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A new study on the optimal prediction of partial
transmission ratios of three-step helical gearboxes
with second-step double gear-sets
VU NGOC PI
Faculty of Mechanical, Maritime and Materials Engineering
Delft University of Technology
Mekelweg 2, 2628 CD Delft
THE NETHERLANDS
v.n.pi@tudelft.nl
Abstract: - This paper presents a new study on the applications of the optimization and regression techniques
for optimal prediction of partial ratios of three-step helical gearboxes with second-step double gear-sets for
getting different objectives including the minimal gearbox length, the minimal gearbox cross section
dimension and the minimal mass of gears. In the study, based on the condition of the moment equilibrium of a
mechanic system including three gear units and their regular resistance condition, three optimization problems
for getting minimal gearbox length, the minimal gearbox cross section dimension and minimal mass of gears
were performed. In addition, explicit models for calculating the partial ratios of the gearboxes were found by
using regression analysis technique. Using these models, the determination of the partial ratios becomes
accurate and simple.
Key-Words: - Gearbox design; Optimal design; Helical gearbox; Transmission ratio.
1 Introduction
In optimal gearbox design, it is known that the
optimal determination of partial transmission ratios
is the most important task. This is because the partial
ratios are main factors which affect the size, the
mass, and the cost of the gearboxes.
Fig. 1 Transmission ratio of second gear step
versus mass of gears
Figure 1 shows the relation between the
transmission ratio of the second gear step and the
total mass of gears (calculated for a two step helical
gearbox with the total transmission ratio is 10, the
torque on the output shaft is 60.000 Nmm, the helix
angle is 120). It follows the figure that there is an
optimal value of the transmission ratio with which
the total mass of gear is minimal. Therefore, optimal
prediction of partial ratios of gearboxes has been
subjected to many studies.
Until now, there have been many studies on the
determination of the partial ratios of the three-step
helical gearboxes. For the gearboxes, the partial
ratios can be determined by two methods including
graph method and modeling method.
In graph method, the partial ratios are predicted
graphically. Using this method, V.N. Kudreavtev [1]
gave a graph for the prediction of the partial ratios
and of the first and the second gear steps (see
Figure 2). The partial ratio of the third step then can
be found by:
1
u2
u
312
=
⋅
h
u
uuu (1)
Where, is the total transmission ratio of the
gearbox. h
u
The graph method was also presented in other
studies such as in the study of Trinh Chat [2] and
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in the study of A.N. Petrovski et al [3].
In modeling method, based on the results of
optimization problems, models for the
determination of the partial ratios are found in
order to get different objectives. Using this
method, Romhild I and Linke H. [4] proposed the
following models for the prediction of the partial
ratios of the first and the second step in order to get
the minimal gearbox mass:
0.609
10.4643=⋅
h
uu (2)
0.262
21.205=⋅
h
uu (3)
Fig. 2 Determination of partial ratios of
three-step helical gearboxes [1]
To find the minimal mass of gears of the
gearboxes, Vu Ngoc Pi et al. [5] carried out a study
on optimal determination the partial ratios for
getting the minimal mass of gears. In the study, the
partial ratios of gear steps 2 and 3 are calculated as
follows:
2
3
10.314=⋅
h
uu
(4)
4
21.33=⋅
h
uu (5)
In both studies [4] and [5], after calculating the
partial ratios 1
u and 2
u, the partial ratio of the third
gear unit 3
u is determined by Equation 1.
Recently, Vu Ngoc Pi [6] introduced a new
study on optimal prediction of the partial
transmission ratios in order to get the minimal cross
section dimension of the gearbox. The partial ratios
of the second and the third gear steps are determined
as follows:
0.4291 0.4294 0.2848
22
20.144 0.1442 0.2853
331
1.0134 cbah
cbaba
Ku
uK
ψ
ψψ
≈⋅⋅⋅ (6)
0.4279 0.428 0.1423
33
30.2854 0.2854 0.1426
221
1.0269 cbah
cbaba
Ku
uK
ψ
ψψ
≈⋅ ⋅⋅
1ba
(7)
In which,
ψ
, 2ba
ψ
and 3ba
ψ
are coefficients
of helical gear face width of the first, the second and
the third gear units, respectively; kc2 and kc3 are
coefficients.
