如何计算两个GPS坐标之间的距离(使用经纬度)?
当前回答
Scala版本
def deg2rad(deg: Double) = deg * Math.PI / 180.0
def rad2deg(rad: Double) = rad / Math.PI * 180.0
def getDistanceMeters(lat1: Double, lon1: Double, lat2: Double, lon2: Double) = {
val theta = lon1 - lon2
val dist = Math.sin(deg2rad(lat1)) * Math.sin(deg2rad(lat2)) + Math.cos(deg2rad(lat1)) *
Math.cos(deg2rad(lat2)) * Math.cos(deg2rad(theta))
Math.abs(
Math.round(
rad2deg(Math.acos(dist)) * 60 * 1.1515 * 1.609344 * 1000)
)
}
其他回答
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
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;
对于java
public static double degreesToRadians(double degrees) {
return degrees * Math.PI / 180;
}
public static double distanceInKmBetweenEarthCoordinates(Location location1, Location location2) {
double earthRadiusKm = 6371;
double dLat = degreesToRadians(location2.getLatitude()-location1.getLatitude());
double dLon = degreesToRadians(location2.getLongitude()-location1.getLongitude());
double lat1 = degreesToRadians(location1.getLatitude());
double lat2 = degreesToRadians(location2.getLatitude());
double a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.sin(dLon/2) * Math.sin(dLon/2) * Math.cos(lat1) * Math.cos(lat2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
return earthRadiusKm * c;
}
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;
计算两个坐标之间的纬度和经度的距离,包括一个Javascript实现。
西部和南部的位置是负的。 记住,分和秒是60度,所以S31 30'是-31.50度。
别忘了把角度转换成弧度。许多语言都有这个功能。或者它是一个简单的计算:弧度=角度* PI / 180。
function degreesToRadians(degrees) {
return degrees * Math.PI / 180;
}
function distanceInKmBetweenEarthCoordinates(lat1, lon1, lat2, lon2) {
var earthRadiusKm = 6371;
var dLat = degreesToRadians(lat2-lat1);
var dLon = degreesToRadians(lon2-lon1);
lat1 = degreesToRadians(lat1);
lat2 = degreesToRadians(lat2);
var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.sin(dLon/2) * Math.sin(dLon/2) * Math.cos(lat1) * Math.cos(lat2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
return earthRadiusKm * c;
}
下面是一些用法的例子:
distanceInKmBetweenEarthCoordinates(0,0,0,0) // Distance between same
// points should be 0
0
distanceInKmBetweenEarthCoordinates(51.5, 0, 38.8, -77.1) // From London
// to Arlington
5918.185064088764