希望有人会回答一个优雅的解决方案,什么都没有,所以这里是我如何做到这一点。
我总是发现这个问题的解决方案很不雅,说,我不知道是否有更好的方法,也没有我努力找到一个。下面是两个函数将返回由函数Math.atan2(yDif, xDif)
返回在范围内的角度得出两个角度之间的最短方向 - Math.PI
到Math.PI
获取最短方向。
// returns true if shortest direction is clockwise else false
function isShortestDirClockwise(ang1, ang2){
if (ang1 < 0) {
if ((ang2 < 0 && ang1 > ang2) || (ang2 >= 0 && ang1 + Math.PI < ang2)) {
return false;
}
} else {
if ((ang2 > 0 && ang1 > ang2) || (ang2 <= 0 && ang1 - Math.PI < ang2)) {
return false;
}
}
return true;
}
找最短的角度
// returns the shortest angle neg angles are counterClockwise positive are clockwise
function getShortestAngle(ang1, ang2){
var cw = true; // clockwise
var ang;
if (ang1 < 0) {
if((ang2 < 0 && ang1 > ang2) || (ang2 >= 0 && ang1 + Math.PI < ang2)) {
cw = false;
}
} else {
if ((ang2 > 0 && ang1 > ang2) || (ang2 <= 0 && ang1 - Math.PI < ang2)) {
cw = false;
}
}
if (cw) {
var ang = ang2 - ang1;
if (ang < 0) {
ang = ang2 + Math.PI * 2 - ang1;
}
return ang;
}
var ang = ang1 - ang2;
if (ang < 0) {
ang = ang1 + Math.PI * 2 - ang2;
}
return -ang;
}
正如我不愿意使用这些功能,我更喜欢下面稍微复杂但更优雅的解决方案。这根据一组3点找出最短角度。中心点以及我希望找到两个角度之间的角度的两点,它总是返回任意两条线之间的最短角度。
// Warning may return NAN if there is no solution (ie one or both points (p1,p2) are at center)
// Also uses Math.hypot check browser compatibility if you wish to use this function or use Math.sqrt(v1.x * v1.x + v1.y * v1.y) instead
function getAngleBetween(center, p1, p2){
var d;
var ang;
// get vectors from center to both points
var v1 = { x : p1.x - center.x, y : p1.y - center.y};
var v2 = { x : p2.x - center.x, y : p2.y - center.y};
// normalise both vectors
d = Math.hypot(v1.x, v1.y);
v1.x /= d;
v1.y /= d;
d = Math.hypot(v2.x, v2.y);
v2.x /= d;
v2.y /= d;
// cross product gets the angle in range -Math.PI/2 to Math.PI/2
ang = Math.asin(v1.x * v2.y - v1.y * v2.x);
// use the cross product of the line perpendicular to the first to find the quadrant
if(-v1.y * v2.y - v1.x * v2.x > 0){
if(ang < 0){
ang = -Math.PI - ang;
}else{
ang = Math.PI - ang;
}
}
return ang;
}
谢谢你的帮助!不过,我会保留这个问题'没有答案',以防万一有人提到你所说的'优雅的解决方案'。 –