The partial ratio of the first gear step u then is
calculated by the following equation: 1
123
h
u
uuu
= (8)
⋅
It is clear that the modeling method is more
accurate and easier to determine the partial ratios
than the graph method. Especially, the method can
be usefully applied in computer programming or
other applications. In addition, it is allowed to have
different objectives when using the modeling
method for determining the partial transmission
ratios.
From previous studies, it is apparent that there
have been many researches on the splitting the total
transmission ratio for three-step helical gearboxes.
However, the optimal determination of the partial
ratios of the helical gearbox with second-step double
gear-sets has not been investigated. In this paper, a
study on optimal calculation of partial ratios for the
gearboxes for getting the minimal gear length, the
minimal gearbox cross section dimension and the
minimal mass of gears are presented.
2 Determination of the gearbox
length
In practice, the length of a three-step helical gearbox
with second-step double gear-sets is decided by the
dimension of L which can be determined as follows
(see Figure 3):
11 23
123
22
ww
ww w
dd
Laaa=++++ (9)
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The center distance of the first gear step can be
calculated by the following equation:
11 21 21 11
121 1
222
ww w w
ww
dd d d
ad
⎛⎞
=+=⋅ +
⎜⎟
⎝⎠
Fig. 3 Calculating schema for three-step helical
gearboxes with second-step double gear-sets
Or we have
21
11
11
2
w
wd
au
⎛⎞
=
⎜
⎝⎠
+
⎟
(10)
Using the same way, the center distances of the
second and the third steps are calculated by:
22
22
11
2
w
wd
au
⎛⎞
=
⎜
⎝⎠
+
⎟
(11)
23
33
11
2
w
wd
au
⎛⎞
=
⎜
⎝⎠
+
⎟
(12)
Where,
1
u, and –the transmission ratios of the
first, the second and the third gear units,
respectively;
2
u3
u
, and are pitch diameters (mm)
of driver gears of the first, the second and the third
gear units, respectively;
11w
d12w
d13w
d
21w
d, and are pitch diameters (mm)
of driven gears of the first, the second and the third
gear units, respectively;
22w
d23w
d
1ba
ψ
, 2ba
ψ
and 3ba
ψ
are coefficients of helical
gear face width of the first, the second and the third
gear units, respectively;
1w
a, and are center distances (mm) of
the first, the second and the third gear units,
respectively.
2w
a3w
a
From Equations 10, 11 and 12 and with the note
that 11 21 1
/
ww
ddu
=
, Equation 9 can be rewritten as
follows:
21 22 23
12
21
11
222
ww w
dd d
Luuu
⎛⎞
⎛⎞ ⎛⎞
3
1
2
=
⋅++ ⋅ ++ ⋅ +
⎜⎟
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠ ⎝⎠
(13)
For the first helical gear unit, the design
equation for pitting resistance is calculated by the
following equation [7]:
[
11 1 1
1111 2
1111
21
H
HMH H
ww
TK u
ZZZ bd u
ε
σσ
+
=≤
