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The Best Glide speed is the indicated airspeed that guarantees the maximum glide distance during an engine out scenario. However, the landing straight ahead option after an engine failure may not be the best option especially if there is nothing but houses in front of you after an engine failure. Sometimes turning back towards the field may be the best option.

The Best Glide speed is actually tied to a best Lift/Drag ratio for a specific angle of attack. When in a banked turn, the pilot must increase the angle of attack to avoid a spiral dive. So this implies that the best glide speed would increase during a banked turn in order to maintain the best L/D angle of attack. My question is how would one calculate the Best Glide speed for an aircraft in a 30 degree bank based off the POH listed Best Glide at Gross Weight (I do realize best glide varies with weight).

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    $\begingroup$ This doesn't quite add up. (A) You say "When in a banked turn, the pilot must increase the angle of attack to avoid a spiral dive." When gliding, why must the pilot increase the AoA? Let's say you want to fly at the AoA for Min Sink Rate. Isn't this the same AoA in wings-level flight as in a banked turn? Same for stall AoA? Same for best L/D AoA? The concept of increasing AoA when banked pertains more to flying with a motor and trying to maintain a constant altitude. (ctd) $\endgroup$
    – quiet flyer
    Mar 18, 2021 at 19:21
  • $\begingroup$ I think your presumptions are incorrect. When turning in an engine out glide you still want to maintain best glide speed, which will correlate to best L/D AOA. However, in order to do so you will need to lower the nose, which will increase sink rate. $\endgroup$
    – Michael Hall
    Mar 18, 2021 at 19:21
  • $\begingroup$ (ctd) Then you say (B) "So this implies that the best glide speed would increase during a banked turn in order to maintain the best L/D angle of attack." --yes, it must increase according to the square root of the change in wing loading . So (B) is true. But (B) doesn't follow logically from (A), because (A) is incorrect. $\endgroup$
    – quiet flyer
    Mar 18, 2021 at 19:23
  • $\begingroup$ I think you can still fix this w/o invalidating the single answer posted so far, because the answer only addresses (B), not (A). $\endgroup$
    – quiet flyer
    Mar 18, 2021 at 19:25
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    $\begingroup$ Another huge complication is that a 180 degree reversal arguably has nothing to do with best L/D at all. You are in no sense trying to maximize forward distance covered. You want to minimize time spent in the reversal and you also want to minimize the sink rate during the reversal. I seem to recall I've seen analyses suggesting a 45-degree banked turn is the way to go, letting the airspeed bleed off (or even pulling the stick aft to help the airspeed bleed off) to near min sink for that bank angle -- but certainly not lower!-- I'd have to look some more to relocate those analyses-- $\endgroup$
    – quiet flyer
    Mar 18, 2021 at 19:31

2 Answers 2

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During a turn the load factor increases. This increases the apparent weight. Thus, during a turn, the best glide speed is selected according to the weight times the load factor.

The load factor at 30 degrees bank is $$n=\frac{1}{\cos{\theta}} \approx 1.2$$

The best glide speed is thus the best glide speed for a weight that is 1.2 times your actual weight. This will be proportional to the square root of the load factor. $$V_{opt,\theta}= \sqrt{\frac{1}{\cos{\theta}}} V_{opt,0^\circ}$$

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  • $\begingroup$ Is this really correct? The load factor is only equal to 1/cos(theta) in level flight where you pull on the yoke to maintain altitude. But presumably one will not be attempting to maintain level flight during engine out. $\endgroup$
    – SMeznaric
    Jun 6 at 15:43
  • $\begingroup$ @SMeznaric you are right about the level flight assumption being likely not valid during engine out conditions. However at small flight path angles, the formula approximates the load factor well. The correct form (assuming constant flight path angle or gamma) is cos(gamma)/cos(theta) $\endgroup$
    – DeltaLima ♦
    Jun 6 at 18:04
  • $\begingroup$ I am not sure that's correct @DeltaLina. If you're not pulling on the yoke and accept a higher descent rate during the turn then your load factor should stay unchanged. $\endgroup$
    – SMeznaric
    Jun 13 at 16:55
  • $\begingroup$ @SMeznaric I am note sure what you are trying to achieve. assuming coordinated flight, the load factor is directly related to the bank angle and the flight path angle. $\endgroup$
    – DeltaLima ♦
    Jun 13 at 17:32
  • $\begingroup$ I can't tell if your final equation is right or not, I am only taking an issue with your statement about the load factor going up in a turn. This is only true in level flight. And the reason it's true in level flight is because you need more lift to counteract gravity in a turn. But if you're not in level flight, you're trimmed out and turning without pulling, you'll increase your descent rate and then this assumption does not hold. The load factor in a turn in general could in general be higher or lower than the normal level flight load factor. $\endgroup$
    – SMeznaric
    Jun 14 at 9:52
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In a descending turn, maintain the same glide speed. Your rate of descent will increase, but you can fly the same speed.

The issue of load factor in a banked turn applies to a level turn. In a descent, fly your glide speed and let the descent rate increase.

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