Module 2-21: Spatial Jacobians

Added: 2026-05-25

[00:00:00]hi i'm scott nockley in this video we're going to look at jacobians for spatial manipulators so by the end of this video you should be able to understand the two methods for determining the jacobian for spatial manipulator so for planar manipulators we previously showed that our velocity of render factor is equal to the jacobian matrix j here times our vector joint rates as well as our torques is equal to the jacobian transpose times our wrench all right now this is the static force case and we showed that the jacobian each column represents one joint of our robot and to determine each column you take the partial derivative of x with respect to theta i right and so typically for a three joint robot this would be a three by three matrix now for the spatial case our formulas are still the same our velocity is equal to the jacobian times q dot and our torque for the static force case is equal to j transpose times f all right the difference is we can't just use this formulation to find our jacobian because we have 3d rotations all right we're not in the plane so we have to look at a different way to find the jacobian all right and as well we can define our velocity in this case of our end effector has six components three translational in the x y and z direction and three rotational which are the angular velocities about the x y and z axes all right and so we have a translational part and we have an angular part as well our wrench f in this case will have three forces in the x y and z direction as well as three moments about the x y and z axes so we're going to look at how we can find this jacobian for the spatial case and there's two approaches for doing that all right so we have two methods for finding the jacobian for spatial manipulators the first method is using joint directions and derivatives and the second is using joint directions and cross products and you can think of this sort as you know basic mechanics approach so let's look at the formulation for each of these all right so if we're using joint directions and derivatives if we were to draw sort of a schematic of our our joints suppose we have z1 here we have z2 over here and then maybe z3 is over here and then suppose we have a spherical wrist so we have just say z4 z5 z6 and then maybe we have our end effector point here all right now for the angular part in terms of the angular velocity of our end effector which is omega it's simply just going to be the summation from i equals 1 to n or whatever number of joints but in this case we have six joints of omega i our joint rate of joint i times the joint direction zi all right and our translational will just be the partial derivative of our x vector with respect to joint i so our jacobian then becomes of our end effector point with respect to some reference frame and again just like for the planar case each column in the jacobian represents one joint of the robot so if our ith joint is a revolute joint our eighth column of our jacobian will have this form all right so we do the partial derivative of x respect to theta i joint i that gives us our translational components and then here's our joint direction zi for joint i which will give us our angular components all right so we use that for revolute joints now say joint j is a prismatic joint we're going to have no angular component here so this will be a zero vector three by one and we'll just simply have our translational joint direction for the uh the translational part all right so we can make a note here in the above i denotes a revlon joint j denotes a prismatic joint and our vector x ref here is just simply our position vector from the origin of frame 0 to the origin of the end effector frame in terms of whatever reference frame we're using all right so this is one approach and in a subsequent video we will show how you use this method to find the jacobian for a spatial robot now let's look at the the second approach all right so in this approach we're using joint directions and cross products all right so you can think of this as you know just using basic mechanics so in this case we'd have say a coordinate frame we can establish this would be our reference frame and we may have a prismatic joint this will be that j say and we could have a revolute joint zi which has a joint rate of omega i and then maybe we have our end effector off of that point all right and the vector from joint i to the end effector we're going to note as r from the origin of frame i to the origin of our end effector frame all right and that's going to be with respect to our reference frame and we can actually make a note these are all with respect to our reference frame okay so now for a revolute joint our velocity of the end effector point due to our angular velocity it's just simply going to be our angular velocity omega i cross our position vector r all right and now omega i vector is just omega i magnitude times our joint direction and we can write out our position vector so it's from the origin of i to the origin of the end effector frame all right and this is the same omega i is the same as theta i dot okay so that is the translational point or component of our velocity on the end effector due to the a our rotational joints all right our revolute joints also obviously provide an angular velocity which is simply omega i which is omega i component or magnitude times zi which is the same as saying theta i dot zi alright and so these are the translational and angular components due to our revolute joint we can also have a prismatic joint all right and so the translational velocity due to our prismatic joint of our end effector which is d dot j it's just going to be the magnitude of our joint rate our translational joint rate times the joint direction all right and then the angular component due to our translational joint well they don't contribute any angular velocity so it's a zero vector all right so again a revlon joint and j is denoting a prismatic joint in this case so we can write this out in matrix form again each column of our jacobian is one joint of our robot so if our ith joint is revolut we have this formulation that's the translational component and then we have our angular component in the bottom part and then if our joint is sorry our jth joint is prismatic it will just be zj here and we'll have a zero vector in the bottom again let's just make a note where i denotes a revlon joint j denotes a prismatic joint now either formulation either this formulation or this formulation will get you your jacobian right those return the same jacobian all right it's just two different approaches and we'll also look at an example of how we use this method to find the jacobian first spatial manipulator in a subsequent video you