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: petrescuflorian@yahoo.com 

**E-mail: petrescuvictoria@yahoo.com 

 
(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 was 

obtained by burning oil is replaced with nuclear 

energy, hydropower, solar energy, wind, and other 

types of unconventional energy, in the conditions 

in which electric motors have been instead of 

internal combustion in public transport, but more 

recently they have entered in the cars world 

(Honda has produced a vehicle that uses a 

compact electric motor and electricity consumed 

by the battery is restored by a system that uses an 

electric generator with hydrogen combustion in 

cells, so we have a car that burns hydrogen, but 

has an electric motor), which is the role and 

prospects which have internal combustion engines 

type 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 atmosphere 

needed our planet by sparks emitted by collecting 

brushes. Immediate consequence is that if we only 

use electric motors in all sectors, we’ll have 

problems with higher ozone shield that protects 

our planet and without which no life could exist 

on Earth. 

Magnetic motors (combined with the 

electromagnetic) are just in the beginning, but 

they 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 generate 

electricity that will fill the battery that will 

actually feed the engine (probably an electric 

motor). 

The Otto engines or those with internal 

combustion in general, will have to adapt to 

hydrogen fuel.  

It is composed of the basic (hydrogen) can 

extract industrially, practically from any item (or 



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

2 

 
 

combination) through nuclear, chemical, photonic 

by radiation, by burning, etc... (Most easily 

hydrogen can be extracted from water by breaking 

up into constituent elements, hydrogen and 

oxygen; by burning hydrogen one obtains water 

again 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 water 

directly on the vehicle, in which case the reservoir 

would feed water (and there were announced 

some successful). 

As a backup, there are trees that can donate a 

fuel oil, which could be planted on the extended 

zone, or directly in the consumer court. With 

many years ago, Professor Melvin Calvin, 

(Berkeley University), discovered that “Euphora” 

tree, a rare species, contained in its trunk a liquid 

that has the same characteristics as raw oil. The 

same professor discovered on the territory of 

Brazil, a tree which contains in its trunk a fuel 

with properties similar to diesel.  

During a journey in Brazil, the natives driven 

him (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 as 

indigenous raw material base for the preparation 

of perfumes or, in concentrated form, as a balm. 

Nobody see that it is a pure fuel that can be used 

directly by diesel engines.  

Calvin said that after he poured the liquid 

extracted from the tree trunk directly into the tank 

of 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 in 

the forests, and the courts of people.  

From a jagged tree is filled about half of the 

tank; one covers the slash and it is not open until 

after six months; it means that having 12 trees in a 

courtyard, a man can fill monthly a tank with the 

new natural diesel fuel. 

In some countries (USA, Brazil, Germany) 

producing alcohol or vegetable oils, for their use 

as fuel. 

In the future, aircraft will use ion engines, 

magnetic, laser or various micro particles 

accelerated. Now, and the life of the jet engine 

begin to end.  

Even in these conditions internal combustion 

engines will be maintained in land vehicles (at 

least), for power, reliability and especially their 

dynamics. Thermal engine efficiency is still low 

and, about one third of the engine power is lost 

just by the distribution mechanism. Mechanical 

efficiency of cam mechanisms was about 4-8%. In 

the past 20 years, managed to increase to about 

14-18%, and now is the time to pick it up again at 

up to 60%. This is the main objective of this 

paper. 

 

 

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 the 

classical valve gear mechanism (see the Figure 1), 

one creates the translating dynamic model, with a 

single degree of freedom (with a single mass), 

with variable internal amortization (see the Figure 

2), 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 the 

form (2). 
 

)()(
0

xkFxyKxcxM       …(2) 
 

 

 

 



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

3 

 
 

M M

k 

kx  F

F(t) c .

cx

xx(t)

K(y-x)
K 

y(t)


camã

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

variable internal amortization. 

 

 

The differential equation, Lagrange, can be 

written in the form (3). 

rm
FFx

dt

dM
xM  

2

1
                     …(3) 

Comparing the two equations, (2 and 3), one 

identifies the coefficients and one obtains the 

resistant force (4), the motor force (5) and the 

coefficient of internal amortization (6). One can 

see that the internal amortization coefficient, c, is 

a variable. 
 

)(
000

xxkxkxkxkFF
r

    …(4) 

)()( xsKxyKF
m

                       …(5) 

dt

dM
c 

2

1
                                                    …(6) 

 

One places the variable coefficient, c, (see the 

relation 6), in the Newton equation (form 1 or 2) 

and obtains the equation (7). 
 

0
)(

2

1
FyKxkKx

dt

dM
xM   …(7) 

 

The reduced mass can be written in the form 

(8), (the reduced mass of the system, reduced at 

the valve). 
 

24
4

21
1

22
325

)()(

)()(

x
J

x
J

x

y
mmmM












                   …(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 valve 

mass; J1 =the inertia mechanical moment of the 

cam; J4 =the inertia mechanical moment of the 

valve rocker; 
2

y  =the tappet velocity, or the 
second movement-low, imposed by the cam’s 

profile; x =the real (dynamic) valve velocity. 

