whereW X_{c}= M_{o}+ L X_{a}+ T X_{b}- D X_{z}

Wweight X_{c}distance from leading edge to center of gravity M_{o}wing moment around the leading edge Llift generated by wing X_{a}distance from leading edge to center of lift Tlift generated by tail X_{b}distance from leading edge to center of tail lift Ddrag generated by wing X_{z}verticle distance between center of gravity and center of drag

Consider that the lift generated by the wing and tail equal the weight:

L = W - T

and substituting

W X_{c}= M_{o}+ (W - T) X_{a}+ T X_{b}- D X_{z}

W (X_{c}- X_{a}) = M_{o}+ T (X_{b}- X_{a}) - D X_{z}

The result above shows that the lift can be dropped from the equation,
and that the tail and CG moment arms can be taken, more conveniently
around the center of lift.
However,
the pitching moment need also be taken around the center of lift.
The *center of lift* is typically located at
1/4 of the mean aerodynamic chord, MAC,
and it is standard practice to measre the pitching moment at this point.
M_{o} is therefore replaced with M_{g}.

whereW X_{g}= M_{g}+ T X_{t}- D X_{z}

X_{g}distance from center of gravity to 1/4 MAC location M_{g}wing moment around the 1/4 MAC location X_{t}distance from center of tail lift to 1/4 MAC location

The moment and force terms can be expanded to consider airspeed:

whereW L_{g}= 1/2 r V^{2}S c C_{m}+ 1/2 r V^{2}S_{t}C_{lt}L_{t}- 1/2 r V^{2}S C_{d}L_{z}

and simplifying

rhoair density divided by viscosity (e.g. 0.002378 slugs / ft^4) Vair speed (e.g. ft per second) Swing area (e.g. ft^2) cmean aerodynamic chord of wing C_{m}wing moment coefficient S_{t}tail area (e.g. ft^2) C_{lt}lift coefficient of tail C_{d}drag coefficient of wing

W L_{g}= 1/2 rho V^{2}{ S c C_{m}+ S_{t}C_{lt}L_{t}- S C_{d}L_{z}}

recognizing that:(W/S) L_{g}/ c = 1/2 rho V^{2}[ C_{m}+ C_{lt}(S_{t}L_{t}) / (S c) - C_{d}L_{z}/ c ]

Also notice that all distances are divided by

W / Swing loading (S_{t}L_{t}) / (S c)tail volume coefficient

where(W/S) x_{g}= 1/2 rho V^{2}[ C_{m}+ C_{lt}(S_{t}x_{t}) / S - C_{d}x_{z}]

x_{g}normalized distance from center of gravity to 1/4 MAC location x_{t}normalized distance from center of tail lift to 1/4 MAC location x_{z}normalized verticle distance between center of gravity and center of drag

x_{g}= 1/2 rho V^{2}/ (W/S) [ C_{m}+ C_{lt}S_{t}x_{t}/ S - C_{d}x_{z}]

- Consider the simplified case of
using an airfoil with a
*C*of zero, and a wing through the fuselage such that x_{m}_{z}is zero.

If the tail*x*_{g}= 1/2 rho V^{2}/ (W/S) [ C_{lt}S_{t}x_{t}/ S ]*incidence*angle is zero, the tail moment is zero and*x*it becomes zero. The plane is neutral. If put into the dive test, it would fly straight into the ground. It is not speed dependent and will fly at any speed, slowing dow due to drag._{g} -
If the tail incidence is negative (negative aoa),
*x*will be positive, making the plane dependent on airspeed in such a way that it is stable. Stability comes from the fact that the CG location causes a constant moment force independent of airspeed, while the tail produces an airspeed dependent moment._{g}In this case, the planes needs to fly at a particular airspeed,

*V*, such that the tail moment offsets the moment cause by the CG. If it flies greater than_{o}*V*, the tail force increases, pitching the plane up, causing it to slow down. Similarly, if the plane flies slower than_{o}*V*, the tail force decreases, allowing the plane to pitch down, and increase speed._{o}A positive tail incidence would require a negative

*x*, resulting in a non-correcting configuration. If the plane were to pitch down and increase speed, the increased speed on the tail would increase the tail moment, causing the tai lto go up, pitching the plane down further._{g} - Now consider the use of a cambered airfoil
with a non-zero
*C*._{m}

The effect of the wing moment requires a neutralizing tail force, which can result in*x*_{g}= 1/2 rho V^{2}/ (W/S) [ C_{m}+ C_{lt}S_{t}x_{t}/ S]*x*being zero. The tail incidence can be adjusted in the previous case requiring a non-zero_{g}*x*producing a non-neutral and stable configuration. However, a non-zero tail incidence produces drag as well as lift._{g} - But when considering the all terms,
we see that
*C*can be offset by_{m}*C*, making it possible to have a tail incidence of zero, and zero drag. However, the downwash produced by the wing must also be considered. (Is it a stabilizing or destabilizing influence?)_{d}x_{z}

Some questions:

- The above only consider moments. What about considering the lift forces on each wing due to aileron deflection about the CG which could be forward of the aerodynamic center? Wouldn't this cause the plane to pitch down?

where1/2 rho V^{2}S c C_{m}= 1/2 rho V^{2}S C_{l}x

Simplifying

xdistance from center of pressure to the 1/4 MAC location

and rearrangingc C_{m}= C_{l}x

The reference can be shifted from the 1/4 MAC location to the leading edge by adding a constantx / c = C_{m}/ C_{l}

It's easy to see that at zero lift, the center-of-pressure is an infinite distance behind the wing. Similarly, at infinite lift, the center-of-pressure can be no further than 1/4 MAC.x / c = 0.25 + C_{m}/ C_{l}

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