creating rotation matrices in opencv coordinate system












1















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







share|improve this question























  • 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
















1















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







share|improve this question























  • 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














1












1








1








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







share|improve this question














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






share|improve this question













share|improve this question











share|improve this question




share|improve this question










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 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

















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












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