n12_spur_gears
page 1 of 5
determine p d (or m) for unknown gear
count the number of teeth N
measure OD [inch] - better for even number of teeth, not as accurate for odd number of teeth
assume full depth involute profile
pd 
N
d
OD  d  2a 
a
1
pd
N
2

pd pd
pd 
N2
OD
p d must be standard valuie from Table 12-2
if not, convert to module
m
d
1

N pd
and check Table 12-3
example
larger Boston gear
smaller Boston gear
N = 64
N = 16
OD= 4.117 inch
OD= 1.117 inch
pd 
N2
 16.03
OD
pd 
N2
 16.11
OD
use p d = 16
use p d = 16
n12_spur_gears
page 2 of 5
design a spur gear reducer for 125 HP at 1000 rpm input with NFS > 2 for 10 years life
P / G ≈ 2.5
C ≈ 6.65 in
choose
NP
choose
= 25°
pd = N / d
NG

pd
F
QV
material
heavy load
d = N / pd
r = N / 2 pd
 = P / G = NG / NP
C = rP + rG = NP / 2 pd + NG / 2 pd = (NP + NG) / 2 pd
~NP
11.40
15.20
19.00
22.80
30.40
pd
3
3
4
5
5
6
8
NP
11
11
15
19
19
23
30
2 pd C = NP + NG = (1+) NP
NG =  NP
NP = 2 pd C / (1+)
pd
3
4
5
6
8
NG = NP
~NG
28.50
38.00
47.50
57.00
76.00
NG
28
29
38
47
48
57
76

2.545
2.636
2.533
2.474
2.526
2.478
2.533
rP
1.833
1.833
1.875
1.900
1.900
1.917
1.875
rG
4.666
4.833
4.750
4.700
4.800
4.750
4.750
C
6.500
6.667
6.625
6.600
6.700
6.667
6.625
Table 12-12
J not available
J not available
all OK for no tip interference using gear_AGMA.xls
pd = 8 are same size gears but smaller teeth than pd = 4
choose
pd = 6
NP = 23
NG = 57
rP = 1.917 in
NP and NG both odd provides hunting tooth effect
 1000 rev  min  2 rad 

P = 

 = 104.72 rad/sec
 min  60 sec  rev 
P=T

 550 ft .lbf
sec

TP = P / P = 125 HP
 104.72 rad  HP. sec
 12 in 

 = 7878 in.lbf
 ft 
n12_spur_gears
T = Wt r
Wt = TP / rP = 4110 lbf
page 3 of 5
Wr = Wt tan= 1917 lbf
 104.72 rad  60 sec  ft 
 = 1004 ft/min
Vt = rP P = 1.917 in 


sec

 min  12 in 
recommend QV ≥ 8
Table 12-7 Norton
choose
QV = 8
choose
F ≈ 12 / pd = 2 in
b 
page 746 Norton 5th ed ( 8/pd < F <16/pd )
Wt p d K a K m
KSK BK I
F J Kv
NP =21
0.31
NG = 55
57
135
JP = 0.32
Eq 12.15us Norton
23
26
0.33
0.32
0.31
0.33
for  = 25°, tip loading, standard addendum
B = (12-QV)2/3 / 4 = 0.6300
Eq 12.17b Norton
A = 50 + 56 ( 1 – B ) = 70.722

A
Kv = 
A V
t

Eq 12.17a Norton
B

 = 0.7920


Eq 12.16us Norton
Km = 1.6
Table 12-16 Norton
Ka = 1.0
assume uniform driver, uniform driven
KS = 1.0
not yet defined by AGMA
KB = 1.0
solid pinion
KI = 1.0
not an idler
 bP 
Table 12-12 Norton

Table 12-17 Norton

Wt p d K a K m
4110 lbf  6 in 1 11.6 111 = 77.84 ksi
KSK BK I 
2 in 0.320.7920
F JP Kv
choose
AISI 4140 nitrided steel
Sfb’ = 39.5 ksi (mean value)
Table 12-20 Norton
n12_spur_gears
Sfb 
KL
Sfb '
KTKR
Eq 12.24 Norton
 1000 rev 
 60 min
N= 
10 years 
 min 
 hour
KL = 1.3558 N -0.0178 = 0.9103
 24 hour

 day
 365 day 

 = 5.256 x 109 revolutions
year


assume commerical use
KT = 1.0
assume T < 250° F
KR = 1.0
assume reliability = 99%
Sfb 
page 4 of 5
Figure 12-24 Norton
Table 12-19 Norton
KL
0.9103
39.5 ksi  = 35.96 ksi
Sfb ' 
11
KTKR
NFS = Sfb / bP = 0.46
change
F = 8 in
new
Km = 1.766
new
 bP 
new
NFS = 1.67
change
Table 12-16 Norton

pd = 4
NP = 15
keep
= 25°
new
Wt = 4202 lbf
QV = 8
NG = 38
rP = 1.875 in
F = 6 in
AISI 4140 nitrided steel
Vt = 981.7 fpm
NP =14
0.34
NG = 35
38
55
JP = 0.35

Wt p d K a K m
4110 lbf  6 in 1 11.766 111 = 21.48 ksi
KSK BK I 
8 in 0.320.7920
F JP Kv
15
17
0.37
0.35
0.34
0.38
for  = 25°, tip loading, standard addendum
same
B = 0.6300
new
KV = 0.7934
new
Km = 1.7
A = 70.722
Table 12-12 Norton
Eq 12.17a and 12.17b Norton
Eq 12.16us Norton
Table 12-16 Norton
n12_spur_gears


Wt p d K a K m
4202 lbf  4 in 1 11.7  111 = 17.15 ksi
KSK BK I 
6 in 0.350.7934
F JP Kv
new
 bP 
new
NFS = 2.10
OK
assume gear load split equally between two ball bearings on each side of pinion
Wr = Wt tan  = 1959 lbf
W  Wt2  Wr2 = 4636 lbf
assume no axial load
P = W / 2 = 2318 lbf
N = 5.256 x 109 revolutions
L10 = 5256
from above
millions of revolutions
C = P (L10)1/ 3 = 40,302 lbf
use two 6328 bearings with C = 44,000 lbf
page 5 of 5