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

tappet-valve, given from the valve rocker, the 

theoretically velocity of the valve, y , (the tappet 

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

where the ratio of transmission, i, is given from 

the formula (10). 

i

y
yy

2

5


                                                    …(9) 

DC

CC
i

0

0
                                                       …(10) 

One can write the following relations (11-16), 

where y’ is the reduced velocity forced at the 

tappet by the cam’s profile. With the relations (10, 

13, 14, 16) the reduced mass (8), can be written in 

the forms (17–19). 
 

'
1

xx                                                      …(11) 

''
2

1
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

DC
J

x
J

x

yi
mmmM








               …(17) 

  
2

1

2

2

0

4

32

2

5

)
'

1
()

'

'
(]

)(

)([

x
J

x

y

DC

J

mmimM





                 …(18) 

2

1

2

5
)
'

1
()

'

'
(*

x
J

x

y
mmM                   ..(19) 

 



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

4 

 
 

It derives dM/d and obtains the relations (20–

22). 

)
'

''

'

''
()

'

'
(2)

'

'
''''(

'

'2

'

)''''''(

'

'2
])

'

'
[(

2

2

2

2

x

x

y

y

x

y

x

y
xy

x

y

x

yxxy

x

y

d

x

y
d













          ...(20) 

 
32

2

'

''
2

'

''

'

2
])

'

1
[(

x

x

x

x

xd

x
d






                  …(21) 

31

2

'

''
2)

'

''

'

''
()

'

'
(*2

x

x
J

x

x

y

y

x

y
m

d

dM



  

                      ...(22) 

The relation (6) can be written in form (23) and 

with relation (22), it’s taking the forms (24–25). 
 





d

dM
c 

2
                                                …(23) 

}
'

''
)

'

''

'

''
()

'

'
(

]
)(

)({[

31

2

2

0

4
32

2

x

x
J

x

x

y

y

x

y

DC

J
mmic



 

         …(24) 

]
'

''
)

'

''

'

''
()

'

'
(*[

31

2

x

x
J

x

x

y

y

x

y
mc     …(25) 

With the notation (26): 

2

0

4

32

2

)(
)(*

DC

J
mmim                  …(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
(

)
'

'
(

FyKxkK
x

x
Jx

x

x

y

y

x

y
mxx

x
J

x
x

y
mmx













                                                                                                   

   …(28) 

 

0

*2

2*22*2

5

2

)(
'

'
''

'')
'

'
()

'

'
(''''

FyKxkK
x

y
ym

x
x

y
m

x

y
xmxm








                                                                                         

                                                                      …(29) 

0

''*2''

5

2

'

'
)( FKy

x

y
ymxkKxm                                                                                          

                                                                      …(30) 

0

*

5

2
)()

'

'
''''( FyKxkK

x

y
ymxm                                                                                              

                                                                      …(31) 

The exact equation (31) can be approximated 

at the form (32) with x’y’. 

 

0

*

5

2
)()''''( FyKxkKymxm                                                                                        

                                                                      …(32) 

 

With the following notations: y=s, y’=s’, 

y’’=s’’, y’’’=s’’’, the equation (32) takes the 

approximate form (33) and the complete equation 

(31) takes the exact form (34). 

0

*

5

2
)()''''( FsKxkKsmxm                                                                                        

                                                                      …(33) 

05

2
)()

'

'
''*''( FsKxkK

x

s
smxm                                                                                         

                                                                      …(34) 

 

 

Solving the Differential Equation by Direct 

Integration and Obtaining the Mother 

Equation 
 

One integrates the equation (31) directly. One 

prepares the equation (31) for the integration. 

First, one writes (31) in form (35) [13]. 
 

I

III

TII

S
x

yym
xm

xkyKxkK






2*
2*

0
)(




           …(35) 

 

The equation (35), can be amplified by x’ and 

one obtains the relation (36). 

III

T

III

S

III

yymxxm

xxkxyKxxkK





2*2*

0
)(


                                                                                                                             

                                                                      …(36) 

 



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

5 

 
 

Now, one replaces the term K.y.x’ 

with
I

y
kK

K
yK 


 , (taken in calculation the 

statically assumption, Fm=Fr) and one obtains the 

form (37). 

III

T

III

S

III

yymxxm

xxkyy
kK

K
xxkK








2*2*

0

2

)(



                                                                                               

                                                                      …(37) 

One integrates directly the equation (37) and 

one obtains the mother equation (38). 

C
y

m
x

m

xxk
y

kK

Kx
kK

TS







2

'

2

'

22
)(

2
2*

2
2*

0

222



                                                                                        

                                                                      …(38) 

With the initial condition, at the =0, y=y’=0 

and x=x’=0, one obtains for the constant of 

integration, C the value 0. In this case the 

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

2

'

2

'

22
)(

2
2*

2
2*

0

222

y
m

x
m

xxk
y

kK

Kx
kK

TS









             …(39) 

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

if one divides it with the 
2

kK 
 . 

