§AERE 261 Homework 5

Sep 25, 2025

§1.

T=D T=D
L=W L=W
TR=WCL/CD=WCDCL T_R=\frac{W}{C_L / C_D}=\frac{W C_D}{C_L}
CL=Lq∞S=Wq∞S C_L=\frac{L}{q_\infty S}=\frac{W}{q_\infty S}
TR=WCDWq∞S=WCDq∞SW=CDq∞S T_R=\frac{W C_D}{\frac{W}{q_\infty S}}=\frac{W C_D q_\infty S}{W}=C_D q_\infty S
CD=CD,0+KCL2=CD,0+K(Wq∞S)2=CD,0+KW2q∞2S2 C_D=C_{D,0}+K C_L^2=C_{D,0}+K \left(\frac{W}{q_\infty S}\right)^2=C_{D,0}+\frac{KW^2}{q_\infty^2 S^2}
TR=(CD,0+KW2q∞2S2)q∞S=q∞SCD,0+q∞SKW2q∞2S2=q∞SCD,0+KW2q∞S T_R=\left(C_{D,0}+\frac{KW^2}{q_\infty^2 S^2}\right) q_\infty S=q_\infty SC_{D,0}+\frac{q_\infty SKW^2}{q_\infty^2 S^2}=q_\infty SC_{D,0}+\frac{KW^2}{q_\infty S}
q∞=12ρ∞V∞2 q_\infty=\frac{1}{2} \rho_\infty V_\infty^2
TR=12ρ∞V∞2SCD,0+KW212ρ∞V∞2S=12ρ∞V∞2SCD,0+2KW2ρ∞V∞2S T_R=\frac{1}{2} \rho_\infty V_\infty^2 SC_{D,0}+\frac{KW^2}{\frac{1}{2} \rho_\infty V_\infty^2 S}=\boxed{\frac{1}{2} \rho_\infty V_\infty^2 SC_{D,0}+\frac{2KW^2}{\rho_\infty V_\infty^2 S}}

§2.

TR=12ρ∞V∞2SCD,0+2KW2ρ∞V∞2S T_R=\frac{1}{2} \rho_\infty V_\infty^2 SC_{D,0}+\frac{2KW^2}{\rho_\infty V_\infty^2 S}
∂TR∂V∞=∂∂V∞(12ρ∞V∞2SCD,0+2KW2ρ∞V∞2S) \frac{\partial T_R}{\partial V_\infty}=\frac{\partial}{\partial V_\infty}\left(\frac{1}{2} \rho_\infty V_\infty^2 SC_{D,0}+\frac{2KW^2}{\rho_\infty V_\infty^2 S}\right)
=∂∂V∞(12ρ∞V∞2SCD,0)+∂∂V∞(2KW2ρ∞V∞2S) =\frac{\partial}{\partial V_\infty}\left(\frac{1}{2} \rho_\infty V_\infty^2 SC_{D,0}\right)+\frac{\partial}{\partial V_\infty}\left(\frac{2KW^2}{\rho_\infty V_\infty^2 S}\right)
∂TR∂V∞=2∗12ρ∞V∞SCD,0+∂∂V∞(2KW2ρ∞V∞2S) \frac{\partial T_R}{\partial V_\infty}=2*\frac{1}{2} \rho_\infty V_\infty SC_{D,0}+\frac{\partial}{\partial V_\infty}\left(\frac{2KW^2}{\rho_\infty V_\infty^2 S}\right)
=ρ∞V∞SCD,0+∂∂V∞(2KW2ρ∞V∞2S) =\rho_\infty V_\infty SC_{D,0}+\frac{\partial}{\partial V_\infty}\left(\frac{2KW^2}{\rho_\infty V_\infty^2 S}\right)
∂TR∂V∞=ρ∞V∞SCD,0+∂∂V∞(2KW2ρ∞SV∞−2) \frac{\partial T_R}{\partial V_\infty}=\rho_\infty V_\infty SC_{D,0}+\frac{\partial}{\partial V_\infty}\left(\frac{2KW^2}{\rho_\infty S}V_\infty^{-2}\right)
=ρ∞V∞SCD,0−2∗2KW2ρ∞SV∞−3 =\rho_\infty V_\infty SC_{D,0}-2*\frac{2KW^2}{\rho_\infty S}V_\infty^{-3}
∂TR∂V∞=ρ∞V∞SCD,0−4KW2ρ∞V∞3S=0 \frac{\partial T_R}{\partial V_\infty}=\rho_\infty V_\infty SC_{D,0}-\frac{4KW^2}{\rho_\infty V_\infty^3 S}=0
ρ∞V∞SCD,0=4KW2ρ∞V∞3S \rho_\infty V_\infty SC_{D,0}=\frac{4KW^2}{\rho_\infty V_\infty^3 S}
ρ∞V∞4SCD,0=4KW2ρ∞S \rho_\infty V_\infty^4 SC_{D,0}=\frac{4KW^2}{\rho_\infty S}
V∞4=4KW2ρ∞2S2CD,0 V_\infty^4=\frac{4KW^2}{\rho_\infty^2 S^2 C_{D,0}}
=4KCD,0(Wρ∞S)2 =\frac{4K}{C_{D,0}}\left(\frac{W}{\rho_\infty S}\right)^2
V∞=(4KCD,0)14(Wρ∞S)24 V_\infty=\left(\frac{4K}{C_{D,0}}\right)^{\frac{1}{4}} \left(\frac{W}{\rho_\infty S}\right)^{\frac{2}{4}}
=4KCD,04Wρ∞S =\sqrt[4]{\frac{4K}{C_{D,0}}} \sqrt{\frac{W}{\rho_\infty S}}
=Wρ∞S4KCD,0 =\sqrt{\frac{W}{\rho_\infty S} \sqrt{\frac{4K}{C_{D,0}}}}
=2Wρ∞SKCD,0 =\boxed{\sqrt{\frac{2W}{\rho_\infty S} \sqrt{\frac{K}{C_{D,0}}}}}

