Tp15

Austin's Test Page

Gear Configuration 1

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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}

This is the angular velocity of the Ring gear in terms of the angular velocity of the Arm and the angular velocity of the Sun gear for the configuration shown.