这是我尝试的代码片段：
final double RADIUS = 6371.01;
double temp = Math.cos(Math.toRadians(latA))
* Math.cos(Math.toRadians(latB))
* Math.cos(Math.toRadians((latB) - (latA)))
+ Math.sin(Math.toRadians(latA))
* Math.sin(Math.toRadians(latB));
return temp * RADIUS * Math.PI / 180;
我正在使用这个公式来获取纬度和经度：
x = Deg + (Min + Sec / 60) / 60)
Dommer提供的Java代码给出的结果略有不准确，但如果处理例如GPS轨迹等大量数据时，这些小误差会不断累积。以下是一个实现了Haversine方法并考虑了两点之间高度差的Java代码。
/**
* Calculate distance between two points in latitude and longitude taking
* into account height difference. If you are not interested in height
* difference pass 0.0. Uses Haversine method as its base.
*
* lat1, lon1 Start point lat2, lon2 End point el1 Start altitude in meters
* el2 End altitude in meters
* @returns Distance in Meters
*/
public static double distance(double lat1, double lat2, double lon1,
double lon2, double el1, double el2) {
final int R = 6371; // Radius of the earth
double latDistance = Math.toRadians(lat2 - lat1);
double lonDistance = Math.toRadians(lon2 - lon1);
double a = Math.sin(latDistance / 2) * Math.sin(latDistance / 2)
+ Math.cos(Math.toRadians(lat1)) * Math.cos(Math.toRadians(lat2))
* Math.sin(lonDistance / 2) * Math.sin(lonDistance / 2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
double distance = R * c * 1000; // convert to meters
double height = el1 - el2;
distance = Math.pow(distance, 2) + Math.pow(height, 2);
return Math.sqrt(distance);
}
这里有一个Java函数可以计算两个经纬度点之间的距离，如下所示，以防它再次消失。
private double distance(double lat1, double lon1, double lat2, double lon2, char unit) {
double theta = lon1 - lon2;
double dist = Math.sin(deg2rad(lat1)) * Math.sin(deg2rad(lat2)) + Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) * Math.cos(deg2rad(theta));
dist = Math.acos(dist);
dist = rad2deg(dist);
dist = dist * 60 * 1.1515;
if (unit == 'K') {
dist = dist * 1.609344;
} else if (unit == 'N') {
dist = dist * 0.8684;
}
return (dist);
}
/*:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::*/
/*:: This function converts decimal degrees to radians :*/
/*:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::*/
private double deg2rad(double deg) {
return (deg * Math.PI / 180.0);
}
/*:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::*/
/*:: This function converts radians to decimal degrees :*/
/*:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::*/
private double rad2deg(double rad) {
return (rad * 180.0 / Math.PI);
}
System.out.println(distance(32.9697, -96.80322, 29.46786, -98.53506, 'M') + " Miles\n");
System.out.println(distance(32.9697, -96.80322, 29.46786, -98.53506, 'K') + " Kilometers\n");
System.out.println(distance(32.9697, -96.80322, 29.46786, -98.53506, 'N') + " Nautical Miles\n");
假设未来读者偶然发现这篇SOF文章。
显然，这个问题是在2010年提出的，现在已经是2019年了。 但它在互联网搜索中很早就出现了。原始问题并没有排除使用第三方库（当我写这篇答案时）。
public double calculateDistanceInMeters(double lat1, double long1, double lat2,
double long2) {
double dist = org.apache.lucene.util.SloppyMath.haversinMeters(lat1, long1, lat2, long2);
return dist;
}
和
<dependency>
<groupId>org.apache.lucene</groupId>
<artifactId>lucene-spatial</artifactId>
<version>8.2.0</version>
</dependency>
https://mvnrepository.com/artifact/org.apache.lucene/lucene-spatial/8.2.0
在深入研究之前，请阅读有关“SloppyMath”的文档！
https://lucene.apache.org/core/8_2_0/core/org/apache/lucene/util/SloppyMath.html
注意：此解决方案只适用于短距离。
我尝试使用dommer发布的一个公式进行应用，并发现它在长距离上表现良好，但在我的数据中，我使用的全部都是非常短的距离，并且dommer的帖子表现非常差。我需要速度，而更复杂的地理计算效果很好，但速度太慢了。所以，在您需要速度并且所有计算都很短（可能小于100米左右）的情况下，我发现这个小估计值非常有效。请注意，它假设世界是平的，因此不要用于长距离，它通过近似给定纬度下的单个经纬度的距离，并返回米制下的勾股距离来工作。
public class FlatEarthDist {
//returns distance in meters
public static double distance(double lat1, double lng1,