§3. (a)

h=7000ft h=7000ft
ρ∞=0.993kg/m3 \rho_\infty=0.993kg/m^3
me=1243kg m_e=1243kg
mf=147kg m_f=147kg
wp=2∗62kg+2∗68kg+30kg=290kg w_p=2*62kg+2*68kg+30kg=290kg
K=0.056 K=0.056
CD,0=0.017 C_{D,0}=0.017
S=22.9m2 S=22.9m^2
g=9.81m/s2 g=9.81m/s^2
We=meg=1243kg∗9.81m/s2=12190N W_e=m_eg=1243kg*9.81m/s^2=12190N
Wf=mfg=147kg∗9.81m/s2=1442N W_f=m_fg=147kg*9.81m/s^2=1442N
Wp=wp=290kg∗9.81m/s2=2845N W_p=w_p=290kg*9.81m/s^2=2845N
W=W0=We+Wp+Wf=12190N+2845N+1442N=16477N W=W_0=W_e+W_p+W_f=12190N+2845N+1442N=16477N
V∞=2Wρ∞SKCD,0 V_\infty=\sqrt{\frac{2W}{\rho_\infty S} \sqrt{\frac{K}{C_{D,0}}}}
=2∗16477N0.993kg/m3∗22.9m20.0560.017 =\sqrt{\frac{2 * 16477N}{0.993kg/m^3 * 22.9m^2} \sqrt{\frac{0.056}{0.017}}}
=51.29m/s =\boxed{51.29m/s}

§3. (b)

(CL/CD)max=14KCD,0 (C_L/C_D)_{max}=\frac{1}{\sqrt{4KC_{D,0}}}
=14∗0.056∗0.017 =\frac{1}{\sqrt{4 * 0.056 * 0.017}}
=16.21 =\boxed{16.21}

§3. (c)

V∞=51.29m/s V_\infty=51.29m/s
q∞=12ρ∞V∞2 q_\infty=\frac{1}{2}\rho_\infty V_\infty^2
=12∗0.993kg/m3∗(51.29m/s)2 =\frac{1}{2} * 0.993kg/m^3 * (51.29m/s)^2
=1306Pa =1306Pa
CL=Wq∞S=16477N1306Pa∗22.9m2=0.5509 C_L=\frac{W}{q_\infty S}=\frac{16477N}{1306Pa * 22.9m^2}=0.5509

§3. (d)

a=0.088deg⁡−1 a = 0.088\deg^{-1}
αL=0=−1.19deg⁡ \alpha_{L=0}=-1.19\deg
CL=0.5509 C_L=0.5509
CL=a(α−αL=0) C_L = a(\alpha-\alpha_{L=0})
CLa=α−αL=0 \frac{C_L}{a} = \alpha-\alpha_{L=0}
α=CLa+αL=0 \alpha=\frac{C_L}{a} + \alpha_{L=0}
=0.55090.088deg⁡−1−1.19deg⁡ =\frac{0.5509}{0.088\deg^{-1}} - 1.19\deg
=5.07deg⁡ =\boxed{5.07\deg}

§3. (e)

CD=CD,0+KCL2 C_D=C_{D,0}+K C_L^2
=0.017+0.056∗(0.5509)2 =0.017 + 0.056 * (0.5509)^2
=0.03400 =0.03400

§3. (f)

(CL/CD)max=CLCD (C_L/C_D)_{max}=\frac{C_L}{C_D}
=0.55090.03400 =\frac{0.5509}{0.03400}
=16.20 =\boxed{16.20}
That's pretty close to the 16.2116.21 that I got in part (b).