We show that – contrary perhaps to many pilots’ intuition – lift in steady climbing flight is smaller than in straight-and-level flight.
Remembering that there are three fundamental forces in flight – weight, thrust, and resultant aerodynamic force (RAF) -, we remind ourselves that lift is defined at the component of RAF which is perpendicular to the flight path and lying in the plane of symmetry (x’z’-plane) of the aircraft. We shall also remember that in unaccelerated flight – as is the case in a constant-airspeed climb at a constant climb rate – all forces must cancel, otherwise there would be an acceleration in some direction.
Now we deal with the problem in the body frame of the aircraft, and for simplicity assume that the longitudinal (x’-)axis is aligned with the flight path. Then along the x’-axis:
$$T=D+W\sin\gamma,$$
where \(\gamma\) is the flight path angle with respect to the horizontal, \(W\) is weight, \(D\) is drag (by definition parallel to the flight path), and \(T\) is thrust.
Along the vertical (z’-)axis of the aircraft:
$$L=W \cos\gamma$$,
where \(L\) denotes lift. Since \(\cos\gamma<1\) for \(0<\gamma<2\pi\), it follows that lift is smaller than in straight-and-level flight. Thrust on the other hand is larger, because it now also opposes a component of weight. The latter is consistent with us having to add power in order to climb at the same airspeed.
Note that at the beginning of a climb lift is momentarily increased by the pilot, as we need an upward acceleration to start climbing. But this is only briefly at the beginning, until a steady climb rate is established.