Compound gear train ratios and how to assemble gears?

[unseenfootage]

Homework Statement

Hello fellow Physics Forumers

My first post and , hopefully not my last. I've been lurking for a long while and the tempo has increased since I started my 1st year engineering course last semester.

So I decided to register today as I have a bit of a puzzle.

We have a project to construct a truck which will be used in a torque/speed test.

First we have to do a speed - torque test of a small 1.5v/3v motor using a gear ratio (output shaft speed/motor speed) of 1/256.

I have the following kit:

http://www.mfacomodrills.com/gearboxes/multiratio.jpg

The motor is 1.5-3.0v with 8200 rpm and 14000 rpm respectively.

The driver gear has 12 teeth.
There are 6 compound gears with 48 outer and 12 inner teeth.
Two shafts , one with pulley
Assembly kit.

My question is how to achieve the gear ratio (output shaft speed/motor speed) of 1/256, with what I have?

Kind regards.

Homework Equations

Gear ratio = (No. Teeth on gear B/No. Teeth on gear A) * (No. Teeth on gear D/No. Teeth on gear C) * (No. Teeth on gear F/ No. Teeth on gear E) * (No. Teeth on gear H/No. Teeth on gear G).Gear A is the driver gear with 12 teeth. The rest are compound gears with 48 outer and 12 inner teeth.

The Attempt at a Solution

48/12 * 48/12 * 48/12 * 48/12
= 4 * 4 * 4 * 4
=256

This appears to be the solution, except I only have two shafts and the assembly plate seems to have only space for two shafts.

Is there a way of assembling it (maybe 2 gears per shaft) which might work?

Kind regards...

Homework Statement

Homework Equations

The Attempt at a Solution

 

[unseenfootage]

Solved, by myself.

I stacked two compound gears on each shaft...One green and one orange.

The orange ones are fixed to the shift and the green ones are free.

So instead of all the gears positioned side by side on four different shafts in a line, I have only two shafts with two gears stacked on each.

Thanks anyway.

Now for the speed torque test.