Your best bet is to iterate over all the lines and find the minimum distance from a point to a line segment. To find the distance from a point to a line segment, you first find the distance from a point to a line by picking arbitrary points P1 and P2 on the line (it … Read more
To give you some more choices of libraries out there: Polyboolean. I never tried this one, but it looks promising: http://www.complex-a5.ru/polyboolean/index.html General Polygon Clipper. This one works very well in practice and does triangulation as well as clipping and holes holes: http://www.cs.man.ac.uk/~toby/alan/software/ My personal recommendation: Use the tesselation from the GLU (OpenGL Utility Library). The … Read more
To flip the triangle you really just need to change the direction of iteration. Instead of going from i = 0 to i < lines you need to go down from i = lines-1 to i >= 0 You also need to change the c to how many spaces and symbols you want to start … Read more
Providing a custom component turned out to be the best solution. I use it like this in my code <Controls:Arc Center=”{Binding Path=PreviousMousePositionPixels}” Stroke=”White” StrokeDashArray=”4 4″ SnapsToDevicePixels=”True” StartAngle=”0″ EndAngle=”{Binding Path=DeltaAngle}” SmallAngle=”True” Radius=”40″ /> SmallAngle when true will render the small angle between the points irrespective of order of StartAngle and EndAngle. When SmallAngle is false the … Read more
It is a little complicated, but you can draw all the objects by the following code: from mpl_toolkits.mplot3d import Axes3D import matplotlib.pyplot as plt import numpy as np from itertools import product, combinations fig = plt.figure() ax = fig.gca(projection=’3d’) ax.set_aspect(“equal”) # draw cube r = [-1, 1] for s, e in combinations(np.array(list(product(r, r, r))), 2): … Read more
There is in fact no real reason to partition the sphere into a regular non-overlapping mesh, try this: partition your sphere into semi-overlapping circles see here for generating uniformly distributed points (your circle centers) Dispersing n points uniformly on a sphere you can identify the points in each circle very fast by a simple dot … Read more
Translated the C function on the site to JavaScript: function intersection(x0, y0, r0, x1, y1, r1) { var a, dx, dy, d, h, rx, ry; var x2, y2; /* dx and dy are the vertical and horizontal distances between * the circle centers. */ dx = x1 – x0; dy = y1 – y0; /* … Read more
Known data: bounding box width W, height H, rotation angle Fi Wanted: coordinates of rotated rectangle vertices. Unknown source rectangle size: w x h Bounding box size for this dimension and rotation angle: H = w * Abs(Sin(Fi)) + h * Abs(Cos(Fi)) W = w * Abs(Cos(Fi)) + h * Abs(Sin(Fi)) denote as = Abs(Sin(Fi)) … Read more
you could use a gradient with stop-opacity to do this. you would add two “middle” stops with opacity 0 and 1 respectively an set the offset of both to the percentage you need. <svg xmlns=”http://www.w3.org/2000/svg” viewBox=”0 0 100 100″ width=”200″ height=”200″> <linearGradient id=”lg” x1=”0.5″ y1=”1″ x2=”0.5″ y2=”0″> <stop offset=”0%” stop-opacity=”1″ stop-color=”royalblue”/> <stop offset=”40%” stop-opacity=”1″ stop-color=”royalblue”/> … Read more
as3.x = centerX + radius * cos(angle) as3.y = centerY + radius * sin(angle) Note that the rotation in the picture linked to is in the “negative direction”. I.e, an increase of the angle, yields a counter-clockwise rotation.