ESFA 2003 – R 10
THE GEAR SYNTHESIS WITH THE BEST EFFICIENCY Relly PETRESCU1, Florian PETRESCU2 Chair GDGI, Bucharest Polytechnic University, ROMANIA, victoriap@emoka.ro 2 Chair TMR, Bucharest Polytechnic University, ROMANIA,
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ABSTRACT The paper presents an original method to determinate the efficiency of the gear. The originality of this method consist in the eliminated of the friction modulus. On analyze the influence of few parameters about the efficiency value. These parameters are: z1 - the number of tooth’s for the primary wheel of gear; z2 - the number of tooth’s for the secondary wheel of gear; α0 the pressure angle normal on the divided circle; β - the inclination angle. With the relations of this presentment, on can synthesizing the gear mechanisms. KEY WORDS: efficiency, gear, constructive parameters, tooth, wheel, circle.
INTRODUCTION In this paper the authors presents an original method to calculate the efficiency of the gear. The originality n © 2002 Victoria PETRESCU 2 consist in the way to determinate the t efficiency of the v2 gear, because on Fτ are not utilized the α1 friction forces of P F m couple (this new way eliminate the v1 rp1 classical method). α1 Fψ K1 On eliminate too the necessity to v12 determinate the rb1 t friction coefficients O1 by different n ω1 experimental 0 1 methods. The determinates Fig. 1. The forces of the gear efficiency by the new method is the same like the classical efficiency, namely the mechanical efficiency of the gear. The model taken in calculation is very simple. It can be fallowed in figure 1.
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1. THE DETERMINATE OF THE MOMENTARY MECHANICAL EFFICIENCY The calculating relations are the next (see the fig. 1): Fτ = Fm . cosα1
(1)
Fψ = Fm . sinα1
(2)
v2 = v1 . cosα1
(3)
v12 = v1 . sinα1
(4)
where: _ _ _ Fm = Fτ + Fψ _ _ _ v1 = v2 + v12 with:
wheel); wheel);
(5) (6)
Fm Fτ Fψ v1
- the motive force (the driving force); - the transmitted force (the useful force); - the slide force (the lost force); - the speed of element 1, or the speed of wheel 1 (the driving
v2
- the speed of element 2, or the speed of wheel 2 (the droves
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) will be writhed with the relation (7): Pc = Pm = Fm . v1
(7)
The useful power (the transmitted power from the profile 1 to the profile 2) will be writhed: Pu = Pτ = Fτ . v2 = Fm .cosα1 .v1 .cosα1
(8)
Pu = Fm .v1 .cos2 α1
(8’)
The lost power will be writhed: Pψ =Fψ .v12 =Fm .sinα1 .v1 .sinα1
(9)
Pψ = Fm . v1 .sin2 α1
(9’)
The momentary efficiency of couple will be determinate directly with the next relation:
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(10)
ηi = Pu / Pc =Pτ /Pm =Fm.v1.cos2α1 / Fm.v1 ηi = cos2 α1
(10’)
The momentary losing coefficient will be writhed: (11)
