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2.1 HW1

2.1.1 Problem 1

calculation notebook in Mathematica appendix.nb

appendix.pdf

Figure 2.1:problem 1 description

2.1.1.1 Part(a)

Figure 2.2:problem 1 part (a)

2.1.1.2 Part(b)

The following diagram shows the calculated aerodynamic dimensions of the wing using the three views.

Figure 2.3:problem 1 part (b)

The wing area $S$ is the mean of the root chord length and the wing tip chord multiplied by the wing span $b$, therefore

$$\begin{align*} S &= \left ( \frac{c_{r}+c_{t}}{2} \right ) b\\ &= \left ( \frac{25+12}{2} \right ) 150\\ &= \SI{2775}{\square \ft } \end{align*}$$

The aspect ratio is

$$\mathcal{A} = \frac{b^2}{S} = \frac{150^2}{2775} = 8.108$$ The taper ratio $\lambda$ is $$\lambda = \frac{c_t}{c_r} = \frac{12}{25} = 0.48$$ To find the length of the mean aerodynamic chord $\bar{c}$, equation (C.3,3) in the textbook was used with $n=0$. In using this equation, $x$, which is the distance from wing tip to the start of the $\bar{c}$ chord was found first using equation (C.3,1) as follows $$\begin{align*} x &= \left (\frac{b}{2}\right ) \left (\frac{1}{3}\right ) \frac{1+2\lambda }{1+\lambda } \tan (\Lambda _0) \\ &= \left (\frac{150}{2}\right ) \left (\frac{1}{3}\right ) \frac{1+2\times 0.48}{1+0.48}\tan (\SI{26}{\degree }) \\ &= \SI{16.148}{\ft } \end{align*}$$

Now equation (C3.3) was used since $x$ is known

$$\begin{align*} \frac{x}{\bar{c}} &= \frac{(1+2\lambda )(1+\lambda )}{8(1+\lambda +\lambda ^{2})} \mathcal{A}\tan (\Lambda _0) \\ \bar{c} &= \left (\frac{x}{\mathcal{A}\tan (\Lambda _0)}\right ) \frac{8(1+\lambda +\lambda ^2)}{(1+2\lambda )(1+\lambda )}\\ &= \frac{16.148}{ 8.108 \tan (\SI{26}{\degree }) } \frac{8(1+0.48+0.48^2)}{(1 + 2 \times 0.48) (1+0.48)}\\ &=\SI{19.2613}{\ft } \end{align*}$$

This value for $\bar{c}$ was verified using figure C.2 on page 261 based on the use of

$$\bar{c} = \frac{2}{3} c_{r} \frac{1+\lambda +\lambda ^{2}}{1+\lambda }$$ The result matched that found using equation (C3.3) above.

2.1.1.3 Part(c)

The location of mean aerodynamic center on the full wing is given by the coordinates $\left (\bar{x},\bar{y},\bar{z}\right )$ in the local frame of reference. For the full wing

$$\bar{y}=0$$ And $$\begin{align*} \bar{x} & = x + \frac{1}{4}\bar{c} \\ & = 16.148 + \left (\frac{1}{4}\right ) 19.261 \\ & = \SI{20.964}{\ft } \end{align*}$$

To obtain $\bar{z}$, (C.1,4) in appendix C was used

$$\begin{equation} \bar{z} = \frac{2}{C_L S} \int _{0}^{\frac{b}{2}} C_{L_{\alpha }}c z\, dy\tag{C.1,4} \end{equation}$$ From the problem $C_{L_{\alpha }}=C_{L}$ as the lift coefficient is uniform. Therefore the above simplifies to $$\begin{equation} \bar{z} = \frac{2}{S}\int _{0}^{\frac{b}{2}}c z\, dy\tag{C.1,4} \end{equation}$$ The value for $c(y)$ in the above integral is given ¹ by the following $$c(y) = \frac{2s}{(1+\lambda )b} \left (1-\frac{2(1-\lambda )}{b}y\right )$$ Given that $z(y)=y\tan (\Gamma )$ where $\Gamma$ is the dihedral angle which is 4° and $S=\SI{2775}{\ft \squared }$ is the wing area, and $\lambda =0.48$, (C.1,4) becomes $$\begin{align*} \bar{z} &= \frac{2}{S} \int _{0}^{\frac{b}{2}} \frac{2s}{(1+\lambda ) b} \left (1- \frac{2 (1-\lambda )}{b} y \right ) y \tan (\Gamma )\, dy\\ &=\frac{2}{2775} \int _{0}^{\frac{150}{2}} \frac{2(2775)}{(1+0.48)150} \left (1-\frac{2(1-0.48)}{150} y \right ) y \tan (\SI{4}{\degree })\,dy\\ & =\SI{2.315}{\ft } \end{align*}$$

2.1.1.4 Appendix for part (c)

This section is extra as it finds the $\{\bar{x},\bar{y},\bar{z}\}$ for half wing, and not the full wing as the problems asks for. This was done to practice the use of appendix C integrals.

Figure 2.4:problem 1 part (e)

In finding $\{\bar{x},\bar{y},\bar{z}\}$ , equations (C.1,2,3,4) in appendix C are used. The expression for $c$ in these integrals is given by

$$c(y) = \frac{2s}{( 1+\lambda ) b}\left (1- \frac{2(1-\lambda )}{b} y \right )$$ From (C1.2) $$\begin{align*} \bar{x} & =\frac{2}{C_{L}S}\int _{0}^{\frac{b}{2}}C_{L_{\alpha }}cx\,dy\\ & =\frac{2}{S}\int _{0}^{\frac{b}{2}}cx\,dy \end{align*}$$

Where $x=\left (\frac{1}{4}\right )c + y\tan (\Gamma )$ as seen in the above diagram. substituting these in the above integral results in

$$\bar{x}=\frac{2}{S}\int _{0}^{\frac{b}{2}}\frac{2s}{\left ( 1+\lambda \right ) b}\left ( 1-\frac{2\left ( 1-\lambda \right ) }{b}y\right ) \left ( \frac{1}{4}\left ( \frac{2s}{\left ( 1+\lambda \right ) b}\left ( 1-\frac{2\left ( 1-\lambda \right ) }{b}y\right ) \right ) +y\tan \left ( \Gamma \right ) \right )\,dy$$ Giving numerical values for all the variables in the above gives $$\bar{x}=\SI{20.9632}{\ft }$$ Similarly for $\bar{y}$ $$\begin{align*} \bar{y} & =\frac{2}{C_{L}S}\int _{0}^{\frac{b}{2}}C_{L_{\alpha }}cy\,dy\\ & =\frac{2}{S}\int _{0}^{\frac{b}{2}}cy\,dy\\ & =\frac{2}{S}\int _{0}^{\frac{b}{2}}\frac{2s}{\left ( 1+\lambda \right ) b}\left ( 1-\frac{2\left ( 1-\lambda \right ) }{b}y\right ) y\,dy \end{align*}$$

Substituting numerical values for all the variables above gives

$$\bar{y}=\SI{33.108}{\ft }$$ This value can also be found based on geometry using the above diagram as follows $$\begin{align*} \tan (\alpha ) &=\frac{\bar{y}}{x}\\ \bar{y} &=x\tan (\SI{90}{\degree }-\Lambda _{0}) \\ &=16.148\tan (\SI{90}{\degree }-\SI{26}{\degree }) \\ &=\SI{33.108}{\ft } \end{align*}$$

And finally for $\bar{z}$

$$\bar{z}=\frac{2}{S}\int _{0}^{\frac{b}{2}}cz\,dy$$ Where $z=y\tan (\Gamma )$ hence the above becomes $$\bar{z}=\frac{2}{S}\int _{0}^{\frac{b}{2}}\left ( 1-\frac{2\left ( 1-\lambda \right ) }{b}y\right ) \left ( y\tan \Gamma \right )\,dy$$ Substituting numerical values for all the variables above gives $$\bar{z}=\SI{2.31514}{\ft }$$ The diagram below was drawn to scale in Mathematica using the actual values found. This diagram shows the aerodynamic center for the full wing as well for the half wing.

