Lift and Drag component question


by NBAJam100
Tags: component, drag, lift
NBAJam100
NBAJam100 is offline
#1
Dec26-08, 02:05 PM
P: 161
Hey guys,

I current picked up Fundamentals of Aerodynamics by John D. Anderson Jr. and am having a little bit of trouble understanding some of the equations i am given. This is the first aero book ive picked up so im not too familiar with aero concepts or terms yet.

On page 20, they break down the Lift and Drag components geometrically. The symbols below are as follows: A= axial force and N=normal force. [tex] \alpha = [/tex] angle of attack.

They have:

[tex]L=N\cos \alpha - A\sin \alpha [/tex]
[tex]D=N\sin \alpha + A\cos \alpha [/tex]

I am having trouble understanding why this is true, and why they put the +/- where they did. I see that [tex]N\cos \alpha [/tex] is equal to L, so why are we subtracting the [tex]A\sin \alpha [/tex] ? Same goes for the equation for D. Im sure im overlooking something very obvious here, but id appreciate someone helping me see it.

Thanks.
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tiny-tim
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#2
Dec26-08, 02:30 PM
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Quote Quote by NBAJam100 View Post
I see that [tex]N\cos \alpha [/tex] is equal to L, so why are we subtracting the [tex]A\sin \alpha [/tex] ?
Hi NBAJam100!

(have an alpha: α )

No, Ncosα is not equal to L, it's only the component of N is the L direction.

cosα -sinα
sinα cosα

is simply the transformation matrix for a rotation (from axes x,y to x',y')

it converts the pair N,D from coordinates fixed in the aeroplane to horizontal and vertical coordinates.
NBAJam100
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#3
Dec26-08, 02:49 PM
P: 161
Quote Quote by tiny-tim View Post
No, Ncosα is not equal to L, it's only the component of N is the L direction.

cosα -sinα
sinα cosα

is simply the transformation matrix for a rotation (from axes x,y to x',y') …

it converts the pair N,D from coordinates fixed in the aeroplane to horizontal and vertical coordinates.
Hey Tiny!

Ahh! now that you point that out i see that is the transformation matrix for CCW rotation.

Now i understand that, but when you say it converts the pair N,D from coordinates fixed in the aeroplane to horizontal and vertical coordinates, do you mean it converts the pair N,A ..... or am i missing something completely?

So regardless, the basis for doing the CCW rotation is rotating the 2 target vectors by the angle of attack (which would make them match up with L and D) hence giving us the components of L and D?

[Edit] Wow! Now that you pointed out at CCW thing that i was neglecting, everything else in the chapter seems to make so much more sense when i look at it!!

Thanks a lot Tiny!

deepthishan
deepthishan is offline
#4
Dec26-08, 04:03 PM
P: 38

Lift and Drag component question


Quote Quote by NBAJam100 View Post
Hey guys,

I current picked up Fundamentals of Aerodynamics by John D. Anderson Jr. and am having a little bit of trouble understanding some of the equations i am given. This is the first aero book ive picked up so im not too familiar with aero concepts or terms yet.

On page 20, they break down the Lift and Drag components geometrically. The symbols below are as follows: A= axial force and N=normal force. [tex] \alpha = [/tex] angle of attack.

They have:

[tex]L=N\cos \alpha - A\sin \alpha [/tex]
[tex]D=N\sin \alpha + A\cos \alpha [/tex]

I am having trouble understanding why this is true, and why they put the +/- where they did. I see that [tex]N\cos \alpha [/tex] is equal to L, so why are we subtracting the [tex]A\sin \alpha [/tex] ? Same goes for the equation for D. Im sure im overlooking something very obvious here, but id appreciate someone helping me see it.

Thanks.

the (rather rough) sketch should help clear up ur doubt-just resolve forces vertically and horizontally
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