Module 2-22: Spatial Jacobians - Examples Part 1

追加: 2026-05-25

[00:00:00]hi i'm scott nockley in this video we're going to look at finding the jacobian for a spatial manipulator so by the end of this video you should be able to derive the jacobian for a spatial manipulator using joint directions and derivatives so for this example we're going to use a spatial robot that's going to have a layout of r parallel r parallel r parallel p which is prismatic perfectly r perpendicular r all right so here's a schematic of the layout and in essence you can see it's a scary robot that has a a pointer group added to it at the end in this example we're going to set our tool frame to be equivalent to frame six all right just to make things a little easier and what we want to find is our jacobian of our end effector with respect to our base frame zero so here's our zero displacement diagram for our that robot and using that zero displacement diagram we can complete our modified dnh parameters as such and you can see because frame six is equivalent to the tool frame we just have a row of zeros here so that homogeneous transform would just be the identity matrix so we don't actually need to calculate that so before we can do or find the jacobian we need to find the forward displacement solution so the first step of that is finding the homogeneous transforms so we have the six of them you just take each row over of our dnh parameters and substitute into the homogeneous transform equation and then for the forward displacement we just multiply each of those matrices out and you'll see actually for the jacobian problem we actually need the joint directions for each joint and we get those from each of these transformers so for joint two we'll get it from t02 from joint three we'll get it from t03 and so on so we need to have all of these so these are the first two here's zero four here's zero five and lastly here's zero six which is our full forward displacement solution now to find the jacobian using joint directions and derivatives right we showed that this is the form that our jacobian takes okay remember each column of the jacobian is one joint all right and if our i joint is a revolute joint our jacobian takes on this form it's the partial derivative of our position vector with respect to the joint variable i that gives us our translational component and then our angular component is just simply the joint direction if the jth joint is prismatic we just need the translational part which is just from the joint direction so it's here and there's no angular component so it's a zero vector here so we use this formulation for revolute joints and we use this formulation for prismatic joints and recall our position vector x here is just going to be our position from our base frame to our end effector frame in terms of whatever reference frame we're using now for this example we're going to use the base frame to be our reference frame all right and so we need our position vector with respect to the base frame and we're going to get that from our forward displacement solution all right there is our position vector of our end of factor with respect to the base frame so let's just highlight that this is our position vector that we're going to use okay so let's write that out on the next page just so we have it so our position vector from our base frame to our end effector frame in terms of our base frame is equal to c1 l1 plus c12 l2 plus c123 s5 ltool that's the x position we have s1 l1 plus s12 l2 plus s123 s5l tool for our x position sorry our y position and then lastly we have our z position which is minus c5 l tool minus d4 okay and just to make a note where did we got that from it came from our forward displacement solution which in this case is from our homogeneous transform t06 okay so now we can start going joint by joint and so the i-th joint is revolut we're going to use this formulation and if the j-th joint is prismatic we're going to use this formulation all right so for joint one we're going to do the partial derivative of our position vector so that vector i just wrote on the previous slide with respect to our joint one variable which is theta one so we need to do the partial derivative of this expression here in terms of theta one and if you do that you're going to get minus s1 l1 minus s12 l2 minus s123 s5l tool we're going to get c1 l1 plus c12 l2 plus c123 s5 l2 and then our last one will be 0 because theta 1 or if you look at our expression here theta 1 there's no theta 1 term in there all right so that's our partial derivative and then the other component we need is we need z1 with respect to our base frame and now where do we get that we go back to our homogeneous transforms here's t01 okay and z1 is this vector here all right that's our z column of our rotation matrix that's our joint direction so we'll just draw that or highlight that in blue here all right so that is our joint direction for joint one so we can write that out and we'll have zero zero one and we can make a note that we got that from t01 all right so that's joint one done we can do the same thing for joint two now all right so we're going to take the partial derivative of our position vector now with respect to theta 2 our second joint variable so we go back to here we're going to do the partial derivative of this now with respect to theta 2.

[00:07:49]and if you do that you're going to get minus s12 l2 minus s 1 2 3 s 5 l tool you're going to get c 1 2 l 2 plus c 1 2 3 s5l tool and again we'll have zero we need z2 now our joint direction with respect to our base frame so where we're going to get that we're going to go to t02 now so let's jump back to that there's t02 here and there is our joint direction so let's highlight that all right and let's go make a note of that on our slide so it's also 0 0 1 which makes sense because joints 1 and 2 are parallel so this is from t02 now all right so that's joint 1 and 2 done and we just keep going so for joint 3 because joint 3 is also revolute we're going to do the partial derivative of our position vector with respect to theta 3 now all right and if you do you're that to get minus s123 s5l tool you get c123 s5l tool you get zero we need join axis 3 now so that's going to be z3 with respect to frame 0. so we need to go to t03 to get that one here's t03 here's our joint direction again it's zero zero one because these first three joints are all parallel to each other all right and so we can go and make a note of that and again where we got them from that was from t 0 3.

[00:10:33]so we're halfway there that's the first three joints done joint 4 all right it's prismatic so all we need is z4 terms our base frame and you guessed it we get that from t04 so let's go to t04 there's t04 and you can see there's our joint direction for that one all right so we can go write that one down and you get zero zero minus one for our joint direction and that's from t04 all right and that's our only prismatic joint the rest are also revolute joints so we're going to use the same formula as we did here except for a different joint so let's just finish that quickly here so for joint 5 it's going to be the partial derivative of our position vector with respect to theta five now if you go do that you're going to get c123 c5l tool you get s123 c5l tool and you're going to get s5l tool we need the z5 joint direction go t05 to get that one all right here's t05 there's our zed our joint direction okay so we can go make a note of that so z5 is s123 minus c123 and zero again we got that from t05 all right and then our final joint is also revolute so we need to do the partial derivative of our position vector respect to our end effector point in terms of theta six if we just jump back and look at our position vector you can see theta 6 doesn't appear in here so the partial derivative is just going to be zero right so we have zero zero and zero all right we still need our joint direction z6 and we get that from t06 all right our forward displacement there's t06 so we need our joint direction which is that so we can write that out we'll have c123 s5 we'll have s123 s5 and we have minus c5 and again we got that from t06 all right so we've actually found the whole jacobian all right each of these is a column all right we just need to now put it in matrix form all right so if you put that all in matrix form i already wrote it out or typed it out this is our jacobian for our robot all right it is a six by six matrix all right because each we have six joints and we have uh so we'll have a column for each joint all right column for each joint joint one joint two joint three joint four which is our prismatic joint joint five and joint six and remember this is the same jacobian that we use for our velocity problem so remember our velocity is just simply our jacobian times our joint rates and then for our static force problem our torques are equal to our jacobian transpose times our wrench okay so this is the jacobian we use in our forward velocity and our inverse force all right just like we did for the the planar case but how we find the jacobian is different