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# Converting X, Y to Lat, Long

Posted on 2009-04-02
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Hi All,

I am creating a golf scorer for a windows mobile 5 device and used some of the posts from the following thread to use for the conversion of x,y to lat and long co ordinates:

http://www.experts-exchange.com/Programming/Languages/Visual_Basic/Q_24135663.html?sfQueryTermInfo=1+10+convert+lat+lon+x+y

The piece of code in particular which i am trying to get working for my cropped image is :
Dim XDim As Integer, yDim As Integer, delX As Decimal, _
tlLon As Double, brLon As Double, delY As Decimal, _
tlLat As Double, brLat As Double
Dim myLattitude As Double, myLon As Double

XDim = 640
yDim = 425

tlLon = -95.97630500793457
brLon = -95.921287536621108

tlLat = 36.060964648518294
brLat = 36.031470597983571

delX = (tlLon - (brLon)) / XDim

txtLon.Text = delX

delY = (tlLat - brLat) / yDim

txtLat.Text = delY
Dim y As Integer = 212
myLattitude = tlLat - y * delY
Dim X As Integer = 132
myLon = tlLon + X * delX
txt1.Text = myLattitude
txt2.Text = myLon

For the life of me, I cannot seem to get it working for any of the course layouts (which are all cropped images)

i will post what my version of the code looks like and as you will find, the lat long co ordinates are not miles off but very wide off the mark in terms of accuracy. If anyone can have a look through the coding and find where i am going wrong i would be eternally grateful!!

the picturebox is a 210x806 dimensioned control

the bitmap has been cropped and 'straightened' but the lat and long of the two corners are for the final image so to speak.

the x and y valye of 58 and 95 respectively is the location of the Green and when i get the co-ordinates from the calculations, its quite a way off.

i will be giving the full 500 points for this as i have spent along time trying to look into this and try to recalculate it but to no avail.

Baz
``````Dim XDim As Integer, yDim As Integer, delX As Decimal, _
tlLon As Double, brLon As Double, delY As Decimal, _
tlLat As Double, brLat As Double
Dim myLattitude As Double, myLon As Double

XDim = 640
yDim = 425

tlLon = -1.136182
brLon = -1.133490

tlLat = 52.668572
brLat = 52.666803

delX = (tlLon - (brLon)) / XDim

txtLon.Text = delX

delY = (tlLat - brLat) / yDim

txtLat.Text = delY
Dim y As Integer = 95
myLattitude = tlLat - y * delY
Dim X As Integer = 58
myLon = tlLon - X * delX
txt1.Text = myLattitude
txt2.Text = myLon
``````
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Question by:baz86
• 8
• 4

Author Comment

ID: 24054841
Correction to code snippet above:
xDim= 210
yDim = 806
0

Author Comment

ID: 24055359
The Lat and Long co-ordinates i should be getting are 52.668575lat and -1.135770long
0

Author Comment

ID: 24059318
The formula works fine for the first hole:
Xdim=210,
Ydim=908,
TLLong = -1.133241666
TLLat  = 52.6674027
BRLong = -1.134216666
BRLat = 52.6703861

But as soon as i enter he details from the database for the second hole (and anyh hole thereafter, the co-ordinates are wrong).
0

Author Comment

ID: 24067024
0

LVL 15

Expert Comment

ID: 24089790
When you say you want to convert x, y to Lat, Lon, you need to chose what projection to use, as mapping from spherical coordinates to planar coordinates isn't as simple as the code you have provided. Basically you're not taking into account the the distance between meridians changes you move away form the equator.

The simplest project to use would be Sinusoidal, which is an equal area projection.
http://en.wikipedia.org/wiki/Sinusoidal_projection
As you're dealing with only a small area, you wouldn't see the kind of deformation you see in the wiki images.

I'll dig out some code I wrote a while ago for generating aviation charts that should do proper mapping for you.
0

Author Comment

ID: 24162002
would you be able to advice how i can alter my code so that i can rotate it and still having the co-ordinates be correct?
0

LVL 15

Expert Comment

ID: 24163372
I may have been a bit hasty saying that you needed a non-linear transform. You're dealing with such small areas that it shouldn't be an issue using the linear transform you're using.

Where is the x, y origin, Top-left corner (like a System.Drawing.Graphics object), or Bottom-left?
From your code, I assume it in the bottom-left corner.

For the coordinates you gave in the question, there's no way that area can fit in an image 210x806 as the aspect ratios are completely different.

For the 2nd set of coordinates and the image shape of 210x908 the aspect ratios are similar.

I'm using Google Earth to get get an idea of what you're looking at, so your images will be slightly different.

Can you upload a sample image, saying what the lat/lon position of a certain point on the image is.
0

Author Comment

ID: 24163452
the origin for the picturebox is top left corner at (0,0)
and the picturebox resizes to the dimensions of the bitmap automatically one the form is loaded......
what i found is that when the image is rotated then the formula doesnt give the correct co-ordinates..ie Pic1
but when the image is not rotated and North is true north, the formula gives the corrrect co-ordinates. ie Pic2
the problem i have with this(Pic2) is that im creating an application for a windows mobile device and having Pic1 would only cause me to have a vertical scrollbar on the panel which i can live with, but Pic2 would have the inconvenience of vertical and horzontal....

