There seem to be some formula and the most common is that:

rate of turn is proportional to TAS / radius.

But the exact relationship seems a little more fraught.

For a circle of circ 'd' (NM) and an aircraft of speed 'v' (Kts) the time taken to complete the circle t = d/v

So the frequency of turning 1/t (per hour) = v (TAS) / d

Multiplying both sides by 360 we get the rate of turn (degrees per hour) = 360 v / d

But d = 2 pi r (radius in NM)

So rate of turn (deg /h) = 360 v / 2 pi r

We don't need to worry about NM because they cancel top and bottom but we can convert the units of time to min

rate of turn (deg / min) = 360 v / 2 x 60 pi r = 3 v / pi r

And again convert units of time to s

rate of turn (deg / s) = 3 v / 60 pi r = v / 20 pi r

From the airline interview example, a std rate turn at 180 kts has a radius of 1 NM

There fore the rate of turn = 180 / 20 pi = 9/pi ~ 3 deg per s.

This all seems to work - but I can't find this formula anywhere else except my derivation.

rate of turn is proportional to TAS / radius.

But the exact relationship seems a little more fraught.

For a circle of circ 'd' (NM) and an aircraft of speed 'v' (Kts) the time taken to complete the circle t = d/v

So the frequency of turning 1/t (per hour) = v (TAS) / d

Multiplying both sides by 360 we get the rate of turn (degrees per hour) = 360 v / d

But d = 2 pi r (radius in NM)

So rate of turn (deg /h) = 360 v / 2 pi r

We don't need to worry about NM because they cancel top and bottom but we can convert the units of time to min

rate of turn (deg / min) = 360 v / 2 x 60 pi r = 3 v / pi r

And again convert units of time to s

rate of turn (deg / s) = 3 v / 60 pi r = v / 20 pi r

From the airline interview example, a std rate turn at 180 kts has a radius of 1 NM

There fore the rate of turn = 180 / 20 pi = 9/pi ~ 3 deg per s.

This all seems to work - but I can't find this formula anywhere else except my derivation.

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