Levai's Gear Configuration 3

# Gear Configuration 3

Here,

*f = 2*

*l = 7*

*a = 1*

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 as well as 5 & 6 are on the same shaft, therefore their ratio is 1, giving us

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

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

(4)\begin{align} \omega_7 & = \omega_1 + \left(\frac{-N_2 N_4 N_6}{N_3 N_5 N_7}\right)\left(\omega_2 - \omega_1\right)\\ \end{align}

page revision: 6, last edited: 11 Sep 2010 15:40