0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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."

The theory of the flow around airfoils is well established, Textbooks would be a better place to find it than research papers, since there isn't likely to be any research effort being done on a well-known standard theory. Beware of web sites, though, since there is a lot of total nonsense about this on the web.

https://en.wikipedia.org/wiki/Lift_(force) is as good a place to start as any - note section 2.4 on "Alternative explanations, misconceptions, and controversies" if you look elsewhere on the web.

$\endgroup$
  • $\begingroup$ Right! ... thanks for that. I'm sure you appreciate that the simple logic that both area presented to the flow and lateral displacement of flow are both poportional to the angle, and that lateral force might be supposed to be proportional to the product of these, is one that is sufficiently compelling to occasion a refutation. I do realise that at least in subsonic flow the fluid doesn't just propel the thing propelled by slamming into it - that the fluid sets up a complete system of flow about the thing propelled. But even so, I did need to hear from soneone explicitly $\endgroup$ – AmbretteOrrisey Nov 6 '18 at 14:46
  • $\begingroup$ & specifically that lateral force is proportional to ̸θ rather than ̸θ². I didn't really think of looking it up under ærofoils, as the kind of fin I am talking about is just a simple flat plate, such as on a dart or arrow, & I think of ærofoils as having a specific cross-sectional shape; so I didn't search under that heading. Nevertheless - the theory of flat plate still does come under it, then. As for my saying academic papers: I really just meant PDF files on the subject in the public domain that can be downloaded, which includes many a textbook. And cranksites - seen 'em alright! $\endgroup$ – AmbretteOrrisey Nov 6 '18 at 14:56
  • $\begingroup$ Flat plates (and pretty much any sensible looking shape) will generate lift quite efficiently. The main reason why aerofoils have fancy shapes is to minimize drag, not to maximize lift. $\endgroup$ – alephzero Nov 6 '18 at 17:15
  • $\begingroup$ Right! ... Hmmm! ... I thought it was to make the speed of the flow greater over the upper surface greater than over the lower surface. $\endgroup$ – AmbretteOrrisey Nov 7 '18 at 0:42
  • $\begingroup$ @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. $\endgroup$ – Mike Dunlavey Nov 7 '18 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.