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--- robots:armadillo:hardware:gearratio [2012/01/11 18:50]
claes created
+++ robots:armadillo:hardware:gearratio [2021/08/14 04:21] (current)
@@ Line 2: / Line 2: @@
 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.
+$$R=\frac{N_b}{N_a}*\frac{N_d}{N_c} $$
-  *$ N_1=22 $
+/*
-  *$ N_2=22 $
+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.
-  *$ N_3=22 $
+$$R_{belt/drivewheel}=\frac{C_{belt}}{C_{drivewheel}}=\frac{D_{belt}}{D_{drivewheel}}=\frac{R_{belt}}{R_{drivewheel}} $$
-  *$ N_4=22 $
+*/
+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>

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