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
  • $\begingroup$ Thanks for the recent additions! I am currently in the process of analysing the new answers and chat discussions. $\endgroup$ Mar 16 at 5:28
  • $\begingroup$ Bear in mind that you can’t keep all other factors constant, eg either airspeed, rate of turn and turn radius can’t all stay the same if you climb or descend. $\endgroup$
    – Frog
    Mar 16 at 21:04

4 Answers 4


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
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    $\begingroup$ @MarkJonesJr. -- why? $\endgroup$ Mar 15 at 1:01
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    $\begingroup$ @MarkJonesJr -- the point is that we are considering steady-state situations, not dynamically changing ones. In the steady-state case of a vertical descent, the wing must be at the zero-lift angle-of-attack, otherwise we would be generating lift which would cause the flight path to curve away from the desired vertical line, just as if we were pulling out of a loop. Unless we want to get off into the weeds with additional trivia like maybe the thrust line isn't parallel to the relative wind at the zero-lift aoa etc but that's not the main point at hand here. $\endgroup$ Mar 16 at 13:23

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

climbing increases the rate of turn while descending decreases it

Vertical speed affects the rolling moment equilibrium as explained here. Climbing will roll the airplane into the turn and descending will roll it out of the turn. However, the changing bank angle should be corrected by the pilot, just as is the sideslip angle which might develop when turning.

However, the opposite is true with regards to the maximum rate of turn!

Turning requires extra lift to add centripetal force, which in turn increases drag and requires extra thrust. Climbing also requires more thrust while descending frees up potential energy which would accelerate the airplane if the engine weren't throttled back and allows to create more lift with full thrust for a tighter turn. Unless your angular velocity is limited by lift and excess thrust remains, the more thrust goes into climbing, the less will be available for turning.

Maximum turn rate diagram

Maximum turn rate diagram for two flap settings: 0° (blue lines) and +10° (red lines). Own work.

Here you see the computed turn rate (Drehrate) over Mach for two flap settings (red and blue lines) and different climb rsp. descent speeds. While limited thrust at 10 m/s climb speed does not leave enough excess thrust to pull even 2 gs (with a turn rate below 4°/s), with -40 m/s climb speed (= 40 m/s sink speed) so much more thrust is freed up that the maximum load factor of 2.66 is easily exceeded and the turn rate reaches 7°/s. I did this calculation for a HALE platform which explains the low g limit. Values above Mach 0.75 are not realistic because I did not properly include transsonic effects.

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    $\begingroup$ This is interesting info but does it really address original question which specified "Assume that the bank angle remains constant, so does the speed (TAS)"? I don't see how the graphic shows how for a given bank angle and TAS, the turn rate will vary according to whether the aircraft is descending, climbing, or flying level. $\endgroup$ Mar 16 at 11:36
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    $\begingroup$ Likewise the arguments you give in the text of the anwer seem focused on total G-load (which is related mainly to bank angle, but also to climb or descent angle) that can be achieved with a given amount of thrust before the aircraft is forced to decelerate due to excess drag. But what if we assume unlimited power is available and want to explore how turn rate varies with climb or descent angle, for a given TAS and bank angle? $\endgroup$ Mar 16 at 11:46
  • $\begingroup$ @quietflyer Acknowledged. Please look at the first paragraph now, I added it to account for unlimited thrust. $\endgroup$ Mar 16 at 17:16
  • $\begingroup$ Perhaps our biggest challenge is that we are having to make a lot of unrealistic assumptions to answer this question. First, unless you are flying a jet fighter, you will be unable to make a sustained speed climbing turn at more than, say, 15 degrees of upward Pitch and you will be climbing into thinner air. Conversely, unless you have dive flaps, you will be unable to make a sustained speed descending turn at more than, say, 15 degrees of downward Pitch. So we are working within a very limited range. To answer this question, we have to remove all realistic elements from our flight models. $\endgroup$ Mar 16 at 19:20
  • $\begingroup$ @PeterKämpf First of all, thank you for taking your valuable time to write an answer! the information you shared is very useful from a practical standpoint. However, it doesn't seem to answer my original question (even after the edit). In particular, it doesn't explain the effect of climb/descent on rate of turn assuming a constant bank angle, VS and TAS, as quiet flyer mentioned above. $\endgroup$ Mar 23 at 11:00

For several years, I have been working on a single-vector version of a Flight Simulator Module which uses speed as the primary vector where Thrust, Drag and Gravity_x_Sin(Pitch) change the magnitude of the Speed Vector and Lift and the Gravity_x_Cos(Pitch) change the direction of the Speed Vector. As seen in this example, Lift acts along the Pitch Plane and Gravity_x_Cos(Pitch) acts along the Vertical Plane.

The OP asked a question which I had also been wondering about. I wanted to both validate my Flight Simulator Module and to compute the Lift settings necessary to hold the Airplane stable in a climbing turn. My initial goal was to simply to mention a few things and hope someone else would come up with a definitive answer.

However, while doing so, I felt that I had stumbled across a vector-based solution when I realized that the horizontal rate of turn is the result of two factors - the horizontal component of the Speed Vector and the horizontal component of the Lift Deflector. In a level banked turn, the Total Lift required equals Gravity/COS(Bank). The vertical component of the Lift equals the Gravity Deflector and the horizonal Lift equals Lift_x_SIN(Bank). Thus the Horizontal Turn Rate equals the (Horizontal Lift/Speed)x(PI/180). The entire equation is:

  • Turn Rate = ((Gravity/COS(Bank))_x_SIN(Bank)/Speed)x(PI/180)

This exactly matches the result given by the "old standard" equation of:

  • Turn Rate = TAN(Bank)x(Gravity/Speed)x(PI/180)

When Pitch changes, two things happen: (1) the Lift required to maintain climbing Flight decreases since the Gravity Deflector is Gravity_x_Cos(Pitch); and (2) the Horizontal Component of the Speed Vector also decreases and equals Speed Vector_x_Cos(Pitch). Since Cos(Pitch) appears in both the numerator and the denominator, my conclusion was that the Turn Rate stays the same. While I initially did not like this answer, because I assumed that the Turn Rate should decrease to zero, I later realized that it is the Turn Radius that is decreasing to zero. (And if Turn Rate did decrease, Turn Radius might remain the same or even increase - an equally unappealing result.)

So I set about to create an example of my flight model which would help determine the correct answer. As can be seen, the latest example appears to confirm my initial conclusion - that the Turn Rate remains identical and that Turn Radius decreases.

This appears to be the answer reached by other respondents - apparently confirmed with some "heroic" real life tests.

I had been hoping to further validate my conclusions with a mathematical explanation of what the Flight Simulation Model is doing (it uses linked rotating objects) - but have not done so yet. In the meantime, I have been able to use this knowledge to compute "autopilot" values for the coefficient of Lift which keep the airplane in a steady climb - which seems to be a further validation. So, at the moment, I am happy with what I have learned and hope that it is useful to others.

Note that this is largely a hypothetical exercise with very little real world application since most aircraft will be unable to maintain a fixed speed at Pitch +/- 15 degrees and because air density almost always decreases as you climb and increases as you descend, both of which affect your turn rate.

Thanks to everyone (especially quiet flyer) for their comments and suggestions!


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