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SQL function to convert UK OS coordinates from easting/northing to longitude and latitude

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sql

geometry

Please can someone post a SQL function to convert easting/northing to longitude/latitude. I know it's incredibly complicated but I haven't found anyone who has documented it in T-SQL.

This javascript code works but I'm having trouble converting it to SQL.

I have 16,000 coordinates and need them all converted to lat/long.

This is what I have so far but it's not getting past the while loop.

DECLARE @east real = 482353, 
        @north real = 213371

DECLARE @a real = 6377563.396, 
        @b real = 6356256.910,
        @F0 real = 0.9996012717,

        @lat0 real = 49*PI()/180, 
        @lon0 real = -2*PI()/180

DECLARE @N0 real = -100000, 
        @E0 real = 400000,
        @e2 real = 1 - (@b*@b)/(@a*@a),
        @n real = (@a-@b)/(@a+@b)

DECLARE @n2 real = @n*@n, 
        @n3 real = @n*@n*@n

DECLARE @lat real = @lat0, 
        @M real = 0

WHILE (@north-@N0-@M >= 0.00001)
BEGIN

    SET @lat = ((@north-@N0-@M)/(@a*@F0)) + @lat

    DECLARE @Ma real = (1 + @n + (5/4)*@n2 + (5/4)*@n3) * (@lat-@lat0),
            @Mb real = (3*@n + 3*@n*@n + (21/8)*@n3) * SIN(@lat-@lat0) * COS(@lat+@lat0),
            @Mc real = ((15/8)*@n2 + (15/8)*@n3) * SIN(2*(@lat-@lat0)) * COS(2*(@lat+@lat0)),
            @Md real = (35/24)*@n3 * SIN(3*(@lat-@lat0)) * COS(3*(@lat+@lat0))

    SET @M = @b * @F0 * (@Ma - @Mb + @Mc - @Md)

END

DECLARE @cosLat real = COS(@lat), 
        @sinLat real = SIN(@lat)

DECLARE @nu real = @a*@F0/sqrt(1-@e2*@sinLat*@sinLat)
DECLARE @rho real = @a*@F0*(1-@e2)/POWER(1-@e2*@sinLat*@sinLat, 1.5)
DECLARE @eta2 real = @nu/@rho-1

DECLARE @tanLat real = tan(@lat)
DECLARE @tan2lat real = @tanLat*@tanLat
DECLARE @tan4lat real = @tan2lat*@tan2lat
DECLARE @tan6lat real = @tan4lat*@tan2lat
DECLARE @secLat real = 1/@cosLat
DECLARE @nu3 real = @nu*@nu*@nu
DECLARE @nu5 real = @nu3*@nu*@nu
DECLARE @nu7 real = @nu5*@nu*@nu
DECLARE @VII real = @tanLat/(2*@rho*@nu)
DECLARE @VIII real = @tanLat/(24*@rho*@nu3)*(5+3*@tan2lat+@eta2-9*@tan2lat*@eta2)
DECLARE @IX real = @tanLat/(720*@rho*@nu5)*(61+90*@tan2lat+45*@tan4lat)
DECLARE @X real = @secLat/@nu
DECLARE @XI real = @secLat/(6*@nu3)*(@nu/@rho+2*@tan2lat)
DECLARE @XII real = @secLat/(120*@nu5)*(5+28*@tan2lat+24*@tan4lat)
DECLARE @XIIA real = @secLat/(5040*@nu7)*(61+662*@tan2lat+1320*@tan4lat+720*@tan6lat)

DECLARE @dE real = (@east-@E0)
DECLARE @dE2 real = @dE*@dE
DECLARE @dE3 real = @dE2*@dE
DECLARE @dE4 real = @dE2*@dE2, 
        @dE5 real = @dE3*@dE2
DECLARE @dE6 real = @dE4*@dE2, 
        @dE7 real = @dE5*@dE2

