# Chapter 13 : Gear Trains¶

## Example 13.1 Page No : 432¶

In [1]:
# Variables:
NA = 975 			#rpm
TA = 20.
TB = 50.
TC = 25
TD = 75
TE = 26
TF = 65

#Solution:
#Calculating the speed of gear F
NF = NA*(TA*TC*TE)/(TB*TD*TF) 			#rpm

#Results:
print " Speed of gear F, NF  =  %d rpm."%(NF)

 Speed of gear F, NF  =  52 rpm.


## Example 13.2 Page No : 433¶

In [2]:
import math
from numpy import linalg

# Variables:
x = 600.
pc = 25. 			#mm
N1 = 360.
N2 = 120. 			#rpm
#Solution:
#Calculating the pitch circle diameters of each gear
#Speed ratio N1/N2  =  d2/d1 or N1*d1-N2*d2  =  0                                .....(i)
#Centre distance between the shafts x  =  1/2*(d1+d2) or d1+d2  =  600*2         .....(ii)
A = [[N1, -N2],[ 1, 1]]
B = [0, 600*2]
V = linalg.solve(A,B)
d1 = V[0] 			#mm
d2 = V[1] 			#mm
#Calculating the number of teeth on the first gear
T1 = round(math.pi*d1/pc)
#Calculating the number of teeth on the second gear
T2 = int(math.pi*d2/pc+1)
#Calculating the pitch circle diameter of the first gear
d1dash = T1*pc/math.pi 			#mm
#Calculating the pitch circle diameter of the second gear
d2dash = T2*pc/math.pi 			#mm
#Calculating the exact distance between the two shafts
xdash = (d1dash+d2dash)/2 			#mm

#Results:
print " The number of teeth on the first and second gear must be %d and %d and their pitch\
circle diameters must be %.2f mm and %.1f mm respectively."%(T1,T2,d1dash,d2dash)
print " The exact distance between the two shafts must be %.2f mm."%(xdash)

 The number of teeth on the first and second gear must be 38 and 114 and their pitch circle diameters must be 302.39 mm and 907.2 mm respectively.
The exact distance between the two shafts must be 604.79 mm.


## Example 13.3 Page No : 435¶

In [4]:
from numpy import linalg
import math

# Variables:
rAD = 12. 			#Speed ratio NA/ND
mA = 3.125          #mm
mB = mA             #mm
mC = 2.5            #mm
mD = mC             #mm
x = 200. 			#mm

#Solution:
#Calculating the speed ratio between the gears A and B and C and D
rAB = math.sqrt(rAD) 			#Speed ratio between the gears A and B
rCD = math.sqrt(rAB) 			#Speed ratio between the gears C and D
#Calculating the ratio of teeth on gear B to gear A
rtBA = rAB 			#Ratio of teeth on gear B to gear A
#Calculating the ratio of teeth on gear D to gear C
rtDC = rCD 			#Ratio of teeth on gear D to gear C
#Calculating the number of teeth on the gears A and B
#Distance between the shafts x  =  mA*TA/2+mB*TB/2 or (mA/2)*TA+(mB/2)*TB  =  x        .....(i)
#Ratio of teeth on gear B to gear A TB/TA  =  math.sqrt(12) or math.sqrt(12)*TA-TB  =  0         .....(ii)
A = [[mA/2, mB/2],[math.sqrt(12) ,-1]]
B = [x, 0]
V = linalg.solve(A,B)
TA = int(V[0])
TB = round(V[1])
#Calculating the number of teeth on the gears C and D
#Dismath.tance between the shafts x  =  mC*TC/2+mD*TD/2 or (mC/2)*TC+(mD/2)*TD  =  x        .....(iii)
#Ratio of teeth on gear D to gear C TD/TC  =  math.sqrt(12) or math.sqrt(12)*TC-TD  =  0         .....(iv)
A = [[mC/2, mD/2],[ math.sqrt(12) ,-1]]
B = [x, 0]
V = linalg.solve(A,B)
TC = round(V[0])
TD = int(V[1])

#Results:
print " Number of teeth on gear A, TA  =  %d."%(TA)
print " Number of teeth on gear B, TB  =  %d."%(TB)
print " Number of teeth on gear C, TC  =  %d."%(TC)
print " Number of teeth on gear D, TD  =  %d."%(TD)

 Number of teeth on gear A, TA  =  28.
Number of teeth on gear B, TB  =  99.
Number of teeth on gear C, TC  =  36.
Number of teeth on gear D, TD  =  124.


## Example 13.4 Page No : 438¶

In [5]:
import math

# Variables:
TA = 36.
TB = 45.
NC = 150. 			#rpm anticlockwise

#Solution:
#Refer Fig. 13.7
#Algebraic method:
#Calculating the speed of gear B when gear A is fixed
NA = 0.
NC = 150. 			#rpm
NB1 = (-TA/TB)*(NA-NC)+NC 			#rpm
#Calculating the speed of gear B when gear A makes 300 rpm clockwise
NA = -300. 			#rpm
NB2 = (-TA/TB)*(NA-NC)+NC 			#rpm

#Results:
print " Speed of gear B when gear A is fixed, NB  =  %d rpm."%(NB1)
print " Speed of gear B when gear A makes 300 rpm clockwise, NB  =  %d rpm."%(NB2)

 Speed of gear B when gear A is fixed, NB  =  270 rpm.
Speed of gear B when gear A makes 300 rpm clockwise, NB  =  510 rpm.


## Example 13.5 Page No : 440¶

In [6]:
import math

# Variables:
TB = 75.
TC = 30.
TD = 90.
NA = 100. 			#rpm clockwise

#Solution:
#Refer Table 13.3
#Calculating the number of teeth on gear E
TE = TC+TD-TB
#Calculating the speed of gear C
y = -100.
x = y*(TB/TE)
NC = y-x*(TD/TC) 			#rpm

#Results:
print " Speed of gear C, NC  =  %d rpm, anticlockwise."%(NC)

 Speed of gear C, NC  =  400 rpm, anticlockwise.


## Example 13.6 Page No : 443¶

In [7]:
import math

# Variables:
TA = 72.
TC = 32.
NEF = 18. 			#Speed of arm EF rpm

#Solution:
#Refer Table 13.5
#Speed of gear C:
y = 18. 			#rpm
x = y*(TA/TC)
NC = x+y 			#Speed of gear C rpm
#Speed of gear B:
#Calculating the number of teeth on gear B
TB = (TA-TC)/2
#Calculating the speed of gear B
NB = y-x*(TC/TB) 			#Speed of gear B rpm

#Solution:
print " Speed of gear C  =  %.1f rpm."%(NC)
print " Speed of gear B  =  %.1f rpm in the opposite direction of arm."%(-NB)

 Speed of gear C  =  58.5 rpm.
Speed of gear B  =  46.8 rpm in the opposite direction of arm.


## Example 13.7 Page No : 444¶

In [9]:
from numpy import linalg
import math

# Variables:
TA = 40.
TD = 90.

#Solution:
#Calculating the number of teeth on gears B and C
#From geometry of the Fig. 13.11 dA+2*dB = dD
#Since the number of teeth are proportional to their pitch circle diameters
TB = (TD-TA)/2
TC = TB
#Refer Table 13.6
#Speed of arm when A makes 1 revolution clockwise and D makes half revolution anticlockwise:
#Calculating the values of x and y
#From the fourth row of the table -x-y  =  -1 or x+y  =  1                            .....(i)
#The gear D makes half revolution anticlockwise i.e. x*(TA/TD)-y  =  1/2            .....(ii)
A = [[1, 1],[TA/TD, -1]]
B = [1, 1./2]
V = linalg.solve(A,B)
x = V[0]
y = V[1]
#Calculating the speed of arm
varm = -y 			#Speed of arm revolutions

#Results:
print " Speed of arm when A makes 1 revolution clockwise and D makes half revolution\
anticlockwise  =  %.2f revolution anticlockwise."%(varm)
#Speed of arm when A makes 1 revolution clockwise and D is stationary:
#Calculating the values of x and y
#From the fourth row of the table -x-y  =  -1 or x+y  =  1                            .....(iii)
#The gear D is stationary i.e. x*(TA/TD)-y  =  0                                    .....(iv)
A = [[1, 1],[ TA/TD, -1]]
B = [1, 0]
V = linalg.solve(A,B)
x = V[0]
y = V[1]
#Calculating the speed of arm
varm = -y 			#Speed of arm revolutions

#Results:
print " Speed of arm when A makes 1 revolution clockwise and D is stationary  =  %.3f revolution\
clockwise."%(-varm)

 Speed of arm when A makes 1 revolution clockwise and D makes half revolution anticlockwise  =  0.04 revolution anticlockwise.
Speed of arm when A makes 1 revolution clockwise and D is stationary  =  0.308 revolution clockwise.


## Example 13.8 Page No : 446¶

In [10]:
import math

# Variables:
TC = 28.
TD = 26.
TE = 18.
TF = TE

#Solution:
#The sketch is as in Fig. 13.12
#Number of teeth on  wheels A and B:
#From geometry dA  =  dC+2*dE and dB  =  dD+2*dF
#Since the number of teeth are proportional to their pitch circle diameters
TA = TC+2*TE
TB = TD+2*TF
#Speed of wheel B when arm G makes 100 rpm clockwise and wheel A is fixed:
#Since the arm G makes 100 rpm clockwise therefore from the fourth row of Table 13.7
y = -100
x = -y
#Calculating the speed of wheel B
NB1 = y+x*(TA/TC)*(TD/TB) 			#Speed of wheel B when arm G makes 100 rpm clockwise and wheel A is fixed rpm
#Speed of wheel B when arm G makes 100 rpm clockwise and wheel A makes 10 rpm counter clockwise:
#Since the arm G makes 100 rpm clockwise therefore from the fourth row of Table 13.7
y = -100
x = 10-y
#Calculating the speed of wheel B
NB2 = y+x*(TA/TC)*(TD/TB) 			#Speed of wheel B when arm G makes 100 rpm clockwise and wheel A makes 10 rpm counter clockwise rpm

#Solution:
print " Number of teeth on  wheel A, TA  =  %d."%(TA)
print " Number of teeth on  wheel B, TB  =  %d."%(TB)
print " Speed of wheel B when arm G makes 100 rpm clockwise and wheel A is fixed  =  %.1f rpm, clockwise."%(-NB1)
print " Speed of wheel B when arm G makes 100 rpm clockwise and wheel A makes 10 rpm counter\
clockwise  =  %.1f rpm, counter clockwise."%(NB2)

 Number of teeth on  wheel A, TA  =  64.
Number of teeth on  wheel B, TB  =  62.
Speed of wheel B when arm G makes 100 rpm clockwise and wheel A is fixed  =  4.1 rpm, clockwise.
Speed of wheel B when arm G makes 100 rpm clockwise and wheel A makes 10 rpm counter clockwise  =  5.4 rpm, counter clockwise.


## Example 13.9 Page No : 447¶

In [1]:
import math

# Variables:
dD = 224.
m = 4. 			#mm

#Solution:
#Refer Table 13.8
#Calculating the values of x and y
y = 1.
x = 5-y
#Calculating the number of teeth on gear D
TD = dD/m
#Calculating the number of teeth on gear B
TB = y/x*TD
#Calculating the number of teeth on gear C
TC = (TD-TB)/2

#Results:
print " Number of teeth on gear D, TD  =  %d."%(TD)
print " Number of teeth on gear B, TB  =  %d."%(TB)
print " Number of teeth on gear C, TC  =  %d."%(TC)

 Number of teeth on gear D, TD  =  56.
Number of teeth on gear B, TB  =  14.
Number of teeth on gear C, TC  =  21.


