問題描述

我正在玩three.js

i′m playing arround with three.js

我想在更大的球體上渲染特定地理坐標(biāo)上的對象，我非常接近解決方案，但我沒有從 lat lon 獲得正確的 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上設(shè)置了一個測試用例，有兩個坐標(biāo)

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

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

紐約和柏林

這是我從緯度和半徑計算 xyz 的函數(shù)

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];
}

但所有嘗試都返回不同的 xy，它們都不正確(z 始終正確).

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

有人可以指導(dǎo)我正確的方式嗎?我不知道可能出了什么問題

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

這就是要玩的小提琴

更新:正在工作的 jsfiddle

推薦答案

不幸的是，我無法進(jìn)一步解釋，但在玩了這個之后，它就像一個魅力:)

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，不是嗎?而且我仍然對一些更短的方程式感興趣

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

工作小提琴

這篇關(guān)于javascript緯度經(jīng)度到地球上的xyz位置(threejs)的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！