Is lift less in a steeper descent than in a shallower descent, and if so, by how much?

For example, if we change our ratio of horizontal distance travelled to vertical distance travelled (through the airmass) from 10:1 (5.7 degree descent angle) to 5:1 (11.3 degree descent angle)-- which, if we hold the airspeed constant, very nearly doubles the sink rate-- have we decreased the lift force, and by how much?

For relatively shallow descent angles (say up to 15 or 20 degrees), does it really make sense to say that if we leave thrust constant, we can increase the descent rate and make the glide path steeper "by reducing lift", meaning the actual lift force as we would measure in Newtons or pounds? Or does the idea of descending by "reducing lift" have some other interpretation that is more accurate or meaningful?

Does the answer depend on whether we are increasing the descent rate and making the glide path steeper by opening spoilers, or by deploying flaps, or by reducing power, or by just putting the nose down and allowing the airspeed to increase, assuming we end up with the same change in glide path in every case?

For example, what really happens, in terms of the lift force, when we open spoilers to "dump lift" and make our descent angle steeper, while holding airspeed constant, without changing thrust? (Assume that the "spoilers" really do decrease the lift coefficient, at the instant that they are deployed.) After we are established on the new flight path, has the lift force really decreased, and if so, by how much? (Specifically, for the change in glide path described above.)

Assume the conventional definitions of lift and drag-- i.e. lift acts perpendicular to the flight path, and drag acts parallel to the flight path. Start by assuming that the thrust line is parallel to the flight path; feel free to elaborate on other variations if you feel that is warranted. Assume we are always talking about the glide path and sink rate through the airmass, i.e. what we'd achieve over the ground if there were no wind-- this is not a "trick question" about effects due to wind.

  • $\begingroup$ This question has a similar title aviation.stackexchange.com/questions/56718/… but it is really about a misconception on a table, plus a few other questions thrown in; the body of the question is not a duplicate. $\endgroup$ May 7, 2021 at 12:20
  • $\begingroup$ This question aviation.stackexchange.com/q/56254/34686, when correctly understood, has similarities to the present question, but the question is posed in a very different way and invites some very different interpretations $\endgroup$ May 7, 2021 at 12:35
  • 2
    $\begingroup$ Maybe I don't know enough, but this seems to be very broad. There are a lot of question marks in there. Maybe try narrowing the focus some? Or at least the question mark count... $\endgroup$
    – FreeMan
    May 7, 2021 at 12:55
  • $\begingroup$ Well, I could truncate it after the second or third paragraph-- $\endgroup$ May 7, 2021 at 12:56
  • $\begingroup$ @FreeMan -- better? $\endgroup$ May 7, 2021 at 12:58

3 Answers 3



in an unaccelerated descent - or ascent -, lift - the force acting perpendicular to the direction of motion - scales with the cosine of the angle of flight from horizontal. Hence, an aircraft in a 90 degree dive will have no lift.

This is independent of airspeed or the method by which lift reduction is achieved.

Here's a graph of lift Vs glide ratio

enter image description here

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Federico
    May 8, 2021 at 15:09
  • $\begingroup$ Um... lift is a force acting perpendicular to the relative airlow, therefore it is still present in a vertical dive, but now it doesn't oppose gravity ..... $\endgroup$
    – Raffles
    May 8, 2021 at 23:57
  • 2
    $\begingroup$ @Raffles No. If it were, the aircraft would accelerate towards the horizon and pitch upeas a result. $\endgroup$ May 9, 2021 at 0:06

you need to specify how that changes Lift/Drag ... Bianfable

That's the answer.

Lift requirement in descent is defined the same way as in a climb:

Cosine angle(from horizon) x weight

The rest of the vertical lift requirement is from the the vertical drag component. (In a climb it is from the vertical thrust component).

When you dump spoilers and flaps, this drastically increases the drag, requiring a higher angle of descent to maintain airspeed.

Flaps increase lift coeficient and increase drag allowing for a slower, steeper descent.

Spoilers decrease lift coefficient and increase drag, allowing for a faster, steeper descent. This very valuable to a glider trying to land under windy conditions.

