# Chapter 10: Toothed Gearing¶

## Example 1, Page 320¶

In [1]:
import math

#Variable declaration
Teeth=48
pitch=.75 #in

#Calculations
D=Teeth*pitch/math.pi

#Result
print "The pitch diameter is %.3f in"%D

The pitch diameter is 11.459 in


## Example 2, Page 320¶

In [2]:
import math

#Variable declaration
T=48#teeth
pd=4#diametral pitch

#Calculations
D=T/pd#pitch diameter
p=math.pi/pd#the circular pitch

#Results
print "The pitch diameter = %.f in\nThe circular pitch = %.4f in"%(D,p)

The pitch diameter = 12 in
The circular pitch = 0.7854 in


## Example 3, Page 321¶

In [3]:
import math

#Variable declaration
T=48
m=6#mm ; module

#Calculations
D=m*T
p=math.pi*m
dia=D/10#cm
P=p*0.0393700787#inches

#Results
print "Pitch diameter = %.1f cm\nCircular pitch = %.4f in"%(dia,P)

Pitch diameter = 28.0 cm
Circular pitch = 0.7421 in


## Example 4, Page 334¶

In [5]:
import math

#Variable declaration
phi=20*math.pi/180

#Calculations
#Solution a)
ar=1
t1=2*ar/math.sin(phi)**2#from equation 10.7
T1=math.ceil(t1)
#Solution b)
aw=1.
t2=2*aw/((1+3*math.sin(phi)**2)**(1./2)-1)#from euation 10.6
T2=math.ceil(t2)
#solution c)
t=1.
T=3.
A=(t/T)*(t/T+2)
t3=2*aw*(t/T)/((1+A*math.sin(phi)**2)**(1./2)-1)#from 10.5
T3=math.ceil(t3)

#Results
print "Smallest number of teeth theoretically required in order to avoid interference on a pinion which is to gear with"\
"\na) A rack , t= %.f\nb) An equal pinion , t=  %.f\nc) A wheel to give a ratio of 3 to 1 , t= %.f"%(T1,T2,T3)

Smallest number of teeth theoretically required in order to avoid interference on a pinion which is to gear with
a) A rack , t= 18
b) An equal pinion , t=  13
c) A wheel to give a ratio of 3 to 1 , t= 15


## Example 5, Page 338¶

In [6]:
import math

#Variable declaration
t=25
phi=20*math.pi/180

#Calculations
#let pitch be 1
R=t/(2*math.pi)#R=t*p/(2*math.pi)
Larc=1.6#1.6*p
#AB=Larc*math.cos(phi)
AB=Larc*math.cos(phi)
Ra=(4.47+13.97)**(1./2)#by simplifying AB+2{(Ra^2-R^2*cos(phi)^2)-R*sin(phi)} and using p =1
#writing p in place of p=1

#Result

Addendum required = 0.32p


## Example 6, Page 338¶

In [7]:
import math

#Variable declaration
#let module be 1
m=1
t1=28
t2=45

#Calculations
r=t1*m/2
R=t2*m/2
ra=r+m
Ra=R+m
phi1=14.5*math.pi/180
#10.8 => AB =(ra^2-r^2*cos(phi)^2)^(1/2)+(Ra^2-R^2*cos(phi)^2)^(1/2)-(r+R)*sin(phi)
#AB=A+B-C
A=m*(ra**2-r**2*math.cos(phi1)**2)**(1./2)
B=m*(Ra**2-R**2*math.cos(phi1)**2)**(1./2)
C=m*(r+R)*math.sin(phi1)
AB=A+B-C
p=math.pi*m
ABp=AB/math.pi
arc1=ABp/math.cos(phi1)#length of arc of contact
phi2=20*math.pi/180
#10.8 => AB =(ra^2-r^2*cos(phi)^2)^(1/2)+(Ra^2-R^2*cos(phi)^2)^(1/2)-(r+R)*sin(phi)
a=m*(ra**2-r**2*math.cos(phi2)**2)**(1./2)
b=m*(Ra**2-R**2*math.cos(phi2)**2)**(1./2)
c=m*(r+R)*math.sin(phi2)
ab=a+b-c
abp=ab/math.pi
arc2=abp/math.cos(phi2)#length of arc of contact

#Results
print "Length of path of contact\nWhen phi = 14.5 degrees = %.3fm\nWhen phi = 20 degrees = %.2fm"\
"\n\nLength of arc of contact\nWhen phi = 14.5 degrees = %.2fp\nWhen phi = 20 degrees = %.3fp"%(AB,ab,arc1,arc2)

Length of path of contact
When phi = 14.5 degrees = 6.092m
When phi = 20 degrees = 4.97m

Length of arc of contact
When phi = 14.5 degrees = 2.00p
When phi = 20 degrees = 1.685p