Your formula differs slightly from the geodetic to ECEF calculation. Refer to the formulas on Dr Math Latitude and Longitude, GPS Conversion and Wikipedia Geodetic to/from ECEF coordinates. This projects the latitude, longitude to a flattened sphere (i.e. the real Earth is not perfectly spherical).
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 = 6378137.0;
var f = 1.0 / 298.257224;
var C = 1.0 / Math.sqrt(cosLat * cosLat + (1 - f) * (1 - f) * sinLat * sinLat);
var S = (1.0 - f) * (1.0 - f) * C;
var h = 0.0;
marker_mesh.position.x = (rad * C + h) * cosLat * cosLon;
marker_mesh.position.y = (rad * C + h) * cosLat * sinLon;
marker_mesh.position.z = (rad * S + h) * sinLat;
In your scenario, because it seems you're gunning for a perfect sphere, you will need to put f = 0.0 and rad = 500.0 instead. This will cause C and S to become 1.0, so, the simplified version of the formula reduces to:
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 = 500.0;
marker_mesh.position.x = rad * cosLat * cosLon;
marker_mesh.position.y = rad * cosLat * sinLon;
marker_mesh.position.z = rad * sinLat;
N.B. I have not validated the syntax of the Java code examples.
与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…