But the lift requirement for stable, linear flight is determined by weight x cos angle to horizon, no matter what combination of drag, lift coefficient, and airspeed one uses.

  • $\begingroup$ @quiet flyer that's ok, I think you are correct that "residual thrust" does not make it more complicated, it is merely subtracted from drag in the glide ratio: D-T = Net Drag. A little thrust flattens the glide (same effect as reducing drag), more and D-T=0 D=T.(level flight). $\endgroup$ May 7, 2021 at 19:50
  • $\begingroup$ Re "Spoilers decrease lift coefficient and increase drag, allowing for a faster, steeper descent. This very valuable to a glider trying to land under windy conditions." -- food for thought-- I'd argue that in actual practice, a glider pilot can always increase speed via the elevator control, and I'd suggest that flaps (especially full-span, deflected near 90 degrees) do a better job of increasing the sink rate at high airspeeds than do spoilers (at least so long as we are still talking about glide angles of 20 degrees or less). $\endgroup$ May 8, 2021 at 17:38
  • $\begingroup$ (But flaps may have a redline for deployment lower than max aircraft redline.) So why do gliders have spoilers instead of flaps? I think there's a great ASE question in there, someone should ask it. Possible answers may include: better control of glide path at relatively low airspeed (e.g during the "float" before touchdown), ease of rapid deployment, ability to create a design that may be deployed at all airspeeds up to redline and prevents the aircraft from exceding redline even when lost in a cloud etc w/ nose straight down ... $\endgroup$ May 8, 2021 at 17:44
  • $\begingroup$ (ctd) , ability to create a design that may be deployed at all airspeeds up to redline and prevents the aircraft from exceding redline even when lost in a cloud etc w/ nose straight down (the latter point is not a universal feature of all glider spoiler designs.) $\endgroup$ May 8, 2021 at 17:45
  • $\begingroup$ @quiet flyer gliders have spoilers to essentially to "shrink the wing", this allows the plane to land at higher airspeed, perhaps less ground effect with spoilers. But yes, massively deflected flaps force a steeper approach and rapidly drop airspeed after the flare. Notice airliners do both, and reverse thrusters too! $\endgroup$ May 8, 2021 at 20:02

Other answers have already pointed out that

  • In a steady-state descent, for the simple case where the thrust vector is considered to act parallel to the flight path, lift =weight * cosine (descent angle).

  • Therefore, the larger the descent angle, the smaller the lift vector must be.

  • However, for reasonably small descent angles-- say 20 degrees or less-- this effect is very small, and (if we assume the thrust line acts parallel to the flight path), lift remains very nearly equal to weight.

For example, at a 10 degree descent angle, which in the context of an instrument approach would be considered absurdly steep, lift is still about 98% of weight. Going from a 10:1 descent ratio (5.7 degree descent angle) to a 5:1 descent ratio (11.3 degree descent angle), we very nearly double the sink rate (if airspeed is constant), but the lift vector only changes from 99.5% of weight to 98.1% of weight, a change of about 1.5% (i.e. the ratio between the two numbers is about 1.015).

Things are different at steeper descent angles, especially well over 30 degrees, where the (drag minus thrust) vector carries a significant part of the aircraft weight-- all of it, in the extreme case of a vertical dive. At steep descent angles, further increasing the descent angle does significantly reduce the magnitude of the lift force. (Also, at very steep descent angles-- above 45 degrees-- increasing the magnitude of the (drag minus thrust) vector actually decreases the sink rate, rather than increasing it.)1

As long as the descent angle is small, when we say we increase the descent angle and sink rate "by reducing lift" -- e.g. by opening spoilers-- it is technically true that the lift force is slightly reduced, but if we are envisioning a significant effect, what we really mean must be something else entirely.

In some cases, we might really mean that we are reducing the lift coefficient. The descent angle is determined by the ratio of Lift to (Drag minus Thrust), which in the power-off case, is exactly equal to the ratio of the lift coefficient to the drag coefficient. This gives some insight into how reducing the lift coefficient as well as increasing the drag coefficient, can help us to descend more steeply.

However, we have to be careful about how we apply this idea as well, for at least a couple of reasons. One, because it gives no insight into how we can descend by deploying flaps, or by reducing thrust. And two, because it gives no insight into any case where we are holding the airspeed constant. Whenever the airspeed is constant, the lift coefficient and the lift force are directly related, so the lift coefficient must be directly related to weight * cosine (descent angle). For shallow descent angles, we've already noted that the value of this expression only decreases slightly even when we make a large increase in the descent angle and sink rate.

So what is really going on when we increase our descent angle and sink rate by deploying spoilers, or deploying flaps, or reducing power, while holding airspeed constant?

If we deploy the flaps, this will have the initial effect of increasing the lift coefficient as well as the drag coefficient. If we don't want to lose airspeed, we'll need to decrease the angle-of-attack of the wing to return to (very nearly) the original lift coefficient. Decreasing the angle-of-attack in this manner slightly reduces the net increase in drag coefficient that would otherwise be created by deploying the flaps, but the drag coefficient is still higher than it was before we deployed the flaps. Unless the airspeed is quite low, we have more drag force and a higher descent angle and a higher sink rate when the flaps are down than when they are up. The higher the airspeed, the more pronounced this effect. When the flaps are deployed to a very large deflection, such as 45 degrees or more, lowering the nose to increase the airspeed can be a very effective way to create a very large increase in the descent angle. Another way to say this is to note that deploying the flaps creates a very large downward shift (toward a higher sink rate) in the far right-hand portion of the "polar curve" of airspeed versus vertical speed.

If we increase the descent angle by opening the spoilers, this will have the initial effect of reducing the lift coefficient as well as increasing the drag coefficient. If we don't want to gain airspeed, we'll need to increase the angle-of-attack of the wing to return to (very nearly) the original lift coefficient. The increase in angle-of-attack also further increases the drag coefficient. The net result on the sink rate and descent angle is somewhat similar to if we had deployed the flaps, except that there is no decrease in stall speed. Rather, the stall speed is raised.

(Note that increases or decreases in angle-of-attack (as measured at the fuselage or the unflapped part of the wing) do not directly correspond to identical increases or decreases in pitch attitude, especially if we're simultaneously making a change in configuration, which also affects the flight path. Note also that for simplicity this answer assumes wing area is constant-- the increased wing area due to deploying Fowler flaps, for example, is just treated as an increase in the lift and drag coefficients. Note also that throughout this answer, for simplicity we treat the thrust vector as acting parallel to the flight path through the airmass.)

At the end of the day, while flying at any given constant airspeed, for shallow descent angles, we increase our descent angle and our descent rate by reducing thrust or by adding drag. The closed vector diagram of Weight, Lift, and (Drag - Thrust) does force a small decrease in the value of the Lift vector whenever we increase the descent angle, but for shallow descent angles, this effect is very small.

Note that if we are operating on the "front side" of the thrust-required curve (i.e. not at extremely low airspeeds), simply moving the control stick forward is another way to add drag, even though the drag coefficient is actually reduced by the decrease in angle-of-attack. What we're really doing here is reducing both the lift and drag coefficients, which leads to a higher airspeed, which (for small descent angles) restores the lift force to (very nearly) its original value, but at cost of a net increase in drag. So the descent angle and sink rate are both increased.


  1. Equation needed, to prove that this statement is true, for a given constraint (e.g. constant lift coefficient? Or constant descent angle?)

Related ASE questions--

Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?

(this answer contains vector diagrams relevant to the present answer, and also includes (near the end) a discussion of effects arising from the thrust line not being parallel to the flight path)

'Gravitational' power vs. engine power -- the variation in the size of the Lift vector as the flight path changes from horizontal to descending comes into play here, exactly as it does in the present answer

Does lift equal weight in a climb?

  • $\begingroup$ Sorry, composing on phone is challenging, finished now-- ! $\endgroup$ May 8, 2021 at 22:30
  • $\begingroup$ Re footnote: I think I worked out the math on this once before but don't have it handy at the moment-- $\endgroup$ May 9, 2021 at 2:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .