Robotics Kinematics and Dynamics/Serial Manipulator Dynamics

Acceleration of a Rigid Body
The linear and angular accelerations are the time derivatives of the linear and angular velocity vectors at any instant:

_a \dot{v} = \dfrac{d \, _a v}{dt} = \lim_{\Delta t \rightarrow 0} \dfrac{_a v(t + \Delta t) - \,_a v(t)}{\Delta t} $$, and:

_a \dot{\omega} = \dfrac{d \, _a \omega}{dt} = \lim_{\Delta t \rightarrow 0} \dfrac{_a \omega(t + \Delta t) - \,_a \omega(t)}{\Delta t} $$

The linear velocity, as seen from a reference frame $$\{a\}$$, of a vector $$q$$, relative to frame $$\{b\}$$ of which the origin coincides with $$\{a\}$$, is given by:

_a v_q = \, ^b _a R \, _b v_q + \, _a \omega_b \times \, ^b _a R \, _b q $$

Differentiating the above expression gives the acceleration of the vector $$q$$:

_a \dot{v}_q = \dfrac{d}{dt} \, ^b _a R \, _b v_q + \, _a \dot{\omega}_b \times \, _a ^b R \, _b q + \, _a \omega_b \times \dfrac{d}{dt} \, ^b _a R \, _b q $$

The equation for the linear velocity may also be written as:

_a v_q = \dfrac{d}{dt} \, ^b _a R \, _b q = \, ^b _a R \, _b v_q + \, _a \omega_b \times \, ^b _a R \, _b q $$

Applying this result to the acceleration leads to:

_a \dot{v}_q = ^b _a R \, _b \dot{v}_q + \, _a \omega_b \times \, ^b _a R \, _b v_q + \, _a \dot{\omega}_b \times \, _a ^b R \, _b q + \, _a \omega_b \times \left( ^b _a R \, _b v_q + \, _a \omega_b \times \, ^b _a R \, _b q \right) $$

In the case the origins of $$\{a\}$$ and $$\{b\}$$ do not coincide, a term for the linear acceleration of $$\{b\}$$, with respect to $$\{a\}$$, is added:

_a \dot{v}_q = \, _a \dot{v}_{b,org} + \, ^b _a R \, _b \dot{v}_q + \, _a \omega_b \times \, ^b _a R \, _b v_q + \, _a \dot{\omega}_b \times \, _a ^b R \, _b q + \, _a \omega_b \times \left( ^b _a R \, _b v_q + \, _a \omega_b \times \, ^b _a R \, _b q \right) $$

For rotational joints, $$_b q$$ is constant, and the above expression simplifies to:

_a \dot{v}_q = \, _a \dot{v}_{b,org} + \, _a \dot{\omega}_b \times \, _a ^b R \, _b q + \, _a \omega_b \times \left( _a \omega_b \times \, ^b _a R \, _b q \right) $$

The angular velocity of a frame $$\{c\}$$, rotating relative to a frame $$\{b\}$$, which in itself is rotating relative to the reference frame $$\{a\}$$, with respect to $$\{a\}$$, is given by:

_a \omega_c = \, _a \omega_b + \, _a ^b R \, _b \omega_c $$

Differentiating leads to:

_a \dot{\omega}_c = \, _a \dot{\omega}_b + \dfrac{d}{dt} \, _a ^b R \, _b \omega_c $$

Replacing the last term with one of the expressions derived earlier:

_a \dot{\omega}_c = \, _a \dot{\omega}_b + \, _a \omega_b \times \, _a ^b R \, _b \omega_c $$

Inertia Tensor
The inertia tensor can be thought of as a generalization of the scalar moment of inertia:

_a I = \begin{pmatrix} I_{xx} & -I_{xy} & -I_{xz} \\ I_{xy} & I_{yy} & -I_{yz} \\ I_{xz} & -I_{yz} & I_{zz} \\ \end{pmatrix} $$

Newton's and Euler's equation
The force $$F$$, acting at the center of mass of a rigid body with total mass$$m$$, causing an acceleration $$\dot{v}_{com}$$, equals:

F = m \dot{v}_{com} $$

In a similar way, the moment $$N$$, causing an angular acceleration $$\dot{\omega}$$, is given by:

N = \, _c I \dot{\omega} + \omega \times \, _c I \omega $$, where $$_c I$$ is the inertia tensor, expressed in a frame $$\{c\}$$ of which the origin coincides with the center of mass of the rigid body.