SET @lat = @lat - @VII*@dE2 + @VIII*@dE4 - @IX*@dE6

DECLARE @lon real = @lon0 + @X*@dE - @XI*@dE3 + @XII*@dE5 - @XIIA*@dE7

SELECT @lon, @lat
like image 323
Mark Clancy Avatar asked Dec 01 '22 04:12

Mark Clancy


2 Answers

I've been struggling with this one for a while. I had a lot of northing/easting points in OSGB36 that have to be converted on the fly on a regular basis. Please note that the UDF below converts northings/eastings in OSGB36 (Ordnance Survey) projection to latitude/longitude in WGS84 projection so they can be used in Google Maps.

/****** Object:  UserDefinedFunction [dbo].[NEtoLL]    Script Date: 09/06/2012 17:06:39 ******/
SET ANSI_NULLS ON
GO

SET QUOTED_IDENTIFIER ON
GO

CREATE FUNCTION [dbo].[NEtoLL] (@East INT, @North INT, @LatOrLng VARCHAR(3)) RETURNS FLOAT AS
BEGIN

--Author: Sandy Motteram
--Date:   06 September 2012

--UDF adapted from javascript at http://www.bdcc.co.uk/LatLngToOSGB.js
--found on page http://mapki.com/wiki/Tools:Snippets

--Instructions:
--Latitude and Longitude are calculated based on BOTH the easting and northing values from the OSGB36
--This UDF takes both easting and northing values in OSGB36 projection and you must specify if a latitude or longitude co-ordinate should be returned.
--IT first converts E/N values to lat and long in OSGB36 projection, then converts those values to lat/lng in WGS84 projection

--Sample values below
--DECLARE @East INT, @North INT, @LatOrLng VARCHAR(3)
--SELECT @East = 529000, @North = 183650 --that combo should be the corner of Camden High St and Delancey St


    DECLARE @Pi              FLOAT
          , @K0              FLOAT
          , @OriginLat       FLOAT
          , @OriginLong      FLOAT
          , @OriginX         FLOAT
          , @OriginY         FLOAT
          , @a               FLOAT
          , @b               FLOAT
          , @e2              FLOAT
          , @ex              FLOAT
          , @n1              FLOAT
          , @n2              FLOAT
          , @n3              FLOAT
          , @OriginNorthings FLOAT
          , @lat             FLOAT
          , @lon             FLOAT
          , @Northing        FLOAT
          , @Easting         FLOAT

    SELECT  @Pi = 3.14159265358979323846
          , @K0 = 0.9996012717 -- grid scale factor on central meridean
          , @OriginLat  = 49.0
          , @OriginLong = -2.0
          , @OriginX =  400000 -- 400 kM
          , @OriginY = -100000 -- 100 kM
          , @a = 6377563.396   -- Airy Spheroid
          , @b = 6356256.910
    /*    , @e2
          , @ex
          , @n1
          , @n2
          , @n3
          , @OriginNorthings*/

    -- compute interim values
    SELECT  @a = @a * @K0
          , @b = @b * @K0

    SET     @n1 = (@a - @b) / (@a + @b)
    SET     @n2 = @n1 * @n1
    SET     @n3 = @n2 * @n1

    SET     @lat = @OriginLat * @Pi / 180.0 -- to radians

    SELECT  @e2 = (@a * @a - @b * @b) / (@a * @a) -- first eccentricity
          , @ex = (@a * @a - @b * @b) / (@b * @b) -- second eccentricity

    SET     @OriginNorthings = @b * @lat + @b * (@n1 * (1.0 + 5.0 * @n1 * (1.0 + @n1) / 4.0) * @lat
          - 3.0 * @n1 * (1.0 + @n1 * (1.0 + 7.0 * @n1 / 8.0)) * SIN(@lat) * COS(@lat)
          + (15.0 * @n1 * (@n1 + @n2) / 8.0) * SIN(2.0 * @lat) * COS(2.0 * @lat)
          - (35.0 * @n3 / 24.0) * SIN(3.0 * @lat) * COS(3.0 * @lat))