Figure 2.5:detailed wing dimensions

(*calculations used in the above*)
chordLength[y_, s_, b_, lambda_] := (2 s)/((1 + lambda) b) (1-(2(1 - lambda))/b y)
Clear[y];
s = 2775;
b = 150;
lambda = 0.48;
c = chordLength[y, s, b, lambda]
    25. (1 - 0.00693333 y)
cBar = 2/s Integrate[c^2, {y, 0, b/2}]
    19.2613
yBar = 2/s Integrate[c y, {y, 0, b/2}]
    33.1081
zBar = 2/s Integrate[c y Tan[4 Degree], {y, 0, b/2}]
    2.31514
xBar = 2/s Integrate[c ((1/4) c + y*Tan[26 Degree]) , {y, 0, b/2}]
    20.9632

2.1.1.5 Part(d)

Figure 2.6:problem 1 part(d)

From the above diagram (not drawn to scale)

$$h_{n}=\frac{9.8151}{19.261}=0.50958$$ From equation (2.3,23)² $$\begin{equation} h_{n} = h_{n_{wb}}+ \frac{a_t}{a} \overline{V}_{H} \left (1-\frac{\partial \epsilon }{\partial \alpha }\right ) -\frac{1}{a}\frac{\partial C_{m_{p}}}{\partial \alpha } \tag{1} \end{equation}$$ Ignoring power plant effects and using $a$ from (2.3,18) given by $$\begin{equation} a=a_{wb}\left (1+\frac{a_{t}}{a_{wb}}\frac{S_{t}}{S}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) \right ) \tag{2} \end{equation}$$ Substituting (2) into (1) and using $\overline{V}_{H}=\frac{l_t}{\bar{c}}\frac{S_{t}}{S}$ results in $$h_{n}=h_{n_{wb}}+\frac{a_{t}}{a_{wb}\left ( 1+\frac{a_{t}}{a_{wb}}\frac{S_{t}}{S}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) \right ) }\frac{l_{t}}{\bar{c}}\frac{S_{t}}{S}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right )$$ Since $a_{t}=a_{wb}$ the above becomes $$h_{n}=h_{n_{wb}}+\frac{S}{S+S_{t}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) }\frac{\bar{l}_{t}}{\bar{c}}\frac{S_{t}}{S}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right )$$ Substituting the following numerical values $S=\SI{2775}{\square \ft },\bar{c}=\SI{19.2613}{\ft }, \bar{l}_{t}=\SI{55}{\ft }, \frac{\partial \epsilon }{\partial \alpha }=0.25$ the above becomes $$0.50958=\frac{1}{4}+\frac{2775}{\left ( 2775+0.75S_{t}\right ) }\frac{55}{19.2613}\frac{0.75S_{t}}{2775}$$ Solving for $S_{t}$ gives the area of tail $$S_{t}=\SI{367}{\square \ft }$$

2.1.2 Problem 2

2.1.2.1 Part(d)

Figure 2.7:problem 2 description

The expression for $C_{m_{\alpha }}$ in the first equation above is given by

$$\begin{equation} C_{m_{\alpha }}=a_{wb}\left ( h-h_{n_{wb}}\right ) -a_{t}V_{H}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) +\frac{\partial C_{m_{p}}}{\partial \alpha } \tag{1} \end{equation}$$ While the expression for $C_{m_{\alpha }}$ in the second equation is given by $$\begin{equation} C_{m_{\alpha }}=a\left ( h-h_{n_{wb}}\right ) -a_{t}\overline{V}_{H}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) +\frac{\partial C_{m_{p}}}{\partial \alpha } \tag{2} \end{equation}$$ The above expressions are given in the class handout on page 32 and 34.

