如何计算两个GPS坐标之间的距离(使用经纬度)?
当前回答
一、关于“面包屑”方法
地球半径在不同的纬度上是不同的。在Haversine算法中必须考虑到这一点。 考虑轴承的变化，它将直线变成拱门(更长的) 考虑到速度变化将把拱门变成螺旋(比拱门更长或更短) 高度变化将使平面螺旋变成3D螺旋(再次变长)。这对丘陵地区非常重要。
下面是考虑#1和#2的C语言函数:
double calcDistanceByHaversine(double rLat1, double rLon1, double rHeading1,
double rLat2, double rLon2, double rHeading2){
double rDLatRad = 0.0;
double rDLonRad = 0.0;
double rLat1Rad = 0.0;
double rLat2Rad = 0.0;
double a = 0.0;
double c = 0.0;
double rResult = 0.0;
double rEarthRadius = 0.0;
double rDHeading = 0.0;
double rDHeadingRad = 0.0;
if ((rLat1 < -90.0) || (rLat1 > 90.0) || (rLat2 < -90.0) || (rLat2 > 90.0)
|| (rLon1 < -180.0) || (rLon1 > 180.0) || (rLon2 < -180.0)
|| (rLon2 > 180.0)) {
return -1;
};
rDLatRad = (rLat2 - rLat1) * DEGREE_TO_RADIANS;
rDLonRad = (rLon2 - rLon1) * DEGREE_TO_RADIANS;
rLat1Rad = rLat1 * DEGREE_TO_RADIANS;
rLat2Rad = rLat2 * DEGREE_TO_RADIANS;
a = sin(rDLatRad / 2) * sin(rDLatRad / 2) + sin(rDLonRad / 2) * sin(
rDLonRad / 2) * cos(rLat1Rad) * cos(rLat2Rad);
if (a == 0.0) {
return 0.0;
}
c = 2 * atan2(sqrt(a), sqrt(1 - a));
rEarthRadius = 6378.1370 - (21.3847 * 90.0 / ((fabs(rLat1) + fabs(rLat2))
/ 2.0));
rResult = rEarthRadius * c;
// Chord to Arc Correction based on Heading changes. Important for routes with many turns and U-turns
if ((rHeading1 >= 0.0) && (rHeading1 < 360.0) && (rHeading2 >= 0.0)
&& (rHeading2 < 360.0)) {
rDHeading = fabs(rHeading1 - rHeading2);
if (rDHeading > 180.0) {
rDHeading -= 180.0;
}
rDHeadingRad = rDHeading * DEGREE_TO_RADIANS;
if (rDHeading > 5.0) {
rResult = rResult * (rDHeadingRad / (2.0 * sin(rDHeadingRad / 2)));
} else {
rResult = rResult / cos(rDHeadingRad);
}
}
return rResult;
}
2有一种更简单的方法，效果很好。
按平均速度。
Trip_distance = Trip_average_speed * Trip_time
由于GPS速度是由多普勒效应检测的，与[Lon,Lat]没有直接关系，如果不是主要的距离计算方法，至少可以考虑作为次要的(备份或校正)。
其他回答
一、关于“面包屑”方法
地球半径在不同的纬度上是不同的。在Haversine算法中必须考虑到这一点。 考虑轴承的变化，它将直线变成拱门(更长的) 考虑到速度变化将把拱门变成螺旋(比拱门更长或更短) 高度变化将使平面螺旋变成3D螺旋(再次变长)。这对丘陵地区非常重要。
下面是考虑#1和#2的C语言函数:
double calcDistanceByHaversine(double rLat1, double rLon1, double rHeading1,
double rLat2, double rLon2, double rHeading2){
double rDLatRad = 0.0;
double rDLonRad = 0.0;
double rLat1Rad = 0.0;
double rLat2Rad = 0.0;
double a = 0.0;
double c = 0.0;
double rResult = 0.0;
double rEarthRadius = 0.0;
double rDHeading = 0.0;
double rDHeadingRad = 0.0;
if ((rLat1 < -90.0) || (rLat1 > 90.0) || (rLat2 < -90.0) || (rLat2 > 90.0)
|| (rLon1 < -180.0) || (rLon1 > 180.0) || (rLon2 < -180.0)
|| (rLon2 > 180.0)) {
return -1;
};
rDLatRad = (rLat2 - rLat1) * DEGREE_TO_RADIANS;
rDLonRad = (rLon2 - rLon1) * DEGREE_TO_RADIANS;
rLat1Rad = rLat1 * DEGREE_TO_RADIANS;
rLat2Rad = rLat2 * DEGREE_TO_RADIANS;
a = sin(rDLatRad / 2) * sin(rDLatRad / 2) + sin(rDLonRad / 2) * sin(
rDLonRad / 2) * cos(rLat1Rad) * cos(rLat2Rad);
if (a == 0.0) {
return 0.0;
}
c = 2 * atan2(sqrt(a), sqrt(1 - a));
rEarthRadius = 6378.1370 - (21.3847 * 90.0 / ((fabs(rLat1) + fabs(rLat2))
/ 2.0));
rResult = rEarthRadius * c;
