Manouvering Speed and weight

1. Dec 1, 2014

Nimeo

Given an aircraft with a fixed MTOW and a maximum load factor of 6 g's. Let's assume that the Manouvering Speed is 120 Knots. Does it change with the actual weight of the airplane? The manouvering speed is the stall speed at the maximum load factor, in this case the maximum speed at six times the weight. If we reduce the take off weight the manouvering speed should remains the same as long as the the stress on the aircraft remains the same. The stresses depends on the force applied on the structure that depend on angle of attack velocity and area the surface immersed in the fluid. So at 120 Knots the lift generated by the wing is the same wether or not the weight of aircraft has changed. WHat have changed is the acceleration, but not the stresses. I'd appreciate any comments and suggestions

2. Dec 6, 2014

Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Dec 29, 2014

montoyas7940

Hmmm, 1 post a month ago... You are probabaly gone but here ya go.
Maneuvering speed increases with weight because there is less reserve angle of attack available as weight increases. Basically the heavier the particular aircraft the sooner the wing will stall thus protecting the structure from over-stress. That is the skeletal explanation. If you are still around and want more detail just ask!

4. Dec 30, 2014

Nimeo

The stress generated by the aerodynamic forces on the aircraft is independent of the weight of the airplane, i.e. : the lift generated by the wing depends only on speed, density surface and angle of attack and it's independent of weight. The only thing that changes when the weight varies it's the acceleration of the aircraft, so if you reduce weight and you pull full aft the stick the lift generated by the wing is the same regardless of the weight of the airplane, you'll have a greater acceleration at a reduced weight so by Newton's second Law d Mv / dt = F the same stresses on the structure...Imagine that you have an airplane designed for 6 g's at an MTOW of 1200 KG with a stall speed of 60 knots. The manouvering speed is 60 X Square Root of Load Factor. Now if you halve the weight you can pull 12 g's without overstressing the structure, but the wing will still generate the same 1200 X 6 = 7200 KG at the same maunevering speed.

5. Jan 1, 2015

montoyas7940

Maybe it would help to discuss just one component of fixed weight such as the engine. Let's pretend the engine mount is designed to withstand 4g.
A lightly loaded plane may be flying at 6 deg. AOA while a heavily loaded plane may need 9 deg. AOA for the same airspeed. Since the wing will always stall at the same AOA, if the plane is lightly loaded there is more reserve AOA available than would be if the plane were heavily loaded. A full scale elevator deflection on the lighter plane will impose more load on the engine mount before the stall (possibly exceeding the 4g limit). So a slower maneuvering speed is imposed for a lightly loaded plane thereby reducing reserve AOA.

You may already know Va (light) =Va (maxgross)[sqrt(lightweight/Maxgross)] I really should learn latex...