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Here is my confusion : I found in my lecture script the following equation for the steady-state climb and descent (wings level)

L = m*g*cos(gamma) , with gamma being the kinematic climb angle (1)
(equation is simplified neglecting the Thrust angle of attack)

Now if an aircraft is flying at the same indicated airspeed (IAS) and the same absolute Vertical Speed (VS) once a climb (positive VS) and once a descent (negative VS), does it need the same lift in both situations ? (Neglect wind influence)

The kinematic climb angle should be equal in magnitude, but with flipped sign. According to this equation

gamma = asin(V_Vs./V_IAS); (2)

After plugging the values into (1) we get the same required Lift. Is therefore the angle of attack (AoA) equal in the two scenarios ?

Also I assumed that:

Theta = AoA + gamma

Thank you for your help and clearing up my confusion :)

Christian Mändle
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  • Hello Christian Mändle, and welcome aviation.stackexchange.com! Your question will be closed as a duplicate of Does lift equal weight in a climb?. The answer of Peter Kämpf addresses lift in descent as well. If there are – DeltaLima Oct 16 '23 at 14:39
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    This question is clearly not the same as the claimed duplicates. This question is about whether the effect is the same in climb and in descent. This is a very reasonable answer because the usual explanation for the lift deficit in climb is that a component of thrust is offsetting the weight -- which is clearly not identically true in descent.

    The short answer to the question is Yes. $L=W, \cos(\theta)$ does not care the sign of theta.

     – Rob McDonald Oct 16 '23 at 16:28
  • Interestingly it is oppositely true in a descent: the vertical component of drag is "offsetting" (some) of the weight – Robert DiGiovanni Oct 16 '23 at 21:23

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