Cam Dynamic Synthesis

8/10/2019 Cam Dynamic Synthesis

1/23

Al-Khwarizmi

Engineering

Journal

Al-Khwarizmi Engineering Journal, Vol. 10, No. 1, P.P. 1- 23 (2014)

Cam Dynamic Synthesis

Florian Ion T. Petrescu* Relly Victoria V. Petrescu***Department of Theory of Mechanisms and Robots/ Polytechnic University of Bucharest / Romania / Europe

** Department of Transport, Traffic & Logistic Department / Polytechnic University of Bucharest/ Romania / Europe

*E-mail: [email protected]

**E-mail: [email protected]

(Received 10 October 2013; accepted 4 February 2014)

Abstract

The paper presents an original method to make the geometric synthesis of the rotary cam and translated tappet with

roll. Classical method uses to the geometric synthesis and the reduced tappet velocity, and in this mode the geometric

classic method become a geometric and kinematic synthesis method. The new geometric synthesis method uses just the

geometric parameters (without velocities), but one utilizes and a condition to realize at the tapped the velocities

predicted by the tapped movement laws imposed by the cam profile. Then, it makes the dynamic analyze for the

imposed cam profile, and one modify the cam profile geometric parameters to determine a good dynamic response

(functionality). In this mode it realizes the dynamic synthesis of the cam, and we obtain a normal functionality.

Keywords: Geometric synthesis, Dynamic synthesis, Rotary cam, Translated tappet with roll.

1. Introduction

In conditions which started to magnetic

motors, oil fuel is decreasing, energy which wasobtained by burning oil is replaced with nuclear

energy, hydropower, solar energy, wind, and othertypes of unconventional energy, in the conditionsin which electric motors have been instead ofinternal combustion in public transport, but morerecently they have entered in the cars world

(Honda has produced a vehicle that uses acompact electric motor and electricity consumedby the battery is restored by a system that uses anelectric generator with hydrogen combustion incells, so we have a car that burns hydrogen, but

has an electric motor), which is the role andprospects which have internal combustion enginestype Otto or Diesel?

Internal combustion engines in four-stroke

(Otto, Diesel) are robust, dynamic, compact,powerful, reliable, economic, autonomous,independent and will be increasingly clean.

Let's look at just remember that any electric

motor that destroy ozone in the atmosphereneeded our planet by sparks emitted by collectingbrushes. Immediate consequence is that if we onlyuse electric motors in all sectors, well haveproblems with higher ozone shield that protects

our planet and without which no life could existon Earth.

Magnetic motors (combined with theelectromagnetic) are just in the beginning, butthey offer us a good perspective, especially in the

aeronautics industry.Probably at the beginning they will not be used

to act as a direct transmission, but will generateelectricity that will fill the battery that willactually feed the engine (probably an electric

motor).The Otto engines or those with internal

combustion in general, will have to adapt tohydrogen fuel.

It is composed of the basic (hydrogen) canextract industrially, practically from any item (or


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Flor ian I on T. Petrescu Al -Khwarizmi Engineer ing Journal, Vol. 10, No. 1, P.P. 1- 23 (2014)

2

combination) through nuclear, chemical, photonic

by radiation, by burning, etc... (Most easilyhydrogen can be extracted from water by breaking

up into constituent elements, hydrogen andoxygen; by burning hydrogen one obtains wateragain that restores a circuit in nature, with no

losses and no pollution).Hydrogen must be stored in reservoirs cell (a

honeycomb) for there is no danger of explosion;the best would be if we could breaking up waterdirectly on the vehicle, in which case the reservoirwould feed water (and there were announcedsome successful).

As a backup, there are trees that can donate afuel oil, which could be planted on the extendedzone, or directly in the consumer court. Withmany years ago, Professor Melvin Calvin,(Berkeley University), discovered that Euphoratree, a rare species, contained in its trunk a liquidthat has the same characteristics as raw oil. Thesame professor discovered on the territory ofBrazil, a tree which contains in its trunk a fuelwith properties similar to diesel.

During a journey in Brazil, the natives drivenhim (Professor Calvin) to a tree called by them

"Copa-Iba".At the time of boring the tree trunk, from it to

begin flow a gold liquid, which was used asindigenous raw material base for the preparationof perfumes or, in concentrated form, as a balm.

Nobody see that it is a pure fuel that can be useddirectly by diesel engines.

Calvin said that after he poured the liquidextracted from the tree trunk directly into the tankof his car (equipped with a diesel), engine

functioned irreproachable.In Brazil the tree is fairly widespread. It could

be adapted in other areas of the world, planted inthe forests, and the courts of people.

From a jagged tree is filled about half of thetank; one covers the slash and it is not open until

after six months; it means that having 12 trees in acourtyard, a man can fill monthly a tank with thenew natural diesel fuel.

In some countries (USA, Brazil, Germany)producing alcohol or vegetable oils, for their useas fuel.

In the future, aircraft will use ion engines,magnetic, laser or various micro particlesaccelerated. Now, and the life of the jet enginebegin to end.

Even in these conditions internal combustionengines will be maintained in land vehicles (at

least), for power, reliability and especially theirdynamics. Thermal engine efficiency is still lowand, about one third of the engine power is lost

just by the distribution mechanism. Mechanical

efficiency of cam mechanisms was about 4-8%. Inthe past 20 years, managed to increase to about

14-18%, and now is the time to pick it up again atup to 60%. This is the main objective of thispaper.

2. Presenting a Dynamic Model, with one

Grade of Freedom, with Variable

Internal Amortization

2.1.Determining the Amortization

Coefficient of the Mechanism

Starting with the kinematical schema of theclassical valve gear mechanism (see the Figure 1),

one creates the translating dynamic model, with asingle degree of freedom (with a single mass),with variable internal amortization (see the Figure2), having the motion equation (1) [3, 13].

The formula (1) is just a Newton equation,where the sum of forces on a single element is 0.

0)( FxcxkxyKxM (1)

5 1

2

3

4

A

B

C

D

C0

O

Fig. 1. The Kinematical schema of the classical

valve gear mechanism.

The Newton equation (1) can be written in theform (2).

)()( 0 xkFxyKxcxM (2)


3/23


3

M M

k

kx F

F(t) c .

cx

xx(t)

K(y-x)K

y(t)

cam

Fig. 2. Dynamic Model with a Single Liberty, with

variable internal amortization.

The differential equation, Lagrange, can bewritten in the form (3).

rm FFxdt

dMxM 2

1 (3)

Comparing the two equations, (2 and 3), oneidentifies the coefficients and one obtains theresistant force (4), the motor force (5) and thecoefficient of internal amortization (6). One cansee that the internal amortization coefficient, c, isa variable.

)( 000 xxkxkxkxkFFr (4)

)()( xsKxyKFm (5)

dt

dMc 2

1 (6)

One places the variable coefficient, c, (see therelation 6), in the Newton equation (form 1 or 2)

and obtains the equation (7).

