For an aircraft in a glide, what—if any—is the optimal bank angle to make a given turn while maximizing glide distance and maintaining published best glide speed?

In other words, at a constant airspeed, which bank angle will produce the greatest turn rate with the smallest loss in altitude. Assume a constant airspeed turn that is already established. I am not asking which bank angle will produce the least reduction in the vertical component of lift.

In envisioning this scenario, I specifically have in mind a single engine aircraft experiencing an engine failure at low altitude. However, the aerodynamic theory may apply to any fixed wing scenario.

Edit: let me emphasize that this question assumes the pilot is maintaining the aircraft's nominal published best glide speed throughout the turn, not adjusting airspeed to some variable theoretical best glide speed. Airspeed should not be a variable.

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    $\begingroup$ this is a very tricky question because so many factors come into play. $\endgroup$
    – rbp
    Commented Jan 10, 2016 at 23:43
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    $\begingroup$ @rbp There are certainly a lot of factors that go into describing the scenario, but I think the solution is probably rather simple. One of my instructors, who was a former Boeing engineer, used to tell me that the answer was 45°. If that is true, I'm looking for the why. $\endgroup$
    – J W
    Commented Jan 11, 2016 at 0:12
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    $\begingroup$ Given the last paragraph, this might be duplicate of aviation.stackexchange.com/q/786/524, since that question does discuss the bank angle and speed to use for the turn. $\endgroup$
    – Jan Hudec
    Commented Jan 11, 2016 at 8:07
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    $\begingroup$ Why are people so eager to close questions on SE Aviation. e.g. This is an excellent question with no obvious answer as far as I can see anywhere on SE Aviation nor on the internet. Yet, some trigger happy fellow wants to close it already as a duplicate. $\endgroup$ Commented Jan 14, 2016 at 6:48
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    $\begingroup$ Please clarify the specific goal. All turns have radius in addition to turn rate, a 180 degree turn does not put you back where you started facing the other direction. A 90 degree turn can be started sooner or later to match the radius with desired exit track. Just eyeballing it you would want a fairly steep initial turn followed by a slow rollout kind of like a chandelle, this would quickly arrest movement in the original direction while sacrificing some glide ratio, then restore the glide ratio as soon as you are moving in the desired general direction. A real answer is knee-deep in calculus $\endgroup$
    – Max Power
    Commented Feb 25, 2023 at 23:53

6 Answers 6


The best bank angle is indeed 45°.

It can be shown relatively easily that that gives the tightest turn for constant angle of attack, where we assume that constant angle of attack results in constant angle of descent too. It would be pretty hard to show for constant speed, and it might not even be exactly true in that case.

However, this can't really be separated from the best speed. And the surprising answer is that the stall speed (which is ~19% ($\sqrt[4]{2}$) higher than in straight flight) gives smallest loss of altitude for turning given number of degrees.

The reason is that the drag (near the stall speed) is proportional to $\frac{1}{v}$, but the turn radius is proportional to $v^2$, so as you slow down, the radius decreases faster then the drag (and therefore vertical speed) increases.


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    $\begingroup$ "on the back of the power curve"? I thought this was an engine out scenario $\endgroup$
    – rbp
    Commented Jan 11, 2016 at 13:13
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    $\begingroup$ @rbp, "back of the power curve" is simply below best glide speed, i.e. in the range where lower speed means higher drag and whether the power to overcome that drag comes from engine or from descending is secondary. $\endgroup$
    – Jan Hudec
    Commented Jan 11, 2016 at 13:25
  • $\begingroup$ i'll tell that to my glider pilot friends. I think you mean "back of the lift curve?" $\endgroup$
    – rbp
    Commented Jan 11, 2016 at 13:26
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    $\begingroup$ @rbp The term "power curve" refers to the fact that the curve is a polynomial -- i.e. Y equals X raised to a power (or negative X, times X raised to a different power, plus a constant, etc.etc.etc.). It has nothing to do with engine power. $\endgroup$ Commented Sep 1, 2019 at 22:06
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    $\begingroup$ @hiddenwindshield wrong on several different levels :) when used to describe a generic function, "power" usually refers to an exponential, not a polynomial. The curve of D/L as a function of airspeed is neither an exponential not a polynomial, and its name does come from "power required for lift to equal weight". That power can come from an engine or from expending gravitational potential energy or any combination. $\endgroup$ Commented Sep 5, 2019 at 4:51

The question as asked is open to interpretation, so I will first rephrase it to have a basis to build upon. Your last paragraph tells me that you want to know the optimum bank angle to get the highest ratio of turn rate to altitude loss in a glide at a given airspeed.

