# Constant turn radius with Δ airspeed and bank

Can someone explain how speed impacts rate of turn and and degree of bank required?

I'm trying to understand this practice problem but can't get the concept: Say you have 2 vehicles travelling at different speeds and carrying out a turn at the same radius. The example states that the faster vehicle will have to bank at a shallower angle and will have a lower rate of turn.

In my mind, the faster vehicle would have to maintain a steeper bank angle to compensate for the higher relative speed, otherwise their radius would increase. If you're moving faster, and banking at a lower angle, wouldn't that mean you're travelling at greater forward distance and less lateral distance thus increasing you're radius?

• I am with you on this question. The faster the object is moving the faster the rate of change will be around the turn. The rate of change of direction has an affect on the centrifugal forces (forces pushing away from axis of rotation). The goal in an aircraft turn is to have the centrifugal and gravity forces being equal resulting in a coordinated turn. – Jeff A Jan 19 at 19:45

You are right on all accounts.

In a steady-state turn, from circular motion, we have, with $$F_c$$ being centripetal force, $$V$$ being airspeed in zero wind, $$R$$ being the radius of turn, and $$m$$ being the aircraft mass:

$$F_c=m\frac{V^2}{R}$$

The centripetal acceleration in a coordinated turn is created from the banked lift vector ($$L$$), therefore, with $$\phi$$ being the bank angle:

$$L\sin\phi=m\frac{V^2}{R}$$

Therefore, the radius of turn is:

$$R=\frac{V^2}{g\tan\phi}$$

and the rate of turn is:

$$\dot{\psi}=\frac{g\tan\phi}{V}$$

As speed increases, you would need an ever larger bank angle in order to achieve the same rate of turn. Of course, there's a limit to how much lift you can get out of the wing at a given speed, especially at higher Mach when transonic buffet becomes the limiting factor.