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| From | Ben Bacarisse <ben.usenet@bsb.me.uk> |
| Newsgroups | comp.programming |
| Subject | Re: Topography contours |
| Date | Thu, 12 Nov 2020 15:57:10 +0000 |
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Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes: > If you have a topology in continuous space that you can sample to > arbitrary resolution, both in space and in height, what is the best > way of producing isomers (contour lines of equal height). I've never heard of the called that before. If you are searching, I think contour lines will be more successful There must be established algorithm for this and if you find one, please post a link. If you can assume your function is continuous and smooth, I thought you could use partial derivatives to find the tangent of the contour at a given point. You could the move some small step in that direction and try again. But to stop errors piling up, you'd have to a search in the neighbourhood of the end point for a close match to the starting value. Given that, maybe just a search at each step is the best bet. If the function is smooth in relation to the step size you want to take, you might get away with a binary search of the points on a circle of radius r (the step size) away from the previous point, with the angles ranging from -pi to +pi and 0 being "ahead" -- the direction of the last line segment in the contour. (It would be easier to describe with a diagram!) And you could, of course, make the step size dependent on how smooth the function is in an particular neighbourhood. -- Ben.
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Topography contours Malcolm McLean <malcolm.arthur.mclean@gmail.com> - 2020-11-11 09:26 -0800
Re: Topography contours Ben Bacarisse <ben.usenet@bsb.me.uk> - 2020-11-12 15:57 +0000
Re: Topography contours Malcolm McLean <malcolm.arthur.mclean@gmail.com> - 2020-11-12 08:24 -0800
Re: Topography contours Ben Bacarisse <ben.usenet@bsb.me.uk> - 2020-11-12 17:44 +0000
Re: Topography contours Malcolm McLean <malcolm.arthur.mclean@gmail.com> - 2020-11-13 05:22 -0800
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