Florian PETRESCU, Relly PETRESCU
THE CAM DESIGN FOR A BETTER EFFICIENCY
Abstract: The paper presents an original method to determine the efficiency of a mechanism with cam and follower. The originality of this method consists in eliminate of the friction modulus. In this paper on analyze three types of cam mechanisms: 1.The mechanism with rotary cam and plate translated follower; 2.The mechanism with rotary cam and translated follower with roll; 3.The mechanism with rotary cam and rocking-follower with roll. In every kind of cam and follower mechanism on utilize a different method for the best efficiency design. Key Words: efficiency, power, cam, follower, roll, force, speed. 1. INTRODUCTION In this paper the authors present an original method to calculate the efficiency of the cam mechanisms. The originality consists in the eliminating of friction forces and friction coefficients. On determine just the mechanical efficiency of cam mechanism. In every kind of cam and follower mechanism on are utilizing a different method for the design with maximal efficiency. In this paper on analyze three kinds of cam and follower mechanisms. 2. DETERMINING THE MOMENTARY MECHANICAL EFFICIENCY OF THE ROTARY CAM AND PLATE TRANSLATED FOLLOWER The consumed motor force, Fc, perpendicular in A on the vector rA, is dividing in two components, [1]: 1. Fm, which represents the useful force, or the motor force reduced to the follower; 2. Fψ, which is the sliding force between the two profiles of cam and follower, (see the picture 1). Pc is the consumed power and Pu represents the useful power. The written relations are the next: Fm = Fc ⋅ sin τ (2.1)
v2 = v1 ⋅ sin τ
(2.2)
Pu = Fm ⋅ v 2 = Fc ⋅ v1 ⋅ sin 2 τ
(2.3)
Pc = Fc ⋅ v1
(2.4)
ηi =
2
Pu Fc ⋅ v1 ⋅ sin τ = = Pc Fc ⋅ v1 2
(2.5)
2
= sin τ = cos δ
sin 2 τ =
s ’2 s ’2 = 2 rA ( r0 + s ) 2 + s ’2
Fψ = Fc ⋅ cos τ v12 = v1 ⋅ cos τ 2
Pψ = Fψ ⋅ v12 = Fc ⋅ v1 ⋅ cos τ
(2.6)
F
& Fc
A rA
δ
v&2
δ
τ
s’
δ
C
v&1
v&12
D
& Fψ
B
& Fm E
s
τ
r0
O
ω
Fig. 1 Forces and speeds to the cam with plate translated follower. Determining the efficiency.
ψi =
Pψ Pc
=
Fc ⋅ v1 ⋅ cos 2 τ = Fc ⋅ v1
2
(2.10)
2
= cos τ = sin δ In the relation number (2.11) on determine the mechanical efficiency:
η = 0.5 ⋅ {1 −
(r0 + sτ M ) ⋅ s ’τ M
} τ M ⋅ [(r0 + sτ M ) 2 + s ’2τ M ]
(2.11)
3. DETERMINING THE MOMENTARY MECHANICAL EFFICIENCY OF THE ROTARY CAM AND TRANSLATED FOLLOWER WITH ROLL The written relations are the next: rB2 = e 2 + (s 0 + s) 2
(3.1)
(2.7)
rB = rB2
(3.2)
(2.8)
e cos α B ≡ sin τ = rB
(3.3)
(2.9)
sin α B ≡ cos τ =
s0 + s rB
(3.4)
DECEMBER 2006 VOLUME 1 NUMBER 2 JIDEG 33
The CAM Design for a Better Efficiency Fu, v 2
Fn, vn δ
Fn, v n
Fi, vi
B rb
F m, v m αA-δ
s
A
Fa, va
s0
B0
rA
rB
A0
n α0 C
x
θB
µ
γ
θA ϕ
αA e r0
O
Fig. 2 Forces and speeds to the cam with translated follower with roll. Determining the efficiency.