0)(

2

1FyKxkKx

dt

dMxM (7)

The reduced mass can be written in the form(8), (the reduced mass of the system, reduced atthe valve).

244

211

22325

)()(

)()(

xJ

xJ

x

ymmmM

(8)

With the following notations:

m2 =the mass of the tappet (of the valve lifter); m3=the mass of the valve push rod; m5 =the valvemass; J1 =the inertia mechanical moment of the

cam; J4 =the inertia mechanical moment of the

valve rocker; 2y =the tappet velocity, or the

second movement-low, imposed by the camsprofile; x =the real (dynamic) valve velocity.

If one notes with i=i25, the ratio of transmission

tappet-valve, given from the valve rocker, thetheoretically velocity of the valve, y , (the tappet

velocity reduced at the valve), takes the form (9),

where the ratio of transmission, i, is given fromthe formula (10).

i

yyy 25

(9)

DC

CCi0

0 (10)

One can write the following relations (11-16),where y is the reduced velocity forced at thetappet by the cams profile. With the relations (10,13, 14, 16) the reduced mass (8), can be written inthe forms (1719).

'1 xx (11)

''21 xx (12)

'1'

212 yiyy (13)

'

1

'1

11

xxx

(14)

DC

y

DC

CC

CC

y

CC

iy

CC

y

CC

y

0

1

0

0

0

1

0

1

0

'

21

0

2

4

''

'.

(15)

'

'1

'

'

010

14

x

y

DCxDC

y

x

..(16)

2

0

4

2

1

2

325

)'

'1()

'

1(

)'

'()(

x

y

DCJ

xJ

x

yimmmM

(17)

2

1

2

2

0

4

32

2

5

)'

1()

'

'(])(

)([

xJ

x

y

DC

J

mmimM

(18)

2

1

2

5 )'

1

()'

'

(* xJx

y

mmM ..(19)


4/23


4

It derives dM/d and obtains the relations (2022).

)'

''

'

''()'

'(2)'

'''''('

'2

'

)''''''(

'

'2])

'

'[(

2

2

2

2

x

x

y

y

x

y

x

yxyx

y

x

yxxy

x

y

d

x

yd

...(20)

32

2

'

''2

'

''

'

2])'

1[(

x

x

x

x

xd

xd

(21)

31

2

'

''2)'

''

'

''()'

'(*2

x

xJ

x

x

y

y

x

ym

d

dM

...(22)

The relation (6) can be written in form (23) andwith relation (22), its taking the forms (2425).

d

dMc 2

(23)

}'

'')'

''

'

''()

'

'(

])(

)({[

31

2

2

0

432

2

x

xJ

x

x

y

y

x

y

DC

Jmmic

(24)

]''')

'''

'''()

''(*[

312

xxJ

xx

yy

xymc (25)

With the notation (26):

2

0

4

32

2

)()(*

DC

Jmmim (26)

2.2.Determining the Movement Equations

With the relations (19, 12, 25, 11) the equation

(2) take the forms (27, 28, 29, 30 and 31) [13]:

0

2 )(''' FyKxkKxcxM (27)

03

''

1

2'

'

''

'

''2*'22''2

'1

''2*2

5

''2

)('

)()'

'()

1(

)'

'(

FyKxkKx

x

Jx

x

x

y

y

x

ymxx

xJ

xx

ymmx

(28)

0

*2

2*22*2

5

2

)('

'''

'')'

'()

'

'(''''

FyKxkKx

yym

xx

ym

x

yxmxm

(29)

0

''*2''

5

2

'

')( FKy

x

yymxkKxm

(30)

0

*

5

2)()

'

'''''( FyKxkKx

yymxm

(31)

The exact equation (31) can be approximated

at the form (32) with xy.

0*

52 )()''''( FyKxkKymxm

(32)

With the following notations: y=s, y=s,y=s, y=s, the equation (32) takes theapproximate form (33) and the complete equation

(31) takes the exact form (34).

0

*

5

2)()''''( FsKxkKsmxm

(33)

052 )()

''''*''( FsKxkK

xssmxm

(34)

Solving the Differential Equation by Direct

Integration and Obtaining the Mother

Equation

One integrates the equation (31) directly. Oneprepares the equation (31) for the integration.First, one writes (31) in form (35) [13].

I

III

TII

Sx

yymxm

xkyKxkK

2*

2*

0)(

(35)

The equation (35), can be amplified by x andone obtains the relation (36).

III

T

III

S

III

yymxxm

xxkxyKxxkK

2*2*

0)(

(36)


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5

Now, one replaces the term K.y.x

with IykK

KyK

, (taken in calculation the

statically assumption, Fm=Fr) and one obtains the

form (37).

III

T

III

S

III

yymxxm

xxkyykK

KxxkK

2*2*

0

2

)(

(37)

One integrates directly the equation (37) and

one obtains the mother equation (38).

Cy

mx

m

xxky

kK

KxkK

TS

2

'

2

'

22)(

22*

22*

0

222

(38)

With the initial condition, at the =0, y=y=0and x=x=0, one obtains for the constant ofintegration, C the value 0. In this case the

equation (38), takes the form (39).

2

'

2

'

22)(

22*

22*

0

222

ym

xm

xxky

kK

KxkK

TS

(39)

The equation (39) can be put in the form (40),

if one divides it with the2

kK .

0)(

'

'2

2

2

22

2*

2

2*

02

ykK

Ky

kK

m

xkK

mx

kK

xkx

T

S

(40)

The mother equation (40), take the form (41),

if one notes: '' ykK

Kx

, (the static

assumption, Fm=Fr).

0')(

)(

)(2

22

**

2

2

2

2

2

02

ykK

mm

kK

K

ykK

Kx

kK

xkx

TS

(41)

Solving the Mother Equation (41) Directly

The equation (41) is a two degree equation in

x; One determines directly, (42-43) and X1,2(44) [13].

22

*

2

2

*

2

22

0 ')(

)(

)(

)()(y

kK

mkK

Km

kK

sKxk TS

(42)

22

*

2

2*

2

22

0 )'()(

)(

)(

)()(Ds

kK

mkK

Km

kK

Kskx TS

(43)

kK

xkX 02,1 (44)

Physically, just the positive solution is valid

(see the relation 45).

kK

xkX

0 (45)

Solving the Mother Equation (41) with

Finished Differences

One can solve the mother equation (41) using

the finished differences [13]. One notes:

XsX (46)

With the notation (46) placed in the mother

equation (41), it obtains the equation (47).

0')(

)(

)(

222)(

22

**

2

2

2

2

2

0022

ykK

mmkK

K

skK

K

XkK

xks

kK

xkXsXs

TS

(47)

The equation (47) is a two degree equation in

X, which can be solved directly with (49) and

X1,2, (50), or transformed in a single degree

equation in X, with (X)20, solved by the

relation (48).