  

0
)(

'

'2

2

2

2
2

2*

2

2*

02





















y
kK

K
y

kK

m

x
kK

m
x

kK

xk
x

T

S





          …(40) 

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

if one notes: '' y
kK

K
x 


 , (the static 

assumption, Fm=Fr). 

0'
)(

)(

)(
2

22

**

2

2

2

2

2

02


















y
kK

mm
kK

K

y
kK

K
x

kK

xk
x

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

m
kK

K
m

kK

sKxk
TS










                                                                                                                                                     

                                                              …(42)                                                                                                                                

22

*

2

2
*

2

22

0 )'(
)(

)(

)(

)()(
Ds

kK

m
kK

K
m

kK

Kskx
TS










                                                                                                                                                                               

                                                               …(43) 
 







kK

xk
X

0

2,1
                             …(44) 

 

Physically, just the positive solution is valid 

(see the relation 45).  

kK

xk
X






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























y
kK

mm
kK

K

s
kK

K

X
kK

xk
s

kK

xk
XsXs

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)
2
0, solved by the 

relation (48). 

 



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

6 

 
 

20

22**
2

0

22

)()(2

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

)1(

kK
kK

xk
s

DsmkKm
kK

K
skKkxskKk

X

TS















                                                                                                                                                                                        

                                                              …(48) 

  

2

22**
2

2

0

222

)(

)'(])([

kK

DsmkKm
kK

K
xksK

TS











                                                                                                                                                                                                       

                                                                 …(49) 

 

)(
0

kK

xk
sX




                             …(50) 

 

 

2.3. Mechanism with Rotary Cam and 

Translated Tappet with Roll 

 

First, one presents an original method to 

determine the efficiency at the mechanism with 

rotary cam and translated follower with roll [5].  

With this occasion it presents and the forces 

and the velocities as well (Figure 3). 

The pressure angle  (Figure 3), is determined 

by relations (1.5-1.6).  

We can write the next forces, speeds and 

powers (1.13-1.18). 

Fm (vm) is perpendicular to the vector rA at A.  

Fm is 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). The 

momentary dynamic efficiency can be obtained 

from relation (1.18). 

 

0
A



A

B







A-

Fn, vn

Fm, vm

Fa, va

Fi, viFn, vn

Fu, v2

B

B0

A0

A

O

x

e

s 0

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

B
r

e
  sincos                                      ...(1.3) 

B

B
r

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

b
A

A

b

A

r

rss

esssr

resssss






cos
sin

)'()(

])'()([)(
sin

0

22

0

22

00

                                                                                  

                                                                    ...(1.10) 

 
























cos
'

)cos(

)'()(

')(
)cos(

22

0

0

A

A

A

A

r

s

esssr

sss

        

                                                                    ...(1.11) 

            


2

cos
'

cos)cos( 
A

A
r

s
           ...(1.12) 









)sin(

)sin(





Ama

Ama

FF

vv
                           ...(1.13) 



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

7 

 
 









)cos(

)cos(





Amn

Amn

FF

vv
                           ...(1.14) 













sin

sin

ni

ni

FF

vv
                                        ...(1.15)  













cos)cos(cos

cos)cos(cos
2

Amnu

Amn

FFF

vvv
             

                                                                    ...(1.16)       









mmc

Ammuu

vFP

vFvFP 
22

2
cos)(cos

                                                                                    

                                                                    ...(1.17) 
































4

2

2

222

22

cos
'

]cos
'

[]cos)[cos(

cos)(cos

A

i

A

Ai

mm

Amm

c

u
i

r

s

r

s

vF

vF

P

P

              

                                                                    ...(1.18)                   

bB
rrr 

00
                                              …(1.19)  

22

0 0
ers

B
                                          …(1.20)  

0

0
cos

B
r

e
                                             …(1.21)         

0

0

0
sin

B
r

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 one 

knows OB, AB, and the angle between them, B 

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

OA and the angle  (AOB) can be determined 

with the relations (1.7-1.8, 1.29-1.31): 
 

BA

bBA

rr

rrr






2
cos

222

                              …(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 
B

r  (1.36) can be deduced from (1.1). 
Then, (1.38) will be obtained from (1.29): 

            

2
sin

B

B
BB

r

re 



                             …(1.34)      

2

0
)(

B

BB
B

rss

rre







                              …(1.35) 












sssrr

sssrr

BB

BB





)(

)(22

0

0
                   …(1.36) 

22

0

0

)(

)(

BB

B
r

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 
A

r  (1.39) from 
expression (1.8): 



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

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

B
r

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 are 

deduced as follows (1.13-1.16): 

 

 

2.6. Determining the Efficiency of the 

Module B 

 
With the relationships (1.17-1.18) we can 

determine the powers and the momentary 

mechanical efficiency [14].   

 

 

Determining the (Dynamic) Transmission 

Function D, for the Module B 

 
The follower’s velocity (1.16) can be written 

into the form (1.45). 

