## Example 13.10 Page No : 448¶

In [13]:
from numpy import linalg
import math

# Variables:
TC = 50.
TD = 20.
TE = 35.
NA = 110. 			#rpm

#Solution:
#Calculating the number of teeth on internal gear G
TG = TC+TD+TE
#Speed of shaft B:
#Calculating the values of x and y
#From the fourth row of Table 13.9
#y-x*(TC/TD)*(TE/TG)  =  0        .....(i)
#Also x+y  =  110 or y+x  =  110                                     .....(ii)
A = [[1, -(TC/TD)*(TE/TG)],[ 1, 1]]
B = [0, 110]
V = linalg.solve(A,B)
x = V[1]
y = V[0]
#Calculating the speed of shaft B
NB = round(+y) 			#Speed of shaft B rpm

#Results:
print " Number of teeth on internal gear G, TG  =  %d."%(TG)
print " Speed of shaft B  =  %d rpm, anticlockwise."%(NB)

 Number of teeth on internal gear G, TG  =  105.
Speed of shaft B  =  50 rpm, anticlockwise.


## Example 13.11 Page No : 450¶

In [14]:
import math

# Variables:
TA = 12.
TB = 30.
TC = 14.
NA = 1.
ND = 5. 			#rps

#Solution:
#Number of teeth on wheels D and E:
#Calculating the number of teeth on wheel E
TE = TA+2*TB
#Calculating the number of teeth on wheel E
TD = TE-(TB-TC)
#Magnitude and direction of angular velocities of arm OP and wheel E:
#Calculating the values of x and y
#From the fourth row of Table 13.10 -x-y  =  -1 or x+y  =  1            .....(i)
#Also x*(TA/TB)*(TC/TD)-y  =  5                                        .....(ii)
A = [[1, 1],[(TA/TB)*(TC/TD) ,-1]]
B = [1, 5]
V = linalg.solve(A,B)
x = V[0]
y = V[1]
#Calculating the angular velocity of arm OP
omegaOP = -y*2*math.pi 			#Angular velocity of arm OP rad/s
#Calculating the angular velocity of wheel E
omegaE = (x*TA/TE-y)*2*math.pi 			#Angular velocity of wheel E rad/s

#Results:
print " Number of teeth on wheel E, TE  =  %d."%(TE)
print " Number of teeth on wheel D, TD  =  %d."%(TD)
print " Angular velocity of arm OP  =  %.3f rad/s, counter clockwise."%(omegaOP)
print " Angular velocity of wheel E  =  %.2f rad/s, counter clockwise."%(omegaE)

 Number of teeth on wheel E, TE  =  72.
Number of teeth on wheel D, TD  =  56.
Angular velocity of arm OP  =  27.989 rad/s, counter clockwise.
Angular velocity of wheel E  =  33.70 rad/s, counter clockwise.


## Example 13.12 Page No : 451¶

In [15]:
import math

# Variables:
TB = 80.
TC = 82.
TD = 28.
NA = 500. 			#rpm

#Solution:
#Calculating the number of teeth on wheel E
TE = TB+TD-TC
#Calculating the values of x and y
y = 800.
x = -y*(TE/TB)*(TC/TD)
#Calculating the speed of shaft F
NF = x+y 			#Speed of shaft F rpm

#Results:
print " Speed of shaft F  =  %d rpm, anticlockwise."%(NF)

 Speed of shaft F  =  38 rpm, anticlockwise.


## Example 13.13 Page No : 452¶

In [11]:
# variables

TA = 100. ;        # Gear A teeth
TC = 101. ;        # Gear C teeth
TD = 99. ;        # Gear D teeth
TP = 20.          # Gear planet teeth
y = 1
x = 0 - y