    SELECT  @northing = @north - @OriginY
         ,  @easting  = @east  - @OriginX

    DECLARE @nu       FLOAT
          , @phid     FLOAT
          , @phid2    FLOAT
          , @t2       FLOAT
          , @t        FLOAT
          , @q2       FLOAT
          , @c        FLOAT
          , @s        FLOAT
          , @nphid    FLOAT
          , @dnphid   FLOAT
          , @nu2      FLOAT
          , @nudivrho FLOAT
          , @invnurho FLOAT
          , @rho      FLOAT
          , @eta2     FLOAT

    /* Evaluate M term: latitude of the northing on the centre meridian */

    SET     @northing = @northing + @OriginNorthings

    SET     @phid  = @northing / (@b*(1.0 + @n1 + 5.0 * (@n2 + @n3) / 4.0)) - 1.0
    SET     @phid2 = @phid + 1.0

    WHILE (ABS(@phid2 - @phid) > 0.000001)
    BEGIN
        SET @phid = @phid2;
        SET @nphid = @b * @phid + @b * (@n1 * (1.0 + 5.0 * @n1 * (1.0 + @n1) / 4.0) * @phid
                   - 3.0 * @n1 * (1.0 + @n1 * (1.0 + 7.0 * @n1 / 8.0)) * SIN(@phid) * COS(@phid)
                   + (15.0 * @n1 * (@n1 + @n2) / 8.0) * SIN(2.0 * @phid) * COS(2.0 * @phid)
                   - (35.0 * @n3 / 24.0) * SIN(3.0 * @phid) * COS(3.0 * @phid))

        SET @dnphid = @b * ((1.0 + @n1 + 5.0 * (@n2 + @n3) / 4.0) - 3.0 * (@n1 + @n2 + 7.0 * @n3 / 8.0) * COS(2.0 * @phid)
                    + (15.0 * (@n2 + @n3) / 4.0) * COS(4 * @phid) - (35.0 * @n3 / 8.0) * COS(6.0 * @phid))

        SET @phid2 = @phid - (@nphid - @northing) / @dnphid
    END

    SELECT @c = COS(@phid)
         , @s = SIN(@phid)
         , @t = TAN(@phid)
    SELECT @t2 = @t * @t
         , @q2 = @easting * @easting

    SET    @nu2 = (@a * @a) / (1.0 - @e2 * @s * @s)
    SET    @nu = SQRT(@nu2)

    SET    @nudivrho = @a * @a * @c * @c / (@b * @b) - @c * @c + 1.0
    SET    @eta2 = @nudivrho - 1
    SET    @rho = @nu / @nudivrho;

    SET    @invnurho = ((1.0 - @e2 * @s * @s) * (1.0 - @e2 * @s * @s)) / (@a * @a * (1.0 - @e2))

    SET    @lat = @phid - @t * @q2 * @invnurho / 2.0 + (@q2 * @q2 * (@t / (24 * @rho * @nu2 * @nu) * (5 + (3 * @t2) + @eta2 - (9 * @t2 * @eta2))))
    SET    @lon = (@easting / (@c * @nu))
                - (@easting * @q2 * ((@nudivrho + 2.0 * @t2) / (6.0 * @nu2)) / (@c * @nu))
                + (@q2 * @q2 * @easting * (5 + (28 * @t2) + (24 * @t2 * @t2)) / (120 * @nu2 * @nu2 * @nu * @c))


    SELECT @lat = @lat * 180.0 / @Pi
         , @lon = @lon * 180.0 / @Pi + @OriginLong


--Now convert the lat and long from OSGB36 to WGS84

    DECLARE @OGlat  FLOAT
          , @OGlon  FLOAT
          , @height FLOAT

    SELECT  @OGlat  = @lat
          , @OGlon  = @lon
          , @height = 24 --London's mean height above sea level is 24 metres. Adjust for other locations.

