javascript纬度经度到地球上的xyz位置(threejs)

问题描述

我正在玩three.js

i´m playing arround with three.js

我想在更大的球体上渲染特定地理坐标上的对象，我非常接近解决方案，但我没有从 lat lon 获得正确的 xyz 位置

i want to render objects on specific geocoordinates on a bigger sphere, i´m pretty near to the solution, but i dont get the correct xyz position from lat lon

我在jsfiddle上设置了一个测试用例，有两个坐标

i have set up a test case on jsfiddle, there are two coordinates

latlons = [[40.7142700,-74.0059700], [52.5243700,13.4105300]];

纽约和柏林

这是我从纬度和半径计算 xyz 的函数

and this is my function to calc xyz from lat lon and radius

function calcPosFromLatLonRad(lat,lon,radius){

// Attempt1
var cosLat = Math.cos(lat * Math.PI / 180.0);
var sinLat = Math.sin(lat * Math.PI / 180.0);
var cosLon = Math.cos(lon * Math.PI / 180.0);
var sinLon = Math.sin(lon * Math.PI / 180.0);
var rad = radius;
y = rad * cosLat * sinLon;
x = rad * cosLat * cosLon;
z = rad * sinLat;

// Attempt2
// x = radius *  Math.sin(lat) * Math.cos(lon)
// y = radius *  Math.sin(lat) * Math.sin(lon)
// z = radius * Math.cos(lat)


// Attempt3
// latitude = lat * Math.PI/180
// longitude = lon * Math.PI/180
// x =  -radius * Math.cos(latitude) * Math.cos(longitude)
// y =  radius * Math.sin(latitude) 
// z =  radius * Math.cos(latitude) * Math.sin(longitude)


// Attempt4
// var phi   = (90-lat)*(Math.PI/180);
// var theta = (lng+180)*(Math.PI/180);
// x = ((rad) * Math.sin(phi)*Math.cos(theta));
// z = ((rad) * Math.sin(phi)*Math.sin(theta));
// y = ((rad) * Math.cos(phi));



   console.log([x,y,z]);
   return [x,y,z];
}

但所有尝试都返回不同的 xy，它们都不正确(z 始终正确).

but all attempts return different xy, and they are all not correct ( z is always correct).

有人可以指导我正确的方式吗?我不知道可能出了什么问题

could someone pleas guide me to the right way ? i have no idea what could be wrong

这就是要玩的小提琴

更新:正在工作的 jsfiddle

推荐答案

不幸的是，我无法进一步解释，但在玩了这个之后，它就像一个魅力:)

unfortunatly i can´t further explain, but after playing around this one works like a charme :)

function calcPosFromLatLonRad(lat,lon,radius){
  
    var phi   = (90-lat)*(Math.PI/180);
    var theta = (lon+180)*(Math.PI/180);

    x = -(radius * Math.sin(phi)*Math.cos(theta));
    z = (radius * Math.sin(phi)*Math.sin(theta));
    y = (radius * Math.cos(phi));
  
    return [x,y,z];

}

是的，那很酷，不是吗?而且我仍然对一些更短的方程式感兴趣

yeah thats pretty cool isnt it ? And i´m still interested into some shorter equation

工作小提琴

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