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Old 28th September 2000 | 14:22
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Valiant
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Question Principles of Flight questions

I'm busily revising for ATPL exams and have got a couple of PoF questions that I cant figure out! I've spent ages looking through the books but to no avail. Can anybody tell me how to work it out.

Question goes:

An aircraft is straight and level at a CL of 0.35 and a 1 degree increase in AoA would increase the CL by 0.079. Following a vertical gust that gives a 2% increase in AoA, what increase in load factor would result.

Thanks very much,

Yours baffled and perplexed!

Valiant
 
Old 28th September 2000 | 17:35
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Dan Dare
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for straight and level flight. Since CL is proportional to angle of attack ( ), then for an uncambered wing and a 2% increase in produces a 2% increase in and a 2% increase in load factor n.

As proof:- when = 0.35 at = 4.43º

Now increase by 2% to get 4.52º. CL = .079 x 4.52 = 0.357.

Thus lift has increase by 2% and load factor has increased by 2%.

For an example with a cambered wing, where + there is insufficient information to complete this example.

 
Old 28th September 2000 | 17:42
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Dan Dare
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Red face

Pants! I carefully wrote out all the equations in Word before pasting to PPRuNe and they don't show! Any advice?

The bulk of my previous post was lots of equations proving that for an uncambered wing the answer is 2%, and that the answer is not determinable with the information given for a cambered wing section.
 
Old 28th September 2000 | 19:52
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Valiant
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Dan

Thanks very much for that, I've understood your explanation about uncambered wings.

I'm still left bewildered though as the answer is apparently 0.45

The question doesnt state whether uncambered or cambered, but there definately isnt any more info in the question if it's assuming cambered.


 
Old 29th September 2000 | 00:33
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John Farley
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Valiant

I suspect a misprint in the question.

If the question asked for the load factor increase due to a 2 degree (not 2 percent) increase in AOA then the load factor increase would be 0.079 x 2 divided by 0.35 which is 0.45.

If there is no misprint then there is no need to quote the the change in lift coefficient with respect to one degree of AOA in the first place (it could be a red herring I suppose)

(Don’t worry about the wing being cambered or not as it makes no difference to the method of calculation, which merely relies on the relationship between AOA and Cl being linear – which it will be on a certificated aircraft operating at those sort of Cls ie not at or past the stall. What varies with/without camber is the AOA for a zero Cl This is zero for an uncambered wing and some positive amount for a cambered one- how much positive depending on the amount of camber)

Happy?

JF
 
Old 29th September 2000 | 01:08
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Dan Dare
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Pipped to the post! Working from the answer back to the question then with 2 degree increse in alpha there would be a .451 increase in load to 1.45g.
 
Old 29th September 2000 | 11:43
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Valiant
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Dan / John

Thanks both very much for your help. The other question I had was along similar lines and I can figure it out now!

You're right about the misprint, I typed % in error - sorry!

BTW, John are you John Farley as in ex BAe / BAC? If so, we met a couple of years ago at Farnborough.

Thanks again

Valiant
 
Old 29th September 2000 | 21:04
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John Farley
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Thumbs up

Valiant

I will admit to BAe and HSA, but not BAC………..

(Sorry I can’t visualise your face right now as I don’t know who you are do I?!)

Glad to help

JF
 
Old 29th September 2000 | 22:02
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Valiant
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I dont think you'll remember me John.

Come to think of it it was actually 4 years ago at the 1996 airshow. If I remember correctly it was for the launch of the Summer air camp scheme with the sponsored Grumman AA5.

Cheers

Valiant
 
Old 30th September 2000 | 22:14
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John Farley
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Well that narrows the field a bit! G-DOEA has lost some of its stickers now. We have 25 organisations sponsoring the scheme these days. Including Lockheed and Boeing...

Cabair still get the kids airborne for us.

Best wishes for the exams

You have my email above this if I can help at all.

JF
 

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