mawingho
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Turning And Induced Drag

Fri Aug 17, 2012 2:05 pm

In the Bank Angle section of the material I am reading, it said "Turning decreases climb performnace due to the additional induced drag".

As I know, Induced Drag is caused by Lift. When turning / rowing, upper wing's lift increase and lower wing's lift decrase which means that the induced drag in total is the same, right?

Then why turning can increase the induced drag?

Hope someone can help me out. Thanks!
 
vikkyvik
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RE: Turning And Induced Drag

Fri Aug 17, 2012 2:30 pm

Quoting mawingho (Thread starter):
As I know, Induced Drag is caused by Lift. When turning / rowing, upper wing's lift increase and lower wing's lift decrase which means that the induced drag in total is the same, right?

No. It is true that the upper and lower wing have slightly different amounts of lift in a turn, but in a standard turn without loss of altitude, they are both ABOVE regular unbanked lift.

When you bank, you tilt the lift vector by the bank angle. Gravity doesn't tilt with it, though, so now you have less vertical lift than gravity, and the airplane will start to descend. So you apply back pressure to maintain altitude, which increases the lift coefficient of the wings, which equals more induced drag.

So for a 30 degree bank, for example, the wings have to produce:

lift = (1/cos30)*weight = 1.154*weight.
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mawingho
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RE: Turning And Induced Drag

Fri Aug 17, 2012 2:36 pm

Quoting vikkyvik (Reply 1):
So for a 30 degree bank, for example, the wings have to produce:

lift = (1/cos30)*weight = 1.154*weight.

Thank you for your reply. However, I am not quite sure how to do this calculation, can you explain more a bit?
 
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Starlionblue
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RE: Turning And Induced Drag

Fri Aug 17, 2012 3:11 pm

What material are you reading? What part of the equation are you having difficulties with?
"There are no stupid questions, but there are a lot of inquisitive idiots." - John Ringo
 
mawingho
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RE: Turning And Induced Drag

Fri Aug 17, 2012 4:18 pm

Why is (1/cos30)*weight, any picture can illustrate this? is it the compoent of weight to make this formula?
 
vikkyvik
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RE: Turning And Induced Drag

Fri Aug 17, 2012 4:25 pm

Quoting mawingho (Reply 2):
Thank you for your reply. However, I am not quite sure how to do this calculation, can you explain more a bit?

Ummm, not really...

The lift generated in a banked level turn is approximately equal to the weight of the aircraft divided by the cosine of the bank angle (where theta is the bank angle):

L = W / cos(theta)

(that's the same equation I wrote before, I just removed the "1/" since it was unnecessary)

See here:

http://en.wikipedia.org/wiki/Bank_angle#Banked_turn_in_aeronautics
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tdscanuck
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RE: Turning And Induced Drag

Fri Aug 17, 2012 7:51 pm

Quoting mawingho (Thread starter):
When turning / rowing, upper wing's lift increase and lower wing's lift decrase which means that the induced drag in total is the same, right?

No. Induced drag is higher on both wings, although it's even higher on the upper wing.

Quoting mawingho (Reply 4):
Why is (1/cos30)*weight, any picture can illustrate this?
http://www.free-online-private-pilot-ground-school.com/images/normal_slipping_skidding.gif

Quoting mawingho (Reply 4):
is it the compoent of weight to make this formula?

Yes. It's the vertical component of lift.

Tom.

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