derivative of the Geometric Jacobian

Hi,

Is there an efficient and easy way to find the \dot{J} based on the actual Jacobian of a manipulator?

I have

J = geometricJacobian(obj.RigidBodyTreeInternal, q, obj.EndEffectorName); 

Thanks in advance

Answer 1

[Leepakshi]

Hi,

No builtin function is provided in MATLAB for direct derivative of Jacobian , but you can compute (\dot{J}) (the time derivative of the Jacobian) either numerically or symbolically: numerically, use the finite difference method in MATLAB by evaluating the Jacobian at (q) and at (q + \dot{q} \cdot dt) and dividing the difference by (dt):

dt = 1e-6; 
J_now = geometricJacobian(obj.RigidBodyTreeInternal, q, obj.EndEffectorName); 
J_next = geometricJacobian(obj.RigidBodyTreeInternal, q + q_dot*dt, obj.EndEffectorName); 
J_dot = (J_next - J_now) / dt; 

Alternatively, you can use symbolic differentiation for analytical insight, for example:

clear all
clc
syms q1(t) q2(t) q3(t) a2 a3

J11 = -sin(q1(t))*(a3*cos(q2(t) + q3(t)) + a2*cos(q2(t)));
dJ11dt = diff(J11, t); % Symbolic time derivative

This allows you to derive (\dot{J}) analytically and then substitute joint values and velocities as needed. Combining both methods gives you flexibility for both simulation and analysis.

Hope this helps!