22

22

2

cos'cos'

cos
'

cos
'

cos)cos(cos

ss

r

s
r

r

s
v

vvv

I

AA

A

AA

A

m

Amn



                                                                  

                                                            …(1.45) 

With the relationships (1.45) and (1.46) we 

determine the transmission function (the dynamic 

modulus), D (1.47): 

 Dsv '
2

                                          …(1.46)    


2

cos
I

A
D                                         …(1.47) 

Expression cos
2
 is known (1.48): 

    
22

0

2

02

)'()(

)(
cos

esss

ss




               …(1.48) 

The expression of the ’A (1.49) is more 

difficult. 

 





































]}')[(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

22

0

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) 

 

 

 



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

9 

 
 

The Dynamics of Distribution Mechanisms 

with Translated Follower with Roll 
 

For the dynamics of the Module B the 

relationships (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

kx
s

y
kK

mm
kK

K

s
kK

kx
s

kK

kKk

X

TS






















                                                                                                          

                                                                   …(1.52)   

][2

)'(

]
)(

[
2

)(

2

0

2

2**

2

2

02

2

2

kK

kx
s

sD
kK

mm
kK

K

s
kK

kx
s

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 B 

for some known movement laws.  

We begin with the classical law SIN (see the 

diagram in Figure 4); A speed rotation n=5500 

[rot/min], for a maxim theoretical displacement of 

the valve h=6 [mm] is used.  

The phase angle 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=65
0
, 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=65
0
, 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 the 

accelerations have the same values as for the 

classical module C for a relaxed phase (75
0
-80

0
). 

In Figure 5 we can see the cam’s profile. It 

uses the profile sin, a rotation speed n=5500 rpm, 

and u=65
0
, 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 dynamic 

analyze of the profile cos, and its profile design 

can be seen in the Figure 7. 

The principal parameters are: 

Law COS, n=5500 rpm, u=65
0
, r0=13 [mm], 

rb=6 [mm], hT=6 [mm], =10.5%. 

 

 
 

Fig. 6. The dynamic analysis of the module B. Law 

COS, n=5500 rpm, u=65
0
, r0=13 [mm], rb=6 [mm], 

hT=6 [mm], =10.5%. 



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

10 

 
 

 
 

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

rpm, u=65
0
, r0=13 [mm], rb=6 [mm], hT=6 [mm].  

 

     

 
 

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

rpm, u=80
0
, 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 are 

diminished, the noises are limited, the effective 

displacement 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 the 

Figure 9 the profile of C4P law is presented.  

 

It starts at the law C4P with n=5500 [rpm], but 

for this law the rotation velocity can increase to 

high values of 30000-40000 [rpm] (see the Figure 

10). 

 

 

 
 

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

C4P1-5, n=40000 rpm. 

 

 

 

 

 

 

 

 



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

11 

 
 

2.8. The New Cam Synthesis 

 
The rotary cam with translated follower with 

roll (Figure 3 or 13), is synthesized dynamic by 

the new next relationships. 

It has exchanged the rotation sense from the 

Figure 1 to Figure 11, from the clockwise to the 

counter-clockwise. 

First, one determines the mass moment of 

inertia (mechanical) of the mechanism, reduced to 

the element of rotation, ie cam (basically using 

kinetic 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

0
0

22

0

2

0

222

0

2

0

222

0

222

0

2222222

2

'
'

'

2

1
22

2

1

8

22

22

2

1

16

1

4

1

2

1

'

'2

'

2

'
2

'

'
2

cossin2

cos2cos

sin2sin

2

1

sm
esss

ssese
rM

sMssMrrrrMJ

hm

e
hh

s

h
se

h
e

rM

hMhsMrrrrMJ

esss

eser

esss

ssr
ssrer

esss

ss
ssr

esss

es
erssrer

sserssrer

ssrrss

rereyxrR

RMJ

Tbc

ccbbc

T
bc

ccbbcm

b

b
bA

b

bbA

bbA

bb

bbAAA

ccama



















                                                                                              

                                                                      ...(2.1) 

 

We considered the law of motion of the tappet 

classic version already used the cosine law (both 

ascending and descending). 

The angular velocity is a function of the cam 

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

Where ωm is the nominal angular velocity of cam 

and express at the distribution mechanisms based 

on the motor shaft speed (2.3). 

2

*

*
2

m
m

J

J
                                                ...(2.2)   

  

60260

2

60
22

nn

n

motor

c
cm













                                     

                                                                      ...(2.3)

  

We start the simulation with a classical law of 

motion, namely the cosine law. To climb cosine 

law system is expressed by the relationships (2.4). 

 



















































































uu

r

uu

r

uu

r

u

h
s

h
as

h
vs

hh
s


































sin
2

'''

cos
2

''

sin
2

'

cos
22

3

3

2

2
                                       

                                                                      ...(2.4) 

Where  takes values from 0 to u.  

Jmax occurs for =u/2. 

With the relation (2.5) is expressed the first 

derivative of the reduced mechanical moment of 

inertia. It is necessary to determine the angular 

acceleration (2.6). 

 

       
    
       
     2/322

0

0

2

0

2

2/322

0

22

00

0

*'

'

'''''

'

''2''

'''2''

esss

sessssssese
rM

esss

essssssse
rM

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 and 

are basically two original equations of motion, 

crucial for mechanical mechanisms. 

