Re: [math-fun] Curve-fitting methods ?
Brent Meeker <meekerdb@verizon.net> wrote:
Keith F. Lynch wrote:
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.
I know. I should have said any three *sides* that obey the triangle inequality. For one thing, speaking of three *points* that obey the triangle inequality makes little sense, as no three points can fail to do so. Suppose all you have are the lengths of three sides of a triangle, and the knowledge that the three vertices of the triangle are equidistant from some "center" point in space, possibly a point at infinity, possibly a point in the same plane with the triangle. You don't know how curved the sides are, except that they all have the same radius of curvature, the curvature such that every point on all the sides is also equidistant from that same center point. Wouldn't you agree that with the information you have, you can't rule out any radius above a certain minimum? I'm not a visual thinker, so I won't guess at whether the locus of possible center points is a line or some other kind of curve. If the three sides of the triangle were equal, I assume it would have to be a line, by symmetry, unless it splits into three curves.
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. :-)
I tried it. Doesn't work in a practical sense as written...just not enough resolution. Maybe I can fix it.
Again, I don't think you have enough information. The law of sines should *always* be satisfied, unless you choose a radius less than that of the circle the three points are on. This is complicated by the fact that you can't assign any definite coordinates for the points or any straight-line distances between them until you know the radius. Any three points in space are on a unique plane and are on a unique circle in that plane. They're also on an infinite number of spheres, with all radii starting with that of the circle. Again, this is complicated by the fact that our points aren't fixed, but move around as the fixed-length sides change curvature, but I think the principle is the same. For extra credit, extend the scenario to negative curvature. See https://en.wikipedia.org/wiki/Hyperbolic_law_of_cosines
You're right. With four points there are four spherical triangles which depend on the radius and which can only be consistent for one radius value. Brent On 9/27/2018 5:46 PM, Keith F. Lynch wrote:
Brent Meeker <meekerdb@verizon.net> wrote:
Keith F. Lynch wrote:
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. I know.
I should have said any three *sides* that obey the triangle inequality. For one thing, speaking of three *points* that obey the triangle inequality makes little sense, as no three points can fail to do so.
Suppose all you have are the lengths of three sides of a triangle, and the knowledge that the three vertices of the triangle are equidistant from some "center" point in space, possibly a point at infinity, possibly a point in the same plane with the triangle. You don't know how curved the sides are, except that they all have the same radius of curvature, the curvature such that every point on all the sides is also equidistant from that same center point. Wouldn't you agree that with the information you have, you can't rule out any radius above a certain minimum?
I'm not a visual thinker, so I won't guess at whether the locus of possible center points is a line or some other kind of curve. If the three sides of the triangle were equal, I assume it would have to be a line, by symmetry, unless it splits into three curves.
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. :-) I tried it. Doesn't work in a practical sense as written...just not enough resolution. Maybe I can fix it. Again, I don't think you have enough information. The law of sines should *always* be satisfied, unless you choose a radius less than that of the circle the three points are on. This is complicated by the fact that you can't assign any definite coordinates for the points or any straight-line distances between them until you know the radius.
Any three points in space are on a unique plane and are on a unique circle in that plane. They're also on an infinite number of spheres, with all radii starting with that of the circle. Again, this is complicated by the fact that our points aren't fixed, but move around as the fixed-length sides change curvature, but I think the principle is the same.
For extra credit, extend the scenario to negative curvature. See https://en.wikipedia.org/wiki/Hyperbolic_law_of_cosines
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Keith F. Lynch