Lift and Drag component question

AI Thread Summary
The discussion centers on understanding the equations for Lift (L) and Drag (D) in "Fundamentals of Aerodynamics" by John D. Anderson Jr. The user is confused about the geometric breakdown of these forces, particularly the roles of the normal force (N) and axial force (A) in the equations L = Ncos(α) - Asin(α) and D = Nsin(α) + Acos(α). A participant clarifies that Ncos(α) represents only the component of N in the lift direction, not equal to L, and explains that the equations utilize a transformation matrix for rotating forces based on the angle of attack (α). This rotation aligns the forces with the horizontal and vertical axes, helping to derive the components of lift and drag. The user expresses gratitude for the clarification, indicating a better understanding of the material.
NBAJam100
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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 I've picked up so I am 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. \alpha = angle of attack.

They have:

L=N\cos \alpha - A\sin \alpha
D=N\sin \alpha + A\cos \alpha

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

Thanks.
 
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NBAJam100 said:
… I see that N\cos \alpha is equal to L, so why are we subtracting the A\sin \alpha ? …

Hi NBAJam100! :smile:

(have an alpha: α :wink:)

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. :smile:
 
tiny-tim said:
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. :smile:

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!
 
Last edited:
NBAJam100 said:
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 I've picked up so I am 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. \alpha = angle of attack.

They have:

L=N\cos \alpha - A\sin \alpha
D=N\sin \alpha + A\cos \alpha

I am having trouble understanding why this is true, and why they put the +/- where they did. I see that N\cos \alpha is equal to L, so why are we subtracting the A\sin \alpha ? Same goes for the equation for D. I am sure I am 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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