# calculations
NC = y + x * TA/TC
ND = y + x * TA/TD

# results
print "revolutions of gear C : %.4f"%NC
print "revolutions of gear D : %.4f "%ND

revolutions of gear C : 0.0099
revolutions of gear D : -0.0101


## Example 13.14 Page No : 453¶

In [3]:
from numpy import linalg
import math

# Variables:
NA = 300. 			#rpm
TD = 40.
TE = 30.
TF = 50.
TG = 80.
TH = 40.
TK = 20.
TL = 30.

#Solution:
#Refer Fig. 13.18 and Table 13.13
#Calculating the speed of wheel E
NE = NA*(TD/TE) 			#rpm
#Calculating the number of teeth on wheel C
TC = TH+TK+TL
#Speed and direction of rotation of shaft B:
#Calculating the values of x and y
#We have -x-y  =  -400 or x+y  =  400                    .....(i)
#Also x*(TH/TK)*(TL/TC)-y  =  0                         .....(ii)
A = [[1, 1],[ (TH/TK)*(TL/TC), -1]]
B = [400, 0]
V = linalg.solve(A,B)
x = V[0]
y = V[1]
#Calculating the speed of wheel F
NF = -y 			#rpm
#Calculating the speed of shaft B
NB = -NF*(TF/TG) 			#Speed of shaft B rpm

#Results:
print " Number of teeth on wheel C, TC  =  %.1f."%(TC)
print " Speed of shaft B  =  %.1f rpm, anticlockwise."%(NB)

 Number of teeth on wheel C, TC  =  90.0.
Speed of shaft B  =  100.0 rpm, anticlockwise.


## Example 13.15 Page No : 455¶

In [17]:
from numpy import linalg
import math

# Variables:
T1 = 80.
T8 = 160.
T4 = 100.
T3 = 120.
T6 = 20.
T7 = 66.

#Solution:
#Refer Fig. 13.19 and Table 13.14
#Calculating the number of teeth on wheel 2
T2 = (T3-T1)/2
#Calculating the values of x and y
#Assuming that wheel 1 makes 1 rps anticlockwise x+y  =  1            .....(i)
#Also y-x*(T1/T3)  =  0 or x*(T1/T3)-y  =  0                           .....(ii)
A = [[1, 1],[ 1, T1/T3]]
B = [1, 0]
V = linalg.solve(A,B)
x = V[0]
y = V[1]
#Calculating the speed of casing C
NC = y 			#Speed of casing C rps
#Calculating the speed of wheel 2
N2 = y-x*(T1/T2) 			#Speed of wheel 2 rps
#Calculating the number of teeth on wheel 5
T5 = (T4-T6)/2
#Calculating the values of x1 and y1
y1 = -2
x1 = (y1-0.4)*(T4/T6)
#Calculating the speed of wheel 6
N6 = x1+y1 			#Speed of wheel 6 rps
#Calculating the values of x2 and y2
y2 = 0.4
x2 = -(14+y2)*(T7/T8)
#Calculating the speed of wheel 8
N8 = x2+y2 			#Speed of wheel 8 rps
#Calculating the velocity ratio of the output shaft B to the input shaft A
vr = N8/1 			#Velocity ratio

#Results:
print " Velocity ratio of the output shaft B to the input shaft A  =  %.2f."%(vr)

 Velocity ratio of the output shaft B to the input shaft A  =  -5.54.


## Example 13.16 Page No : 459¶

In [18]:
import math

# Variables:
TA = 40.
TB = 30.
TC = 50.
NX = 100.
NA = NX 			    #rpm
Narm = 100. 			#Speed of armrpm

#Solution:
#Refer Fig. 13.22 and Table 13.18
#Calculating the values of x and y
y = +100
x = -100-y
#Calculating the speed of the driven shaft
NY = y-x*(TA/TB) 			#rpm

#Results:
print " Speed of the driven shaft, NY  =  %.1f rpm, anticlockwise."%(NY)

 Speed of the driven shaft, NY  =  366.7 rpm, anticlockwise.


