Levai's Gear Configuration 4

Gear Configuration D

flickr:4977934654

Here,

a = 1
f = 2
l = 7

from which, we get

(1)
\begin{align} \frac {\omega_{7/1}}{\omega_{2/1}}& =\frac {\omega_7-\omega_1}{\omega_2-\omega_1}\\ \end{align}
(2)
\begin{split} \frac {\omega_{7/1}}{\omega_{2/1}}& =\left(\frac {N_2}{N_3}\right)\left(\frac {N_3}{N_4}\right)\left(\frac{-N_4}{N_5}\right)\left(\frac{N_5}{N_6}\right)\left(\frac{N_6}{N_7}\right)\\ \end{split}

Gears 3 & 4 in addition to 5 & 6 are located on the same shaft, resulting in a ration of 1

(3)
\begin{align} \frac {\omega_7-\omega_1}{\omega_2-\omega_1}& =\left(\frac {-N_2N_4N_6}{N_3N_5N_7}\right) \end{align}

therefore from (1) and the expression above, we get

(4)
\begin{align} \omega_7 & =\left(\frac {-N_2N_4N_6}{N_3N_5N_7}\right)\left(\omega_2 - \omega_1\right) + \omega_1\\ \end{align}