creating rotation matrices in opencv coordinate system
I have been looking at creating rotation matrices from a direction vector in the OpenCV coordinate space and there is one thing that has me slightly confused. Here is an OpenCV rotation matrix, that I got from an opencv function (let us call this matrix r):
array([[-0.4136457 , -0.19724711, 0.88881427],
[-0.57926765, 0.810177 , -0.08978985],
[-0.70238609, -0.55200255, -0.44938511]])
Now, starting from the z-axes unit vector, I am trying to recreate this vector. So, I have the following code:
def basis(v):
v = v / np.linalg.norm(v)
if v[0] > 0.9:
b1 = np.asarray([0.0, 1.0, 0.0])
else:
b1 = np.asarray([1.0, 0.0, 0.0])
b1 -= v * np.dot(b1, v)
b1 *= np.reciprocal(np.linalg.norm(b1))
b2 = np.cross(v, b1)
return b1, b2, v
I can call this function as:
x, y, z = basis(r[:, 2])
Then I compute the rotation matrix as:
avg = np.asarray([[x[0], y[0], z[0]],
[x[1], y[1], z[1]],
[x[2], y[2], z[2]]])
Now running this code returns:
array([[ 0.4582676 , 0. , 0.88881427],
[ 0.17414826, -0.98061724, -0.08978985],
[ 0.8715866 , 0.19593324, -0.44938511]])
So, the signs along the x and y-axes are flipped.
Now, in my basis function, if I change the line
b1 = np.asarray([1.0, 0.0, 0.0]) to b1 = np.asarray([-1.0, 0.0, 0.0]). It returns with the correct sign like:
array([[-0.4582676 , 0. , 0.88881427],
[-0.17414826, 0.98061724, -0.08978985],
[-0.8715866 , -0.19593324, -0.44938511]])
I am guessing this has something to do with the handedness as the opencv origin is at the top left corner and the y axes is increasing in the downward direction rather than upward but the thing that has me confused is why does chaging the sign of the x coordinate of the unit vector makes a difference? I was expecting to have to change the other condition i.e. b1 = np.asarray([0.0, 1.0, 0.0]) to have the sign flip in the y coordinate.
opencv geometry coordinates coordinate-transformation
asked Nov 26 '18 at 10:48
LucaLuca
3,25552782
add a comment |
I have been looking at creating rotation matrices from a direction vector in the OpenCV coordinate space and there is one thing that has me slightly confused. Here is an OpenCV rotation matrix, that I got from an opencv function (let us call this matrix r):
array([[-0.4136457 , -0.19724711, 0.88881427],
[-0.57926765, 0.810177 , -0.08978985],
[-0.70238609, -0.55200255, -0.44938511]])
Now, starting from the z-axes unit vector, I am trying to recreate this vector. So, I have the following code:
def basis(v):
v = v / np.linalg.norm(v)
if v[0] > 0.9:
b1 = np.asarray([0.0, 1.0, 0.0])
else:
b1 = np.asarray([1.0, 0.0, 0.0])
b1 -= v * np.dot(b1, v)
b1 *= np.reciprocal(np.linalg.norm(b1))
b2 = np.cross(v, b1)
return b1, b2, v
I can call this function as:
x, y, z = basis(r[:, 2])
Then I compute the rotation matrix as:
avg = np.asarray([[x[0], y[0], z[0]],
[x[1], y[1], z[1]],
[x[2], y[2], z[2]]])
Now running this code returns:
array([[ 0.4582676 , 0. , 0.88881427],
[ 0.17414826, -0.98061724, -0.08978985],
[ 0.8715866 , 0.19593324, -0.44938511]])
So, the signs along the x and y-axes are flipped.
Now, in my basis function, if I change the line
b1 = np.asarray([1.0, 0.0, 0.0]) to b1 = np.asarray([-1.0, 0.0, 0.0]). It returns with the correct sign like:
array([[-0.4582676 , 0. , 0.88881427],
[-0.17414826, 0.98061724, -0.08978985],
[-0.8715866 , -0.19593324, -0.44938511]])
I am guessing this has something to do with the handedness as the opencv origin is at the top left corner and the y axes is increasing in the downward direction rather than upward but the thing that has me confused is why does chaging the sign of the x coordinate of the unit vector makes a difference? I was expecting to have to change the other condition i.e. b1 = np.asarray([0.0, 1.0, 0.0]) to have the sign flip in the y coordinate.
opencv geometry coordinates coordinate-transformation
asked Nov 26 '18 at 10:48
LucaLuca
3,25552782
There is an infinite number of rotation matrices with a given z-axis. You chose one of them. Whatever created your reference matrixrsimply chose another (either because it had more information about the underlying problem or randomly).
