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 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. 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 $$R=\frac{N_b}{N_a}*\frac{N_d}{N_c}$$

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

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

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