返回两个圆之间的 x,y 交点的 JavaScript 函数?
I got the (x,y) center location of two circles and their radius but I need to find their intersection points (marked with red) using JavaScript.
I think the best explanation as far as the math is concerned is found here (Intersection of two circles), but I don't really understand the math so I'm not able to implement it.
For example d = ||P1 - P0|| , what do the || stand for? Does it mean that the resulting number is always a positive?
And also P2 = P0 + a ( P1 - P0 ) / d , aren't the P's here something like (10, 50)? But doing (10,50)+13 in JavaScript gives you 63, so it just ignores the first number, so what's suppose to happen? Should the outcome be (23,63) here or? And also the P1-P0 part or (40,30)-(10,60), how do you express that in JavaScript?
解决方案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;
/* Determine the straight-line distance between the centers. */
d = Math.sqrt((dy*dy) + (dx*dx));
/* Check for solvability. */
if (d > (r0 + r1)) {
/* no solution. circles do not intersect. */
return false;
}
if (d < Math.abs(r0 - r1)) {
/* no solution. one circle is contained in the other */
return false;
}
/* 'point 2' is the point where the line through the circle
* intersection points crosses the line between the circle
* centers.
*/
/* Determine the distance from point 0 to point 2. */
a = ((r0*r0) - (r1*r1) + (d*d)) / (2.0 * d) ;
/* Determine the coordinates of point 2. */
x2 = x0 + (dx * a/d);
y2 = y0 + (dy * a/d);
/* Determine the distance from point 2 to either of the
* intersection points.
*/
h = Math.sqrt((r0*r0) - (a*a));
/* Now determine the offsets of the intersection points from
* point 2.
*/
rx = -dy * (h/d);
ry = dx * (h/d);
/* Determine the absolute intersection points. */
var xi = x2 + rx;
var xi_prime = x2 - rx;
var yi = y2 + ry;
var yi_prime = y2 - ry;
return [xi, xi_prime, yi, yi_prime];
}
相关文章