A slight confusion about a particular step in Euclidean Transformations (Computer Vision)
On page 392 of Computer vision by Simon Prince (http://web4.cs.ucl.ac.uk/staff/s.prince/book/book.pdf), equation 15.2 has the following expression for the Euclidean transformation in homogeneous coordinates:
$$
lambda
begin{bmatrix}
x\y\1
end{bmatrix} =
begin{bmatrix}
phi_x&gamma&delta_x\
0 &phi_y&delta_y\
0&0&1
end{bmatrix}
begin{bmatrix}
w_{11}&w_{12}&tau_x\
w_{21}&w_{22}&tau_y\
0&0&D
end{bmatrix}
begin{bmatrix} u\v\1end{bmatrix}
$$
In what follows, he switches the position of $1$ and $D$:
$$
lambda
begin{bmatrix}
x\y\1
end{bmatrix} =
begin{bmatrix}
phi_x&gamma&delta_x\
0 &phi_y&delta_y\
0&0&D
end{bmatrix}
begin{bmatrix}
w_{11}&w_{12}&tau_x\
w_{21}&w_{22}&tau_y\
0&0&1
end{bmatrix}
begin{bmatrix} u\v\1end{bmatrix}
$$
with the comment that the last equation is unchanged.
My question is this: the other two equations do seem to change when you make this switch, so firstly, why can we make this switch? Secondly, is there a geometric explanation underlying this switch?
Edit:
Here are the multiplied out matrices:
$$
begin{bmatrix}
phi_x&gamma&delta_x\
0 &phi_y&delta_y\
0&0&1
end{bmatrix}
begin{bmatrix}
w_{11}&w_{12}&tau_x\
w_{21}&w_{22}&tau_y\
0&0&D
end{bmatrix} =
begin{bmatrix}
phi_x w_{11} + gamma w_{21}&phi_x w_{12} + gamma w_{22}&phi_x tau_x + gamma tau_y + Ddelta_x\
phi_y w_{21}& phi_y w_{22}&phi_y tau_y + Ddelta_y\
0&0&D
end{bmatrix}
$$
$$
begin{bmatrix}
phi_x&gamma&delta_x\
0 &phi_y&delta_y\
0&0&D
end{bmatrix}
begin{bmatrix}
w_{11}&w_{12}&tau_x\
w_{21}&w_{22}&tau_y\
0&0&1
end{bmatrix} =
begin{bmatrix}
phi_x w_{11} + gamma w_{21}&phi_x w_{12} + gamma w_{22}&phi_x tau_x + gamma tau_y + delta_x\
phi_y w_{21}& phi_y w_{22}&phi_y tau_y + delta_y\
0&0&D
end{bmatrix}
$$
As can be seen in the last column, the first two equations have changed.
Edit 2:
I have checked with my lecturer and he thinks that there might be an error in the book.
machine-learning rigid-transformation
asked Nov 30 '18 at 2:23
Sean LeeSean Lee
1578
add a comment |
On page 392 of Computer vision by Simon Prince (http://web4.cs.ucl.ac.uk/staff/s.prince/book/book.pdf), equation 15.2 has the following expression for the Euclidean transformation in homogeneous coordinates:
$$
lambda
begin{bmatrix}
x\y\1
end{bmatrix} =
begin{bmatrix}
phi_x&gamma&delta_x\
0 &phi_y&delta_y\
0&0&1
end{bmatrix}
begin{bmatrix}
w_{11}&w_{12}&tau_x\
w_{21}&w_{22}&tau_y\
0&0&D
end{bmatrix}
begin{bmatrix} u\v\1end{bmatrix}
$$
In what follows, he switches the position of $1$ and $D$:
$$
lambda
begin{bmatrix}
x\y\1
end{bmatrix} =
begin{bmatrix}
phi_x&gamma&delta_x\
0 &phi_y&delta_y\
0&0&D
end{bmatrix}
begin{bmatrix}
w_{11}&w_{12}&tau_x\
w_{21}&w_{22}&tau_y\
0&0&1
end{bmatrix}
begin{bmatrix} u\v\1end{bmatrix}
$$
with the comment that the last equation is unchanged.
My question is this: the other two equations do seem to change when you make this switch, so firstly, why can we make this switch? Secondly, is there a geometric explanation underlying this switch?
Edit:
Here are the multiplied out matrices:
$$
begin{bmatrix}
phi_x&gamma&delta_x\
0 &phi_y&delta_y\
0&0&1
end{bmatrix}
begin{bmatrix}
w_{11}&w_{12}&tau_x\
w_{21}&w_{22}&tau_y\
0&0&D
end{bmatrix} =
begin{bmatrix}
phi_x w_{11} + gamma w_{21}&phi_x w_{12} + gamma w_{22}&phi_x tau_x + gamma tau_y + Ddelta_x\
phi_y w_{21}& phi_y w_{22}&phi_y tau_y + Ddelta_y\
0&0&D
end{bmatrix}
$$
$$
begin{bmatrix}
phi_x&gamma&delta_x\
0 &phi_y&delta_y\
0&0&D
end{bmatrix}
begin{bmatrix}
w_{11}&w_{12}&tau_x\
w_{21}&w_{22}&tau_y\
0&0&1
end{bmatrix} =
begin{bmatrix}
phi_x w_{11} + gamma w_{21}&phi_x w_{12} + gamma w_{22}&phi_x tau_x + gamma tau_y + delta_x\
phi_y w_{21}& phi_y w_{22}&phi_y tau_y + delta_y\
0&0&D
end{bmatrix}
$$
As can be seen in the last column, the first two equations have changed.
Edit 2:
I have checked with my lecturer and he thinks that there might be an error in the book.
machine-learning rigid-transformation
asked Nov 30 '18 at 2:23
Sean LeeSean Lee
1578
Multiply it out for yourself!
– amd
Nov 30 '18 at 2:42
@amd Question has been edited - I am still unable to see why we can switch the positions of the 1 and D given that the first two equations are not equivalent. Is there a particular reason why we don't need to be concerned about the switching of 1 and D?
– Sean Lee
Nov 30 '18 at 10:56
add a comment |
On page 392 of Computer vision by Simon Prince (http://web4.cs.ucl.ac.uk/staff/s.prince/book/book.pdf), equation 15.2 has the following expression for the Euclidean transformation in homogeneous coordinates:
$$
lambda
begin{bmatrix}
x\y\1
end{bmatrix} =
begin{bmatrix}
phi_x&gamma&delta_x\
0 &phi_y&delta_y\
0&0&1
end{bmatrix}
begin{bmatrix}
w_{11}&w_{12}&tau_x\
w_{21}&w_{22}&tau_y\
0&0&D
end{bmatrix}
begin{bmatrix} u\v\1end{bmatrix}
$$
In what follows, he switches the position of $1$ and $D$:
$$
lambda
begin{bmatrix}
x\y\1
end{bmatrix} =
begin{bmatrix}
phi_x&gamma&delta_x\
0 &phi_y&delta_y\
0&0&D
end{bmatrix}
begin{bmatrix}
w_{11}&w_{12}&tau_x\
w_{21}&w_{22}&tau_y\
0&0&1
end{bmatrix}
begin{bmatrix} u\v\1end{bmatrix}
$$
with the comment that the last equation is unchanged.
My question is this: the other two equations do seem to change when you make this switch, so firstly, why can we make this switch? Secondly, is there a geometric explanation underlying this switch?
Edit:
Here are the multiplied out matrices:
$$
begin{bmatrix}
phi_x&gamma&delta_x\
0 &phi_y&delta_y\
0&0&1
end{bmatrix}
begin{bmatrix}
w_{11}&w_{12}&tau_x\
w_{21}&w_{22}&tau_y\
0&0&D
end{bmatrix} =
begin{bmatrix}
phi_x w_{11} + gamma w_{21}&phi_x w_{12} + gamma w_{22}&phi_x tau_x + gamma tau_y + Ddelta_x\
phi_y w_{21}& phi_y w_{22}&phi_y tau_y + Ddelta_y\
0&0&D
end{bmatrix}
$$
$$
begin{bmatrix}
phi_x&gamma&delta_x\
0 &phi_y&delta_y\
0&0&D
end{bmatrix}
begin{bmatrix}
w_{11}&w_{12}&tau_x\
w_{21}&w_{22}&tau_y\
0&0&1
end{bmatrix} =
begin{bmatrix}
phi_x w_{11} + gamma w_{21}&phi_x w_{12} + gamma w_{22}&phi_x tau_x + gamma tau_y + delta_x\
phi_y w_{21}& phi_y w_{22}&phi_y tau_y + delta_y\
0&0&D
end{bmatrix}
$$
As can be seen in the last column, the first two equations have changed.
