3 Added more steps, removed extraneous comment about lifted actions. edited Jan 29 '16 at 23:34 RLH 13633 bronze badges The Jacobian in that equation is from the joint velocity to the "spatial velocity" of the end effector. The spatial velocity of an object is a somewhat unintuitive concept: it is the velocity of a frame rigidly attached to the end effector but currently coincident with the origin frame. It may help to think of the rigid body as extending to cover the whole space, and you're measuring the velocity by standing at the origin and looking at what is happening to the end effector where you are, instead of at the end of the arm. In the example given, the origin is at the same location as the first joint. This means that rotating the joint will sweep the end effector in a circle, but that the frame at the origin will only pivot around the vertical axis; this is encoded by the first column of the Jacobian being [0 0 0 0 0 1]'. Rotating joints two and three will pull the origin-overlapping frame away from the origin, and hence have translational components. Edit:To see how this works in action, note that the Jacobian that you asked about from MLS can be simplified to $$J_{st,SCARA}^{s} = \begin{bmatrix} 0 & \phantom{-}y_{2} & \phantom{-}y_{3} & 0\\ 0 & -x_{2} & - x_{3} & 0 \\ 0 &0&0&1 \\ 0& 0& 0&0\\ 0& 0& 0&0\\ 1&1&1 &0 \end{bmatrix},$$ in which the first three columns encode the velocity at the origin of an object rotating around the corresponding axis and the fourth column gives the velocity at the origin of an object sliding up the fourth axis. You can convert the spatial-velocity Jacobian into a world-frame Jacobian by incorporating the Jacobian from base-frame motion to motion at the end effector's current position and orientation. For the SCARA arm, this works out fairly cleanly, with the only difference between the two frame velocities being the "cross product" term that accounts for the extra motion of the end effector sweeping around the base, $$J_{\text{world, SCARA}} = \begin{bmatrix} % \begin{pmatrix} 1 &&\\ &1&\\ && 1 \end{pmatrix} % & % \begin{pmatrix} 0 & & -y\\ & 0 & \phantom{-}x \\ & & \phantom{-}0 \end{pmatrix} \\ \begin{pmatrix} 0&& \\ &0&\\ &&0 \end{pmatrix} & \begin{pmatrix} 1 & & \\ & 1 & \\ && 1 \end{pmatrix} \end{bmatrix} J_{st,\text{SCARA}}$$$$J^{w}_{st, SCARA} = \begin{bmatrix} % \begin{pmatrix} 1 &&\\ &1&\\ && 1 \end{pmatrix} % & % \begin{pmatrix} 0 & & -y_{4}\\ & 0 & \phantom{-}x_{4} \\ & & \phantom{-}0 \end{pmatrix} \\ \begin{pmatrix} 0&& \\ &0&\\ &&0 \end{pmatrix} & \begin{pmatrix} 1 & & \\ & 1 & \\ && 1 \end{pmatrix} \end{bmatrix} J^{s}_{st,\text{SCARA}}.$$ This matrix is often called the "Jacobianproduct evaluates to $$J_{st,SCARA}^{w} = \begin{bmatrix} -y_{4} & -(y_{4}-y_{2}) & -(y_{4}-y_{3}) & 0\\ \phantom{-}x_{4} & \phantom{-}x_{4}-x_{2} & \phantom{-}x_{4}- x_{3} & 0 \\ 0 &0&0&1 \\ 0& 0& 0&0\\ 0& 0& 0&0\\ 1&1&1 &0 \end{bmatrix},$$ which matches what we would expect to see: Each of the right group action" orrotary joints contributes to the "Right lifted action"end effector velocity by the cross product between its rotational velocity and the vector from the joint to the end effector (note that $$x_{4}-x_{3}$$ and $$y_{4}-y_{3}$$ are both zero). If you had rotation === In the general case, where the rotations are not only around other axesthe $$z$$ axis, you would see correspondingwant to use the full form of the matrix mapping between the spatial to world Jacobians, $$J_{w} = \begin{bmatrix} % \begin{pmatrix} 1 &&\\ &1&\\ && 1 \end{pmatrix} % & % \begin{pmatrix} \phantom{-}0 & \phantom{-}z & -y\\ -z& \phantom{-}0 & \phantom{-}x \\ \phantom{-}y& -x & \phantom{-}0 \end{pmatrix} \\ \begin{pmatrix} 0&& \\ &0&\\ &&0 \end{pmatrix} & \begin{pmatrix} 1 & & \\ & 1 & \\ && 1 \end{pmatrix} \end{bmatrix} J^{s},$$ which encodes the cross-product product terms infor all three rotation axes when the upper-right blockend effector is at $$(x,y,z)$$ relative to the base frame. The Jacobian in that equation is from the joint velocity to the "spatial velocity" of the end effector. The spatial velocity of an object is a somewhat unintuitive concept: it is the velocity of a frame rigidly attached to the end effector but currently coincident with the origin frame. It may help to think of the rigid body as extending to cover the whole space, and you're measuring the velocity by standing at the origin and looking at what is happening to the end effector where you are, instead of at the end of the arm. In the example given, the origin is at the same location as the first joint. This means that rotating the joint will sweep the end effector in a circle, but that the frame at the origin will only pivot around the vertical axis; this is encoded by the first column of the Jacobian being [0 0 0 0 0 1]'. Rotating joints two and three will pull the origin-overlapping frame away from the origin, and hence have translational components. Edit: You can convert the spatial-velocity Jacobian into a world-frame Jacobian by incorporating the Jacobian from base-frame motion to motion at the end effector's current position and orientation. For the SCARA arm, this works out fairly cleanly, with the only difference between the two frame velocities being the "cross product" term that accounts for the extra motion of the end effector sweeping around the base, $$J_{\text{world, SCARA}} = \begin{bmatrix} % \begin{pmatrix} 1 &&\\ &1&\\ && 1 \end{pmatrix} % & % \begin{pmatrix} 0 & & -y\\ & 0 & \phantom{-}x \\ & & \phantom{-}0 \end{pmatrix} \\ \begin{pmatrix} 0&& \\ &0&\\ &&0 \end{pmatrix} & \begin{pmatrix} 1 & & \\ & 1 & \\ && 1 \end{pmatrix} \end{bmatrix} J_{st,\text{SCARA}}$$ This matrix is often called the "Jacobian of the right group action" or the "Right lifted action". If you had rotation around other axes, you would see corresponding cross-product terms in the upper-right block. The Jacobian in that equation is from the joint velocity to the "spatial velocity" of the end effector. The spatial velocity of an object is a somewhat unintuitive concept: it is the velocity of a frame rigidly attached to the end effector but currently coincident with the origin frame. It may help to think of the rigid body as extending to cover the whole space, and you're measuring the velocity by standing at the origin and looking at what is happening to the end effector where you are, instead of at the end of the arm. In the example given, the origin is at the same location as the first joint. This means that rotating the joint will sweep the end effector in a circle, but that the frame at the origin will only pivot around the vertical axis; this is encoded by the first column of the Jacobian being [0 0 0 0 0 1]'. Rotating joints two and three will pull the origin-overlapping frame away from the origin, and hence have translational components. To see how this works in action, note that the Jacobian that you asked about from MLS can be simplified to $$J_{st,SCARA}^{s} = \begin{bmatrix} 0 & \phantom{-}y_{2} & \phantom{-}y_{3} & 0\\ 0 & -x_{2} & - x_{3} & 0 \\ 0 &0&0&1 \\ 0& 0& 0&0\\ 0& 0& 0&0\\ 1&1&1 &0 \end{bmatrix},$$ in which the first three columns encode the velocity at the origin of an object rotating around the corresponding axis and the fourth column gives the velocity at the origin of an object sliding up the fourth axis. You can convert the spatial-velocity Jacobian into a world-frame Jacobian by incorporating the Jacobian from base-frame motion to motion at the end effector's current position and orientation. For the SCARA arm, this works out fairly cleanly, with the only difference between the two frame velocities being the "cross product" term that accounts for the extra motion of the end effector sweeping around the base, $$J^{w}_{st, SCARA} = \begin{bmatrix} % \begin{pmatrix} 1 &&\\ &1&\\ && 1 \end{pmatrix} % & % \begin{pmatrix} 0 & & -y_{4}\\ & 0 & \phantom{-}x_{4} \\ & & \phantom{-}0 \end{pmatrix} \\ \begin{pmatrix} 0&& \\ &0&\\ &&0 \end{pmatrix} & \begin{pmatrix} 1 & & \\ & 1 & \\ && 1 \end{pmatrix} \end{bmatrix} J^{s}_{st,\text{SCARA}}.$$ This product evaluates to $$J_{st,SCARA}^{w} = \begin{bmatrix} -y_{4} & -(y_{4}-y_{2}) & -(y_{4}-y_{3}) & 0\\ \phantom{-}x_{4} & \phantom{-}x_{4}-x_{2} & \phantom{-}x_{4}- x_{3} & 0 \\ 0 &0&0&1 \\ 0& 0& 0&0\\ 0& 0& 0&0\\ 1&1&1 &0 \end{bmatrix},$$ which matches what we would expect to see: Each of the rotary joints contributes to the end effector velocity by the cross product between its rotational velocity and the vector from the joint to the end effector (note that $$x_{4}-x_{3}$$ and $$y_{4}-y_{3}$$ are both zero). === In the general case, where the rotations are not only around the $$z$$ axis, you would want to use the full form of the matrix mapping between the spatial to world Jacobians, $$J_{w} = \begin{bmatrix} % \begin{pmatrix} 1 &&\\ &1&\\ && 1 \end{pmatrix} % & % \begin{pmatrix} \phantom{-}0 & \phantom{-}z & -y\\ -z& \phantom{-}0 & \phantom{-}x \\ \phantom{-}y& -x & \phantom{-}0 \end{pmatrix} \\ \begin{pmatrix} 0&& \\ &0&\\ &&0 \end{pmatrix} & \begin{pmatrix} 1 & & \\ & 1 & \\ && 1 \end{pmatrix} \end{bmatrix} J^{s},$$ which encodes the cross product terms for all three rotation axes when the end effector is at $$(x,y,z)$$ relative to the base frame. 2 added 977 characters in body edited Jan 28 '16 at 21:13 RLH 13633 bronze badges The Jacobian in that equation is from the joint velocity to the "spatial velocity" of the end effector. The spatial velocity of an object is a somewhat unintuitive concept: it is the velocity of a frame rigidly attached to the end effector but currently coincident with the origin frame. It may help to think of the rigid body as extending to cover the whole space, and you're measuring the velocity by standing at the origin and looking at what is happening to the end effector where you are, instead of at the end of the arm. In the example given, the origin is at the same location as the first joint. This means that rotating the joint will sweep the end effector in a circle, but that the frame at the origin will only pivot around the vertical axis; this is encoded by the first column of the Jacobian being [0 0 0 0 0 1]'. Rotating joints two and three will pull the origin-overlapping frame away from the origin, and hence have translational components. Edit: You can convert the spatial-velocity Jacobian into a world-frame Jacobian by incorporating the Jacobian from base-frame motion to motion at the end effector's current position and orientation. For the SCARA arm, this works out fairly cleanly, with the only difference between the two frame velocities being the "cross product" term that accounts for the extra motion of the end effector sweeping around the base, $$J_{\text{world, SCARA}} = \begin{bmatrix} % \begin{pmatrix} 1 &&\\ &1&\\ && 1 \end{pmatrix} % & % \begin{pmatrix} 0 & & -y\\ & 0 & \phantom{-}x \\ & & \phantom{-}0 \end{pmatrix} \\ \begin{pmatrix} 0&& \\ &0&\\ &&0 \end{pmatrix} & \begin{pmatrix} 1 & & \\ & 1 & \\ && 1 \end{pmatrix} \end{bmatrix} J_{st,\text{SCARA}}$$ This matrix is often called the "Jacobian of the right group action" or the "Right lifted action". If you had rotation around other axes, you would see corresponding cross-product terms in the upper-right block. The Jacobian in that equation is from the joint velocity to the "spatial velocity" of the end effector. The spatial velocity of an object is a somewhat unintuitive concept: it is the velocity of a frame rigidly attached to the end effector but currently coincident with the origin frame. It may help to think of the rigid body as extending to cover the whole space, and you're measuring the velocity by standing at the origin and looking at what is happening to the end effector where you are, instead of at the end of the arm. In the example given, the origin is at the same location as the first joint. This means that rotating the joint will sweep the end effector in a circle, but that the frame at the origin will only pivot around the vertical axis; this is encoded by the first column of the Jacobian being [0 0 0 0 0 1]'. Rotating joints two and three will pull the origin-overlapping frame away from the origin, and hence have translational components. The Jacobian in that equation is from the joint velocity to the "spatial velocity" of the end effector. The spatial velocity of an object is a somewhat unintuitive concept: it is the velocity of a frame rigidly attached to the end effector but currently coincident with the origin frame. It may help to think of the rigid body as extending to cover the whole space, and you're measuring the velocity by standing at the origin and looking at what is happening to the end effector where you are, instead of at the end of the arm. In the example given, the origin is at the same location as the first joint. This means that rotating the joint will sweep the end effector in a circle, but that the frame at the origin will only pivot around the vertical axis; this is encoded by the first column of the Jacobian being [0 0 0 0 0 1]'. Rotating joints two and three will pull the origin-overlapping frame away from the origin, and hence have translational components. Edit: You can convert the spatial-velocity Jacobian into a world-frame Jacobian by incorporating the Jacobian from base-frame motion to motion at the end effector's current position and orientation. For the SCARA arm, this works out fairly cleanly, with the only difference between the two frame velocities being the "cross product" term that accounts for the extra motion of the end effector sweeping around the base, $$J_{\text{world, SCARA}} = \begin{bmatrix} % \begin{pmatrix} 1 &&\\ &1&\\ && 1 \end{pmatrix} % & % \begin{pmatrix} 0 & & -y\\ & 0 & \phantom{-}x \\ & & \phantom{-}0 \end{pmatrix} \\ \begin{pmatrix} 0&& \\ &0&\\ &&0 \end{pmatrix} & \begin{pmatrix} 1 & & \\ & 1 & \\ && 1 \end{pmatrix} \end{bmatrix} J_{st,\text{SCARA}}$$ This matrix is often called the "Jacobian of the right group action" or the "Right lifted action". If you had rotation around other axes, you would see corresponding cross-product terms in the upper-right block. 1 answered Jan 27 '16 at 6:59 RLH 13633 bronze badges The Jacobian in that equation is from the joint velocity to the "spatial velocity" of the end effector. The spatial velocity of an object is a somewhat unintuitive concept: it is the velocity of a frame rigidly attached to the end effector but currently coincident with the origin frame. It may help to think of the rigid body as extending to cover the whole space, and you're measuring the velocity by standing at the origin and looking at what is happening to the end effector where you are, instead of at the end of the arm. In the example given, the origin is at the same location as the first joint. This means that rotating the joint will sweep the end effector in a circle, but that the frame at the origin will only pivot around the vertical axis; this is encoded by the first column of the Jacobian being [0 0 0 0 0 1]'. Rotating joints two and three will pull the origin-overlapping frame away from the origin, and hence have translational components.