Untitled-4-copy.bmp
test1unmoved.bmp
0

Author Comment

ID: 24163463
apologies for the enlarged pictures!!! didn't realise them to be so large
0

Author Comment

ID: 24163603
an example of this:

for Pic1...point(107,87) on a 210X860 bitmap is the bright white bunker near the green...this results in the lat long of (52.6686080837,-1.1348202380) which places the mark 60 yards to the right.
for Pic 2...point (84,98) on a 413X678 is the bright white bunker right near the green....this results in the lat long of (52.6685142094,-1.1354876101) which is correct.

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LVL 15

Accepted Solution

oobayly earned 2000 total points
ID: 24169675
Ah, that'd be your problem. Your transforming, rotating & scaling your image coordinates to geoditic coordinates. That'll always be tricky.

First, you need to be able to determine through what angle the image has been rotated. This is relatively easy as you're dealing with a small area, so you can use the Rhumb bearing between points.

The easiest way to do all this is to use the System.Draw.Drawing2D.Matrix. You tell it what transforms to do, and then just feed it an array of points to transform.

For all this to work, you need to pick 2 distinct points on the image, ideally top-left & bottom-right as the further they are apart the better. ie.for the example below I've picked:
Lat, Lon: (x, y)
52.668572, -1.136182: (0, 0)
52.666803, -1.13349: (88, 666)

Then use the code below to create a Transformation matrix. I've also uploaded the image I've based the bounds on
``````public static void Test() {
Matrix imgToGeo = Transform(
new PointF(-1.136182f, 52.668572f),
new PointF(-1.13349f, 52.666803f),
new Point(0, 0),
new Point(88, 666)
);
Matrix geoToImg = imgToGeo.Clone();
geoToImg.Invert();

// Create a set of points inside the image
PointF[] pOrig = new PointF[] {
new PointF(0, 0),
new PointF(68, 79) ,
new PointF(88, 666)
};

// Transform the image points to geoditic
PointF[] pGeo = (PointF[])pOrig.Clone();
imgToGeo.TransformPoints(pGeo);

// Transform the geoditic points back to image points
PointF[] pImage = (PointF[])pGeo.Clone();
geoToImg.TransformPoints(pImage);
}

private static void Rhumb(PointF p1, PointF p2, out double distance, out double heading) {
// Average Radius of the earth in metres
double R = 6371000;

p1.X = (float)(p1.X * Math.PI / 180);
p1.Y = (float)(p1.Y * Math.PI / 180);
p2.X = (float)(p2.X * Math.PI / 180);
p2.Y = (float)(p2.Y * Math.PI / 180);

/* Phi = Latitude
* Lambda = Longitude
*/
double dPhi = (p2.Y - p1.Y);
double dLambda = (p2.X - p1.X);

double dX = dLambda;
double dY = Math.Log(
Math.Tan((Math.PI / 4) + ((double)p2.Y / 2))
/ Math.Tan((Math.PI / 4) + ((double)p1.Y / 2))
);

double q;
if (p1.Y == p2.Y) {
q = Math.Cos(p1.Y);
} else {
q = dPhi / dY;
}

// Distance between points in metres
distance = Math.Sqrt(
Math.Pow(dPhi, 2) + Math.Pow(q * dLambda, 2)
) * R;

// Heading from p1 -> p2 in degrees
heading = ((Math.Atan2(dX, dY) * 180 / Math.PI) + 360) % 360;
}

private static Matrix Transform(PointF geoTL, PointF geoBR, Point imgTL, Point imgBR ) {
/* Calculate the angle that image has been rotated
* Can use Rhumb bearing between 2 geoditic points as the scale
* is small.
* Using rhumb bearing will cause inacuracies over larger scales
*/
float rotate;
{
// Rhumb bearing
double d, thetaRhumb;
Rhumb(geoTL, geoBR, out d, out thetaRhumb);

// Angle from Top-left -> bottom-right on the image
double thetaImage = (Math.Atan2(imgBR.X - imgTL.X, imgBR.Y - imgTL.Y) * 180 / Math.PI);

rotate = (float)(180 - thetaImage - thetaRhumb);
}

/* Create Transformation matrix
* 1). Translate image point to origin
* 2). Rotate the the point about the origin
*     Negative, as we're rotating back
*/
Matrix m = new Matrix();
m.Translate(-imgTL.X, -imgTL.Y);
m.Rotate(-rotate, MatrixOrder.Append);

/* Calcualate the scale
* 1). Get the transformed & rotated value for Point2
* 2). Use this new point to determine the X & Y axis scales
*/
float sX, sY;
{
PointF[] pTemp = new PointF[] { new PointF(imgBR.X, imgBR.Y) };
m.TransformPoints(pTemp);
sX = (float)((geoBR.X - geoTL.X) / pTemp[0].X);
sY = (float)((geoBR.Y - geoTL.Y) / pTemp[0].Y);
}

/* Final part of the transformation matrix
* 1). Set the X & Y scaling
* 2). Translate transformed point back to geoditic coordinate
*/
m.Scale(sX, sY, MatrixOrder.Append);
m.Translate(geoTL.X, geoTL.Y, MatrixOrder.Append);

return m;
}
``````
Foo.jpg
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ID: 24241559
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