    DECLARE @deg2rad  FLOAT
          , @rad2deg  FLOAT
          , @radOGlat FLOAT
          , @radOGlon FLOAT

    SELECT  @deg2rad = @Pi / 180
          , @rad2deg = 180 / @Pi

    --first off convert to radians
    SELECT  @radOGlat = @OGlat * @deg2rad
          , @radOGlon = @OGlon * @deg2rad
    --these are the values for WGS84(GRS80) to OSGB36(Airy) 

    DECLARE @a2       FLOAT
          , @h        FLOAT
          , @xp       FLOAT
          , @yp       FLOAT
          , @zp       FLOAT
          , @xr       FLOAT
          , @yr       FLOAT
          , @zr       FLOAT
          , @sf       FLOAT
          , @e        FLOAT
          , @v        FLOAT
          , @x        FLOAT
          , @y        FLOAT
          , @z        FLOAT
          , @xrot     FLOAT
          , @yrot     FLOAT
          , @zrot     FLOAT
          , @hx       FLOAT
          , @hy       FLOAT
          , @hz       FLOAT
          , @newLon   FLOAT
          , @newLat   FLOAT
          , @p        FLOAT
          , @errvalue FLOAT
          , @lat0     FLOAT

    SELECT  @a2 = 6378137             -- WGS84_AXIS
          , @e2 = 0.00669438037928458 -- WGS84_ECCENTRIC
          , @h  = @height             -- height above datum (from $GPGGA sentence)
          , @a  = 6377563.396         -- OSGB_AXIS
          , @e  = 0.0066705397616     -- OSGB_ECCENTRIC
          , @xp = 446.448
          , @yp = -125.157
          , @zp = 542.06
          , @xr = 0.1502
          , @yr = 0.247
          , @zr = 0.8421
          , @s  = -20.4894

    -- convert to cartesian; lat, lon are in radians
    SET @sf = @s * 0.000001
    SET @v = @a / (sqrt(1 - (@e * (SIN(@radOGlat) * SIN(@radOGlat)))))
    SET @x = (@v + @h) * COS(@radOGlat) * COS(@radOGlon)
    SET @y = (@v + @h) * COS(@radOGlat) * SIN(@radOGlon)
    SET @z = ((1 - @e) * @v + @h) * SIN(@radOGlat)

    -- transform cartesian
    SET @xrot = (@xr / 3600) * @deg2rad
    SET @yrot = (@yr / 3600) * @deg2rad
    SET @zrot = (@zr / 3600) * @deg2rad
    SET @hx = @x + (@x * @sf) - (@y * @zrot) + (@z * @yrot) + @xp
    SET @hy = (@x * @zrot) + @y + (@y * @sf) - (@z * @xrot) + @yp
    SET @hz = (-1 * @x * @yrot) + (@y * @xrot) + @z + (@z * @sf) + @zp

    -- Convert back to lat, lon
    SET @newLon = ATAN(@hy / @hx)
    SET @p = SQRT((@hx * @hx) + (@hy * @hy))
    SET @newLat = ATAN(@hz / (@p * (1 - @e2)))
    SET @v = @a2 / (SQRT(1 - @e2 * (SIN(@newLat) * SIN(@newLat))))
    SET @errvalue = 1.0;
    SET @lat0 = 0
    WHILE (@errvalue > 0.001)
    BEGIN
        SET @lat0 = ATAN((@hz + @e2 * @v * SIN(@newLat)) / @p)
        SET @errvalue = ABS(@lat0 - @newLat)
        SET @newLat = @lat0
    END

    --convert back to degrees
    SET @newLat = @newLat * @rad2deg
    SET @newLon = @newLon * @rad2deg

    DECLARE @ReturnMe FLOAT
    SET @ReturnMe = 0

    IF @LatOrLng = 'Lat'
        SET @ReturnMe = @newLat
    IF @LatOrLng = 'Lng'
        SET @ReturnMe = @newLon

