如何计算距离另一点一定距离的点的纬度？
在地图上画一个圆我有一个以米为单位的中心GLatLng（A）和半径（r）。如何计算距离另一点一定距离的点的纬度？
这里有一个图：
-----------
--/ \--
-/ \-
/ \
/ \
/ r \
| *-------------*
\ A /B
\ /
\ /
-\ /-
--\ /--
-----------
如何计算的GLatLng位置B？假设r与赤道平行。
使用GLatLng.distanceFrom（）方法获得A和B时的半径是微不足道的 - 但以相反的方式进行。似乎我需要做一些更重的数学。
我们需要给出一个轴承和从源点移动的距离时，返回目标点的方法。幸运的是，Chris Veness在Calculate distance, bearing and more between Latitude/Longitude points上有一个很好的JavaScript实现。
下已经适应与google.maps.LatLng类工作：
var pointA = new google.maps.LatLng(25.48, -71.26);
var radiusInKm = 10;
var pointB = pointA.destinationPoint(90, radiusInKm);
下面是使用Google Maps API v3一个完整的例子：
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="content-type" content="text/html; charset=UTF-8"/>
<title>Google Maps Geometry</title>
<script src="http://maps.google.com/maps/api/js?sensor=false"
type="text/javascript"></script>
</head>
<body>
<div id="map" style="width: 400px; height: 300px"></div>
<script type="text/javascript">
Number.prototype.toRad = function() {
return this * Math.PI/180;
}
Number.prototype.toDeg = function() {
return this * 180/Math.PI;
}
google.maps.LatLng.prototype.destinationPoint = function(brng, dist) {
dist = dist/6371;
brng = brng.toRad();
var lat1 = this.lat().toRad(), lon1 = this.lng().toRad();
var lat2 = Math.asin(Math.sin(lat1) * Math.cos(dist) +
Math.cos(lat1) * Math.sin(dist) * Math.cos(brng));
var lon2 = lon1 + Math.atan2(Math.sin(brng) * Math.sin(dist) *
Math.cos(lat1),
Math.cos(dist) - Math.sin(lat1) *
Math.sin(lat2));
if (isNaN(lat2) || isNaN(lon2)) return null;
return new google.maps.LatLng(lat2.toDeg(), lon2.toDeg());
}
var pointA = new google.maps.LatLng(40.70, -74.00); // Circle center
var radius = 10; // 10km
var mapOpt = {
mapTypeId: google.maps.MapTypeId.TERRAIN,
center: pointA,
zoom: 10
};
var map = new google.maps.Map(document.getElementById("map"), mapOpt);
// Draw the circle
new google.maps.Circle({
center: pointA,
radius: radius * 1000, // Convert to meters
fillColor: '#FF0000',
fillOpacity: 0.2,
map: map
});
// Show marker at circle center
new google.maps.Marker({
position: pointA,
map: map
});
// Show marker at destination point
new google.maps.Marker({
position: pointA.destinationPoint(90, radius),
map: map
});
</script>
</body>
</html>
Number.prototype.toRad = function() {
return this * Math.PI/180;
}
Number.prototype.toDeg = function() {
return this * 180/Math.PI;
}
google.maps.LatLng.prototype.destinationPoint = function(brng, dist) {
dist = dist/6371;
brng = brng.toRad();
var lat1 = this.lat().toRad(), lon1 = this.lng().toRad();
var lat2 = Math.asin(Math.sin(lat1) * Math.cos(dist) +
Math.cos(lat1) * Math.sin(dist) * Math.cos(brng));
var lon2 = lon1 + Math.atan2(Math.sin(brng) * Math.sin(dist) *
Math.cos(lat1),
Math.cos(dist) - Math.sin(lat1) *
Math.sin(lat2));
if (isNaN(lat2) || isNaN(lon2)) return null;
return new google.maps.LatLng(lat2.toDeg(), lon2.toDeg());
}
只需将如下使用
屏幕截图：
UPDATE：
在回答Paul's下面的评论，这是当圆圈环绕的一极会发生什么。
绘制pointA北极附近，与千公里半径：
var pointA = new google.maps.LatLng(85, 0); // Close to north pole
var radius = 1000; // 1000km
截图供pointA.destinationPoint(90, radius)：
那么这总是会让目的地位于东方吗？ – Nirmal 2010-04-14 12:06:36
@Nirmal：不，它取决于你传递给'destinationPoint（）'的第一个参数。 90为东，但您可以使用任何轴承，从0 =北开始，顺时针方向移动。 – 2010-04-14 12:09:48
谢谢，那很完美。 – 2010-04-14 12:31:28
的这个问题的答案，更可以在这里找到：http://www.edwilliams.org/avform.htm
我在页面上看到很多acos，sin，tan等。但我会坚持丹尼尔的回答。感谢您的帮助。 – 2010-04-14 12:32:29
@保罗，谢谢你这样一个不错的链接... – aProgrammer 2011-09-16 09:01:45
链接现在被打破了。 – 2018-01-27 17:18:02
如果你是2纬度/经度点之间的距离后，穿过地球表面，那么你可以在这里找到javascript：
http://www.movable-type.co.uk/scripts/latlong-vincenty.html
这是Android的API用于在android.location.Location::distanceTo
您可以轻松地从JavaScript将代码转换成Java相同的公式。
如果要计算目标点给出的起点，方位和距离， 那么你需要这个方法：
http://www.movable-type.co.uk/scripts/latlong-vincenty-direct.html
下面是Java中的公式：
public class LatLngUtils {
/**
* @param lat1
* Initial latitude
* @param lon1
* Initial longitude
* @param lat2
* destination latitude
* @param lon2
* destination longitude
* @param results
* To be populated with the distance, initial bearing and final
* bearing
*/
public static void computeDistanceAndBearing(double lat1, double lon1,
double lat2, double lon2, double results[]) {
// Based on http://www.ngs.noaa.gov/PUBS_LIB/inverse.pdf
// using the "Inverse Formula" (section 4)
int MAXITERS = 20;
// Convert lat/long to radians
lat1 *= Math.PI/180.0;
lat2 *= Math.PI/180.0;
lon1 *= Math.PI/180.0;
lon2 *= Math.PI/180.0;
double a = 6378137.0; // WGS84 major axis
double b = 6356752.3142; // WGS84 semi-major axis
double f = (a - b)/a;
double aSqMinusBSqOverBSq = (a * a - b * b)/(b * b);
double L = lon2 - lon1;
double A = 0.0;
double U1 = Math.atan((1.0 - f) * Math.tan(lat1));
double U2 = Math.atan((1.0 - f) * Math.tan(lat2));
double cosU1 = Math.cos(U1);
double cosU2 = Math.cos(U2);
double sinU1 = Math.sin(U1);
double sinU2 = Math.sin(U2);
double cosU1cosU2 = cosU1 * cosU2;
double sinU1sinU2 = sinU1 * sinU2;
double sigma = 0.0;
double deltaSigma = 0.0;
double cosSqAlpha = 0.0;
double cos2SM = 0.0;
double cosSigma = 0.0;
double sinSigma = 0.0;
double cosLambda = 0.0;
double sinLambda = 0.0;
double lambda = L; // initial guess
for (int iter = 0; iter < MAXITERS; iter++) {
double lambdaOrig = lambda;
cosLambda = Math.cos(lambda);
sinLambda = Math.sin(lambda);
double t1 = cosU2 * sinLambda;
double t2 = cosU1 * sinU2 - sinU1 * cosU2 * cosLambda;
double sinSqSigma = t1 * t1 + t2 * t2; // (14)
sinSigma = Math.sqrt(sinSqSigma);
cosSigma = sinU1sinU2 + cosU1cosU2 * cosLambda; // (15)
sigma = Math.atan2(sinSigma, cosSigma); // (16)
double sinAlpha = (sinSigma == 0) ? 0.0 : cosU1cosU2 * sinLambda
/sinSigma; // (17)
cosSqAlpha = 1.0 - sinAlpha * sinAlpha;
cos2SM = (cosSqAlpha == 0) ? 0.0 : cosSigma - 2.0 * sinU1sinU2
/cosSqAlpha; // (18)
double uSquared = cosSqAlpha * aSqMinusBSqOverBSq; // defn
A = 1 + (uSquared/16384.0) * // (3)
(4096.0 + uSquared * (-768 + uSquared * (320.0 - 175.0 * uSquared)));
double B = (uSquared/1024.0) * // (4)
(256.0 + uSquared * (-128.0 + uSquared * (74.0 - 47.0 * uSquared)));
double C = (f/16.0) * cosSqAlpha * (4.0 + f * (4.0 - 3.0 * cosSqAlpha)); // (10)
double cos2SMSq = cos2SM * cos2SM;
deltaSigma = B
* sinSigma
* // (6)
(cos2SM + (B/4.0)
* (cosSigma * (-1.0 + 2.0 * cos2SMSq) - (B/6.0) * cos2SM
* (-3.0 + 4.0 * sinSigma * sinSigma)
* (-3.0 + 4.0 * cos2SMSq)));
lambda = L
+ (1.0 - C)
* f
* sinAlpha
* (sigma + C * sinSigma
* (cos2SM + C * cosSigma * (-1.0 + 2.0 * cos2SM * cos2SM))); // (11)
double delta = (lambda - lambdaOrig)/lambda;
if (Math.abs(delta) < 1.0e-12) {
break;
}
}
double distance = (b * A * (sigma - deltaSigma));
results[0] = distance;
if (results.length > 1) {
double initialBearing = Math.atan2(cosU2 * sinLambda, cosU1 * sinU2
- sinU1 * cosU2 * cosLambda);
initialBearing *= 180.0/Math.PI;
results[1] = initialBearing;
if (results.length > 2) {
double finalBearing = Math.atan2(cosU1 * sinLambda, -sinU1 * cosU2
+ cosU1 * sinU2 * cosLambda);
finalBearing *= 180.0/Math.PI;
results[2] = finalBearing;
}
}
}
/*
* Vincenty Direct Solution of Geodesics on the Ellipsoid (c) Chris Veness
* 2005-2012
*
