Derivative of a Jacobian matrix

Question

I have an RRR planar robot:

Its forward kinematics transform is:

$$ {}^{0}T_3 = \\ \left[\begin{array}{cccc} \cos\!\left(\mathrm{\theta_1} + \mathrm{\theta_2} + \mathrm{\theta_3}\right) & - \sin\!\left(\mathrm{\theta_1} + \mathrm{\theta_2} + \mathrm{\theta_3}\right) & 0 & \mathrm{l_2}\, \cos\!\left(\mathrm{\theta_1} + \mathrm{\theta_2}\right) + \mathrm{l_1}\, \cos\!\left(\mathrm{\theta_1}\right) + \mathrm{l_3}\, \cos\!\left(\mathrm{\theta_1} + \mathrm{\theta_2} + \mathrm{\theta_3}\right)\\ \sin\!\left(\mathrm{\theta_1} + \mathrm{\theta_2} + \mathrm{\theta_3}\right) & \cos\!\left(\mathrm{\theta_1} + \mathrm{\theta_2} + \mathrm{\theta_3}\right) & 0 & \mathrm{l_2}\, \sin\!\left(\mathrm{\theta_1} + \mathrm{\theta_2}\right) + \mathrm{l_1}\, \sin\!\left(\mathrm{\theta_1}\right) + \mathrm{l_3}\, \sin\!\left(\mathrm{\theta_1} + \mathrm{\theta_2} + \mathrm{\theta_3}\right)\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right] $$

With the joint parameters $q = \left[\begin{array}{ccc} \mathrm{\theta_1} & \mathrm{\theta_2} & \mathrm{\theta_3} \end{array}\right]^T$ and the end-effector position $X = \left[\begin{array}{ccc} \mathrm{x} & \mathrm{y} & \mathrm{\theta} \end{array}\right]^T$. $\theta$ is constrained to 0. (Also the image is misleading, $\theta$ is actually $\theta_1 + \theta_2 + \theta_3$.)

The jacobian matrix is

$$ J = \\ \left[\begin{array}{ccc} - \mathrm{l_2}\, \sin\!\left(\mathrm{\theta_1} + \mathrm{\theta_2}\right) - \mathrm{l_1}\, \sin\!\left(\mathrm{\theta_1}\right) - \mathrm{l_3}\, \sin\!\left(\mathrm{\theta_1} + \mathrm{\theta_2} + \mathrm{\theta_3}\right) & - \mathrm{l_2}\, \sin\!\left(\mathrm{\theta_1} + \mathrm{\theta_2}\right) - \mathrm{l_3}\, \sin\!\left(\mathrm{\theta_1} + \mathrm{\theta_2} + \mathrm{\theta_3}\right) & - \mathrm{l_3}\, \sin\!\left(\mathrm{\theta_1} + \mathrm{\theta_2} + \mathrm{\theta_3}\right)\\ \mathrm{l_2}\, \cos\!\left(\mathrm{\theta_1} + \mathrm{\theta_2}\right) + \mathrm{l_1}\, \cos\!\left(\mathrm{\theta_1}\right) + \mathrm{l_3}\, \cos\!\left(\mathrm{\theta_1} + \mathrm{\theta_2} + \mathrm{\theta_3}\right) & \mathrm{l_2}\, \cos\!\left(\mathrm{\theta_1} + \mathrm{\theta_2}\right) + \mathrm{l_3}\, \cos\!\left(\mathrm{\theta_1} + \mathrm{\theta_2} + \mathrm{\theta_3}\right) & \mathrm{l_3}\, \cos\!\left(\mathrm{\theta_1} + \mathrm{\theta_2} + \mathrm{\theta_3}\right)\\ 1 & 1 & 1 \end{array}\right] $$

I'm trying to find $\ddot{q}$:

$$ [\ddot{q}] = J^{-1}(q) \cdot \left([\ddot{X}] - \dot{J}(q) \cdot J^{-1}(q) \cdot [\dot{X}] \right) $$

My question is: how can I find $\dot{J}$? What is it?

score 2 · Accepted Answer · answered Nov 06 '16 at 01:49

2

Look at example 283 and its derivation here. You just take the time derivative of each element of J, paying particular attention to the chain rule.

answered Nov 06 '16 at 01:49

SteveO

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I used to use Matlab to compute it. It is messy. – CroCo Nov 07 '16 at 01:28
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@SteveO could you update or correct the link, it gives 404 error. – 50k4 Feb 28 '19 at 13:16

score 2 · Answer 2 · answered Nov 07 '16 at 02:04

2

You can use Matlab to compute $\dot{J}$. For example, to compute $\dot{J}_{11}$, we can use the following code

clear all
clc
syms T1 T2 T3 l1 l2 l3 t

T1(t) = symfun(sym('T1(t)'), t);
T2(t) = symfun(sym('T2(t)'), t);
T3(t) = symfun(sym('T3(t)'), t);

J11 = -l2*sin(T1(t)+T2(t)) - l1*sin(T1(t)) -l3*sin(T1(t)+T2(t)+T3(t));

dJ11dt = diff(J11,t)

which yields

- l3*cos(T1(t) + T2(t) + T3(t))*(diff(T1(t), t) + diff(T2(t), t) + diff(T3(t), t)) - l1*cos(T1(t))*diff(T1(t), t) - l2*cos(T1(t) + T2(t))*(diff(T1(t), t) + diff(T2(t), t))

Or

$$ \dot{J}_{11} = -l_3\cos(\theta_1 + \theta_2 + \theta_3)(\dot{\theta}_1+\dot{\theta}_2+\dot{\theta}_3) - l_1\cos(\theta_1) \dot{\theta}_1 - l_2\cos(\theta_1+\theta_2)(\dot{\theta}_1+\dot{\theta}_2) $$

Now it is easy to compute the rest.

answered Nov 07 '16 at 02:04

CroCo

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This is similar to what I did:
`j_p = sym('j_p', [3 3]);

% for each row

for a=1:3

% for each column

for b=1:3
```
   sum = sym(0);  

   % for each ddl  

   for i=1:3  

       sum = sum + (diff(j(a,b),q(i)))*q_p(i);  

   end  

   j_p(a,b) = sum;  
```
end

end`

(Edit: Hmm that's not very readable. Here)
– christophebedard Nov 07 '16 at 02:18
@ChrisVolkoff, so what is exactly your question? – CroCo Nov 07 '16 at 02:33
I mean that's what I ended up doing after @SteveO's answer. – christophebedard Nov 07 '16 at 02:34

Derivative of a Jacobian matrix

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