    RETURN @ReturnMe
END
GO
like image 89
SiO2y Avatar answered Dec 03 '22 18:12

SiO2y


I ended up using the following javascript functions to convert the values. I know it's not a SQL solution but it did the job for me.

function OSGridToLatLong(E, N) {

  var a = 6377563.396, b = 6356256.910;              // Airy 1830 major & minor semi-axes
  var F0 = 0.9996012717;                             // NatGrid scale factor on central meridian
  var lat0 = 49*Math.PI/180, lon0 = -2*Math.PI/180;  // NatGrid true origin
  var N0 = -100000, E0 = 400000;                     // northing & easting of true origin, metres
  var e2 = 1 - (b*b)/(a*a);                          // eccentricity squared
  var n = (a-b)/(a+b), n2 = n*n, n3 = n*n*n;

  var lat=lat0, M=0;
  do {
    lat = (N-N0-M)/(a*F0) + lat;

    var Ma = (1 + n + (5/4)*n2 + (5/4)*n3) * (lat-lat0);
    var Mb = (3*n + 3*n*n + (21/8)*n3) * Math.sin(lat-lat0) * Math.cos(lat+lat0);
    var Mc = ((15/8)*n2 + (15/8)*n3) * Math.sin(2*(lat-lat0)) * Math.cos(2*(lat+lat0));
    var Md = (35/24)*n3 * Math.sin(3*(lat-lat0)) * Math.cos(3*(lat+lat0));
    M = b * F0 * (Ma - Mb + Mc - Md);                // meridional arc

  } while (N-N0-M >= 0.00001);  // ie until < 0.01mm

  var cosLat = Math.cos(lat), sinLat = Math.sin(lat);
  var nu = a*F0/Math.sqrt(1-e2*sinLat*sinLat);              // transverse radius of curvature
  var rho = a*F0*(1-e2)/Math.pow(1-e2*sinLat*sinLat, 1.5);  // meridional radius of curvature
  var eta2 = nu/rho-1;

  var tanLat = Math.tan(lat);
  var tan2lat = tanLat*tanLat, tan4lat = tan2lat*tan2lat, tan6lat = tan4lat*tan2lat;
  var secLat = 1/cosLat;
  var nu3 = nu*nu*nu, nu5 = nu3*nu*nu, nu7 = nu5*nu*nu;
  var VII = tanLat/(2*rho*nu);
  var VIII = tanLat/(24*rho*nu3)*(5+3*tan2lat+eta2-9*tan2lat*eta2);
  var IX = tanLat/(720*rho*nu5)*(61+90*tan2lat+45*tan4lat);
  var X = secLat/nu;
  var XI = secLat/(6*nu3)*(nu/rho+2*tan2lat);
  var XII = secLat/(120*nu5)*(5+28*tan2lat+24*tan4lat);
  var XIIA = secLat/(5040*nu7)*(61+662*tan2lat+1320*tan4lat+720*tan6lat);

  var dE = (E-E0), dE2 = dE*dE, dE3 = dE2*dE, dE4 = dE2*dE2, dE5 = dE3*dE2, dE6 = dE4*dE2, dE7 = dE5*dE2;
  lat = lat - VII*dE2 + VIII*dE4 - IX*dE6;
  var lon = lon0 + X*dE - XI*dE3 + XII*dE5 - XIIA*dE7;


  return {

    longitude: lon.toDeg(),
    latitude: lat.toDeg()

  };

}

Number.prototype.toRad = function() {  // convert degrees to radians
  return this * Math.PI / 180;
}
Number.prototype.toDeg = function() {  // convert radians to degrees (signed)
  return this * 180 / Math.PI;
}
Number.prototype.padLZ = function(w) {
  var n = this.toString();
  for (var i=0; i<w-n.length; i++) n = '0' + n;
  return n;
}
like image 23
Mark Clancy Avatar answered Dec 03 '22 17:12

Mark Clancy