gear’s design for the mechanism’s highest efficiency

7/31/2019 GEARS DESIGN FOR THE MECHANISMS HIGHEST EFFICIENCY

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International Conference on Engineering Graphics and Design - Bucharest,Romania 2005

GEARS DESIGN FOR THE MECHANISMS HIGHEST EFFICIENCY

Relly PETRESCU, Florian PETRESCU

Abstract: The paper presents an original method to determine the efficiency of the gear. The originality

of this method relies on the eliminated friction modulus. The paper is analyzing the influence of a few

parameters concerning gear efficiency. These parameters are: z1 - the number of teeth for the primary

wheel of gear; z2 - the number of teeth of the secondary wheel of gear; 0 - the normal pressure angle

on the divided circle; - the inclination angle. With the relations presented in this paper, one can

synthesize the gears mechanisms.

Key Words: efficiency, gear, constructive parameters, teeth, outside circle, wheel.

1. INTRODUCTION

In this paper the authors present an original method

for calculating the efficiency of the gear.

The originality consists in the way of determination

of the gears efficiency, because one hasnt used thefriction forces of couple (this new way eliminates the

classical method). One eliminates the necessity of

determining the friction coefficients by different

experimental methods as well. The efficiency

determinates by the new method is the same like the

classical efficiency, namely the mechanical efficiency of

the gear.

The considerate model is very simple. It can be

followed in figure 1.

2. DETERMINING THE MOMENTARY

MECHANICAL EFFICIENCY

The calculating relations [1,2], are the next (see the

fig. 1):

+=

+=

=

=

=

=

1221

1112

112

1

1

sin

cos

sin

cos

vvv

FFF

vv

vv

FF

FF

m

m

m

(1)

with: Fm - the motive force (the driving force);

- the transmitted force (the useful force);F

F - the slide force (the lost force);

v1 - the speed of element 1, or the speed of wheel

1 (the driving wheel);

v2 - the speed of element 2, or the speed of wheel

2 (the driven wheel);

v12 - the relative speed of the wheel 1 in relation

with the wheel 2 (this is a sliding speed).

The consumed power (in this case the driving

power):

1vFPP mmc = (2)

The useful power (the transmitted power from the

profile 1 to the profile 2) will be written:

P

O1

K1

n

n

t

t

1

1

rb1

rp1

Fm

F

F

v1

v12

v2

1

1

0

2 2002 Victoria PETRESCU

Fig. 1. The forces of the gear

12

12 cos == vFvFPP mu (3)

The lost power will be written:

12

112 sin == vFvFP m (4)

The momentary efficiency of couple will be

calculated directly with the next relation:

=

==

12

1

12

1

cos

cos

i

m

m

mc

ui vF

vF

P

P

P

P

(5)

The momentary losing coefficient [3], will be written:

=+=+

=

==

1sincos

sinsin

12

12

12

1

12

1

ii

m

m

mi

vF

vF

P

P

(6)

One can easily see that the sum of the momentary

efficiency and the momentary losing coefficient is 1:Now one can determine the geometrical elements of

gear. These elements will be used in determining the

couple efficiency, .

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3. THE GEOMETRICAL ELEMENTS OF THE

GEAR

One can determine the next geometrical elements of

the external gear, [1,2], (for the right teeth, =0):The radius of the basic circle of wheel 1 (of the

driving wheel), (7):

011 cos2

1= zmrb (7)

The radius of the outside circle of wheel 1 (8):

)2(2

)2(2

1111 +=+= z

mmzmra (8)

One determines now the maximum pressure angle of

the gear (9):

2

cos

)2(2

1

cos2

1

cos

1

01

1

01

1

11

+

=

+

=

==

z

z

zm

zm

r

r

a

bM

(9)

And now one determines the same parameters for the

wheel 2, the radius of basic circle (10) and the radius of

the outside circle (11) for the wheel 2:

022 cos21 = zmrb (10)

)2(2

22 += zm

ra (11)

Now one can determine the minimum pressure angle

of the external gear (12, 13):

++

+=

=+

+=

=+=

=

]44sin

sin)[(2

cos)2(

2sin)(

2

1

)(

2022

2

021

022

22

2

021

22

22021

11

zz

zzm

zz

mzzm

rrtgrrN

r

Ntg

babb

bm

(12)

)cos/(]44sin

sin)[(

012022

2

0211

++

+=

zzz

zztg m(13)

Now one can determine, for the external gear, theminimum (13) and the maximum (9) pressure angle for

the right teeth.

For the external gear with bended teeth (0) one

uses the relations (14, 15 and 16):

cos

0tgtg t = (14)

t

t

tm

z

zz

zztg

cos

cos]4

cos4

cos

sin

cos

sin)[(

1

2

2

222

211

++

+=

(15)

2cos

cos

cos

cos1

1

1

+

=

z

z t

M (16)

For the internal gear with bended teeth (0) oneuses the relations (14 with 17, 18-A or 19, 20-B):

A. When the driving wheel 1, has external teeth:

t

t

tm

z

zz

zztg

cos

cos]4

cos4

cos

sin

cos

sin)[(

1

2

2

222

211

+

+=

(17)

2cos

cos

cos

cos1

1

1

+

=

z

z t

M (18)

B. When the driving wheel 1, have internal teeth:

t

t

tM

z

zz

zztg

cos

cos]4

cos4

cos

sin

cos

sin)[(

1

2

2

222

211

++

+=

(19)

2cos

cos

cos

cos1

1

1

=

z

z t

m (20)

4. DETERMINING THE EFFICIENCY

The efficiency of the gear will be calculated through

the integration of momentary efficiency on all sections of

gearing movement, namely from the minimum pressure

angle to the maximum pressure angle (21), [1, 2]:

5.0)(4

)2sin()2sin(

]2

)2sin()2sin(

[2

1

])2sin(2

1[

2

1

cos11 2

+

=

=+

=

=+

=

=

=

mM

mM

mM

a

i

M

m

M

m

M

m

dd

(21)

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Table 1.

Determining the efficiency of the

gears right teeth for i12effective= - 4

5. THE CALCULATED EFFICIENCY OF THE

GEAR

We shall now see four tables with the calculated

efficiency depending on the input parameters and once

we proceed with the results we will draw some

conclusions.

The input parameters are:

z1 = the number of teeth for the driving wheel 1;

z2 = the number of teeth for the driven wheel 2, or the

ratio of transmission, i (i12=-z2/z1);0 = the pressure angle normal on the divided circle;

= the bend angle.

We begin with the right teeth (the toothed gear), with

i=-4, once for z1 we shall take successively different

values, rising from 8 teeth.

One can see that for 8 teeth of the driving wheel the

standard pressure angle, 0=200, is to small to be used

(one obtains a minimum pressure angle, m, negative

and this fact is not admitted!).

In the second table we shall diminish (in module) the

value for the ratio of transmission, i, from 4 to 2.

One willl see how for a lower value of the number of

teeth of the wheel 1, the standard pressure angle (0=200)

is to small and it will be necessary to increase it to a

minimum value.

Table 2.

Determining the efficiency of the

gears right teeth for i12effective= - 2i12effective= - 4 right teeth

z1 =8 z2 =32

0 =200 ? 0 =29

0 0 =350

m = -16.220 ? m = 0.7159

0 m = 11.13030

M=41.25740 M=45.5974

0 M=49.05600

=0.8111 =0.7308z1 =10 z2 =40

0 =200

? 0 =260 0 =30

0

m = -9.890 ? m = 1.3077

0 m = 8.22170

M=38.45680 M=41.4966

0 M=43.80600

=0.8375 =0.7882z1 =18 z2 =72

0 =190 0 =20

0 0 =300

m = 0.98600 m =2.7358

0 m =18.28300

M=31.68300

M=32.25050

M=38.79220

=0.90105 =0.8918 =0.7660z1 =30 z2 =120

0 =150 0 =20

0 0 =300

m = 1.50660 m =9.5367

0 m =23.12250

M=25.10180 M=28.2414

0 M=35.71810

=0.9345 =0.8882 =0.7566z1 =90 z2 =360

0 =80 ? 0 =9

0 0 =200

m =-0.16380 ? m =1.5838

0 m =16.49990

M=14.36370 M=14.9354

0 M=23.18120

=0.9750 =0.8839

i12effective= - 2 right teeth

z1 =8 z2 =16

0 =200 ? 0 =28

0 0 =350

m=-12.650 ? m = 0.9149

0 m =12.29330

M=41.25740 M=45.0606

0 M=49.05590

=0.8141 =0.7236z1 =10 z2 =20

0 =200

? 0 =250 0 =30

0

m = -7.130

? m = 1.3330

0 m = 9.41060

M=38.45680 M=40.9522

0 M=43.80600

=0.8411 =0.7817z1 =18 z2 =36

0 =180

0 =200

0 =300

m = 0.67560 m =3.9233

0 m =18.69350

M=31.13510 M=32.2505

0 M=38.79220

=0.9052 =0.8874 =0.7633z1 =90 z2 =180

0 =80 0 =20

0 0 =300

m =0.52270 m =16.5667

0 m =27.78250

M=14.36370 M=23.1812

0 M=32.09170

=0.9785 =0.8836 =0.7507

For example, if z1=8, the necessary minimum value is0=29

0 for an i=-4 (see the table 1) and 0=280 for an i=-

2 (see the table 2). If z1=10, the necessary minimum

pressure angle is 0=260 for i=-4 (see the table 1) and

0=250 for i=-2 (see the table 2).

