Useful Notes and Equations
For a comprehensive and thorough summary of the theory, check MuChenSun’s wonderful note here.
My own notes:
Useful Equations:
Textbook Exercises Attempts
Exercise 4.2 The RRRP SCARA robot of Figure 4.12 is shown in its zero position. Determine the end-effector zero position configuration M, the screw axes \(\mathscr{S}_{i}\) in {0}, and the screw axes \(\mathscr{B}_{i}\) in {b}. For \(l_{0} = l_{1} = l_{2} = 1\) and the joint variable values \(\theta = (0, \frac{\pi}{2}, -\frac{\pi}{2},1)\), use both the FKinSpace and the FKinBody functions to find the end-effector configuration \(T\in \text{SE}(3)\). Confirm that they agree with each other.
%% Exercise 4.2
% function T = FKinSpace(M, Slist, thetalist)
M = [1 0 0 0;
0 1 0 2;
0 0 1 1;
0 0 0 1];
Slist = [[0; 0; 1; 0; 0; 0], ...
[0; 0; 1; 1; 0; 0], ...
[0; 0; 1; 2; 0; 0], ...
[0; 0; 0; 0; 0; 1]];
Blist = [[0; 0; 1;-2; 0; 0], ...
[0; 0; 1;-1; 0; 0], ...
[0; 0; 1; 0; 0; 0], ...
[0; 0; 0; 0; 0; 1]];
thetalist =[0; pi/2; -pi/2; 1];
T_Slist = FKinSpace(M, Slist, thetalist);
T_Blist = FKinBody(M, Blist, thetalist);
============================================
>> T_Slist
T_Slist =
1 0 0 -1
0 1 0 1
0 0 1 2
0 0 0 1
>> T_Blist
T_Blist =
1 0 0 -1
0 1 0 1
0 0 1 2
0 0 0 1
Exercise 4.4 to 4.6 are already attempted in my notes
Exercise 4.7 The PRRRRR spatial open chain of Figure 4.13 is shown in its zero position. Determine the end-effector zero position configuration M, the screw axes \(\mathscr{S}_{i}\) in {0}, and the screw axes \(\mathscr{B}_{i}\) in {b}.
Exercise 4.8 The spatial RRRRPR open chain of Figure 4.14 is shown in its zero position, with fixed and end-effector frames chosen as indicated. Determine the end-effector zero position configuration M, the screw axes \(\mathscr{S}_{i}\) in {0}, and the screw axes \(\mathscr{B}_{i}\) in {b}.
Exercise 4.9 The spatial RRPPRR open chain of Figure 4.15 is shown in its zero position. Determine the end-effector zero position configuration M, the screw axes \(\mathscr{S}_{i}\) in {0}, and the screw axes \(\mathscr{B}_{i}\) in {b}.
Exercise 4.10 The URRPR spatial open chain of Figure 4.16 is shown in its zero position. Determine the end-effector zero position configuration M, the screw axes \(\mathscr{S}_{i}\) in {0}, and the screw axes \(\mathscr{B}_{i}\) in {b}.
Exercise 4.11 The spatial RPRRR open chain of Figure 4.17 is shown in its zero position. Determine the end-effector zero position configuration M, the screw axes \(\mathscr{S}_{i}\) in {0}, and the screw axes \(\mathscr{B}_{i}\) in {b}.
Exercise 4.12 The RRPRRR spatial open chain of Figure 4.18 is shown in its zero position (all joints lie on the same plane). Determine the end-effector zero position configuration M, the screw axes \(\mathscr{S}_{i}\) in {0}, and the screw axes \(\mathscr{B}_{i}\) in {b}. Setting \(\theta_{5}=\pi\) and all other joint variables to zero, find \(T_{06}\) and \(T_{60}\).
