如何计算两个GPS坐标之间的距离(使用经纬度)?
当前回答
飞镖版本
Haversine Algorithm。
import 'dart:math';
class GeoUtils {
static double _degreesToRadians(degrees) {
return degrees * pi / 180;
}
static double 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 = sin(dLat/2) * sin(dLat/2) +
sin(dLon/2) * sin(dLon/2) * cos(lat1) * cos(lat2);
var c = 2 * atan2(sqrt(a), sqrt(1-a));
return earthRadiusKm * c;
}
}
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
飞镖版本
Haversine Algorithm。
import 'dart:math';
class GeoUtils {
static double _degreesToRadians(degrees) {
return degrees * pi / 180;
}
static double 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 = sin(dLat/2) * sin(dLat/2) +
sin(dLon/2) * sin(dLon/2) * cos(lat1) * cos(lat2);
var c = 2 * atan2(sqrt(a), sqrt(1-a));
return earthRadiusKm * c;
}
}
在SQL Server 2008中使用地理类型非常容易做到这一点。
SELECT geography::Point(lat1, lon1, 4326).STDistance(geography::Point(lat2, lon2, 4326))
-- computes distance in meters using eliptical model, accurate to the mm
4326是WGS84椭球地球模型的SRID
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.
下面是Kotlin的一个变种:
import kotlin.math.*
class HaversineAlgorithm {
companion object {
private const val MEAN_EARTH_RADIUS = 6371.008
private const val D2R = Math.PI / 180.0
}
private fun haversineInKm(lat1: Double, lon1: Double, lat2: Double, lon2: Double): Double {
val lonDiff = (lon2 - lon1) * D2R
val latDiff = (lat2 - lat1) * D2R
val latSin = sin(latDiff / 2.0)
val lonSin = sin(lonDiff / 2.0)
val a = latSin * latSin + (cos(lat1 * D2R) * cos(lat2 * D2R) * lonSin * lonSin)
val c = 2.0 * atan2(sqrt(a), sqrt(1.0 - a))
return MEAN_EARTH_RADIUS * c
}
}
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