# How to properly use the lift-curve slope for divergence calculations?

I was reading Wright and Cooper's writings on static aeroelasticity and I think I'm missing something. To calculate the acting lift, the equation $$L = q S a_1 \theta_0$$ is used, where $$q$$ is the dynamic pressure, $$S$$ is the wing area, $$a_1$$ is the lift-curve slope ($$\frac{\delta C_L}{\delta \alpha}$$) and $$\theta_0$$ is the initial incidence angle.

However, the way I see it, that only equals to the traditional $$L = qS C_L$$ when/if I have a curve where both $$C_L$$ and $$\alpha$$ are zero at the same time, since the slope indicates the rate of my $$C_L$$ variation but not the starting point.

So the question is: is the Wright and Cooper's formula only supposed to calculate the increase in lift generated or am I reading it wrong?

I'll try and show a few examples of where that is used, so it may clarify.

In divergence studies, to calculate the total lift:

$$L_{total} = L_{rigid} + L_{elastic} = q S \frac{\delta C_L}{\delta \alpha} (\alpha_0 + \theta)$$

Where $$\alpha_0$$ is the original angle of attack and $$\theta$$ is the twist angle due to aerodynamic loading. The way I see it, this should be equal to $$\large L_{total} = qS C_{L_{\alpha+\theta}}$$, but it isn't. I'm not sure if it's not supposed to, or if I'm seeing it wrong.

In Wright and Cooper's book, to calculate the pitching moment on an airfoil.

Here again, they calculate lift as $$L = q c a_1 \theta_0$$, and that is not equal to $$L = qc C_L$$, because the admitted $$C_L$$ values will be lower.

• Without being certain, a possibility is that the value $M= q\;e\ c^2 a_1\theta_0$, is indeed the moment only from the initial position of the airfoil, which can be then later used to calculate the $\theta$ (or $\delta \theta$ if you like). Then the new total (rigid+ elastic) can be used in an iterative process until convergence. – NMech Apr 20 at 9:23