The problem asks to show that these two expression are the same. Starting from (2) in order to show it can be rewritten as (1). For this purpose, the following two definitions are used

$$\begin{align} a & =a_{wb}\left ( 1+\frac{a_{t}}{a_{wb}}\frac{S_{t}}{S}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) \right ) \tag{3}\\ \overline{V}_{H} & =\frac{\bar{l}_{t}}{\bar{c}}\frac{S_{t}}{S}\nonumber \end{align}$$

Since $\bar{l}_{t}=l_{t}+\left ( h-h_{n_{wb}}\right ) \bar{c}$ the above becomes

$$\begin{align} \overline{V}_{H} & =\frac{l_{t}+\left ( h-h_{n_{wb}}\right ) \bar{c}}{\bar{c}}\frac{S_{t}}{S}\nonumber \\ & =\frac{l_{t}}{\bar{c}}\frac{S_{t}}{S}+\left ( h-h_{n_{wb}}\right ) \frac{S_{t}}{S}\tag{4} \end{align}$$

Substituting Eqs (3,4) into Eq (2) gives

$$\begin{align*} C_{m_{\alpha }} & =a_{wb}\left ( 1+\frac{a_{t}}{a_{wb}}\frac{S_{t}}{S}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) \right ) \left ( h-h_{n_{wb}}\right ) -a_{t}\left [ \frac{l_{t}}{\bar{c}}\frac{S_{t}}{S}+\left ( h-h_{n_{wb}}\right ) \frac{S_{t}}{S}\right ] \left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) +\frac{\partial C_{m_{p}}}{\partial \alpha }\\ & =\left ( a_{wb}+a_{t}\frac{S_{t}}{S}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) \right ) \left ( h-h_{n_{wb}}\right ) -\left [ a_{t}\frac{l_{t}}{\bar{c}}\frac{S_{t}}{S}+a_{t}\left ( h-h_{n_{wb}}\right ) \frac{S_{t}}{S}\right ] \left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) +\frac{\partial C_{m_{p}}}{\partial \alpha }\\ & =a_{wb}\left ( h-h_{n_{wb}}\right ) +\overbrace{a_{t}\frac{S_{t}}{S}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) \left ( h-h_{n_{wb}}\right ) }-a_{t}\frac{l_{t}}{\bar{c}}\frac{S_{t}}{S}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) -\overbrace{a_{t}\left ( h-h_{n_{wb}}\right ) \frac{S_{t}}{S}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) }+\frac{\partial C_{m_{p}}}{\partial \alpha } \end{align*}$$

The second term and the fourth term in the above cancel each others resulting in

$$C_{m_{\alpha }}=a_{wb}\left ( h-h_{n_{wb}}\right ) -a_{t}\overset{V_{H}}{\overbrace{\frac{l_{t}}{\bar{c}}\frac{S_{t}}{S}}}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) +\frac{\partial C_{m_{p}}}{\partial \alpha }$$ In the above $\frac{l_{t}}{\bar{c}}\frac{S_{t}}{S}$ is $V_{H}$ hence the above becomes $$C_{m_{\alpha }}=a_{wb}\left ( h-h_{n_{wb}}\right ) -a_{t}V_{H}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) +\frac{\partial C_{m_{p}}}{\partial \alpha }$$ Comparing the above to (1), it can be seen it is same as (2).

