====== Armadillo - Gear Ratio ======
The gear ratio $R$ is inversely proportional to ratio of the number of teeth, $N_a$ and $N_b$, of the two gears.
$$R=\frac{N_b}{N_a} $$
The beltmodules uses the same concept as shown in the picture below, from [[http://www.dynamicscience.com.au/tester/solutions/hydraulicus/gears1compound.htm|dynamicscience.com.au]] where the gear train of four gears are connected, and the gears "B" and "C" are fixed together on the same shaft.
{{ :robots:armadillo:hardware:gearscompound1.gif?nolink& |}}
From this it is possible to derive the gear ratio formula for the cogs in the beltmodule.
$$R=\frac{N_b}{N_a}*\frac{N_d}{N_c} $$

/*
To get the correct gear-ratio the ratio between the drivewheel and the belt also needs to be taken in to account. The belt and drivewheel can be seen as 2 cirkles, the ratio is the the ratio between circumference , diameter or radius.
$$R_{belt/drivewheel}=\frac{C_{belt}}{C_{drivewheel}}=\frac{D_{belt}}{D_{drivewheel}}=\frac{R_{belt}}{R_{drivewheel}} $$

*/

The complete solution for the gear-ratio in the beltmodule is
$$R=\frac{N_b}{N_a}*\frac{N_d}{N_c}$$

===== Gears =====
The beltmodule has the following gears from the motor to the drivewheel
  *$ N_a=22 $
  *$ N_b=80 $
  *$ N_c=17 $
  *$ N_d=45 $

which gives the following ratio

$$R=\frac{N_b}{N_a}*\frac{N_d}{N_c}=\frac{80}{22}*\frac{45}{17}=\frac{3600}{374}=\frac{1800}{187}=9,625 $$

From the divewheel to the motor the ratio is

$$G=\frac{N_a}{N_b}*\frac{N_c}{N_d}=\frac{22}{80}*\frac{17}{45}=\frac{374}{3600}=\frac{187}{1800}=0,10389$$

<file matlab gearing.m>
close all; clear all;

%no of teeth on the cogs
Na=22; %Small belt cog on the motor
Nb=80; %large belt cog on driveshaft 1
Nc=17; %small chain cog on driveshaft 1
Nd=45; %large chain cog on driveshaft 2

R = Nb/Na*Nd/Nc
G = Na/Nb*Nc/Nd
</file>