double lat2, double lng2){
double a = (lat1-lat2)*FlatEarthDist.distPerLat(lat1);
double b = (lng1-lng2)*FlatEarthDist.distPerLng(lat1);
return Math.sqrt(a*a+b*b);
}
private static double distPerLng(double lat){
return 0.0003121092*Math.pow(lat, 4)
+0.0101182384*Math.pow(lat, 3)
-17.2385140059*lat*lat
+5.5485277537*lat+111301.967182595;
}
private static double distPerLat(double lat){
return -0.000000487305676*Math.pow(lat, 4)
-0.0033668574*Math.pow(lat, 3)
+0.4601181791*lat*lat
-1.4558127346*lat+110579.25662316;
}
}
虽然已经提供了许多很好的答案，但我发现一些性能方面的不足之处，所以我提供了一个着眼于性能的版本。每个常量都是预先计算的，并引入了x、y变量，以避免重复计算相同的值。希望能对你有所帮助。
private static final double r2d = 180.0D / 3.141592653589793D;
private static final double d2r = 3.141592653589793D / 180.0D;
private static final double d2km = 111189.57696D * r2d;
public static double meters(double lt1, double ln1, double lt2, double ln2) {
double x = lt1 * d2r;
double y = lt2 * d2r;
return Math.acos( Math.sin(x) * Math.sin(y) + Math.cos(x) * Math.cos(y) * Math.cos(d2r * (ln1 - ln2))) * d2km;
}
这是一个包含各种球面计算的JavaScript示例页面。页面上的第一个示例应该可以满足您的需求。
http://www.movable-type.co.uk/scripts/latlong.html
以下是JavaScript代码：
var R = 6371; // km
var dLat = (lat2-lat1).toRad();
var dLon = (lon2-lon1).toRad();
var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(lat1.toRad()) * Math.cos(lat2.toRad()) *
Math.sin(dLon/2) * Math.sin(dLon/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;
其中 'd' 将保存距离。
package distanceAlgorithm;
public class CalDistance {
public static void main(String[] args) {
// TODO Auto-generated method stub
CalDistance obj=new CalDistance();
/*obj.distance(38.898556, -77.037852, 38.897147, -77.043934);*/
System.out.println(obj.distance(38.898556, -77.037852, 38.897147, -77.043934, "M") + " Miles\n");
System.out.println(obj.distance(38.898556, -77.037852, 38.897147, -77.043934, "K") + " Kilometers\n");
System.out.println(obj.distance(32.9697, -96.80322, 29.46786, -98.53506, "N") + " Nautical Miles\n");
}
public double distance(double lat1, double lon1, double lat2, double lon2, String sr) {
double theta = lon1 - lon2;
double dist = Math.sin(deg2rad(lat1)) * Math.sin(deg2rad(lat2)) + Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) * Math.cos(deg2rad(theta));
dist = Math.acos(dist);
dist = rad2deg(dist);
dist = dist * 60 * 1.1515;
if (sr.equals("K")) {
dist = dist * 1.609344;
} else if (sr.equals("N")) {
dist = dist * 0.8684;
}
return (dist);
}
public double deg2rad(double deg) {
return (deg * Math.PI / 180.0);
}
public double rad2deg(double rad) {
return (rad * 180.0 / Math.PI);
}
}
public static double distance(double lat1, double lat2, double lon1,
double lon2, double el1, double el2) {
final int R = 6371; // Radius of the earth
double latDistance = Math.toRadians(lat2 - lat1);
double lonDistance = Math.toRadians(lon2 - lon1);
double a = Math.sin(latDistance / 2) * Math.sin(latDistance / 2)
+ Math.cos(Math.toRadians(lat1)) * Math.cos(Math.toRadians(lat2))
* Math.sin(lonDistance / 2) * Math.sin(lonDistance / 2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
double distance = R * c * 1000; // convert to meters
double height = el1 - el2;
distance = Math.pow(distance, 2) + Math.pow(height, 2);
return Math.sqrt(distance);
}
public static double distanceBetweenLocations(Location l1, Location l2) {
if(l1.hasAltitude() && l2.hasAltitude()) {
return distance(l1.getLatitude(), l2.getLatitude(), l1.getLongitude(), l2.getLongitude(), l1.getAltitude(), l2.getAltitude());
}
return l1.distanceTo(l2);
}
distance函数相同，但我创建了一个小的包装函数，它接受2个Location对象。由于这样，只有在两个位置实际上都有海拔时才使用distance函数，因为有时它们没有。这可能会导致奇怪的结果（如果位置不知道其海拔高度，则返回0）。在这种情况下，我将退回到经典的distanceTo函数。