3rd Gear Solution

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

First equation,

f = 5
a = 3
l = 7

from which, we get

(1)
\begin{align} \frac {\omega_{7/3}}{\omega_{5/3}}& =\frac {\omega_7-\omega_3}{\omega_5-\omega_3}\\ \end{align}
(2)
\begin{split} \frac {\omega_{7/3}}{\omega_{5/3}}& =\left(\frac {N_8}{N_7}\right)\left(\frac {-N_5}{N_8}\right)\\ & =\left(\frac {-N_5}{N_7}\right)\\ \end{split}

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

(3)
\begin{align} \omega_7 & =\left(\frac {-N_5}{N_7}\right)\left(\omega_5 - \omega_3\right) + \omega_3\\ \end{align}

simplifying, we get

(4)
\begin{align} \omega_7 & =\left(\frac {-N_5}{N_7}\right)\omega_5\\ + \left( 1 + \frac{N_5}{N_7}\right)\omega_3\\ \end{align}

Second equation,

f = 6
a = 7
l = 3

from which, we get

(5)
\begin{align} \frac {\omega_{6/7}}{\omega_{3/7}}& =\frac {\omega_6-\omega_7}{\omega_3-\omega_7}\\ \end{align}
(6)
\begin{split} \frac {\omega_{6/7}}{\omega_{3/7}}& =\left(\frac {-N_2}{N_6}\right)\left(\frac {N_3}{N_2}\right)\\ & =\left(\frac {-N_3}{N_6}\right)\\ \end{split}

therefore from (5) and the expression above, we get

(7)
\begin{align} \omega_7 & =\left(\frac {N_3}{N_6}\right)\left(\omega_3 - \omega_7\right) + \omega_6\\ \end{align}

because $\omega_6$ is fixed, it equals 0, and simplifying, we get

(8)
\begin{align} \omega_7 & =\left(\frac{1}{\left(\frac {N_6}{N_3}\right) + 1}\right)\omega_3\\ \end{align}

Now we have two equations for $\omega_7$, (4) and (8), which we need to set equal to one another to solve for$\omega_3$.

(9)
\begin{align} \left(\frac {-N_5}{N_7}\right)\omega_5\\ + \left( 1 + \frac{N_5}{N_7}\right)\omega_3\\ & =\left(\frac{1}{\left(\frac {N_6}{N_3}\right) + 1}\right)\omega_3\\ \end{align}
(10)
\begin{align} \omega_3 & =\frac{\left(\frac {N_5}{N_7}\right)\omega_5}{\left( 1 + \frac{N_5}{N_7}\right)-\left(\frac{1}{\left(\frac {N_6}{N_3}\right) + 1}\right)} \end{align}

Now we can use $\omega_3$ to solve for $\omega_4$, which is the output of the transmission.

Lastly, equation three,

f = 5
a = 4
l = 3

from which, we get

(11)
\begin{align} \frac {\omega_{3/4}}{\omega_{5/4}}& =\frac {\omega_3-\omega_4}{\omega_5-\omega_4}\\ \end{align}
(12)
\begin{split} \frac {\omega_{3/4}}{\omega_{5/4}}& =\left(\frac {N_1}{N_3}\right)\left(\frac {-N_5}{N_3}\right)\\ & =\left(\frac {-N_5}{N_3}\right)\\ \end{split}

therefore from (11) and the expression above, we get

(13)
\begin{align} \omega_4 & =\left(\frac {N_5}{N_3}\right)\left(\omega_5 - \omega_4\right) + \omega_3\\ \end{align}

simplifying, we get

(14)
\begin{align} \omega_4 & =\left(\frac{1}{\left(\frac {N_3}{N_5}\right) + 1}\right)\omega_5\\ + \left(\frac{1}{\left(\frac {N_5}{N_3}\right) + 1}\right)\omega_3\\ \end{align}
page revision: 34, last edited: 09 Sep 2010 17:53