I was part of a discussion about the differences between all engine climb gradients and one engine inoperative (OEI) climb gradients, and a colleague suggested that climb gradient can be calculated as

$G = \frac{T-D}{W}$


  • $G$ is Climb Gradient as a percent,
  • $T$ is Thrust,
  • $D$ is Drag, and
  • $W$ is Weight.

This derivation was never really explained, and it doesn't really make sense to me. He continues by stating that lift can be assumed to equal weight for small climb angles, so that his equation becomes

$G = \frac{T}{W} -\frac{D}{L}$

Where $L$ is Lift.

For some analyses, I can see using the approximation $W = L$, but since that is essentially assuming your climb gradient is 0, it doesn't seem prudent to use that assumption to calculate the climb gradient itself.

Has anyone seen these equations before? Or is there a piece to this that I'm missing that someone could explain to me?

  • $\begingroup$ Lift exactly equals weight at any climb angle so long as the aircraft is in an unaccelerated climb. $\endgroup$ Commented Aug 28, 2014 at 22:07

1 Answer 1


Assuming that thrust is entirely in the direction of travel and the flight path angle is constant, the Lift is equal to the weight times the cosine of the flight path angle:

$L = W\cos(\gamma)$

For small $\gamma$, $\cos(\gamma) \approx 1$

E.g. for a flight path angle of 10 degrees, the error introduced by the approximation is ~ 1.5%

enter image description here

For unaccelerated climb, the sum of all forces in the along path direction cancel each other out:

$ T-D-W\sin(\gamma) = 0$

The gradient is $\tan (\gamma)$.

Again, for small angles, $\sin(\gamma)\approx\tan(\gamma)$

From there it is a small step to see that the calculation suggested by your colleague is acceptable for unaccelerated climbs at small flight path angles.

  • $\begingroup$ Ah yes, the $sin(γ)≈tan(γ)$ part is the bit I had forgotten, I follow it now. Thanks for you help. $\endgroup$ Commented Aug 26, 2014 at 15:58
  • $\begingroup$ Excellent answer. However i didn't understand how you calculated the equation for unaccelerated climb. Could you elaborate on that a bit? Thanks in advance. $\endgroup$
    – KeyC0de
    Commented Aug 29, 2015 at 16:46
  • 1
    $\begingroup$ @RestlessC0bra Thrust is a force in the direction of travel, drag acts in the opposite direction. There is also a component of weight ( -sin(gamma) W) that counteracts the thrust. If there is no acceleration all these force must be in balance, hence their sum is zero. $\endgroup$
    – DeltaLima
    Commented Aug 29, 2015 at 20:07
  • $\begingroup$ If the flight is climbing without acceleration, with a particular value of γ (gamma), we have a particular ratio (T-D)/W. What happens to aircraft if suddenly increasing or decreasing T (thrust) ? $\endgroup$ Commented Jun 8, 2016 at 9:25
  • $\begingroup$ @d.pensopositivo Suddenly there will be an acceleration in the flight path direction which causes the speed to change. Since the line of thrust is often not though the centre of mass of the aircraft, it will also cause a pitch moment, causing the pitch angle to change. As a secondary effect, both the change of speed and the change of pitch angle will affect the lift and therefore the flight path angle with change, unless some form of pitch control will maintain the flight path angle. $\endgroup$
    – DeltaLima
    Commented Jun 8, 2016 at 9:32

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