Question

如上图所示，假设有一条从中心发出的光线在矩形边缘碰撞。所以我想计算它碰撞的点，这样我就可以计算出边缘点和中心点之间的距离。

我们知道什么？
- 矩形的宽度和高度
- 学位与b度（如上图所示）
- centerX，centerY

Answer 1

让中心为（0,0），光线角度为phi。伪代码：

c = Cos(phi)
s = Sin(phi)
if Width * Abs(s) < Height * Abs(c) then
   x = Sign(c) * Width / 2
   y = Tan(phi) * x
else
   y = Sign(s) * Height / 2
   x = CoTan(phi) * y

Answer 2

我终于构建了一个函数来查找矩形边缘点的坐标。

function mine(rect,angle){ 
/* rect is object like this {x:number,y:number,width:number,height:number} */
/* angle is an angle(number) between -Math.PI and Math.PI (in radians) */
if(angle==0){
    return{
        x:rect.x+rect.width,
        y:rect.y+rect.height/2
    }
}
/* cp is the center point of the rectangle */
var cp = { 
    x:rect.x + rect.width/2,
    y:rect.y + rect.height/2
};

/*dp is the point of the top right diagonal of the rectangle*/
var dp = {
    x:rect.x + rect.width,
    y:rect.y
}

var dx = dp.x - cp.x;   //distanceX between dp and cp
var dy = dp.y - cp.y;   //distanceY between dp and cp

/* calculate the angle of diagonal then abs the angle*/
var diagonalAngle = Math.abs(Math.atan2(dy,dx)); // now we can know from which side the edgePoint is

var distanceX;
var distanceY;
var signX;  //a vector
var signY = Math.abs(angle) / angle;    //a vector
angle = Math.abs(angle);
if(angle<=Math.PI/2){ //that means that the edge point is somewhere on the half right side of the rectangle
    signX = 1;
    if(angle<diagonalAngle){ //the edge point is on the right side of the rectangle
        distanceX = rect.width/2; // this number is the adjacent of the right triangle
        var distance = distanceX/Math.cos(angle); // this number is the hypotenuse width of the right triangle. it is calulated using: cos(angle) = opposite / hypothenuse then hyp = opposite / cos(angle)
        distanceY = Math.sqrt(distance*distance - distanceX*distanceX); // theorem of pythagoras
    }
    else{ //the edge point is on the top or the bottom of the rectangle
        distanceY = rect.height/2;
        var distance = distanceY/Math.sin(angle); // this number is the hypotenuse width of the right triangle. it is calulated using: sin(angle) = opposite / hypothenuse then hyp = opposite / sin(angle)
        distanceX = Math.sqrt(distance*distance - distanceY*distanceY);
    }
}
else{ //that means that the edge point is somewhere on the half left side of the rectangle
    signX = -1;
    var angle2 = Math.PI - angle;
    if(angle2<diagonalAngle){//the edge point is on the left side of the rectangle
        distanceX = rect.width/2;
        var distance = distanceX/Math.cos(angle);
        distanceY = Math.sqrt(distance*distance - distanceX*distanceX);
    }
    else{// the edge point is somewhere on the top or on the bottom
        distanceY = rect.height/2;
        var distance = distanceY/Math.sin(angle);
        distanceX = Math.sqrt(distance*distance - distanceY*distanceY);
    }
}
return {
    x:cp.x+distanceX*signX,
    y:cp.y+distanceY*signY
};
/*By ABOUD*/
}

我测试了我的函数，如果您需要了解它是如何工作的，请访问 my page 并查看源代码。

javascript中心与矩形边缘上任意点之间的距离

2 个答案: