Why is there no lift component when calculating rate of climb? I read this post about how to calculate the rate of climb
How does climb-rate vary with density/pressure altitude?
So in the image, all the forces are drawn. I am wondering, why the rate of climb does not depend on the lift force (or not directly (?) ).
So the forces acting in the vertical direction are: a small drag component, a big lift component, a weight component and a thrust component.
A steady climb would mean, that all forces are in balance - so no acceleration in any direction. But on the other hand in on post it is said &quot;Calculate how much power is needed to overcome drag, and any excess can be used for climbing&quot; - so when there is more force available in one direction - there should be an acceleration in this direction?
I am a bit confused ... is the rate of climb = velocity in the body fixed airframe (so where the airplane is flying to) ? Or is this the vertical velocity in z-direction?
Thank you for your support says
Helmut
Power and force are different things. Power is force times velocity.

The calculation is done in the reference frame of the air mass. In the reference frame of the aircraft none of the forces are doing work, and the engine power all goes to accelerating the air, but we don't have formulas for that.

The lift is, by definition, orthogonal to the flight path, so it does not do any work. That's why it does not enter the equations directly (it does indirectly due to reduction of induced drag at high flight path angles).

The “power needed to overcome drag” is needed for overcoming the aerodynamic drag. In climb the “aft” direction is tilted own, so part of gravity is added to drag and the “excess can be used for climbing” means the engine is able to produce the higher thrust needed to balance the sum of them.

Rate of climb is vertical velocity, i.e. velocity in the $z$ direction. But for that derivation to work, you need the velocity in the $x$ (horizontal) direction too.