Module 2-23: Spatial Jacobians - Examples Part 2

Added: 2026-05-25

[00:00:00]hi i'm scott knockbee in this video we're going to look at finding the jacobian for 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 cross products so in this example we're going to use the same manipulator that we did for the previous video where we were doing joint directions and derivatives to find the jacobian so we're just going to examine the same robot but this time we're going to use joint directions and cross products to find our jacobian so recall this is a 6 degree of freedom robot it's layout is r parallel r parallel r parallel p perpendicular r perpendicular r all right we're choosing that the tool frame is equivalent to frame six we have our zero displacement diagram and we have our modified dnh parameters based on that zero displacement diagram again here are our homogeneous transforms that we find by just taking our dna primers and plugging them into the homogeneous transform equation and like before we need to find the forward displacement solution so we have t02 t03 zero four zero five and then finally zero six which is the full forward displacement solution for that that robot now for the joint directions and cross product method we previously showed that this is the form of the jacobian so if the i joint is revolute we take the joint direction zi cross the vector from the origin of frame i to the origin of our end effector frame all right and then the bottom is just our joint direction if the jth joint is prismatic we just need the joint direction okay and for this example just like the previous one we're going to use that the reference frame is the base frame all right but we could we could choose other frames but we're going to do the same reference frame for this example all right so in order to use this method for our revolute joints we need to know this r vector so let's look at that in a bit more detail all right so we require the position vector terms of our base frame from the origin of frame i to the origin of the end effector frame all right and what we know is we know the position vector of from our base frame to our end effector we know that from our forward displacement solution as well we know the vectors from the base frame to each frame i all right and that's from our our transforms t01 t02 t03 and so on okay so we can use those two vectors to actually calculate r so let's draw this out so say we have frame 0 here we have frame i here and we have our end effector frame over here so we know this vector all right that's just p from the origin of the base frame to the origin of frame i all right and as well we know this vector which is p from the origin of the base frame to the origin of the end effector frame and we're trying to find this frame or sorry this vector which is r from the origin of frame i to the origin of the end effector frame again these are all in terms of our our base frame so in this case our vector r is just simply going to be our vector p from the base frame to the end effector frame minus p from the base frame to frame i so we can use this to calculate our r vector for each of our revolute joints all right we're just going to take the overall position of the end effector with respect to the base frame minus the position of frame i with respect to the base frame okay so let's start looking at each joint in turn so joint one is a revolute joint so the first thing we need to do is we need to calculate r from the origin of frame one to the order in the end effector frame all right so that's just going to be p from the origin of our base frame to the origin of our end effector frame minus p from the origin of our base frame to the origin of frame one so we get this vector from our forward displacement solution so if we jump back this is that vector here so we can make a note of that that this is p from the origin of the base frame to the origin at the end of vector frame in terms of frame zero so we're going to keep using that one so we'll write that in and you'll see we have c1l1 plus c12l2 plus c123 s5l tool and then we have s1 l1 that's s12 l2 plus s123 s5l tool and then we have minus c5 ltool minus d4 all right so that here is this vector here now we need to go look at transform t01 to find this position vector so we jump to that t01 and you can see the position vector is zero all right so we know that this is zero so for this case our r vector is just equal to our p vector all right so now that we have our r vector we need to do our cross product of our joint direction said one cross r okay z1 in this case we're going to get from t01 so let's go to t01 there's z1 so it's going to be 0 0 1.

[00:08:45]cross our r vector i'm not going to write that all out okay but if you do the cross product you'll get minus s1 l1 minus s12 l12 and minus s123 s5 l tool you'll then get c1 l1 plus c12 l2 plus c123 s5l tool and you'll get zero so we have the two pieces of information we need we have z cross r here so we need that that's the first part of our jacobian for that column and then our z direction which i'm not going to rewrite again but it's right here all right so we have all the information we have need for joint one let's look at joint let's look at joint two so here we're gonna have r02 now frame two to the end effector frame it's going to equal our p vector from the base frame to our end effector frame minus our p vector from the base frame to frame two all right this is the same p vector we just used before i'm not going to keep writing that out because it's a big mess but we know that this one we're going to get from t02 so let's go to t02 all right and there's our position vector there so we can put that in so this is going to be minus l1 c1 l1 s1 and 0.

