Cesium 已知一点经纬度和距离求另一个点的经纬度

整篇参考了：GIS算法–已知一点经纬度，方位角，距离求另一点

大神用的算法：
大圆算法：http://www.movable-type.co.uk/scripts/latlong.html

使用方法：直接传值，返回的就是另一个点的经纬度。

说明：本方法没有考虑高程问题。

/** * * @param {*} lon 经度 * @param {*} lat 纬度 * @param {*} angle 角度 (0~360度) * @param {*} distance 距离 (米) * */
function get_another_point(lon, lat, angle, distance) {
    
	// WGS84坐标系
	var a = 6378137;	// 赤道半径
	var b = 6356752.3142;	// 短半径
	var f = 1 / 298.257223563;	// 扁率

	var alpha1 = angle * (Math.PI / 180)
	var sinAlpha1 = Math.sin(alpha1);
	var cosAlpha1 = Math.cos(alpha1);
	var tanU1 = (1 - f) * Math.tan(lat * (Math.PI / 180));
	var cosU1 = 1 / Math.sqrt((1 + tanU1 * tanU1)), sinU1 = tanU1 * cosU1;
	var sigma1 = Math.atan2(tanU1, cosAlpha1);
	var sinAlpha = cosU1 * sinAlpha1;
	var cosSqAlpha = 1 - sinAlpha * sinAlpha;
	var uSq = cosSqAlpha * (a * a - b * b) / (b * b);
	var A = 1 + uSq / 16384 * (4096 + uSq * (-768 + uSq * (320 - 175 * uSq)));
	var B = uSq / 1024 * (256 + uSq * (-128 + uSq * (74 - 47 * uSq)));
	var sigma = distance / (b * A), sigmaP = 2 * Math.PI;
	while (Math.abs(sigma - sigmaP) > 1e-12) {
    
		var cos2SigmaM = Math.cos(2 * sigma1 + sigma);
		var sinSigma = Math.sin(sigma);
		var cosSigma = Math.cos(sigma);
		var deltaSigma = B * sinSigma * (cos2SigmaM + B / 4 * (cosSigma * (-1 + 2 * cos2SigmaM * cos2SigmaM) -
			B / 6 * cos2SigmaM * (-3 + 4 * sinSigma * sinSigma) * (-3 + 4 * cos2SigmaM * cos2SigmaM)));
		sigmaP = sigma;
		sigma = distance / (b * A) + deltaSigma;
	}

	var tmp = sinU1 * sinSigma - cosU1 * cosSigma * cosAlpha1;
	var lat2 = Math.atan2(sinU1 * cosSigma + cosU1 * sinSigma * cosAlpha1,
		(1 - f) * Math.sqrt(sinAlpha * sinAlpha + tmp * tmp));
	var lambda = Math.atan2(sinSigma * sinAlpha1, cosU1 * cosSigma - sinU1 * sinSigma * cosAlpha1);
	var C = f / 16 * cosSqAlpha * (4 + f * (4 - 3 * cosSqAlpha));
	var L = lambda - (1 - C) * f * sinAlpha *
		(sigma + C * sinSigma * (cos2SigmaM + C * cosSigma * (-1 + 2 * cos2SigmaM * cos2SigmaM)));

	var revAz = Math.atan2(sinAlpha, -tmp); // final bearing

	return {
     lon: Number(lon) + L * (180 / Math.PI), lat: lat2 * (180 / Math.PI) };
}