]
1
(14)
From (14) we have:
[]
()
[]
()
2
21
1111
11 2
11111
21 H
ww
HMH
bd u
TuKZZZ
ε
σ
⋅⋅
=⋅+ ⋅⋅⋅ (15)
Where, bw1 and dw11 are calculated by the
following equations:
()
1111
111 1
2
ba w
wbaw du
ba
ψ
ψ
⋅
⋅+
=⋅= (16)
21
11 1
w
wd
du
= (17)
Substituting Equation 16 and 17 into Equation
15 we get
[]
[
]
3
12101
11 2
1
4
ba w
dK
Tu
ψ
⋅⋅
=⋅ (18)
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Where
[]
[]
()
2
1
01 2
1111
H
HMH
KKZZZ
ε
σ
=⋅⋅⋅ (19)
From (18) the pitch diameter can be
calculated by 21w
d
[]
[]
1/3
2
11 1
21 101
4
wba
Tu
dK
ψ
⎛⎞
=⎜
⎜
⎝⎠
⎟
⎟
(20)
Calculating in the same manner, we have the
following equations for the second and the third gear
units:
[]
[]
1/3
2
12 2
22 202
4
wba
Tu
dK
ψ
⎛⎞
=⎜
⎜
⎝⎠
⎟
⎟
(21)
[]
[]
1/3
2
13 3
23 303
4
wba
Tu
dK
ψ
⎛⎞
=⎜
⎜
⎝⎠
⎟
⎟
(22)
In the above equations:
[]
[]
()
2
2
02 2
2222
ε
σ
=⋅⋅⋅
H
HMH
KKZZZ (23)
[] []
()
2
3
03 2
3333
ε
σ
=⋅⋅⋅
H
HMH
KKZZZ (24)
Where,
1
M
Z
, 2
M
Z
, 3
M
Z
are coefficients which
consider the effects of the gear material when
calculate the pitting resistance of the first, the second
and the third gear units, respectively;
1
H
Z
,2
H
Z
,3
H
Z
are coefficients which consider
the effects contact surface shape when calculate the
pitting resistance of the first, the second and the
third gear units, respectively;
1
ε
Z
,2
ε
Z
, 3
ε
Z
are coefficients which consider
the effects of the contact ratio when calculate the
pitting resistance of the first, the second and the
third gear units, respectively;
1
σ
H
- contact stresses of the first gear unit
(N/mm2);
[
]
1H
σ
,
[
]
2
σ
Hand 3
σ
⎡
⎤
⎣
⎦
H are allowable contact
stresses (N/mm2) of the first, the second and the
third gear units, respectively.
From the condition of the moment equilibrium
of the first gear step and the regular resistance
condition of the system we have:
[
]
[]
3
123
11 11
r
rbrt o
T
Tuuu
TT 3
η
η
=
=⋅⋅⋅ ⋅ (25)
In which,
brt
η
-helical gear transmission efficiency; brt
η
is
from 0.96 to 0.98 [7];
o
η
-transmission efficiency of a pair of rolling
bearing ; o
η
is from 0.99 to 0.995 [7].
Choosing 97.0
=
brt
η
, 992.0=
o
η
and
substituting them into Equation 24 we get
[]
[
]
11 123
0.8909 r
T
Tuuu
=
⋅
⋅⋅ (26)
Substituting (26) into (20) we have
[]
[]
1/3
1
21 10123
4.4898 r
wba
Tu
dKuu
ψ
⎛⎞
⋅⋅
=⎜
⎜
⋅⋅⋅
⎝⎠
⎟
⎟
(27)
For the second step, the following equation can
be given
[
]
[]
2
23
12 12
22
r
rbrt o
T
Tuu
TT 2
η
η
=
=⋅⋅ ⋅
⋅⋅ (28)
As it was done with the first step, by choosing
97.0
=
brt
η
and 992.0
=
o
η
Equation 28 can be
rewritten as follows:
[]
[
]
12 23
1.8518r
T
Tuu
=
⋅
⋅ (29)
Where,
[
]
11
T,
[
]
12
T -torque (Nmm) on the driver shaft
of the first and the second gear units, respectively;
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[]
r
T –torque (Nmm) on the out put shaft.
