Robotics Kinematics Review: Fast Version for Everything

Notes for ME5421

Force of one joint

$$ \begin{aligned} & \sum F = 0 \\ & i.e.\ f_i - f_{i+1} = 0 \\ & \sum T_{\text{torques about origin i-1}} = 0 \\ & i.e.\ n_i - n_{i+1} + (p_i - p_{i-1}) \times - f_{i+1} = 0 \end{aligned} $$ Another view: $$ \begin{aligned} f_i &= f_{i+1} \\ n_i &= n_{i+1} + (p_i - p_{i-1}) \times f_{i+1} \end{aligned} $$ If robot arm is rigid, the force needed in the joint is shown. Where z indicate the direction and n or f indicate the force. $$ \begin{aligned} T_i &= n_i^T z_{i-1} \text{ (rotational)} \\ T_i &= f_i^T z_{i-1} \text{ (translational)} \end{aligned} $$

Force and Jacobian

$\tau$ is the force the every joint. shape: (n, 1) $$ \begin{aligned} \tau &= J^T F \\ \end{aligned} $$ Singularity: J in not full rank.

Force transformation

Let $^AF_B$ be the force and moment experienced at B, $^AF_C$ be the force and moment excerted on C. B C are linked by a rigid body. Then: $$ \begin{aligned} ^Af_B &= {}^Af_C \\ ^An_B &= {}^An_C + (^Ap_B - {}^Ap_C) \times {}^Af_C \\ \begin{bmatrix} {}^Af_B \\ ^An_B \end{bmatrix} &= \begin{bmatrix} I & 0 \\ \hat{{}^AR_B{}^Bp_C} & I \end{bmatrix} \begin{bmatrix} {}^Af_C \\ {}^An_C \end{bmatrix} \\ \\ \begin{bmatrix} {}^Bf_B \\ ^Bn_B \end{bmatrix} &= \begin{bmatrix} {}^BR_C & 0 \\ {{}^Bp_C} \times {}^BR_C & {}^BR_C \end{bmatrix} \begin{bmatrix} {}^Cf_C \\ {}^Cn_C \end{bmatrix} \\ \end{aligned} $$ > Note: $ \begin{bmatrix} {}^Af_C \\ {}^An_C \end{bmatrix} = \begin{bmatrix} {}^AR_C & 0 \\ 0 & {}^AR_C \end{bmatrix} \begin{bmatrix} {}^Cf_C \\ {}^Cn_C \end{bmatrix} \\\\ $