_{L_{\delta_e}} = \eta \tau \frac{S_t}{S} C_{L_{\alpha_t}}
Cmδe=−ητVHCLαt C_{m_{\delta_e}} = - \eta \tau V_H C_{L_{\alpha_t}}
Derivation:
CLαt=CLδeητSSt C_{L_{\alpha_t}} = \frac{C_{L_{\delta_e}}}{\eta \tau} \frac{S}{S_t}
Cmδe=−ητltStcˉSCLδeητSSt C_{m_{\delta_e}} = - \cancel{\eta \tau} \frac{l_t \cancel{S_t}}{\bar{c} \cancel{S}} \frac{C_{L_{\delta_e}}}{\cancel{\eta \tau}} \frac{\cancel{S}}{\cancel{S_t}}
Cmδe=−ltcˉCLδe \boxed{C_{m_{\delta_e}} = - \frac{l_t}{\bar{c}} C_{L_{\delta_e}}}

§d.

Cmδe=−ltcˉCLδe C_{m_{\delta_e}} = - \frac{l_t}{\bar{c}} C_{L_{\delta_e}}
lt=−CmδeCLδecˉ l_t = - \frac{C_{m_{\delta_e}}}{C_{L_{\delta_e}}} \bar{c}
lt=−−0.9230.3555.7ft=14.82ft l_t = - \frac{-0.923}{0.355} 5.7ft = \boxed{14.82ft}
Coming from part (b), I would have expected these values to be much closer than what I got. The actual value is 16ft16ft, so it's in the ballpark, but not exactly right.

§3.

dCmdCL=−0.15 \frac{d C_m}{dC_L} = -0.15
Cm(CL=0)=0.08 C_m(C_L = 0) = 0.08
Xcg/cˉ=h=0.3 X_{cg} / \bar c = h = 0.3
hn=h−(dCm/dCL)=0.3−(−0.15)=0.45 h_n = h - (d C_m / d C_L) = 0.3 - (-0.15) = \boxed{0.45}

§4.

W=2750lb W = 2750lb
S=180ft2 S = 180ft^2
cg=0.25cˉ  ⟹  h=0.25 cg = 0.25 \bar c \implies h = 0.25

§a.

Two points following the line for δe=−15°\delta_e = -15\degree:
(CL,Cm)0=(0.0,0.15) (C_L, C_m)_0 = (0.0, 0.15)
(CL,Cm)1=(1.0,0.0) (C_L, C_m)_1 = (1.0, 0.0)
dCmdCL=y1−y0x1−x0=0.0−0.151.0−0.0=−0.15 \frac{d C_m}{d C_L} = \frac{y_1 - y_0}{x_1 - x_0} = \frac{0.0 - 0.15}{1.0 - 0.0} = -0.15
hn=h−(dCm/dCL)=0.25−(−0.15)=0.4 h_n = h - (d C_m / d C_L) = 0.25 - (-0.15) = \boxed{0.4}

§b.

V=125ft/s V = 125ft/s
ρ=2.3769∗10−3slug/ft3 \rho = 2.3769 * 10^{-3} slug/ft^3
q=12ρV2=12∗2.3769∗10−3slug/ft3∗(125ft/s)2=0.129psi q = \frac{1}{2} \rho V^2 = \frac{1}{2} * 2.3769 * 10^{-3} slug/ft^3 * (125ft/s)^2 = 0.129psi
W=L=CLqS W = L = C_L q S
CL=WqS=2750lb0.129psi∗180ft2=0.8224 C_L = \frac{W}{q S} = \frac{2750lb}{0.129psi * 180ft^2} = \boxed{0.8224}
For the CLC_L to be 0.82240.8224 and CmC_m to be 00, we would need:
δe≈−12° \boxed{\delta_e \approx -12\degree}