Mixing
calculating new 3D position on sphere with angular velocity vector
$begingroup$
I feel like this is actually pretty simple but still could not find any solutions so far...
I'm trying to calculate the movement of a point in a rigid rod with the equation
$ dot P = [ v + omega times (P - P_{center} ) ] $ ,
with $ v $ being the translational velocity of the center of the rod $P_{center}$, $ omega $ the rotational/angular velocity around it and $P$ the position of the point. I derive both velocities from a force and momentum equilibrium ( $ F=m cdot a $ and $ M = J cdot omega $ ).
Neglecting the transitional movement and focusing on the rotation, my question is how to get $P_1$ in a 3D environment after a given time step $Delta t$ when all vectors are given and in a mutual cartesian coordinate system (x,y,z), especially $omega$.
Since my simulation is time discrete I cannot simply multiply $dot P_1$ with $Delta t$ (which would result in an elongation of the rod). Hence I want to calculate the new position analytically and then reassign the new position to my point. I read a lot about Eulerian angles, but the angles I get from integrating $omega$ refer to a rotation around fixed axes x,y and z and cannot be used for the Eulerian method (...I think).
Thanks in advance for any help.
vectors 3d rotations spheres angle
asked Jan 14 at 18:49
BlueFireBlueFire
133
$endgroup$
add a comment |
$begingroup$
I feel like this is actually pretty simple but still could not find any solutions so far...
I'm trying to calculate the movement of a point in a rigid rod with the equation
$ dot P = [ v + omega times (P - P_{center} ) ] $ ,
with $ v $ being the translational velocity of the center of the rod $P_{center}$, $ omega $ the rotational/angular velocity around it and $P$ the position of the point. I derive both velocities from a force and momentum equilibrium ( $ F=m cdot a $ and $ M = J cdot omega $ ).
Neglecting the transitional movement and focusing on the rotation, my question is how to get $P_1$ in a 3D environment after a given time step $Delta t$ when all vectors are given and in a mutual cartesian coordinate system (x,y,z), especially $omega$.
Since my simulation is time discrete I cannot simply multiply $dot P_1$ with $Delta t$ (which would result in an elongation of the rod). Hence I want to calculate the new position analytically and then reassign the new position to my point. I read a lot about Eulerian angles, but the angles I get from integrating $omega$ refer to a rotation around fixed axes x,y and z and cannot be used for the Eulerian method (...I think).
Thanks in advance for any help.
vectors 3d rotations spheres angle
asked Jan 14 at 18:49
BlueFireBlueFire
133
$endgroup$
add a comment |
$begingroup$
I feel like this is actually pretty simple but still could not find any solutions so far...
I'm trying to calculate the movement of a point in a rigid rod with the equation
$ dot P = [ v + omega times (P - P_{center} ) ] $ ,
with $ v $ being the translational velocity of the center of the rod $P_{center}$, $ omega $ the rotational/angular velocity around it and $P$ the position of the point. I derive both velocities from a force and momentum equilibrium ( $ F=m cdot a $ and $ M = J cdot omega $ ).
Neglecting the transitional movement and focusing on the rotation, my question is how to get $P_1$ in a 3D environment after a given time step $Delta t$ when all vectors are given and in a mutual cartesian coordinate system (x,y,z), especially $omega$.
Since my simulation is time discrete I cannot simply multiply $dot P_1$ with $Delta t$ (which would result in an elongation of the rod). Hence I want to calculate the new position analytically and then reassign the new position to my point. I read a lot about Eulerian angles, but the angles I get from integrating $omega$ refer to a rotation around fixed axes x,y and z and cannot be used for the Eulerian method (...I think).
Thanks in advance for any help.
vectors 3d rotations spheres angle
asked Jan 14 at 18:49
BlueFireBlueFire
133
$endgroup$
I feel like this is actually pretty simple but still could not find any solutions so far...
I'm trying to calculate the movement of a point in a rigid rod with the equation
$ dot P = [ v + omega times (P - P_{center} ) ] $ ,
with $ v $ being the translational velocity of the center of the rod $P_{center}$, $ omega $ the rotational/angular velocity around it and $P$ the position of the point. I derive both velocities from a force and momentum equilibrium ( $ F=m cdot a $ and $ M = J cdot omega $ ).
