It turns out that Wikipedia does have the best answer: // h = 1..12, m = 0..59 static double angle(int h, int m) { double hAngle = 0.5D * (h * 60 + m); double mAngle = 6 * m; double angle = Math.abs(hAngle – mAngle); angle = Math.min(angle, 360 – angle); return angle; } … Read more
If you can, do the picking on the CPU by calculating a ray from the eye through the mouse pointer and intersect it with your models. If this isn’t an option I would go with some type of ID rendering. Assign each object you want to pick a unique color, render the objects with these … Read more
I think it’s easiest to think of your problem as an intersection problem: what is the parameter location (s,t) where the point p intersects the arbitrary 2D bilinear surface defined by p0, p1, p2 and p3. The approach I’ll take to solving this problem is similar to tspauld’s suggestion. Start with two equations in terms … Read more
If y = 1 / (1 + exp(-x)) then its inverse is x = ln(y / (1 – y))
SVD can be understood from a geometric sense for square matrices as a transformation on a vector. Consider a square n x n matrix M multiplying a vector v to produce an output vector w: w = M*v The singular value decomposition M is the product of three matrices M=U*S*V, so w=U*S*V*v. U and V … Read more
The following resources survey some popular numerical methods for inverse kinematics problems: Samuel R. Buss. Introduction to Inverse Kinematics with Jacobian Transpose, Pseudoinverse and Damped Least Squares methods Bill Baxter. Fast Numerical Methods for Inverse Kinematics Chris Welman. Inverse Kinematics and Geometric Constraints for Articulated Figure Manipulation Buss’s survey may be particularly interesting, because it … Read more
I know this is 2 years old, but I’ve recently been looking around for the same problem and I don’t see an elegant solution without ifs posted in here, so here it goes: shortest_angle=((((end – start) % 360) + 540) % 360) – 180; return shortest_angle * amount; that’s it ps: of course, % is … Read more
Let P0, P1, P2 be the control points, and Pc be your fixed point you want the curve to pass through. Then the Bezier curve is defined by P(t) = P0*t^2 + P1*2*t*(1-t) + P2*(1-t)^2 …where t goes from zero to 1. There are an infinite number of answers to your question, since it might … Read more
Proof: Point D is on a line CD perpendicular to AB, and of course D belongs to AB. Write down the Dot product of the two vectors CD.AB = 0, and express the fact D belongs to AB as D=A+t(B-A). We end up with 3 equations: Dx=Ax+t(Bx-Ax) Dy=Ay+t(By-Ay) (Dx-Cx)(Bx-Ax)+(Dy-Cy)(By-Ay)=0 Subtitute the first two equations in … Read more
For each edge in both polygons, check if it can be used as a separating line. If so, you are done: No intersection. If no separation line was found, you have an intersection. /// Checks if the two polygons are intersecting. bool IsPolygonsIntersecting(Polygon a, Polygon b) { foreach (var polygon in new[] { a, b … Read more