I know that I can find an upper bound for the bank angle for a plane, if I know the maximum sustained load factor of that plane because the roll angle and the load factor $n_z$ in a turn are directly related: $$tan (\varphi) = \sqrt{n_z^2 - 1} \;\; \text{or} \;\; \varphi = arctan\left(\sqrt{n_z^2 - 1}\right)$$ And also I known that the bank angle would not be more than 30° for civilian and 75° and more for military aircraft.

But my question is :

How can one get the minimum possible bank angle an airplane can use to perform a turn with a given radius before reaching the stall speed?

Also I want to know if the stall speed is known, can we use the formula :

$$ W = V/R $$

to calculate the angular rate of turn (of course considering the level turn)?


3 Answers 3


Given a radius $R$ and a velocity $V$, the bank angle can be computed by:

$\phi = \textrm{atan}\left(\frac{V^2}{R\cdot g }\right)$,

where $g$ is the acceleration by gravity. This shows that the lower the speed becomes, the lower the bank angle becomes. The lowest attainable speed is the stall speed, however we have to take into account the load factor during the turn.

The load factor $n$ during the turn is:

$n=\frac{1}{\cos \phi}$

Due to the increased load factor, the stall speed increases in the turn:

$V_{stall,turn} = V_{stall} \cdot \sqrt n $

The minimum bank angle for a turn with radius R will occur when the speed is equal to the stall speed for that angle.

Solving this mathematically for $\phi$ results in:

$\phi_{min} = \textrm{atan}\left(\frac{V_{stall}^2}{\sqrt{R^2g^2-V_{stall}^4}}\right)$

Note that is the theoretical approach; in practice you would not fly at the stall speed, especially not during turns.

Units are all in SI, you have to apply conversion factors if you want to use aviation units.


  • $V_{stall}$ = 50 m/s (approx. 97 knots)
  • $R$ = 400 m
  • $\phi_{min}$ = ~39.6 degrees
  • $V$ = ~57 m/s (approx. 111 knots)
  • $\begingroup$ can i know the usual bank angle of the military and the civilian planes. I mean the usual bank angle and hence the velocity, to do a complete turn, or it is not possible to know that information ? $\endgroup$
    – AAEM
    Sep 12, 2018 at 11:03
  • 1
    $\begingroup$ @Ahmed I don't understand what you mean. There are three related variables in simple turn mechanics: velocity, bank-angle and turn radius. If you know two of them, you can calculate the third. For typical aircraft operations, the bank angle and the velocity are controlled by the pilot, the turn radius is the result. For passenger aircraft a maximum bank angle of 25 or 30 degrees is often used. In other cases the radius of the turn is given (see fix radius turn) and the bank angle is the result. $\endgroup$
    – DeltaLima
    Sep 12, 2018 at 11:15
  • $\begingroup$ ok i got it, i want to ask you again about the equation, it seems that we can not put bank angle = 90 degree(in coordinated level turn), am i right? $\endgroup$
    – AAEM
    Sep 12, 2018 at 11:22
  • 1
    $\begingroup$ @Ahmed That is right. If the wings roll 90 degrees, the lift vector is fully horizontal. That means there is no vertical component any more to counteract the gravity force. This means that the aircraft will accelerate downwards, and therefore it is no longer a level turn. $\endgroup$
    – DeltaLima
    Sep 12, 2018 at 11:25
  • $\begingroup$ Still you do not have a given v, or g, dependent on v. What one could do is have r variable for chosen speeds, plot on graph, and extrapolate to v stall? In other words, it may not be possible to turn plane in given radius safely (unless you know hammerhead!). $\endgroup$ Sep 13, 2018 at 14:13

TL;DR: just assume that you are flying at your stall speed and compute the bank angle from there.

Please do not fly at your stall speed, especially not in a turn, you will stall and could have serious problems

Assuming you are flying at your stall speed $V_{stall}$, and you want to have a turn with radius $R$, you will need to have a turn rate of

$$\dot\psi = \frac{V_{stall}}{R}$$

and a bank angle of

$$\phi = atan \left( \frac{V_{stall}^2}{g \cdot R} \right) $$

(where $g$ is the gravity acceleration) will let you fly in a coordinated way.

All terms are in SI units.

  • $\begingroup$ @Ahmed you cannot fly below stall speed, so that's the angle. yes, each aircraft will have a different stall speed, and thus a different "minimum angle", but the answer does not change. $\endgroup$
    – Federico
    Sep 12, 2018 at 8:36
  • 3
    $\begingroup$ Stall speed depends on load factor, so in a coordinated level turn the stall speed will be higher than in straight and level flight. That should be factored in... $\endgroup$ Sep 12, 2018 at 9:10
  • $\begingroup$ @CptReynolds indeed. See my answer. $\endgroup$
    – DeltaLima
    Sep 12, 2018 at 9:29
  • $\begingroup$ @DeltaLima Perfect! $\endgroup$ Sep 12, 2018 at 12:59
  • 1
    $\begingroup$ "you could have serious problems": well, that a nice understatement! Or a different way to say that you could die. $\endgroup$
    – kebs
    Sep 13, 2018 at 5:27

How can I compute the minimum possible bank angle for a turn with a given radius?

No computation is needed; the answer is zero degrees. Apply rudder as needed to turn, while cross-controlling with the ailerons as needed to keep the bank angle at zero.

(Note: if you are trying to achieve an extremely low turn radius in relation to your airspeed, you might "run out of rudder" before achieving your goal. To say whether or not this will happen, requires more information than is contained in the question. If you are thinking in the ballpark of the kind of turn radius that could be accomplished with thirty degrees of bank or less, as your question seems to indicate, this will probably not be an issue.)

(Did you intend to specify that the turn should be "coordinated", i.e. non-skidding?)

I know that I can find an upper bound for the bank angle for a plane,


And also I known that the bank angle would not be more than 30° for civilian and 75° and more for military aircraft.

The last quoted sentence above is unfounded. There's no basis for those limits.


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