Jtam.dvi


JOURNAL OF THEORETICAL

AND APPLIED MECHANICS

45, 2, pp. 379-403, Warsaw 2007

DYNAMIC COMPENSATION OF DYNAMIC FORCES IN

TWO PLANES FOR THE RIGID ROTOR

Tadeusz Majewski

Departamento de Ingeniera Mecánica e Industrial, Universidad de las Américas-Puebla, Puebla,

México and Faculty of Mechatronical Engineering, Warsaw University of Technology, Poland

e-mail: tadeusz.majewski@udlap.mx

Radoslaw Domagalski

Warsaw University of Technology, Poland

Marco Meraz Melo

Departamento Metalmecánica, Instituto Tecnológico de Puebla, Puebla, México

The paper presents a method of automatic compensation of dynamic forces
actingonarigidrotorsasa resultof theirunbalance. It isdoneby freeelements
located in two planes. The free elements rotate together with the rotor and
generate forces that canbeopposedto the rotorunbalance.Thepaperpresents
physical andmathematical models. It is shown for what positions of the free
elements they can compensate the unbalance. Numerical simulations show
that the elements status goes to these positions. The vibration forces were
defined and it was pointed out that they push the free elements to these
positions. The vibration forces take zero values in ball positions for which the
system is balanced.They are positions of the equilibrium.The stability of the
free elements in these positions is discussed. Ranges of the rotor velocity in
which the system balances itself are defined. The paper presents a simulation
of the behavior of the system during the balancing for different unbalance
situation. The influence of resistance of the free elements on the efficiency of
the method is verified. The theoretical results are verified during laboratory
experiments.

Key words: rotor, vibration, self-organizing system

1. Introduction

A rotor has static and dynamic unbalances. The first one happens when the
center of mass of the rotor is not on the axis of rotation and the second one
exists when the axis of rotation is not one of the principal axes of inertia of



380 T. Majewski et al.

the rotor. The unbalance of the rotor generates vibration and dynamic forces
that act on the rotor and its bearings. They can shorten life of the rotor
or they can destroy it. Therefore, all rotors are balanced before they can be
used. Special stands are used to balance rotating systems. They measure the
rotor unbalance, its position, calculate what counterbalanced masses should
be added or remove from the rotor, define the positions where they should
be given, and execute the operation (Breal-Kjear, 1973; Harris, 1988). With
short rotors, such as grinding wheels or disks, it is enough to balance them
in one plane. Longer, rigid rotors have to be balanced at least in two planes
(Breal-Kjear, 1973). Elastic rotors have to be balanced in more planes. It
all depends on the range of the rotor speed and rotor natural frequencies.
The rotor unbalance can change due to its wear or thermal deformation, and
therefore the rotor has to be balanced from time to time.

For rotors in which the distribution of mass is changing, for each start or
during operation (e.g. washing machine, centrifuge, grinding wheel) this me-
thod can not be used. In these situations we can use self-balancing method.
Thismethod was proposed byThearle (1950) for balancing in one plane. The
author developed this method (Majewski, 1976, 1978) and checked in what
situations it can be used and defined its efficiency. Sokolowska (1981) investi-
gated the possibility of compensating dynamic forces for an object rotating
about a fixed point. Sokolowska showed that only a part of the rotor unbalan-
ce could be compensated by free elements. In papers Alfriend (1974), Chang
and Chou (1991), the authors used a liquid for the stabilization of gyroscopic
motion. This kind of balancing is also important for machines that run the
risk of failure, e.g. aircraft turbines or compressors when one of the blades
brakes. Such break-downs can result in a terrible accident.

If the rotor has static and dynamic unbalance, then there is a combination
of a centrifugal force Fo and a moment Mo that rotate together with the
rotor. These loadings can also be presented as two oblique forces acting in
two arbitrary planes perpendicular to the axis of rotation. So, to dynamically
balance the rotor, it is necessary to create two opposing forces. These forces
can be generated by free elements, e.g. balls or rollers located in two planes.
They should generate a force and a moment that are always opposed to the
unbalance. First results of this research were given inMajewski (1980, 1988).

Themethodhaspositive andnegative features, and it is necessary to define
them so that later engineers might decide in what situations this method
can be used in practice. The physical and mathematical models will define
properties of the method and for which situations it can be used.

Theaimof thepaper is to determine the possibility of balancing, define the
velocity range of the rotor for which the free elements compensate dynamic
forces or increase them,and the timeof reaction. If the free elements are able to
balance the rotor, then it is necessary todefine the efficiency of themethodand



Dynamic compensation of dynamic forces... 381

check the influence of each parameter on it. The physical and mathematical
model will allow verification of the stability of the dynamic system.The result
of research should be verified in laboratory tests.