Neglecting the transitional movement and focusing on the rotation, my question is how to get $P_1$ in a 3D environment after a given time step $Delta t$ when all vectors are given and in a mutual cartesian coordinate system (x,y,z), especially $omega$.
Since my simulation is time discrete I cannot simply multiply $dot P_1$ with $Delta t$ (which would result in an elongation of the rod). Hence I want to calculate the new position analytically and then reassign the new position to my point. I read a lot about Eulerian angles, but the angles I get from integrating $omega$ refer to a rotation around fixed axes x,y and z and cannot be used for the Eulerian method (...I think).
Thanks in advance for any help.
vectors 3d rotations spheres angle
vectors 3d rotations spheres angle
asked Jan 14 at 18:49
BlueFireBlueFire
133
asked Jan 14 at 18:49
BlueFireBlueFire
133
asked Jan 14 at 18:49
BlueFireBlueFire
133
asked Jan 14 at 18:49
BlueFireBlueFire
133
asked Jan 14 at 18:49
BlueFireBlueFire
133
133
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Roughly speaking, you would have to use a procedure as follows. First you need to define the current orientation of your rod (i.e. body) and represent it in some form (quaternions are preferred). Assuming that you choose quaternions you can call the current orientation $q_0$. The problem then is to determine the next orientation after $Delta t$, which we will call $q_1$. The basic update rule is as follows:
1) Get the quaternion derivative as $dot{q} = frac{1}{2}q_0 omega$ where $omega$ is in body-fixed coordinates. Also $omega$ is a quaternion with a real component as 0, and the vector part equal to the components of the angular velocity. In other words $omega = [0,, omega_xmathbf{i},, omega_ymathbf{j},, omega_zmathbf{k}]$
2) Update the quaternion using $q_1 = q_0 + dot{q}Delta t$. (or use some other more sophisticated integration technique, i.e. Runge-Kutta).
3) Finally normalize the result $q_1 leftarrow frac{q_1}{|q_1|}$
answered Jan 15 at 2:51
TpofofnTpofofn
3,6171427
$endgroup$
$begingroup$
I unfortunately have not worked with quaternions yet. I dug a little into it though after your answer and am sure your way works. But since the other solution sofar works too, I spared the trouble trying to implement it. Thank you anyway, and I will get back to this as soon as my code starts crashing again :-P
$endgroup$
– BlueFire
Jan 17 at 11:22
add a comment |
$begingroup$
...it took some time but I found a simple answer :-)
Knowing the angular velocity is sufficient: It points in the direction of the rotation axis and its magnitude multiplied with the time step reveals the angle by which the particle has been rotated around said axis during the timestep. With these two parameters I can calculate a rotation matrix that projects my particle onto its new position. Mathematically speaking:
=> $vert vec omega vert cdot Delta t = theta $ and $ frac{vec omega}{vert vec omega vert} = vec n $
=> the conversion into a matrix can be found on the Wikipedia article on rotation matrices: $ R = Matrix(vec n, theta) $
=> $P_1 = R cdot P_0 $
answered Jan 17 at 11:19
BlueFireBlueFire
133
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Roughly speaking, you would have to use a procedure as follows. First you need to define the current orientation of your rod (i.e. body) and represent it in some form (quaternions are preferred). Assuming that you choose quaternions you can call the current orientation $q_0$. The problem then is to determine the next orientation after $Delta t$, which we will call $q_1$. The basic update rule is as follows:
1) Get the quaternion derivative as $dot{q} = frac{1}{2}q_0 omega$ where $omega$ is in body-fixed coordinates. Also $omega$ is a quaternion with a real component as 0, and the vector part equal to the components of the angular velocity. In other words $omega = [0,, omega_xmathbf{i},, omega_ymathbf{j},, omega_zmathbf{k}]$
2) Update the quaternion using $q_1 = q_0 + dot{q}Delta t$. (or use some other more sophisticated integration technique, i.e. Runge-Kutta).
