PostGIS中具有纬度/长度SRID的实际（大圆）距离？

Question

我在PostGIS数据库中使用lat / long SRID（-4326）。我希望以有效的方式找到给定点的最近点。我试过做

ORDER BY    ST_Distance(point, ST_GeomFromText(?,-4326))

这给了我48个州的好结果，但在阿拉斯加，它给了我垃圾。有没有办法在PostGIS中进行真正的距离计算，或者我是否需要提供合理大小的缓冲区然后计算大圆距离并在之后的代码中对结果进行排序？

Answer 1

您正在寻找ST_distance_sphere（point，point）或st_distance_spheroid（point，point）。

请参阅：

http://postgis.refractions.net/documentation/manual- 1.3 / ch06.html＃distance_sphere http://postgis.refractions.net/documentation/manual-1.3/ch06 html的＃distance_spheroid

这通常被称为测地距离或大地距离......虽然这两个术语的含义略有不同，但它们往往可以互换使用。

或者，您可以投影数据并使用标准的st_distance函数...这仅适用于短距离（使用UTM或状态平面）或者所有距离都相对于一个或两个点（等距投影）。

Answer 2

PostGIS 1.5使用lat longs和meter处理真正的地球距离。它意识到纬度/长度本质上是角度的并且具有360度线

Answer 3

这是来自SQL Server，而且我使用Haversine可能会遇到可能受到阿拉斯加问题影响的快速距离（可能会偏离一英里）：

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((@lat2-@lat1)/2.0E)) + 
    (cos(@lat1) * cos(@lat2) * square(sin((@lon2-@lon1)/2.0E))) )

set @distance =
    @radius * ( 2.0E *asin(case when 1.0E < @a then 1.0E else @a end ) )

return @distance

end

Vicenty很慢，但准确到1毫米以内（我只发现了一个javascript imp）：

/*
 * 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;
}