Kapitel V: Velocity Kinematics and Statics

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 5.1 A wheel of unit radius is rolling to the right at a rate of 1 rad/s (see Figure 5.14; the dashed circle shows the wheel at \(t = 0\)).

(a) Find the spatial twist \(\mathscr{V}_{s}(t)\) as a function of \(t\).

(b) Find the linear velocity of the {b}-frame origin expressed in {s}-frame coordinates.

Exercise 5.2 The 3R planar open chain of Figure 5.15(a) is shown in its zero position.

(a) Suppose that the last link must apply a wrench corresponding to a force of 5 N in the \(\hat{x}^{s}\)-direction of the {s} frame, with zero component in the \(\hat{y}_{s}\)-direction and zero moment about the \(\hat{z}_{s}\)-axis. What torques should be applied at each joint?

(b) Suppose that the last link must apply a force of 5 N in the yˆs-direction, with zero components in the other wrench directions. What torques should be applied at each joint?

N.B. in figure part (a), the robot is not in its zero position!!!!

(a) Since \(\tau = J_{s}^{T}(\theta) \cdot \mathscr{F}_{s}\), we need to first find SList and thetalist and then use the function JacobianSpace to find \(J_{s}^{T}(\theta)\). \(\mathscr{F}_{s}\) can be found by observation. The following MATLAB code is self-explanatory.

%% Exercise 5.2 

%  Part (a)
F_s = [0; 0; 0; 5; 0; 0];
Slist = [[0; 0; 1; 0; 0; 0], ...
         [0; 0; 1; 0;-1; 0], ...
         [0; 0; 1; 0;-2; 0]];
thetalist =[0; pi/4; 0];
J_s_T = transpose(JacobianSpace(Slist, thetalist));
tau = J_s_T * F_s;

%  Part (b)
F_s_b = [0; 0; 0; 0; 5; 0];
tau_b = J_s_T * F_s_b;
===================================================
>> tau_a

tau_a =

         0
         0
    3.5355

>> tau_b

tau_b =

         0
   -5.0000
   -8.5355

Exercise 5.3 Answer the following questions for the 4R planar open chain of Figure 5.15(b).

(a) For the forward kinematics of the form \(T(\theta) = e^{[\mathscr{S}_{1}]\theta_{1}} e^{[\mathscr{S}_{2}]\theta_{2}} e^{[\mathscr{S}_{3}]\theta_{3}} e^{[\mathscr{S}_{4}]\theta_{4}} M\) write down \(M \in \text{SE}(2)\) and each \(\mathscr{S}_{i} = (\omega_{zi}, v_{xi}, v_{yi}) \in \mathbb{R}^{3}\).

(b) Write down the body Jacobian.

(c) Suppose that the chain is in static equilibrium at the configuration \(\theta_{1}=\theta_{2}=0, \theta_{3}=\pi/2, \theta_{4}=-\pi/2\) and that a force \(f = (10,10,0)\) and a moment \(m = (0,0,10)\) are applied to the tip (both f and m are expressed with respect to the fixed frame). What are the torques experienced at each joint?

(d) Under the same conditions as (c), suppose that a force \(f = (-10,10,0)\) and a moment \(m = (0,0,-10)\), also expressed in the fixed frame, are applied to the tip. What are the torques experienced at each joint?

(e) Find all kinematic singularities for this chain.

(a) \(M = \begin{bmatrix} 1 & 0 & L_{1} + L_{2} + L_{3} + L_{4} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \in \text{SE}(2)\), \(\mathscr{S} = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & -L_{1} & -(L_{1}+L_{2}) & -(L_{1}+L_{2}+L_{3}) \\ \end{bmatrix} \in \mathbb{R}^{3}\).
(b) I am too lazy to do that by hand. One of the online quizzes of this Coursera MOOC actually gives you the answer…
Because I am using MATLAB to do this course, and MATLAB is quite impotent on symbolic manipulation, part (c) and (d) are skipped.
(e) When \(\theta_{2}=\theta_{3}=\theta_{4}=n\pi, \,\text{where}\,n\in \mathbb{Z}^{+}\).

Exercise 5.7 The RRP robot in Figure 5.19 is shown in its zero position.

(a) Write down the screw axes in the space frame. Evaluate the forward kinematics when \(\theta = (90^{\circ}, 90^{\circ}, 1)\). Hand-draw or use a computer to show the arm and the end-e↵ector frame in this configuration. Obtain the space Jacobian \(J_{s}\) for this configuration.

