Environmental Engineering Reference
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at this point and ignore the Z-coordinate. The focal point of the camera is offset by -s 1x and
-s 1y at horizontal and vertical directions respectively from the lower left corner of the screen.
Then the position of the corner is s 1 =[s 1x , s 1y ] T . The lower right corner is s 2 =[s 2x , s 2y ] T
=[L H +s 1x , s 1y ] T where L H is the width of the screen. Likewise, the upper right corner is
s 3 =[s 3x , s 3y ] T =[L H +s 1x , L V +s 1y ] T where L V is the height of the screen.
2.2 Pointer location using stereovision
Assuming the optical axis of the left camera is rotated from X-axis by angle θ 1 , we can obtain
the rotation matrix of the camera from screen-based coordinate system to camera-based
coordinate system as
cos
sin
1
1
A
.
(1)
1
sin
cos
1
1
Likewise, assuming the focal point of the right camera is located at d 2 =[d 2x , d 2y ] T with angle
θ 2 , the transformation (rotation and translation) matrix is
cos
sin
d
2
2
2
x
A
sin
cos
d
.
(2)
2
2
2
2
x
0
0
1
Please note that we implied the use of homogeneous coordinates in Eq. 2. We will change
back and forth between [x, y] T and [x, y, 1] T in calculations and in expressions whenever
necessary. Therefore any point P=[P x , P y ] T in the screen can be transformed to
P
cos
P
sin
P

x
1
y
1
1
x
y
P
A P
 

(3)
1
1
P
P
sin
P
cos

1
x
1
y
1
at camera 1 and
P
P
cos
P
sin

x
2
y
2
2
x
P
A P
(4)

1
2
P
P
sin
P
cos

2
y
x
2
y
2
at camera 2. We assume a pinpoint camera model with the focal length as λ 1 .
For camera 1, the projection of the point P on the image plane can be easily measured from
the image as N 1 =n 1 *c px1 where n 1 is the pixel position at the image plane and c px1 is the size
of each pixel for camera 1. Hence, we can easily obtain the following relationship using
similar triangles:
P
sin
P
cos
x
1
y
1
1
(5)
P
cos
P
sin
N
x
1
y
1
1
Likewise, we can find similar result for camera 2 as:
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