Al III-lea Seminar National de Mecanisme, Craiova, 2008
1
THE DYNAMICS OF CAM GEARS AT THE MODULE B (WITH TRANSLATED FOLLOWER WITH ROLL) Relly Victoria PETRESCU, Adriana COMĂNESCU, Florian Ion PETRESCU, Ovidiu ANTONESCU Abstract: The paper presents shortly an original method in determining the dynamic of the mechanisms with rotation cam and translated follower with roll. First, one presents the dynamics kinematics. Then one makes the dynamic analyze of few models, for some movement laws, imposed at the follower, by the designed cam profile.
1 Introduction The paper proposes an original dynamic model of cam gear and translated follower with roll. 2 The dynamic of distribution mechanism with translated follower with roll 2.1 Generalities The angle α0 defines the basic position of the vector, rB 0 , in the OCB0 triangle having a straight angle (1-4): s e rB0 = r0 + rb (1) s 0 = rB20 − e 2 (2) cos α 0 = (3) sin α 0 = 0 (4) rB0 rB0
The pressure angle, δ, between the normal n (which pass by the contact point A) and a vertical line, can be calculated with the relations (5-7). s0 + s s '−e cos δ = sin δ = (5) (6) ( s 0 + s ) 2 + ( s '−e) 2 ( s 0 + s ) 2 + ( s '−e) 2
s '−e s0 + s The vector rA can be determined with the relations (8-9): tgδ =
rA2 = (e + rb ⋅ sin δ ) 2 + ( s 0 + s − rb ⋅ cos δ ) 2
(7)
(8)
Al III-lea Seminar National de Mecanisme, Craiova, 2008
2
rA = (e + rb ⋅ sin δ ) 2 + ( s 0 + s − rb ⋅ cos δ ) 2
(9)
Fn, vn
Fu, v2 δ Fn, vn
Fi, vi
B rb s
Fm, vm α A-δ
A
Fa, va rB rA
B0 rb n
s0
C
A0
x
θB
μ
γ
θA ϕ
α0 αA e
r0
O
Fig. 1. Mechanism with rotate cam and translated follower with roll
One can calculate αA (10-11): e + rb ⋅ sin δ (10) cos α A = rA
sin α A =
s 0 + s − rb ⋅ cos δ rA
(11)
2.2. The relations to design the profile
γ = α A −α0 cos γ = cos α A ⋅ cos α 0 + sin α A ⋅ sin α 0 sin γ = sin α A ⋅ cos α 0 − cos α A ⋅ sin α 0 θA =ϕ −γ cos θ A = cos ϕ ⋅ cos γ + sin ϕ ⋅ sin γ sin θ A = sin ϕ ⋅ cos γ − sin γ ⋅ cos ϕ
(12) (13) (14) (15) (16) (17)
Al III-lea Seminar National de Mecanisme, Craiova, 2008
3
2.3. The exactly kinematics of B Module
From the triangle OCB (fig. 1) one determines the length rB (OB) and the complementary angles αB (COB) AND τ (CBO). rB2 = e 2 + ( s 0 + s) 2
rB = rB2 (19) s +s e cos α B ≡ sin τ = (20) sin α B ≡ cos τ = 0 (21) rB rB From the general triangle OAB, where one knows OB, AB, and the angle between them, B (ABO, which is the sum of τ with δ), one can determine the length OA and the angle μ (AOB): cos(δ + τ ) = cos δ ⋅ cos τ − sin δ ⋅ sin τ (22) 2 2 2 rA = rB + rb − 2 ⋅ rb ⋅ rB ⋅ cos(δ + τ ) (23)
cos μ =
(18)
rA2 + rB2 − rb2 2 ⋅ rA ⋅ rB
(24)
rb ⋅ sin(δ + τ ) rA With αB and μ one can determine now αA and α& A : αA =αB − μ (27) α& A = α& B − μ& From (20) one obtains α& B (32), (see 29-32) where r&B (31) obtained from (18). Then, one obtains (33) from (24): e ⋅ r& e ⋅ rB ⋅ r&B (29) α& B = − sin α B ⋅ α& B = − 2 B rB ( s 0 + s ) ⋅ rB2