## Example 13.17 Page No : 460¶

In [19]:
from numpy import linalg
import math

# Variables:
TB = 20.
TC = 80.
TD = 80.
TE = 30.
TF = 32.
NB = 1000. 			#rpm

#Solution:
#Refer Fig. 13.23 and Table 13.19
#Speed of the output shaft when gear C is fixed:
#Calculating the values of x and y
#From the fourth row of the table y-x*(TB/TC)  =  0                .....(i)
#Also x+y  =  +1000 or y+x  =  1000                                 .....(ii)
A = [[1, -TB/TC],[ 1, 1]]
B = [0, 1000]
V = linalg.solve(A,B)
x = V[1]
y = V[0]
#Calculating the speed of output shaft
NF1 = y-x*(TB/TD)*(TE/TF) 			#Speed of the output shaft when gear C is fixed rpm
#Speed of the output shaft when gear C is rotated at 10 rpm counter clockwise:
#Calculating the values of x and y
#From the fourth row of te table  y-x*(TB/TC)  =  +10                .....(iii)
#Also x+y  =  +1000 or y+x  =  1000                                 .....(iv)
A = [[1, -TB/TC],[1, 1]]
B = [10, 1000]
V = linalg.solve(A,B)
x = V[1]
y = V[0]
#Calculating the speed of output shaft
NF2 = y-x*(TB/TD)*(TE/TF) 			#Speed of the output shaft when gear C is rotated at 10 rpm counter clockwise rpm

#Results:
print " Speed of the output shaft when gear C is fixed  =  %.1f rpm, counter clockwise."%(NF1)
print " Speed of the output shaft when gear C is rotated at 10 rpm counter clockwise  =  %.1f rpm, \
counter clockwise."%(NF2)

 Speed of the output shaft when gear C is fixed  =  12.5 rpm, counter clockwise.
Speed of the output shaft when gear C is rotated at 10 rpm counter clockwise  =  22.4 rpm,  counter clockwise.


## Example 13.18 Page No : 461¶

In [20]:
import math

# Variables:
TA = 10.
TB = 60.
NA = 1000.
NQ = 210.
ND = NQ 			#rpm

#Solution:
#Refer Fig. 13.24 and Table 13.20
#Calculating the speed of crown gear B
NB = NA*(TA/TB) 			#rpm
#Calculating the values of x and y
y = 200.
x = y-210.
#Calculating the speed of road wheel attached to axle P
NC = x+y 			#Speed of road wheel attached to axle P rpm

#Results:
print " Speed of road wheel attached to axle P  =  %d rpm."%(NC)

 Speed of road wheel attached to axle P  =  190 rpm.


## Example 13.19 Page No : 463¶

In [5]:
from numpy import linalg
import math

# Variables:
TA = 15.
TB = 20.
TC = 15.
NA = 1000. 			#rpm
Tm = 100. 			#Torque developed by motor N-m

#Solution:
#Refer Fig. 13.26 and Table 13.21
#Calculating the number of teeth on gears E and D
TE = TA+2*TB
TD = TE-(TB-TC)
#Speed of the machine shaft:
#From the fourth row of the table x+y  =  1000 or y+x  =  1000        .....(i)
#Also y-x*(TA/TE)  =  0                                              .....(ii)
A = [[1, 1],[1, -TA/TE]]
B = [1000, 0]
V = linalg.solve(A,B)
y = round(V[0])
x = round(V[1])
#Calculating the speed of machine shaft
ND = y-x*(TA/TB)*(TC/TD) 			#rpm
#Calculating the torque exerted on the machine shaft
Ts = Tm*NA/ND 			#Torque exerted on the machine shaft N-m

#Results:
print " Speed of machine shaft, ND  =  %.2f rpm, anticlockwise."%(ND)
print " Torque exerted on the machine shaft  =  %.f N-m."%(Ts)

 Speed of machine shaft, ND  =  37.15 rpm, anticlockwise.
Torque exerted on the machine shaft  =  2692 N-m.


## Example 13.20 Page No : 465¶

In [23]:
import math

# Variables:
Ts = 100 			#Torque on the sun wheel N-m
r = 5 			#Ratio of speeds of gear S to C NS/NC
#Refer Fig. 13.27 and Table 13.22
#Number of teeth on different wheels:
#Calculating the values of x and y
y = 1.
x = 5-y

#Calculating the number of teeth on wheel E
TS = 16.
TE = 4*TS
#Calculating the number of teeth on wheel P
TP = (TE-TS)/2
#Torque necessary to keep the internal gear stationary:
Tc = Ts*r 			#Torque on CN-m
#Caluclating the torque necessary to keep the internal gear stationary
Ti = Tc-Ts 			#Torque necessary to keep the internal gear stationary N-m

#Results:
print " Number of teeth on different wheels, TE  =  %d."%(TE)
print " Torque necessary to keep the internal gear stationary  =  %d N-m."%(Ti)

 Number of teeth on different wheels, TE  =  64.
Torque necessary to keep the internal gear stationary  =  400 N-m.


