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--- robots:armadillo:hardware:gearratio [2012/03/05 11:56]
claes [Armadillo - Gear Ratio]
+++ robots:armadillo:hardware:gearratio [2021/08/14 04:21] (current)
@@ Line 7: / Line 7: @@
 $$R=\frac{N_b}{N_a}*\frac{N_d}{N_c} $$
+/*
 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
-$$R=\frac{N_b}{N_a}*\frac{N_d}{N_c}*R_{belt/drivewheel} $$
+$$R=\frac{N_b}{N_a}*\frac{N_d}{N_c}$$
 ===== Gears =====
-The beltmodule has the following gears
+The beltmodule has the following gears from the motor to the drivewheel
   *$ N_a=22 $
   *$ N_b=80 $
-  *$ N_c=10 $
+  *$ N_c=17 $
-  *$ N_d=20 $
+  *$ N_d=45 $
 which gives the following ratio
-$$R=\frac{N_b}{N_a}*\frac{N_d}{N_c}= \frac{80}{22}*\frac{20}{10}= \frac{160}{22}=7,27 $$
+$$R=\frac{N_b}{N_a}*\frac{N_d}{N_c}=\frac{80}{22}*\frac{45}{17}=\frac{3600}{374}=\frac{1800}{187}=9,625 $$
+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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