Question

我想在两个圆圈的交叉区域随机x / y坐标

What the script does

我现在的问题是我完成了大部分工作，并且它在浏览器中使用Canvas进行演示。但仅限2D坐标。我正在使用谷歌地图，我想输入两个具有Lat Lng位置和半径的圆圈。结果应该是我在2D上计算的Lat / Lang。但我的数学以现实世界坐标结束..

所以circle1.x和.y例如.lat和.lng。

/******************************************************************
    Generate New Coord From Intersect
        Generate random position point on cntersect area of two circles.
*******************************************************************/
var circle1 = {x : 0, y : 0, radius : 10};
var circle2 = {x : 0, y : 0, radius : 10};
var ctx;
var p2 = {x : 0, y : 0};
var p3 = {x : 0, y : 0};
var t3 = {x : 0, y : 0};
var t6 = {x : 0, y : 0};
var t7 = {x : 0, y : 0};
function GenerateNewCoordFromIntersect(circle1, circle2) {
    var c = document.getElementById('canvasID');
    c.width = 500;
    c.height = 500;
    ctx = c.getContext('2d');
    drawCircle(circle1.x,circle1.y,circle1.radius,"red");
    drawCircle(circle2.x,circle2.y,circle2.radius,"blue");

    var distance = Math.sqrt(Math.pow(circle2.x-circle1.x,2) + Math.pow(circle2.y-circle1.y,2));
    // then there are no solutions, the circles are separate.
    if (distance > circle1.radius +  circle2.radius ) {
        return;
    }
    // then there are no solutions because one circle is contained within the other.
    if (distance < circle1.radius -  circle2.radius ) {
        return;
    }
    // then the circles are coincident and there are an infinite number of solutions.
    if (distance == 0 && circle1.radius == circle2.radius) {
        return;
    }
    // random if take sector of circle 1 or 2
    if (Math.random() > 0.5) {
        var newcircle1 = JSON.parse(JSON.stringify(circle1));
        var newcircle2 = JSON.parse(JSON.stringify(circle2));
        circle1 = newcircle2;
        circle2 = newcircle1;
    }
    // calc a
    a = ((Math.pow(circle1.radius,2) - Math.pow(circle2.radius,2) + Math.pow(distance,2))) / (2 * distance);
    // calc height
    h = Math.sqrt(Math.pow(circle1.radius,2) - Math.pow(a,2));
    // calc middle point of intersect
    p2.x = circle1.x + a * (circle2.x - circle1.x) / distance;
    p2.y = circle1.y + a * (circle2.y - circle1.y) / distance;
    // calc upper radius intersect point
    p3.x = p2.x + h * (circle2.y - circle1.y) / distance;
    p3.y = p2.y - h * (circle2.x - circle1.x) / distance;
    // random height for random point position
    var randNumber = Math.random() / 2 + Math.random() * 0.5; 
    var radiusOfH = (h*2);
    var randh = (randNumber) * radiusOfH - h;
    // calc random Hypotenuse
    var hypDistance = Math.abs(Math.sqrt(Math.pow(a,2) + Math.pow(randh,2)));
    var randomHyp = (circle1.radius - hypDistance) + hypDistance;
    // random point on line of middlepoint
    t3.x = p2.x +  ((randh) * (circle2.y - circle1.y)) / distance;
    t3.y = p2.y - (randh) * (circle2.x - circle1.x) / distance;
    // angle calc
    var winkel = Math.atan(randh / a);
    var newA = Math.cos(winkel) * randomHyp; //(randomHyp);
    var newH = Math.sin(winkel) * randomHyp;//newA ;
    t6.x = circle1.x + newA * (circle2.x - circle1.x) / distance;
    t6.y = circle1.y + newA * (circle2.y - circle1.y) / distance;
    t7.x = t6.x + newH * (circle2.y - circle1.y) / distance;
    t7.y = t6.y - newH * (circle2.x - circle1.x) / distance;
    randNumber = Math.random();
    var xDist = (t7.x - t3.x) * (randNumber);
    var yDist = (t7.y - t3.y) * (randNumber);
    var rx = t3.x + xDist;
    var ry = t3.y + yDist;
    drawCircle(rx,ry,2,"blue");
}
function drawCircle(x, y, r, fill) {
    ctx.beginPath();
    ctx.arc(x,y,r,0,2*Math.PI);
    ctx.strokeStyle = fill;
    ctx.stroke();
}

Answer 1

您需要知道它用于将纬度/经度坐标转换为2d平面的x / y坐标的map projection。

您可以找到Google Maps API here的具体信息。

将X / Y圆弧坐标转换为功能性Lat / Lng

1 个答案: