In the past, I used to have this idea that if all other factors remain constant, then climbing increases the rate of turn while descending decreases it. This idea was most likely planted into my mind from my past flight simulator experience.

Assume that the bank angle remains constant, so does the speed (TAS), and the aircraft is in a steady climb/descent (constant vertical speed). Will the rate of turn be any different from what it would be in level flight?

It is known that lift is less than the weight of the aircraft in a climb/descent, since thrust/drag are bearing the remaining weight. It is also known that the horizontal component of lift is what provides the centripetal force for a turn to take place. Does this have any affect on the rate of turn? what other factors affect it? or is the rate of turn unaffected?

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    $\begingroup$ Your first sentence (for a student pilot) indicates that ROT increases in a climb because airspeed is dropping. Very important to keep an eye on airspeed. Realisticly, we can start with climbs and descents of less than 10 degrees. Cos 10 is 0.985 so lift requirement does not change much but sin 10 is 0.173. This is the percentage of your aircraft weight as gravity "thrust" in a dive and what the engine must lift in a climb. Climbing or descending turns can be done by adding or subtracting power. Try a few in the simulator, let us know how you do. $\endgroup$ Dec 10, 2022 at 10:07
  • $\begingroup$ That's most likely how that idea got into my mind. rate of turn depends on TAS; in my simulator (X-plane), in climb after takeoff, I have to use relatively small bank angles for rate 1 turn, because my TAS is low. When I'm descending, I'm perhaps at 20~30k feet, and so the TAS is much higher. So for the same bank angle, I get unacceptably low rates of turn. I understood this concept some time ago while studying flight mechanics, but that's when I had this new problem: What if the TAS remained constant? Will the rate still vary in climb/descent, compared to level flight? $\endgroup$ Dec 10, 2022 at 10:36
  • $\begingroup$ This new problem that I had could not be answered through simulator - even if it's physics was perfectly identical to real life. For any appreciable changes in cos ∅, you need extremely large values of ∅ (angle of climb/descent), and at those extreme values, your TAS is changing all the time; it's practically impossible to maintain a constant TAS. So looks like this question cannot be answered by experimenting in simulator. $\endgroup$ Dec 10, 2022 at 10:47
  • $\begingroup$ But I now see much better why airliners use Mach (TAS) and AoA while I putter around using IAS. $\endgroup$ Dec 10, 2022 at 12:10

2 Answers 2


This is one of those situations where it is helpful to extrapolate to the most extreme possible cases.

Imagine an aircraft in a descending turn, with the descent path incrementally getting steeper and steeper. I.e. the glide ratio is incrementally getting poorer and poorer. The flight path is incrementally getting closer and closer to a vertical rolling dive, with the direction of the roll being toward the lower wingtip.

In less extreme cases, when the flight path is below horizontal, but not yet vertical, note that a non-zero roll rate is needed just to maintain a constant bank angle, with the direction of roll being toward the low wingtip. If the roll rate decreases below the required value, the bank angle will decrease.

At the extreme limit case where the flight path truly is aimed vertically downward, the wing is generating no lift, and the aircraft's weight is entirely supported by the drag vector. In this case turn radius is zero, and the bank angle is undefined, and the roll rate is no longer constrained by the dynamics of the turn. The pilot can set the ailerons to any position he or she desires, and the roll rate will be limited only by the aerodynamic "damping" effect in the roll axis.

Note that if the airspeed is stipulated to be constant, then as the flight path gets progressively more vertical, the component of the airspeed vector which is tangential to the circle of the turn-- i.e. the horizontal component of the airspeed vector-- gets progressively smaller. This means that a progressively smaller centripetal force component is needed to obtain a given turn radius, and also that a given centripetal force component will drive a progressively tighter (smaller) turn radius.

Of course the situation is complicated by the fact that the centripetal force component is not constant, because as the flight path gets progressively steeper, a progressively larger portion of the aircraft weight is borne by the drag vector rather than lift vector, so the lift vector gets smaller. Still, the net effect is that due to the decreasing turn radius, the turn rate will increase as the flight path gets steeper, given the assumption of constant airspeed and constant bank angle.

Essentially all the same logic applies to a climbing turn as well. The main differences are that 1) the climbing turn, it is the thrust vector rather than the drag vector that supports some of the aircraft weight, and 2) in the climbing turn, the direction of roll required to maintain a constant bank angle is toward the high wing tip, not the low wingtip, so if the roll rate falls below the required value, the bank angle will increase, and 3) the extreme limit case is a vertical rolling climb with the direction of roll being toward the wingtip that was the higher wingtip when the flight path was less vertical. The end result will be the same-- the turn rate will be higher when the flight path is more vertical than when the flight path is entirely horizontal, given the assumptions of a constant airspeed and a constant bank angle.

In the real world, of course, the idea of using one given constant airspeed for a wide variety of trajectories ranging from vertical rolling climbs to ordinary horizontal turns to vertical rolling dives doesn't make a lot of sense.

  • $\begingroup$ Indeed considering the extreme cases is how I understand flight mechanics; I couldn't wrap my head around the extreme cases in this particular situation, and your answer really helped me with that. However, I have a question. You state that the net effect of the two factors (reduced forward component of velocity and reduced horizontal component of lift) favours the former, causing the turn radius to decrease. I couldn't agree with that statement, since I believe that both factors must change in equal proportion, cancelling the effects of each other. (Continued...) $\endgroup$ Dec 10, 2022 at 7:24
  • $\begingroup$ Let ∅ denote the angle of climb or descent. We know that the forward component of velocity is a function of cos ∅. We also know that lift (and therefore it's horizontal component) is also a function of cos ∅. So we know that these two factors change proportionally. Then how does one outweigh the other? Thank you for reading. $\endgroup$ Dec 10, 2022 at 7:30
  • $\begingroup$ @AdityaSharma -- the answer admittedly isn't fully complete yet. It may take me quite some time to post a revised version $\endgroup$ Dec 10, 2022 at 13:23
  • $\begingroup$ No issues! just wanted to adress that I messed up somewhere: The forward velocity (forward TAS) is a function of cos ∅, but the radius of turn is proportional to the SQUARE of forward velocity; I completely forgot about that. $\endgroup$ Dec 11, 2022 at 0:26

Does climbing or descending affect rate of turn

Many people new to flying will only pull the yoke to climb and push to descend, resulting in an decrease in indicated airspeed while climbing and an increase while descending.

Since rate of turn is proportional to bank angle and inversely proportional to speed, what is going on here? With higher AoA, the same centripetal force is created for a lower airspeed, resulting in a higher rate of turn. Opposite for lower AoA. All fine and good, as long as you do not stall in your climbing turn, or exceed V manuvering in your descending turn.

Fast forward to the airliner simulator, 30,000 feet. "Unacceptably slow rate of turn". At higher altitudes, air density Rho comes into play in a much more significant way. TAS becomes increasingly greater than IAS as density decreases. If IAS is kept constant, TAS increases and rate of turn decreases with altitude, and while descending, TAS decreases and rate of turn increases (within the constraints of less than +/- 10 degrees from the horizon).

However, if TAS is kept constant, we can model our climbing/descending turn as a cylinder rather than a circle. Flying at the same TAS, more distance must be covered if the cylinder radius remains the same as the horizontal circle.

What follows is an amazing group of cosines canceling each other out:

  1. Cosine angle of climb/descent × Weight = Lift component
    Lift is less than weight in a climb/descent.
    less centripetal force

  2. Cosine angle of climb/descent × TAS = horizontal TAS
    if the radius of the horizontal circle component of the cylinder remains the same, horizontal TAS is slower
    rate of turn (around the horizontal circle) is lower

But what of the rolling component as pitch to the horizon increases? In a practical application of climbing in the range +/- 10 degrees, it may not factor in significantly$^1$.

$^1$ this would be revisited if radius of turn decreased

  • $\begingroup$ I think I didn't make my point clear. I know that the rate of turn varies with TAS, and so I'm interested in the case of a constant TAS. Will the rate of turn still vary in a climb/descent, compared to in level flight? $\endgroup$ Dec 11, 2022 at 0:23
  • $\begingroup$ @AdityaSharma some ROT reading as for how airliners turn. They seem to limit bank angle. to 25 degrees, and just take more time turning at higher TAS. Good to work with you. I learned some as well. $\endgroup$ Dec 11, 2022 at 1:27
  • $\begingroup$ Thank you for that link, that's some interesting information! $\endgroup$ Dec 11, 2022 at 4:51
  • $\begingroup$ Just noticed something in your second point of the cosine section. Wouldn't the radius of the cylinder reduce, if the forward component of TAS is made to reduce, while keeping the same net TAS? quiet flyer pointed this out, and I found it reasonable (for instance, as the flight profile tends to vertical climb/descent, the turn radius tends to zero). $\endgroup$ Dec 11, 2022 at 9:00
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    $\begingroup$ @AdityaSharma well, with all the progress, there is certainly no right or wrong here. At +/- 10 degrees, cosine values are little changed, so ROT should be only slightly changed. One for the test pilots to confirm. Thanks. Robert. $\endgroup$ Dec 11, 2022 at 15:08

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