Page Comparison

...

Image Modified

...

Derivation

This derivation is based off equation 2:

$a e^{j\theta_2} + b e^{j\theta_3} - c_x - c_y = 0$ Image Modified

...

(Eq. 2)

Where the derivative with respect to time is:

$aj\omega_2e^{j\theta_2} + bj\omega_3e^{j\theta_3} - \dot{c}_x = 0$ Image Modified

...

(Eq. 9)

The claw linkage velocities are similarly the derivative of equation 4:

$de^{j\theta_4} - fe^{j\theta_5} - c_x - c_y = 0$ Image Modified

...

(Eq. 4)

$dj\omega_4e^{j\theta_4} - fj\omega_5e^{j\theta_5} - \dot{c}_x = 0$ Image Modified

...

(Eq. 10)

I again use an equation solver on equation 9 to find the velocity of the slider, 'c_x, and then use the equation solver on equation 10 to find ω

...

₄ and ω_5.

Again, because the rest of the system is parallel to links 4 and 5, the end effectors have the same angular velocities.

Image Removed

...

Using the values from Table 1 and ω₂ = 1 rad/s