6/23


6

20

22**2

0

22

)()(2

)'(])([)(2)2(

)1(

kKkK

xks

DsmkKmkK

KskKkxskKk

X

TS

(48)

2

22**2

2

0

222

)(

)'(])([

kK

DsmkKmkK

KxksK TS

(49)

)( 0

kK

xksX

(50)

2.3.Mechanism with Rotary Cam and

Translated Tappet with Roll

First, one presents an original method todetermine the efficiency at the mechanism withrotary cam and translated follower with roll [5].

With this occasion it presents and the forcesand the velocities as well (Figure 3).

The pressure angle (Figure 3), is determinedby relations (1.5-1.6).

We can write the next forces, speeds andpowers (1.13-1.18).

Fm(vm) is perpendicular to the vector rAat A.Fmis divided into Fa(the sliding force) and Fn

(the normal force). Fn is divided too, into Fi (the

bending force) and Fu (the useful force). Themomentary dynamic efficiency can be obtained

from relation (1.18).

0A

A

B

A-

Fn, vn

Fm, vm

Fa, va

Fi, viFn, vn

Fu, v2

B

B0

A0

A

O

x

e

s0

rb

r0

rA

rB

s

n

C

rb

Fig. 3. Forces and velocities at the cam with

translated follower with roll.

The written relations are the following.

2

0

22

B s)(ser ... (1.1)

2

BB rr ...(1.2)

B

Bre sincos ...(1.3)

B

Br

ss 0cossin ...(1.4)

22

0

0

)'()(cos

esss

ss

...(1.5)

22

0 )'()(

'sin

esss

es

...(1.6)

sinsincoscos)cos( ...(1.7)

)cos(2

)cos()sin(

222

2

0

22

BbbBA

bbA

rrrrr

rssrer

...(1.8)

A

b

A

A

b

A

r

re

esssr

esressse

sin

cos

)'()(

)'()'()(cos

22

0

22

0

...(1.9)

A

bA

A

b

A

r

rss

esssr

resssss

cossin

)'()(

])'()([)(sin

0

22

0

22

00

...(1.10)

cos'

)cos(

)'()(

')()cos( 22

0

0

A

A

A

A

r

s

esssr

sss

...(1.11)

2cos'

cos)cos( A

Ar

s ...(1.12)

)sin(

)sin(

Ama

Ama

FF

vv ...(1.13)


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7

)cos(

)cos(

Amn

Amn

FF

vv ...(1.14)

sin

sin

ni

ni

FF

vv ...(1.15)

cos)cos(cos

cos)cos(cos2

Amnu

Amn

FFF

vvv

...(1.16)

mmc

Ammuu

vFP

vFvFP 222 cos)(cos

...(1.17)

4

2

2

222

22

cos'

]cos'[]cos)[cos(

cos)(cos

A

i

A

Ai

mm

Amm

c

ui

r

s

r

s

vF

vF

P

P

...(1.18)

bB rrr 00 (1.19)

22

00

ers B (1.20)

0

0cosBr

e (1.21)

0

0

0sinBr

s (1.22)

2.4.The Relations to Design the Profile

Now one determines the profile of the cam(relations 1.23-1.28).

0 A (1.23)

00 sinsincoscoscos AA

(1.24)

00 sincoscossinsin AA (1.25)

A (1.26)

sinsincoscoscos A (1.27)

cossincossinsin A (1.28)

2.5.The Exact Kinematics of B Module

From the triangle OCB (fig. 3) the length rB

(OB) and the complementary angles B (COB)

and (CBO) are determined by the relation (1.1-1.4).

From the general triangle OAB, where oneknows OB, AB, and the angle between them, B

(ABO, which is the sum of with ), the length

OA and the angle (AOB) can be determinedwith the relations (1.7-1.8, 1.29-1.31):

BA

bBA

rr

rrr

2

cos222

(1.29)

cossincossin)sin( ...(1.30)

)sin(sin A

b

r

r (1.31)

With B and we can deduce now A and

A (the relations 1.32-1.33):

BA (1.32)

BA (1.33)

From (1.3) one obtains B (1.37), (see 1.34-

1.37) where Br (1.36) can be deduced from (1.1).

Then, (1.38) will be obtained from (1.29):

2sin

B

B

BBr

re

(1.34)

2

0 )( B

BB

Brss

rre

(1.35)

sssrr

sssrr

BB

BB

)(

)(22

0

0 (1.36)

22

0

0

)(

)(

BB

Br

se

rss

ssse

(1.37)

BBAABA

BABA

rrrrrr

rrrr

22sin2

cos2cos2

(1.38)

From (1.38) one writes (1.43), but it is

necessary to obtain first Ar (1.39) fromexpression (1.8):


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8

)()sin(2

)cos(222

Bb

BbBBAA

rr

rrrrrr

(1.39)

To solve (1.39) we need the derivatives

(1.40 and 1.41) and (1.42).

22

0

0

)'()(

)'(')('''

esss

essess

(1.40)

' (1.41)

2

B

Br

se

(1.42)

Now we can determine

(1.43), A

(1.33)and A

(1.44):

sin

coscos

BA

BBAABABA

rr

rrrrrrrr

(1.43)

AA (1.44)

We write cos A and sin A (1.9-1.10):

Further, we can obtain the expression cos(A-

) (1.11), and cos(A-).cos(1.12).

Finally the forces and the velocities arededuced as follows (1.13-1.16):

2.6.Determining the Efficiency of the

Module B

With the relationships (1.17-1.18) we candetermine the powers and the momentarymechanical efficiency [14].

Determining the (Dynamic) Transmission

Function D, for the Module B

The followers velocity (1.16) can be writteninto the form (1.45).

22

22

2

cos'cos'

cos'

cos'

cos)cos(cos

ss

r

sr

r

sv

vvv

IAA

A

AA

A

m

Amn

(1.45)

With the relationships (1.45) and (1.46) we

determine the transmission function (the dynamicmodulus), D (1.47):

Dsv '2 (1.46)

2cos IAD (1.47)

Expression cos2is known (1.48):

22

0

2

02

)'()(

)(cos

esss

ss

(1.48)

The expression of the A (1.49) is moredifficult.

]}')[(2

)'()(

])/{[(

/])'()/[(]})'(

)()'(')(''[

)'()(])'(

){[(])'()(

')[(

22

0

22

0

222

0

22

0

2

2

00

22

0

2

2

0

22

0

22

0

seessr

esss

ress

essses

ssesssssr

essses

ssesss

rseess

b

b

b

b

I

A

(1.49)

We will determine by its expressions (1.50-1.51):

22

0

220

22

0

22

0

22

0

)'()(

]')[(

)'()(

)'()(])[(

cos

esssrr

seessr

esssrr

esssess

BA

b

BA

(1.50)

22

0

0

)'()(

')(sin

esssrr

sssr

BA

b

(1.51)


9/23


9

The Dynamics of Distribution Mechanisms

with Translated Follower with Roll

For the dynamics of the Module B therelationships (49-50) are used in the forms (1.52-1.54), where D is determined from (1.47).

