Derivation of Aerodynamic Forces

[VyRianS]

Good day all:

I'm having trouble deriving the forces for Lift and Drag in the attached diagram.

The text I have (Fundamentals of Aerodynamics - John Anderson) states the solution to be:
L = N cosα - A sinα
D = N sinα + A cosα

Can anyone guide me through this? I think the cofunction identity is used somewhere:
sin(90 - x) = cos x
cos(90 - x) = sin x

But that's as far as I've gotten.

Help is appreciated, thanks!

 

[Samy_A]

Good day all:

I'm having trouble deriving the forces for Lift and Drag in the attached diagram.

The text I have (Fundamentals of Aerodynamics - John Anderson) states the solution to be:
L = N cosα - A sinα
D = N sinα + A cosα

Can anyone guide me through this? I think the cofunction identity is used somewhere:
sin(90 - x) = cos x
cos(90 - x) = sin x

But that's as far as I've gotten.

Help is appreciated, thanks!

For $L$:
$\cos (\alpha + x)=\frac{L}{R}$
So $L=R\cos(\alpha + x)=R\cos x \cos \alpha -R\sin x \sin \alpha \ \ \ \ (1)$
Now $\cos x=\frac{N}{R}$ and $\sin x=\frac{A}{R}\ \ \ \ (2)$.
Then (1) becomes: $L=N \cos \alpha -A\sin \alpha$.

Similarly for $D$:
$D=R \cos y= R \cos(\frac{\pi}{2}-\alpha -x)=R\sin(\alpha + x)=R\cos x \sin \alpha + R\sin x \cos \alpha$.
Using (2) again, you get $D=N\sin \alpha + A\cos \alpha$.

 

[NatSusan]

Hi there,

I'm not an expert in aerodynamics, but I'll try my best to help you with this problem. From my understanding, the lift and drag forces in aerodynamics are perpendicular to each other. In the attached diagram, the lift force (L) is the force acting upwards and perpendicular to the direction of airflow, while the drag force (D) is the force acting parallel to the direction of airflow.

To derive the equations for lift and drag, we can use the trigonometric identities that you mentioned. Let's start with the lift force:

L = N cosα - A sinα

In this equation, N is the normal force (the force acting perpendicular to the surface), and α is the angle between the normal force and the direction of airflow. Now, if we look at the diagram, we can see that the normal force (N) is equal to the weight of the object (W), which is acting downwards. So, we can rewrite the equation as:

L = W cosα - A sinα

Next, we can use the identity sin(90 - x) = cos x to rewrite the cosα term:

L = W sin(90 - α) - A sinα

Now, since the lift force (L) is acting upwards, the sign in front of the second term should be positive. This is why we have a positive sign before the A sinα term in the original equation.

Moving on to the drag force, we can use a similar approach:

D = N sinα + A cosα

Again, we can replace N with W and use the identity cos(90 - x) = sin x to get:

D = W sinα + A sin(90 - α)

Since the drag force (D) is acting parallel to the direction of airflow, it should have a negative sign in front of the second term. This is why we have a negative sign before the A sin(90 - α) term in the original equation.

I hope this helps to clarify the equations for lift and drag. Let me know if you have any further questions. Good luck with your studies!