问题
Hello I need to make for school a program in Javascript that says if circles had an collision. It doesn't need to be shown graphical.
I gave it a try but my code doesn't seem to work.
Hope you could help me out with a script.
Here's my code what I produced.
function collision (p1x, p1y, r1, p2x, p2y, r2) {
var a;
var x;
var y;
a = r1 + r2;
x = p1x - p2x;
y = p1y - p2y;
if (a > (x*x) + (y*y)) {
return true;
} else {
return false;
}
}
var collision = collision(5, 500, 10, 1000, 1500, 1500);
alert(collision);
回答1:
Your check should be if (a > Math.sqrt((x*x) + (y*y)))
http://cgp.wikidot.com/circle-to-circle-collision-detection
So the complete code is
function collision(p1x, p1y, r1, p2x, p2y, r2) {
var a;
var x;
var y;
a = r1 + r2;
x = p1x - p2x;
y = p1y - p2y;
if (a > Math.sqrt((x * x) + (y * y))) {
return true;
} else {
return false;
}
}
var collision = collision(5, 500, 10, 1000, 1500, 1500);
console.log(collision);
and for a less computational implementation (using es7 syntax for the snippet) use
const checkCollision = (p1x, p1y, r1, p2x, p2y, r2) => ((r1 + r2) ** 2 > (p1x - p2x) ** 2 + (p1y - p2y) ** 2)
var collision = checkCollision(5, 500, 10, 1000, 1500, 1500);
console.log(collision);
as Darek Rossman shows in his answer.
https://stackoverflow.com/a/8331460/128165
回答2:
In your if statement, try this instead:
if ( a * a > (x * x + y * y) ) {
...
} else {
...
}
回答3:
The length of a triangle having sides dx
and dy
(i.e. the distance between points (x1, y1)
and (x2, y2)
where dx = x2 - x1
and dy = y2 - y1
) is equal to:
sqrt(dx^2 + dy^2)
So you probably want:
if(a > Math.sqrt(x*x + y*y)) {
来源:https://stackoverflow.com/questions/8331243/circle-collision-javascript