ray and ellipsoid intersection accuracy improvement
ray and ellipsoid intersection accuracy improvement
ray and ellipsoid intersection accuracy improvement
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
what about some bitmap+vector approach like this: obtain bounding box of point cloud area coverage Do this if it is not already known. It should be simple O(N) cycle through all points. create map[N][N] of the area It is a ‘bitmap’ of the area for easy data density computation. Just create projection from area(x,y) -> … Read more
Try this code which makes use of a cross product: public bool isLeft(Point a, Point b, Point c){ return ((b.X – a.X)*(c.Y – a.Y) – (b.Y – a.Y)*(c.X – a.X)) > 0; } Where a = line point 1; b = line point 2; c = point to check against. If the formula is equal … Read more
I thought I might briefly mention my own polygon clipping and offsetting library – Clipper. While Clipper is primarily designed for polygon clipping operations, it does polygon offsetting too. The library is open source freeware written in Delphi, C++ and C#. It has a very unencumbered Boost license allowing it to be used in both … Read more
Here’s the answer I found: Just to make the definition complete, in the Cartesian coordinate system: the x-axis goes through long,lat (0,0), so longitude 0 meets the equator; the y-axis goes through (0,90); and the z-axis goes through the poles. The conversion is: x = R * cos(lat) * cos(lon) y = R * cos(lat) … Read more
Despite the title to your question, I think you’re actually looking for the minimum dissection into rectangles of a rectilinear polygon. (Jason’s links are about minimum covers by rectangles, which is quite a different problem.) David Eppstein discusses this problem in section 3 of his 2010 survey article Graph-Theoretic Solutions to Computational Geometry Problems, and … Read more
Taking E is the starting point of the ray, L is the end point of the ray, C is the center of sphere you’re testing against r is the radius of that sphere Compute: d = L – E ( Direction vector of ray, from start to end ) f = E – C ( … Read more
In general, x and y must satisfy (x – center_x)² + (y – center_y)² < radius². Please note that points that satisfy the above equation with < replaced by == are considered the points on the circle, and the points that satisfy the above equation with < replaced by > are considered the outside the … Read more
if (RectA.Left < RectB.Right && RectA.Right > RectB.Left && RectA.Top > RectB.Bottom && RectA.Bottom < RectB.Top ) or, using Cartesian coordinates (With X1 being left coord, X2 being right coord, increasing from left to right and Y1 being Top coord, and Y2 being Bottom coord, increasing from bottom to top — if this is not … Read more