# How could I find 𝐶𝐿𝛼 lift coefficient slope?

I'm trying to find ๐ถ๐ฟ๐ผ lift coefficient slope through the below diagram, could anyone explain for me please, how could I find it?

by the way, I depend on this equation ๐ถ๐ฟ = ๐ถ๐ฟ๐ผ (๐ผโ๐ผ0)

• You can eyeball it from the plot, it is the slope of your curve, rise over run. Your cL goes up .1 for every 4 degrees, so cLalpha is (.1)/4 – MikeY May 26 '19 at 14:19
• @MikeY Thank you so much sir – MohammedALNasrawi May 27 '19 at 5:00

Basic maths:

$$\alpha_0$$ = point where the line crosses the x-axis = -4

$$C_L(\alpha)$$ is linear, so take any two points, e.g. (-4,0) and (16,0.5)

$$\Delta y$$ = 0.5 - 0 = 0.5; $$\Delta x = 16 -(-4) = 16 + 4 = 20$$

$$C_L = \frac{0.5}{20} * (\alpha + 4) = 0.025 * (\alpha + 4)$$

Check:

• $$\alpha = -4: C_L = 0$$
• $$\alpha = 16: C_L = 0.025 * (16 + 4) = 0.5$$
• Thank you so much sir. – MohammedALNasrawi May 26 '19 at 7:27

The lift coefficient slope is the ratio of a small increment of CL to the corresponding small increment of AOA. Since your curve is reduced to a straight line the result of the ratio would be the same either for small increments ratio or for large increments ratio. To calculate the slope ahead of stall you may go from alpha = -4 to alpha = 16; the corresponding CL goes from 0 to 0.5 So the slope will be (0.5-0)/(16-(-4))=0.5/20 If you go from alpha =0 to alpha = 16, CL will go from 0.1 to 0.5 and the ratio is: (0.5-0.1)/(16-0). Both examples will give you the same result as you may check.

• Thank you so much sir. – MohammedALNasrawi May 26 '19 at 7:27