Spoiler: Since steeper bank angles require more lift, and aircraft with better L/D are more efficient in producing lift, the optimum bank angle depends on the aerodynamic qualities of the aircraft.

What is given

  • Glider or powered aircraft with inoperative engine. The polar and the weight are known and do not change over time.
  • Airspeed. This will result in a restricted optimum - the absolute best bank angle will require a suitable speed.

What can be changed

  • Bank angle $\varphi$ (obviously - you are asking for this)
  • Lift $L$ (again, obviously. You want to stay airborne)


First I need to formulate the ratio of turn rate over height loss. This then needs to be derived with respect to the bank angle and set to zero. To have a derivable polar, I use the quadratic polar where $c_D = c_{D0}\cdot\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$.

I further assume a coordinated turn, so we can define the lift and drag equations. Drag is compensated by selecting a suitable glide path angle $\gamma$ in order to convert potential into kinetic energy to keep the speed constant. The angular velocity $\Omega$ in a turn with the radius $R$ is $$\Omega = \frac{v}{R} = \frac{g\cdot tan\varphi}{v} = \frac{g\cdot \sqrt{n_z^2-1}}{v}$$ Height loss over time is vertical speed $v_z$, and this can be calculated from speed $v$ and flight path angle $\gamma$: $$v_z = v\cdot sin\gamma$$ Since $v$ is given and constant, we can rephrase the problem as a maximization of turn rate over flight path angle or sink speed. This is equivalent to the smallest height loss for a given azimuth change. $$\frac{\Omega}{v_z} = \frac{g\cdot tan\varphi}{sin\gamma}$$

Before deriving this, we need to express $\gamma$ in terms of $\varphi$. If we had the liberty to adjust speed, we could directly solve for the optimum bank angle at optimum L/D. Now, however, speed is fixed and L/D is what the airplane produces at the required lift. Since for gliders $sin\gamma = \frac{c_D}{c_L}$, we can write: $$\frac{\Omega}{v_z} = \frac{g\cdot tan\varphi\cdot c_L}{c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}} = \frac{g\cdot sin\varphi\cdot \frac{m\cdot g}{q\cdot S}}{c_{D0}\cdot cos^2\varphi + \frac{\left(\frac{m\cdot g}{q\cdot S}\right)^2}{\pi\cdot AR\cdot\epsilon}}$$ with $c_L = \frac{m\cdot g}{q\cdot S\cdot cos\varphi}$. Since the dynamic pressure $q$ is constant, we can now derive with respect to the bank angle. With the chain rule we get a fraction, and since it will be set to zero, it is enough to look for the condition when the numerator is zero: $$g\cdot cos\varphi\cdot \frac{m\cdot g}{q\cdot S}\cdot\left({c_{D0}\cdot cos^2\varphi + \frac{\left(\frac{m\cdot g}{q\cdot S}\right)^2}{\pi\cdot AR\cdot\epsilon}}\right) = g\cdot sin\varphi\cdot \frac{m\cdot g}{q\cdot S}\cdot 2\cdot c_{D0}\cdot sin\varphi\cdot cos\varphi$$ $$\frac{\left(\frac{m\cdot g}{q\cdot S}\right)^2}{c_{D0}\cdot\pi\cdot AR\cdot\epsilon} = 2\cdot sin^2\varphi - cos^2\varphi = \frac{1}{2} - \frac{3}{2}cos2\varphi$$ $$\varphi = \frac{1}{2}\cdot arccos\left(\frac{1}{3} - \frac{2\cdot\left(\frac{m\cdot g}{q\cdot S}\right)^2}{3\cdot c_{D0}\cdot\pi\cdot AR\cdot\epsilon}\right)$$ This does not obviously look wrong, but I could very well have screwed up on the path to the result. If you plug in the numbers for an airplane you know, you can check whether the result makes sense. At least, with too low airspeed you get a negative argument for the cosine which mathematically means a roll angle of >90° and can be interpreted as too slow for that turn.


