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 Turning And Induced Drag
 mawingho From Hong Kong, joined Jul 2012, 41 posts, RR: 0Posted Fri Aug 17 2012 07:05:56 UTC (1 year 8 months 3 days 11 hours ago) and read 3868 times:

 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 From United States of America, joined Jul 2003, 9393 posts, RR: 27 Reply 1, posted Fri Aug 17 2012 07:30:40 UTC (1 year 8 months 3 days 11 hours ago) and read 3855 times:

 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.

 "Two and a Half Men" was filmed in front of a live ostrich.
 mawingho From Hong Kong, joined Jul 2012, 41 posts, RR: 0 Reply 2, posted Fri Aug 17 2012 07:36:46 UTC (1 year 8 months 3 days 11 hours ago) and read 3854 times:

 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?

 Starlionblue From Greenland, joined Feb 2004, 16908 posts, RR: 67 Reply 3, posted Fri Aug 17 2012 08:11:38 UTC (1 year 8 months 3 days 10 hours ago) and read 3845 times:

 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."
 mawingho From Hong Kong, joined Jul 2012, 41 posts, RR: 0 Reply 4, posted Fri Aug 17 2012 09:18:31 UTC (1 year 8 months 3 days 9 hours ago) and read 3818 times:

 Why is (1/cos30)*weight, any picture can illustrate this? is it the compoent of weight to make this formula?
 vikkyvik From United States of America, joined Jul 2003, 9393 posts, RR: 27 Reply 5, posted Fri Aug 17 2012 09:25:32 UTC (1 year 8 months 3 days 9 hours ago) and read 3809 times:

 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

 "Two and a Half Men" was filmed in front of a live ostrich.
 tdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80 Reply 6, posted Fri Aug 17 2012 12:51:59 UTC (1 year 8 months 3 days 5 hours ago) and read 3753 times:

 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?

 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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