][2

'

])(

[2

)(

2

0

2

2**

2

2

02

2

2

kK

kxs

ykK

mmkK

K

skK

kxs

kK

kKk

X

TS

(1.52)

][2

)'(

])(

[2

)(

2

0

2

2**

2

2

02

2

2

kK

kxs

sDkK

mmkK

K

skK

kxs

kK

kKk

X

TS

(1.53)

XsX (1.54)

2.7.The Dynamic Analysis of the Module B

It presents now the dynamics of the module Bfor some knownmovement laws.

We begin with the classical law SIN (see thediagram in Figure 4); A speed rotation n=5500[rot/min], for a maxim theoretical displacement of

the valve h=6 [mm] is used.Thephaseangle is u=c=65 [degree]; the ray

of the basic circle is r0=13 [mm].For the ray of the roll the value rb=13 [mm]

has been adopted.

Fig. 4. The dynamic analysis of the module B. The

law SIN, n=550 rpm, u=650

, r0=13 [mm], rb=13[mm], hT=6 [mm], e=0 [mm],k=30 [N/mm], and

x0=20 [mm].

Fig.5. The profile SIN at the module B. n=5500 rpm

u=650, r0=13 [mm], rb=13 [mm], hT=6 [mm].

The dynamics are better than for the classical

module C. For a phase angle of just 65 degrees theaccelerations have the same values as for theclassical module C for arelaxedphase (750-800).

In Figure 5 we can see the cams profile. Ituses the profile sin, a rotation speed n=5500 rpm,

and u=650, r0=13 [mm], rb=13 [mm], hT=6 [mm].

The law COS can be seen in the Figures 6 and

7. In the Figure 6 is presented the dynamicanalyze of the profile cos, and its profile designcan be seen in the Figure 7.

The principal parameters are:

Law COS, n=5500 rpm, u=650, r0=13 [mm],

rb=6 [mm], hT=6 [mm], =10.5%.

Fig. 6. The dynamic analysis of the module B. LawCOS, n=5500 rpm, u=65

0, r0=13 [mm], rb=6 [mm],

hT=6 [mm], =10.5%.


10/23


10

Fig. 7. The profile COS at the module B, n=5500

rpm, u=650

, r0=13 [mm], rb=6 [mm], hT=6 [mm].

Fig. 8. The dynamic analysis. Law C4P1-0, n=5500

rpm, u=800, r0=13 [mm], rb=6 [mm], hT=6 [mm].

In the figure 8 the law C4P, created by the

author, is analyzed dynamic. The vibrations arediminished, the noises are limited, the effectivedisplacement of the valve is increased, smax=5.37

[mm].

Fig. 9. The profile C4P of the module B.

The efficiency has a good value =8.6%. In theFigure 9 the profile of C4P law is presented.

It starts at the law C4P with n=5500 [rpm], butfor this law the rotation velocity can increase tohigh values of 30000-40000 [rpm] (see the Figure10).

Fig. 10. The dynamic analysis of the module B. Law

C4P1-5, n=40000 rpm.


11/23


11

2.8.The New Cam Synthesis

The rotary cam with translated follower withroll (Figure 3 or 13), is synthesized dynamic bythe new next relationships.

It has exchanged the rotation sense from the

Figure 1 to Figure 11, from the clockwise to thecounter-clockwise.

First, one determines the mass moment ofinertia (mechanical) of the mechanism, reduced tothe element of rotation, ie cam (basically usingkinetic energy conservation, system 2.1).

2

22

0

2

0

2

2

00

2

0

2*

2

0

22

2

0

2

0

2

0

2

0

2

00

22

0

*

22

0

22

0

2

02

0

222

22

0

00

22

0

2

0

222

0

2

0

222

0

222

0

2222222

2

''

'

2

122

2

1

8

22

22

2

1

16

1

4

1

2

1

'

'2

'

2

'2

'

'2

cossin2

cos2cos

sin2sin

2

1

smesss

sseserM

sMssMrrrrMJ

hm

ehh

s

hse

he

rM

hMhsMrrrrMJ

esss

eser

esss

ssrssrer

esss

ssssr

esss

eserssrer

sserssrer

ssrrss

rereyxrR

RMJ

Tbc

ccbbc

Tbc

ccbbcm

b

bbA

b

bbA

bbA

bb

bbAAA

ccama

...(2.1)

We considered the law of motion of the tappetclassic version already used the cosine law (bothascending and descending).

The angular velocity is a function of the cam

position () but also its rotation speed (2.2).

Where mis the nominal angular velocity of camand express at the distribution mechanisms basedon the motor shaft speed (2.3).

2

*

*2

m

m

J

J ...(2.2)

60260

2

6022

nn

n

motor

ccm

...(2.3)

We start the simulation with a classical law ofmotion, namely the cosine law. To climb cosinelaw system is expressed by the relationships (2.4).

uu

r

uu

r

uu

r

u

hs

has

hvs

hhs

sin2

'''

cos2

''

sin2

'

cos22

3

3

2

2

...(2.4)

Where takes values from 0 to u.

Jmaxoccurs for =u/2.With the relation (2.5) is expressed the first

derivative of the reduced mechanical moment ofinertia. It is necessary to determine the angularacceleration (2.6).

2/3220

0

2

0

2

2/322

0

22

00

0

*'

'

'''''

'

''2''

'''2''

esss

sesssssseserM

esss

essssssserM

ssmssMssMJ

bc

bc

Tcc

...(2.5)

Differentiating the formula (2.2), against time,is obtained the angular acceleration expression(2.6).

*

*'2

2 J

J

...(2.6)

Relations (2.2) and (2.6) a general nature andare basically two original equations of motion,crucial for mechanical mechanisms.

For a rotary cam and translated tappet with rollmechanism (without valve), dynamic movementtappet is expressed by equation (2.7), which is an


12/23


12

original dynamic equation deduced for the

distribution mechanism (50) and now bycanceling the valve mass, will customize and

reaching form below (2.7).

kK

xkskK

skKxksKkksmkK

sx

T

02

0

2222

)(2

)(2)2(')(

...(2.7)

Where x is the dynamic movement of thepusher, while s is its normal, kinematicsmovement. K is the spring constant of the system,

and k is the spring constant of the tappet spring.It note, with x0the tappet spring preload, with

mT the mass of the tappet, with the angularrotation speed of the cam (or camshaft), where sis the first derivative in function of of the tappet

movement, s. Differentiating twice successively,the expression (2.7) in the angle , we obtain areduced tappet speed (equation 2.8), and reducedtappet acceleration (2.9).

2

02

0

0

22

0

2222

2

''

'

'2'22'''2

)(2)2(')(

kK

kxskK

Msx

sNkK

kxs

skKkxsskKkssmkKM

skKxksKkksmkKN

T

T

(2.8)

3

02

00

0

22

22

0

0

22

0

2222

2

'2''

''''

''2'''22

''''''2

'

'2'22'''2

)(2)2(')(

kK

kxskK

sMkK

kxssN

kK

kxsO

sx

skKxksssKkk

sssmkKO

sNkK

kxs

skKkxsskKkssmkKM

skKxksKkksmkKN

T

T

T

(2.9)

Further the acceleration of the tappet can be

determined directly real (dynamic) using therelation (2.10).