When the number of teeth of the wheel 1 increases,

one can decrease the normal pressure angle, 0. Oneshall see that for z1=90 one can take less for the normal

pressure angle (for the pressure angle of reference),

0=80.

In the table 3 on increase the module of i, value

(for the ratio of transmission), from 2 to 6.

In the table 4, the teeth are bended (0). The modulei, take now the value 2.

6. CONCLUSION

The efficiency (of the gear) increases when the

number of teeth for the driving wheel 1, z1, increases too

and when the pressure angle, 0, diminishes; z2 or i12 are

not that much influenced about the efficiency value;

One can easily see that for the value 0=200, the

efficiency takes roughly the value 0.89 for any values

of the others parameters (this justifies the choice of this

value, 0=200, for the standard pressure angle of

reference).But the better efficiency may be obtained only for an

0200.

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Table 3.

Determining the efficiency of the gears

right teeth for i12effective= - 6

But the pressure angle of reference, 0, can be

decreased the same time the number of teeth for the

driving wheel 1, z1, increases, to increase the gears

efficiency;

Contrary, when we desire to create a gear with a low

z1 (for a less gauge), it will be necessary to increase the

0 value, for maintaining a positive value form (in thiscase the gear efficiency will be diminished);

When increases, the efficiency, , increases too, but

the growth is insignificant.

The module of the gear, m, has not any influence on

the gears efficiency value.

When 0 is diminished one can take a higher normal

module, for increasing the addendum of teeth, but the

increase of the m at the same time with the increase of

the z1 can lead to a greater gauge.

The gears efficiency, , is really a function of0 and

z1: =f(0,z1); m and M are just the intermmediate

parameters.For a good projection of the gear, its necessary a z1

and a z2 greater than 30-60; but this condition may

increase the gauge of mechanism.

Table 4.

The determination of the gears

parameters in bend teeth for i=-4

i12effective= - 4 bend teeth Table 4

=150

z1 =8 z2 =32

0 =200

? 0 =300 0 =35

0

m=-16.8360? m = 1.1265

0 m = 9.44550

M=41.08340 M=46.2592

0 M=49.29530

=0.8046 =0.7390

z1 =18 z2 =72

0 =190 0 =20

0 0 =300

m =0.327150 m =2.0283

0 m =17.18400

M=31.71800 M=32.32020 M=39.18030

=0.9029 =0.8938 =0.7702

z1 =30 z2 =120

0 =150 0 =20

0 0 =300

m =1.02690 m =8.8602

0 m =22.15500

M=25.13440 M=28.4591

0 M=36.25180

=0.9357 =0.8899 =0.7593

z1 =90 z2 =360

0=90

0=200

0=300

m =1.31870 m =15.8944

0 m =26.94030

M=14.96480 M=23.6366

0 M=32.82620

=0.9754 =0.8845 =0.7513

i12effective= - 6 right teeth

z1 =8 z2 =48

0 =200 ? 0 =30

0 0 =350

m=-17.860? m = 1.7784

0 m =10.6600

M=41.25740 M=46.1462

0 M=49.05590

=0.8026 =0.7337z1 =10 z2 =60

0 =200 ? 0 =26

0 0 =300

m=-11.120

? m =0.6054

0 m = 7.73910

M=38.45680 M=41.4966

0 M=43.80600

=0.8403 =0.7908z1 =18 z2 =108

0 =190 0 =200 0 =300

m =0.42940 m =2.2449

0 m =18.12800

M=31.68300 M=32.2505

0 M=38.79220

=0.9028 =0.8935 =0.7670z1 =30 z2 =180

0 =150 0 =20

0 0 =300

m =1.09220 m =9.3414

0 m =23.06660

M=25.10180 M=28.2414

0 M=35.71810

=0.9356 =0.8891 =0.7570z1 =90 z2 =540

0 =90 0 =200 0 =300

m =1.36450 m =16.4763

0 m =27.75830

M=14.93540 M=23.1812

0 M=32.09170

=0.9754 =0.8841 =0.7509

7. REFERENCES

[1] Petrescu, V., Petrescu, I. Randamentul cuplei

superioare de la angrenajele cu roi dinate cu axe fixe.

PRASIC 02, Braov, Romania, 2002, Vol. I, p. 333-

338.[2] Petrescu, R., Petrescu, F. The gear synthesis with the

best efficiency. ESFA03, Bucharest, 2003, Vol. 2, p.

63-70.

[3] Pelecudi, Chr., .a., Mecanisme. E.D.P., Bucureti,

1985.

Authors: Drd. Eng. Florian-Ion PETRESCU, proffesor-

asisstent, Univ. POLITEHNICA Bucureti, chair TMR,

phone: 021. 4029632;

Dr. Eng. Relly-Victoria PETRESCU, lecturer,

Univ. POLITEHNICA Bucureti, chair GDGI, phone:

0722.529.840.

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gear’s design for the mechanism’s highest efficiency

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