我有一个网格,我在其中放置一个对象,我想围绕一个固定的单元格(cell3)旋转.该对象包含坐标,如：
activeObject = {
coords: {
cell1: {
x: 0,y: 0
},cell2: {
x: 1,y: 1
},cell3: {
x: 2,y: 2
},cell4: {
x: 2,y: 3
},cell5: {
x: 3,}
}
输出：
我想围绕cell3旋转这个对象,例如使用x：2,y：2,不使用某种(基本)三角函数对每个单元格位置进行硬编码.我意识到我必须检查每个细胞与细胞3的距离以及方向.但是我不知道如何进行计算,因为我对三角学不太了解.新的活动对象将是：
activeObject = {
coords: {
cell1: {
x: 4,cell2: {
x: 4,cell4: {
x: 1,cell5: {
x: 1,}
}
输出：
最佳答案
一些基本的想法
>如果枢轴不是原点,则必须对变换进行一些修正.
>如果一个点位于(1,2),则围绕原点旋转
> 90°CW：变为(2,-1)
> 90°CCW：变为(-2,1)
>点(x,y)的结论
> 90°CW：成为(y,-x)
> 90°CCW：变为(-y,x)
>至少应用枢轴修正
一些数学被发现here.
function Point(x,y) {
this.x = x;
this.y = y;
}
Point.prototype.rotateCW = function (c) {
// x' = x cos phi + y sin phi \ formula with pivot at (0,0)
// y' = -x sin phi + y cos phi /
// phi = 90° insert phi
// cos 90° = 0 sin 90° = 1 calculate cos and sin
// x' = y \ formula with pivot at (0,0)
// y' = -x /
// x' = (cy - y) + cx \ formula with different pivot needs correction
// y' = -(cx - x) + cy /
// y' = -cx + x + cy /
return new Point(
c.x + c.y - this.y,c.y - c.x + this.x
);
}
Point.prototype.rotateCCW = function (c) {
// source: https://en.wikipedia.org/wiki/Rotation_(mathematics)#Two_dimensions
// x' = x cos phi + y sin phi \ formula with pivot at (0,0)
// y' = -x sin phi + y cos phi /
// phi = -90°
// cos -90° = 0 sin -90° = -1
// x' = -y \ formula with pivot at (0,0)
// y' = x /
// x' = -(cy - y) + cx \ formula with different pivot needs correction
// x' = -cy + y + cx /
// y' = (cx - x) + cy /
return new Point(
c.x - c.y + this.y,c.y + c.x - this.x
);
}
var activeObject = {
coords: {
cell1: new Point(0,0),cell2: new Point(1,1),cell3: new Point(2,cell4: new Point(2,3),cell5: new Point(3,}
},pivot = new Point(2,rotated = { coords: {} };
Object.keys(activeObject.coords).forEach(function (k) {
rotated.coords[k] = activeObject.coords[k].rotateCW(pivot);
});
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