DETERMINING THE DYNAMIC EFFICIENCY OF GEARS, Relly Victoria Petrescu, Florian Ion T. Petrescu

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mm IFToMM ARoTMM

THE NINTH IFToMM INTERNATIONAL SYMPOSIUM ON THEORY OF MACHINES AND MECHANISMS BUCHAREST, ROMANIA, SEPTEMBER 1 - 4, 2005 SYROM 2005

DETERMINING THE DYNAMIC EFFICIENCY OF GEARS Relly Victoria PETRESCU Florian Ion PETRESCU

Department of Descriptive Geometry and Engineering Graphics, University “POLITEHNICA” of Bucharest, 313 Splaiul Independentei, 77206 Bucharest, ROMANIA E-mail: victoriap@emoka.ro Department of Mechanisms and Robots, University “POLITEHNICA” of Bucharest, 313 Splaiul Independentei, 77206 Bucharest, ROMANIA E-mail: tiberiuionro@yahoo.com

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. Today, the gears are present every where, in the mechanical’s world (In vehicle’s industries, in electronics and electro-technique equipments, in energetically industries, etc…). Optimizing this mechanism (the gears mechanism), one can improve the functionality of the transmissions with gears.

Keywords: 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 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. Precisely one determines the dynamics efficiency, but at the gears transmissions, the dynamics efficiency is the same like the mechanical efficiency; this is a greater advantage of the gears transmissions.

2. DETERMINING THE MOMENTARY DYNAMIC (MECHANICAL) EFFICIENCY The calculating relations [2, 3], are the next (1-21), (see the fig. 1): ⎧ Fτ = Fm ⋅ cos α1 ⎪ ⎪ Fψ = Fm ⋅ sin α1 ⎪v = v ⋅ cos α ⎪ 2 1 1 ⎨ = ⋅ v v sin α 12 1 1 ⎪ ⎪F = F + F τ ψ ⎪ m ⎪⎩v1 = v2 + v12

with: Fm - the motive force (the driving force); Fτ - the transmitted force (the useful force); Fψ - the slide force (the lost force);  v1 - the velocity of element 1, or the speed of wheel 1 (the driving wheel);

(1)


v2 - the velocity 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).

n

2

© 2002 Victoria PETRESCU The Copyright-Law Of March, 01, 1989 U.S. Copyright Office Library of Congress Washington, DC 20559-6000 202-707-3000

t

v2 Fτ Fm

α1

v1

1 P rp1 α1

rb1

v12 t

K1

O1 0

ω1

n

Fig. 1. The forces of the gear The consumed power (in this case the driving power): Pc ≡ Pm = Fm ⋅ v1

(2)

The useful power (the transmitted power from the profile 1 to the profile 2) will be written: Pu ≡ Pτ = Fτ ⋅ v2 = Fm ⋅ v1 ⋅ cos2 α1

(3)

Pψ = Fψ ⋅ v12 = Fm ⋅ v1 ⋅ sin 2 α1

(4)

The lost power will be written:

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

(5)

The momentary losing coefficient [1], 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.

3. THE GEOMETRICAL ELEMENTS OF THE GEAR One can determine the next geometrical elements of the external gear, [2,3], (for the right teeth, β=0): The radius of the basic circle of wheel 1 (of the driving wheel), (7): 1 ⋅ m ⋅ z1 ⋅ cos α 0 2

(7)

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

(8)

rb1 =

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

One determines now the maximum pressure angle of the gear (9): 1 ⋅ m ⋅ z1 ⋅ cos α 0 rb1 2 z ⋅ cos α 0 cos α1M = = = 1 1 ra1 z1 + 2 ⋅ m ⋅ ( z1 + 2) 2

(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: 1 rb 2 = ⋅ m ⋅ z2 ⋅ cos α 0 (10) 2 m ra 2 = ⋅ ( z2 + 2) (11) 2 Now one can determine the minimum pressure angle of the external gear (12, 13): N ⎧ ⎪tgα1m = r b1 ⎪ ⎪ m 1 2 2 2 2 2 ⎨ N = (rb1 + rb 2 ) ⋅ tgα 0 − ra 2 − rb 2 = ⋅ m ⋅ ( z1 + z2 ) ⋅ sin α 0 − ⋅ ( z2 + 2) − z2 ⋅ cos α 0 = 2 2 ⎪ ⎪ m 2 2 ⎪= 2 ⋅ [( z1 + z2 ) ⋅ sin α 0 − z2 ⋅ sin α 0 + 4 ⋅ z2 + 4 ] ⎩ tgα1m = [( z1 + z2 ) ⋅ sin α 0 − z22 ⋅ sin 2 α 0 + 4 ⋅ z2 + 4 ] /( z1 ⋅ cos α 0 )

(12)

(13)

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 uses the relations (14, 15 and 16): tgα t =

tgα1m = [( z1 + z2 ) ⋅

tgα 0 cos β

sin α t sin 2 α t z cos β − z22 ⋅ + 4 ⋅ 2 + 4]⋅ cos β cos β z1 ⋅ cos α t cos 2 β

cos α1M

z1 ⋅ cos α t cos β = z1 +2 cos β

(14) (15)

(16)

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


A. When the driving wheel 1, has external teeth: z cos β sin α t sin 2 α t + z22 ⋅ − 4 ⋅ 2 + 4]⋅ 2 z1 ⋅ cos α t cos β cos β cos β

tgα1m = [( z1 − z2 ) ⋅

cos α1M

(17)

z1 ⋅ cos α t cos β = z1 +2 cos β

(18)

B. When the driving wheel 1, have internal teeth: tgα1M = [( z1 − z2 ) ⋅

z cos β sin α t sin 2 α t + z22 ⋅ + 4 ⋅ 2 + 4]⋅ 2 β z cos β cos cos β 1 ⋅ cos α t cos α1m

