I’m not even sure if this is a safe/allowed procedure at controlled airports so it is purely hypothetical.

Assuming there are calm winds at the field, is there a way to calculate the distance your plane must be from the centerline (see “x”) of the runway while on downwind, so that once you start your turn to base and final thereafter, in a constant rate turn (assume 3 deg./sec.), that you will end up rolling out on the centerline of the runway?

Speed changes (slows) by 27 knots from the beginning of the turn to the end of the turn.

Essentially, if I wanted to execute an 180 degree turn at 3 deg. per second, while slowing the airplane from traffic pattern speed (90 kts) to final approach speed (63 kts), how far away would I need to be from the runway while on downwind before beginning the maneuver? The goal is to be able to use this on any plane given their own specific traffic pattern speed and final approach to landing speed. enter image description here

  • $\begingroup$ This really isn't a standard pattern, why would you do that? You don't want to be turning and making configuration changes while turning, it's one of a few reasons you don't make a semicircular approach. $\endgroup$
    – Ron Beyer
    Commented Aug 9, 2019 at 1:07
  • 3
    $\begingroup$ Although it is not common in the civilian world, I believe procedures like this are quite common in the military. Specifically the Navy. $\endgroup$
    – Austin S
    Commented Aug 9, 2019 at 19:22
  • $\begingroup$ FWIW, there was a study planned on this, although I never heard about the results. As per another comment, this is common in the military world, including civilian pilots who fly ex-military aircraft. $\endgroup$
    – Pondlife
    Commented Aug 9, 2019 at 23:50

2 Answers 2


Turning at 3 deg/sec, a 180-degree turn will take 60 seconds. If the entire turn is performed at 63 knots, the aircraft will travel 1.05 nautical on a half circle (assuming zero wind). The radius of the circle will be the circumference, 2.10 divided by 2*pi = 0.3342 nm. Therefore, x = 2*0.3342 = 0.6684 nm. If starting at 90 knots and slowing down to 63 knots, the trajectory will not be a a semi-circle but rather a section of a spiral. The starting radius will be 0.4775 and the final radius 0.3342. In this case, x will be approximately 0.8117 nautical miles. To get an exact value for x, an integration must be performed.

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    $\begingroup$ I'd like to second your point on having the approach speed being achieved at the abeam (position A on the diagram). You may, at your option, want to add that the military constant turn approach is a constant airspeed approach with the approach speed being stable at the abeam position ... in part for the reason that you cite (why make it more complicated than it needs to be ....) $\endgroup$ Commented Jan 29, 2020 at 17:04

This sounds like a power-off 180. Your numbers may be off based on the fact that you are going into your descent with absolutely no power available. You would be better off squaring off your approach with a downwind to base to final configuration with 90 degree turns. If you feel like you have too much lateral distance from the runway for all three legs done properly, you may try a downwind to diagonal base to diagonal final. This can be accomplished either with a shallow turn, then a steep turn, then a shallow turn. Or, the better option of side-slip, steep turns, side-slip.


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