3 回答
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TA贡献1797条经验 获得超6个赞
基于关于stackoverflow的另一个问题,我得到了此代码。这将以米为单位计算结果,而不是以英里为单位:)
public static float distFrom(float lat1, float lng1, float lat2, float lng2) {
double earthRadius = 6371000; //meters
double dLat = Math.toRadians(lat2-lat1);
double dLng = Math.toRadians(lng2-lng1);
double a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(Math.toRadians(lat1)) * Math.cos(Math.toRadians(lat2)) *
Math.sin(dLng/2) * Math.sin(dLng/2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
float dist = (float) (earthRadius * c);
return dist;
}
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TA贡献1895条经验 获得超7个赞
您可以将Java Geodesy Library用于GPS,它使用Vincenty的公式,该公式考虑了地球表面的曲率。
实现是这样的:
import org.gavaghan.geodesy.*;
...
GeodeticCalculator geoCalc = new GeodeticCalculator();
Ellipsoid reference = Ellipsoid.WGS84;
GlobalPosition pointA = new GlobalPosition(latitude, longitude, 0.0); // Point A
GlobalPosition userPos = new GlobalPosition(userLat, userLon, 0.0); // Point B
double distance = geoCalc.calculateGeodeticCurve(reference, userPos, pointA).getEllipsoidalDistance(); // Distance between Point A and Point B
结果距离以米为单位。
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TA贡献1827条经验 获得超8个赞
在C ++中,是这样完成的:
#define LOCAL_PI 3.1415926535897932385
double ToRadians(double degrees)
{
double radians = degrees * LOCAL_PI / 180;
return radians;
}
double DirectDistance(double lat1, double lng1, double lat2, double lng2)
{
double earthRadius = 3958.75;
double dLat = ToRadians(lat2-lat1);
double dLng = ToRadians(lng2-lng1);
double a = sin(dLat/2) * sin(dLat/2) +
cos(ToRadians(lat1)) * cos(ToRadians(lat2)) *
sin(dLng/2) * sin(dLng/2);
double c = 2 * atan2(sqrt(a), sqrt(1-a));
double dist = earthRadius * c;
double meterConversion = 1609.00;
return dist * meterConversion;
}
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