Now we got a similar question but with both speed and roll angle as variables. Obviously, now we need to derive both with respect to speed and roll angle. But it is more fun to plot the results over these two as a contour plot. I just had to do this since several answers here claim that the optimum angle is 45°. Equally obviously, this is too simplistic.

First the math: I start from the same equations as above and added a term for wind ($w_z$) which adds rising or sinking air mass to the problem. $$∆h = \frac{\pi\cdot v}{g\cdot tan\phi}\cdot(v_z+w_z) = \frac{\pi\cdot v}{g\cdot\sqrt{n_z^2-1}}\cdot\left(\frac{v\cdot c_D}{c_L}+w_z\right)$$ Expressing the lift coefficient as $$c_L = \frac{2\cdot n_z\cdot m\cdot g}{\rho\cdot S\cdot v^2}$$ brings us to $$∆h = \frac{\pi}{g^2\cdot\sqrt{n_z^2-1}}\cdot \left(\frac{\rho\cdot S\cdot v^4\cdot c_{D0}}{2\cdot n_z\cdot m} + \frac{2\cdot n_z\cdot m\cdot g^2}{\pi\cdot\rho\cdot S\cdot AR\cdot\epsilon} + w_z\cdot v\cdot g\right)$$ Nomenclature:
$\kern4mm g\kern6mm$gravitational acceleration
$\kern4mm n_z\kern4mm$vertical load factor
$\kern4mm \rho\kern6mm$air density
$\kern4mm S\kern5mm$wing area
$\kern4mm v\kern6mm$flight speed
$\kern4mm c_{D0}\kern2mm$zero-lift drag coefficient
$\kern4mm m\kern5mm$aircraft mass
$\kern4mm AR\kern1mm$wing aspect ratio
$\kern4mm \epsilon\kern6mm$Oswald factor

The figure below is the result plotted in R. Since I need to read the full matrix of values for the contour plot, the area of low speed and high bank angle is filled with the result of a strict penalty function, so please disregard the values to the right and below the red line.

Contour plot of height losses a A320 type aircraft in a 180° turn

Contour plot of height losses a A320 type aircraft in a 180° turn at sea level and MTOW (78 tons), no wind. X is bank angle in degrees and Y is flight speed in m/s. Own work.

As you can see, the minimum (approx. 170 m) is achieved right before stall at a high bank angle and speed. Unfortunately, you need the aerobatic version of the A320 to fly this safely.

  • $\begingroup$ Naive question. What constrains your acos numerator from becoming negative? Are those physically meaningless cases? Else your expression yields negatve bank angles? $\endgroup$ Commented Jan 13, 2016 at 8:34
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    $\begingroup$ @curious_cat: Negative bank angles show that the speed is not sufficient to create the required lift even in straight flight. The constraint is actually the speed, and this results in an upper bound for the lift coefficient. $\endgroup$ Commented Jan 13, 2016 at 11:09
  • $\begingroup$ At the stage where you define the term "smallest height loss for a given azimuth change", why cannot you cancel velocity & then that term becomes simply 1 / (R sin glide_angle) Since glide angle is known you must minimize R to maximize the term "smallest height loss for a given azimuth change". But R will be v^2/ (g tan bank_angle ). So to minimize R maximize tan bank_angle & hence maximize bank_angle. So I might recommend bank as much as you can without stalling. Am I making a blunder? $\endgroup$ Commented Jan 13, 2016 at 18:32
  • $\begingroup$ I have added another answer giving a detailed derivation of what I mean. Would love your critique on that if you could. Thanks! $\endgroup$ Commented Jan 14, 2016 at 6:02
  • $\begingroup$ @curious_cat: Steepest bank and AoA close to stall is the correct answer for the highest absolute turn rate. The smallest altitude loss is at minimum power loss, when $\frac{c_L^3}{c_D^2}$ is at its maximum. The best ratio of turn rate to altitude loss must be somewhere between both optima. Since speed is fixed, the end result depends on speed (and is certainly not always 45°). $\endgroup$ Commented Jan 14, 2016 at 20:45

The best bank angle for a gliding aircraft to use to optimize both turn rate and sink rate can be generalized as 45°.