The pressure angle, δ, is determined by the relations (3.5-3.6): s0 + s cos δ = (3.5) ( s 0 + s ) 2 + ( s ’−e) 2
s ’−e
sin δ =
( s 0 + s ) 2 + ( s ’−e) 2
cos(δ + τ ) = cos δ ⋅ cosτ − sin δ ⋅ sin τ
rA2
=
rB2
+
rb2
− 2 ⋅ rb ⋅ rB ⋅ cos(δ + τ )
va = vm ⋅ sin(α A − δ ) Fa = Fm ⋅ sin(α A − δ )
(3.13)
vn = vm ⋅ cos(α A − δ ) Fn = Fm ⋅ cos(α A − δ )
(3.14)
vi = vn ⋅ sin δ Fi = Fn ⋅ sin δ
(3.15)
(3.8)
v2 = vn ⋅ cos δ = vm ⋅ cos(α A − δ ) ⋅ cos δ Fu = Fn ⋅ cos δ = Fm ⋅ cos(α A − δ ) ⋅ cos δ
(3.16)
(3.9)
Pu = Fu ⋅ v2 = Fm ⋅ vm ⋅ cos 2 (α A − δ ) ⋅ cos 2 δ (3.17) Pc = Fm ⋅ vm (3.18)
(3.6) (3.7)
cos α A = =
e ⋅ ( s 0 + s ) 2 + ( s ’−e) 2 + rb ⋅ ( s ’−e) rA ⋅ ( s 0 + s ) 2 + ( s ’−e) 2
The momentary mechanical efficiency can be obtained by the relation (3.19):
sin α A = =
( s 0 + s ) ⋅ [ ( s 0 + s ) 2 + ( s ’−e) 2 − rb ]
(3.10)
ηi =
rA ⋅ ( s 0 + s ) 2 + ( s ’−e) 2
cos(α A − δ ) = ( s0 + s ) ⋅ s ’
s’ = = ⋅ cos δ 2 2 r A rA ⋅ ( s0 + s ) + ( s ’−e) s’ cos(α A − δ ) ⋅ cos δ = ⋅ cos 2 δ rA
(3.11)
=
Pu = Pc
Fm ⋅ vm ⋅ cos 2 (α A − δ ) ⋅ cos 2 δ Fm ⋅ vm
= cos 2 (α A − δ ) ⋅ cos 2 δ = (3.12)
On can write the next forces and speeds (see the picture 2): Fm, vm, are perpendicular on the vector rA in A. Fm is dividing in Fa (the sliding force) and Fn (the normal force). Fn is dividing too in Fi (the bending force) and Fu (the useful force).
34 DECEMBER 2006 VOLUME 1 NUMBER 2 JIDEG
= (3.19)
2
= [cos(α A − δ ) ⋅ cos δ ] = =[
s’ s ’2 ⋅ cos 2 δ ]2 = 2 ⋅ cos 4 δ rA rA
4. DETERMINING THE MOMENTARY MECHANICAL EFFICIENCY OF THE ROTARY CAM AND ROCKING FOLLOWER WITH ROLL The written relations are the next:
The CAM Design for a Better Efficiency b 2 + d 2 − (r0 + rb ) 2 2⋅b⋅d ψ 2 =ψ +ψ 0 RAD = cosψ 0 =
= d 2 + b 2 (1 − ψ ’) 2 − 2bd (1 − ψ ’) cosψ 2 d ⋅ cosψ 2 + b ⋅ ψ ’−b RAD d ⋅ sin ψ 2 cos δ = RAD rB2 = b 2 + d 2 − 2 ⋅ b ⋅ d ⋅ cosψ 2
sin δ =
2
rB2
cos α B =
d + −b 2 ⋅ d ⋅ rB
sin α B =
b ⋅ sin ψ 2 rB
2
sin(δ + ψ 2 ) = sin δ cosψ 2 + sin ψ 2 cos δ cos(δ + ψ 2 ) = cos δ cosψ 2 − sin ψ 2 sin δ π B = δ +ψ 2 + α B − 2 cos B = sin(δ + ψ 2 + α B ) sin B = − cos(δ + ψ 2 + α B ) cos B = sin(δ + ψ 2 ) ⋅ cos α B + + sin α B ⋅ cos(δ + ψ 2 ) sin B = sin(δ + ψ 2 ) ⋅ sin α B − − cos α B ⋅ cos(δ + ψ 2 )
rA2 = rB2 + rb2 − 2 ⋅ rb ⋅ rB ⋅ cos B
(4.1) (4.2) (4.3) (4.4) (4.5) (4.6) (4.7) (4.8) (4.9) (4.10) (4.11) (4.12) (4.13) (4.14) (4.15) (4.16)
r 2 + rB2 − rb2 cos µ = A 2 ⋅ rA ⋅ rB