Substituting (29) into (21) we can have as
follows: 22w
d
[]
[]
1/3
2
22 2023
2.1601 r
wba
Tu
dKu
ψ
⎛⎞
⋅⋅
=⎜
⎜
⋅⋅
⎝⎠
⎟
⎟
(30)
Using the same way, the pitch diameter for the
third step is determined by the following
equation: 23w
d
[]
[]
1/3
3
23 303
4.1571 r
wba
Tu
dK
ψ
⎛⎞
⋅⋅
=⎜
⎜
⋅
⎝⎠
⎟
⎟
)
3
(31)
Substituting (27), (30) and (31) into Equation
13 with the note that , the length of
the gearbox can be rewritten as follows:
(
12
/
h
uuuu=⋅
[]
1/3 1/3
23
22
01 123
1/3 1/3
3
2
223 2 33 3
4.4898 2
11
2
4.1571
2.1601 11
1
rh
h
ba
ba c ba c
Tuuu
LKu
uu
u
u
Ku u K u
ψ
ψψ
⎡
⎛⎞
⎛⎞
⎛⎞
⋅⋅⋅
⎢
=⋅
⎜⎟
⎜⎟
⎜⎟
⎜⎟
⎜⎟
⎢
⎡⎤ ⋅⋅ ⎝⎠
⎣⎦ ⎝⎠
⎝⎠
⎣
⎤
⎛⎞⎛⎞⎛
⎛⎞ ⋅
⋅2
++
⎞
⎥
+⋅++⋅
⎜⎟⎜⎟⎜
⎜⎟
⋅⋅ ⋅ +
⎟
⎥
⎝⎠
⎝⎠⎝⎠⎝
⎦
⎠
(32)
Where, 02
201
cK
KK
⎡
⎣
=⎡
⎣⎦
⎤
⎦
⎤
and 03
301
cK
KK
⎡
⎣
=⎡
⎣⎦
⎤
⎦
⎤
are
coefficients.
3 Determination of the dimension of
the gearbox cross section
The cross section of a three-step helical gearbox
with second-step double gear-sets is decided by the
dimension of A which is determined as follows (see
Figure 2):
A
Lh=⋅ (33)
Where
L-the length of the gearbox (mm) which is
determined by Equation 32.
h - the height of the gearbox (mm); h is
determined by the maximal value among pitch
diameters , and or it can be predicted
by the following equation:
21w
d22w
d23w
d
(
21 22 23
max , ,
www
hddd=
From (27), (30) and (31), Equation 33 can be
rewritten as follows:
3
2
22
123 2 23 3 3
4.4898 4.1571
2.1601
max ; ;
ψψ ψ
⎛⎞
⋅
⋅
⋅
=⎜⎟
⋅⋅ ⋅ ⋅ ⋅
⎝⎠
h
ba ba c ba c
uu
u
huu k u k (35)
4 Determination of the mass of gears
For a three-step helical gearbox with second-step
double gear-sets, the mass of gears can be
determined as follows:
12
2a
GG G G
3
=
+⋅ + (36)
Where, ,and are the mass of gears of
unit 1, unit 2a and unit 3, respectively (see Figure 2).
can be determined as follows:
1
G2a
G3
G
1
G
22
11 1 1 2 1
1()
4
ww
dbeeu
G
πρ
⋅
⋅⋅⋅+⋅
= (37)
In which,
ρ
-the density of the gear material (kg/m3);
e1, e2 -volume coefficients of gear 1 and 2,
respectively; the volume coefficient of a gear is ratio
of the actual volume to the theoretical volume of the
gear. For a helical gear unit, we can have e1=1;
e2=0.6 [2].