2.1.3 Problem 3

Figure 2.8:problem 3 description

2.1.3.1 Part(a)

$C_{m_{0}}$ is given by (2.3,22) on page 32 of the textbook.

$$\begin{equation} C_{m_{0}}=C_{m_{ac_{wb}}}+a_{t}\bar{V}_{H}\left ( \epsilon _{0}+i_{t}\right ) \left [ 1-\frac{a_{t}}{a}\frac{S_{t}}{S}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) \right ] \tag{0} \end{equation}$$ Where (using SI units) $$\begin{align*} \overline{V}_{H} &=\frac{\bar{l}_{t}}{\bar{c}}\frac{S_{t}}{S}\\ & =\frac{38.84}{15.61}\frac{0.0342}{0.139}\\ & =0.6122 \end{align*}$$

And

$$\begin{align*} a &= a_{wb}\left ( 1+\frac{a_{t}}{a_{wb}}\frac{S_{t}}{S}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) \right ) \\ & =0.077\left ( 1+\frac{0.064}{0.077}\frac{0.0342}{0.139}\left ( 1-0.3\right ) \right ) \\ & = \SI{0.08802}{\per \deg } \end{align*}$$

Using the numerical values given by (0) the above becomes

$$\begin{align} C_{m_{0}} & =-0.018+0.064\left ( 0.6122\right ) \left ( 0.72+i_{t}\right ) \left ( 1-\frac{0.064}{0.08802}\frac{0.0342}{0.139}\left ( 1-0.3\right ) \right ) \nonumber \\ & =0.03427\,i_{t}+0.006677 \tag{1} \end{align}$$

Hence

$$\begin{align*} C_{m_{0}} & >0\\ 0.03427\,i_{t}+0.006677\, & >0\\ i_{t} & >\frac{-0.006677}{0.03427}\\ & >\SI{-0.19484}{\degree } \end{align*}$$

$C_{m_{\alpha }}$ is given by

$$C_{m_{\alpha }}=a\left ( h-h_{n_{wb}}\right ) -a_{t}\overline{V}_{H}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) +\overset{0}{\overbrace{\frac{\partial C_{mp}}{\partial \alpha }}}$$ Hence $$\begin{align*} C_{m_{\alpha }} &= 0.08802\left ( h-h_{n_{wb}}\right ) -a_{t}\left ( 0.6122\right ) \left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) \\ & =0.08802\left ( h-0.25\right ) -0.064\left ( 0.6122\right ) \left ( 1-0.3\right ) \end{align*}$$

Therefore

$$\begin{equation} C_{m_{\alpha }}=0.08802\,h-0.04943 \tag{2} \end{equation}$$ Hence $$\begin{align*} C_{m_{\alpha }} & <0\\ 0.08802h-0.04943 & <0\\ h & <\frac{0.04943}{0.08802}<0.562 \end{align*}$$

2.1.3.2 Part(b)

$$C_{m}=C_{m_{0}}+C_{m_{\alpha }}\alpha$$ But at trim $C_{m}=0$, hence at trim the above becomes $$\begin{equation} C_{m_{0}}+C_{m_{\alpha }}\alpha _{trim}=0 \tag{3} \end{equation}$$ We can find $\alpha _{trim}$ since $\left ( C_{L}\right ) _{trim}=a\alpha _{trim}$ and we know $a$ which is $C_{L_{\alpha }}$ from part (a). Hence we just need to find $C_{L}$ at trim. But $$\left ( C_{L}\right ) _{trim}=\frac{L}{\frac{1}{2}\rho V^{2}S}=\frac{W}{\frac{1}{2}\rho V^{2}S}$$ where at trim the lift $L$ is equal to the weight of the aircraft $W$. Therefore, since $\rho =\SI{1.225}{\kg \per \meter \cubed }$, $V=\SI{123}{\meter \per \second }$ and at trim $L=W=mg=22680(9.8)$, and the scaled wing area is $S=\left (0.139\right )25^{2}=\SI{86.875}{\meter \squared }$ then the above becomes $$\begin{align*} \left ( C_{L}\right ) _{trim} & =\frac{22680\left ( 9.8\right ) }{\frac{1}{2}\left ( 1.225\right ) \left ( 123^{2}\right ) \left ( 86.875\right ) }\\ & =0.27609 \end{align*}$$