For a rotary cam and translated tappet with roll 

mechanism (without valve), dynamic movement 

tappet is expressed by equation (2.7), which is an 



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

12 

 
 

original dynamic equation deduced for the 

distribution mechanism (50) and now by 

canceling the valve mass, will customize and 

reaching form below (2.7). 




















kK

xk
skK

skKxksKkksmkK

sx

T

02

0

2222

)(2

)(2)2(')(                                                                                                

                                                                      ...(2.7)                                                                                                                                                  

Where x is the dynamic movement of the 

pusher, while s is its normal, kinematics 

movement. K is the spring constant of the system, 

and k is the spring constant of the tappet spring.  

It note, with x0 the tappet spring preload, with 

mT the mass of the tappet, with ω the angular 

rotation speed of the cam (or camshaft), where s’ 

is the first derivative in function of  of the tappet 

movement, s. Differentiating twice successively, 

the expression (2.7) in the angle , we obtain a 

reduced tappet speed (equation 2.8), and reduced 

tappet acceleration (2.9).  

 

      

 













































2

02

0

0

22

0

2222

2

''

'

'2'22'''2

)(2)2(')(

kK

kx
skK

M
sx

sN
kK

kx
s

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

kx
skK

sM
kK

kx
ssN

kK

kx
sO

sx

skKxksssKkk

sssmkKO

sN
kK

kx
s

skKkxsskKkssmkKM

skKxksKkksmkKN

T

T

T







                                                                     …(2.9) 

Further the acceleration of the tappet can be 

determined directly real (dynamic) using the 

relation (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 the 

corresponding acceleration is obtained (see Figure 

11). It then summarizes the corresponding cam 

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

 

 

 
 

Fig. 11. Dynamic diagram to the rotary cam with 

translated follower with roll. 

 

 

    

    









































































coscossinsin

sincoscossin

cossin

sincos

cos

sin

0

0

0

bbC

bbC

TTC

TTC

bT

bT

rssrey

rssrex

yxy

yxx

rssy

rex

                                                                                                                                                                                   

                                                                    ...(2.11) 



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

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 Cam 
and the Translated Follower with Roll 

 
Now, we shall see the geometry of a rotary 

cam with translated follower with roll (Figure 13). 

The cam rotation sense is positive (trigonometric). 

We can make the geometrical synthesis of the 

cam profile with the help of the cinematics of the 

mechanism. One uses as well the reduced speed, 

s’. OA=r=rA; r
2
=rA

2 

It establishes a system fixed Cartesian, xOy = 

xfOyf, and a mobil Cartesian system, xOy = 

xmOym fixed 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 Ai
0
 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 can 

determine the coordinates of the point A from the 

tappet (2.12), and from the cam (2.13). 

 











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)                                

                                                                                  

f


m





A

0

i
A

0
A

O

f
x

f
y

m
x

0
B

B



e

s

0
s



r

0
r

0
r

b
r

b
r



B
r

0

 
 

Fig. 13. The geometry of the rotary cam with 

translated follower with roll. 

 

 

One uses and the next relationships (2.14), 

where the pressure angle  was obtained with the 

classic 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 normal 

force, and Fa, a force of slipping. Fn is divided, as 

well, in two components: FT is the transmitted (the 

utile) force, and FR is a radial force which bend 

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



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

14 

 
 

 

 

   

    




























































































cossin2

cossin

;;
2

cos

;
22

;

coscoscoscos

coscoscoscos

coscoscos

coscoscos

cos

cos

0

2

0

22

2

0

222

2

0

2
222

222

sserssrerr

rssreyxr

sser
rr

rrr
A

AA

vF

vF

vF

vF

P

P

vvv

FFF

vv

FF

bbA

bbAfAf

B

b

Bb

mm

mm

mm

TT

c

u
i

mnT

mnT

mn

mn

                                                                                             
                                                              … (2.15) 

f


m





A

0

i
A

0
A

O

f
x

f
y

m
x

0
B

B


e

s

0
s



r

0
r

0
r

b
r

b
r


B

r

0

T
F

R
F

n
F

n
F

m
F

a
F





 

 
Fig. 14. Forces and velocities of the rotary cam with 

translated follower with roll. 

 

 

3.3. New Geometro-Kinematics Synthesis 
 

For a good work one proposes to make a new 

geometric and kinematics synthesis of the cam 

profile, using some new relationships for the 

pressure angle delta (2.16). The new synthesis 

relations already presented (2.12 and 2.13) will 

use 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 new 

method): 

 
 

     

 

 

     

 

     

 
































































































][2

'2'4'4
arccossinsin

][2

'2'4'4
arccos

cosarccos

][2

'2'4'4
cos

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]; 

  



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

15 

 
 

3.4. Demonstration (Explication) 
 

The original relationships (2.16) have been 

deduced by the expressions (2.17). 

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. 