(4.17)
r sin µ = b ⋅ sin B rA
(4.18)
αA =αB + µ cos α A = cos α B cos µ − sin α B sin µ sin α A = sin α B cos µ + cos α B sin µ α = π − α A −ψ 2 − δ cos α = − cos(ψ 2 + δ + α A ) = = sin(ψ 2 + δ ) ⋅ sin α A − − cos(ψ 2 + δ ) ⋅ cos α A ψ ’⋅b cos α = ⋅ cos δ rA ψ ’⋅b cos α ⋅ cos δ = ⋅ cos 2 δ rA
(4.19) (4.20) (4.21) (4.22)
Fa = Fm ⋅ sin α va = vm ⋅ sin α Fn = Fm ⋅ cos α vn = vm ⋅ cos α F = F ⋅ sin δ n c v = v ⋅ sin δ n c Fu = Fn ⋅ cos δ = Fm ⋅ cos α ⋅ cos δ v2 = vn ⋅ cos δ = vm ⋅ cos α ⋅ cos δ 2 2 Pu = Fu ⋅ v2 = Fm ⋅ vm ⋅ cos α ⋅ cos δ Pc = Fm ⋅ vm P η i = u = cos 2 α ⋅ cos 2 δ = Pc
ψ ’⋅b = (cos α ⋅ cos δ ) 2 = ( ⋅ cos 2 δ ) 2 = rA =
(4.26)
(4.27)
ψ ’2 ⋅b 2 ⋅ cos 4 δ r A2
sin(δ + ψ 2 ) = sin δ ⋅ cosψ 2 + d − b ⋅ cosψ 2 ⋅ (1 − ψ ’) RAD rb rb sin µ = ⋅ sin B = ⋅ rA rA + cos δ ⋅ sin ψ 2 =
(4.28)
(4.29)
⋅ [sin(δ + ψ 2 ) ⋅ sin α B − cos(δ + ψ 2 ) ⋅ cos α B ] cos µ =
rB − rb ⋅ cos B rB rb = − ⋅ rA rA rA
(4.30)
⋅ [sin(δ + ψ 2 ) ⋅ cos α B + cos(δ + ψ 2 ) ⋅ sin α B ]
cos(δ + ψ 2 ) = cos δ ⋅ cosψ 2 − b ⋅ sin ψ 2 ⋅ (1 − ψ ’) RAD cos α A = cos α B ⋅ cos µ − sin α B ⋅ sin µ = − sin δ ⋅ sin ψ 2 =
=
(4.31)
rB r ⋅ cos α B − b ⋅ [sin(δ + ψ 2 ) ⋅ cos 2 α B rA rA
+ cos(δ + ψ 2 ) ⋅ sin α B ⋅ cos α B + + sin(δ + ψ 2 ) ⋅ sin 2 α B −
(4.23)
− cos(δ + ψ 2 ) ⋅ sin α B ⋅ cos α B ] = =
1 ⋅ [rB ⋅ cos α B − rb ⋅ sin(δ + ψ 2 )] = rA
=
1 ⋅ [d − b ⋅ cosψ 2 − rb ⋅ sin(δ + ψ 2 )] = rA
=
1 ⋅ [d − b ⋅ cosψ 2 − rA
(4.24) (4.25)
Forces and speeds are writhing in the relations (4.26) and the efficiency is writhen in the relation (4.27): On demonstrate now the mode of deduction for the relation (4.24). On can see now a very difficult algorithm for the obtained of this relation (4.24):
− rb ⋅
(4.32)
d − b ⋅ cosψ 2 ⋅ (1 − ψ ’) ] RAD
DECEMBER 2006 VOLUME 1 NUMBER 2 JIDEG 35
The CAM Design for a Better Efficiency picture 3): Fm, vm, are perpendicular on the vector rA in A. Fm is dividing in Fa (the sliding force) and Fn (the normal force). Fn is dividing too in Fc (the compressed force) and Fu (the useful force). For the mechanisms, with rotary cam and diverse kind of followers, on must utilize different methods for realizing the design with maximal efficiency to every type of follower. cos Îą = sin(Ďˆ 2 + δ ) â‹… sin Îą A − cos(Ďˆ 2 + δ ) â‹… cos Îą A =
Fig. 3 Forces and speeds to the cam with rocking follower with roll. Determining the efficiency.
