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# Right, lets see whos clever enough to answer this:

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Right, lets see whos clever enough to answer this:

physics
Guest Dec 20, 2014

#1
+1037
+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 NaClO3. 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
Sort:

#1
+1037
+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 NaClO3. 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

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