计算两条线的交点 [英] calculating the point of intersection of two lines

问题描述

我有动态生成的动画线，我想在线撞到另一条线时检测到它们。我正在尝试实现一些基本的线性代数来获得线的方程，然后求解x，y，但结果是不稳定的。在这一点上，我只测试两条线，这意味着我应该得到一个交叉点，但我得到两条线。我只是想确保我的数学运算正常，我应该在别处寻找问题。

I have dynamically generated lines that animate and I want to detect when a lines hits another. I'm trying to implement some basic linear algebra to obtain the equation of the lines and then solving for x,y, but the results are erratic. At this point I'm testing only with two lines which means I should be getting one point of intersection, but I get two. I just want to make sure my math is ok and that I should be looking elsewhere for the problem.

function collision(boid1, boid2) {
    var x1 = boid1.initialX, y1 = boid1.initialY, x2 = boid1.x, y2 = boid1.y, x3 = boid2.initialX, y3 = boid2.initialY, x4 = boid2.x, y4 = boid2.y;

      slope1 = (y1 - y2)/(x1 - x2);
      slope2 = (y3 - y4)/(x3- x4);

      //console.log("slope1:"+slope1);
  //console.log('x2:'+x2+' y2:'+y2);

    if(slope1 != slope2){
        var b1 = getB(slope1,x1,y1);
        var b2 = getB(slope2,x3,y3);

        if(slope2 >= 0){
            u = slope1 - slope2;
        }else{
            u = slope1 + slope2;
        }

        if(b1 >= 0){
            z = b2 - b1;
        }else{
            z = b2 + b1;
        }

        pointX = z / u;

        pointY = (slope1*pointX)+b1;

        pointYOther = (slope2*pointX)+b2;

            console.log("pointx:"+pointX+" pointy:"+pointY+" othery:"+pointYOther);
            //return true;
            context.beginPath();
      context.arc(pointX, pointY, 2, 0, 2 * Math.PI, false);
      context.fillStyle = 'green';
      context.fill();
      context.lineWidth = 1;
      context.strokeStyle = '#003300';
      context.stroke();



    }

 return false;


}




function getB(slope,x,y){

    var y = y, x = x, m = slope;

    a = m*x;

    if(a>=0){
        b = y - a;
    }else{

        b = y + a;
    }

    return b;
}

编辑：

问题是我得到两个不同的交叉点值。应该只有一个，这让我相信我的计算是错误的。是的，x2，y2，x4，y4都在移动，但它们有一个设定的角度，一致的斜率确认了这一点。

The problem is that I'm getting two different values for the point of intersection. There should only be one, which leads me to believe my calculations are wrong. Yes, x2,y2,x4,y4 are all moving, but they have a set angle and the consistent slopes confirm that.

推荐答案

当将'found-x'插回其中一个方程时，您不需要在添加/减去y-交叉之间交替：

You don't need to alternate between adding/subtracting y-intersects when plugging 'found-x' back into one of the equations:

(function () {
    window.linear = {
        slope: function (x1, y1, x2, y2) {
            if (x1 == x2) return false;
            return (y1 - y2) / (x1 - x2);
        },
        yInt: function (x1, y1, x2, y2) {
            if (x1 === x2) return y1 === 0 ? 0 : false;
            if (y1 === y2) return y1;
            return y1 - this.slope(x1, y1, x2, y2) * x1 ;
        },
        getXInt: function (x1, y1, x2, y2) {
            var slope;
            if (y1 === y2) return x1 == 0 ? 0 : false;
            if (x1 === x2) return x1;
            return (-1 * ((slope = this.slope(x1, y1, x2, y2)) * x1 - y1)) / slope;
        },
        getIntersection: function (x11, y11, x12, y12, x21, y21, x22, y22) {
            var slope1, slope2, yint1, yint2, intx, inty;
            if (x11 == x21 && y11 == y21) return [x11, y11];
            if (x12 == x22 && y12 == y22) return [x12, y22];

            slope1 = this.slope(x11, y11, x12, y12);
            slope2 = this.slope(x21, y21, x22, y22);
            if (slope1 === slope2) return false;

            yint1 = this.yInt(x11, y11, x12, y12);
            yint2 = this.yInt(x21, y21, x22, y22);
            if (yint1 === yint2) return yint1 === false ? false : [0, yint1];

            if (slope1 === false) return [y21, slope2 * y21 + yint2];
            if (slope2 === false) return [y11, slope1 * y11 + yint1];
            intx = (slope1 * x11 + yint1 - yint2)/ slope2;
            return [intx, slope1 * intx + yint1];
        }
    }
}());

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