이 SQL 서버에서, 그리고 내가 (마일으로 해제 할 수 있습니다) 당신의 알래스카 문제로 고통 수있는 말도 안되게 빠른 거리 하버 사인을 사용
ALTER function [dbo].[getCoordinateDistance]
(
@Latitude1 decimal(16,12),
@Longitude1 decimal(16,12),
@Latitude2 decimal(16,12),
@Longitude2 decimal(16,12)
)
returns decimal(16,12)
as
/*
fUNCTION: getCoordinateDistance
Computes the Great Circle distance in kilometers
between two points on the Earth using the
Haversine formula distance calculation.
Input Parameters:
@Longitude1 - Longitude in degrees of point 1
@Latitude1 - Latitude in degrees of point 1
@Longitude2 - Longitude in degrees of point 2
@Latitude2 - Latitude in degrees of point 2
*/
begin
declare @radius decimal(16,12)
declare @lon1 decimal(16,12)
declare @lon2 decimal(16,12)
declare @lat1 decimal(16,12)
declare @lat2 decimal(16,12)
declare @a decimal(16,12)
declare @distance decimal(16,12)
-- Sets average radius of Earth in Kilometers
set @radius = 6366.70701949371
-- Convert degrees to radians
set @lon1 = radians(@Longitude1)
set @lon2 = radians(@Longitude2)
set @lat1 = radians(@Latitude1)
set @lat2 = radians(@Latitude2)
set @a = sqrt(square(sin((@[email protected])/2.0E)) +
(cos(@lat1) * cos(@lat2) * square(sin((@[email protected])/2.0E))))
set @distance =
@radius * (2.0E *asin(case when 1.0E < @a then 1.0E else @a end))
return @distance
end
Vicenty이 내에 정확한 느리지 만, 1mm (그리고 나는 단지 그것의 꼬마 도깨비 자바 스크립트를 발견) :
/*
* Calculate geodesic distance (in m) between two points specified by latitude/longitude (in numeric degrees)
* using Vincenty inverse formula for ellipsoids
*/
function distVincenty(lat1, lon1, lat2, lon2) {
var a = 6378137, b = 6356752.3142, f = 1/298.257223563; // WGS-84 ellipsiod
var L = (lon2-lon1).toRad();
var U1 = Math.atan((1-f) * Math.tan(lat1.toRad()));
var U2 = Math.atan((1-f) * Math.tan(lat2.toRad()));
var sinU1 = Math.sin(U1), cosU1 = Math.cos(U1);
var sinU2 = Math.sin(U2), cosU2 = Math.cos(U2);
var lambda = L, lambdaP = 2*Math.PI;
var iterLimit = 20;
while (Math.abs(lambda-lambdaP) > 1e-12 && --iterLimit>0) {
var sinLambda = Math.sin(lambda), cosLambda = Math.cos(lambda);
var sinSigma = Math.sqrt((cosU2*sinLambda) * (cosU2*sinLambda) +
(cosU1*sinU2-sinU1*cosU2*cosLambda) * (cosU1*sinU2-sinU1*cosU2*cosLambda));
if (sinSigma==0) return 0; // co-incident points
var cosSigma = sinU1*sinU2 + cosU1*cosU2*cosLambda;
var sigma = Math.atan2(sinSigma, cosSigma);
var sinAlpha = cosU1 * cosU2 * sinLambda/sinSigma;
var cosSqAlpha = 1 - sinAlpha*sinAlpha;
var cos2SigmaM = cosSigma - 2*sinU1*sinU2/cosSqAlpha;
if (isNaN(cos2SigmaM)) cos2SigmaM = 0; // equatorial line: cosSqAlpha=0 (§6)
var C = f/16*cosSqAlpha*(4+f*(4-3*cosSqAlpha));
lambdaP = lambda;
lambda = L + (1-C) * f * sinAlpha *
(sigma + C*sinSigma*(cos2SigmaM+C*cosSigma*(-1+2*cos2SigmaM*cos2SigmaM)));
}
if (iterLimit==0) return NaN // formula failed to converge
var uSq = cosSqAlpha * (a*a - b*b)/(b*b);
var A = 1 + uSq/16384*(4096+uSq*(-768+uSq*(320-175*uSq)));
var B = uSq/1024 * (256+uSq*(-128+uSq*(74-47*uSq)));
var deltaSigma = B*sinSigma*(cos2SigmaM+B/4*(cosSigma*(-1+2*cos2SigmaM*cos2SigmaM)-
B/6*cos2SigmaM*(-3+4*sinSigma*sinSigma)*(-3+4*cos2SigmaM*cos2SigmaM)));
var s = b*A*(sigma-deltaSigma);
s = s.toFixed(3); // round to 1mm precision
return s;
}
유감 스럽지만 Debian Stable (1.1.6-2)과 함께 제공되는 PostGIS 버전에는 없습니다. 백 포트 찾기를 시작할 시간입니다. –