Two circles intersect if, and only if, the distance between their centers is between the sum and the difference of their radii. Given two circles (x0, y0, R0) and (x1, y1, R1), the formula is as follows: ABS(R0 – R1) <= SQRT((x0 – x1)^2 + (y0 – y1)^2) <= (R0 + R1) Squaring both sides … Read more
You should be able to exploit the fact that the matrix is affine to speed things up over a full inverse. Namely, if your matrix looks like this A = [ M b ] [ 0 1 ] where A is 4×4, M is 3×3, b is 3×1, and the bottom row is (0,0,0,1), then … Read more
Let me try to explain it a little better. I need to export a model from Blender, in which the z axis faces up, into OpenGL, where the y axis faces up. For every coordinate (x, y, z) it’s simple; just swap the y and z values: (x, z, y). Because I have swapped the … Read more
Starting point you know (x1, x2), end point is (x1 + l * cos(ang), y1 + l * sin(ang)) where l is the length and ang is the angle.
I think the above answer deserves a few words of explanation here nevertheless. A short note in advance: Arithmetic expressions in Prolog are just terms (“Everything is a term in Prolog”), which are not evaluated automatically. (If you have a Lisp background, think of quoted lists). So 3 + 4 is just the same as … Read more
As you’ve discovered, both Quadratic curves and Cubic Bezier curves just connect 2 points with a curve. Since the Cubic curve has more control points, it is more flexible in the path it takes between those 2 points. For example, let’s say you want to draw this letter “R”: Start drawing with the “non-curvey” parts … Read more
The quaternion representation of rotation is a variation on axis and angle. So if you rotate by r radians around axis x, y, z, then your quaternion q is: q[0] = cos(r/2); q[1] = sin(r/2)*x; q[2] = sin(r/2)*y; q[3] = sin(r/2)*z; If you want to create a quaternion that only rotates around the y axis, … Read more
yes it does… return num*-1; or simply return -num;
I wrote an article about finding the bounding coordinates: http://JanMatuschek.de/LatitudeLongitudeBoundingCoordinates The article explains the formulae and also provides a Java implementation. (It also shows why IronMan’s formula for the min/max longitude is inaccurate.)
using this referance to calculate Angle: private double angleFromCoordinate(double lat1, double long1, double lat2, double long2) { double dLon = (long2 – long1); double y = Math.sin(dLon) * Math.cos(lat2); double x = Math.cos(lat1) * Math.sin(lat2) – Math.sin(lat1) * Math.cos(lat2) * Math.cos(dLon); double brng = Math.atan2(y, x); brng = Math.toDegrees(brng); brng = (brng + 360) % … Read more