1. Coordinate Transformation

2. How to transform coordinates from one system to another. In this situation we have earth coordinates on the left and digitizer coordinates on the right. So we are transforming between locations on the globe (i.e. geographic coordinates) and the coordinate system of the digitizing table (i.e. millimeters).

3. What we want We would like to click on the digitizing “puck” and get a UTM coordinate placed in our GIS. In order to do this we need to calculate the “transformation parameters” between the UTM coordinate system and the coordinate system of the Digitizing Tablet. To align these systems we go through a translation, a scaling and a rotation.

4. Translation N E x’ y’ TxTx TyTy x’ = Tx + x y’ = Ty + y x y

5. Scaling y’ N E x’’ = S N ” * x’ y’’ = S E ” * y’ y’’ x’’ x’

6. Rotation N E N = x a ’’ sin  + y a ’’cos  E = x a ’’ cos  – y a ’’sin  y’’ x’’ E a,N a x’’a y’’a  x a ’’ cos  y a ’’ sin  y a ’’ cos  x a ’’ sin 

7. The transformation equations The final transformation equations are: –N = a 1 + a 2 x + a 3 y –E = b 1 + b 2 x + b 3 y –Where: the a 1 and b 1 contain the translation parameters and a 2, a 3, b 2, and b 3 contain the scale and rotation parameters. a 1, a 2,a 3, b 1, b 2, and b 3 are the unknown coefficients that need to be solved in order to transform from the (x,y) coordinate system to the (N,E) coordinate system.

8. Solving for the Unknown coefficients To solve for the unknowns we need to select a least four coordinates that can be physically identified in both of the coordinate systems. Road intersections are often used as well as geographic features such as the tip of a peninsula. (N 1,E 1 ) (N 2,E 2 ) (N 3,E 3 ) (N 4,E 4 ) (x 1,y 1 ) (x 2,y 2 ) (x 3,y 3 ) (x 4,y 4 )

9. Solving for the Unknown coefficients (cont) With four points we can write 8 equations one for each x and y coordinate. These equations are solved simultaneously to yield the 6 coefficients. By plugging these coefficients back into the equations we can solver for the residuals ( v y, v x ) which gives us the accuracy of our transformation. –N 1 = a 1 + a 2 x 1 + a 3 y 1 N 1 = a 1 + a 2 x 1 + a 3 y 1 + v y1 –E 1 = b 1 + b 2 x 1 + b 3 y 1 E 1 = b 1 + b 2 x 1 + b 3 y 1 + v x1 –N 2 = a 1 + a 2 x 2 + a 3 y 2 N 2 = a 1 + a 2 x 2 + a 3 y 2 + v y2 –E 2 = b 1 + b 2 x 2 + b 3 y 2 Least Squares solution E 2 = b 1 + b 2 x 2 + b 3 y 2 + v x2 –N 3 = a 1 + a 2 x 3 + a 3 y 3 N 3 = a 1 + a 2 x 3 + a 3 y 3 + v y3 –E 3 = b 1 + b 2 x 3 + b 3 y 3 Plug in coefficients to solve for residuals E 3 = b 1 + b 2 x 3 + b 3 y 3 + v x3 –N 4 = a 1 + a 2 x 4 + a 3 y 4 N 4 = a 1 + a 2 x 4 + a 3 y 4 + v y4 –E 4 = b 1 + b 2 x 4 + b 3 y 4 E 4 = b 1 + b 2 x 4 + b 3 y 4 + v x4

10. How it works. Now we can transform our digitizer coordinates into the coordinate system of the map. N1 = a1 + a2(23.4567) + a3(33.9875) E1 = b1 + b2(23.4567) + b3 (33.9875) X = 23.4567 Y = 33.9875 b1 = 488,524.790 b2 = 150.0567 b3 = 234.0678 a1 = 4,831,480.600 a2 = 560.546 a3 = 465.347 N1 = 4,831,480.60 + 560.546(23.4567) + 465.347(33.9875) E1 = 488,524.79 + 150.0567(23.4567) + 234.0678 (33.9875) N1 = 4,871,656. E1 = 500,000.