(b) Write down the screw axes in the end-effector body frame. Evaluate the forward kinematics when \(\theta = (90^{\circ}, 90^{\circ}, 1)\) and confirm that you get the same result as in part (a). Obtain the body Jacobian \(J_{b}\) for this configuration.

See the code below:

%% Exercise 5.7

% Part (a)
SList = [[0; 0; 1; 0; 0; 0], ...
         [1; 0; 0; 0; 2; 0], ...
         [0; 0; 0; 0; 1; 0]];
thetalist = [pi/2; pi/2; 1];
M = [-1 0 0 0;
      0 0 1 3;
      0 1 0 2;
      0 0 0 1];
% T = FKinSpace(M, Slist, thetalist)
T_sb = FKinSpace(M, SList, thetalist);
% Js = JacobianSpace(Slist, thetalist)
J_s = JacobianSpace(SList, thetalist);

=======================================

>> T_sb

T_sb =

   -0.0000    1.0000   -0.0000   -0.0000
   -1.0000   -0.0000    0.0000    0.0000
         0    0.0000    1.0000    6.0000
         0         0         0    1.0000

>> J_s

J_s =

         0    0.0000         0
         0    1.0000         0
    1.0000         0         0
         0   -2.0000   -0.0000
         0    0.0000    0.0000
         0         0    1.0000
      
=======================================

% Part (b)
BList = [[0; 1; 0; 3; 0; 0], ...
        [-1; 0; 0; 0; 3; 0], ...
         [0; 0; 0; 0; 0; 1]];
T_bs = FKinBody(M, BList, thetalist);
J_b = JacobianBody(BList, thetalist);

=======================================

>> T_bs

T_bs =

   -0.0000    1.0000   -0.0000   -0.0000
   -1.0000   -0.0000    0.0000    0.0000
         0    0.0000    1.0000    6.0000
         0         0         0    1.0000

>> J_b

J_b =

         0   -1.0000         0
    0.0000         0         0
    1.0000         0         0
    0.0000         0         0
         0    4.0000         0
         0         0    1.0000

Exercise 5.26 The kinematics of the 7R WAM robot are given in Section 4.1.3.

(a) Give the numerical body Jacobian \(J_{b}\) when all joint angles are \(\pi/2\). Separate the Jacobian matrix into an angular-velocity portion \(J_{\omega}\) (the joint rates act on the angular velocity) and a linear-velocity portion \(J_{v}\) (the joint rates act on the linear velocity).

(b) For this configuration, calculate the directions and lengths of the principal semi-axes of the three-dimensional angular-velocity manipulability ellipsoid (based on \(J_{\omega}\)) and the directions and lengths of the principal semi-axes of the three-dimensional linear-velocity manipulability ellipsoid (based on \(J_{v}\)).

(a) By observation (also on Coursera quizzes), we have

\[J_{\omega} = \begin{bmatrix} 0 & -1 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 1 \\ \end{bmatrix}\] \[J_{v} = \begin{bmatrix} -0.105 & 0 & 0.006 & -0.045 & 0 & 0.006 & 0 \\ -0.889 & 0.006 & 0 & -0.844 & 0.006 & 0 & 0 \\ 0 & -0.105 & 0.889 & 0 & 0 & 0 & 0 \\ \end{bmatrix}\]

(b) See code below

%% Exercise 5.26
% Part (b)
J_b = [0 -1 0 0 -1 0 0;
       0 0 1 0 0 1 0;
       1 0 0 1 0 0 1;
       -0.105 0 0.006 -0.045 0 0.006 0;
       -0.889 0.006 0 -0.844 0.006 0 0;
       0 -0.105 0.889 0 0 0 0];
J_w = J_b(1:3,:);
J_v = J_b(4:6,:);
A_w = J_w * transpose(J_w);
[V_w D_w] = eig(A_w);
A_v = J_v * transpose(J_v);
[V_v D_v] = eig(A_v);

========================================

>> V_w

V_w =

     1     0     0
     0     1     0
     0     0     1

>> D_w

D_w =

     2     0     0
     0     2     0
     0     0     3

The length of the principal semi-axis is the sqrt of the eigenvalue. Therefore the length are \(\sqrt{2},\,\sqrt{2},\,\sqrt{3}\) resepectively. The direction of the semi-axis are V_w as shown above.

>> V_v

V_v =

   -0.9962    0.0067   -0.0872
    0.0872   -0.0004   -0.9962
    0.0067    1.0000    0.0002
    
>> D_v

D_v =

    0.0016         0         0
         0    0.8014         0
         0         0    1.5142

Again, the length are the sqrt of D_v, and the direction are V_v.