3) Finally normalize the result $q_1 leftarrow frac{q_1}{|q_1|}$
answered Jan 15 at 2:51
TpofofnTpofofn
3,6171427
$endgroup$
$begingroup$
I unfortunately have not worked with quaternions yet. I dug a little into it though after your answer and am sure your way works. But since the other solution sofar works too, I spared the trouble trying to implement it. Thank you anyway, and I will get back to this as soon as my code starts crashing again :-P
$endgroup$
– BlueFire
Jan 17 at 11:22
add a comment |
$begingroup$
Roughly speaking, you would have to use a procedure as follows. First you need to define the current orientation of your rod (i.e. body) and represent it in some form (quaternions are preferred). Assuming that you choose quaternions you can call the current orientation $q_0$. The problem then is to determine the next orientation after $Delta t$, which we will call $q_1$. The basic update rule is as follows:
1) Get the quaternion derivative as $dot{q} = frac{1}{2}q_0 omega$ where $omega$ is in body-fixed coordinates. Also $omega$ is a quaternion with a real component as 0, and the vector part equal to the components of the angular velocity. In other words $omega = [0,, omega_xmathbf{i},, omega_ymathbf{j},, omega_zmathbf{k}]$
2) Update the quaternion using $q_1 = q_0 + dot{q}Delta t$. (or use some other more sophisticated integration technique, i.e. Runge-Kutta).
3) Finally normalize the result $q_1 leftarrow frac{q_1}{|q_1|}$
answered Jan 15 at 2:51
TpofofnTpofofn
3,6171427
$endgroup$
$begingroup$
I unfortunately have not worked with quaternions yet. I dug a little into it though after your answer and am sure your way works. But since the other solution sofar works too, I spared the trouble trying to implement it. Thank you anyway, and I will get back to this as soon as my code starts crashing again :-P
$endgroup$
– BlueFire
Jan 17 at 11:22
add a comment |
$begingroup$
Roughly speaking, you would have to use a procedure as follows. First you need to define the current orientation of your rod (i.e. body) and represent it in some form (quaternions are preferred). Assuming that you choose quaternions you can call the current orientation $q_0$. The problem then is to determine the next orientation after $Delta t$, which we will call $q_1$. The basic update rule is as follows:
1) Get the quaternion derivative as $dot{q} = frac{1}{2}q_0 omega$ where $omega$ is in body-fixed coordinates. Also $omega$ is a quaternion with a real component as 0, and the vector part equal to the components of the angular velocity. In other words $omega = [0,, omega_xmathbf{i},, omega_ymathbf{j},, omega_zmathbf{k}]$
2) Update the quaternion using $q_1 = q_0 + dot{q}Delta t$. (or use some other more sophisticated integration technique, i.e. Runge-Kutta).
3) Finally normalize the result $q_1 leftarrow frac{q_1}{|q_1|}$
answered Jan 15 at 2:51
TpofofnTpofofn
3,6171427
$endgroup$
Roughly speaking, you would have to use a procedure as follows. First you need to define the current orientation of your rod (i.e. body) and represent it in some form (quaternions are preferred). Assuming that you choose quaternions you can call the current orientation $q_0$. The problem then is to determine the next orientation after $Delta t$, which we will call $q_1$. The basic update rule is as follows:
1) Get the quaternion derivative as $dot{q} = frac{1}{2}q_0 omega$ where $omega$ is in body-fixed coordinates. Also $omega$ is a quaternion with a real component as 0, and the vector part equal to the components of the angular velocity. In other words $omega = [0,, omega_xmathbf{i},, omega_ymathbf{j},, omega_zmathbf{k}]$
2) Update the quaternion using $q_1 = q_0 + dot{q}Delta t$. (or use some other more sophisticated integration technique, i.e. Runge-Kutta).
