I have plotted this CL^2 vs CD graph for a glide test performed in a flight simulator. I want to find the Oswald Efficiency Factor, e from it and I assume I can do that from its gradient. With a wing Aspect Ratio of 5.93, Equating the gradient (19.824) to 1/(𝝅eAR) and rearranging for e gives me a value of 0.00271 which is really strange as I was expecting it to be in the range 0.80-0.95. Also the gradient looks reasonable too. What am I doing wrong here?CL^2 vs CD


1 Answer 1


You have the relationship between ${C_L}^2$ and $C_D$ the wrong way around. It is

$$ C_D = {C_D}_0 + \frac{{C_L}^2}{\pi\,e\,\Lambda} $$

so the gradient should be

$$ \frac{1}{19.824} = 0.0504 = \frac{1}{\pi\,e\,\Lambda} $$

which resolves to

$$ e = \frac{1}{\pi \cdot 0.0504 \cdot 5.93} = 1.064 $$

which is at least in the ballpark.

If you determine the slope $dC_D/d{C_L}^2$ the other way around you may find that it is slightly larger than $0.0504$ and will give you an efficiency factor in the $e = 0.9..0.95$ range.

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    $\begingroup$ -1 The efficiency factor cannot ever be greater than 1. This is just as wrong or moreso as .00271. $\endgroup$
    – Jim
    Commented Apr 5, 2022 at 19:15
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    $\begingroup$ If the slope is e.g. 0.052 instead of 0.0504, the Oswald factor will be slightly below 1.0, so in the correct range. But thanks for explaining the downvote. $\endgroup$ Commented Apr 5, 2022 at 19:33
  • $\begingroup$ Thanks a lot for this. Missed out on a simple manipulation. And yes, I did adjust the slope slightly and got the Oswald value as 0.97ish so it's reasonable. $\endgroup$ Commented Apr 6, 2022 at 22:57

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