Question

html5画布中矩形的旋转以弧度形式存储。为了找出后续鼠标是否在任何给定矩形内，我将鼠标x和y平移到矩形的旋转原点，将旋转的反转应用于鼠标坐标，然后将鼠标坐标平移回来。

这根本不起作用 - 鼠标坐标没有按预期变换（也就是说，当在旋转的矩形的可见边界内单击时，不在原始矩形的边界内），并且对矩形进行测试＆＃ 39;边界是失败的。鼠标点击的检测仅在矩形的最中心区域内起作用。请参阅下面的代码段，告诉我你是否能看到这里的错误。

 // Our origin of rotation is the center of the rectangle
 // Our rectangle has its upper-left corner defined by x,y, its width
 // defined in w, height in h, and rotation(in radians) in r.  
var originX = this.x + this.w/2, originY = this.y + this.h/2, r = -this.r;

 // Perform origin translation
mouseX -= originX, mouseY -= originY;
// Rotate mouse coordinates by opposite of rectangle rotation
mouseX = mouseX * Math.cos(r) - mouseY * Math.sin(r);
mouseY = mouseY * Math.cos(r) + mouseX * Math.sin(r);
// Reverse translation
mouseX += originX, mouseY += originY;

// Bounds Check
if ((this.x <= mouseX) && (this.x + this.w >= mouseX) && (this.y <= mouseY) && (this.y + this.h >= mouseY)){
    return true;
}

Answer 1

经过一些进一步的工作后，得出了以下解决方案，我认为我会在这里转录给未来可能需要它的人：

// translate mouse point values to origin
    var dx = mouseX - originX, dy = mouseY - originY;
    // distance between the point and the center of the rectangle
    var h1 = Math.sqrt(dx*dx + dy*dy);
    var currA = Math.atan2(dy,dx);
    // Angle of point rotated around origin of rectangle in opposition
    var newA = currA - this.r;
    // New position of mouse point when rotated
    var x2 = Math.cos(newA) * h1;
    var y2 = Math.sin(newA) * h1;
    // Check relative to center of rectangle
    if (x2 > -0.5 * this.w && x2 < 0.5 * this.w && y2 > -0.5 * this.h && y2 < 0.5 * this.h){
        return true;
    }

HTML5 Canvas中旋转矩形内的鼠标位置

1 个答案: