To start with something familiar: here's a flight polar. As we learned, the best glide speed is found by starting at the 0,0 point, and drawing a line tangent to the sink rate line. The intersection point gives the airspeed for the best glide. In headwind or sink, the polar moves and so does the tangent line, producing different best glide airspeed.

flight polar (credit nashvillecfi.com and Tom Fleming)

And here is a "turning polar" (from BGA webinar "Thermal Centering"). How is it constructed?

Turning polar (credit BGA)

It looks to me that, to draw each small polar (for 20°, 30°, 40°, 50°, 60° bank), the bank angle is fixed and all possible pitch angles are "flown"; the resulting turn radius and sinking speed produce the point on the plot.

But how is that "optimum speed" found, on each small polar (e.g. 43kts for 30° bank)? This will probably also answer the question why does the overall plot of the optimum speed (dotted line) looks what it looks.


2 Answers 2


It looks to me that, to draw each small polar (for 20°, 30°, 40°, 50°, 60° bank), the bank angle is fixed and all possible pitch angles are "flown"; the resulting turn radius and sinking speed produce the point on the plot.

Correct, except for the "all possible" part. Note that the right-hand end of each of the bank angle curves corresponds to the angle-of-attack that would give the minimum sink rate with the wings level. Moving further left on each of the bank angle curves, we are increasing the angle-of-attack (moving closer to the stall angle-of-attack), which increases the sink rate. So we are only looking at a very small part of the glider's overall flight envelope in terms of the possible angles-of-attack, compared to the glide polar shown in the upper figure in the question.

But how is that "optimum speed" found, on each small polar (e.g. 43kts for 30° bank)?

Imagine that rather than drawing only curves for 5 bank angles between 20 and 60 degrees, curves were drawn for every possible bank angle in between. For any given turn radius, the dashed line is drawn through whatever airspeed on whatever bank angle curve gives the lowest possible sink rate.

Note that the diagram indicates that at steep bank angles-- which result in high airspeeds due to the high wing loading-- there is an advantage to flying at an angle-of-attack higher (closer to stall) than the one that would generate the minimum sink rate for that bank angle, if the end goal is to minimize the sink rate for a given turn radius.1

It would be interesting to know whether the curves for the various bank angles were generated by simply "adjusting" the wings-level glide polar for the increased wing loading, or by actual measurements. In turning flight, the "relative wind" is actually curved in all three dimensions-- for example, a string trailing from the tail would stream in a different direction than a string attached to the front of a probe projecting forward from the nose, due in part to the continual nose-up, tail-down pitch rotation of the aircraft in the turn.2 This has aerodynamic consequences, such as an apparent decrease in the decalage angle between the wing and tail, requiring more up-elevator (or less down-elevator) to maintain any given angle-of-attack when the aircraft is steeply banked and turning tightly, than when the wings are level. These aerodynamic changes associated with turning flight affect drag, and sink rate.3 Therefore the L/D ratio does vary to some degree as the bank angle is increased and the angle-of-attack of the wing is held constant. These aerodynamic consequences associated with turning flight are most pronounced in slow-flying aircraft, where the radius of the turn is not orders of magnitude larger than the physical dimensions of the aircraft.


  1. The reason for this is that if the glider were constrained to fly at the min-sink-rate angle of attack rather than a higher angle-of-attack, a higher bank angle would be required to achieve a given turn radius.

  2. This is a bit tricky-- we're really trying to describe the direction of the undisturbed relative wind, referenced to various different points on the aircraft. We're not including the downwash from the wing, or any other effects caused by the physical interaction between the aircraft and the airflow. This change in direction of the undisturbed relative wind also causes changes in the direction of the actual airflow experienced at various points on the aircraft.

  3. The ideal fuselage shape for minimizing drag in thermalling flight would be bent like a banana, convex side down, curved to conform to the curving flight path and relative wind.

  • 1
    $\begingroup$ The polars in the question look like they have been copy-pasted from a straight flight polar and fit so the results look OK. In truth, the steeper polars also look more compressed in X direction. $\endgroup$ Aug 18, 2022 at 15:02
  • $\begingroup$ @PeterKämpf - interesting observation-- $\endgroup$ Aug 18, 2022 at 15:08

I think it's much more intuitive to draw this "turn polar" in terms of airspeed instead of turn radius. Note that bank angle directly relates to centripetal acceleration, which together with airspeed directly results in a turn radius. So instead of "all possible pitch angles", the diagram is constructed using all possible airspeeds.

Optimum speed in this context is minimum altitude loss per unit of angle. Turn rate is equal to centripetal acceleration divided by airspeed, so inversely proportional to turn radius. So instead of a straight line tangent, you draw a hyperbola that just touches your polar to get the optimum speed.

I assume that the fact that a single hyperbola touches all bank angle polars is due to the same reason that a single tangent line touches all different mass lines in a normal polar: a different load factor simply produces the same result at a different air speed. This is probably why they plotted turn radius instead of airspeed. I can't at the moment give a difinitive reason however.

  • $\begingroup$ But the dashed line is not touching all the bank angles curves at the point corresponding to the same angle-of-attack (and therefore the same L/D, if we ignore the additional nuance explored in the final paragraph of my answer.). It's not clear to me whether or not this answer is consistent with this observation. $\endgroup$ Aug 18, 2022 at 14:29
  • $\begingroup$ @quietflyer agreed. Some more thinking is required... :) $\endgroup$
    – Sanchises
    Aug 18, 2022 at 14:41

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