The reason for this is that 45° is the point at which the vertical component of lift equals the horizontal component of lift.

In other words, a bank angle of 45° will produce the greatest centripetal turning force (horizontal lift) while maintaining the best sink rate (as a function of the vertical component of lift). A lesser bank angle will yield a better sink rate, but produce a lesser turn rate that decreases at a greater rate than the sink rate improves. Conversely, a greater bank angle will produce a better turn rate, but yield a greater sink rate that increases at a greater rate than the sink rate improves.

This phenomenon is purely a function of bank angle, wholly independent of other design or load factors, and therefore holds true for all fixed wing aircraft.

Edit: This may be a nominal answer, not accounting for minor variations in the L/D curves, but it satisfies my operational needs as a pilot experiencing an emergency where I will be maintaining "best glide" throughout a turn and my 45° bank is probably +/- 5°.

  • 4
    $\begingroup$ That's totally fine, answering your own question is actually encouraged $\endgroup$
    – Pondlife
    Commented Jan 11, 2016 at 3:48
  • $\begingroup$ The highest turn rate regardless of sink speed can be obtained at maximum lift (i.e. right before stall), and the minimum altitude loss at the point when $\frac{c_L^3}{c_D^2}$ is at its maximum. The correct answer must be somewhere between both points and certainly depends on the fixed speed (which determines what lift reserves are available for turning). 45° will only be right at one specific speed. $\endgroup$ Commented Jan 14, 2016 at 20:57

There is no general solution, at least without further assumptions.

What I have up to now

  • The problem depends on both bank angle and airspeed

  • We can calculate the angular velocity (turn rate) from bank angle and airspeed

$$\omega = \frac{g \cdot tan~\theta}{v_H}$$

  • We can look up the vertical velocity from the (bank angle corrected) polar curve

$$v_V = \frac{f(v_H \cdot \sqrt{cos~\theta~~})}{\sqrt{cos~\theta~~}}$$

  • We can calculate the ratio $\frac{\omega}{v_V}$ that the asker wants to maximize for any airspeed and bank angle.

Without any knowledge of the polar in analytical form or assumptions we're stuck here.


Let $\theta$, $a_N$, $a_H$, $a_V$ be bank angle, normal, horizontal and vertical acceleration. Their relationships are:

$$a_V = a_N \cdot cos~\theta$$ $$a_H = a_N \cdot sin~\theta$$

with $a_V = 1g$

$$a_N = \frac{1}{cos~\theta} \cdot 1g$$

$$a_H = a_N \cdot sin~\theta = \frac{sin~\theta}{cos~\theta} \cdot 1g = g \cdot tan~\theta$$

Let $r$ and $\omega$ be turn radius and angular velocity. Let $v_H$ be the airspeed. The horizontal acceleration is the centripetal acceleration in our turn, so:

$$\frac{v_H^2}{r} = g \cdot tan~\theta$$

The angular velocity (turn rate) is:

$$\omega = \frac{v_H}{r} = \frac{g \cdot tan~\theta}{v_H}$$

Let $v_V$ be the sink rate which is a function $f$ of the airspeed. The function $f$ is usually given as polar curve. For load factors other than $1g$, we have to scale it by the square root of the load factor $k$.

$$k = \frac{1}{\sqrt{cos~\theta~~}}$$

$$v_V = k \cdot f \left( \frac{v_H}{k} \right)$$


Wouldn’t the mathematical problem look something like this?

$$V = L\cos\theta$$

$$H = L\sin\theta$$

$$R_S = k_1 (W - V)$$

$$R_T = k_2 H$$

Maximize $R_T / R_S$ with respect to $\theta$ given constant $L, k_1, k_2, W$


  • $\theta$ is Bank Angle
  • $L$ is Lift
  • $H, V$ are Horizontal & Vertical Components of Lift
  • $W$ is Weight
  • $R_S$ is Sink Rate
  • $R_T$ is Turn Rate
  • $k_1, k_2$ are positive constants

The Math:

$R_T / R_S$ evaluates to

$$\frac{\frac{k_2 L}{k_1}\sin\theta}{W - L \sin\theta}$$

Maximise this expression with respect to $\theta$.