ψi =Pψ /Pm = Fm.v1.sin2α1 / Fm.v1 = sin2α1
On can see easily that the sum of the momentary efficiency and the momentary losing coefficient will be always 1: ηi +ψi = cos2α1 +sin2α1 =1
(12)
And now on shall determinate the geometrical elements of gear. These elements will be utilized to the determination of the efficiency of the couple, η. 2. THE GEOMETRICAL ELEMENTS OF GEAR On shall be determinate the next geometrical elements of gear: The radius of the basic circle of wheel 1 (of the driving wheel): rb1 = 1/2.m.z1 .cosα0
(13)
The radius of the head circle of wheel 1: ra1 = 1/2.(m.z1 +2.m)=m/2.(z1 +2)
(14)
On determinate now the maximum of pressure angle of the gear: cosα1M =rb1 /ra1 =1/2.m.z1 .cosα0 /[1/2.m.(z1 +2)]=z1 .cosα0 /(z1 +2)
(15)
And now on determinate the same parameters for the wheel 2, the radius of basic circle and the radius of the head circle for the wheel 2: rb2 = 1/2.m.z2 .cosα0
(16)
ra2 = 1/2.(m.z2 +2.m)=m/2.(z2 +2)
(17)
Now on can determinate the minimum of pressure angle of the gear: tgα1m =[(rb1 +rb2 ).tgα0 -sqrt(r2a2 -r2b2 )]/rb1=N/rb1 N=1/2.m.(z1 +z2 ).sinα0 -sqrt[(m/2)2 .(z2 +2)2 -(m/2)2 .z22 .cos2α0 ] (19)
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(18)
N=m/2.[(z1 +z2 ).sinα0 -sqrt(z22 .sin2 α0 +4.z2 +4)]
(20)
With the relation (20) the relation (18) become the relation (21): (21)
tgα1m =[(z1 +z2 ).sinα0 –sqrt(z22.sin2α0 +4.z2 +4)]/(z1 .cosα0)
Now on can determinate the minimum (22) and the maximum (23) for the pressure angle: α1m =atan{[(z1 +z2 ).sinα0 -sqrt(z22.sin2α0 +4.z2 +4)]/(z1 .cosα0 )} (22) α1M = acos[z1 .cosα0 /(z1 +2)]
(23)
3. DETERMINATING THE EFFICIENCY The efficiency of gear will be determinate through the integration of momentary efficiency on all section of gearing movement, namely from the minimum pressure angle to the maximum pressure angle: αM η = 1 / ∆α . ∫ ηi . dα αm αM η = 1 / ∆α . ∫ [cos2α .dα] αm αM η = 1/(2.∆α). ∫ [2.cos2α.dα] αm αM η = 1 / (2.∆α). ∫ { [cos(2.α)+1].dα} αm αM η = 1/(2.∆α). [1/2.sin(2.α)+α] αm
(24)
(25)
(26)
(27)
(28)
η=1/(2.∆α).{1/2.[sin(2.αM)-sin(2.αm)]+∆α}
(29)
η= [sin(2.αM) - sin(2.αm)] / [4 . (αM - αm)] + 0.5
(30)
Finally the efficiency of gear will be calculated with the relation (30), utilizing for the pressure angle the relation (22) for the minimum and the relation (23) for the maximum, in case of utilizing the gear with right teeth (β=0) and in case of utilizing the gear with bended teeth (β≠0) on shall be use for the determination of the extreme pressure angles the relations (32) and (33).
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αt =atan(tgα0 /cosβ)
(31)
αm=atan{[(z1+z2)/cosβ.sinαt -sqrt(z22/cos2β.sin2αt+4.z2 /cosβ+4)].cosβ/(z1.cosαt )} (32) αM = acos[z1 /cosβ.cosαt /(z1 /cosβ+2)]
(33)
Continuing to see some calculated relations on shall see how the efficiency is varying with the input parameters.