From Equation 37, with is
diameter coefficient of the first helical gear step, the
mass of gears of unit 1 can be calculated as follows:
11
/
bd w w
bd
ψ
=11
32
111 1
1(1 0.6 )
4
bd w
du
G
πρψ
⋅
⋅⋅⋅+⋅
= (38)
For the first helical gear unit, the allowable
torque on the driving shaft 1 can be predicted by the
following equation [7]:
[]
3
1111 01
11 1
[]
2( 1)
bd w
duK
Tu
ψ
⋅⋅⋅
=⋅+
(39)
It follows Equation 39 that:
311
111 101
2( 1)[ ]
[]
bd w uT
duK 1
ψ
⋅
+⋅
⋅= ⋅ (40)
)
(34)
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Substituting (40) into (38) the mass of gears of
e first step is determined as follows
th
2
11 1 1
101 1
2[](1)(10.6
4[ ]
Tu u
GKu
πρ
⋅⋅⋅ ⋅ +⋅+ ⋅
=⋅⋅
)
(41)
of e second gear unit and the third
gear unit :
By calculating in the same way, the following
equations were found for the calculation of the mass
of gears th 2a
G
3
G
2
12 2 2
202 2
2[](1)(10.6
4[ ]
aTu u
GKu
πρ
⋅⋅⋅ ⋅ +⋅+ ⋅
=⋅⋅
)
(42)
2
13 3 3
303 3
2[](1)(10.6
4[ ]
Tu u
GKu
πρ
⋅⋅⋅ ⋅ +⋅+ ⋅
=⋅⋅
)
(43)
In the above equations,
[
]
01
K
,
[
]
02
K and
[
]
03
K
are determined by Equation 19, 23 and 24,
resp
regular resistance condition of
e system we have:
ectively.
Based on the condition of the moment
equilibrium of the mechanic system which includes
the gear units and the
th
[]
[]
33
123
11
T11
r
rbrt o
T
Tuuu
T
η
η
==⋅⋅⋅⋅
(44)
As it was done in Section 3, by choosing
97.0=
brt
η
, 992.0=
o
η
and substituting them into
quation 44 we have
E
[] []
11 123
0.8909 r
T
Tuuu
=⋅⋅ ⋅ (45)
to (41) the following
quation can be obtained:
Substituting (45) in
e
2
11
12
01 1 2 3
2[](1)(10.6
4 [ ] 0.8909
r
Tu u
GKu
πρ
⋅⋅⋅ +⋅+ ⋅
=⋅
⋅⋅
)
uu⋅⋅
(46)
For the second helical gear unit it can be written
s:
a
[
]
[]
22
23
12 12
22
r
rbrt o
T
Tuu
η
TT
η
=
= ⋅
⋅
With
⋅⋅
⋅ (47)
97.0
=
brt
η
and 992.0=
o
η
Equation 47
becomes
[
[] ]
12 23
1.8518r
T
uu
=
T
⋅
⋅ (48)
From Equations 48 and 42 we get the mass of
gear of the second unit:
2
22
22
02
4 [ ] 1.8518
r
aKuu
2 3
2[](1)(10.6)Tu u
G
πρ
⋅
⋅⋅ +⋅+ ⋅
In exactly similar manner, we can derive the
m ss of gears of unit 3 as follows:
=⋅
⋅⋅⋅
(49)
a
2
33
32
3
4 [ ] 0 9622
r
Ku
=⋅
⋅⋅
(50)
Substituting (46), (49) and (50) int
03
(1)(10.6)
2[] .
uu
T
G
πρ
+⋅+ ⋅
⋅⋅⋅
o (28) with
the note that
(
)
12
/
h
uuuu
3
=
⋅, the mass of gears of
the gearbox can be rewritten as follows:
()
222
23 23
2
01
2
233
22
)(1
(
1.8518 .9622
22
223 33
()10.6/
2[]
4[ ] 0.8909
(1 0.6)
1)(10.6)
0
hh
r
h
CC
uuu uuu
T
GKu
uu
uu
kuu u
πρ
⎧⎡⎤
+⋅ ⋅+ ⋅ ⋅
⋅⋅⋅ ⎪⎣⎦
k
=
+
⎨
⋅⋅
⎪
⎩
⎫
+⋅+ ⋅
+ ⋅ ⎪
+⎬
⋅⋅⋅ ⋅ ⎪
⎭
(51)
1
]
+⋅
+⋅
where, 2020
[]/[
C
kKK
=
and 1
]
.