sin(δ + τ ) = sin δ ⋅ cos τ + sin τ ⋅ cos δ
(25)
2 ⋅ rB ⋅ r&B = 2 ⋅ ( s 0 + s ) ⋅ s&
α& B =
sin μ =
rB ⋅ r&B = ( s 0 + s ) ⋅ s&
e ⋅ ( s 0 + s ) ⋅ s& ( s 0 + s) ⋅ r
2 B
=
e ⋅ s& rB2
(26)
(28) can be (30)
(31) (32)
2 ⋅ r&A ⋅ rB ⋅ cos μ + 2 ⋅ rA ⋅ r&B ⋅ cos μ − 2 ⋅ rA ⋅ rB ⋅ sin μ ⋅ μ& = 2 ⋅ rA ⋅ r&A + 2 ⋅ rB ⋅ r&B (33) From (33) one writes μ& (38), but one needs first to obtain r&A (34) from the expression (23): 2 ⋅ rA ⋅ r&A = 2 ⋅ rB ⋅ r&B − 2 ⋅ rb ⋅ r&B ⋅ cos(δ + τ ) (34) + 2 ⋅ r ⋅ r ⋅ sin(δ + τ ) ⋅ (δ& + τ&) b
B
To solve (34) one needs the derivatives δ& and τ& . From (7) one obtains (35 and 36). τ& takes the form (37):
Al III-lea Seminar National de Mecanisme, Craiova, 2008
4
δ '=
s ' '⋅( s 0 + e) − s '⋅( s '−e) e ⋅ s& (35) δ& = δ '⋅ω (36) τ& = −α& B = − 2 2 2 ( s 0 + s ) + ( s '−e) rB Now one can determine μ& (38), α& A (28) and θ&A (39): r& ⋅ r ⋅ cos μ + rA ⋅ r&B ⋅ cos μ − rA ⋅ r&A − rB ⋅ r&B μ& = A B rA ⋅ rB ⋅ sin μ & θ A = ϕ& − γ& = ω − α& A One writes cos α A and sin α A (40-41):
cos α A =
sin α A =
(43):
e ⋅ ( s 0 + s ) 2 + ( s '−e) 2 + rb ⋅ ( s '−e)
(37)
(38) (39)
(40)
rA ⋅ ( s 0 + s ) 2 + ( s '−e) 2
( s 0 + s ) ⋅ [ ( s 0 + s ) 2 + ( s '−e) 2 − rb ] rA ⋅ ( s 0 + s ) 2 + ( s '−e) 2
(41)
Now, one can obtain the expression cos(αA-δ) (42), and cos(αA-δ).cosδ cos(α A − δ ) =
( s0 + s) ⋅ s' rA ⋅ ( s 0 + s ) + ( s '−e) 2
2
=
s' ⋅ cos δ rA
s' ⋅ cos 2 δ rA Now one writes the forces and the velocities (48-50): v a = v m ⋅ sin(α A − δ ) v n = v m ⋅ cos(α A − δ ) (44) Fa = Fm ⋅ sin(α A − δ ) Fn = Fm ⋅ cos(α A − δ ) v i = v n ⋅ sin δ v 2 = v n ⋅ cos δ = v m ⋅ cos(α A − δ ) ⋅ cos δ (46) Fi = Fn ⋅ sin δ Fu = Fn ⋅ cos δ = Fm ⋅ cos(α A − δ ) ⋅ cos δ cos(α A − δ ) ⋅ cos δ =
(42) (43)
(45) (47)
2.4. Determining the efficiency at the Module B
Pu = Fu ⋅ v 2 = Fm ⋅ v m ⋅ cos 2 (α A − δ ) ⋅ cos 2 δ
ηi =
(48)
Pc = Fm ⋅ v m
Pu Fm ⋅ v m ⋅ cos 2 (α A − δ ) ⋅ cos 2 δ = = cos 2 (α A − δ ) ⋅ cos 2 δ = Pc Fm ⋅ v m
s' s' 2 = [cos(α A − δ ) ⋅ cos δ ] = [ ⋅ cos 2 δ ] 2 = 2 ⋅ cos 4 δ rA rA 2
(49)
(50)
Al III-lea Seminar National de Mecanisme, Craiova, 2008
5
2.5. Determining the transmission function D, at the Module B
The follower velocity (47) can be puted in the form (51): s' v 2 = v n ⋅ cos δ = v m ⋅ cos(α A − δ ) ⋅ cos δ = v m ⋅ ⋅ cos 2 δ = rA
(51) s' 2 2 I 2 & & = rA ⋅ θ A ⋅ ⋅ cos δ = θ A ⋅ s '⋅ cos δ = θ A ⋅ ω ⋅ s '⋅ cos δ rA With the relations (51) and (52) one determine the transmission function (the dynamic modulus), D (53): v 2 = s'⋅D ⋅ ω (52) D = θ AI ⋅ cos 2 δ (53) 2 The expression cos δ is knew (54): ( s 0 + s) 2 cos 2 δ = (54) ( s 0 + s ) 2 + ( s '−e) 2 The expression of the θ’A is more difficult (55):
θ AI = [( s 0 + s ) 2 + e 2 − e ⋅ s '− rb ⋅ ( s 0 + s) 2 + ( s'−e) 2 ] ⋅ {[( s 0 + s ) 2 + ( s '−e) 2 ] ⋅ ( s 0 + s ) 2 + ( s '−e) 2 + rb ⋅ [ s ' '⋅( s 0 + s ) − s '⋅( s '−e) − ( s 0 + s ) 2 − ( s '−e) 2 ]} /
(55)