using $a=\SI{0.08802}{\degree }$ which was found from part (a), the angle of attack at trim $\alpha _{trim}$ is now found

$$\alpha _{trim}=\frac{\left ( C_{L}\right )_{trim}}{a}=\frac{0.2761}{0.08802}=\SI{3.1367}{\degree }$$ Now that $\alpha _{trim}$ is found, then equation (3) is used to find the following equation $$\begin{align*} C_{m_{0}}+C_{m_{\alpha }}\alpha _{trim} & =0\\ \overset{C_{m_{0}}\text{ from part (a)}}{\overbrace{\left ( 0.03427\,i_{t}+0.006677\right ) }}+\,\overset{C_{m_{\alpha }}\text{ from part(a)}}{\overbrace{\left ( 0.08802h-0.04943\,\right ) }}3.1367 & =0\\ 0.27609h+0.03427i_{t}-0.14837 & =0 \end{align*}$$

Solving for $i_{t}$ as a function of $h$ gives

$$\begin{align*} i_{t} & =\frac{0.1484-0.2761\,h}{0.0343}\\ & =4.3294-8.0563\,h \end{align*}$$

The following is a plot in a small region around $i_{t}=\SI{-0.19}{\degree }$ and $h=0.56$

Figure 2.9:problem 3 part b

Figure 2.10:Plot for $h=0\cdots 1$ showing location $i_{t}=\SI{-0.19}{\degree }$ and $h=0.56$

For static stability, $i_{t}>\SI{-0.19}{\degree }$ and $h<0.562$ as was obtained above. This is the value of the above line to the left of the shown small point and above the point, which is the limit of static stability.

2.1.4 Problem 4

Figure 2.11:problem 4 description

2.1.4.1 Part(a)

The aspect ratio is (SI units are used)

$$\mathcal{A=}\frac{b^{2}}{S}=\frac{50.29^{2}}{353}=7.1645$$ The taper ratio $\lambda$ is $$\lambda =\frac{c_{t}}{c_{r}}=\frac{2.68}{11.37}=0.23571$$ Using table C.1 in appendix C of the textbook, page 359 $$\begin{align*} \bar{c} & =\frac{2}{3}c_{r}\frac{1+\lambda +\lambda ^{2}}{1+\lambda }\\ & =\frac{2}{3}(11.37) \frac{1+0.236+0.236^{2}}{1+0.236}\\ & =\SI{7.921}{\meter } \end{align*}$$

2.1.4.2 Part(b)

For this part, figure B.1-2 on page 322 was used. This figure is shown below

$$a_{w}=C_{L_{w_{\alpha }}}=\frac{\partial C_{L_{w}}}{\partial \alpha }$$ In the above figure, $\beta$ is the Prandtl-Glauert compressibility factor and $\kappa =\frac{\beta C_{l_{\alpha }}}{2\pi }$where $C_{l_{\alpha }}$ is the 2D airfoil lift-curve slope and $C_{L_{w}}$.

The half chord sweep angle is $\Lambda _{\frac{1}{2}}=\SI{22}{\degree }$. The problem asks to use the expression in the figure inset to find $C_{L_{\alpha }}$

$$\begin{align*} \frac{C_{L_{\alpha }}}{\mathcal{A}} & =\frac{2\pi }{2+\sqrt{\frac{\mathcal{A}^{2}\beta ^{2}}{\kappa ^{2}}\left ( 1+\frac{\tan \left ( \Lambda _{\frac{1}{2}}\right ) ^{2}}{\beta ^{2}}\right ) +4}}\\ \frac{C_{L_{\alpha }}}{7.1645} & =\frac{2\pi }{2+\sqrt{\frac{\left ( 7.1645^{2}\right ) \left ( 1^{2}\right ) }{1^{2}}\left ( 1+\frac{\tan \left ( 22^{0}\right ) ^{2}}{1^{2}}\right ) +4}}\\ & =0.63247 \end{align*}$$