artial ratios for
From timal problem for
determining the minimal gearbox length can be
expressed as follows:
The objective function is:
3030
[]/[
C
kKK=
5 Optimization problems and results
5.1 For finding the optimal p
getting the minimal gearbox length
5.1.1 Optimization problem
Equations 32, the op
23
n ( ; ; )
h
Lfuuumi
=
(52)
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With the following constraints:
22min2 uu ≤≤
(53)
maxmin hhh uuu ≤≤
umax
3min 3 3max
uuu≤≤
2min 2 2maxccc
KKK≤≤
3min 3 3maxccc
KKK≤≤
1min 1 1max
ψ
ψψ
≤≤
ba ba ba
max22min2 bababa
ψ
ψ
ψ
≤≤
3min 3 3max
ψ
ψψ
≤≤
b
ogram was b
the above optimizat
th
a ba ba
A computer pr uilt in order to solve
ion problem. The used data in
e program were: 2c
K
nd 3c
a
K
are from 1 to 1.3,
1ba
ψ
, 2ba
ψ
and 3ba
ψ
are fro 0.25 to 0.4 [7], and
are from 1 to 9 [1]; is from 30 to 120.
whe ing he following values of
coef
m 2
u
3
u h
u
5.1.2 Results and discussions
Figure 4 describes the relation between the partial
transmission ratios and the total transmission ratio
n calculat with t
ficients: 2c
K
=1.2, 3c
K
=1.3, 1ba
ψ
=0.25,
2ba
ψ
=0.3 and 3ba
ψ
=0.35.
It is observed that with the increase of the total
ratio h
uthe partial ratios increase. In addition, it is
found that with the increase of the total ratio, the
increase of partial ratio of the first step 1
u is mu h
larger than that of the third step 3
u. It can be
explained that with the increase of the total ratio h
u,
the torque on the output shaft r
Tis much larger than
that on the driver shaft of the first gear unit 11
T. For
that reason, the increase of the partial ratio should
be m
c
und
for finding the optimal values of partial ratios u2 and
u3 of the second and the third step, respectively:
3
u
uch smaller than that of 1
u in order to reduce
the gearbox length.
To find the models for determination of the
partial ratios, regression analysis was carried out
based on the results of the optimization program.
The following regression models then were fo
()
()
0.4386 0.2649
22
20.1328 0.3061
1
33
1.8452 Cba h
ba
Cba
Ku
uK
ψ
ψ
ψ
⋅
≈⋅ ⋅
⋅ (54)
()
()
0.4386 0.1219
33
30.2974 0.1412
1
22
1.0689 Cba h
ba
Cba
Ku
uK
ψ
ψ
ψ
⋅
≈⋅ ⋅
⋅ (55)
Fig. 4: Partial transmission ratios
. The icients of
for the models for the second step and
the
partial transmission ratio of the first gear
step 1 then can be calculated by the following
equation:
versus the total transmission ratio
The above regression models fit quite well with
the calculated data coeff
determination
third step were 20.9999R= and 20.9998R=,
respectively.
Equations 54 and 55 can be used for the
prediction of the partial transmission ratios u2 and u3
of gear steps 2 and 3, respectively. After finding u2
and u3, the
u
123
uu
h
u
u=
⋅
(56)
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5.2 For finding the optimal partial ratios for
From mal problem
for finding the minimal cross section dimension of
the gearbox can be exp
The objective function is:
n ( ; ; )
h
getting the minimal cross section dimension
of the gearbox
5.2.1 Optimization problem
Equations 33, 32 and 35, the opti
ressed as follows:
23
mi
A
fu u u= (57)
With the following constraints:
(58)
maxmin hhh uuu ≤≤
max22min2 uuu ≤≤
3min 3 3max
uuu≤≤
2min 2 2maxccc
KKK≤≤
3min 3 3maxccc
KKK≤≤
1min 1 1max
ψ
ψψ
≤≤
ba ba ba
max22min2 bababa
ψ
ψ
ψ
≤≤
3min 3 3max
ψ
ψψ
≤≤
For lvin
co
were used in the program
ba ba ba
so g the above optimization problem a
mputer program was built. The following data
:2c
K
and 3c
K
are from 1 to
1.3, 1ba
ψ
, 2ba
ψ
and 3ba
ψ
are from
2
uand 3
u are from 1 to 9 [1]; h
u is 0.25 to 0.4 [7],
from 30 to 120.