 

 

   

 

 
 

 

   

     

       
 

     

 





















































































































































][2

'2'4'4
cos

][2

'4'2'2
cos

0'cos'2cos

cos'2cos'coscos

cos'cos1cos

coscossin'
cossin

cos

'

cossincossincossinsin

sin
cos

'
sin

cos

'

sinsin

sin
cos

'

sin
2

cos
2

coscos

22
;;

2

coscos'coscos

22

0

22

00

2

0

22

0

22

0

2
22

0

2

02

222

0

422

0

242242

0

22

0

22

0

2

0
0

0

ess

sesesssssss

ess

essssessesss

ssessess

seesssss

esss

esss
r

ess

r

s

r

e

r

ss
BBB

B
r

s
B

r

r

r

s

B
r

r
A

A
r

s

AAA

AA

rsrs

BB

BB

BA

B

A

A

B

A

AA










































 

                                                                   ...(2.17) 

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. The synthesis was 

made using the natural geometro-kinematics 

parameters (of cam mechanism). It follows the 

proper functioning dynamics. We will optimize 

and 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), cosine 

law. 

The synthesis of the cam profile can be made 

with the relationships (3.1) when the cam rotates 

clockwise 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, u is 

/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 by 

integrating the instantaneous efficiency 

throughout the climb and descent) is 0.39 or 

=39%. The dynamic diagram can be seen in the 

fig. 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 technological 

knowledge. 

 



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

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 the 

Fig. 18). 

The r0 (the radius of the base circle of the cam) 

is 0.015 [m]. The h (the maximum displacement 

of the tappet) is 0.010 [m]. The angle of lift, u is 

/6 [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. 18. 

 

 
 

Fig. 18. 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]. 
 

 

 

 

The obtained mechanical yield (obtained by 

integrating the instantaneous efficiency 

throughout the climb and descent) is 0.428 or 

=43%. The dynamic diagram can be seen in the 

fig. 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 technological 

knowledge. 

 

 

 

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 three 

times (nc=n/6), we can reduce and the preload of 

the valve spring (x0=5[cm]); see the dynamic 

diagram in the Fig. 20. However, in this case, the 

cam 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 50 100 150 200

xpp [m/s2]

 
 

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]. 



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

17 

 
 

 

 
 

Fig. 21. 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]. 

 

 

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 the 

tappet) is 0.009 [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. 22. 

 

 
 

Fig. 22. 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]. 

 

 

The obtained mechanical yield (obtained by 

integrating the instantaneous efficiency 

throughout the climb and descent) is 0.538 or 

=54%. The dynamic diagram can be seen in the 

fig. 23 (the dynamic setting are not normal). 

Valve spring preload 30 cm no longer poses 

today. Instead, achieve a long arc very-very hard 

(k=1600000[N/m]), require special technological 

knowledge. 

 

 
 

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] 

 

 



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

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 of 

the valve spring, k=15000[N/m]; see the dynamic 

diagram 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], can 

increase performance even further, if the size 

tappet race take a greater value (h=12[mm]). The 

r0 (the radius of the base circle of the cam) is 

0.013 [m].  

The h (the maximum displacement of the 

tappet) is 0.012 [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. 

 

 
 

Fig. 26. The cosine profile at the cam with 

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

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

 

For correct operation it is necessary to 

decrease the speed of the camshaft four times, and 

all four times multiplication of the cam profile. 

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]. The elastic constant of the 

valve spring is k=1500000[N/m]. See the dynamic 

diagram in the Fig. 27. However, in this case, the 

cam profile should be fourfold. The obtained 

mechanical yield is 0.60 or =60%. 
 



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

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 50 100 150 200 250 300

xpp [m/s2]

-2000

-1500

-1000

-500

0

500

0 50 100 150 200

xpp [m/s2]

 

Fig. 27. The dynamic diagram at the cosine fourfold 
profile 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 valve 

K the elastically constant of the 
system 

k the elastically constant of the valve 
spring 

dt

dM
c 

2

1

 

the coefficient of the system’s 

amortization 

F0 the elastically force which 
compressing the valve spring 

r
F  the resistant force 

m
F    the motor force 

x the effective displacement of the 
valve 

x0 the valve (tappet) spring preload 
x’ the reduced valve (tappet) speed 
x’’ the reduced valve (tappet) 

acceleration 
ys  the theoretical displacement of the 

tappet reduced at the valve, imposed 

by the cam’s profile  

y’s’ the velocity of the theoretical 
displacement of the tappet reduced 

at the valve, imposed by the cam’s 

profile 

mT=m2 the mass of the tappet (of the valve 
lifter) 

m3  the mass of the valve push rod 
m5 the valve mass 

J1=Jc the inertia mechanical moment of 

the cam 

J4 the inertia mechanical moment of 

the valve rocker  

2
y  the tappet velocity, or the second 

movement-low, imposed by the 

cam’s profile  

x  the real (dynamic) valve velocity 

x  the real (dynamic) valve 
acceleration 

i=i25 the ratio of transmission tappet-

valve, given from the valve rocker 
y  the tappet velocity reduced at the 

valve 

D the dynamic transmission function 

(the dynamic transmission 

coefficient) 

