Question on descnding turn (load factor and stall speed)
09-23-2008, 07:29 AM
#1
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Question on descnding turn (load factor and stall speed)
Hi all,
My friend and I were having a discussion with respect to the following and I'd be most grateful if anyone could provide a solid answer?
I maintain that in a 45-degree bank angle descending turn the load factor, and hence the stall speed, is lower than in a 45-degree bank angle level flight (constant altitude) turn. Is that the case and if not, why not?!
Thanks for your help.
EiNY
My friend and I were having a discussion with respect to the following and I'd be most grateful if anyone could provide a solid answer?
I maintain that in a 45-degree bank angle descending turn the load factor, and hence the stall speed, is lower than in a 45-degree bank angle level flight (constant altitude) turn. Is that the case and if not, why not?!
Thanks for your help.
EiNY
09-23-2008, 08:10 AM
#2
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you are correct. In a constant bank LEVEL turn, load factor is increased. when load factor is increased, stall speed is increased proportional to the square root of the load factor. When in a descending turn you are taking some of that loading off the wing. remember, lift is OPPOSING gravity(weight,load). Think about it as letting gravity do its work by making your airplane go towards the ground, whereas if you stayed level the wings would actually have to work harder against gravity to keep you there...load factor.
09-23-2008, 12:33 PM
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Originally Posted by pagey
you are correct. In a constant bank LEVEL turn, load factor is increased. when load factor is increased, stall speed is increased proportional to the square root of the load factor. When in a descending turn you are taking some of that loading off the wing. remember, lift is OPPOSING gravity(weight,load). Think about it as letting gravity do its work by making your airplane go towards the ground, whereas if you stayed level the wings would actually have to work harder against gravity to keep you there...load factor.
As we know, in straight and level flight, you have a load factor of one.
If you pull back on the stick to start a climb the load factor increases above one as long as your rate of climb is increasing. Once you are established in a constant rate climb the load factor goes back to one.
If you push forward on the stick to start a descent the load factor will go below one, but as soon as you are established in a constant rate descent then the load factor returns to one.
The thing that confuses many people is that in a constant rate climb or descent lift is equal to weight. It is only during changes in rate that lift is not equal to weight.
During a constant rate, constant speed climb or descent it is thrust that is not equal to drag.
Joe
09-23-2008, 02:22 PM
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Joe gets the cookie on this one. If the airplane is in a constant speed descent or climb, then weight = gravity and the wings are holding 1g plus the amount of loading used to keep the turn coming. As we know turns require horizontal force and some wing loading, but the loading is the same whether descending or climbing for a given speed and turn rate.
If there is no turn going on, then the wing is loaded with exactly 1g in either descent or climb. Do not to confuse the forces in equilibrium steady flight with the momentary flux of forces associated with the beginning of an ascent, descent, or turn. At that time there is a momentary flux in the force equation as the forces rebalance due to the addition or removal of thrust or the change in angle of attack. In a descent the airplane starts to derive forward force in place of thrust by trading potential energy for kinetic, and in a climb it increases potential energy by the thrust derived from the prop. In either case as long as the airplane is not accelerating or decelerating the g forces are constant and they are in equilibrium.
Ref: FAA-H-8083-25 Pilot's Handbook of Aeronautical Knowledge.
If there is no turn going on, then the wing is loaded with exactly 1g in either descent or climb. Do not to confuse the forces in equilibrium steady flight with the momentary flux of forces associated with the beginning of an ascent, descent, or turn. At that time there is a momentary flux in the force equation as the forces rebalance due to the addition or removal of thrust or the change in angle of attack. In a descent the airplane starts to derive forward force in place of thrust by trading potential energy for kinetic, and in a climb it increases potential energy by the thrust derived from the prop. In either case as long as the airplane is not accelerating or decelerating the g forces are constant and they are in equilibrium.
Ref: FAA-H-8083-25 Pilot's Handbook of Aeronautical Knowledge.
Last edited by Cubdriver; 09-23-2008 at 02:36 PM.
09-23-2008, 02:53 PM
#6
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You would momentarily have a lift factor that is less than 1 if you roll into the bank without adding backpressure. As the vertical component of lift decreases, gravity pulls the airplane down, causing the initially lowered load factor. But, as mentioned before, the load will stabilize at 1 if the turn and descent are constant rate or speed.
09-23-2008, 06:20 PM
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Originally Posted by the King
You would momentarily have a lift factor that is less than 1 if you roll into the bank without adding backpressure. As the vertical component of lift decreases, gravity pulls the airplane down, causing the initially lowered load factor. But, as mentioned before, the load will stabilize at 1 if the turn and descent are constant rate or speed.
If you are in a 60 degree bank with a constant vertical speed, (up, down or zero) your load factor will be 2g. The vertical component of lift remains 1.0g, but the load factor increases wih the bank angle.
Note: This statement does not apply to those few aircraft (such as a light weight F-16 with the big engine) that are capable of constant rate vertical climbs, because the load factor (perpendicular to the wings) would be zero, and the bank angle in a vertical climb is undefined.
Joe
