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For a glider in a steady bank at angle $\mu$, to a good approximation, the sink rate has a factor $1/\sqrt{\cos \mu}$ relative to the sink rate at zero bank angle.

What are the dynamics of the sink rate during a roll, for instance, when rolling from a bank angle of zero to a bank angle of $\mu \sim 20^{\circ}$ in one second?

I'm not looking for a very precise, quantitative answer but rather some qualitative idea of the physical effects that underlie the dynamics of the sink at such roll rates for a model plane.

Any references or guidance will be helpful, thanks!

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  • $\begingroup$ Would you be able to take the average of the 2 angles (start and finish)? so, the average of 0 and 20 degrees? That would be 10 degrees. so your sink rate would be using that same formula but instead of mu being 20, it would be 10. $\endgroup$
    – alex
    Apr 6 '17 at 9:26
  • $\begingroup$ @alex the bank rate influence is non-linear, as seen from the 1/sqrt(cos) equation, so applying it to the average of the input (the bank angle) does not provide the average of the output $\endgroup$
    – Federico ♦
    Apr 6 '17 at 10:06
  • $\begingroup$ @Federico Clearly, it is not, but -intuitively- I believe that the form of the curve lift/bank angle is probably concave during the first half of the roll and convex during the second. Thus, the non-linearity may (roughly) cancel out... and a linear rate might be OK for a first approximation... $\endgroup$
    – xxavier
    Apr 6 '17 at 13:21
  • $\begingroup$ @xxavier "I believe that the form of the curve lift/bank angle is probably concave during the first half of the roll and convex during the second" it is not $\endgroup$
    – Federico ♦
    Apr 6 '17 at 13:25
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    $\begingroup$ @xxavier The lift will not necessarily be zero when inverted. It may be negative, or something else depending on angle of attack. $\endgroup$
    – DJClayworth
    Apr 6 '17 at 15:19
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As always, it depends. Besides the steady-state drag sources of lift creation and friction, you need to look at the drag created by roll acceleration and manoeuvring, including adverse yaw.

If the glider starts the roll at a bank angle which is sufficient to reach a steady roll rate by the time it rolls through 0°, the roll acceleration term becomes zero. However, now you have deflected ailerons and a deflected rudder to counter adverse yaw, which all add their own drag. How much depends on the roll rate and the needed deflection.

On top, now the local incidence of the wing shows a discontinuity at the spanwise station where the ailerons begin. This will create a less than optimum lift distribution over span, so the induced drag of the rolling glider is higher than that of the same glider in a steady glide. Again, the magnitude depends on the roll rate, and the aileron placement and deflection angle.

I would not be surprised if the sink rate in a brisk roll increases by 50% over that in a steady glide at the same speed, even with zero sideslip. If adverse yaw is not properly countered by the rudder, the additional drag from the resulting sideslip angle can easily double the sink rate.

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  • $\begingroup$ Thanks! We have a good amount of data using a model plane. The change in sink rate is about 50% for a turn from $20^{\circ}$ to $40^{\circ}$ as you guessed but the sink rate changes in both directions - it increases when we move the bank angle farther from 0 and decreases when we move it towards 0. The sign suggests this is due to unbalanced lift. This change is not affected when we have a rudder mix in our rolls so it is not due to sideslip nor is it affected when we soften our aileron deflection. We can test if it's due to roll acceleration; a reference for this will be greatly appreciated. $\endgroup$
    – un_anonymous
    Apr 10 '17 at 8:32
  • $\begingroup$ @un_anonymous: Sorry - I do not know of a reference how roll acceleration will increase drag. But this should be straightforward to calculate once you know the roll inertia of the plane and the drag increase of a roll command. $\endgroup$
    – Peter Kämpf
    Apr 10 '17 at 20:34

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