ωm  
 

the nominal (average) angular 

velocity of cam 

ω=1=  
 

the angular (real) rotation speed of 

the cam (or camshaft) 

4  the angular rotation speed of the 
valve rocker 

 The pressure angle  

   The additional pressure angle 

rb   roll radius 

r0  basic radius 

e   horizontally misalignment 

s0 vertical misalignment 

BA
rr ,  position vectors 

J*    the reduced mechanical moment of 

inertia 

J*’  the first derivative of the reduced 

mechanical moment of inertia 

J*m the average reduced moment of 

inertia 

 
 

 

 

 

 

 

 

 

 

 

 

 

 



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

20 

 
 

4. Discussion 
 

In this paper one presents an original method 

to determine the dynamic parameters at the 

camshaft (the distribution mechanisms). It makes 

the synthesis, of the rotary cam and tappet with 

translational motion with roll, with a great 

precision.  

The authors introduce a new pressure angle, 

alpha, and a new method to determine the two 

pressure angles, alpha and delta. 

The presented method is the most elegant and 

direct method to determine the kinematics and 

dynamic parameters. 

The dynamic synthesis can generate a cam 

profile which will work without vibrations. 

Processes robotization increasingly determine 

and influence the emergence of new industries, 

applications in specific environmental conditions, 

approach new types of technological operations, 

handling of objects in outer space, leading 

teleoperator in disciplines such as medicine, 

robots that covers a whole larger service benefits 

our society, modern and computerized. In this 

context, this paper seeks to contribute to the 

scientific and technical applications in dynamic 

analysis 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 tappet 

cam 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] a 

study on dynamics of roller gear cam system, 

considering clearances. 

In 2002 D. Taraza synthesized analyzes the 

influence of the cam profile, the variation of the 

angular velocity distribution tree, and the 

parameters of power load consumption and 

emissions of internal combustion engine [16]. 

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

R.V. Petrescu present a synthesis method of rotary 

cam 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 small 

efficiency for about 150 years; this fact affects the 

total yield of the internal heat engines. Much of 

the mechanical energy of an engine is lost through 

the mechanism of distribution. Multi-years the 

yield of the distribution mechanisms was only 4-

8%. In the past 20 years it has managed a lift up to 

the value of 14-18%; car pollution has decreased 

and people have better breathing again. 

Meanwhile the number of vehicles has tripled and 

the pollution increased again.  

Now, it’s the time when we must try again to 

grow the yield of the distribution mechanisms.  

The paper presents an original method to 

increase the efficiency of a mechanism with cam 

and follower, used at the distribution mechanisms. 

This paper treats only one module: the 

mechanism with rotary cam and translated follower 

with roll (the modern module B). 

At the classical module C we can increase again 

the yield to about 30%. The growth is difficult. 

Dimensional parameters of the cam must be 

optimized; optimization and synthesis of the cam 

profile are made dynamic, and it must set the 

elastic (dynamic) parameters of the valve (tappet) 

spring: k and x0.  

The law used is not as important as the module 

used, sizes and settings used. We take the classical 

law cosine; dimensioning the radius cam, lift 

height, and angle of lift.  

To grow the cam yield again we must leave the 

classic 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%, and 

we can consider for the moment that we have gain 

importance, since we work with the cam and 

tappet mechanisms. 

If we more increase h, or decrease the angle 

u, then is tapering cam profile very much. We 

must stop now at a yield value, =60%. 



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

21 

 
 

It can synthesize high-speed cam, or high-

performance camshafts. 

 

 

6. References 
 
[1] Angelas J., Lopez-Cajun C., Optimal 

synthesis of cam mechanisms with oscillating 

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Theory 23,(1988), Nr. 1., p. 1-6., 1988. 

[2] Antonescu, P., Petrescu, F., Antonescu, O., 
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Disc-Cam Profile,  In VIII-th International 

Conference on the Theory of  Machines and 

Mechanisms, Liberec, Czech Republic, p. 51-

56, 2000. 

[3] Antonescu, P., Oprean, M., Petrescu, F.I., 
Analiza dinamică a mecanismelor de 

distribuţie cu came. În al VII-lea Simpozion 

Naţional de Roboţi Industriali, MERO'87, 

Bucureşti, 1987, Vol. III, p. 126-133. 

[4] Bagepalli, B.S., a.o., Generalized Modeling 
of Dynamic Cam-follower Pairs in 

Mechanisms, Journal of Mechanical Design, 

June 1991, Vol. 113, Issue 2, p. 102-109. 

[5] Chang, Z., a.o., A study on dynamics of 
roller gear cam system considering 

clearances, Mechanism and Machine Theory, 

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

[6] Chang, Z., a.o., Effects of clearance on 
dynamics of parallel indexing cam 

mechanism, ICIRA’11 Proceedings of the 4th 

international conference on Intelligent 

Robotics and Applications – Volume, Part I, 

2011, p. 270-280. 

