# Lateral Force on a Fin

I am wondering what the lateral force is on the fin of, say, an arrow, (or other befinned missile) as a function of the angle through which it is turned ... or more particularly what it's dependence on the angle is. By a naïve argument it could be held to be a ̸θ² dependence, as both the area that the fin presents to the airflow and the angle through which the flow is turned are proportional to ̸θ. But this is not necessarily so, I might imagine, as the the interaction of the flow with the fin is not simply that of its impacting a target of a certain size, but rather of the entire flow for some distance around the fin being altogether reshapen. I have not been able to find a straightforward answer to this through a web 'conjuration' of academic papers - but I suspect that whatever the solution is, the Schwarz-Kristoffel transformation should play a major rôle in it.

And in addition, the solution to the equation of motion, given some intial rate-of-turning as boundary-condition, becomes a really quite appalling one with an elliptic integral in it (although I'm sure it can be simplified: the Wolfram™ online integrator is notorious for returning solutions in an extremely raw form!) ... but nature doesn't care about that, of course!

Intuitively there is a certain repugnance in the idea of the lateral (& therefore restoring force) having θ² dependence, as such dependence might be held to admit of rather large swings of the missile about orthobaticity, by reason of its thereby having a significant latitude in which to swing without the restoring force even beginning significantly to increase.

But maybe they do! I can't honestly say I have often closely observed an arrow in flight.

Aerodynamically, a fin works the same way as a wing. For small angles of attack $$\theta$$ to the airflow, the "lift force" is proportional to $$\theta$$. For large angles of attack where the airflow may stall, etc, there is no simple "formula."
• @AmbretteOrrisey: This answer is right. What makes airfoils work is that they have sharp trailing edges and they impose a circulation (change the direction of the airflow), resulting in a momentum flux (lift coefficient) proportional to $\theta$ up to the stall angle. The stall angle depends on many things. – Mike Dunlavey Nov 7 '18 at 13:22