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robots:armadillo:hardware:gearratio [2012/03/05 11:42]
claes [Armadillo - Gear Ratio]
robots:armadillo:hardware:gearratio [2021/08/14 04:21] (current)
Line 4: Line 4:
 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. 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& |}} {{ :robots:armadillo:hardware:gearscompound1.gif?nolink& |}}
-From this it is possible to derive the gear ratio formula+From this it is possible to derive the gear ratio formula for the cogs in the beltmodule.
 $$R=\frac{N_b}{N_a}*\frac{N_d}{N_c} $$ $$R=\frac{N_b}{N_a}*\frac{N_d}{N_c} $$
  
 +/*
 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. 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 ===== ===== Gears =====
-The beltmodule has the following gears+The beltmodule has the following gears from the motor to the drivewheel
   *$ N_a=22 $   *$ N_a=22 $
   *$ N_b=80 $   *$ N_b=80 $
-  *$ N_c=10 +  *$ N_c=17 
-  *$ N_d=20 $+  *$ N_d=45 $ 
 which gives the following ratio 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>
robots/armadillo/hardware/gearratio.1330944122.txt.gz · Last modified: 2021/08/14 04:20 (external edit)