GetCoords and Mercator Projection ?

[alfalf99]

GetCoords and Mercator Projection ?

[neveldo]

Hi, I think this is because the tutorial is for the maps using equi-rectangular projection only, it can be applied for other projections. For the mercator, you will have to define the proper algorithm to convert the coordinates.

Feel free to take a look at this pull-request that includes a map with a mercator projection to see how the getCoords() function is built : #169

[alfalf99]

Thank you.

I have tried to translate the mercator getcoords function (from the link provided) in VBA.
Coordinates of my south america submap are hardcoded in the VBA script.

But when I test RIO DE JANEIRO, lat, long = (-43.23, -22.9)
I have the result :
x= -0,753617936081045
y = -0,689338045886754
which is wrong, but perhaps I have missed something :(
If it returns x, y I would expect values that are based on dim of my map: width = 510 and height : 792,25.
so something like that : x= 400 and y = 600

Below the VBA code :
`Option Explicit

'// Width of the map, in pixel
Public Const WWwidth As Double = 1008.77
'// Height of the map, in pixel
Public Const WWheight As Double = 651.44
'// Longitude of the left side of the map, in degree
'leftLongitude: -169.6,
'// Longitude of the right side of the map, in degree
'rightLongitude: 190.25,
'// Latitude of the top of the map, in degree
'topLatitude: 83.68,
'// Latitude of the bottom of the map, in degree
'bottomLatitude: -55.55,

Public Sub teste()

Call getCoords(-43.23, -22.9) 'RIO DE JANEIRO lat, lon

End Sub

'*
' * Get x,y point from lat,lon coordinate
' *
' * Principle:
' * (lat, lon) are inputs
' * Projection(lat, lon) = (x', y')
' * Transformation(x', y') = (x, y)
' *
' * Source: http://jkwiens.com/2009/01/23/miller-projection/
' *
' * @param lat latitude value in degree
' * @param lon longitude value in degree
' * @return {x, y} coordinate in pixel
' *
Sub getCoords(lat As Double, lon As Double)

Dim xLeftPrime As Double, xRightPrime As Double, yTopPrime As Double, yBottomPrime As Double
Dim leftLongitude As Double, rightLongitude As Double, topLatitude As Double, bottomLatitude As Double

Dim xPrime As Double, yPrime As Double
Dim x As Double, y As Double

'---------------INPUT  MAP INFO MANUALLY
xLeftPrime = 0
bottomLatitude = -86.66

xRightPrime = 510.9827 '-- sub map width
rightLongitude = -32.16

yTopPrime = 0
topLatitude = 13.5

yBottomPrime = 792.25 '-- sub map height
bottomLatitude = -58
'----------------------------------------


'Project map boundaries with projection (only once for performance)
'If (xLeftPrime Is Empty) Then xLeftPrime = projectLongitude(leftLongitude)
'If (xRightPrime Is Empty) Then xRightPrime = projectLongitude(rightLongitude)
'If (yTopPrime Is Empty) Then yTopPrime = projectLatitude(topLatitude)
'If (yBottomPrime Is Empty) Then yBottomPrime = projectLatitude(bottomLatitude)

'Compute x' and y' (projection-specific value)
xPrime = projectLongitude(lon)
yPrime = projectLatitude(lat)

'Compute x and y
x = (xPrime - xLeftPrime) * (WWwidth / (xRightPrime - xLeftPrime))
y = (yTopPrime - yPrime) * (WWheight / (yTopPrime - yBottomPrime))

Debug.Print x
Debug.Print y
'return {x: x, y: y}

End Sub

' * Transform a longitude coordinate into projection-specific x' coordinate
' * Note: this is not in pixel
' *
' * @param lon longitude value in degree
' * @return x' projection-specific value
' *
Public Function projectLongitude(lon As Double)

'Compute longitude in radian
projectLongitude = lon * Excel.WorksheetFunction.Pi / 180
End Function

'*
' * Transform a latitude coordinate into projection-specific y' coordinate
' * Note: this is not in pixel
' *
' * @param lat latitude value in degree
' * @return y' projection-specific value
' *
Public Function projectLatitude(lat As Double)
'Compute latitude in radian and get its SINUS
Dim latRadSinus As Double
latRadSinus = Math.Sin(lat * Excel.WorksheetFunction.Pi / 180)
projectLatitude = 0.5 * Math.Log((1 + latRadSinus) / (1 - latRadSinus))
End Function
`