## Example 13.21 Page No : 466¶

In [9]:
import math
from numpy import linalg

# Variables:
TA = 14.
TC = 100.
r = 98./41 			#TE/TD
PA = 1.85*1000 			#W
NA = 1200. 			#rpm
TB = 43

#Solution:
#Refer Fig. 13.28 and Table 13.23
#Calculating the number of teeth on wheel B TB = (TC-TA)/2
#Calculating the values of x and y
#From the fourth row of the table  -y+x*(TA/TC)  =  0 or x*(TA/TC)-y  =  0    .....(i)
#Also x-y  =  1200 or x+y  =  -1200                                         .....(ii)
A = [[TA/TC, -1],[ 1, 1]]
B = [0, -1200]
V = linalg.solve(A,B)
x = V[0]
y = V[1]
#Calculating the speed of gear E
NE = round(-y+x*(TA/TB)*(1./r)) 			#rpm
#Fixing torque required at C:
#Calculating the torque on A
Ta = PA*60./(2*math.pi*NA) 			#Torque on A N-m
#Calculating the torque on E
Te = PA*60./(2*math.pi*NE) 			#Torque on E
#Calculating the fixing torque required at C
Tc = Te-Ta 			#Fixing torque at C N-m

#Results:
print " Speed and direction of rotation of gear E, NE  =  %d rpm, anticlockwise."%(NE)
print " Fixing torque required at C  =  %.1f N-m."%(Tc)

 Speed and direction of rotation of gear E, NE  =  4 rpm, anticlockwise.
Fixing torque required at C  =  4401.8 N-m.


## Example 13.22 Page No : 468¶

In [25]:
from numpy import linalg
import math

# Variables:
TB = 15.
TA = 60.
TC = 20.
omegaY = 740.
P = 130*1000. 			#W

#Solution:
#Refer Fig. 13.29 and Table 13.24
#Calculating the number of teeth on wheel D
TD = TA-(TC+TB)
#Calculating the values of x and y
#From the fourth row of the table y-x*(TD/TC)*(TB/TA)  =  740            .....(i)
#Also x+y  =  0 or y+x  =  0                                              .....(ii)
A = [[1, -(TD/TC)*(TB/TA)],[ 1, 1]]
B = [740, 0]
V = linalg.solve(A,B)
x = V[1]
y = V[0]
#Calculating the speed of shaft X
#Holding torque on wheel D:
#Calculating the torque on A
Ta = P/omegaA 			#Torque on A N-m
#Calculating the torque on X
Tx = P/omegaX 			#Torque on X N-m
#Calculating the holding torque on wheel D
Td = Tx-Ta 			#Holding torque on wheel D N-m

#Results:
print " Speed of shaft X, omegaX  =  %.1f rad/s."%(omegaX)
print " Holding torque on wheel D  =  %.1f N-m."%(Td)

 Speed of shaft X, omegaX  =  563.8 rad/s.
Holding torque on wheel D  =  54.9 N-m.


## Example 13.23 Page No : 469¶

In [26]:
from numpy import linalg
import math

# Variables:
TP = 144.
TQ = 120.
TR = 120.
TX = 36.
TY = 24.
TZ = 30.
NI = 1500. 			#rpm
P = 7.5*1000 			#W
eta = 0.8

#Solution:
#Refer Fig. 13.30 and Table 13.25
#Calculating the values of x and y
#From the fourth row of the table x+y  =  -1500        .....(i)
#Also y-x*(TZ/TR)  =  0 or -x*(TZ/TR)+y  =  0           .....(ii)
A = [[1, 1],[-TZ/TR, 1]]
B = [-1500, 0]
V = linalg.solve(A,B)
x = V[0]
y = V[1]

