Question

我正在寻找一种计算两条线的交点的方法，我有一个javascript方法可以正常使用样本x和y

//Start of linear part
    window.linear = {
        getSlope: function(x1, y1, x2, y2) {
            if (x1 == x2) return false;
            return (y1 - y2) / (x1 - x2);
        },
        getYInt: function(x1, y1, x2, y2) {
            if (x1 === x2) return y1 === 0 ? 0 : false;
            if (y1 === y2) return y1;
            return y1 - this.getSlope(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.getSlope(x1, y1, x2, y2)) * x1 - y1)) / slope;
        },
        getInt: function(x11, y11, x12, y12, x21, y21, x22, y22) {
            var slope1, slope2, yint1, yint2, intx, inty;

            // check if either of the points are the same. if so that's our intersection
            if (x11 == x21 && y11 == y21) return [x11, y11];
            if (x12 == x22 && y12 == y22) return [x12, y22];

            // get slope: (y1 - y1) / (x1 - x2)
            slope1 = this.getSlope(x11, y11, x12, y12);
            slope2 = this.getSlope(x21, y21, x22, y22);

            // if both lines have the same slope, they are paralell; never touch.
            if (slope1 === slope2) return false;

            // get y-intersection: y - slope * x
            yint1 = this.getYInt(x11, y11, x12, y12);
            yint2 = this.getYInt(x21, y21, x22, y22);

            // check to see if both lines have the same yintcept, and if so, return the point
            if (yint1 === yint2) return yint1 === false ? false : [0, yint1];

            // if one of the lines doesn't have a slope:
            if (slope1 === false) return [y21, slope2 * y21 + yint2];
            if (slope2 === false) return [y11, slope1 * y11 + yint1];

            //if both lines have a slop and y intercept, calulate the x-intercept:
            // (slope1 * x1 + yint1 - yint2) / slope2;
            intx = (slope1 * x11 + yint1 - yint2) / slope2;

            // calculate the y-intercept, and return an array:
            return [intx, slope1 * intx + yint1];
        }
    }
 // END OF linear

这就是它。但是当我从铯中添加一个样本点时，它给出了错误的答案。我认为这是因为光辉和地理对话。任何人都知道如何解决它？

这是我的样本行的两点

var start=new Cesium.Cartesian3(0.7007890932991216, -1.3246378751475405, 0)
var end=new Cesium.Cartesian3(0.7007917595851312, -1.3246388181033268, 0)

这是另一条线

var st=new Cesium.Cartesian3(0.7007909619371219, -1.3246425889783457, 0)
var en=new Cesium.Cartesian3(0.7007881826635102, -1.3246415011312433, 0)

这就是我使用它的方式

var  e = linear.getInt(start.x,start.y,end.x,end.y,st.x,st.y,en.x, en.y)||["n/a","n/a"];

它给了我错误的观点...我认为我必须进行一些审判，因为积分是光芒四射的......是否可以帮助我？

铯 - 两条线交叉点

0 个答案: