这是我的尝试,它只是我的代码片段:
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;
我正在使用这些公式来获取纬度和经度:
x = Deg + (Min + Sec / 60) / 60)
答案 0 :(得分:167)
以上Dommer给出的Java代码给出了稍微不正确的结果,但如果您正在处理GPS轨道,则会出现小错误。以下是Java中Haversine方法的实现,它还考虑了两点之间的高度差异。
/**
* 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);
}
答案 1 :(得分:72)
这是Java function that calculates the distance between two lat/long points。
修改强>
我找到了another reference to the code。
并在下面发布,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");
答案 2 :(得分:11)
注意:此解决方案仅适用于短距离。
我尝试使用dommer发布的公式用于应用程序,发现它适用于长距离,但在我的数据中我使用的距离非常短,而且dommer的帖子做得非常差。我需要速度,更复杂的地理计算效果很好,但速度太慢。因此,在您需要速度的情况下,您所做的所有计算都很短(可能<100米左右)。我发现这个小近似值很好。它假设世界是平心的,所以不要长距离使用它,它通过近似给定纬度的单个纬度和经度的距离并以米为单位返回毕达哥拉斯距离。
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;
}
}
答案 3 :(得分:4)
这是一个包含各种球形计算的javascript示例的页面。页面上的第一个应该为您提供所需的信息。
http://www.movable-type.co.uk/scripts/latlong.html
这是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;
'd'将保持距离。
答案 4 :(得分:4)
偶然发现这篇SOF文章的读者。
很明显,这个问题是在2010年和现在的2019年提出的。 但是它出现在互联网搜索的早期。最初的问题并没有减少对第三方库的使用(当我撰写此答案时)。
Foo
和
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;
}
https://mvnrepository.com/artifact/org.apache.lucene/lucene-spatial/8.2.0
在潜水之前,请阅读有关“ SloppyMath”的文档!
https://lucene.apache.org/core/8_2_0/core/org/apache/lucene/util/SloppyMath.html
答案 5 :(得分:1)
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);
}
}
答案 6 :(得分:1)
@David George的答案略有升级:
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);
}
距离函数是相同的,但是我创建了一个小型包装函数,该函数需要2个 Location 对象。因此,如果两个位置实际上都具有高度,我只会使用距离功能,因为有时它们没有。并可能导致奇怪的结果(如果位置不知道其高度0将返回)。在这种情况下,我将使用经典的 distanceTo 函数。
答案 7 :(得分:0)
此维基百科article提供了公式和示例。该文本是德文,但计算不言自明。
答案 8 :(得分:0)
提供了很多很好的答案,但是我发现了一些性能缺陷,所以让我提供一个考虑性能的版本。预先计算每个常数,并引入x,y变量以避免两次计算相同的值。希望对您有帮助
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) {
final double x = lt1 * d2r;
final 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;
}