enter image description hereAs controller, I always wonder when should I instruct aircraft to turn. The aircraft does not turn at right angles as soon as instructed, but rather has turning radius. So, I tried to find the formula but failed.

If speed of aircraft is 300 kts, and he is instructed to make 90 degree, how long distance the aircraft need to turn? I make image, and I wonder the formula for 'x'

  • $\begingroup$ Is that 300 kts airspeed or groundspeed? At what altitude does the aircraft make the turn? $\endgroup$ – DeltaLima Aug 19 '19 at 7:19
  • $\begingroup$ I didn't consider those things as factor. For exeample, altitude is 5,000ft and ground speed. And I also want to know the fomula for 30degree or other random degree case. Thanks. $\endgroup$ – Min Aug 19 '19 at 12:04

I was taught the following rule of thumb for providing a final vector to final, in the USA, using the STARS radar system which (when using ADS-B inputs) has a one-second update time:

Set the Predicted Track Line to 30 seconds. Vector the aircraft to a 90º base-to-final. When the PTL reaches the extended runway final, issue the Position-Turn-Altitude-Clearance. The aircraft should roll out on a good heading to intercept final approach course.

This is obviously at low altitudes and relatively low speeds, 230ish knots or below. Theoretically (because the method uses the aircraft's speed instead of a fixed distance) this method should work for any aircraft going any speed, but that may not always pan out. In particular, some military aircraft will be flying rather quickly on the base leg and slow down as soon as they get their dogleg turn-to-final, which could end up leaving them too far from final for a legal intercept.


Standard turn rate is 360° in 2 minutes so a 90° turn will take half a minute (30 s). The distance covered will be $ s = \frac {v}{60 \times 2} = \frac {v}{120} $ (nautical miles).

The number you are looking for is the radius of the arc. This is given by $ s \times \frac {2}{\pi} $ (nautical miles).

So the radius is given by $ r = \frac {v}{120}\times \frac {2}{\pi} = 0.0053 v$. This is approximately $ \frac {v}{200} $.

So for 300 kt the start turning would be $ \frac {300}{200} = 1.5 $ nautical miles from rollout.

I'm an engineer, not a pilot or controller, so wait for confirmation of my maths by someone who knows what they're talking about.

  • 2
    $\begingroup$ I think your math is correct, but an aircraft traveling that fast under IFR will generally make half standard rate turns to avoid excessive angles of bank. The rule of thumb we used for a half standard radius was 1.0% of airspeed. So, 3nm at 300kias, 2.5nm at 250kias, etc. Avoids algebra while flying! ;) $\endgroup$ – Michael Hall Aug 18 '19 at 16:16
  • $\begingroup$ That makes sense. You can post it as an answer. Thank you. $\endgroup$ – Transistor Aug 18 '19 at 16:30
  • $\begingroup$ Thanks for answer. If the degree is 30, Could I get the lead point too? $\endgroup$ – Min Aug 19 '19 at 12:07
  • $\begingroup$ @Transistor, I don't have time to verify against your formulas, but feel free to incorporate into your answer to make it more complete. $\endgroup$ – Michael Hall Aug 19 '19 at 17:29
  • $\begingroup$ @Min, I will assume you are asking degrees of turn and not angle of bank. I am not sure if the math above supports this, but personally I would just divide by 3. For example, if I was doing 300 knots and using my rule of thumb of 3nm for a 90 degree turn, I would lead a 30 degree turn by about a mile. I would also recommend you confer with fellow controllers for advice on this matter. $\endgroup$ – Michael Hall Aug 19 '19 at 17:39

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