🚀 go-pugleaf

Matthias Weiss wrote:
> 
> 
> Suppose you have your two points P and Q. Without loss of generality,
> you can put the radius from the center C to P on the x-axis, and move
> Q around freely. Hence you have P=(x1,  0), Q=(x2, y2). You are also
> given the arc length L.
> 

That's a nice trick and I think it'll even work for the program I'm
working on. 

> The center of your circle has coordinates C = (x0,0). You can assume
> that x0 < x1, therefore the radius R=x1-x0.
> 
> In polar coordinates, with angular parameter t,
> 
> y2 = R sin(t2)
> 
> for a certain, as yet unknown angle t2.
> Hence
> 
> sin(t2) =
> y2/R
> (*)
> 
> You don't know how big R is, but that's no biggie, as
> 
> R =
> L/t2
> (**)
> 
> which gives you with (*)
> 
> sin(t2) = y2*t2/L,
> 
> or
> 
> sin(t2)/t2 =
> y2/L                                                            (#)
> 
> the right hand side of which contains only known quantities.

It's unbelievable but I understand!
> 
> (Notice that when y2 = L, ie. Q is directly on top of P, we have
>  sin(t2) = t2    =>   t2=0   =>   R=infinity, as one would assume!)

Or rather R is undefined/can be anything.

> 
> Now R can be computed from (**), and so, if desired, the center C.
> 
> The whole computation can be done with P not on the x-axis,
> but the formulae get a bit more cumbersome.
> 
I think I'll skip that, if possible....

> You may compute t2 from eqn (#) using, for example, Newton
> iteration. You can also try the following: Solve for t2
> 
> t2 = L/y2 sin(t2),
> 
> use an initial guess for t2 (in radians!!!), compute the right-hand
> side,
> plug the thus obtained t2 back into the right-hand side, and see if
> you get convergence.
> 
> Notice also that, since  x>sin(x)  for all x>0, you must have L>y2,
> again
> as one would expect.
> 
> Cheerio,
> Matthias Weiss
> mmshiro@math.lsa.umich.edu

Thanks a lot.

Remco de Korte