GEAR’S DESIGN FOR THE MECHANISM’S HIGHEST EFFICIENCY Relly PETRESCU, Florian PETRESCU

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

GEAR’S DESIGN FOR THE MECHANISM’S 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; α0 - the normal pressure angle on the divided circle; β - the inclination angle. With the relations presented in this paper, one can synthesize the gear’s mechanisms.

Key Words: efficiency, gear, constructive parameters, teeth, outside circle, wheel. profile 1 to the profile 2) will be written:

1. INTRODUCTION

n

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 gear’s efficiency, because one hasn’t used the friction 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.

Fτ Fm

α1

v1

P rp1 α1

Fψ t

K1

rb1

v12 O1 0

n

ω1 1

Fig. 1. The forces of the gear

The calculating relations [1,2], are the next (see the fig. 1):

Pu ≡ Pτ = Fτ ⋅ v 2 = Fm ⋅ v1 ⋅ cos 2 α 1

(3)

The lost power will be written: Pψ = Fψ ⋅ v12 = Fm ⋅ v1 ⋅ sin 2 α 1

(4)

(1) The momentary efficiency of couple will be calculated directly with the next relation: ⎧ Pu P F ⋅ v ⋅ cos 2 α1 ≡ τ = m 1 ⎪ηi = Pc Pm Fm ⋅ v1 ⎨ ⎪ 2 ⎩ηi = cos α1

with: Fm - the motive force (the driving force); Fτ  - the transmitted force (the useful force); Fψ - 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): Pc ≡ Pm = Fm ⋅ v1

t

v2

2. DETERMINING THE MOMENTARY MECHANICAL EFFICIENCY

⎧ Fτ = Fm ⋅ cos α1 ⎪ ⎪ Fψ = Fm ⋅ sin α1 ⎪v = v ⋅ cos α ⎪ 2 1 1 ⎨ = ⋅ sin α v v 1 1 ⎪ 12 ⎪F = F + F τ ψ ⎪ m ⎩⎪v1 = v2 + v12

© 2002 Victoria PETRESCU

2

(5)

The momentary losing coefficient [3], will be written: ⎧ Pψ F ⋅ v ⋅ sin 2 α1 = m 1 = sin 2 α1 ⎪ψ i = Pm Fm ⋅ v1 ⎨ ⎪ 2 2 ⎩ηi + ψ i = cos α1 + sin α1 = 1

(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, η.

(2)

The useful power (the transmitted power from the

257

uses the relations (14, 15 and 16):

3. THE GEOMETRICAL ELEMENTS OF THE GEAR

tgα t =

One can determine the next geometrical elements of the external gear, [1,2], (for the right teeth, β=0): The radius of the basic circle of wheel 1 (of the driving wheel), (7): rb1 =

1 ⋅ m ⋅ z1 ⋅ cos α 0 2

tgα 1m = [( z1 + z 2 ) ⋅ z 22

(7)

The radius of the outside circle of wheel 1 (8): ra1 =

1 m ⋅ (m ⋅ z1 + 2 ⋅ m) = ⋅ ( z1 + 2) 2 2

(10)

ra 2

(11)

cos α 1M

tgα 1m = [( z1 + z 2 ) ⋅ sin α 0 − − z 22 ⋅ sin 2 α 0 + 4 ⋅ z 2 + 4 ] /( z1 ⋅ cos α 0 )

z1 ⋅ cos α t cos β = z1 +2 cos β

(18)

B. When the driving wheel 1, have internal teeth: sin α t tgα 1M = [( z1 − z 2 ) ⋅ + cos β z 22

Now one can determine the minimum pressure angle of the external gear (12, 13): N ⎧ ⎪tgα1m = r b1 ⎪ ⎪ 2 2 ⎪ N = (rb1 + rb 2 ) ⋅ tgα 0 − ra 2 − rb 2 = ⎪ 1 m ⎪⎪= ⋅ m ⋅ ( z1 + z2 ) ⋅ sin α 0 − ⋅ 2 2 ⎨ ⎪ 2 2 2 ⎪⋅ ( z2 + 2) − z2 ⋅ cos α 0 = ⎪ m ⎪= ⋅ [( z1 + z2 ) ⋅ sin α 0 − ⎪ 2 ⎪ 2 2 ⎩⎪− z2 ⋅ sin α 0 + 4 ⋅ z2 + 4 ]

(16)

A. When the driving wheel 1, has external teeth: sin α t tgα 1m = [( z1 − z 2 ) ⋅ + cos β (17) 2 z2 cos β 2 sin α t z2 ⋅ − 4⋅ + 4]⋅ cos β z1 ⋅ cos α t cos 2 β

