计算GPS WGS_84 两点的距离 更加细腻的算法

上面的一篇博客,两点距离比较粗糙,现贴一篇更加细腻的算法

wgs84PointsDistance = function(fromPoint, toPoint) {
	if(arguments.length != 2) {
		return 0;
	}
	lon1 = fromPoint.x;
	lat1 = fromPoint.y;
	lon2 = toPoint.x;
	lat2 = toPoint.y;
	var a = 6378137, b = 6356752.3142, f = 1 / 298.257223563;
	var L = (lon2 - lon1).toRad();
	var U1 = Math.atan((1 - f) * Math.tan(lat1.toRad()));
	var U2 = Math.atan((1 - f) * Math.tan(lat2.toRad()));
	var sinU1 = Math.sin(U1), cosU1 = Math.cos(U1);
	var sinU2 = Math.sin(U2), cosU2 = Math.cos(U2);
	var lambda = L, lambdaP, iterLimit = 100;
  	do {
		var sinLambda = Math.sin(lambda), cosLambda = Math.cos(lambda);
		var sinSigma = Math.sqrt((cosU2 * sinLambda) * (cosU2 * sinLambda) + (cosU1 * sinU2 - sinU1 * cosU2 * cosLambda) * (cosU1 * sinU2 - sinU1 * cosU2 * cosLambda));
		if(sinSigma == 0)
			return 0;
		var cosSigma = sinU1 * sinU2 + cosU1 * cosU2 * cosLambda;
		var sigma = Math.atan2(sinSigma, cosSigma);
		var sinAlpha = cosU1 * cosU2 * sinLambda / sinSigma;
		var cosSqAlpha = 1 - sinAlpha * sinAlpha;
		var cos2SigmaM = cosSigma - 2 * sinU1 * sinU2 / cosSqAlpha;
		if(isNaN(cos2SigmaM))
			cos2SigmaM = 0;
		var C = f / 16 * cosSqAlpha * (4 + f * (4 - 3 * cosSqAlpha));
		lambdaP = lambda;
		lambda = L + (1 - C) * f * sinAlpha * (sigma + C * sinSigma * (cos2SigmaM + C * cosSigma * (-1 + 2 * cos2SigmaM * cos2SigmaM)));
    } while (Math.abs(lambda-lambdaP) > (1e-12) && --iterLimit>0);
	if(iterLimit == 0) {
		return NaN
	}
	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 deltaSigma = B * sinSigma * (cos2SigmaM + B / 4 * (cosSigma * (-1 + 2 * cos2SigmaM * cos2SigmaM) - B / 6 * cos2SigmaM * (-3 + 4 * sinSigma * sinSigma) * (-3 + 4 * cos2SigmaM * cos2SigmaM)));
	var s = b * A * (sigma - deltaSigma);
	var fwdAz = Math.atan2(cosU2 * sinLambda, cosU1 * sinU2 - sinU1 * cosU2 * cosLambda);
	var revAz = Math.atan2(cosU1 * sinLambda, -sinU1 * cosU2 + cosU1 * sinU2 * cosLambda);
	return s;
}

  

posted @ 2012-12-05 18:07  HPhone  阅读(5349)  评论(6编辑  收藏  举报