''' 2 xxx ...(2.10)

3. New Dynamic Synthesis

Give the following parameters:r0=0.013 [m]; rb=0.005 [m]; h=0.008 [m]; e=0.01

[m]; x0=0.03 [m]; u=/2; c=/2; K=5000000[N/m]; k=20000 [N/m]; mT=0.1 [kg]; Mc=0.2[kg]; nmotor=5500 [rot/min].

To sum up dynamically based on a computer

program, you can vary the input data until thecorresponding acceleration is obtained (see Figure11). It then summarizes the corresponding cam

profile (Figure 12) using the relations (2.11).

Fig. 11. Dynamic diagram to the rotary cam with


coscossinsin

sincoscossin

cossin

sincos

cos

sin

0

0

0

bbC

bbC

TTC

TTC

bT

bT

rssrey

rssrex

yxy

yxx

rssy

rex

...(2.11)


13/23


13

Fig. 12. The cam profile to the rotary cam with

translated follower with roll; rb=0.003 [m]; e=0.003

[m]; h=0.006 [m]; r0=0.013 [m]; 0=/2 [rad];.

3.1.

The New Geometry of the Rotary Camand the Translated Follower with Roll

Now, we shall see the geometry of a rotarycam with translated follower with roll (Figure 13).The cam rotation sense is positive (trigonometric).

We can make the geometrical synthesis of thecam profile with the help of the cinematics of themechanism. One uses as well the reduced speed,s. OA=r=rA; r

2=rA2

It establishes a system fixed Cartesian, xOy =xfOyf, and a mobil Cartesian system, xOy =

xmOymfixed with the cam.From the lower position 0, the tappet, pushed

by cam, uplifts to a general position, when the

cam rotates with the angle. The contact point A,

go from Ai0 to A

0(on the cam), and to A (on the

tappet). The position angle of the point A from the

tappet is f, and from the cam is m. We candetermine the coordinates of the point A from thetappet (2.12), and from the cam (2.13).

ffAbfAT

ffAb

f

AT

rrrssyy

rrrexx

sinsincos

coscossin

0

...(2.12)

sinsincoscossincos

cossincossinsinsin

sincoscossinsincos

sinsincoscoscoscos

0

0

bbTT

fffmA

m

Ac

bbTT

fffmA

m

Ac

rerssxy

rrrryy

rssreyx

rrrrxx

(2.13)

f

m

A

0

iA

0AO

fx

fy

mx

0B

B

e

s

0s

r

0r

0r

br

br

Br

0

Fig. 13. The geometry of the rotary cam with


One uses and the next relationships (2.14),

where the pressure angle was obtained with theclassic Antonescu P. method [2].

22

0

22

0

0

22

00

'

'sin

'cos

esss

es

esss

ss

errs b

...(2.14)

3.2.Determining the Forces, the Velocities

and the Efficiency

The driving force Fm, perpendicular on r in A,is divided in two components: Fn, the normalforce, and Fa, a force of slipping. Fn is divided, aswell, in two components: FTis the transmitted (theutile) force, and FR is a radial force which bend

the tappet (see 2.15, and the Figure 14).


14/23


14

cossin2

cossin

;;2

cos

;22

;

coscoscoscos

coscoscoscos

coscoscos

coscoscos

cos

cos

0

2

0

22

2

0

222

2

0

2222

222

sserssrerr

rssreyxr

sserrr

rrrA

AA

vF

vF

vF

vF

P

P

vvv

FFF

vv

FF

bbA

bbAfAf

B

b

Bb

mm

mm

mm

TT

c

ui

mnT

mnT

mn

mn

(2.15)

f

m

A

0

iA

0A

O

fx

fy

mx

0B

B

e

s

0s

r

0r

0r

br

br

Br

0

TF

RF

nF

nF

mF

aF

Fig.14. Forces and velocities of the rotary cam withtranslated follower with roll.

3.3.New Geometro-Kinematics Synthesis

For a good work one proposes to make a newgeometric and kinematics synthesis of the camprofile, using some new relationships for thepressure angle delta (2.16). The new synthesisrelations already presented (2.12 and 2.13) willuse delta pressure angle, deduced now with new

relationship (2.16).

ffAb

f

AT

ffAb

f

AT

rrrssyy

rrrexx

sinsincos

coscossin

0

...(2.12)

sinsincoscossincos

cossincossinsinsin

sincoscossinsincos

sinsincoscoscoscos

0

0

bbTT

fffmA

m

Ac

bbTT

fffmA

m

Ac

rerssxy

rrrryy

rssreyx

rrrrxx

(2.13)

One uses and the next relationships (where the

pressure angle was obtained with the newmethod):

][2

'2'4'4arccossinsin

][2

'2'4'4arccos

cosarccos

][2

'2'4'4cos

22

0

22

00

2

0

22

0

22

00

2

0

22

0

22

00

2

0

22

00

ess

sesesssssss

ess

sesesssssss

ess

sesesssssss

errs b

...(2.16)

The new profile can be seen in the Figure 15.

Fig. 15. The new cam profile to the rotary cam with

translated follower with roll; rb=0.003 [m]; e=0.003

[m]; h=0.006 [m]; r0=0.013 [m]; 0=/2 [rad];


15/23


15

3.4.Demonstration (Explication)

The original relationships (2.16) have beendeduced by the expressions (2.17).

Classical method uses to the geometricsynthesis and the reduced tappet velocity, and in

this mode the geometric classic method become ageometric and kinematic synthesis method. Thenew geometric synthesis method uses just thegeometric parameters (without velocities), but oneutilizes and a condition to realize at the tapped thevelocities predicted by the tapped movement lawsimposed by the cam profile.

][2

'2'4'4cos

][2

'4'2'2cos

0'cos'2cos

cos'2cos'coscos

cos'cos1cos

coscossin'cossin

cos

'

cossincossincossinsin

sincos

'sin

cos

'

sinsin

sincos

'

sin

2

cos

2

coscos

22;;

2

coscos'coscos

22

0

22

00

2

0

22

0

22

0

222

0

2

02

222

0

422

0

242242

0

22

0

22

0

2

00

0

ess

sesesssssss

ess

essssessesss

ssessess

seesssss

esss

esssr

ess

r

s

r

e

r

ssBBB

Br

sB

r

r

r

s

Br

rA

Ar

s

AAA

AA

rsrs

BB

BB

BA

B

A

A

B

A

AA

...(2.17)

Then, it makes the dynamic analyze for theimposed cam profile, and one modify the cam

profile geometric parameters to determine a gooddynamic response (functionality). In this mode it

realizes the dynamic synthesis of the cam, and weobtain a normal functionality. The synthesis wasmade using the natural geometro-kinematicsparameters (of cam mechanism). It follows the

proper functioning dynamics. We will optimizeand the couple cam-pusher efficiency. Forces,velocities and accelerations are also limited.

3.5.Increasing the mechanical efficiency at

the Rotary Cam and Translated

Follower with Roll

The used law is the classical law (2.4), cosinelaw.

The synthesis of the cam profile can be madewith the relationships (3.1) when the cam rotatesclockwise and with the expressions from the

system (3.2) when the cam rotates

counterclockwise (trigonometric).

sinsincoscos

sincoscossin

0

0

bbC

bbC

rerssy

rssrex

(3.1)

sinsincoscos

sincoscossin

0

0

bbc

bbc

rerssy

rssrex

...(3.2)

The r0(the radius of the base circle of the cam)is 0.013 [m]. The h (the maximum displacement

of the tappet) is 0.020 [m]. The angle of lift, uis/3 [rad]. The radius of the tappet roll is rb=0.002[m]. The misalignment is e=0 [m]. The cosine

profile can be seen in the fig. 16.

Fig. 16. The cosine profile at the cam with

translated follower with roll; r0=13[mm],

h=20[mm], u=/3[rad], rb=2[mm], e=0[mm].

The obtained mechanical yield (obtained byintegrating the instantaneous efficiencythroughout the climb and descent) is 0.39 or

=39%. The dynamic diagram can be seen in thefig. 17 (the dynamic setting are partial normal).Valve spring preload 9 cm no longer poses today.

Instead, achieve a long arc very hard(k=500000[N/m]), require special technologicalknowledge.


16/23


16

Fig. 17. The dynamic diagram at the cosine

profile at the cam with translated follower with

roll; r0=13[mm]; h=20[mm]; u=/3[rad];

rb=2[mm]; e=0[mm]; n=5500[rpm]; x0=9[cm];

k=500[kN/m]

It tries increase the yield [8-9, 15-16]; angle of

climb is halved u=/6[rad] (see the profile in theFig. 18).

The r0(the radius of the base circle of the cam)is 0.015 [m]. The h (the maximum displacementof the tappet) is 0.010 [m]. The angle of lift, uis

/6 [rad]. The radius of the tappet roll is rb=0.002[m]. The misalignment is e=0 [m]. The cosineprofile can be seen in the Fig. 18.



h=10[mm], u=/6[rad], rb=2[mm], e=0[mm].

The obtained mechanical yield (obtained by

integrating the instantaneous efficiencythroughout the climb and descent) is 0.428 or

=43%. The dynamic diagram can be seen in thefig. 19 (the dynamic setting are not normal).Valve spring preload 20 cm no longer poses

today. Instead, achieve a long arc very-very hard(k=1500000[N/m]), require special technologicalknowledge.

Fig. 19. The dynamic diagram at the cosine profile

at the cam with translated follower with roll;

r0=15[mm]; h=10[mm]; u=/6[rad]; rb=2[mm];

e=0[mm]; n=5500[rpm]; x0=20[cm]; k=1500[kN/m]

Camshaft runs at a shaft speed halved (nc=n/2).

If we more reduce camshaft speed by threetimes (nc=n/6), we can reduce and the preload ofthe valve spring (x0=5[cm]); see the dynamicdiagram in the Fig. 20. However, in this case, thecam profile should be tripled (see the Fig. 21).

-1400

-1200

-1000

-800

-600

-400

-200

0

200

400

600

0 50 100 150 200 250 300 350 400

xpp [m/s2]

xpp [m/s2]

-1400

-1200

-1000

-800

-600

-400

-200

0

200

400

600

0 50 100 150 200 250 300

xpp [m/s2]

-1400

-1200

-1000

-800

-600

-400

-200

0

200

400

600

0 50 100 150

Fig. 20. The dynamic diagram at the cosine tripled

profile at the cam with translated follower with

roll; r0=15[mm]; h=10[mm]; u=/6[rad];rb=2[mm]; e=0[mm]; n=5500[rpm]; x0=5[cm];

k=1500[kN/m].


17/23


17

Fig. 21. The cosine tripled profile at the cam withtranslated follower with roll; r0=15[mm],

h=10[mm], u=/6[rad], rb=2[mm], e=0[mm].

It tries increase the yield again; angle of climb

is reduced to the value u=/8[rad].

The r0(the radius of the base circle of the cam)is 0.013 [m].

The h (the maximum displacement of thetappet) is 0.009 [m].

The angle of lift, uis /8 [rad].The radius of the tappet roll is rb=0.002 [m].The misalignment is e=0 [m]. The cosine

profile can be seen in the fig. 22.


translated follower with roll; r0=13[mm], h=9[mm],

u=/8[rad], rb=2[mm], e=0[mm].

The obtained mechanical yield (obtained byintegrating the instantaneous efficiency

throughout the climb and descent) is 0.538 or=54%. The dynamic diagram can be seen in thefig. 23 (the dynamic setting are not normal).Valve spring preload 30 cm no longer posestoday. Instead, achieve a long arc very-very hard(k=1600000[N/m]), require special technologicalknowledge.

Fig. 23. The dynamic diagram at the cosine profile

at the cam with translated follower with roll;

r0=13[mm]; h=9[mm]; u=/8[rad]; rb=2[mm];

e=0[mm]; n=5000[rpm]; x0=30[cm]; k=1600[kN/m]


18/23


18

Camshaft runs at a shaft speed halved (nc=n/2).

If we more reduce camshaft speed by four times(nc=n/8), we can reduce and the preload of the

valve spring, x0=9[cm] and the elastic constant ofthe valve spring, k=15000[N/m]; see the dynamicdiagram in the Fig. 24. However, in this case, the

cam profile should be fourfold (see the Fig. 25).

-4000

-3000

-2000

-1000

0

1000

2000

3000

4000

0 50 100 150 200 250 300 350 400

xpp [m/s2]

xpp [m/s2]

-4000

-3000

-2000

-1000

0

1000

2000

3000

4000

0 50 100 150 200 250 300 350 400

xpp [m/s2]

-4000

-3000

-2000

-1000

0

1000

2000

3000

4000

0 50 100 150 200

xpp [m/s2]

-4000

-3000

-2000

-1000

0

1000

2000

3000

4000

0 50 100 150 200 250 300

xpp [m/s2]

Fig. 24. The dynamic diagram at the cosine fourfold

profile at the cam with translated follower with roll;

r0=13[mm]; h=9[mm]; u=/8[rad]; rb=2[mm];

e=0[mm]; n=5000[rpm]; x0=9[cm]; k=15[kN/m]

Fig. 25. The cosine fourfold profile at the cam with

translated follower with roll; r0=13[mm], h=9[mm],

u=/8[rad], rb=2[mm], e=0[mm].

With the same angle of climb u=/8[rad], canincrease performance even further, if the sizetappet race take a greater value (h=12[mm]). Ther0 (the radius of the base circle of the cam) is0.013 [m].

The h (the maximum displacement of the

tappet) is 0.012 [m]. The angle of lift, u is /8[rad]. The radius of the tappet roll is rb=0.002 [m].The misalignment is e=0 [m]. The cosine profile

can be seen in the fig. 26.



h=12[mm], u=/8[rad], rb=2[mm], e=0[mm].