Table 1 The calculated efficiency for the right teeth
4. THE CALCULATED EFFICIENCY
i12effective= - 4
right teeth
z1 =8 α0 =200 ? αm = -16.220 ? αM=41.25740
z2 =32 α0 =290 αm = 0.71590 αM=45.59740 η=0.8111
Table 1 α0 =350 αm = 11.13030 αM=49.05600 η=0.7308
We shall see now four table with the calculated efficiency in function of the input parameters and once the proceed from results we shall take some z1 =10 z2 =40 0 conclusions. α0 =20 ? α0 =260 α0 =300 0 0 The input parameters are: αm = -9.89 ? αm = 1.3077 αm = 8.22170 z1 = the number of tooth’s for the αM=38.45680 αM=41.49660 αM=43.80600 driving wheel 1; η=0.8375 η=0.7882 z2 = the number of tooth’s for the droves wheel 2, or the ratio of z1 =18 z2 =72 0 transmission, i (i12=-z2/z1); α0 =19 α0 =200 α0 =300 0 0 α0 = the pressure angle normal on αm = 0.9860 αm =2.7358 αm =18.28300 the divided circle; αM=31.68300 αM=32.25050 αM=38.79220 β = the bend angle. η=0.90105 η=0.8918 η=0.7660 We begin with the right teeth (the toothed gear), with i=-4, once for z1 we z1 =30 z2 =120 shall take successively diverse values, α0 =150 α0 =200 α0 =300 in growth (in rise), starting from 8 αm = 1.50660 αm =9.53670 αm =23.12250 tooth’s. αM=25.10180 αM=28.24140 αM=35.71810 On can see that for 8 tooth’s of the η=0.9345 η=0.8882 η=0.7566 driving wheel the standard pressure angle, α0=200, is to little for to can it be z1 =90 z2 =360 utilized (on obtain a minimum pressure α =80 ? α0 =90 α0 =200 0 angle, αm, negative and this fact is not α =-0.16380 ? α =1.58380 αm =16.49990 m m admitted!). αM=14.36370 αM=14.93540 αM=23.18120 In the second table we shall η=0.9750 η=0.8839 diminish (in module) the value for the ratio of transmission, i, from 4 to 2. On shall see how for the littler values to the number of tooth’s to the wheel 1,
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Table 2
Table 3 The calculated efficiency
The calculated efficiency
i12effective= - 6
right teeth
z1 =8 α0 =200 ? αm=-17.860 ? αM=41.25740
z2 =48 α0 =300 αm = 1.77840 αM=46.14620 η=0.8026
α0 =350 αm =10.6600 αM=49.05590 η=0.7337
z1 =10 α0 =200 ? αm=-11.120 ? αM=38.45680
z2 =60 α0 =260 αm =0.60540 αM=41.49660 η=0.8403
α0 =300 αm = 7.73910 αM=43.80600 η=0.7908
α0 =300 αm =18.69350 αM=38.79220 η=0.7633
z1 =18 α0 =190 αm =0.42940 αM=31.68300 η=0.9028
z2 =108 α0 =200 αm =2.24490 αM=32.25050 η=0.8935
α0 =300 αm =18.12800 αM=38.79220 η=0.7670
z2 =60 α0 =200 αm =10.04160 αM=28.24140 η=0.8859
α0 =300 αm =23.27740 αM=35.71810 η=0.7555
z1 =30 α0 =150 αm =1.09220 αM=25.10180 η=0.9356
z2 =180 α0 =200 αm =9.34140 αM=28.24140 η=0.8891
α0 =300 αm =23.06660 αM=35.71810 η=0.7570
z2 =180 α0 =200 αm =16.56670 αM=23.18120 η=0.8836
α0 =300 αm =27.78250 αM=32.09170 η=0.7507
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
i12effective= - 2
right teeth
Table 2
z1 =8 α0 =200 ? αm=-12.650 ? αM=41.25740
z2 =16 α0 =280 αm = 0.91490 αM=45.06060 η=0.8141
z1 =10 α0 =200 ? αm = -7.130 ? αM=38.45680
z2 =20 α0 =250 αm = 1.33300 αM=40.95220 η=0.8411
α0 =30 αm = 9.41060 αM=43.80600 η=0.7817
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
z1 =30 α0 =140 αm =0.88450 αM=24.54270 η=0.9388 z1 =90 α0 =80 αm =0.52270 αM=14.36370 η=0.9785
α0 =350 αm =12.29330 αM=49.05590 η=0.7236
0
Table 3
the standard pressure angle (α0=200) is to little and on shall 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 the tooth’s to the wheel 1 increase, we can decrease the normal pressure angle, α0. On can see that for z1=90 we shall take a very little value for the normal pressure angle (for the pressure angle of reference), α0=80. In the table 3 on increase the module value for i (for ratio of transmission), from 2 to 6. Table 4
The calculated efficiency for the bended teeth
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The table 4 keep for i the initial value (i=4), but now we shall operate with bended teeth (β≠0). For this table (for this calculation) we shall take for β a normal value, β=150. The bending of the teeth are not so much influence to the value of the gear efficiency. For this reason we do not changed again the bending of the teeth (β).
i12effective= - 4
bend teeth β=15
z1 =8
Table 4
0
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
0
?
z1 =10
z2 =40 α0 =260
α0 =300
αm=-10.5630 ?