2. Description of the system

Figure 1 presents a sketch of the system. Themass mn11 represents the static
unbalance and the two equalmasses mn21,mn22 represent the dynamic unba-
lance of the rotor. The bearings D1 and D2 of the rigid rotor are supported
elastically. The elastic and damping properties of the supports are described
by parameters kjx, kjy, and njx, njy, (j = 1,2 indicates the number of the
support). The positions of the bearings D1 and D2 are defined by the coor-
dinates zj. At the ends of the rotor there are two drums in which free balls

Fig. 1. A system for automatic balancing

or rollers are placed. The position of the drum along the rotor axis is defined
by the distance zi from themass center of the system. The center of the i-th
free element moves along a circular path with the radius Ri. The radius of
the ball or roller is ri. It means that the radius of the drum is Ri+ ri. The
free elements roll without slipping in the drum. In their motion, with respect
to the drums, the free elements have to overcome the viscous and rolling resi-
stance. The static unbalance of the rotor is presented by the principle vector
−→
P o =

−→
Meω2 and principle moment

−→
Mo =

−→
Mdω2. They are perpendicular to

the axis of rotation and they spin with the rotor. The coordinate system xyz
is fixed to the rotor and spins with it, while the frame XYZ is fixed. It was
assumed that the rotor is symmetrical. The mass moment of inertia of the
rotor with respect to the axis of rotation is A and the moment with respect
to the axis perpendicular to it is B. The displacement of the rotor is defined
with respect to the fixed frame XYZ.



382 T. Majewski et al.

3. Equations of motion

Vibrations of the rotor are definedby two components x, y of thedisplacement
of the mass center in the frame XYZ and two angles Θ, Ψ describing the
position of the rotor axis with respect to the axis X and Y . The position
of i-th free element with respect to the coordinate system fixed to the rotor
is described by the angle αi(t), i = 1, . . . ,N (N is the number of the free
elements) – Fig.2.

Fig. 2. The coordinate system for the free element

The equation of motion of the rotor and the free elements were obtained
fromLagrange’s equations.Thekinetic andpotential energyof the systemwere
defined as a function of generalized coordinates q = [x,y,Θ,Ψ]⊤ describing
the vibration of the rotor with respect to the fixed frame XYZ, the angles
α1, . . . ,αN of the free element with respect to the rotor, and their velocities.
It was shown in previous papers (Majewski, 1976, 1978) that it is enough
to model the rotor unbalance as two generalized forces; the centrifugal force
Po from the static unbalance Me and the moment Mo given by the dynamic
unbalance Md.Between thevector of the static anddynamicunbalance there is
anangle ε. For any rotor, themagnitudeof theunbalance and its locationwith
respect to the rotor are unknowns. The rotor spins with a constant speed ω.
Equations of the rotor

Mẍ+
(

N
∑

i=1

mizi
)

Θ̈+
(

2
∑

j=1

njx
)

ẋ+
(

2
∑

j=1

njxzj
)

Θ̇+
(

2
∑

j=1

kjx
)

x+
(

2
∑

j=1

kjxzj
)

Θ=

=Meω2cosωt+
N
∑

i=1

miRi[(ω+ α̇i)
2cosβi+ α̈i sinβi]



Dynamic compensation of dynamic forces... 383

Mÿ−
(

N
∑

i=1

mizi
)

Φ̈+
(

2
∑

j=1

njy
)

ẏ−
(

2
∑

j=1

njyzj
)

Φ̇+
(

2
∑

j=1

kjy
)

y−
(

2
∑

j=1

kjyzj
)

Φ=

=Meω2 sinωt+
N
∑

i=1

miRi[(ω+ α̇i)
2 sinβi− α̈icos(ωt+αi)]

−
(

N
∑

i=1

mizi
)

ÿ+
[

B+
N
∑

i=1

mi(z
2
i +R

2
i cos

2βi)
]

Φ̈−
1

2

(

N
∑

i=1

miR
2
i sin2βi

)

Θ̈+

−
(

2
∑

j=1

njxzj
)

ẏ+
[

2
∑

j=1

njyz
2
j +

N
∑

i=1

miR
2
i(ω+ α̇i)sin2βi

]

Φ̇+ (3.1)

+
[

Aω+2
N
∑

i=1

miR
2
i(ω+ α̇i)sin

2βi
]

Θ̇+
(

2
∑

j=1

kjyzj
)

y+
(

2
∑

j=1

kjyz
2
j

)

Φ=

=Mdω2 sin(ωt−ε)+
N
∑

i=1

miRizj[−(ω+ α̇i)
2 sinβi+ α̈icosβi]

(

N
∑

i=1

mizi
)

ẍ−
1

2

(

N
∑

i=1

miR
2
i sin2βi

)

Φ̈+
[

B+
N
∑

i=1

mi(z
2
i +R

2
i cos

2βi)
]

Θ̈+

+
(

2
∑

j=1

njxzj
)

ẋ−
[

Aω+2
N
∑

i=1

miR
2
i(ω+ α̇i)cos

2βi
]

Φ̇+

+
[

2
∑

j=1

njxz
2
j −

N
∑

i=1

miR
2
i(ω+ α̇i)sin2βi

]

Θ̈+
(

2
∑

j=1

kjxzj
)

x+
(

2
∑

j=1

kjxz
2
j

)

Θ=

=Mdω2 sin(ωt−ε)+
N
∑

i=1

miRizj[(ω+ α̇i)
2cosβi+ α̈i sinβi]

where βi =ωt+αi.