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: 1 ⋅ m ⋅ z 2 ⋅ cos α 0 2 m = ⋅ ( z 2 + 2) 2

z1 ⋅ cos α t cos β = z1 +2 cos β

(15)

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

(9)

rb 2 =

sin α t − cos β

z cos β ⋅ + 4 ⋅ 2 + 4]⋅ 2 cos β z1 ⋅ cos α t cos β

(8)

rb1 = ra1

1 ⋅ m ⋅ z1 ⋅ cos α 0 z ⋅ cos α 0 = 2 = 1 1 z1 + 2 ⋅ m ⋅ ( z1 + 2 ) 2

(14)

sin 2 α t

cos α 1M

One determines now the maximum pressure angle of the gear (9): cos α 1M =

tgα 0 cos β

sin 2 α t

z cos β ⋅ + 4 ⋅ 2 + 4]⋅ 2 cos β z1 ⋅ cos α t cos β

cos α 1m

z1 ⋅ cos α t cos β = z1 −2 cos β

(19)

(20)

4. DETERMINING THE EFFICIENCY (12)

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]: η =

(13)

=

1 ⋅ ∆α

αM

∫

η i ⋅ dα =

am

1 ∆α

αM

∫ cos α

2

α ⋅ dα

m

αM 1 1 ⋅ [ ⋅ sin( 2 ⋅ α ) + α ] = 2 ⋅ ∆α 2 αm

sin( 2α M ) − sin( 2α m ) 1 + ∆α ] = [ 2 2 ⋅ ∆α sin( 2 ⋅ α M ) − sin( 2 ⋅ α m ) + 0 .5 = 4 ⋅ (α M − α m ) =

Now one can determine, for the external gear, the minimum (13) and the maximum (9) pressure angle for the right teeth.

For the external gear with bended teeth (β≠0) one

258

(21)

Table 1.

Table 2.

Determining the efficiency of the gear’s right teeth for i12effective= - 4

i12effective= - 4 z1 =8 α0 =200 ? αm = -16.220 ? αM=41.25740 z1 =10 α0 =200 ? αm = -9.890 ? αM=38.45680 z1 =18 α0 =190 αm = 0.98600 αM=31.68300 η=0.90105 z1 =30 α0 =150 αm = 1.50660 αM=25.10180 η=0.9345 z1 =90 α0 =80 ? αm =-0.16380 ? αM=14.36370

right teeth z2 =32 α0 =290 αm = 0.71590 αM=45.59740 η=0.8111 z2 =40 α0 =260 αm = 1.30770 αM=41.49660 η=0.8375 z2 =72 α0 =200 αm =2.73580 αM=32.25050 η=0.8918 z2 =120 α0 =200 αm =9.53670 αM=28.24140 η=0.8882 z2 =360 α0 =90 αm =1.58380 αM=14.93540 η=0.9750

Determining the efficiency of the gear’s right teeth for i12effective= - 2

i12effective= - 2 z1 =8 α0 =200 ? αm=-12.650 ? αM=41.25740

0

α0 =35 αm = 11.13030 αM=49.05600 η=0.7308 α0 =300 αm = 8.22170 αM=43.80600 η=0.7882

z1 =10 α0 =200 ? αm = -7.130 ? αM=38.45680

α0 =300 αm =18.28300 αM=38.79220 η=0.7660

z1 =18 α0 =180 αm = 0.67560 αM=31.13510 η=0.9052 z1 =90 α0 =80 αm =0.52270 αM=14.36370 η=0.9785

α0 =300 αm =23.12250 αM=35.71810 η=0.7566 α0 =200 αm =16.49990 αM=23.18120 η=0.8839

right teeth z2 =16 α0 =280 αm = 0.91490 αM=45.06060 η=0.8141 z2 =20 α0 =250 αm = 1.33300 αM=40.95220 η=0.8411 z2 =36 α0 =200 αm =3.92330 αM=32.25050 η=0.8874 z2 =180 α0 =200 αm =16.56670 αM=23.18120 η=0.8836

α0 =350 αm =12.29330 αM=49.05590 η=0.7236 α0 =300 αm = 9.41060 αM=43.80600 η=0.7817 α0 =300 αm =18.69350 αM=38.79220 η=0.7633 α0 =300 αm =27.78250 αM=32.09170 η=0.7507

For example, if z1=8, the necessary minimum value is α0=290 for an i=-4 (see the table 1) and α0=280 for an i=2 (see the table 2). If z1=10, the necessary minimum pressure angle is α0=260 for i=-4 (see the table 1) and α0=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, α0. One shall see that for z1=90 one can take less for the normal pressure angle (for the pressure angle of reference), α0=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 (β≠0). The module i, take now the value 2.

