What is the relation between airspeed and rate of turn?

I am reading about aircraft maneuvering and the end of section check contains this question:

During a turn, if you reduce the airspeed and maintain the angle of bank, the result is a _____________ rate of turn.

The correct answer is "higher" which is confusing me.

• If I reduce the airspeed, won't it take longer to complete the turn, thus "lower" rate of turn?
• I can understand that the turn might be smoother and more coordinated if I reduce the airspeed, but how can the rate of turn be higher?
• What am I missing here?
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Look here for the physics behind it. With speed the turn radius increases, so it takes longer to fly the circle. Rate of turn is inverse to speed. – Peter Kämpf Mar 28 at 6:11

Independent of airspeed, angle of bank gives you the "G" force in a level coordinated turn. A 60 degree bank = 2 G's, at any speed.

At low speed, a given G force will give you a greater heading change than it would at a high speed. Driving your car at 10 mph you can complete a U-turn, 180 degrees of heading change, in a couple seconds without pinning passengers against the door, but at highway speeds, even a hard turn that does impose big side-loads will take much longer to give you that much heading change.

Related to all this is turn radius: at low speed, your turn radius for any given loading (bank angle) is smaller than it would be at a higher speed with that same loading/bank angle. Thus, even though you're going slower, you have less distance (circumference of the circle) to cover, and so you are able to complete the circle (or one 360th of it) more quickly.

((For the pedantic ones, everything here assumes level, coordinated turns. Change those assumptions enough, things can be different, but those aren't particularly instructive cases until the base case is understood.))

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Ah got it. Thanks for the examples - really put things in perspective! – atlantis Mar 28 at 3:50
Excellent analogy. Never thought to explain it like that before. – Burhan Khalid Mar 28 at 6:10

The rate of turn is inversely proportional to the (True) airspeed. For an aircraft in a level, coordinated turn, the rate of turn is given by

$\mathrm{Rate\ of\ turn} = \frac{1091 \tan\theta}{V}$

where

• Rate of turn is in degrees per second,
• $\theta$ is the bank angle in degrees, and
• $V$ is the TAS in knots.

So, as the airspeed decreases, the rate of turn increases and vice versa—as long as the angle of bank is kept constant.

Image from cfinotebook.net

For an aircraft in a coordinated turn (i.e., no skidding or slipping), the vertical component of lift is equal to the weight, while the horizontal component is equal to the centrifugal force.

Forces in a turn, image from Instrument Flying Handbook

As the aircraft turns, if the airspeed increases with the bank angle held constant, the radius of turn increases with the square of the speed ($r = \frac{V^{2}}{11.26 \tan\theta}\ \mathrm{ft}$). Hence, the distance traveled during the turn increases as the square of the speed. Even though the aircraft is flying faster, the distance to be flown increases faster than the speed. As a result, the time taken to complete the turn is increased and the turn rate decreases.

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The cfinotebook image was super-helpful. Thank you for the effort! Marking Ralph's answer as the car and highway u-turn makes it very intuitive to remember this. – atlantis Mar 28 at 4:00