如何计算两个GPS坐标之间的距离(使用经纬度)?
当前回答
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。
其他回答
对于任何寻找Delphi/Pascal版本的人:
function GreatCircleDistance(const Lat1, Long1, Lat2, Long2: Double): Double;
var
Lat1Rad, Long1Rad, Lat2Rad, Long2Rad: Double;
const
EARTH_RADIUS_KM = 6378;
begin
Lat1Rad := DegToRad(Lat1);
Long1Rad := DegToRad(Long1);
Lat2Rad := DegToRad(Lat2);
Long2Rad := DegToRad(Long2);
Result := EARTH_RADIUS_KM * ArcCos(Cos(Lat1Rad) * Cos(Lat2Rad) * Cos(Long1Rad - Long2Rad) + Sin(Lat1Rad) * Sin(Lat2Rad));
end;
我对这个代码没有任何功劳,我最初是在一个公共论坛上发现Gary William发布的。
如果你需要更准确的数据,可以看看这个。
Vincenty's formulae are two related iterative methods used in geodesy to calculate the distance between two points on the surface of a spheroid, developed by Thaddeus Vincenty (1975a) They are based on the assumption that the figure of the Earth is an oblate spheroid, and hence are more accurate than methods such as great-circle distance which assume a spherical Earth. The first (direct) method computes the location of a point which is a given distance and azimuth (direction) from another point. The second (inverse) method computes the geographical distance and azimuth between two given points. They have been widely used in geodesy because they are accurate to within 0.5 mm (0.020″) on the Earth ellipsoid.
这是“Henry Vilinskiy”为MySQL和km改编的版本:
CREATE FUNCTION `CalculateDistanceInKm`(
fromLatitude float,
fromLongitude float,
toLatitude float,
toLongitude float
) RETURNS float
BEGIN
declare distance float;
select
6367 * ACOS(
round(
COS(RADIANS(90-fromLatitude)) *
COS(RADIANS(90-toLatitude)) +
SIN(RADIANS(90-fromLatitude)) *
SIN(RADIANS(90-toLatitude)) *
COS(RADIANS(fromLongitude-toLongitude))
,15)
)
into distance;
return round(distance,3);
END;
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
打印稿版本
export const degreeToRadian = (degree: number) => {
return degree * Math.PI / 180;
}
export const distanceBetweenEarthCoordinatesInKm = (lat1: number, lon1: number, lat2: number, lon2: number) => {
const earthRadiusInKm = 6371;
const dLat = degreeToRadian(lat2 - lat1);
const dLon = degreeToRadian(lon2 - lon1);
lat1 = degreeToRadian(lat1);
lat2 = degreeToRadian(lat2);
const a = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.sin(dLon / 2) * Math.sin(dLon / 2) * Math.cos(lat1) * Math.cos(lat2);
const c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
return earthRadiusInKm * c;
}
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