Transcript Document

Coordinate Transformation
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).
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.
Translation
x’ = Tx + x
y’ = Ty + y
N
Tx
Ty
E
Scaling
x’’ = SN” * x’
y’’ = SE” * y’
N
E
Rotation
N
xa’’ cos q
ya’’ sin q
N = xa’’ sinq + ya’’cosq
E = xa’’ cosq – ya’’sinq
Ea,Na
ya’’ cos q
xa’’ sin q
E
The transformation equations
• The final transformation equations are:
– N = a1 + a2x + a3y
– E = b1 + b2x + b3y
– Where: the a1 and b1 contain the translation
parameters and a2, a3, b2, and b3 contain the scale
and rotation parameters.
• a1, a2,a3, b1, b2, and b3 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.
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.
(N1,E1)
(N2,E2)
(x2,y2)
(x1,y1)
(x3,y3)
(N3,E3)
(N4,E4)
(x4,y4)
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 (vy, vx) which gives us the accuracy of
our transformation.
–
–
–
–
–
–
–
–
N1 = a1 + a2x1 + a3y1
E1 = b1 + b2x1 + b3y1
N2 = a1 + a2x2 + a3y2
E2 = b1 + b2x2 + b3y2
N3 = a1 + a2x3 + a3y3
E3 = b1 + b2x3 + b3y3
N4 = a1 + a2x4 + a3y4
E4 = b1 + b2x4 + b3y4
Least Squares solution
Plug in coefficients to solve for residuals
N1 = a1 + a2x1 + a3y1 +
E1 = b1 + b2x1 + b3y1 +
N2 = a1 + a2x2 + a3y2 +
E2 = b1 + b2x2 + b3y2 +
N3 = a1 + a2x3 + a3y3 +
E3 = b1 + b2x3 + b3y3 +
N4 = a1 + a2x4 + a3y4 +
E4 = b1 + b2x4 + b3y4 +
vy1
vx1
vy2
vx2
vy3
vx3
vy4
vx4
How it works.
Now we can transform our digitizer coordinates into the coordinate system of the map.
a1 = 4,831,480.600
a2 =
560.546
a3 =
465.347
b1 =
b2 =
b3 =
488,524.790
150.0567
234.0678
X = 23.4567
Y = 33.9875
N1 = a1 + a2(23.4567) + a3(33.9875)
E1 = b1 + b2(23.4567) + b3 (33.9875)
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.