The be the p
and the total transmission ratio is described in Figure
5. T
5.2.2 Results and discussions
relation tween artial transmission ratios
he following data were used in the calculation:
2c
K
=1.2, 3c
K
=1.3, 1ba
ψ
=0.25, 2ba
ψ
=0.3 and
3ba
ψ
=0.35.
As in the optimization problem for getting th
minimal gearbox length (see Subsection 5.1.2), with
the increase o the total transmission ratio h
uthe
partial transmission ratios increase. Also, while the
partial transmission rat of the first step 1
u
increases very much, the partial transmission ratio of
the third step 3
u nearly does not change when the
total ratio increases. T is is because with the
increase of the total ratio h
u, the torque on the output
shaft r
Tis much larger than that on the driver shaft
of the first gear unit 11
T. Therefore, the partial
trans
e
f
io
h
of the transmission ratio of a helical
ear set [7]. Consequently, the objective of getting
the minimal gearbox cross section dimension should
be used in cases that the total transmission ratio is
less than 90.
mission ratio 3
ushould be much smaller than
the partial transmission ratio 1
u in order to reduce
the gearbox size as well as the gearbox cross section
dimension.
It is observed that when the total transmission
ratio is beyond about 90, the partial transmission
ratio of the first gear unit is larger than 9 (see Figure
5) - the limit
g
Fig. 5: Partial transmission ratios
versus the total transmission ratio
From the calculated results of the optimization
program, the following regression models were
found to determine the optimal values of partial
ratios u2 and u3 of the second and the third step,
respectively:
()
()
0.285
0.43
22
20.1443 1
33
1.3655 Cba h
ba
Cba
Ku
uK
ψ
ψ
ψ
⋅⎛⎞
≈⋅ ⋅
⎜⎟
⋅⎝⎠
(59)
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236
Issue 11, Volume 2, November 2007
()
()
0.1424
0.428
33
30.2852 1
22
0.8409 Cba h
ba
Cba
Ku
uK
ψ
ψ
ψ
⋅⎛⎞
≈⋅ ⋅
⎜⎟
⋅⎝⎠ (60)
odels for t cond step step
of the first step u1 then can be determined as follows:
The above models fit very well with the
calculated data. The coefficients of determination for
the m he se and the third
were 20.9995R= and 20.9998R=, respectively.
Equations 59 and 60 are used to calculate the
transmission ratios u2 and u3 of steps 2 and 3,
respectively. From u2 and u3, the transmission ratio
123
h
u
uuu
=⋅ (61)
l ratios for
oObjective function:
(62)
With the following constraints:
(63)
fro are from 1 to 1.3 [7] and
m 0.
5.3 For finding the optimal partia
getting the minimal mass of gears
5.3.2 Optimization problem
From Equation 51, the optimal problem for
determining the partial transmission ratio in order to
get the minimal mass of the gears is described as
llows:
f
23
min ( ; ; )
h
Gfuuu=
maxmin hhh uuu ≤≤
max22min2 uuu ≤≤
3min 3 3max
uuu≤≤
2min 2 2maxCCC
kkk≤≤
3min 3 3maxCCC
kkk≤≤
For performing the above optimization
problem, a computer program was built. The data
used in the program were follows: 2
uand 3
u were
m 1 to 9 [1], 2C
kand 3C
k
h was frou 30 to 18
5.3.2 Results and discussions
Figure 6 shows the relation between the partial
transmission ratios and the total transmission ratio
when the coefficients 2c
K
and 3c
K
equal 1.1.