#Calculating the values of x1 and y1
#We have y1-x1*(TY/TQ)  =  y                           .....(iii)
#Also x1+y1  =  x+y or y1+x1  =  x+y                    .....(iv)
A = [[1, -TY/TQ],[ 1, 1]]
B = [y, x+y]
V = linalg.solve(A , B)
x1 = V[1]
y1 = V[0]
#Speed and direction of the driven shaft O and the wheel P:
#Calculating the speed of shaft O
NO = y1 			#rpm
#Calculating the speed of wheel P
NP = y1+x1*(TY/TQ)*(TX/TP) 			#rpm
#Torque tending to rotate the fixed wheel R:
#Calculating the torque on shaft I
T1 = P*60/(2*math.pi*NI) 			#N-m
#Calculating the torque on shaft O
T2 = eta*P*60/(2*math.pi*(-NO)) 			#N-m
#Calculating the torque tending to rotate the fixed wheel R
T = T2-T1 			#Torque tending to rotate the fixed wheel R N-m

#Results:
print " Speed of the driven shaft O, NO  =  %d rpm, clockwise."%(-NO)
print " Speed of the wheel P, NP  =  %d rpm, clockwise."%(-NP)
print " Torque tending to rotate the fixed wheel R  =  %.2f N-m."%(T)

 Speed of the driven shaft O, NO  =  500 rpm, clockwise.
Speed of the wheel P, NP  =  550 rpm, clockwise.
Torque tending to rotate the fixed wheel R  =  66.85 N-m.


## Example 13.24 Page No : 471¶

In [6]:
from numpy import linalg
import math

# Variables:
TA = 34.
TB = 120.
TC = 150.
TD = 38.
TE = 50.
PX = 7.5*1000 			#W
NX = 500. 			#rpm
m = 3.5 			#mm

#Solution:
#Refer Fig. 13.31 and Table 13.27
#Output torque of shaft Y:
#Calculating the values of x and y
#From the fourth row of the table x+y  =  500 or y+x  =  500        .....(i)
#Alsoy-x*(TA/TC)  =  0                                            .....(ii)
A = [[1, 1],[ 1, -TA/TC]]
B = [500, 0]
V = linalg.solve(A, B)
y = round(V[0],1) 			#rpm
x = round(V[1],1) 			#rpm
#Calculating the speed of output shaft Y
NY = y-x*(TA/TB)*(TD/TE) 			#rpm
#Calculating the speed of wheel E
NE = NY 			#rpm
#Calculating the input power assuming 100 per cent efficiency
PY = PX 			#W
#Calculating the output torque of shaft Y
Ty = PY*60/(2*math.pi*NY*1000) 			#Output torque on shaft Y kN-m
#Tangential force between wheels D and E:
#Calculating the pitch circle radius of wheel E
rE = m*TE/(2*1000) 			#m
#Calculating the tangential force between wheels D and E
FtDE = Ty/rE 			#Tangential force between wheels D and E kN
#Tangential force between wheels B and C:
#Calculating the input torque on shaft X
Tx = PX*60/(2*math.pi*NX) 			#Input torque on shaft X N-m
#Calculating the fixing torque on the fixed wheel C
Tf = Ty-Tx/1000 			#Fixing torque on the fixed wheelC kN-m
#Calculating the pitch circle radius of wheel C
rC = m*TC/(2*1000) 			#m
#Calculating the tangential forces between wheels B and C
FtBC = Tf/rC 			#kN

#Results:
print " Output torque of shaft Y  =  %.3f kN-m."%(Ty)
print " Tangential force between wheels D and E  =  %.1f kN."%(FtDE)
print " Tangential force between wheels B and C  =  %.f kN."%(FtBC)

# note : answers are slightly different because of solve function and rounding off errors.

 Output torque of shaft Y  =  15.468 kN-m.
Tangential force between wheels D and E  =  176.8 kN.
Tangential force between wheels B and C  =  58 kN.