2nd Gear Solution

Second Gear refers to Figure 9-47 in Norton.

In this configuration, clutch C2 is engaged and Brake B2 is also engaged.
The gear train will be solved with 5 as the input and 4 as the output.

There will be two equations for this case. The first, last and arm for each will be defined as

f = 5
l = 7
a = 3

and

f = 5
l = 3
a = 4

The first step is to solve each equation for omega 3.

(1)
\begin{align} \frac {\omega_{7/3}}{\omega_{5/3}}& =\frac {\omega_7-\omega_3}{\omega_5-\omega_3}& = \left(\frac{-N_5}{N_7}\right)\\ \end{align}

from (1)

(2)
\begin{align} (\omega_{7}-\omega_{3})=(\omega_{5}-\omega_{3})&\left(\frac{-N_5}{N_7}\right) \end{align}

simplifying (2), the following expression for omega 3 is obtained

(3)
\begin{align} \omega_{3}=\frac{\omega_{7}+\left(\frac {N_5}{N_7}\right)\omega_{5}}{\left(1-\frac{N_5}{N_7}\right)} \end{align}

but with B2 engaged,

(4)
\begin{align} \omega_{7} = 0 \end{align}

so (3) becomes

(5)
\begin{align} \omega_{3}=\frac{N_5}{N_7+N_5}*\omega_5 \end{align}

Now, solving the second equation for omega 3

(6)
\begin{align} \frac {\omega_{3/4}}{\omega_{5/4}}& =\frac {\omega_3-\omega_4}{\omega_5-\omega_4}& = \left(\frac{-N_5}{N_3}\right)\\ \end{align}

from (4)

(7)
\begin{align} (\omega_{3}-\omega_{4})=(\omega_{5}-\omega_{4})&\left(\frac {-N_5}{N_3}\right) \end{align}

Solving for omega 3,

(8)
\begin{align} \omega_{3}=\omega_{5}&\left(\frac{-N_5}{N_3}\right)+\left(1+\frac{N_5}{N_3}\right)&\omega_{4}\\ \end{align}

Substituting (5) into (8)

(9)
\begin{align} \omega_{3}=\frac{N_5}{N_7+N_5}*\omega_5=-\omega_{5}&\left(\frac{N_5}{N_3}\right)+\left(1+\frac{N_5}{N_3}\right)&\omega_{4}\\ \end{align}

solving for the output…

(10)
\begin{align} \omega_{4}=\frac{\left(\frac{N_5}{N_7+N_6}\right)+\frac{N_5}{N_3}}{\left(1+\frac{N_5}{N_3}\right)}*\omega_5 \end{align}
page revision: 40, last edited: 24 Sep 2010 19:03