Reading this paper, it is actually a pretty simple calculation. The trick is to calculate the signed volume of a tetrahedron – based on your triangle and topped off at the origin. The sign of the volume comes from whether your triangle is pointing in the direction of the origin. (The normal of the triangle … Read more
Quaternion q; vector a = crossproduct(v1, v2); q.xyz = a; q.w = sqrt((v1.Length ^ 2) * (v2.Length ^ 2)) + dotproduct(v1, v2); Don’t forget to normalize q. Richard is right about there not being a unique rotation, but the above should give the “shortest arc,” which is probably what you need.
Instead of cylinder to sphere mapping I would do a sphere triangulation… I would first start with 2 hexagons each start at pole and end on the equator or do the half only and mirror the other when all done… then recursively subdivide the triangles so divide lines to half and change the mid point … Read more
Let’s say you want to scale a range [min,max] to [a,b]. You’re looking for a (continuous) function that satisfies f(min) = a f(max) = b In your case, a would be 1 and b would be 30, but let’s start with something simpler and try to map [min,max] into the range [0,1]. Putting min into … Read more
I understand the original poster was looking for a simple solution, but unfortunately there really is no simple solution. Nevertheless, I’ve recently created an open-source freeware clipping library (written in Delphi, C++ and C#) which clips all kinds of polygons (including self-intersecting ones). This library is pretty simple to use: http://sourceforge.net/projects/polyclipping/ .
Some of the suggested methods will fail in the case of a non-convex polygon, such as a crescent. Here’s a simple one that will work with non-convex polygons (it’ll even work with a self-intersecting polygon like a figure-eight, telling you whether it’s mostly clockwise). Sum over the edges, (x2 − x1)(y2 + y1). If the … Read more
Editor’s note: this is not actually what compilers do, and gives the wrong answer for large positive integers ending with 9, starting with div10(1073741829) = 107374183 not 107374182. It is exact for smaller inputs, though, which may be sufficient for some uses. Compilers (including MSVC) do use fixed-point multiplicative inverses for constant divisors, but they … Read more
How to generate a random point within a circle of radius R: r = R * sqrt(random()) theta = random() * 2 * PI (Assuming random() gives a value between 0 and 1 uniformly) If you want to convert this to Cartesian coordinates, you can do x = centerX + r * cos(theta) y = … Read more
Just a clarification Although the previous answers are right whenever you try to spot the randomness of a pseudo-random variable or its multiplication, you should be aware that while Random() is usually uniformly distributed, Random() * Random() is not. Example This is a uniform random distribution sample simulated through a pseudo-random variable: BarChart[BinCounts[RandomReal[{0, 1}, 50000], … Read more
It’s all a matter of adequate storage and algorithms to treat numbers as smaller parts. Let’s assume you have a compiler in which an int can only be 0 through 99 and you want to handle numbers up to 999999 (we’ll only worry about positive numbers here to keep it simple). You do that by … Read more