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We know from other questions and answers that airplanes and gliders in particular can have their performance described in terms of glide polar and Lift-to-Drag ratio.

As it appears from the images in the first linked answer (included below), the two are connected to each other.

Despite my research I couldn't find a source on how to derive one curve given the other (the polar given the ratio or the ratio given the polar).

Is it possible to do so? I have heard from friends that are glider pilots that the glide polar curve can be described as a parabola, but I haven't found any reference.

L/D ratio

Glide Polar Images from this question

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  • $\begingroup$ @PeterKämpf I only changed the tags, not the title, so I plead not guilty :-) $\endgroup$
    – Pondlife
    Feb 19, 2015 at 19:08
  • $\begingroup$ @Pondlife: Right, that was fooot. No problem, enjoyed it. $\endgroup$
    – Peter Kämpf
    Feb 19, 2015 at 19:10
  • $\begingroup$ @fooot: That made me rewrite my answer. The new title needs a fitting answer. $\endgroup$
    – Peter Kämpf
    Feb 19, 2015 at 19:10
  • $\begingroup$ @PeterKämpf We're trying to improve titles and it's good to see it eliciting improved answers as well. Hopefully it doesn't change the actual question too much. Thanks for adding to your answer. $\endgroup$
    – fooot
    Feb 19, 2015 at 20:14

2 Answers 2

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Let's look at what exactly is shown by each curve:

  • The glide ratio curve plots glide ratio (horizontal distance divided by vertical distance) against airspeed.
  • The polar curve plots vertical speed against airspeed.

The x axis (airspeed) is the same for both plots, but the y axis is different. To convert one curve to the other, we need to convert glide ratio to vertical speed and vice versa.

Note: I'm going to make a slight simplification here and assume that airspeed is the same as horizontal speed. This is not true when climbing or descending. A more accurate answer will require some trigonometry to calculate horizontal speed from airspeed. But the error introduced by this simplification is very small.

  • Vertical speed is simply airspeed times glide ratio.
  • Glide ratio is simply vertical speed divided by airspeed.

So, to make one curve out of the other:

  • To plot the polar curve, take a glide ratio curve and multiply it by the x coordinate (thus converting glide ratios to vertical speeds). Conceptually, as you travel outward on the x axis, you are magnifying the curve by x.
  • To plot a glide ratio curve, take the polar curve and divide it by the x coordinate (thus converting vertical speeds to glide ratios).

You mentioned parabolas. The polar curve is a "parabola" only in the vaguest sense that it is shaped like a parabola. It is not a precise mathematical parabola. Its exact shape is determined by extremely complicated aerodynamic factors.

All this being said, if your airplane has a published polar curve, use it, instead of deriving your own. Its values were measured and verified during flight test and you're better off using those than something you produced yourself using other data.

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    $\begingroup$ The parabola approximation is often used in glide computers due to its simplicity. You can find a number of parameters for several gliders here. $\endgroup$
    – yankeekilo
    Apr 28, 2014 at 14:14
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The curves in the top chart are gradients. The bottom chart lists these gradients over flight speed. Say you have in the top chart the y-value L/D = 36 at an x-value of 36 m/s, you do this:

Glide polar plot with example for glide ratio calculation

Every point in the bottom diagram can be constructed by drawing a line (red in the example above) from the origin of the coordinate system with the gradient given by the y-value. Where it reaches the corresponding x-value, you get one point of the blue curve in the bottom diagram. You will need to do this for many x-y pairs to get a full polar curve. I used m/s on both axes to make the procedure more transparent.

The parabola is not so bad for a first-order approximation. If we assume that drag is composed of friction drag and induced drag, we can express this as

$$c_D = c_{D0} + \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$

where $c_D$ is the drag coefficient, $c_{D0}$ is the drag coefficient at zero lift (caused mainly by friction drag), $c_L$ is the lift coefficient, $\pi$ is 3.14159…, AR is the aspect ratio of the wing and $\epsilon$ is the Oswald factor (which mainly describes how well the lift is distributed over the span of the wing. Use 0.98 for gliders and 0.7 - 0.8 for other aircraft).

If you plot this, it is indeed a parabola, and it fits with measured drag polars quite well. The model breaks down beyond the upper and lower stall angles of attack when flow separation causes the lift slope to become nonlinear. If you want to re-create the DG plots in your question, you should use the equation above and keep $c_L$ constant at the $c_{L max}$ for plotting, but calculate $c_D$ with the linearly increasing $c_L$, so the induced drag continues to grow even when the wing stalls. This gives a very good approximation even beyond the stall angle of attack.

Otto Lilienthal was the first pioneer of manned flight who measured the lift and drag of airfoils and wings, and he published the results in a polar diagram. That is why we still call these plots polars today, even when we use Carthesian coordinate systems.

To arrive at speeds, you need to add wing loading $\frac{W}{S} = \frac{m\cdot g}{S}$ and air density $\rho$ like this: $$v = \sqrt{\frac{2\cdot m\cdot g}{\rho\cdot S\cdot c_L}}$$ For the sink speed, things get much easier if we assume that the cosine of the glide path angle $\gamma$ is 1. Then we can write: $$v_z = v\cdot \frac{c_{D0}}{c_L} + v\cdot \frac{c_L}{\pi\cdot AR\cdot\epsilon}$$ Tabulate $c_L$ in your favorite spreadsheet, compute the speeds like that and plot the result. Make sure you restrict $c_L$ for plotting as explained above! Let me know how close the result is.

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  • $\begingroup$ thank you! what if we do not assume $cos(\gamma) = 1$? how does the formula look like? $\endgroup$
    – Federico
    Apr 29, 2014 at 6:40
  • $\begingroup$ @Federico: Instead of $c_L$ you get $\sqrt{c_L^2 + c_D^2}$ in the denominator. The sink speed is $v_z = v \cdot \frac{c_D}{\sqrt{c_L^2 + c_D^2}}$. Please use the formula for $c_D$ on the top of my answer to expand that further. Typing this in a comment will become impossible to read. Also, the parabolic polar is an approximation, and in order to improve accuracy, maybe the improvement should start there. $\endgroup$
    – Peter Kämpf
    Apr 29, 2014 at 8:34
  • $\begingroup$ It is clear enough, thank you very much! $\endgroup$
    – Federico
    Apr 29, 2014 at 8:37
  • $\begingroup$ Thank you for answering a nagging question of mine about why those plots are called polars when they don't normally look like the polar diagrams I'm used to. $\endgroup$
    – PJNoes
    Feb 7, 2017 at 14:37

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