PS. When I do the math I get the $\theta$ that maximizes turn rate per sink rate as 90 degrees bank angle.

Obviously, I’m either messing up my math or my model. I must be wrong. Perhaps my blunder was to take lift $L$ as a constant? I suppose $L$ will change with bank too?

Also, I guess the stall characteristics should matter? Maybe that additional constraint would say bank at max angle that won’t stall you?

PS. This is just back of the envelope estimation. I’m probably being naive by not considering the complexities of the issue.

  • $\begingroup$ @JonathanWalters Could be. It might still disappear from the final expression. $\endgroup$ Commented Jan 11, 2016 at 3:03
  • $\begingroup$ Weird, I'm getting 90 degrees as the angle that maximizes the ration of turn rate to loss of altitude. I must be doing something wrong. $\endgroup$ Commented Jan 11, 2016 at 3:23
  • $\begingroup$ Lift is definitely not constant, because the vertical component must remain equal to weight, since there is no vertical acceleration throughout the manoeuvre. $\endgroup$
    – Jan Hudec
    Commented Jan 11, 2016 at 7:29
  • $\begingroup$ You can't compute the sink rate from lift. Sink rate is equal to drag (times speed). (Induced) drag is needed in this equation. $\endgroup$
    – Jan Hudec
    Commented Jan 11, 2016 at 7:33
  • $\begingroup$ Note that to assume drag is constant, you need to actually increase speed, because the best glide speed is actually best glide angle of attack and increases in the turn (by ~19% (⁴√2) for 45° bank). $\endgroup$
    – Jan Hudec
    Commented Jan 11, 2016 at 7:51

I'm adding another answer which looks way too simplistic but I cannot spot why it is wrong.

Here goes:

The term we want to maximize is "smallest height loss for a given azimuth change" & this can be shown to be the following borrowing @PeterKampfs logic:

$$ \frac{\omega}{v_z}=\frac{1}{R * sin \phi}$$

where $\phi$ is the glide angle.

Since glide angle is known you must minimize $R$ to maximize the term "smallest height loss for a given azimuth change". But $$R=\frac{v^2}{g*tan \theta}$$

So to minimize R maximize $tan \theta$ & hence maximize $\theta$. So I might recommend bank as much as you can.

But there's an additional constraint introduced by the fact that you must not stall (obviously). You must use maximum $\theta$ but not above $\theta_{stall}$

Let the un-banked stall speed be $v_{stall}$. Under a banked turn of $\theta$ the stall speed increases to:


where $n$ is the load factor.

$$n=\frac{L}{mg}=\frac{1}{cos \theta}$$

Therefore in a banked turn of $\theta$ the stall velocity will be:

$$v_{stall}^{banked}=\frac{v_{stall}}{\sqrt{cos \theta}}$$

Hence the max angle of bank would be the one you should use and that will be:

$$\theta=arccos \left( \left(\frac{v_{stall}}{v}\right)^2 \right) $$

  • $\begingroup$ @JonathanWalters Indeed. I'm using constant airspeed $v$. Where am I going wrong? Sorry if I'm not seeing the obvious. $\endgroup$ Commented Jan 14, 2016 at 13:47
  • $\begingroup$ @JonathanWalters No worries! But what do you feel about it? Can you spot any other flaws? If it is indeed correct I'm curious how it meshes with the other derivations given by Peter Kampf etc., $\endgroup$ Commented Jan 14, 2016 at 15:54
  • $\begingroup$ @curious_cat: Since speed is fixed, $c_L$ varies with bank angle and we cannot assume that the glide angle is known. The airplane is somewhere on the polar, and not necessarily at the best $\frac{c_L}{c_D}$. Your result gives the highest possible turn rate regardless of altitude loss. $\endgroup$ Commented Jan 14, 2016 at 20:50

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