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); α0 = the pressure angle normal on the divided circle; β = 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, α0=200, is to small to be used (one obtains a minimum pressure angle, αm, 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 (α0=200) is to small and it will be necessary to increase it to a minimum value.

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, α0, diminishes; z2 or i12 are not that much influenced about the efficiency value; One can easily see that for the value α0=200, the efficiency takes roughly the value η≈0.89 for any values of the others parameters (this justifies the choice of this value, α0=200, for the standard pressure angle of reference).

But the better efficiency may be obtained only for an α0≠200.

259

Table 3.

Determining the efficiency of the gear’s right teeth for i12effective= - 6

i12effective= - 4

i12effective= - 6 z1 =8

right teeth z2 =48

α0 =200 ? αm=-17.860 ? αM=41.25740

α0 =350 αm =10.6600 αM=49.05590 η=0.7337

z1 =18

α0 =300 αm = 1.77840 αM=46.14620 η=0.8026 z2 =60 α0 =260 αm =0.60540 αM=41.49660 η=0.8403 z2 =108

α0 =190 αm =0.42940 αM=31.68300 η=0.9028 z1 =30 α0 =150 αm =1.09220 αM=25.10180 η=0.9356 z1 =90

α0 =200 αm =2.24490 αM=32.25050 η=0.8935 z2 =180 α0 =200 αm =9.34140 αM=28.24140 η=0.8891 z2 =540

α0 =90 αm =1.36450 αM=14.93540 η=0.9754

α0 =200 αm =16.47630 αM=23.18120 η=0.8841

α0 =300 αm =27.75830 αM=32.09170 η=0.7509

z1 =10 α0 =200 ? αm=-11.120 ? αM=38.45680

Table 4.

The determination of the gear’s parameters in bend teeth for i=-4

bend teeth β=15

z1 =8

z2 =32 α0 =300

α0 =350

αm=-16.8360 ?

αm = 1.12650

αm = 9.44550

αM=41.08340

αM=46.25920

αM=49.29530

η=0.8046

η=0.7390

α0 =20

α0 =300 αm = 7.73910 αM=43.80600 η=0.7908

Table 4

0

0

?

z1 =18

z2 =72 α0 =200

α0 =300

αm =0.327150

αm =2.02830

αm =17.18400

α0 =300 αm =18.12800 αM=38.79220 η=0.7670

αM=31.71800

αM=32.32020

αM=39.18030

η=0.9029

η=0.8938

η=0.7702

z1 =30

z2 =120

α0 =300 αm =23.06660 αM=35.71810 η=0.7570

α0 =19

α0 =200

α0 =300

αm =1.02690

αm =8.86020

αm =22.15500

αM=25.13440

αM=28.45910

αM=36.25180

η=0.9357

η=0.8899

η=0.7593

z1 =90

z2 =360

0

α0 =200

α0 =300

αm =1.31870

αm =15.89440

αm =26.94030

αM=14.96480

αM=23.63660

αM=32.82620

η=0.9754

η=0.8845

η=0.7513

α0 =15

α0 =9

But the pressure angle of reference, α0, 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 α0 value, for maintaining a positive value for αm (in this case the gear efficiency will be diminished); When β increases, the efficiency, η, increases too, but the growth is insignificant.

The module of the gear, m, has not any influence on the gear’s efficiency value.

When α0 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, η, is really a function of α0 and z1: η=f(α0,z1); αm and αM are just the intermmediate parameters.

For a good projection of the gear, it’s necessary a z1 and a z2 greater than 30-60; but this condition may increase the gauge of mechanism.

0

0

7. REFERENCES [1] Petrescu, V., Petrescu, I. Randamentul cuplei superioare de la angrenajele cu roţi dinţate cu axe fixe.

PRASIC’ 02, Braşov, Romania, 2002, Vol. I, p. 333338.

[2] Petrescu, R., Petrescu, F. The gear synthesis with the best efficiency. ESFA’03, Bucharest, 2003, Vol. 2, p.

63-70.

[3] Pelecudi, Chr., ş.a., Mecanisme. E.D.P., Bucureşti, 1985.

Authors: Drd. Eng. Florian-Ion PETRESCU, proffesorasisstent, Univ. POLITEHNICA Bucureşti, chair TMR, phone: 021. 4029632; Dr. Eng. Relly-Victoria PETRESCU, lecturer, Univ. POLITEHNICA Bucureşti, chair GDGI, phone: 0722.529.840.

260