Hence

$$\begin{align*} C_{L_{\alpha }} & = 7.165 \times 0.633\\ & =\SI{4.531}{\per \radian }\\ & =\SI{0.079}{\per \deg } \end{align*}$$

The angle $C_{L}$ makes with the horizontal is $\arctan (4.531) = \SI{1.354}{\radian } = \SI{77.556}{\degree }$

Figure 2.12:plot for problem 4 part b

Therefore

$$a_{w}=\SI{4.531}{\per \radian }$$

2.1.4.3 Part (c)

The lift curve slope of the aircraft $a$ is given by

$$\begin{equation} a=a_{wb}\left ( 1+\frac{a_{t}}{a_{wb}}\frac{S_{t}}{S}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right ) \right ) \tag{1} \end{equation}$$ Using $a_{t}=0.068$ deg$^{-1}$, $a_{wb}=a_{w}=0.079$ deg$^{-1}$ and $$\begin{align*} \frac{\partial \epsilon }{\partial \alpha } & =\frac{2a_{w}}{\pi \mathcal{A}}\\ & =\frac{2(4.531)}{7.165\pi }\\ & =0.403 \end{align*}$$

From Eq (1) we find

$$\begin{align*} a &= 0.079 \left ( 1 + \left (\frac{0.068}{0.079}\right ) \left (\frac{80.83}{353}\right ) (1-0.403) \right ) \\ &=\SI{0.0883}{\per \deg } \\ &=\SI{5.059}{\per \radian } \end{align*}$$

2.1.4.4 Part(d)

$$C_{m_{\alpha }}=a\left ( h-h_{n_{wb}}\right ) -a_{t}\frac{\bar{l}_{t}}{\bar{c}}\frac{S_{t}}{S}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right )$$ The problem says that $\bar{l}_{t}=l_{t}$. This implies that the distance between aerodynamic center (a.c.) and the center of gravity (c.g.) of the aircraft is zero. This means $\left ( h-h_{n_{wb}}\right ) =0$. Therefore $$C_{m_{\alpha }}=-a_{t}\frac{\bar{l}_{t}}{\bar{c}}\frac{S_{t}}{S}\left ( 1-\frac{\partial \epsilon }{\partial \alpha }\right )$$ Using $\bar{c}=\SI{7.921}{\meter }, \bar{l}_{t}=\SI{28.04}{\meter }, S_{t}=\SI{80.83}{\meter \squared }, S=\SI{353}{\meter \squared }$ and using $\frac{\partial \epsilon }{\partial \alpha }=0.403$ and using $a_{t}=\SI{0.068}{\per \deg }$ found from part(c), then the above gives $$\begin{align*} C_{m_{\alpha }} & =-0.068\frac{28.04}{7.921}\frac{80.83}{353}(1-0.403) \\ & = \SI{-0.0329}{\per \deg } \\ & =\SI{-1.885}{\per \radian } \end{align*}$$

Since $C_{m_{\alpha }}<0$ then the airplane is statically stable.

2.1.4.5 Part(e)

From Fig 2.29, $\frac{\partial C_{m}}{\partial \delta _{e}}$ can be estimated using $C_{m}=0.25$ and the corresponding line for $\delta _{e}=\SI{0}{\degree }$ and using $C_{m}=0.125$ and its corresponding line for $\delta _{e}=\SI{5}{\degree }$. This gives

$$\begin{align*} \frac{dC_{m}}{d\delta _{e}} & =\frac{0.25-0.125}{-0.50}\\ C_{m_{\delta _{e}}} & =\SI{-0.025}{\per \deg } \end{align*}$$