[7] Dan, K., a.o., Research on Dynamic Behavior 
Simulation Technology for Cam-Drive 

Mechanism 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-linkage 

Mechanical Press, Key Engineering 

Materials Journal, Vol. 474-476 (2011), p. 

803-806. 

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

Mechanics 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 Machine 

Theory 14 (1979), nr. 5., p. 327-340, 1979. 

[11] Hain K., Optimization of a cam mechanism 
to give goode transmissibility maximal 

output angle of swing and minimal 

acceleration. Journal of Mechanisms 6 

(1971), Nr. 4., p.419-434. 

[12] Norton, R.L., a.o., Effect of Valve-Cam 
Ramps 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 Ninth 

IFToMM International Symposium on 

Theory of Machines and Mechanisms, 

SYROM 2005, Bucharest, Romania, 2005, 

Vol. I, p. 123-128. 

[14] Petrescu F.I., ş.a., Cams Dynamic 
Efficiency 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 Response 
of Cam-Follower Mechanism, SAE Technic 

Paper, 2009, paper number 2009-01-1416. 

[16] Taraza, D., "Accuracy Limits of IMEP 
Determination from Crankshaft Speed 

Measurements," SAE Transactions, Journal 

of Engines 111, p. 689-697, 2002. 

 

 
 

 

 

 

 

 

 

 
 

 

 

 

 

 

 

 



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

22 

 
 

 

AUTHORS’ INFORMATION 
 
1
Dr. Eng. Florian Ion T. Petrescu, Senior Lecturer 

at UPB (Bucharest Polytechnic University), TMR 

(Theory of Mechanisms and Robots) department. 
2
Dr. Eng. Relly Victoria V. Petrescu, Senior 

Lecturer at UPB (Bucharest Polytechnic 

University), TTL (Transport, Traffic and 

Logistics) department. 

 

 
 

1. Ph.D. Eng. Florian Ion T. 
PETRESCUSenior Lecturer at UPB (Bucharest 

Polytechnic University), Theory of Mechanisms 

and Robots department, 

 

Date of birth: March.28.1958; Higher education: 

Polytechnic University of Bucharest, Faculty of 

Transport, Road Vehicles Department, graduated 

in 1982, with overall average 9.63;  

 

Doctoral Thesis: "Theoretical and Applied 

Contributions About the Dynamic of Planar 

Mechanisms 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. PETRESCU 
Senior Lecturer at UPB (Bucharest Polytechnic 

University), Transport, Traffic and Logistics 

department, 

 

Citizenship: Romanian; 

Date of birth: March.13.1958; 

Higher education: Polytechnic University of 

Bucharest, Faculty of Transport, Road Vehicles 

Department, graduated in 1982, with overall 

average 9.50; 

 

Doctoral Thesis: "Contributions to analysis and 

synthesis of mechanisms with bars and sprocket". 

 

Expert in Industrial Design, Engineering 

Mechanical Design, Engines Design, Mechanical 

Transmissions, Projective and descriptive 

geometry, Technical drawing, CAD, Automotive 

engineering, Vehicles, Transportations. 

 

Association: 

Member ARoTMM, IFToMM, SIAR, FISITA, 

SRR, SORGING, AGIR. 
 

 

 

 

 

 

 

 

 
 

 

 

 

 
 

 

 

 

 

 

 

 

 

 

 



 Florian Ion T. Petrescu                                            (2014) 65- 82، صفحة 1، العدد10دمجلة الخوارزمي الهندسية المجل 

 
 

 
 

 التركيب الديناميكي للحدبات

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

  *  اوربا/ رومانيا / جامعة بوخارست / قسم نظريات المكائن واالنسان االلي 
اوربا/ رومانيا / جامعة بوخارست / قسم النقل والمواصالت   ** 

petrescuflorian@yahoo.com : االتلكتروني البريد*  
petrescuvictoria@yahoo.com : االلكتروني البريد**  

  
 

 
 الخالصة

 
يستخدم األسلوب التقليدي في وصف . لتركيب الهندسي للحدبات الدوارة وانتقال التابع الملتصق اثناء االندفاع يقدم هذا البحث طريقة جديدة لوصف ا     

ركيب تستخدم طريقة الت. التركيب الهندسي وتقليل سرعة التابع المندفع، وفي هذا الوضع فان األسلوب الكالسيكي يصبح كطريقة تركيب هندسية وحركية 

، ولكن يمكن استخدام ومعاينة سرعة التابع المندفع المتحققة والتي يمكن تنبأها من خالل قوانين الحركة (بدون السرعات)الهندسية الجديدة المحددات الهندسية 

حيث يمكن من .اميكي الجيد في العمل والتعديل على محدداتها لحساب التصرف الدين عندها يتم التحليل الديناميكي لشكل الحدبة.   النتجة من شكل الحدبة

 .خالل هذا الوضع ادراك التركيب الديناميكي االمثل للحصول على االداء االعتياد
 

 

 

mailto:petrescuflorian@yahoo.com*البريد
mailto:petrescuflorian@yahoo.com*البريد
mailto:petrescuvictoria@yahoo.com**البريد
mailto:petrescuvictoria@yahoo.com**البريد