# Extrapolating Oswald Efficiency Factor from CL^2 vs CD Graph

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

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

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

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.

• -1 The efficiency factor cannot ever be greater than 1. This is just as wrong or moreso as .00271.
– Jim
Apr 5, 2022 at 19:15
• 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. Apr 5, 2022 at 19:33
• 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. Apr 6, 2022 at 22:57