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Given DH parameters of a serial manipulator, the forward kinematics gives me

T0n = T01*T12*T23*...*Tn-1n , n is the DOF of manipulator.

I can calculate the Jacobian matrix J for the end-effector by taking first order partial directives.

Then I know the relation between joint speed and end-effector speed

eeSpeed = J*jointSpeed or jointSpeed = J_inv*eeSpeed

Now my question is how to calculate the Jacobian for joints 1 to n-1, and how to derive the relationship between joint speed and speed of any point on a robot link?

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2 Answers 2

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Well, the answer is quite straight forward. With the given equation $ T^n_0 = \ldots $ you calculate the transformation from the $n^\mathrm{th}$ coordinate frame (attached to the end effector) to the inertial coordinate frame.

Just imagine you want to know the Jacobian from the joint $n-1$, you do the exact same procedure. Calculate $T^{n-1}_0 = \ldots $ and calculate the first partial derivative.

You can do that for each coordinate frame, or point you wish for.

In other words, the Jacobian method is just the use of the chain rule from differentiation. Given the translational velocity $v=\frac{\mathrm{d}}{\mathrm{d}t}p$ of any point $p$. Any point $p$, expressed in the inertial frame depends on the joint configuration $q(t)$, so $p(q(t))$. Therefore you can apply the chain rule $$ \frac{\mathrm{d}}{\mathrm{d}t}p = \frac{\partial}{\partial q} p \frac{\mathrm{d}}{\mathrm{d}t}q ~.$$ The partial derivative $\frac{\partial}{\partial q} p $ is called the Jacobian of the point $p$. If $p$ describes the end effector position, the Jacobian is called the manipulator jacobian.

Note: $p$, $v$ and $q$ are usually vectors, therefore the partial derivative yields a matrix.

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I believe you refer to first order partial derivatives. It is as easy as to take the forward kinematics from T0n , where n is the point to where you want to take the partial jacobian matrix and do the same of the first order partial derivatives.

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