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Last edited by gonurvia; 03-30-2017 at 04:25 AM.
cross-posted here don't know if there are more: http://www.msofficeforums.com/excel-...oint-lies.html Note that cross-posting is not necessarily a bad thing, but you can create a lot of bad feelings on forums like this (and risk getting your topic locked) if you do not include links to your cross-posts.
Do you have any more descriptions of this algorithm? The mathworks link really only shows how convenient it is to call the "inpolyhedron" function in MATLAB as part of a larger procedure that plots a surface and filters several random points for those that are inside the polyhedron, with no real discussion of the actual code or algorithm inpolyhedron uses.
Other results of my search suggest that there is a lot of "linear algebra" in this (dot products, cross products, etc.). Can we assume that you are familiar with these linear algebra topics? This could be a useful review, though it stops at the much simpler "convex polyhedron" example: https://www.cs.oberlin.edu/~bob/cs35...orGeometry.pdf
Originally Posted by shg
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Last edited by gonurvia; 03-30-2017 at 04:26 AM.
Inside 3D is what I’m looking for, except I want it to consider only x and y coord. (2D surface, as in Inside function) when calculating if its inside/outside and the Z coordinate to know if its wither above or below. Let me set an example. Say I have the following 3D plane;
(0,0,0)
(0,4,1)
(4,4,3)
(4,0,1)
(0,0,0)
and a point with (2,2,8)
In this example, Inside function would tell me its inside (only x&y coord. taken into account), while Inside3D would say its outside, as it is designed for it. How could I get it to tell me it is Inside and, in this example, ABOVE the plane??
This sounds like a different problem from what was originally requested.
Here's what I see:
1) Is the point above the plane's projection in the XY plane.
1a) Is minx<=x<=maxx -- In this case, is 0<=2<=4 -- and Is miny<=y<=maxy -- in this case is 0<=2<=4. Note that this plane is a rectangle and the projection onto the XY plane is also a rectangle -- Will that always be the case?? Because this gets more complicated for non-rectangles.
2) Is the point above, on, or below the plane.
2a) It is obvious in this case that z is above the plane because z>=maxz. Likewise, if z<=minz, then it would be obvious that the point is below the plane.
2b) The tricky part is when minz<=z<=maxz. In those cases, yoiu probably need to regress on the points that define the plane to get zplane=f(x,y). then compute zplane from f, and compare it to z. If z>zplane, then it is above.
Does that sound like the right kind of algorithm?
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