#1**+10 **

$$\text {The lift equation } \ C_L \;=\; C_L_\alpha \; + \; \alpha \;+ \; C_{L_0}

\\

\\ C_L_\alpha \; = \; \dfrac {0.052 -0}{5-(-1.5)}\rightarrow C_L_\alpha \; = \; 0.08 / Deg. \rightarrow \\\

\ C_{L_0} \; = \; (0.08)(1.5)\; = \; 0.12 \\\

\text {The moment equation } C_M \; = \; C_M_\alpha \; + \; C_{M_0} \\

\ C_M_\alpha \; = \; \frac {0.05 -(-0.01)}{7.88-1}\; = \; 0.0087 / Deg.\rightarrow \\\

\ C_{M_0} \; = \; (0.0087)(-1) \; +0.01 \; = \; -0.0187 \\\

\text{Use results from above in equation below for } C_M \text{ as a function of } C_L \\

\ C_M (\alpha) \; = \; \Bigg (\dfrac {C_M_ \alpha}{C_L_\alpha}\Bigg ) \; C_L \; - \; \Bigg ( \Bigg(\dfrac {C_M_ \alpha C_{L_0}}{C_L_ \alpha} \Bigg ) \; - \; C_{M_0} \Bigg ) \rightarrow \\

\ C_M (\alpha) = \; \Bigg(\dfrac {(0.0081)}{0.08} \Bigg ) \; C_L \; - \; \Bigg ( \Bigg(\dfrac {(0.0081)(0.012)}{0.08} \Bigg ) - (-0.0187) \Bigg ) \\\\

\ C_M (\alpha) = \; 0.1087 C_L \; -\; 0.0318 \rightarrow

\\

\ C_M \; = \; 0.0318

\\\\

\text { The aerodynamic center is } \ 0.035 \; + \; 0.0318 \; = \; 0.0668 \\$$

**This aircraft is not stable. The center of gravity is aft of the neutral point (based on**

**$$C_{M_0} \; = \; (-0.0187)\\ \ \\$$**

This may be “normal” for a Harrier type jet.

More likely I have been breathing too much Ozone from the Special NaClO_{3}.** This may not be correct. Mayday! Mayday! Mayday!** Someone needs to apply the dumbness correction factor(s) to compensate for the remnants of my CDD.

**Alan can solve this before his morning tea finishes brewing.**

Nauseated Dec 21, 2014

#1**+10 **

Best Answer

$$\text {The lift equation } \ C_L \;=\; C_L_\alpha \; + \; \alpha \;+ \; C_{L_0}

\\

\\ C_L_\alpha \; = \; \dfrac {0.052 -0}{5-(-1.5)}\rightarrow C_L_\alpha \; = \; 0.08 / Deg. \rightarrow \\\

\ C_{L_0} \; = \; (0.08)(1.5)\; = \; 0.12 \\\

\text {The moment equation } C_M \; = \; C_M_\alpha \; + \; C_{M_0} \\

\ C_M_\alpha \; = \; \frac {0.05 -(-0.01)}{7.88-1}\; = \; 0.0087 / Deg.\rightarrow \\\

\ C_{M_0} \; = \; (0.0087)(-1) \; +0.01 \; = \; -0.0187 \\\

\text{Use results from above in equation below for } C_M \text{ as a function of } C_L \\

\ C_M (\alpha) \; = \; \Bigg (\dfrac {C_M_ \alpha}{C_L_\alpha}\Bigg ) \; C_L \; - \; \Bigg ( \Bigg(\dfrac {C_M_ \alpha C_{L_0}}{C_L_ \alpha} \Bigg ) \; - \; C_{M_0} \Bigg ) \rightarrow \\

\ C_M (\alpha) = \; \Bigg(\dfrac {(0.0081)}{0.08} \Bigg ) \; C_L \; - \; \Bigg ( \Bigg(\dfrac {(0.0081)(0.012)}{0.08} \Bigg ) - (-0.0187) \Bigg ) \\\\

\ C_M (\alpha) = \; 0.1087 C_L \; -\; 0.0318 \rightarrow

\\

\ C_M \; = \; 0.0318

\\\\

\text { The aerodynamic center is } \ 0.035 \; + \; 0.0318 \; = \; 0.0668 \\$$

**This aircraft is not stable. The center of gravity is aft of the neutral point (based on**

**$$C_{M_0} \; = \; (-0.0187)\\ \ \\$$**

This may be “normal” for a Harrier type jet.

More likely I have been breathing too much Ozone from the Special NaClO_{3}.** This may not be correct. Mayday! Mayday! Mayday!** Someone needs to apply the dumbness correction factor(s) to compensate for the remnants of my CDD.

**Alan can solve this before his morning tea finishes brewing.**

Nauseated Dec 21, 2014