oldveedubs

Like the title says...I'm just wondering. I'm watching Modern Marvels : B2 and I was thinking about a BWB aircraft (flying wing) and its ability to create any lift in a 90 degree bank. Lemme know your thoughts...

oldveedubs

Bump...where are my aeronautical engineers on this?!

aerospacepilot

Here is the best explanation I can give you. There is not a yes or no answer, but I think you will understand after you read this.

When an aircraft is flying straight and level, the lift is being used to counteract the weight. As the aircraft begins to bank, the lift vector is still defined as perpendicular to the wing, regardless! At a modest bank angle, most of this lift is still being used to counteract weight, however, some of the lift is being used to pull the aircraft through the turn. Lift is the force that is causing you to turn. As the bank angle increases, more lift is being used to pull the aircraft through the turn, while less and less is being used to counteract weight.
When broken into components at a bank angle of theta, the amount of lift counteracting weight is proportional to cos(theta) while the amount of lift pulling the aircraft through the turn is proportional to sin(theta).

For example:
At a 30 degree bank, 86.6% of the original lift force is still holding the aircraft up (counteracting weight). To maintain level flight, slight back pressure must be applied. This is because you must increase the angle of attack to maintain 100% of the lift force to remain level. 50% of the original lift force is pulling the aircraft through the turn. You will notice these two numbers (86.6 and 50) do not add up to 100. No I am not going crazy. This is because they are components. If you use the pythageoran theorem, you will notice that the magnitude is still equal to 100% of the total lift.
If you now increase to a 55 degree bank, only 57.3% of the original lift force is being used to counteract weight, while 81.9% of the original lift force is being used to pull the aircraft through the turn. This is why you turn faster at 55 degrees bank, but you must apply a lot of back pressure to increase the angle of attack, to increase the lift, to maintain a constant altitude.

As you go to 85 degrees of bank, only 8.7% of the original lift is counteracting weight. At 90 degrees, 0% of the lift is being used to counteract weight. No lift is able to hold the aircraft at a constant altitude. Lift is still being produced (sort of). It is pulling the aircraft through the turn, not holding it in the air.

Two things to add. If you try to put a Cessna-172 into an almost 90 degree bank, it will stall. You would have to apply so much back pressure on the elevator that you would exceed the stalling angle of attack of the airplane. You still produce lift when the wing stalls, but it drops dramatically as I am sure you know.

Some aircraft can maintain level flight in a 90 degree bank (mostly military aircraft). How do you ask? Well, the force from lift is not holding it in the air. It is the centripetal acceleration of the aircraft that is keeping it up. Imagine a ball attached to the end of a long string. Swing it slowly, and the ball will stay low. However, swing it at a fast enough speed, and you can keep the ball level with your eyes. This is centripetal acceleration that is keeping the ball in the air. This is what allows some aircraft to bank at 90 degrees. In order to achieve this, they must be powerful, and must be able to stall at a higher than normal angle of attack.


So, the answer to your question is:
Technically, an aircraft can produce lift at 90 degree angle of bank. None of it is being used to counteract weight (and keep the aircraft in the air), all of it is being used to pull it through the turn. A cessna-172 probably will not produce lift in a 90 degree turn because the angle of attack would be so great it would stall. Some military aircraft will be able to bank at 90 degrees, but it is not lift that is holding them at a constant altitude.

A flying wing aircraft will not be able to maintain a constant altitude at 90 degrees bank because of its wing generating lift to counteract weight. It will only be able to do this if it is powerful enough, fast enough, and stalls at a high enough angle of attack so centripetal acceleration can keep it at a constant altitude.

I hope you were able to understand this. If you have any more questions, feel free to ask!

Cubdriver

Structural damage occurs above Va and below that it stalls rather than tears off the wings. I assure you it would be one or the other of these in a 90-degree turn, depending on its speed, and this is why above 60 degrees is unlawful for normal-category aircraft.

Military jets can bank at very high angles because they can withstand more G's. They cannot do true 90 banking however, because at this bank angle the N-loading is infinite. Therefore they cannot fly sideways, at least as a theoretical ultimate, although they can get very close.

Of course if they are willing to descend, or trade off some climb rate, they can bank to any amount required and this allows rolls and other acrobatics.

Around 86 degrees in non-descending flight the N loading is is something like 12 or 14, which most military jets can withstand. I do not know about pilots, I would think 8 or 9 G's is about it for them, but if the airplane is built so strongly that it can withstand a full 90 degree bank in nondescending flight, there eventually will still have to be driftdown due to gravity because there is no upward component of lift coming from the wings or jets. It is like a rocket that flies horizontally; it eventually arcs back to earth unless there is an upward component of thrust from the rocket motor.

The rudder could be turned to produce some lift, but I assume the question is that lift is not produced from the rudder or engine. If rudder were deflected enough the body of the aircraft will produce some lift, and none of the lift would be from the main wings and the airplane would fly that way. I doubt the B-3 can generate any such lift, because it is all wing. Also, at that angle force from the thrusting engines being tilted upwards would come into the equation and I did not think these alternative sources of force were part of the question.