# 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 "Calculate how much power is needed to overcome drag, and any excess can be used for climbing" - 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

• Related- may help -- aviation.stackexchange.com/questions/40921/… . Note that a steeper climb angle is associated with a smaller lift force, not a larger lift force. This link should also help answer your question about whether the excess thrust causes an acceleration. This link makes the simplifying assumption of constant air density, IAS, and TAS-- to take things to another level and account for the acceleration required to increase the TAS as the aircraft climbs with constant IAS, see aviation.stackexchange.com/a/48855/34686 . Commented Feb 24, 2021 at 5:10
• There are several different questions here. You may be able to delete some of it without invalidating an existing answer, and re-post that content separately. (Perhaps it's too late now-- ) Commented Feb 24, 2021 at 6:14

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.

• I agree with most of the answer, but not with the 3rd bullet. In climb the (aerodynamic) drag does not increase. But the weight vector has a component along the path. The steeper the climb, the larger the component of weight in the along path direction. The extra power means the engine is able to produce the higher thrust needed to balance the drag + component of weight. # Commented Feb 24, 2021 at 10:19
• Not often discussed, but in a climb aerodrag may decrease due to lower lift requirement (less efficient prop must take up slack), enough to be included in calcs. Commented Feb 24, 2021 at 16:18
• @RobertDiGiovanni -- looks like second bullet point already has a reference to that. Although it's not only induced drag that is being reduced. If we hold angle-of-attack and L/D constant, and we increase power and increase climb angle, we actually end up with a reduced airspeed, and therefore reduced Lift and Drag. That's the only way that the force vector diagram can still balance, even though more of the Weight is opposed by Thrust rather than by Lift in the steeper climb. Also covered in aviation.stackexchange.com/questions/40921/… Commented Feb 24, 2021 at 21:36
• We certainly would take caution to reduce speed in a climb, but the point has interest in that prop efficiency increases with lower airspeed, as would holding AoA optimal. Commented Feb 25, 2021 at 2:27