3) Finally normalize the result $q_1 leftarrow frac{q_1}{|q_1|}$
answered Jan 15 at 2:51
TpofofnTpofofn
3,6171427
answered Jan 15 at 2:51
TpofofnTpofofn
3,6171427
answered Jan 15 at 2:51
TpofofnTpofofn
3,6171427
answered Jan 15 at 2:51
TpofofnTpofofn
3,6171427
3,6171427
$begingroup$
I unfortunately have not worked with quaternions yet. I dug a little into it though after your answer and am sure your way works. But since the other solution sofar works too, I spared the trouble trying to implement it. Thank you anyway, and I will get back to this as soon as my code starts crashing again :-P
$endgroup$
– BlueFire
Jan 17 at 11:22
add a comment |
$begingroup$
I unfortunately have not worked with quaternions yet. I dug a little into it though after your answer and am sure your way works. But since the other solution sofar works too, I spared the trouble trying to implement it. Thank you anyway, and I will get back to this as soon as my code starts crashing again :-P
$endgroup$
– BlueFire
Jan 17 at 11:22
$begingroup$
I unfortunately have not worked with quaternions yet. I dug a little into it though after your answer and am sure your way works. But since the other solution sofar works too, I spared the trouble trying to implement it. Thank you anyway, and I will get back to this as soon as my code starts crashing again :-P
$endgroup$
– BlueFire
Jan 17 at 11:22
$begingroup$
I unfortunately have not worked with quaternions yet. I dug a little into it though after your answer and am sure your way works. But since the other solution sofar works too, I spared the trouble trying to implement it. Thank you anyway, and I will get back to this as soon as my code starts crashing again :-P
$endgroup$
– BlueFire
Jan 17 at 11:22
add a comment |
$begingroup$
...it took some time but I found a simple answer :-)
Knowing the angular velocity is sufficient: It points in the direction of the rotation axis and its magnitude multiplied with the time step reveals the angle by which the particle has been rotated around said axis during the timestep. With these two parameters I can calculate a rotation matrix that projects my particle onto its new position. Mathematically speaking:
=> $vert vec omega vert cdot Delta t = theta $ and $ frac{vec omega}{vert vec omega vert} = vec n $
=> the conversion into a matrix can be found on the Wikipedia article on rotation matrices: $ R = Matrix(vec n, theta) $
=> $P_1 = R cdot P_0 $
answered Jan 17 at 11:19
BlueFireBlueFire
133
$endgroup$
add a comment |
$begingroup$
...it took some time but I found a simple answer :-)
Knowing the angular velocity is sufficient: It points in the direction of the rotation axis and its magnitude multiplied with the time step reveals the angle by which the particle has been rotated around said axis during the timestep. With these two parameters I can calculate a rotation matrix that projects my particle onto its new position. Mathematically speaking:
=> $vert vec omega vert cdot Delta t = theta $ and $ frac{vec omega}{vert vec omega vert} = vec n $
=> the conversion into a matrix can be found on the Wikipedia article on rotation matrices: $ R = Matrix(vec n, theta) $
=> $P_1 = R cdot P_0 $
answered Jan 17 at 11:19
BlueFireBlueFire
133
$endgroup$
add a comment |
$begingroup$
...it took some time but I found a simple answer :-)
Knowing the angular velocity is sufficient: It points in the direction of the rotation axis and its magnitude multiplied with the time step reveals the angle by which the particle has been rotated around said axis during the timestep. With these two parameters I can calculate a rotation matrix that projects my particle onto its new position. Mathematically speaking:
=> $vert vec omega vert cdot Delta t = theta $ and $ frac{vec omega}{vert vec omega vert} = vec n $
=> the conversion into a matrix can be found on the Wikipedia article on rotation matrices: $ R = Matrix(vec n, theta) $
=> $P_1 = R cdot P_0 $
answered Jan 17 at 11:19
BlueFireBlueFire
133
$endgroup$
...it took some time but I found a simple answer :-)
Knowing the angular velocity is sufficient: It points in the direction of the rotation axis and its magnitude multiplied with the time step reveals the angle by which the particle has been rotated around said axis during the timestep. With these two parameters I can calculate a rotation matrix that projects my particle onto its new position. Mathematically speaking:
=> $vert vec omega vert cdot Delta t = theta $ and $ frac{vec omega}{vert vec omega vert} = vec n $
=> the conversion into a matrix can be found on the Wikipedia article on rotation matrices: $ R = Matrix(vec n, theta) $
=> $P_1 = R cdot P_0 $
answered Jan 17 at 11:19
BlueFireBlueFire
133
answered Jan 17 at 11:19
BlueFireBlueFire
133
answered Jan 17 at 11:19
BlueFireBlueFire
133
answered Jan 17 at 11:19
BlueFireBlueFire
133
133
add a comment |
add a comment |
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