Determining 2D Triangle Intersection in a Cartesian Coordinate System

Question

I am writing a simple graphics editor where a user can draw a triangle (in either clockwise or counterclockwise orientation) and can then select the triangle and drag it around.

My issue comes down to finding whether or not the mouse cursor coordinates intersect with the triangle. Because I am using GLFW, my window space is defined on a Cartesian coordinate system where x/y have range [-1, 1].

This causes an issue when I try to determine whether I have an intersection using barycentric coordinates (or any other method found here)

My current approach is as follows:

double xpos, ypos;
glfwGetCursorPos(window, &xpos, &ypos);

// Get the size of the window
int width, height;
glfwGetWindowSize(window, &width, &height);

// Convert screen position to world coordinates
xworld = ((xpos / double(width)) * 2) - 1;
yworld = (((height - 1 - ypos) / double(height)) * 2) - 1; // NOTE: y axis is flipped in glfw

double area = 0.5 * (-p1y * p2x + p0y * (-p1x + p2x) + p0x * (p1y - p2y) + p1x * p2y);

double s = 1 / (2 * area) * (p0y * p2x - p0x * p2y + (p2y - p0y) * xworld + (p0x - p2x) * yworld),
       t = 1 / (2 * area) * (p0x * p1y - p0y * p1x + (p0y - p1y) * xworld + (p1x - p0x) * yworld);
if (s > 0 && t > 0 && 1 - s - t > 0)
    return true;
return false;

Which kinda works if triangle in the first quadrant and oriented counter-clockwise, however, it also recognizes as intersection if to the left of the triangle.

Thanks for any help.

EDIT: Issue was a typo I had in my code (wrong value for a vertice in my triangle) Ended up using the calculation of area approach to detect intersection.

Compare with this: https://stackoverflow.com/questions/48719711/ — MBo, Oct 05 '18 at 18:59

score 0 · Answer 1 · answered Oct 06 '18 at 08:00

If you don't know the winding order of your triangle, you can check if mouse cursor position is to the left of every edge and if it is to the right of every edge. If one of this is true, so the mouse cursor is inside the triangle indeed.

Luckily, in the case of a triangle, any 2-combination of its vertices yield its edge. So the problem amounts to calculate six cross products.

But also, after the user has finished drawing of a triangle you can save triangle's vertices with a certain winding order, then you can do only three cross products.

Determining 2D Triangle Intersection in a Cartesian Coordinate System

1 Answers1