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Posted 23 Jul 2007

Mapping Images on Spherical Surfaces Using C#

, 15 Jul 2016
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Mapping images on spherical surfaces using C#


This article describes how to map a flat 2D image (JPG, BMP, or GIF) on a sphere by using basic algebra.

The process is very simple where the x axis of the image will be mapped on sphere longitudes and the y axis of the image will be mapped on sphere latitudes.

The process of mapping is similar to proportion equations x-x0/y-y0 = px-x0/py-y0

public static double MapCoordinate(double i1, double i2, double w1,
    double w2, double p)
    return ((p - i1) / (i2 - i1)) * (w2 - w1) + w1;

Screenshot - worldmap4.gif
Original image

Screenshot - mapping.png
Resulting image


A Sphere Can Be Represented by Spherical Coordinates in R3

  • radius
  • phi (latitude angle)
  • theta (longitude angle)

Image 2

  • Where radius is a constant, phi=[-PI/2,PI/2], and theta=[0,2*PI]


To Find the Cartesian Coordinates from Spherical Coordinates

  • x = radius * sin(phi) * cos(theta)
  • y = radius * sin(phi) * sin(theta)
  • z = radius * cos(theta)
double phi0 = 0.0;
double phi1 = Math.PI;
double theta0 = 0.0;
double theta1 = 2.0*Math.PI;

The Code

At first we code the image loading

System.Drawing.Image image1 = new Bitmap(Server.MapPath(
Bitmap imgBitmap = new Bitmap(image1);

Now we make a loop through the 2 dimensions of the image, map phi and theta angles from image coordinates, get the cartesian 3D coordinates from phi and theta, provide some rotation to the obtained 3D points and plot them with respective image color:

for (int i = 0; i < imgBitmap.Width; i++)
     for (int j = 0; j < imgBitmap.Height; j++)
          // map the angles from image coordinates
          double theta = Algebra.MapCoordinate(0.0, imgBitmap.Width - 1,
              theta1, theta0, i);
          double phi = Algebra.MapCoordinate( 0.0, imgBitmap.Height - 1,phi0,
              phi1, j);
          // find the cartesian coordinates
          double x = radius * Math.Sin(phi) * Math.Cos(theta);
          double y = radius * Math.Sin(phi) * Math.Sin(theta);
          double z = radius * Math.Cos(phi);
          // apply rotation around X and Y axis to reposition the sphere
          RotX(1.5, ref y, ref z);
          RotY(-2.5, ref x, ref z);
          // plot only positive points
          if (z > 0)
               Color color = imgBitmap.GetPixel(i, j);
               Brush brs = new SolidBrush(color);
               int ix = (int)x + 100;
               int iy = (int)y + 100;
               graphics.FillRectangle(brs, ix, iy, 1, 1);

The Rotation Functions [almost forgot]

Actually I made a 3D Math class, but here you will need only these functions

public static void RotX(double angle, ref double y, ref double z)
     double y1 = y * System.Math.Cos(angle) - z * System.Math.Sin(angle);
     double z1 = y * System.Math.Sin(angle) + z * System.Math.Cos(angle);
     y = y1;
     z = z1;
public static void RotY(double angle, ref double x, ref double z)
     double x1 = x * System.Math.Cos(angle) - z * System.Math.Sin(angle);
     double z1 = x * System.Math.Sin(angle) + z * System.Math.Cos(angle);
     x = x1;
     z = z1;
public static void RotZ(double angle, ref double x, ref double y)
     double x1 = x * System.Math.Cos(angle) - y * System.Math.Sin(angle);
     double y1 = x * System.Math.Sin(angle) + y * System.Math.Cos(angle);
     x = x1;
     y = y1;

See sample


This article, along with any associated source code and files, is licensed under The GNU General Public License (GPLv3)


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Comments and Discussions

Jokesweet Pin
Ben Daniel23-Jul-07 19:10
memberBen Daniel23-Jul-07 19:10 
AnswerRe: sweet Pin
andalmeida24-Jul-07 4:53
memberandalmeida24-Jul-07 4:53 
GeneralRe: sweet Pin
N.L. Neilson2-Aug-07 12:01
memberN.L. Neilson2-Aug-07 12:01 
GeneralRe: sweet Pin
andalmeida2-Aug-07 12:03
memberandalmeida2-Aug-07 12:03 
GeneralRe: sweet Pin
andalmeida2-Aug-07 12:17
memberandalmeida2-Aug-07 12:17 
GeneralRe: sweet Pin
N.L. Neilson2-Aug-07 12:51
memberN.L. Neilson2-Aug-07 12:51 
GeneralRe: sweet Pin
andalmeida2-Aug-07 12:51
memberandalmeida2-Aug-07 12:51 
GeneralRe: sweet Pin
N.L. Neilson2-Aug-07 14:48
memberN.L. Neilson2-Aug-07 14:48 
Yes, this will draw an ellipsoid, or a close approximation.
The distance using Math.Cos(theta) is close enough for most purposes. For an actual drawing of the Earth, as a spheroid or an ellipsoid per WGS84, it is visually indistinguishable.
The problem comes into play when when a distance over the drawing is measured from two points, where an icon or other image is placed, tracking, etc.

The GIS, WMS images are quite accurate. To get the data FROM a pixel on the screen and working backwards to get the lat-lon the error is compounded.

In the Vincenty Formula, and others, getting the accurate "cos" which is found by iteration until the desired accuracy is obtained.

The latitude measured for a point on the Earth and the GIS data is normal to the surface of the ellipsoid, it is not the same as on a sphere except at the poles and the equator.

You answered my original question on placing a 2D image on an ellipse.
The link given in your first reply may be able to modify the approx. to a more accurate placement. Trimble's GPS Survey software will do this as others will (I think from the data read) but of course it is not open source.


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