As in Sections 5.1 and 5.2, with the increase of
the total ratio the partial ratio of the first step
increases significantly. In contrast, the partial
transmission ratio of the third step increases very
slowly. The reason of that can be explained as in the
above sections.
h
u1
u
3
u
From the results of the optimization program,
regression analysis was carried out and the following
models were found for determining the optimal
values of the partial ratios of the second and the
third step:
Fig. 6: Partial transmission ratios versus the
total transmission ratio
0.3827 0.2721
2
20.089
3
1.5598 Ch
C
k
u
k
≈⋅⋅
u
(64)
0.2984 0.0814
3
30.2107
2
2.0604 Ch
C
k
u
k
≈⋅⋅
u
is calculated by the following equation:
(65)
As in Subsection 5.1.2 and 5.2.2, the regression
models (64) and (65) fit quite well with the
calculated data. The coefficients of determination for
the above models were and ,
respectively.
20.9998R=20.9978R=
Equations 19 and 20 are used to determine the
transmission partial ratios and
2
u3
u of the second
and the third helical gear units. After finding
and ,
2
u3
u the transmission ratio of the first gear unit
1
u
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Issue 11, Volume 2, November 2007
123
h
u
uuu
=⋅ (66)
Conclusion that the minimal gearbox length,
udreavtev; I.A. Gierzaves; E.G.
[2]
[3]
[4]
[5]
[6]
[7] Uyen, Design and calculus of
6
It can be concluded
the minimal gearbox cross section dimension and the
minimal mass of gears of a three-step helical
gearbox with second-step double gear-sets can be
obtained by optimal splitting the total transmission
ratio of the gearbox.
The objective of finding the optimal partial
ratios in order to get the minimal gearbox cross
section dimension must be used for the gearboxes
with the total transmission ratios is less than 90.
Models for prediction of the optimal partial
ratios of the gearboxes in order to get the minimal
gearbox length, the minimal gearbox cross section
dimension and the minimal mass of gears have been
found.
Using explicit models, the partial ratios of the
gearboxes can be calculated accurately and simply
by. Further researches on the optimal determination
of the partial transmission ratios of helical gearboxes
with four and more steps should be carried out.
References:
[1] V.N. K
Glukharev, Design and calculus of gearboxes (in
Russian), Mashinostroenie Publishing, Sankt
Petersburg, 1971.
Trinh Chat, Optimal calculation the total
transmission ratio of helical gear units (in
Vietnamese), Scientific Conference of Hanoi
University of Technology, 1996, pp. 74-79.
A.N. Petrovski, B.A. Sapiro, N.K. Saphonova,
About optimal problem for multi-step gearboxes
(in Russian), Vestnik Mashinostroenie, No. 10,
1987, pp. 13-14.
Romhild I. , Linke H., Gezielte Auslegung Von
Zahnradgetrieben mit minimaler Masse auf der
Basis neuer Berechnungsverfahren, Konstruktion
44, 1992, pp. 229- 236.
Vu Ngoc Pi, Nguyen Dang Binh, Vu Quy Dac,
Phan Quang The, Optimal calculation of total
transmission ratio of three-step helical gearboxes
for minimum mass of gears (In Vietnamese),
Journal of Science and Technology of Technical
Engineering Universities, Vol. 55, 2006, pp. 91-
93.
Vu Ngoc Pi, A new study on optimal calculation
of partial transmission ratios of three-step helical
reducers for getting minimal cross section
dimension, 2nd WSEAS International Conference
on Computer Engineering and Applications
(CEA’08), Acapulco, Mexico, January 25-27,
2008, pp. 290-293.
Trinh Chat, Le Van
Mechanical Transmissions (in Vietnamese),
Educational Republishing House, Hanoi, 1998.
WSEAS TRANSACTIONS on
APPLIED and THEORETICAL MECHANICS
Vu Ngoc Pi
ISSN: 1991-8747
238
Issue 11, Volume 2, November 2007