rB r â‹… sin Îą B − b â‹… [cos(δ + Ďˆ 2 ) â‹… sin 2 Îą B rA rA
+ sin(δ + Ďˆ 2 ) â‹… sin Îą B â‹… cos Îą B −
r r = B â‹… sin Îą B − b â‹… cos(δ + Ďˆ 2 ) = rA rA =
(4.33)
b r â‹… sinĎˆ 2 − b â‹… cos(δ + Ďˆ 2 ) = rA rA
d − b â‹… cosĎˆ 2 â‹… (1 − Ďˆ ’) (4.34) RAD b â‹… sinĎˆ 2 â‹… (1 − Ďˆ ’) cos(Ďˆ 2 + δ ) = (4.35) RAD 1 r â‹… b â‹… sinĎˆ 2 â‹… (1 − Ďˆ ’) sin Îą A = â‹… [b â‹… sinĎˆ 2 − b ] (4.36) rA RAD sin(Ďˆ 2 + δ ) =
r â‹… b â‹… cosĎˆ 2 â‹… (1 − Ďˆ ’) − rb â‹… d + b ] RAD
(4.38)
rb â‹… b 2 â‹… sinĎˆ 2 â‹… cosĎˆ 2 â‹… (1 − Ďˆ ’) 2 + RAD r â‹… b â‹… d â‹… sinĎˆ 2 â‹… (1 − Ďˆ ’) + b ]= RAD b â‹… Ďˆ ’⋅d â‹… sinĎˆ 2 b â‹… Ďˆ ’ d â‹… sinĎˆ 2 b â‹… Ďˆ ’ = = â‹… = â‹… cos δ rA â‹… RAD rA RAD rA
5. CONCLUSION
1 b â‹… sinĎˆ 2 â‹… (1 − Ďˆ ’) = â‹… [b â‹… sinĎˆ 2 − rb â‹… ] rA RAD
1 cos Îą A = â‹… [d − b â‹… cosĎˆ 2 + rA
rb â‹… b â‹… d â‹… sinĎˆ 2 â‹… (1 − Ďˆ ’) + RAD r â‹… b 2 â‹… sinĎˆ 2 â‹… cosĎˆ 2 â‹… (1 − Ďˆ ’) 2 + b − RAD − b â‹… d â‹… sinĎˆ 2 â‹… (1 − Ďˆ ’) + b 2 â‹… sinĎˆ 2 â‹… cosĎˆ 2 â‹… (1 − Ďˆ ’) − −
− sin(δ + Ďˆ 2 ) â‹… sin Îą B â‹… cosÎą B + + cos(δ + Ďˆ 2 ) â‹… cos 2 Îą B ] =
1 â‹… [b â‹… d â‹… sinĎˆ 2 − b 2 â‹… sinĎˆ 2 â‹… cosĎˆ 2 â‹… (1 − Ďˆ ’) − rA â‹… RAD
−
sin Îą A = sin Îą B â‹… cos Âľ + cos Îą B â‹… sin Âľ = =
=
(4.37)
In figure number three, on can see the forces and the speeds of the mechanism with rotary cam and rocking follower with roll. The cam and the follower are represented in two positions, successively. The distance between the two rotary centers is noted by d. The radius of follower is b. The movement laws are known: Ďˆ, Ďˆâ€™, Ďˆâ€™â€™, Ďˆâ€™â€™â€™.
On can write the next forces and speeds (see the 36 DECEMBER 2006  VOLUME 1  NUMBER 2 JIDEG
The follower with roll, make input-force, to be divided in more components. This is the motive for that, the dynamic and the precisely-kinematics of mechanism with rotary cam and follower with roll, are more different and difficult. 6. REFERENCES [1] Petrescu, R., Petrescu, F. The gear synthesis with the best efficiency. ESFA’ 03, Bucharest, Romania, 2003, Vol. 2, pp. 63-70. [2] Antonescu P., Oprean M., Petrescu, Fl., La projection de la came oscillante chez les mechanismes a distribution variable. CONAT MATMA’ 85, BUDúRY Romania, 1985. Authors: Eng. Florian-Ion PETRESCU, associate professor, University POLITEHNICA of Bucureúti; Eng. Relly-Victoria PETRESCU, Ph.D., lecturer, University POLITEHNICA of Bucureúti.