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)
- By
uniform scalingthey 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):
- Another question is how to decompose matrix
mto rotation, scaling and translation matrices, i.e.m = R*T*S?
python opencv computer-vision affinetransform
asked Nov 10 at 14:52
mrgloom
4,985958125
add a comment |
up vote
1
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)
- By
uniform scalingthey 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):
- Another question is how to decompose matrix
mto rotation, scaling and translation matrices, i.e.m = R*T*S?
python opencv computer-vision affinetransform
asked Nov 10 at 14:52
mrgloom
4,985958125
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
1
Actually in code there is4 degrees of freedomgithub.com/opencv/opencv/blob/… github.com/opencv/opencv/blob/… , so this is a mistake in documentation.
– mrgloom
Nov 14 at 15:40
add a comment |
up vote
1
down vote
favorite
up vote
1
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)
- By
uniform scalingthey 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):
- Another question is how to decompose matrix
mto rotation, scaling and translation matrices, i.e.m = R*T*S?
python opencv computer-vision affinetransform
asked Nov 10 at 14:52
mrgloom
4,985958125
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)
- By
uniform scalingthey 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):
- Another question is how to decompose matrix
mto rotation, scaling and translation matrices, i.e.m = R*T*S?
python opencv computer-vision affinetransform
python opencv computer-vision affinetransform
asked Nov 10 at 14:52
mrgloom
4,985958125
asked Nov 10 at 14:52
mrgloom
4,985958125
asked Nov 10 at 14:52
mrgloom
4,985958125
asked Nov 10 at 14:52
mrgloom
4,985958125
asked Nov 10 at 14:52
mrgloom
4,985958125
4,985958125
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
1
Actually in code there is4 degrees of freedomgithub.com/opencv/opencv/blob/… github.com/opencv/opencv/blob/… , so this is a mistake in documentation.
– mrgloom
Nov 14 at 15:40
add a comment |
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
1
Actually in code there is4 degrees of freedomgithub.com/opencv/opencv/blob/… github.com/opencv/opencv/blob/… , so this is a mistake in documentation.
– mrgloom
Nov 14 at 15:40
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
1
1
Actually in code there is
4 degrees of freedom github.com/opencv/opencv/blob/… github.com/opencv/opencv/blob/… , so this is a mistake in documentation.– mrgloom
Nov 14 at 15:40
Actually in code there is
4 degrees of freedom github.com/opencv/opencv/blob/… github.com/opencv/opencv/blob/… , so this is a mistake in documentation.– mrgloom
Nov 14 at 15:40
add a comment |
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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
1
Actually in code there is
4 degrees of freedomgithub.com/opencv/opencv/blob/… github.com/opencv/opencv/blob/… , so this is a mistake in documentation.– mrgloom
Nov 14 at 15:40