Setting \(\theta_{5}=\pi\) and all other joint variables to zero, we have \(\begin{align} \begin{split} T_{06} &= e^{[\mathscr{S}_{1}]\theta_{1}} e^{[\mathscr{S}_{2}]\theta_{2}} e^{[\mathscr{S}_{3}]\theta_{3}} e^{[\mathscr{S}_{4}]\theta_{4}} e^{[\mathscr{S}_{5}]\theta_{5}} M\\ &= I I I I e^{[\mathscr{S}_{5}]\theta_{5}} M\\ &= \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 0 &-1 & 5L \\ 0 &-1 & 0 & 5L \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & L \\ 0 & 1 & 0 & (4+\sqrt{2}) L \\ 0 & 0 & 1 & -\sqrt{2} L \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \\ \end{split} \end{align}\)
%% Exercise 4.12
e_s5_theta5 = [-1 0 0 0;
0 0 -1 5;
0 -1 0 5;
0 0 0 1];
M = [1 0 0 1;
0 1 0 4+sqrt(2);
0 0 1 -sqrt(2);
0 0 0 1];
T_06 = e_s5_theta5 * M;
T_60 = inv(T_06);
==========================
>> T_06
T_06 =
-1.0000 0 0 -1.0000
0 0 -1.0000 6.4142
0 -1.0000 0 -0.4142
0 0 0 1.0000
>> T_60
T_60 =
-1.0000 0 0 -1.0000
0 0 -1.0000 -0.4142
0 -1.0000 0 6.4142
0 0 0 1.0000
Exercise 4.13 The spatial RRRPRR open chain of Figure 4.19 is shown in its zero position. Determine the end-effector zero position configuration M, the screw axes \(\mathscr{S}_{i}\) in {0}, and the screw axes \(\mathscr{B}_{i}\) in {b}.
Exercise 4.14 The RPH robot of Figure 4.20 is shown in its zero position. Determine the end-effector zero position configuration M, the screw axes \(\mathscr{S}_{i}\) in {s}, and the screw axes \(\mathscr{B}_{i}\) in {b}. Use both the FKinSpace and the FKinBody functions to find the end-effector configuration \(T\in \text{SE}(3)\) when \(\theta = (\pi/2, 3, \pi)\). Confirm that the results agree.
%% Exercise 4.14
Slist = [[0; 0; 1; 4; 0; 0], ...
[0; 0; 0; 0; 1; 0], ...
[0; 0;-1;-6; 0; -0.1]];
M = [-1 0 0 0;
0 1 0 6;
0 0 -1 2;
0 0 0 1];
thetalist = [pi/2; 3; pi];
Blist = [[0; 0;-1; 2; 0; 0], ...
[0; 0; 0; 0; 1; 0], ...
[0; 0; 1;-6; 0; 0.1]];
T_03_s = FKinSpace(M, Slist, thetalist);
T_03_b = FKinBody(M, Blist, thetalist);
=========================================
>> T_03_s
T_03_s =
-0.0000 1.0000 0 -5.0000
1.0000 0.0000 0 4.0000
0 0 -1.0000 1.6858
0 0 0 1.0000
>> T_03_b
T_03_b =
-0.0000 1.0000 0 7.0000
1.0000 0.0000 0 4.0000
0 0 -1.0000 1.6858
0 0 0 1.0000
Question! Still don’t know why \(T_{03,s}\) and \(T_{03,b}\) aren’t equal.
Exercise 4.17 Figure 4.22 shows a snake robot with end-effectors at each end. Reference frames {\(b_{1}\)} and {\(b_{2}\)} are attached to the two end-effectors, as shown.
- (a) Suppose that end-effector 1 is grasping a tree (which can be thought of as “ground”) and end-effector 2 is free to move. Assume that the robot is in its zero configuration. Then \(T_{b_{1}b_{2}} \in \text{SE}(3)\) can be expressed in the following product of exponentials form: \(T_{b_{1}b_{2}} = e^{\mathscr{S}_{1}\theta_{1}} e^{\mathscr{S}_{2}\theta_{2}} \dots e^{\mathscr{S}_{5}\theta_{5}} M\). Find \(S_{3}\), \(S_{5}\), and \(M\).
- (b) Don’t know how to do.