For correct operation it is necessary todecrease the speed of the camshaft four times, andall four times multiplication of the cam profile.Camshaft runs at a shaft speed halved (nc=n/2). Ifwe more reduce camshaft speed by four times

(nc=n/8), we can reduce and the preload of thevalve spring, x0=9[cm]. The elastic constant of thevalve spring is k=1500000[N/m]. See the dynamicdiagram in the Fig. 27. However, in this case, thecam profile should be fourfold. The obtained

mechanical yield is 0.60 or =60%.


19/23


19

-2000

-1500

-1000

-500

0

500

0 50 100 150 200 250 300 350 400

xpp [m/s2]

xpp [m/s2]

-2000

-1500

-1000

-500

0

500

0 50 100 150 200 250 300 350 400

xpp [m/s2]

-2000

-1500

-1000

-500

0

500

0 50 100 150 200 250 30

xpp [m/s2]

-2000

-1500

-1000

-500

0

500

0 50 100 150 200

xpp

Fig.27. The dynamic diagram at the cosine fourfoldprofile at the cam with translated follower with roll;

r0=13[mm]; h=12[mm]; u=/8[rad]; rb=2[mm];

e=0[mm]; n=5000[rpm]; x0=9[cm]; k=1500[kN/m]

For now is necessary to stop here.If we increase h, or decrease the angle u, then

is tapering cam profile very much. We must stop

now at a yield value, =60%.

Nomenclature

M the mass of the mechanism, reduced

at the valveK the elastically constant of the

systemk the elastically constant of the valve

spring

dt

dMc 2

1

the coefficient of the systemsamortization

F0 the elastically force whichcompressing the valve spring

rF the resistant force

mF the motor forcex the effective displacement of the

valvex0 the valve (tappet) spring preloadx the reduced valve (tappet) speedx the reduced valve (tappet)

acceleration

ys the theoretical displacement of thetappet reduced at the valve, imposedby the cams profile

ys the velocity of the theoretical

displacement of the tappet reducedat the valve, imposed by the camsprofile

mT=m2 the mass of the tappet (of the valvelifter)

m3 the mass of the valve push rod

m5 the valve massJ1=Jc the inertia mechanical moment of

the cam

J4 the inertia mechanical moment ofthe valve rocker

2y the tappet velocity, or the secondmovement-low, imposed by thecams profile

x the real (dynamic) valve velocity

x the real (dynamic) valveacceleration

i=i25 the ratio of transmission tappet-valve, given from the valve rocker

y the tappet velocity reduced at the

valveD the dynamic transmission function(the dynamic transmissioncoefficient)

m the nominal (average) angularvelocity of cam

=1= the angular (real) rotation speed ofthe cam (or camshaft)

4 the angular rotation speed of thevalve rocker

The pressure angle

The additional pressure angle

rb roll radiusr0 basic radius

e horizontally misalignments0 vertical misalignment

BA rr , position vectors

J* the reduced mechanical moment ofinertia

J* the first derivative of the reducedmechanical moment of inertia

J*m the average reduced moment of

inertia


20/23


20

4. Discussion

In this paper one presents an original methodto determine the dynamic parameters at thecamshaft (the distribution mechanisms). It makesthe synthesis, of the rotary cam and tappet with

translational motion with roll, with a greatprecision.

The authors introduce a new pressure angle,alpha, and a new method to determine the twopressure angles, alpha and delta.

The presented method is the most elegant anddirect method to determine the kinematics anddynamic parameters.

The dynamic synthesis can generate a camprofile which will work without vibrations.

Processes robotization increasingly determineand influence the emergence of new industries,

applications in specific environmental conditions,approach new types of technological operations,handling of objects in outer space, leadingteleoperator in disciplines such as medicine,robots that covers a whole larger service benefitsour society, modern and computerized. In thiscontext, this paper seeks to contribute to thescientific and technical applications in dynamicanalysis and synthesis of cam mechanisms.

In 1971 K. Hain proposes an optimization

method to cam mechanism to achieve the

optimum output transmission angle (maximum)and minimum acceleration [11].In 1979 F. Giordana investigates the influence

of measurement errors in kinematic analysis of

cam [10].In 1985 P. Antonescu presents an analytical

method for the synthesis mechanism flat tappetcam and tappet rocker mechanism [2].

In 1987 F.I. Petrescu presents a new dynamic

model with general applications [3].In 1988 J. Angelas and C. Lopez-Cajun

presents optimal synthesis mechanism oscillating

flat tappet cam and [1].In 1991 B.S. Bagepalli presents a generalized

model of dynamic cam-follower pairs [4].In 1999 R.L. Norton studying the effect of

valve-cam ramps on the valve-train dynamics[12].

Z. Chang presents in 2001 [5] and 2011 [6] astudy on dynamics of roller gear cam system,considering clearances.

In 2002 D. Taraza synthesized analyzes theinfluence of the cam profile, the variation of theangular velocity distribution tree, and the

parameters of power load consumption andemissions of internal combustion engine [16].

In 2005 [13] and 2008 [14], F.I. Petrescu and

R.V. Petrescu present a synthesis method of rotarycam profile, and translational or rotary tappet, flat

or with roll, to obtain high yields output.In 2009 K. Dan makes some research on

dynamic behavior simulation technology for cam-

drive mechanism in single-cylinder engines [7].In 2009 M. Satyanarayana makes a dynamic

experiment in cam-follower mechanism [15].In 2011 Z. Ge makes the design and dynamic

analysis of the cam mechanisms [8-9].

5. Conclusions

The distribution mechanisms work with smallefficiency for about 150 years; this fact affects thetotal yield of the internal heat engines. Much ofthe mechanical energy of an engine is lost throughthe mechanism of distribution. Multi-years theyield of the distribution mechanisms was only 4-8%. In the past 20 years it has managed a lift up tothe value of 14-18%; car pollution has decreasedand people have better breathing again.Meanwhile the number of vehicles has tripled andthe pollution increased again.

Now, its the time when we must try again togrow the yield of the distribution mechanisms.

The paper presents an original method to

increase the efficiency of a mechanism with camand follower, used at the distribution mechanisms.This paper treats only one module: the

mechanism with rotary cam and translated followerwith roll (the modern module B).

At the classical module C we can increase againthe yield to about 30%. The growth is difficult.Dimensional parameters of the cam must be

optimized; optimization and synthesis of the camprofile are made dynamic, and it must set theelastic (dynamic) parameters of the valve (tappet)spring: k and x0.

The law used is not as important as the moduleused, sizes and settings used. We take the classicallaw cosine; dimensioning the radius cam, liftheight, and angle of lift.

To grow the cam yield again we must leave theclassic module C and take the modern module B.In this way the efficiency can be as high as 60%.

Yields went increased from 4% to 60%, andwe can consider for the moment that we have gainimportance, since we work with the cam andtappet mechanisms.

If we more increase h, or decrease the angle

u, then is tapering cam profile very much. Wemust stop now at a yield value, =60%.


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It can synthesize high-speed cam, or high-

performance camshafts.

