Re: [math-fun] Curve-fitting methods ?
Brent Meeker <meekerdb@verizon.net> wrote:
What you do is iterate. You can take three points and using spherical geometry determine a radius and a center.
But any three points that obey the triangle inequality are compatible with any radius of curvature whatsoever. (Except for a radius less than that which makes them coplanar with the center of the circle they're on.)
No they aren't. The spherical law of cosines is cos a = cos b cos c + sin b sin c cos A where lower case are edges and upper case are interior angles. So you guess some radius R and solve for A and by permutation also B and C. Then you see whether the law of sines is satisfied. sin A/sin a = sin B/ sin b = sin C/sin c In general it won't be but one side, say (sin a/sin b) depends on R while the other side (sin A/sin B) doesn't. So you adjust R and iterate. I guess I should try to this see if you really works. :-) Brent On 9/26/2018 5:25 PM, Keith F. Lynch wrote:
Brent Meeker <meekerdb@verizon.net> wrote:
What you do is iterate. You can take three points and using spherical geometry determine a radius and a center. But any three points that obey the triangle inequality are compatible with any radius of curvature whatsoever. (Except for a radius less than that which makes them coplanar with the center of the circle they're on.)
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I tried it. Doesn't work in a practical sense as written...just not enough resolution. Maybe I can fix it. Brent On 9/26/2018 7:09 PM, Brent Meeker wrote:
No they aren't. The spherical law of cosines is
cos a = cos b cos c + sin b sin c cos A
where lower case are edges and upper case are interior angles. So you guess some radius R and solve for A and by permutation also B and C. Then you see whether the law of sines is satisfied.
sin A/sin a = sin B/ sin b = sin C/sin c
In general it won't be but one side, say (sin a/sin b) depends on R while the other side (sin A/sin B) doesn't. So you adjust R and iterate.
I guess I should try to this see if you really works. :-)
Brent
On 9/26/2018 5:25 PM, Keith F. Lynch wrote:
Brent Meeker <meekerdb@verizon.net> wrote:
What you do is iterate. You can take three points and using spherical geometry determine a radius and a center. But any three points that obey the triangle inequality are compatible with any radius of curvature whatsoever. (Except for a radius less than that which makes them coplanar with the center of the circle they're on.)
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