# 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 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 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 at 22:57