Reverse Gear Solution

To solve 3rd gear, we will need to solve 3 equations.

First equation,

f = 6
a = 7
l = 3

from which, we get

(1)
\begin{align} \frac {\omega_{3/7}}{\omega_{6/7}}& =\frac {\omega_3-\omega_7}{\omega_6-\omega_7}\\ \end{align}
(2)
\begin{split} \frac {\omega_{3/7}}{\omega_{6/7}}& =\left(\frac {-N_6}{N_3}\right)\\ \end{split}

Since $\omega_3$ = 0, (2) Becomes

(3)
\begin{split} \frac {\omega_7}{\omega_{6/7}}& =\left(\frac {N_6}{N_3}\right)\\ \end{split}

simplifying, we get

(4)
\begin{align} \omega_7 & =\left(\frac {N_6\omega_6}{N_3+N_6}\right)\\ \end{align}

Second equation,

f = 7
a = 3
l = 5

Because $\omega_3$ = 0, we get

(5)
\begin{align} \frac {\omega_{5/3}}{\omega_{7/3}}& =\frac {\omega_5}{\omega_7}\\ \end{align}
(6)
\begin{split} \frac {\omega_5}{\omega_{7}}& =\left(\frac {-N_7}{N_5}\right)\\ \end{split}

therefore from (6) we get

(7)
\begin{align} \omega_5 & =\left(\frac {-N_7\omega_7}{N_5}\right) \end{align}

Plugging in $\omega_7$ we get:

(8)
\begin{align} \omega_5 & =\left(\frac{-N_7N_6\omega_6}{N_5\left(N_3+N_6\right) }\right)\\ \end{align}

Lastly, equation three,

f = 5
a = 4
l = 3

Because $\omega_3$ = 0, we get

(9)
\begin{align} \frac {-\omega_4}{\omega_{5/4}}& =\frac {-\omega_4}{\omega_5-\omega_4}\\ \end{align}
(10)
\begin{split} \frac {-\omega_4}{\omega_{5/4}}& =\left(\frac {-N_5}{N_3}\right)\\ \end{split}

therefore from (9) and (10) and the expression above, we get

(11)
\begin{align} \omega_4 & =\left(\frac {N_5}{N_3}\right)\left(\omega_5 - \omega_4\right) \\ \end{align}

simplifying, we get

(12)
\begin{align} \omega_4 & =\left(\frac{N_5}{\left(N_3+N_4\right) }\right)\omega_5\\ \end{align}

Substituting $\omega_5$ = 0, we get:

(13)
\begin{align} \omega_4 & =\left(\frac{N_5}{\left(N_3+N_4\right) }\right)\left(\frac{-N_7N_6\omega_6}{N_5\left(N_3+N_6\right) }\right)\\ \end{align}

Therefore, our solution is:

(14)
\begin{align} \omega_4 & =\left(\frac{-N_7N_6}{\left(N_3+N_4\right)\left(N_3+N_6\right) }\right)\omega_6\\ \end{align}

With $\omega_6$ as our input.