The efficiency of a Weston differential pulley block can be determined using the following equation.
$$E=\frac{FR \ (100)}{MR}$$
As this is the case, in order to increase the efficiency the biggest possible force ratio value is needed and the smallest possible movement ratio is needed. \begin{align} FR & = \frac{Ouput \ Force}{Input \ Force} \\ & = \frac{2}{1-\frac{r_{output}}{r_{input}}} \end{align}
$$MR = \frac{Input \ Movement}{Output \ Movement}$$
To work out the radius of a cog the equation is $r = chain \ pitch \ (\frac{1}{2sin(\frac{\pi}{teeth})})$ but as the chain pitch is constant in a Weston differentail pulley block, we can just use the number of teeth on each cog as the values for $r_{output}$ and $r_{input}$.
With this is mind the I decided to try it out in this simple set of factors.
Input Cog: 25 teeth
Output Cog: 20 teeth
Chain Pitch: 20mm
Input Cog Pulled: 1m
\begin{align} Revolutions & = \frac{1000}{20(25)} \\ & = 2 \end{align}
\begin{align} Output \ Movement & = 2(20)(20) \\ & = 0.8m \end{align}
$$MR = 1.25$$
\begin{align} FR & = \frac{2}{1-\frac{20}{25}} \\ & = 10 \end{align}
So the efficiency would be $\frac{10(100)}{1.25}$ which is $800\%$.
Am I correct in assuming I went wrong somewhere as the efficiency is so high? And where did I go wrong?
Yes your force ratio and movement ratio are both wrong.
Firstly the equation for force ratio is actually the equation for movement ratio. And the little r's are NOT the number of teeth on the cogs, they are the RADII of those cogs. The equation is derived from effort movement / force movement which is 2R / (R-r) after the pi's are cancelled out, and can be rearranged into the form you give it in above, but I repeat it is the movement ratio not the force ratio.
Force ratio is Load Force divided by Effort Force.
So quite a lot to think about!