用纬度经度计算两点之间的距离，我做错了什么?

Here's my try, it's just a snippet of my code:

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;

I am using this formulae to get the latitude and longitude:

我用这个公式来求纬度和经度:

x = Deg + (Min + Sec / 60) / 60)

6 个解决方案

#1

The Java code given by Dommer above gives slightly incorrect results but the small errors add up if you are processing say a GPS track. Here is an implementation of the Haversine method in Java which also takes into account height differences between two points.

上面Dommer给出的Java代码给出了稍微不正确的结果，但是如果您正在处理GPS跟踪，那么这些小错误就会累积起来。下面是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);
}

#2

Here's a Java function that calculates the distance between two lat/long points.

这是一个Java函数，计算两个lat/长点之间的距离。

Edit

编辑

I found another reference to the code.

我找到了另一个对代码的引用。

And, posted below, just in case it disappears again.

还有，贴在下面，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");

#3

Note: this solution only works for short distances.

注意:此解决方案只适用于短距离。

I tried to use dommer's posted formula for an application and found it did well for long distances but in my data I was using all very short distances, and dommer's post did very poorly. I needed speed, and the more complex geo calcs worked well but were too slow. So, in the case that you need speed and all the calculations you're making are short (maybe < 100m or so). I found this little approximation to work great. it assumes the world is flat mind you, so don't use it for long distances, it works by approximating the distance of a single Latitude and Longitude at the given Latitude and returning the Pythagorean distance in meters.

我尝试使用dommer的发布公式来应用程序，发现它在长距离上做得很好，但是在我的数据中，我使用的都是非常短的距离，dommer的帖子做得非常糟糕。我需要速度，更复杂的地球资源系统运行得很好，但速度太慢了。所以，在需要速度的情况下，所有的计算都很短(可能小于100米左右)。我发现这个小近似很有效。它假设世界是平的，所以不要用它来做长距离的旅行，它的工作原理是在给定的纬度和经度上近似一个纬度，然后返回以米为单位的毕达哥拉斯距离。

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;
    }
}

#4

Here is a page with javascript examples for various spherical calculations. The very first one on the page should give you what you need.

这里有一个包含各种球面计算的javascript示例的页面。这一页上的第一个应该给你你所需要的。

http://www.movable-type.co.uk/scripts/latlong.html

http://www.movable-type.co.uk/scripts/latlong.html

Here is the Javascript code

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;

Where 'd' will hold the distance.

d会保持距离。

#5

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);
        }


    }

#6

This wikipedia article provides the formulae and an example. The text is in german, but the calculations speak for themselves.

这篇*的文章提供了公式和一个例子。文本是德文的，但计算结果不言自明。