– Nico Schertler
Nov 26 '18 at 17:08
add a comment |
I have been looking at creating rotation matrices from a direction vector in the OpenCV coordinate space and there is one thing that has me slightly confused. Here is an OpenCV rotation matrix, that I got from an opencv function (let us call this matrix r):
array([[-0.4136457 , -0.19724711, 0.88881427],
[-0.57926765, 0.810177 , -0.08978985],
[-0.70238609, -0.55200255, -0.44938511]])
Now, starting from the z-axes unit vector, I am trying to recreate this vector. So, I have the following code:
def basis(v):
v = v / np.linalg.norm(v)
if v[0] > 0.9:
b1 = np.asarray([0.0, 1.0, 0.0])
else:
b1 = np.asarray([1.0, 0.0, 0.0])
b1 -= v * np.dot(b1, v)
b1 *= np.reciprocal(np.linalg.norm(b1))
b2 = np.cross(v, b1)
return b1, b2, v
I can call this function as:
x, y, z = basis(r[:, 2])
Then I compute the rotation matrix as:
avg = np.asarray([[x[0], y[0], z[0]],
[x[1], y[1], z[1]],
[x[2], y[2], z[2]]])
Now running this code returns:
array([[ 0.4582676 , 0. , 0.88881427],
[ 0.17414826, -0.98061724, -0.08978985],
[ 0.8715866 , 0.19593324, -0.44938511]])
So, the signs along the x and y-axes are flipped.
Now, in my basis function, if I change the line
b1 = np.asarray([1.0, 0.0, 0.0]) to b1 = np.asarray([-1.0, 0.0, 0.0]). It returns with the correct sign like:
array([[-0.4582676 , 0. , 0.88881427],
[-0.17414826, 0.98061724, -0.08978985],
[-0.8715866 , -0.19593324, -0.44938511]])
I am guessing this has something to do with the handedness as the opencv origin is at the top left corner and the y axes is increasing in the downward direction rather than upward but the thing that has me confused is why does chaging the sign of the x coordinate of the unit vector makes a difference? I was expecting to have to change the other condition i.e. b1 = np.asarray([0.0, 1.0, 0.0]) to have the sign flip in the y coordinate.
opencv geometry coordinates coordinate-transformation
asked Nov 26 '18 at 10:48
LucaLuca
3,25552782
I have been looking at creating rotation matrices from a direction vector in the OpenCV coordinate space and there is one thing that has me slightly confused. Here is an OpenCV rotation matrix, that I got from an opencv function (let us call this matrix r):
array([[-0.4136457 , -0.19724711, 0.88881427],
[-0.57926765, 0.810177 , -0.08978985],
[-0.70238609, -0.55200255, -0.44938511]])
Now, starting from the z-axes unit vector, I am trying to recreate this vector. So, I have the following code:
def basis(v):
v = v / np.linalg.norm(v)
if v[0] > 0.9:
b1 = np.asarray([0.0, 1.0, 0.0])
else:
b1 = np.asarray([1.0, 0.0, 0.0])
b1 -= v * np.dot(b1, v)
b1 *= np.reciprocal(np.linalg.norm(b1))
b2 = np.cross(v, b1)
return b1, b2, v
I can call this function as:
x, y, z = basis(r[:, 2])
Then I compute the rotation matrix as:
avg = np.asarray([[x[0], y[0], z[0]],
[x[1], y[1], z[1]],
[x[2], y[2], z[2]]])
Now running this code returns:
array([[ 0.4582676 , 0. , 0.88881427],
[ 0.17414826, -0.98061724, -0.08978985],
[ 0.8715866 , 0.19593324, -0.44938511]])
So, the signs along the x and y-axes are flipped.
Now, in my basis function, if I change the line
b1 = np.asarray([1.0, 0.0, 0.0]) to b1 = np.asarray([-1.0, 0.0, 0.0]). It returns with the correct sign like:
array([[-0.4582676 , 0. , 0.88881427],
[-0.17414826, 0.98061724, -0.08978985],
[-0.8715866 , -0.19593324, -0.44938511]])
I am guessing this has something to do with the handedness as the opencv origin is at the top left corner and the y axes is increasing in the downward direction rather than upward but the thing that has me confused is why does chaging the sign of the x coordinate of the unit vector makes a difference? I was expecting to have to change the other condition i.e. b1 = np.asarray([0.0, 1.0, 0.0]) to have the sign flip in the y coordinate.
opencv geometry coordinates coordinate-transformation
opencv geometry coordinates coordinate-transformation
asked Nov 26 '18 at 10:48
LucaLuca
3,25552782
asked Nov 26 '18 at 10:48
LucaLuca
3,25552782
asked Nov 26 '18 at 10:48
LucaLuca
3,25552782
asked Nov 26 '18 at 10:48
LucaLuca
3,25552782
asked Nov 26 '18 at 10:48
LucaLuca
3,25552782
3,25552782
There is an infinite number of rotation matrices with a given z-axis. You chose one of them. Whatever created your reference matrixrsimply chose another (either because it had more information about the underlying problem or randomly).
– Nico Schertler
Nov 26 '18 at 17:08
add a comment |
There is an infinite number of rotation matrices with a given z-axis. You chose one of them. Whatever created your reference matrixrsimply chose another (either because it had more information about the underlying problem or randomly).
– Nico Schertler
Nov 26 '18 at 17:08
There is an infinite number of rotation matrices with a given z-axis. You chose one of them. Whatever created your reference matrix
r simply chose another (either because it had more information about the underlying problem or randomly).– Nico Schertler
Nov 26 '18 at 17:08
There is an infinite number of rotation matrices with a given z-axis. You chose one of them. Whatever created your reference matrix
r simply chose another (either because it had more information about the underlying problem or randomly).– Nico Schertler
Nov 26 '18 at 17:08
add a comment |
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There is an infinite number of rotation matrices with a given z-axis. You chose one of them. Whatever created your reference matrix
rsimply chose another (either because it had more information about the underlying problem or randomly).– Nico Schertler
Nov 26 '18 at 17:08