cv2.estimateRigidTransform with fullAffine=False











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According to documentation cv2.estimateRigidTransform have parameter fullAffine:




fullAffine – If true, the function finds an optimal affine
transformation with no additional restrictions (6 degrees of freedom).
Otherwise, the class of transformations to choose from is limited to
combinations of translation, rotation, and uniform scaling (5 degrees
of freedom).




I don't understand what is meant by 5 degrees of freedom, as I understand translation, rotation, and uniform scaling can be done with 4 variables (some more info here http://nghiaho.com/?p=2208)




  1. By uniform scaling they mean that x and y scale will be the same?


I have tried



print('cv2.__version__', cv2.__version__)
m = cv2.estimateRigidTransform(_prev_pts, _curr_pts, fullAffine=False)
print('m.shape', m.shape)
print('m',m)


Output:



cv2.__version__ 3.4.3
m.shape (2, 3)
m [[ 1.00165841e+00 -2.10742695e-04 4.28874117e+00]
[ 2.10742695e-04 1.00165841e+00 1.23242652e+00]]


Output looks like stated in documentation(4 unique values):



enter image description here




  1. Another question is how to decompose matrix m to rotation, scaling and translation matrices, i.e. m = R*T*S?










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  • I too think it should be only 4 dof. Decomposition dhould be: angle = asin(a12), scale=a11/cos(angle) and translation = (b1,b2) not sure whether angle is missing a factor -1 though...
    – Micka
    Nov 10 at 17:43















up vote
0
down vote

favorite












According to documentation cv2.estimateRigidTransform have parameter fullAffine:




fullAffine – If true, the function finds an optimal affine
transformation with no additional restrictions (6 degrees of freedom).
Otherwise, the class of transformations to choose from is limited to
combinations of translation, rotation, and uniform scaling (5 degrees
of freedom).




I don't understand what is meant by 5 degrees of freedom, as I understand translation, rotation, and uniform scaling can be done with 4 variables (some more info here http://nghiaho.com/?p=2208)




  1. By uniform scaling they mean that x and y scale will be the same?


I have tried



print('cv2.__version__', cv2.__version__)
m = cv2.estimateRigidTransform(_prev_pts, _curr_pts, fullAffine=False)
print('m.shape', m.shape)
print('m',m)


Output:



cv2.__version__ 3.4.3
m.shape (2, 3)
m [[ 1.00165841e+00 -2.10742695e-04 4.28874117e+00]
[ 2.10742695e-04 1.00165841e+00 1.23242652e+00]]


Output looks like stated in documentation(4 unique values):



enter image description here




  1. Another question is how to decompose matrix m to rotation, scaling and translation matrices, i.e. m = R*T*S?










share|improve this question






















  • I too think it should be only 4 dof. Decomposition dhould be: angle = asin(a12), scale=a11/cos(angle) and translation = (b1,b2) not sure whether angle is missing a factor -1 though...
    – Micka
    Nov 10 at 17:43













up vote
0
down vote

favorite









up vote
0
down vote

favorite











According to documentation cv2.estimateRigidTransform have parameter fullAffine:




fullAffine – If true, the function finds an optimal affine
transformation with no additional restrictions (6 degrees of freedom).
Otherwise, the class of transformations to choose from is limited to
combinations of translation, rotation, and uniform scaling (5 degrees
of freedom).




I don't understand what is meant by 5 degrees of freedom, as I understand translation, rotation, and uniform scaling can be done with 4 variables (some more info here http://nghiaho.com/?p=2208)




  1. By uniform scaling they mean that x and y scale will be the same?


I have tried



print('cv2.__version__', cv2.__version__)
m = cv2.estimateRigidTransform(_prev_pts, _curr_pts, fullAffine=False)
print('m.shape', m.shape)
print('m',m)


Output:



cv2.__version__ 3.4.3
m.shape (2, 3)
m [[ 1.00165841e+00 -2.10742695e-04 4.28874117e+00]
[ 2.10742695e-04 1.00165841e+00 1.23242652e+00]]


Output looks like stated in documentation(4 unique values):



enter image description here




  1. Another question is how to decompose matrix m to rotation, scaling and translation matrices, i.e. m = R*T*S?










share|improve this question













According to documentation cv2.estimateRigidTransform have parameter fullAffine:




fullAffine – If true, the function finds an optimal affine
transformation with no additional restrictions (6 degrees of freedom).
Otherwise, the class of transformations to choose from is limited to
combinations of translation, rotation, and uniform scaling (5 degrees
of freedom).




I don't understand what is meant by 5 degrees of freedom, as I understand translation, rotation, and uniform scaling can be done with 4 variables (some more info here http://nghiaho.com/?p=2208)




  1. By uniform scaling they mean that x and y scale will be the same?


I have tried



print('cv2.__version__', cv2.__version__)
m = cv2.estimateRigidTransform(_prev_pts, _curr_pts, fullAffine=False)
print('m.shape', m.shape)
print('m',m)


Output:



cv2.__version__ 3.4.3
m.shape (2, 3)
m [[ 1.00165841e+00 -2.10742695e-04 4.28874117e+00]
[ 2.10742695e-04 1.00165841e+00 1.23242652e+00]]


Output looks like stated in documentation(4 unique values):



enter image description here




  1. Another question is how to decompose matrix m to rotation, scaling and translation matrices, i.e. m = R*T*S?







python opencv computer-vision affinetransform






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asked Nov 10 at 14:52









mrgloom

4,848954122




4,848954122












  • I too think it should be only 4 dof. Decomposition dhould be: angle = asin(a12), scale=a11/cos(angle) and translation = (b1,b2) not sure whether angle is missing a factor -1 though...
    – Micka
    Nov 10 at 17:43


















  • I too think it should be only 4 dof. Decomposition dhould be: angle = asin(a12), scale=a11/cos(angle) and translation = (b1,b2) not sure whether angle is missing a factor -1 though...
    – Micka
    Nov 10 at 17:43
















I too think it should be only 4 dof. Decomposition dhould be: angle = asin(a12), scale=a11/cos(angle) and translation = (b1,b2) not sure whether angle is missing a factor -1 though...
– Micka
Nov 10 at 17:43




I too think it should be only 4 dof. Decomposition dhould be: angle = asin(a12), scale=a11/cos(angle) and translation = (b1,b2) not sure whether angle is missing a factor -1 though...
– Micka
Nov 10 at 17:43

















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