# What is the relation of stall speed to the position of the center of pressure with respect to the centre of gravity?

Would the centre of pressure being aft or forward of the centre of gravity (CG), reduce or increase the stall speed in straight and level flight? I know it is a weird question, I just cannot figure out if a centre of pressure being ahead of the CG (rare) would increase or decrease the stall speed, or even change it at all.

The comment written by wbeard52 is correct however this does still not explain exactly why the stall speed is increased. As an airframe stalls at an angle of attack, which will result in CLmax.

If the CG is moved to a more forward position, more downward force will be required on the horizontal stabilator to counteract the higher moment of the lift generated by the wings. This will increase the apparent weight. As (apparent) weight is increased stall speed will increase. All other parameters are fixed, in this situation.

Stall speed formula : V = √( 2 W g / ρ S CLmax )

This is a loaded question as it can possibly deal with aircraft stability, maneuverability and controllability.

Figure - 5-23 FAA Pilot Handbook of Aeronautical Knowledge

Take a look at the picture above. For the forces to balance, there has to be one force in the opposite direction in between two other forces. CG (center of gravity) and T (tail-down force) are used to oppose CL (center of lift). The distance between CG and CL and CL and T are important as they provide the characteristics of stability, maneuverability and controllability.

A stall occurs when the wings reaches its critical angle of attack. At greater weights, the wings will have to produce greater lift. The T force is a weight that has to be countered by CL. If CL is closer to CG, less T is required and the stall speed will decrease. If CL is farther from CG, there is a need for more T and the stall speed will increase. Now, as pilots, we don't talk about CL moving but changing CG. If the airplane has a forward CG, there is a lot of length between CG and CL and a lot of T is required increasing the stall speed. If the airplane has an aft CG, the length between CG and CL is very small and T doesn't isn't required as much. As a result, an aft CG reduces stall speeds.

If CL is forward of CG for an airplane like what is pictured above, the aircraft will be unstable. There are two opposing forces on one side helping the other force cause a rotation. Try it.

Take a pencil between your two hands. On one hand, put your thumb and pointer finger on it. With the other hand use your pointer finger to push on the pencil in the middle. You should have forces just like in the picture above. Slowly move your two pointer fingers together with the pencil in the middle. At some point, the thumb will no longer be required to maintain stability. Without using your thumb now, move your pointer fingers away from each other. The pencil should rotate (very easily) out of your hand.

This is what would happen if the CL is forward of CG.

There are some airplane where the CL is forward of CG but to solve the stability issue, there is a canard which provide a nose-up force to keep the nose from rotating. I have read, that airplanes with canards are difficult to stall due to the fact the canard will stall before the main wings.

• "If CL is forward of CG […] the aircraft will be unstable." No, not necessarily. Only less stable, but still fine for regular flying. Jan 29, 2020 at 23:33

First of all, as answered in this question, centre of pressure of the whole aircraft and centre of gravity must coincide for quasi-steady flight. Since the 1G stall speed, by definition of Part 23 and 25 regulations, is essentially quasi-steady, this holds true.

While the maximum wing lift is a fixed number for a given Mach and Reynolds number, the total aircraft CLmax can vary due to tail lift contribution. As the CG moves forward, CLmax decreases due to increased tail down force for trim. The following relationship, found in Advisory Circular 25-7C, roughly captures the influence of CG on maximum lift coefficient, which you can use to compute the stall speed:

$$C_{L_{MAX_{CG2}}} = C_{L_{MAX_{CG1}}}[1 + \frac{\bar{c}}{l_t}(CG_2- CG_1)]$$

where $$\bar{c}$$ is the mean aerodynamic chord (MAC), $$l_t$$ is the tail arm, and CG is expressed in percent MAC/100.