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robots:armadillo:hardware:gearratio

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robots:armadillo:hardware:gearratio [2012/03/05 16:06]
claes [Gears] 		robots:armadillo:hardware:gearratio [2021/08/14 04:21] (current)
Line 7: Line 7:
 $$R=\frac{N_b}{N_a}*\frac{N_d}{N_c} $$ $$R=\frac{N_b}{N_a}*\frac{N_d}{N_c} $$
  
 +/*
 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.
 $$R_{belt/drivewheel}=\frac{C_{belt}}{C_{drivewheel}}=\frac{D_{belt}}{D_{drivewheel}}=\frac{R_{belt}}{R_{drivewheel}} $$ $$R_{belt/drivewheel}=\frac{C_{belt}}{C_{drivewheel}}=\frac{D_{belt}}{D_{drivewheel}}=\frac{R_{belt}}{R_{drivewheel}} $$
 +
 +*/
  
 The complete solution for the gear-ratio in the beltmodule is 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 ===== ===== 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=17 $   *$ N_c=17 $
   *$ N_d=45 $   *$ N_d=45 $
-The drive wheel - all in m. +
-  * $ C_d=0.4084 $ +
-  * $ D_d=0.13 $ +
-  * $ R_d=0.065 $ +
-The Belt - all in m. +
-  * $ C_b=3.12 $ +
-  * $ D_b=0.9931 $ +
-  * $ R_b=0.4966 $+
 which gives the following ratio which gives the following ratio
-$$R_{belt/drivewheel}=\frac{3.12}{0.4048}=\frac{0.9966}{0.13}=\frac{0.4966}{0.065}= 7.6394$$ 
-$$R=\frac{N_b}{N_a}*\frac{N_d}{N_c}*R_{belt/drivewheel}=\frac{80}{22}*\frac{45}{17}*7.6394=\frac{1800}{187}*7.6394=73.5347 $$ 
  
-<file matlab garing.m>+$$R=\frac{N_b}{N_a}*\frac{N_d}{N_c}=\frac{80}{22}*\frac{45}{17}=\frac{3600}{374}=\frac{1800}{187}=9,625 $$ 
 + 
 +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; close all; clear all;
  
Line 40: Line 40:
 Nd=45; %large chain cog on driveshaft 2 Nd=45; %large chain cog on driveshaft 2
  
-%drive wheel +R = Nb/Na*Nd/Nc 
-Dd = 0.13;  %diameter in meters +G = Na/Nb*Nc/Nd
-Rd = Dd/2; % Radius in meters +
-Od = Dd*pi; % circumference +
- +
-%Belt +
-Ob = 3.12;  % circumference in meters +
-Db = Ob/pi; % diameter in meters +
-Rb = Db/2;  % Radius in meters +
- +
-%relationship - %This is the one needed +
-RelO = Ob/Od; +
-RelD = Db/Dd; +
-RelR = Rb/Rd; +
- +
-Rel = (RelO+RelD+RelR)/3 +
-%Rel_O = Od/Ob; +
-%Rel_D = Dd/Db; +
-%Rel_R = Rd/Rb; +
- +
-R = Nb/Na*Nd/Nc*Rel+
 </file> </file>
robots/armadillo/hardware/gearratio.1330959994.txt.gz · Last modified: 2021/08/14 04:20 (external edit)

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