In solving for $h_{n}$, figure 2.30 was used. The slope for the $h=0.35$ line is $-\frac{8}{0.78}=-10.256$ and the slope for the $h=0.25$ line is $-\frac{8}{0.6}=-13.333$. Using

$$\begin{equation} \left ( \frac{d\delta _{e}}{dC_{L}}\right )_{trim}=-\frac{C_{L_{\alpha }}}{\Delta }\left ( h-h_{n}\right ) \tag{1} \end{equation}$$ Where $\Delta =C_{L_{\alpha }}C_{m_{\delta _{e}}}-C_{L_{\delta _{e}}}C_{m_{\alpha }}$. Evaluating Eq (1) for the two given values of $h$ results in two equations $$\begin{align} -10.256 & =-\frac{C_{L_{\alpha }}}{\Delta }\left ( 0.35-h_{n}\right ) \tag{2}\\ -13.333 & =-\frac{C_{L_{\alpha }}}{\Delta }\left ( 0.25-h_{n}\right ) \tag{3} \end{align}$$

From (2) $\frac{C_{L_{\alpha }}}{\Delta }=\frac{10.256}{\left ( 0.35-h_{n}\right ) }$, substituting this in (3) gives

$$\begin{align*} -13.333 & =-\frac{10.256}{\left ( 0.35-h_{n}\right ) }\left (0.25-h_{n}\right ) \\ -12.25\left ( 0.35-h_{n}\right ) & =-9.494\left (0.25-h_{n}\right ) \\ h_{n} & =0.6833 \end{align*}$$

$\allowbreak c_{m_{\alpha }}$ at $h=0.3$ is now found. Since

$$c_{m_{\alpha }}=a\left ( h-h_{n}\right )$$ Where $a$ is found from part(c) as 0.088 296 deg⁻¹ and $h_{n}=0.6833$ therefore $$\begin{align*} c_{m_{\alpha }} & =0.088296\left ( 0.3-0.6833\right ) \\ & =\SI{-0.03384}{\per \deg }\\ & =\SI{-1.9389}{\per \radian } \end{align*}$$

$c_{m_{\alpha }}<0$ indicates static stability.

2.1.5 Problem 5

Figure 2.13:problem 5 description

2.1.5.1 Part(a)

$C_{L}=\frac{L}{\frac{1}{2}\rho V^{2}S}$ and at trim $L=mg$. Using $\rho =\SI{1.225}{\kg \per \meter \cubed }$ and $S=\SI{16.21}{\meter \squared }$ Values for $C_{L}$ for the different $V_{E}$ values given in the table and the corresponding mass are calculated and plotted against the angle $i_{t}$

Figure 2.14:table and plot for part a problem 5

2.1.5.2 Part(b)

The three segments are first fitted each to a straight line giving the following plot. The fitted straight lines found by fitting³ are $\{-3.78307+5.3984x,-4.02058+9.2621x,-3.76524+12.6932x\}$ where $y=i_{t}\approx -3.85$ is the intercept angle in degrees where $C_{L}=0$ for all three lines

Figure 2.15:problem 5 part b plot

2.1.5.3 Part(c)

It was not clear if one should use the speed effect here and plot $\frac{di_{t}}{dV_{trim}}$ against $h$ to find the intersection or to use slope given by $\frac{di_{t}}{d_{C_{L_{trim}}}}$ against $h$ to find the intersection on the x-axis in order to determine $h_{n}$.

The second approach is used below since that is what figure 2.31 used which was referred to in the problem above. Therefore, the slope of each of the above lines is found for each $h$. This results in the following table

slope	$h$ meter (c.g. measured from tip of aircraft)
$5.4$	$2.385$
$9.26$	$2.205$
$12.69$	$2.043$

The data above gives three points. They are plotted and the intersection with $h$ axis is found. This intersection is $h_{n}$. Below is the plot of the data in the above table

Figure 2.16:problem 5 part c (1)

Figure 2.17:Line extended to the x-axis

From the above diagram

$$\begin{align*} h_{n} & =2.65 \hspace{2pt}\text{meter}\\ & =265 \hspace{2pt}\text{cm}\\ & =8.694 \hspace{2pt}\text{ft} \end{align*}$$

2.1.6 HW 1 key solution

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