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Given a lat/lon, how do I find the precise position on the tile?

Given a latitude and longitude, I can find the correct tile number - but how do I find the correct position (in pixel) on the tile?

tile latitude calulation longitude position

This question is marked "community wiki".

asked 07 Sep '10, 08:14

petschge
8.3k●21●73●98
accept rate: 21%

closed 17 Feb '11, 14:03

(17 Feb '11, 14:37) emj

4 Answers:

When you calulate the tile number according to the wiki you will actually get a floating point number.

The integer part indicates which tile you are (or should be) looking at.

The fractional part indicates the position within the tile. As a tile is 256 pixel wide, multiplying the fractional part with 256 will give you the pixel position from the top left.

answered 07 Sep '10, 08:16

petschge
8.3k●21●73●98
accept rate: 21%

edited 07 Sep '10, 10:19

Gnonthgol ♦
13.8k●16●103●198

Can you show a detailed example? for example we have lat : 36 lon: 6 we create square earth map with real scale in autocad how we can find the position x,y of the coordinates in the map? i don't see an example or a formula in the wiki

(14 Oct '21, 06:27) Seghier

1	If you just want to see the position on the map, you can use mlat and mlon to get a marker. answered 07 Sep '10, 13:39 scai ♦ 33.3k●21●309●459 accept rate: 23% what you mean by mlat and mlon? where is the formula? (14 Oct '21, 06:29) Seghier

I've written the following conversion functions to convert from Lat/Long coordinates to OSM pixel x,y coordinates for a given zoom level.

Friend Sub LatLongToPixelXYOSM(ByVal latitude As Double, ByVal longitude As Double, ByVal zoomLevel As Integer, ByRef pixelX As Integer, ByRef pixelY As Integer)
        Dim MinLatitude = -85.05112878
        Dim MaxLatitude = 85.05112878
        Dim MinLongitude = -180
        Dim MaxLongitude = 180
        Dim mapSize = Math.Pow(2, zoomLevel) * 256

        latitude = Clip(latitude, MinLatitude, MaxLatitude)
        longitude = Clip(longitude, MinLongitude, MaxLongitude)

        Dim p As PointF = New Point()
        p.X = CSng((longitude + 180.0) / 360.0 * (1 << zoomLevel))
        p.Y = CSng((1.0 - Math.Log(Math.Tan(latitude * Math.PI / 180.0) + 1.0 / Math.Cos(toRadians(latitude))) / Math.PI) / 2.0 * (1 << zoomLevel))

        Dim tilex As Integer = CInt(Math.Truncate(p.X))
        Dim tiley As Integer = CInt(Math.Truncate(p.Y))
        pixelX = ClipByRange((tilex * 256) + ((p.X - tilex) * 256), mapSize - 1)
        pixelY = ClipByRange((tiley * 256) + ((p.Y - tiley) * 256), mapSize - 1)
    End Sub

Friend Sub PixelXYToLatLongOSM(ByVal pixelX As Integer, ByVal pixelY As Integer, ByVal zoomLevel As Integer, ByRef latitude As Double, ByRef longitude As Double)
        Dim mapSize = Math.Pow(2, zoomLevel) * 256
        Dim tileX As Integer = Math.Truncate(pixelX / 256)
        Dim tileY As Integer = Math.Truncate(pixelY / 256)

        Dim p As PointF = New Point()
        Dim n As Double = Math.PI - ((2.0 * Math.PI * (ClipByRange(pixelY, mapSize - 1) / 256)) / Math.Pow(2.0, zoomLevel))

        longitude = CSng(((ClipByRange(pixelX, mapSize - 1) / 256) / Math.Pow(2.0, zoomLevel) * 360.0) - 180.0)
        latitude = CSng(180.0 / Math.PI * Math.Atan(Math.Sinh(n)))
    End Sub

Private Function ClipByRange(ByVal n As Double, ByVal range As Double)
        Return n Mod range
End Function

Private Function Clip(ByVal n As Double, ByVal minValue As Double, ByVal maxValue As Double)
      Return Math.Min(Math.Max(n, minValue), maxValue)
End Function

answered 14 Nov '11, 10:27

johnmeilak
25●1
accept rate: 0%

0	In the editor, press CTRL+SHIFT+L to display the Location Panel. answered 09 Jan '23, 11:19 joper34 11●1●1●3 accept rate: 0% edited 09 Jan '23, 11:24