αm =0.23550
αm = 6.91880
αM=38.34740
αM=41.571390
αM=43.99650
η=0.8412
η=0.7937
α0 =20
0
?
CONCLUSIONS z1 =18
z2 =72
After the analyzing α0 =190 α0 =200 α0 =300 of the four table on can 0 αm =2.02830 αm =17.18400 take the next αm =0.32715 αM=31.71800 αM=32.32020 αM=39.18030 conclusions: - the efficiency η=0.9029 η=0.8938 η=0.7702 (of the gear) increase when z2 =120 the number of z1 =30 0 α0 =200 α0 =300 tooth’s for the α0 =15 driving wheel αm =1.02690 αm =8.86020 αm =22.15500 1, z1, increase α =25.13440 αM=28.45910 αM=36.25180 M too and that η=0.8899 η=0.7593 when the η=0.9357 pressure angle, α0, z1 =90 z2 =360 diminish; 0 α0 =9 α0 =200 α0 =300 - z2 or i12 are not 0 αm =15.89440 αm =26.94030 so much αm =1.3187 influence about αM=14.96480 αM=23.63660 αM=32.82620 on the η=0.9754 η=0.8845 η=0.7513 efficiency value; - on can see easily that for the value α0=200, the efficiency take roughly the value η≈0.89 for any values of the others parameters (this justify the choice of this value α0=200 for the standard pressure angle of reference). But the pressure angle of reference, α0, can be decreased in the same time with the increase of the number of tooth’s for the driving wheel 1, z1, in reason to increasing the gear efficiency, η; contrary, when we desire to realize an gear with a little z1 (for a
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-
-
little gauge), it will be necessary to increasing the α0 value, for maintaining a positive value for αm (in this case the gear efficiency will be diminished); the module of the gear, m, have not any influence about on the gear efficiency value; when α0 is diminished on can take a bigger normal module, for increasing the addendum of tooth, but the increase of the m in the same time with the increase of the z1 can lead to a greater gauge; when β increase, the efficiency, η, increase too but the growth is insignificant; the gear efficiency, η, is really a function of α0 and z1: η=f(α0, z1); αm and αM are just the intermediate parameters.
BONUS: THE RELATIONS TO CALCULATING THE EFFICIENCY FOR THE INTERNAL GEAR On calculate the values for the extreme angles, αm and αM with the relations (35), (36) respective (39), (38) and then on utilize the same relation (30) like in the external gear case. A. THE DRIVING WHEEL 1, HAVE EXTERNAL TEETH tgαt=tgα0/cosβ
(34)
tgα1m=[(z1 - z2)/cosβ.sinαt + (z22/cos2β.sin2αt - 4.z2/cosβ + 4)1/2]/[z1/cosβ.cosαt] cosα1M=z1/cosβ.cosαt/(z1/cosβ + 2)
(35) (36)
B. THE DRIVING WHEEL 1, HAVE INTERNAL TEETH tgαt=tgα0/cosβ
(37)
tgα1M=[(z1 - z2)/cosβ.sinαt + (z22/cos2β.sin2αt + 4.z2/cosβ + 4)1/2]/[z1/cosβ.cosαt] cosα1m=z1/cosβ.cosαt/(z1/cosβ - 2)
(38) (39)
REFERENCES [1] Petrescu, V., Petrescu, I., Antonescu, O. Randamentul cuplei superioare de la angrenajele cu roţi dinţate cu axe fixe. PRASIC’ 02, Braşov, Romania, 2002, Vol. I, p. 333-338. [2] Petrescu, I., Petrescu, V., Ocnărescu, C. The cam synthesis with maximal efficiency. PRASIC’ 02, Braşov, Romania, 2002, Vol. I, p. 339-344. [3] Antonescu, P., Petrescu, R., Adîr, G., Antonescu, O. Mecanisme cu roţi dinţate. Editura Printech, Bucureşti, 1999. [4] Pelecudi, Ch., ş.a. Mecanisme. EDP, Bucureşti, 1985.
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