[00:11:06]so we subtract that and we get l2 c12 plus actually let's just write that c12 l2 just to be consistent plus c123 s5 l tool then we'll have s12 l2 plus s123 s5 l tool and we'll have minus c5 ltool minus d4 okay we now need to do our cross product so we need to do z2 across our r vector z2 we're going to get from t02 let's go to ct02 and there's that two there in blue okay zero zero one so we're gonna have zero zero one for z2 our r vector and again i won't write it out but it's right above here it's right here and if you do that cross product you're going to get minus s12 l2 minus s123 s5 ltool you get c12 l2 plus c123 s5 l2 and zero all right and so those are the two parts of the revelate joint for the jacobian we need we need our cross product here result and we need our joint direction let's look at joint three which is also a revolute joint so we need r03 to the end effector which is going to be p from the origin to the base end effector frame from the base frame to the end effector frame minus p from the origin of the base frame to the origin of frame three so we're gonna again use this vector we already know we need to go get this vector so that's gonna come from t03 so if we go to t03 here that's our position vector so we can go put that in so it's going to be c1 l1 plus c12 l2 it's going to be s1 l1 plus s12 l2 n0 subtract those two vectors and you're left with c123 s5l tool s123 s5l tool and minus c5 l tool minus d4 all right now we can do our cross product which in this case will be z3 cross r03 to the end effector we get z3 from t03 all right so there's z3 here in blue zero zero one again so zero zero one across our r vector all right and if you do that cross product you'll get minus s s123 s5l tool you get c123 s5 ltool and you get zero all right so that's joint three done joint four is prismatic which is pretty easy all we need is the joint direction so we just need z4 expected the base frame so we're gonna get that from t04 all right and that's our joint direction there zero zero minus one so let's just go write that down so we have it all right two more joints to go and they're both revolut so we need to find our r vector first so r05 the vector from frame five to the end effector frame it's gonna be the vector from our base frame to the end effector frame minus the vector from the base frame to frame five okay so we need to go get this from sorry we need to get this from t05 so this is our position okay so we can write that out it's going to be minus c1 l1 plus c12 l2 it's gonna be s1 l1 plus s12 l2 and then it's gonna be minus d4 so we subtract those two vectors and we're left with c123 s5 ltool s123 s5 ltool and minus c5 ltool okay we can do our cross product now which is z5 across our r vector let's add five we're going to get from t05 all right so here's our z5 joint direction so we can write that out it's going to be s123 minus c123 and zero cross r all right and if you do that cross product you'll get c123 c5l tool there you get c123 c5l tool you get s123 c5 l tool and you get s5 ltool okay and now we need one more joint joint six it's our revolute joint as well so r06 to the end effector frame it's going to be p0 from the base frame to the end effector frame minus p from the base frame to frame six well these are the same vector because our end effector frame is frame six for this example so this product will be zero our cross product all right it's also going to be 0 but let's just write in z6 so we have that information so we can complete our jacobian all right so z6 we get from t06 so this is z6 here so it would be c123 s5 s123 s5 minus c5 across the r vector which we just said was zero so obviously that cross product is going to be zero all right but now we have all the information we have the jacobian calculated all right so this is for joint six this is the the first three elements and the z is the the second three elements same for joint five these are the first three elements of that column and these are the second three and so on so if you were to then write this out in matrix form this will be your answer okay and remember each column is one joint so this is joint 1 joint 2 joint 3 joint 4 which is our prismatic joint joint five and joint six and if you compare this answer to the example from the the answer from the previous example where we use cross pro sorry joint directions and derivatives you see we get the same answer which is what you would expect all right and keep in mind again that this jacobian is what we use for our forward velocity solution okay where we have v it's equal to our jacobian times our vector joint rates q dot as well as our inverse force problem where torques are is equal to j transpose f all right so just a different way of finding our jacobian you can use either method all right and you will get the same answer