(19)

z1 ⋅ cos α t cos β = z1 −2 cos β

(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, the relation (21), [2, 3]: η =

1 ⋅ Δα

αM

η i ⋅ dα =

am

1 Δα

αM

cos 2 α ⋅ d α =

αm

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

(21)

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

Table 1. Determining the efficiency of the gear’s right teeth for i12effective= - 4 i12effective= - 4 z1 =8 α0 =200 ? αm = -16.220 ? αM=41.25740

right teeth z2 =32 α0 =290 αm = 0.71590 αM=45.59740 η=0.8111

α0 =350 αm = 11.13030 αM=49.05600 η=0.7308

z1 =10 α0 =200 ? αm = -9.890 ? αM=38.45680

z2 =40 α0 =260 αm = 1.30770 αM=41.49660 η=0.8375

α0 =300 αm = 8.22170 αM=43.80600 η=0.7882

z1 =18 α0 =190 αm = 0.98600 αM=31.68300 η=0.90105

z2 =72 α0 =200 αm =2.73580 αM=32.25050 η=0.8918

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

z1 =30 α0 =150 αm = 1.50660 αM=25.10180 η=0.9345

z2 =120 α0 =200 αm =9.53670 αM=28.24140 η=0.8882

α0 =300 αm =23.12250 αM=35.71810 η=0.7566

z1 =90 α0 =80 ? αm =-0.16380 ? αM=14.36370

z2 =360 α0 =90 αm =1.58380 αM=14.93540 η=0.9750

α0 =200 αm =16.49990 αM=23.18120 η=0.8839


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. Table 2. Determining the efficiency of the gear’s right teeth for i12effective= - 2 i12effective= - 2 z1 =8 α0 =200 ? αm=-12.650 ? αM=41.25740

right teeth z2 =16 α0 =280 αm = 0.91490 αM=45.06060 η=0.8141

α0 =350 αm =12.29330 αM=49.05590 η=0.7236

z1 =18 α0 =180 αm = 0.67560 αM=31.13510 η=0.9052

z2 =36 α0 =200 αm =3.92330 αM=32.25050 η=0.8874

α0 =300 αm =18.69350 αM=38.79220 η=0.7633

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

z2 =20 α0 =250 αm = 1.33300 αM=40.95220 η=0.8411

α0 =300 αm = 9.41060 αM=43.80600 η=0.7817

z1 =90 α0 =80 αm =0.52270 αM=14.36370 η=0.9785

z2 =180 α0 =200 αm =16.56670 αM=23.18120 η=0.8836

α0 =300 αm =27.78250 αM=32.09170 η=0.7507

Table 3. Determining the efficiency of the gear’s right teeth for i12effective= - 6 i12effective= - 6 z1 =8

right teeth z2 =48

0

α0 =20 ? αm=-17.860 ? αM=41.25740

α0 =30 αm = 1.77840 αM=46.14620 η=0.8026

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

z2 =60 α0 =260 αm =0.60540 αM=41.49660 η=0.8403

0

z1 =18

z2 =108

α0 =35 αm =10.6600 αM=49.05590 η=0.7337

0

α0 =19 αm =0.42940 αM=31.68300 η=0.9028

α0 =200 αm =2.24490 αM=32.25050 η=0.8935

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

α0 =30 αm = 7.73910 αM=43.80600 η=0.7908

z1 =90 α0 =90 αm =1.36450 αM=14.93540 η=0.9754

z2 =540 α0 =200 αm =16.47630 αM=23.18120 η=0.8841

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

0

0

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 will 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. 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 one increases 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. .


Table 4. The determination of the gear’s parameters in bend teeth for i=-4 i12effective= - 4 z1 =8 α0 =200 ? αm=-16.8360 ? αM=41.08340

z1 =18 α0 =190 αm =0.327150 αM=31.71800 η=0.9029

bend teeth β=150 z2 =32 α0 =300 αm = 1.12650 αM=46.25920 η=0.8046

Table 4

α0 =350 αm = 9.44550 αM=49.29530 η=0.7390

z1 =30 α0 =150 αm =1.02690 αM=25.13440 η=0.9357

z2 =120 α0 =200 αm =8.86020 αM=28.45910 η=0.8899

α0 =300 αm =22.15500 αM=36.25180 η=0.7593

z2 =72 α0 =200 αm =2.02830 αM=32.32020 η=0.8938

α0 =30 αm =17.18400 αM=39.18030 η=0.7702

z1 =90 α0 =90 αm =1.31870 αM=14.96480 η=0.9754

z2 =360 α0 =200 αm =15.89440 αM=23.63660 η=0.8845

α0 =300 αm =26.94030 αM=32.82620 η=0.7513

0

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 so much influence 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). The better efficiency may be obtained only for an α0≠200. 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. In this paper, one determines precisely, the dynamics-efficiency, but at the gears transmissions, the dynamics efficiency is the same like the mechanical efficiency; this is a greater advantage of the gears transmissions. This advantage, specifically of the gear’s mechanisms, may be found at the cams mechanisms with plate followers too.

REFERENCES 1. Pelecudi, Chr., ş.a., Mecanisme. E.D.P., Bucureşti, 1985. 2. Petrescu, V., Petrescu, I., Randamentul cuplei superioare de la angrenajele cu roţi dinţate cu axe fixe, In: The Proceedings of 7th National Symposium PRASIC, Braşov, vol. I, pp. 333-338, 2002. 3. Petrescu, R., Petrescu, F., The gear synthesis with the best efficiency, In: The Proceedings of ESFA’03, Bucharest, vol. 2, pp. 63-70, 2003.


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