6. References

[1]

Angelas J., Lopez-Cajun C., Optimalsynthesis of cam mechanisms with oscillatingflat-face followers. Mechanism and MachineTheory 23,(1988), Nr. 1., p. 1-6., 1988.

[2] Antonescu, P., Petrescu, F., Antonescu, O.,Contributions to the Synthesis of The RotaryDisc-Cam Profile, In VIII-th InternationalConference on the Theory of Machines andMechanisms, Liberec, Czech Republic, p. 51-56, 2000.

[3] Antonescu, P., Oprean, M., Petrescu, F.I.,Analiza dinamic a mecanismelor dedistribuie cu came. n al VII-lea SimpozionNaional de Roboi Industriali, MERO'87,Bucureti, 1987, Vol. III, p. 126-133.

[4] Bagepalli, B.S., a.o., Generalized Modelingof Dynamic Cam-follower Pairs inMechanisms, Journal of Mechanical Design,June 1991, Vol. 113, Issue 2, p. 102-109.

[5] Chang, Z., a.o., A study on dynamics ofroller gear cam system consideringclearances, Mechanism and Machine Theory,

January 2001, Vol. 36, N. 1, p. 143-152.

[6]

Chang, Z., a.o., Effects of clearance ondynamics of parallel indexing cammechanism, ICIRA11 Proceedings of the 4thinternational conference on Intelligent

Robotics and Applications Volume, Part I,2011, p. 270-280.

[7] Dan, K., a.o., Research on Dynamic BehaviorSimulation Technology for Cam-DriveMechanism in Single-cylinder Engines, SAE

Technical Paper, 2009, paper number 2009-32-0089.

[8] Ge, Z., a.o., Mechanism Design amd

Dynamic Analysis of Hybrid Cam-linkageMechanical Press, Key EngineeringMaterials Journal, Vol. 474-476 (2011), p.803-806.

[9] Ge, Z., a.o., CAD/CAM/CAE for the Parallel

Indexing Cam Mechanisms, AppliedMechanics and Materials Journal, Vol. 44-47

(2011), p. 475-479.[10] Giordana F., s.a., On the influence of

measurement errors in the Kinematic

analysis of cam. Mechanism and MachineTheory 14 (1979), nr. 5., p. 327-340, 1979.

[11] Hain K., Optimization of a cam mechanismto give goode transmissibility maximaloutput angle of swing and minimalacceleration. Journal of Mechanisms 6(1971), Nr. 4., p.419-434.

[12]Norton, R.L., a.o., Effect of Valve-CamRamps on Valve Train Dynamics, SAE,International Congress & Exposition, 1999,Paper Number 1999-01-0801.

[13] Petrescu, F.I., Petrescu, R.V. Contributions

at the dynamics of cams. In the NinthIFToMM International Symposium onTheory of Machines and Mechanisms,SYROM 2005, Bucharest, Romania, 2005,Vol. I, p. 123-128.

[14] Petrescu F.I., .a., Cams DynamicEfficiency Determination. In New Trends

in Mechanisms, Ed. Academica -Greifswald, 2008, I.S.B.N. 978-3-940237-10-1, p. 49-56.

[15] Satyanarayana, M., a.o., Dynamic Responseof Cam-Follower Mechanism, SAE Technic

Paper, 2009, paper number 2009-01-1416.[16] Taraza, D., "Accuracy Limits of IMEP

Determination from Crankshaft SpeedMeasurements," SAE Transactions, Journalof Engines 111, p. 689-697, 2002.


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AUTHORSINFORMATION

1Dr. Eng. Florian Ion T. Petrescu, Senior Lecturerat UPB (Bucharest Polytechnic University), TMR(Theory of Mechanisms and Robots) department.2Dr. Eng. Relly Victoria V. Petrescu, SeniorLecturer at UPB (Bucharest PolytechnicUniversity), TTL (Transport, Traffic andLogistics) department.

1. Ph.D. Eng. Florian Ion T.PETRESCUSenior Lecturer at UPB (BucharestPolytechnic University), Theory of Mechanismsand Robots department,

Date of birth: March.28.1958; Higher education:Polytechnic University of Bucharest, Faculty ofTransport, Road Vehicles Department, graduated

in 1982, with overall average 9.63;

Doctoral Thesis: "Theoretical and AppliedContributions About the Dynamic of PlanarMechanisms with Superior Joints".

Expert in: Industrial Design, Mechanical Design,Engines Design, Mechanical Transmissions,Dynamics, Vibrations, Mechanisms, Machines,Robots.

Association:

Member ARoTMM, IFToMM, SIAR, FISITA,SRR, AGIR. Member of Board of SRRB(Romanian Society of Robotics).

2. Ph.D. Eng. Relly Victoria V. PETRESCUSenior Lecturer at UPB (Bucharest PolytechnicUniversity), Transport, Traffic and Logisticsdepartment,

Citizenship: Romanian;

Date of birth: March.13.1958;Higher education: Polytechnic University ofBucharest, Faculty of Transport, Road Vehicles

Department, graduated in 1982, with overallaverage 9.50;

Doctoral Thesis: "Contributions to analysis andsynthesis of mechanisms with bars and sprocket".

Expert in Industrial Design, EngineeringMechanical Design, Engines Design, Mechanical

Transmissions, Projective and descriptivegeometry, Technical drawing, CAD, Automotiveengineering, Vehicles, Transportations.

Association:Member ARoTMM, IFToMM, SIAR, FISITA,

SRR, SORGING, AGIR.


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Florian Ion T. Petrescu* Relly Victoria V. Petrescu** / / / *

/ / / **

[email protected] *:

[email protected] **:

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mailto:[email protected]*%D8%A7%D9%84%D8%A8%D8%B1%D9%8A%D8%AFmailto:[email protected]*%D8%A7%D9%84%D8%A8%D8%B1%D9%8A%D8%AFmailto:[email protected]*%D8%A7%D9%84%D8%A8%D8%B1%D9%8A%D8%AFmailto:[email protected]*%D8%A7%D9%84%D8%A8%D8%B1%D9%8A%D8%AFmailto:[email protected]**%D8%A7%D9%84%D8%A8%D8%B1%D9%8A%D8%AFmailto:[email protected]**%D8%A7%D9%84%D8%A8%D8%B1%D9%8A%D8%AFmailto:[email protected]**%D8%A7%D9%84%D8%A8%D8%B1%D9%8A%D8%AFmailto:[email protected]**%D8%A7%D9%84%D8%A8%D8%B1%D9%8A%D8%AFmailto:[email protected]**%D8%A7%D9%84%D8%A8%D8%B1%D9%8A%D8%AFmailto:[email protected]**%D8%A7%D9%84%D8%A8%D8%B1%D9%8A%D8%AFmailto:[email protected]*%D8%A7%D9%84%D8%A8%D8%B1%D9%8A%D8%AFmailto:[email protected]*%D8%A7%D9%84%D8%A8%D8%B1%D9%8A%D8%AF

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Cam Dynamic Synthesis

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