使用经度计算两点之间的距离?

问题描述

这是我的尝试，这只是我的代码的一部分:

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

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;

我正在使用以下公式获取纬度和经度:

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

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

推荐答案

上面Dommer给出的Java代码给出了稍微不正确的结果，但是如果您要处理GPS轨迹，则小错误加起来.这是Java中Haversine方法的实现，该方法还考虑了两点之间的高度差.

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.

/**
 * 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);
}

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