// Chord to Arc Correction based on Heading changes. Important for routes with many turns and U-turns
if ((rHeading1 >= 0.0) && (rHeading1 < 360.0) && (rHeading2 >= 0.0)
&& (rHeading2 < 360.0)) {
rDHeading = fabs(rHeading1 - rHeading2);
if (rDHeading > 180.0) {
rDHeading -= 180.0;
}
rDHeadingRad = rDHeading * DEGREE_TO_RADIANS;
if (rDHeading > 5.0) {
rResult = rResult * (rDHeadingRad / (2.0 * sin(rDHeadingRad / 2)));
} else {
rResult = rResult / cos(rDHeadingRad);
}
}
return rResult;
}
2有一种更简单的方法，效果很好。
按平均速度。
Trip_distance = Trip_average_speed * Trip_time
由于GPS速度是由多普勒效应检测的，与[Lon,Lat]没有直接关系，如果不是主要的距离计算方法，至少可以考虑作为次要的(备份或校正)。
c#版本的Haversine
double _eQuatorialEarthRadius = 6378.1370D;
double _d2r = (Math.PI / 180D);
private int HaversineInM(double lat1, double long1, double lat2, double long2)
{
return (int)(1000D * HaversineInKM(lat1, long1, lat2, long2));
}
private double HaversineInKM(double lat1, double long1, double lat2, double long2)
{
double dlong = (long2 - long1) * _d2r;
double dlat = (lat2 - lat1) * _d2r;
double a = Math.Pow(Math.Sin(dlat / 2D), 2D) + Math.Cos(lat1 * _d2r) * Math.Cos(lat2 * _d2r) * Math.Pow(Math.Sin(dlong / 2D), 2D);
double c = 2D * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1D - a));
double d = _eQuatorialEarthRadius * c;
return d;
}
这里有一个。net小提琴，所以你可以用你自己的Lat/ long测试它。
PHP版本:
(删除所有deg2rad()如果您的坐标已经是弧度。)
$R = 6371; // km
$dLat = deg2rad($lat2-$lat1);
$dLon = deg2rad($lon2-$lon1);
$lat1 = deg2rad($lat1);
$lat2 = deg2rad($lat2);
$a = sin($dLat/2) * sin($dLat/2) +
sin($dLon/2) * sin($dLon/2) * cos($lat1) * cos($lat2);
$c = 2 * atan2(sqrt($a), sqrt(1-$a));
$d = $R * $c;
这段Lua代码改编自维基百科和Robert Lipe的GPSbabel工具:
local EARTH_RAD = 6378137.0
-- earth's radius in meters (official geoid datum, not 20,000km / pi)
local radmiles = EARTH_RAD*100.0/2.54/12.0/5280.0;
-- earth's radius in miles
local multipliers = {
radians = 1, miles = radmiles, mi = radmiles, feet = radmiles * 5280,
meters = EARTH_RAD, m = EARTH_RAD, km = EARTH_RAD / 1000,
degrees = 360 / (2 * math.pi), min = 60 * 360 / (2 * math.pi)
}
function gcdist(pt1, pt2, units) -- return distance in radians or given units
--- this formula works best for points close together or antipodal
--- rounding error strikes when distance is one-quarter Earth's circumference
--- (ref: wikipedia Great-circle distance)
if not pt1.radians then pt1 = rad(pt1) end
if not pt2.radians then pt2 = rad(pt2) end
local sdlat = sin((pt1.lat - pt2.lat) / 2.0);
local sdlon = sin((pt1.lon - pt2.lon) / 2.0);
local res = sqrt(sdlat * sdlat + cos(pt1.lat) * cos(pt2.lat) * sdlon * sdlon);
res = res > 1 and 1 or res < -1 and -1 or res
res = 2 * asin(res);
if units then return res * assert(multipliers[units])
else return res
end
end
在Python中，你可以使用geopy库使用WGS84椭球来计算测地线距离:
from geopy.distance import geodesic
newport_ri = (41.49008, -71.312796)
cleveland_oh = (41.499498, -81.695391)
print(geodesic(newport_ri, cleveland_oh).km)