Edit 2:
I have checked with my lecturer and he thinks that there might be an error in the book.
machine-learning rigid-transformation
asked Nov 30 '18 at 2:23
Sean LeeSean Lee
1578
On page 392 of Computer vision by Simon Prince (http://web4.cs.ucl.ac.uk/staff/s.prince/book/book.pdf), equation 15.2 has the following expression for the Euclidean transformation in homogeneous coordinates:
$$
lambda
begin{bmatrix}
x\y\1
end{bmatrix} =
begin{bmatrix}
phi_x&gamma&delta_x\
0 &phi_y&delta_y\
0&0&1
end{bmatrix}
begin{bmatrix}
w_{11}&w_{12}&tau_x\
w_{21}&w_{22}&tau_y\
0&0&D
end{bmatrix}
begin{bmatrix} u\v\1end{bmatrix}
$$
In what follows, he switches the position of $1$ and $D$:
$$
lambda
begin{bmatrix}
x\y\1
end{bmatrix} =
begin{bmatrix}
phi_x&gamma&delta_x\
0 &phi_y&delta_y\
0&0&D
end{bmatrix}
begin{bmatrix}
w_{11}&w_{12}&tau_x\
w_{21}&w_{22}&tau_y\
0&0&1
end{bmatrix}
begin{bmatrix} u\v\1end{bmatrix}
$$
with the comment that the last equation is unchanged.
My question is this: the other two equations do seem to change when you make this switch, so firstly, why can we make this switch? Secondly, is there a geometric explanation underlying this switch?
Edit:
Here are the multiplied out matrices:
$$
begin{bmatrix}
phi_x&gamma&delta_x\
0 &phi_y&delta_y\
0&0&1
end{bmatrix}
begin{bmatrix}
w_{11}&w_{12}&tau_x\
w_{21}&w_{22}&tau_y\
0&0&D
end{bmatrix} =
begin{bmatrix}
phi_x w_{11} + gamma w_{21}&phi_x w_{12} + gamma w_{22}&phi_x tau_x + gamma tau_y + Ddelta_x\
phi_y w_{21}& phi_y w_{22}&phi_y tau_y + Ddelta_y\
0&0&D
end{bmatrix}
$$
$$
begin{bmatrix}
phi_x&gamma&delta_x\
0 &phi_y&delta_y\
0&0&D
end{bmatrix}
begin{bmatrix}
w_{11}&w_{12}&tau_x\
w_{21}&w_{22}&tau_y\
0&0&1
end{bmatrix} =
begin{bmatrix}
phi_x w_{11} + gamma w_{21}&phi_x w_{12} + gamma w_{22}&phi_x tau_x + gamma tau_y + delta_x\
phi_y w_{21}& phi_y w_{22}&phi_y tau_y + delta_y\
0&0&D
end{bmatrix}
$$
As can be seen in the last column, the first two equations have changed.
Edit 2:
I have checked with my lecturer and he thinks that there might be an error in the book.
machine-learning rigid-transformation
machine-learning rigid-transformation
asked Nov 30 '18 at 2:23
Sean LeeSean Lee
1578
asked Nov 30 '18 at 2:23
Sean LeeSean Lee
1578
edited Nov 30 '18 at 23:27
Sean Lee
asked Nov 30 '18 at 2:23
Sean LeeSean Lee
1578
asked Nov 30 '18 at 2:23
Sean LeeSean Lee
1578
asked Nov 30 '18 at 2:23
Sean LeeSean Lee
1578
1578
Multiply it out for yourself!
– amd
Nov 30 '18 at 2:42
@amd Question has been edited - I am still unable to see why we can switch the positions of the 1 and D given that the first two equations are not equivalent. Is there a particular reason why we don't need to be concerned about the switching of 1 and D?
– Sean Lee
Nov 30 '18 at 10:56
add a comment |
Multiply it out for yourself!
– amd
Nov 30 '18 at 2:42
@amd Question has been edited - I am still unable to see why we can switch the positions of the 1 and D given that the first two equations are not equivalent. Is there a particular reason why we don't need to be concerned about the switching of 1 and D?
– Sean Lee
Nov 30 '18 at 10:56
Multiply it out for yourself!
– amd
Nov 30 '18 at 2:42
Multiply it out for yourself!
– amd
Nov 30 '18 at 2:42
@amd Question has been edited - I am still unable to see why we can switch the positions of the 1 and D given that the first two equations are not equivalent. Is there a particular reason why we don't need to be concerned about the switching of 1 and D?
– Sean Lee
Nov 30 '18 at 10:56
@amd Question has been edited - I am still unable to see why we can switch the positions of the 1 and D given that the first two equations are not equivalent. Is there a particular reason why we don't need to be concerned about the switching of 1 and D?
– Sean Lee
Nov 30 '18 at 10:56
add a comment |
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Multiply it out for yourself!
– amd
Nov 30 '18 at 2:42
@amd Question has been edited - I am still unable to see why we can switch the positions of the 1 and D given that the first two equations are not equivalent. Is there a particular reason why we don't need to be concerned about the switching of 1 and D?
– Sean Lee
Nov 30 '18 at 10:56