🚀 go-pugleaf

Remco de Korte wrote:

> When the radius is variable as well as the angle between the two lines
> connecting the center of the circle with both points the length of the
> arc can be a constant.

Ah...of course you are right. I was thinking of the SHORTER arc joining
the two points, but of course you can flip the whole thing over. Try
this for size:

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.

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.

(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!)

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.

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