#### Introduction

These notes consider the moment forces around the pitch axis of a typical aircraft. The analysis results in the CG location based on normalized dimensions, wing loading, tail volume coefficient, and flight speed.

#### Analysis

A sailplane is affected by lift, weight, drag, tail-force, and the pitching moment of a cambered airfoil. Unlike a powered aircraft, a sailplane is not affected by the thrust of an engine. Each force generates a torque, or moment, around some reference point. The moment is simply the force multiplied by the distance between the force and reference. In quilibrium, or straight and level flight, the sum of these moments must be zero. Their sum is:
W Xc = Mo + L Xa + T Xb - D Xz
where
 W weight Xc distance from leading edge to center of gravity Mo wing moment around the leading edge L lift generated by wing Xa distance from leading edge to center of lift T lift generated by tail Xb distance from leading edge to center of tail lift D drag generated by wing Xz verticle distance between center of gravity and center of drag

#### Eliminating Lift

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

L = W - T

and substituting

W Xc = Mo + (W - T) Xa + T Xb - D Xz
W (Xc - Xa) = Mo + T (Xb - Xa) - D Xz

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. Mo is therefore replaced with Mg.

W Xg = Mg + T Xt - D Xz
where
 Xg distance from center of gravity to 1/4 MAC location Mg wing moment around the 1/4 MAC location Xt distance from center of tail lift to 1/4 MAC location

#### Affect of Airspeed

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

W Lg = 1/2 r V2 S c Cm + 1/2 r V2 St Clt Lt - 1/2 r V2 S Cd Lz
where
 rho air density divided by viscosity (e.g. 0.002378 slugs / ft^4) V air speed (e.g. ft per second) S wing area (e.g. ft^2) c mean aerodynamic chord of wing Cm wing moment coefficient St tail area (e.g. ft^2) Clt lift coefficient of tail Cd drag coefficient of wing
and simplifying
W Lg = 1/2 rho V2 { S c Cm + St Clt Lt - S Cd Lz }

#### Normalizing Dimensions

By dividing each term by S c, the dimensions in the above equations can be made relative to the MAC of the wing. In addition, the weight can effectively be replaced by wing loading. :
(W/S) Lg / c = 1/2 rho V2 [ Cm + Clt (St Lt) / (S c) - Cd Lz / c ]
recognizing that:
 W / S wing loading (St Lt) / (S c) tail volume coefficient
Also notice that all distances are divided by c and can be represented by normalized values:
(W/S) xg = 1/2 rho V2 [ Cm + Clt (St xt) / S - Cd xz ]
where
 xg normalized distance from center of gravity to 1/4 MAC location xt normalized distance from center of tail lift to 1/4 MAC location xz normalized verticle distance between center of gravity and center of drag

#### CG Location

Finally, dividing both sides by the wing loading isolates the CG location, referenced to the wing's 1/4 MAC location:
xg = 1/2 rho V2 / (W/S) [ Cm + Clt St xt / S - Cd xz ]

#### Observations

Some observations:
• Consider the simplified case of using an airfoil with a Cm of zero, and a wing through the fuselage such that xz is zero.
xg = 1/2 rho V2 / (W/S) [ Clt St xt / S ]
If the tail incidence angle is zero, the tail moment is zero and xg 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.

• If the tail incidence is negative (negative aoa), xg 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.

In this case, the planes needs to fly at a particular airspeed, Vo, such that the tail moment offsets the moment cause by the CG. If it flies greater than Vo, the tail force increases, pitching the plane up, causing it to slow down. Similarly, if the plane flies slower than Vo, the tail force decreases, allowing the plane to pitch down, and increase speed.

A positive tail incidence would require a negative xg, 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.

• Now consider the use of a cambered airfoil with a non-zero Cm.
xg = 1/2 rho V2 / (W/S) [ Cm + Clt St xt / S]
The effect of the wing moment requires a neutralizing tail force, which can result in xg being zero. The tail incidence can be adjusted in the previous case requiring a non-zero xg producing a non-neutral and stable configuration. However, a non-zero tail incidence produces drag as well as lift.

• But when considering the all terms, we see that Cm can be offset by Cd xz, 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?)

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?

#### Dynamic Behavior

How does the plane react to a pertubation of its angle of attack?

#### Moment vs Center-of-Pressure

The following show the relationship between the moment coefficient and center-of-pressure
1/2 rho V2 S c Cm = 1/2 rho V2 S Cl x
where
 x distance from center of pressure to the 1/4 MAC location
Simplifying
c Cm = Cl x
and rearranging
x / c = Cm / Cl
The reference can be shifted from the 1/4 MAC location to the leading edge by adding a constant
x / c = 0.25 + Cm / Cl
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.