[( s 0 + s ) 2 + ( s '−e) 2 ] /{[( s 0 + s ) 2 + e 2 + rb2 ] ⋅ ⋅ ( s 0 + s ) 2 + ( s '−e) 2 − 2 ⋅ rb ⋅ [( s 0 + s ) 2 + e 2 − e ⋅ s ' ]} One determine μ by its expressions (56-57): cos μ =
[( s 0 + s ) 2 + e 2 ] ⋅ ( s 0 + s ) 2 + ( s '−e) 2 − rb ⋅ [( s 0 + s ) 2 + e 2 − e ⋅ s ' ] rA ⋅ rB ⋅ ( s 0 + s ) 2 + ( s '−e) 2
sin μ =
rb ⋅ ( s 0 + s ) ⋅ s ' rA ⋅ rB ⋅ ( s 0 + s ) 2 + ( s '−e) 2
(56)
(57)
2.6. The dynamics of the Module B
One uses for the dynamics of the Module B the relations (58-60): K2 [ ⋅ m S* + mT* ] ⋅ ω 2 2 2 (K + k ) k + 2kK 2 2kx 0 ⋅s + ⋅s+ ⋅ y' 2 2 K k + K k + (K + k ) ΔX = − (58) kx 0 2 ⋅ [s + ] K +k
Al III-lea Seminar National de Mecanisme, Craiova, 2008
6
K2 ⋅ m S* + mT* ] ⋅ ω 2 2 2 (K + k ) k + 2kK 2 2kx 0 ⋅s + ⋅s+ ⋅ ( D ⋅ s' ) 2 2 K +k K +k (K + k ) ΔX = − (59) kx 0 2 ⋅ [s + ] K +k X = s + ΔX (60) [
2.7. The dynamics analyze at the module B
One presents now the dynamics of the module B for some knew movement laws. One starting with the classical law SIN (see the diagram from picture number 2); One use a speed rotation n=5500 [rot/min], for a maxim theoretical displacement at valve h=6 [mm]. The faze angle is ϕu=ϕc=65 [grad]; the ray of the basic circle is r0=13 [mm]. For the ray of the roll has been adopted the value rb=13 [mm]. Analiza dinamicã la cama rotativã cu tachet translant cu rolã
8000
amax =6400
6000
s max =5.81
x0=20 [mm]
4000
hs =6 [mm]
2000 0 0
50
-2000 -4000
n=5500[rot/min] ϕu=65 [grad] k=30 [N/mm] r0=13 [mm]
100
hT=6 [mm] i=1;η=11.5% rb=13 [mm] e=6 [mm] 150 legea: sin-0 y=x-sin(2π x)/(2π )
amin= -3000
s*k[mm] k=
Fig. 2. The dynamic analyze at the module B. The law sin, n=550 [rpm], ϕu=65 [grad], r0=13 [mm], rb=13 [mm], hT=6 [mm], e=0 [mm], k=30 [N/mm], and x0=20 [mm].
a[m/s2] 880.53
Al III-lea Seminar National de Mecanisme, Craiova, 2008
7
PROFIL Camã rotativã cu tachet translant cu rolã
yC [mm]
20 ϕ u= 65[grad]
15
ϕ c= 65[grad]
10
r0 = 13[mm]
5
rb = 13[mm] e= 6[mm] hT = 6[mm] 20 Legea SIN
0 -20
-10
-5
0
10 ω
-10 -15
Suportã o turatie n=5500[rot/min]
Fig. 3. The profile SIN at the module B. n=5500 [rot/min] ϕu=65 [grad], r0=13 [mm], rb=13 [mm], hT=6 [mm].
The dynamic is better than of the classical module C. For a faze angle of just 65 degree one has the same values of accelerations that the classical module C had them at phase relaxed 75-80 grade. In the pictures number three one can see the cam profile. The law cos can be seen in the pictures 4 and 5.
5000
Analiza dinamicã la cama rotativã cu tachet translant cu rolã amax =4300
4000
s max =5.74
n=5500[rot/min] ϕu=65 [grad] k=30 [N/mm] r0=13 [mm]
3000
x0=30 [mm]
2000
hs =6 [mm]
1000 0 -1000 0
50
-2000 -3000
100
hT=6 [mm] i=1;η=10.5% rb=6 [mm] 150e=0 [mm] legea: cos-0 y=.5-.5cos(π x)
amin= -2000
s*k[mm] k=
a[m/s2] 601.01
Fig. 4. The dynamic analyze at the module B. Law COS, n=5500 [rot/min], ϕu=65 [grad], r0=13 [mm], rb=6 [mm], hT=6 [mm], η=10.5%.
