Does angle of attack affect affect load factor?

Question

I have been reading FAA Part 23 recently, and I have some confusion regarding the flight envelope of an aircraft. In some references, the maneuvering speed of an aircraft is written as follows:

It says that the lift produced is equal to weight times the maximum positive load factor. When defining the load factor, some references quote that it is simply the acceleration the aircraft is experiencing, and some more references quote that it is simply the lift divided by the weight of the aircraft. After further reading, I have found that you can find this load factor by considering the force body diagram during a steady level turn. Let us set the AoA as alpha, roll angle as phi, and also consider that the wind x axis coincides with the local horizon x axis. When I formulate the force equilibrium of the aircraft with respect to the local horizon axes, I get the following equations:

If I multiply the former with sin(phi) and the latter with cos(phi), and substract the former with the latter, I get:

With n equal to 1/cos(phi) according to some references

That is still fine since it match the reference that I read. But my problem starts when I substitute n = 1/cos(phi) to the force equilibrium equation on the z axis, I get:

If we see the equation above, it doesn't match the lift used when calculating the maneuvering speed, nor does it match the definition of load factor equals L/W. After looking at the equation repeatedly, I conclude that the equation would match the definition of load factor that I have read if the AoA is zero. However, I then realize that the lift coefficient used is the maximum lift coefficient which certainly happens when the AoA is not zero. I'm still having trouble understanding this specific topic, I would really appreciate if someone can clarify this to me.

Thanks in advance

Answer 1

Yes angle of attack affects load factor! With a load factor > 1 during manoeuvres, a higher angle of attack allows the C$_L$ to be high enough to create the extra lift. At an AoA of zero there is very little lift created.

Usually, thrust is not considered in the gravity axis. In jet the L/D > 10, $\alpha$ < 15°, so disregarding the vertical component of thrust yields an error of 3% or so.

Answer 2

A dissection of the given formula yields the familiar lift equation:

L = Density × Area × Coefficient of Lift × V$^2$

L = Weight × max load factor (placarded + and - in G)

Angle of attack at Clift max would be the stall AOA, where coefficient of lift is greatest.

Using the placarded G forces as a limit, one can see that below a certain V, AoA will exceed stall angle before the G limit is reached. This is why one can manuever at lower speeds without overly stressing the airframe, but one must be careful not to stall the aircraft.

Above this "manuevering speed", an AoA less than stall is sufficient to produce enough G forces to exceed the G limit.

AoA is built into coefficient of lift in a generally linear manner. Actual data for the airfoil of your aircraft (coefficient of lift vs AoA and stall AoA) is available at airfoiltools. The load limits and manuvering speeds should be in the POH manual of your aircraft.