* from: Vincenty direct formula - T Vincenty, "Direct and Inverse Solutions
* of Geodesics on the Ellipsoid with application of nested equations", Survey
* Review, vol XXII no 176, 1975 http://www.ngs.noaa.gov/PUBS_LIB/inverse.pdf
*/
/**
* Calculates destination point and final bearing given given start point,
* bearing & distance, using Vincenty inverse formula for ellipsoids
*
* @param lat1
* start point latitude
* @param lon1
* start point longitude
* @param brng
* initial bearing in decimal degrees
* @param dist
* distance along bearing in metres
* @returns an array of the desination point coordinates and the final bearing
*/
public static void computeDestinationAndBearing(double lat1, double lon1,
double brng, double dist, double results[]) {
double a = 6378137, b = 6356752.3142, f = 1/298.257223563; // WGS-84
// ellipsiod
double s = dist;
double alpha1 = toRad(brng);
double sinAlpha1 = Math.sin(alpha1);
double cosAlpha1 = Math.cos(alpha1);
double tanU1 = (1 - f) * Math.tan(toRad(lat1));
double cosU1 = 1/Math.sqrt((1 + tanU1 * tanU1)), sinU1 = tanU1 * cosU1;
double sigma1 = Math.atan2(tanU1, cosAlpha1);
double sinAlpha = cosU1 * sinAlpha1;
double cosSqAlpha = 1 - sinAlpha * sinAlpha;
double uSq = cosSqAlpha * (a * a - b * b)/(b * b);
double A = 1 + uSq/16384
* (4096 + uSq * (-768 + uSq * (320 - 175 * uSq)));
double B = uSq/1024 * (256 + uSq * (-128 + uSq * (74 - 47 * uSq)));
double sinSigma = 0, cosSigma = 0, deltaSigma = 0, cos2SigmaM = 0;
double sigma = s/(b * A), sigmaP = 2 * Math.PI;
while (Math.abs(sigma - sigmaP) > 1e-12) {
cos2SigmaM = Math.cos(2 * sigma1 + sigma);
sinSigma = Math.sin(sigma);
cosSigma = Math.cos(sigma);
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 = s/(b * A) + deltaSigma;
}
double tmp = sinU1 * sinSigma - cosU1 * cosSigma * cosAlpha1;
double lat2 = Math.atan2(sinU1 * cosSigma + cosU1 * sinSigma * cosAlpha1,
(1 - f) * Math.sqrt(sinAlpha * sinAlpha + tmp * tmp));
double lambda = Math.atan2(sinSigma * sinAlpha1, cosU1 * cosSigma - sinU1
* sinSigma * cosAlpha1);
double C = f/16 * cosSqAlpha * (4 + f * (4 - 3 * cosSqAlpha));
double L = lambda
- (1 - C)
* f
* sinAlpha
* (sigma + C * sinSigma
* (cos2SigmaM + C * cosSigma * (-1 + 2 * cos2SigmaM * cos2SigmaM)));
double lon2 = (toRad(lon1) + L + 3 * Math.PI) % (2 * Math.PI) - Math.PI; // normalise
// to
// -180...+180
double revAz = Math.atan2(sinAlpha, -tmp); // final bearing, if required
results[0] = toDegrees(lat2);
results[1] = toDegrees(lon2);
results[2] = toDegrees(revAz);
}
private static double toRad(double angle) {
return angle * Math.PI/180;
}
private static double toDegrees(double radians) {
return radians * 180/Math.PI;
}
}
请注意：Vincenty的公式精确到0.5mm或0.000015“（！）以内，正在使用的椭球体。基于球形模型的计算，如（更简单）Haversine，精确到0.3％左右。所以以前的JavaScript解决方案可能是大多数人需要的。 – 2012-12-13 19:39:19
用于许多测地计算的Javascript（直接&反问题，面积计算等）。可在
http://geographiclib.sourceforge.net/scripts/geographiclib.js
示例用法是
http://geographiclib.sourceforge.net/scripts/geod-calc.html
Google地图的界面处所示
http://geographiclib.sourceforge.net/scripts/geod-google.html
这包括绘制测地线（蓝色提供），测地圆（绿色）和测地线包络（红色）。
为了从另一个给定的方位和距离计算纬度，经度点，你可以使用google's JavaScript实现：
var pointA = new google.maps.LatLng(25.48, -71.26);
var distance = 10; // 10 metres
var bearing 90; // 90 degrees
var pointB = google.maps.geometry.spherical.computeOffset(pointA, distance, bearing);
见https://developers.google.com/maps/documentation/javascript/reference#spherical 对于文档
@Rene ：使我的答案适应GMaps API v2应该很简单。我相信这只是用'GLatLng'替换'google.maps.LatLng'的问题。如果您发现任何困难，请告诉我。 – 2010-04-14 12:20:57
谢谢，这里没有任何困难:) – 2010-04-14 12:34:31