8
Al III-lea Seminar National de Mecanisme, Craiova, 2008
PROFIL Camã rotativã cu tachet translant cu rolã
yC [mm]
20 ϕ u = 65[grad]
15
ϕ c= 65[grad]
10
r0= 13[mm] rb = 6[mm] e= 0[mm] hT = 6[mm] 20 Legea COS
5 0 -20
-10
-5
0
10 ω
-10 -15
Suportã o turatie n=5500[rot/min]
Fig. 5. The profile COS at the module B, n=5500 [rot/min], ϕu=65 [grad], r0=13 [mm], rb=6 [mm], hT=6 [mm].
14000
Analiza dinamicã la cama rotativã cu tachet translant cu rolã a max=13000
12000 10000
s max= 5.37
n=5500[rot/min] ϕu=80 [grad] k=50 [N/mm] r0=13 [mm] x0=50 [mm]
8000
hs =6 [mm]
6000
hT=6 [mm] i=1;η=8.6%
4000
r b=6 [mm] e=0 [mm] legea: C4P1-0 2 y=2x-x
2000 0 -2000 0
50
100 150 a = -600 min
200
s*k[mm] k=
a[m/s2] 1896.75
Fig. 6. The dynamic analyze. Law C4P1-0, n=5500 [rot/min], ϕu=80 [grad], r0=13 [mm], rb=6 [mm], hT=6 [mm].
In the pictures number 6 one analyze dynamic the law C4P, created by the authors. The vibrations are diminished, the noises are limited, the effective
Al III-lea Seminar National de Mecanisme, Craiova, 2008
9
displacement of the valve is increased, smax=5.37 [mm]. The efficiency has a good value η=8.6%. In the pictures number 7 one presents the profile of C4P law. One starts at the law C4P with n=5500 [rpm], but this law can up the rotation velocity at a value of 30000-40000 [rpm] (see picture number 8). PROFIL Camã rotativã cu tachet translant cu rolã
yC [mm]
25 ϕu= 80[grad]
20
ϕc = 80[grad] r0= 13[mm]
15
rb = 3[mm] e= 0[mm] hT= 6[mm] Legea C4P1-0
10 5 0 -20
-10
-5
0
10
-10
20
ω
-15 Suportã o turatie n=5500[rot/min]
-20
Fig. 7. The profile C4P at the module B.
120000 100000 80000
Analiza dinamicã la cama rotativã cu tachet n=40000[rot/min] translant cu rolã ϕu=80 [grad] amax =94000 k=400 [N/mm] r0=13 [mm] s max =3.88 x0=150 [mm]
60000
hs =10 [mm]
40000 20000 0 -20000 0 -40000
50
100 amin= -33000
150
200
hT=10 [mm] i=1;η=14.4% rb=6 [mm] e=0 [mm] legea: C4P1-5 2 y=2x-x
s*k[mm] k=
a[m/s2] 19371.43
. Fig. 8. The dynamic analyze of the module B. Law C4P1-5, n=40000 [rot/min].
Al III-lea Seminar National de Mecanisme, Craiova, 2008
10
3. Conclusion
One can speak about an advantage of the module B than the classical module C. With the module B, (when the follower is with roll) one can obtain an up rotation velocity with a superior efficiency. References 1- Petrescu F.I., Petrescu R.V., Contributions at the dynamics of cams. In the Ninth IFToMM International Sympozium on Theory of Machines and Mechanisms, SYROM 2005, Bucharest, Romania, 2005, Vol. I, p. 123-128. Relly Victoria PETRESCU Universitatea Politehnica din Bucureşti, Departamentul GDGI Splaiul Independenţei 313, Bucureşti, Sector 6, cod 060042 petrescuvictoria@yahoo.com Adriana COMĂNESCU Universitatea Politehnica din Bucureşti, Departamentul TMR Splaiul Independenţei 313, Bucureşti, Sector 6, cod 060042 adrianacomanescu@yahoo.com Florian Ion PETRESCU Universitatea Politehnica din Bucureşti, Departamentul TMR Splaiul Independenţei 313, Bucureşti, Sector 6, cod 060042 petrescuflorian@yahoo.com Ovidiu ANTONESCU Universitatea Politehnica din Bucureşti, Departamentul TMR Splaiul Independenţei 313, Bucureşti, Sector 6, cod 060042 oval33@hotmail.com