Themass of the system is M =Mw+
∑N
i=1mi, where Mw means the total

mass of the rotor and
∑N
i=1mi is the total mass of all free elements.

The differential equation of motion of the free element with respect to the
rotor is (i=1, . . . ,N)

Iiα̈i =miRi[ẍsinβi− ÿcosβi+ Φ̈zicosβi+ Θ̈zi sinβi]−Fi (3.2)

where N is the total number of the free elements and Ii is the moment of
inertia of the i-th free element reduced to the center of the drum. For the free
element that rolls inside the drum without slipping Ii =miR

2
i + Iic(Ri/ri)

2.



384 T. Majewski et al.

The force Fi which is opposed to motion of the free element consists of the
viscous Fiv and the rolling resistance Fir

Fi =Fiv +Fir Fiv =niR
2
i α̇i

(3.3)

Fir ∼=
1

ri
mi(ω+ α̇i)

2fiR
2
i

where ni is the coefficient of viscous resistance, fi is the coefficient of the
rolling resistance for the free element, and ri is the radius of the ball or roller.
Some of the components in Eqs. (3.1) can be neglected because they are

very small with respect to the others. The simulation of the behavior of the
system shows that it is possible. Now Eqs. (3.1) can be written in a matrix
form

Mq̈+(Cd+Cg)q̇+Kq=Qo(ωt)+
N
∑

i=1

Qi(ωt,αi, α̇i) (3.4)

where Qo(ωt) is the harmonic excitation from the unbalance, Qi(ωt,αi) is
the reaction of one free element on the rotor. The components of the first one
depend on the speed of the rotor and the static and dynamic unbalance of the
rotor.The components of Qi(ωt,αi) are a functionof the angle of rotation and
the positions of the free elementswith respect to the rotor.Thematrices of the
inertia of the rotor M, damping Cd, and the stiffness K have the dimension
4×4. The inertial forces generated by the accelerations α̈i are very small with
respect to the free element centrifugal forces (Gawlak and Majewski, 1991;
Majewski, 1976, 1978) and can be neglected in Eqs. (3.1), thus

Qo =ω
2[Mecosωt,Mesinωt,Mdcos(ωt−ε),Mdsin(ωt−ε)]⊤

(3.5)

Qi =miRi(ω+ α̇i)
2[cosβi,sinβi,zi sinβi,−zicosβi]

⊤

Equation (3.2) can be written in the form

Iiα̈i =miRiBiq̈−Fi =P i−Fi i=1, . . . ,N (3.6)

where

P i =miRiBiq̈ Bi = [Biccosωt−Bis sinωt]

Bic = [sinαi,−cosαi,zicosαi,zi sinαi]

Bis = [−cosαi,−sinαi,zi sinαi,−zicosαi]

Equations (3.4) and (3.6) describe the behavior of the rotor and the free ele-
ments during the balancing. They will be used for further analysis of the
system dynamics.



Dynamic compensation of dynamic forces... 385

4. Possibility of balancing

As a result of rotor vibrations there are inertial forces P1, . . . ,PN that push
the free elements to new positions

lim
t→∞
α(t)=αif (4.1)

The system will be completely balanced with the free elements in these new
positions when for any moment of time the rotor excitation is zero. In these
positions, the free elements have to fulfill the following conditions

Qo(t)+
N
∑

i=1

Qi(t,αif)≡0 (4.2)

Conditions (4.2) represent the resultant force acting on the rotor. It can be
written as

[

Qco+
N
∑

i=1

Qci(αif)
]

cosωt−
[

Qso+
N
∑

i=1

Qsi(αiif)
]

sinωt≡0 (4.3)

or
Qccosωt−Qs sinωt≡0

When the coefficients in front of the time functions are zero

Qco+
N
∑

i=1

Qci(αif)=0 Qso+
N
∑

i=1

Qsi(αif)=0 (4.4)

then condition (4.2) occurs.
From this condition, we obtain

Me+
N
∑

i=1

miRicosαif =0
N
∑

i=1

miRi sinαif =0

(4.5)

Md−
N
∑

i=1

miRizi sin(αif −ε)= 0
N
∑

i=1

miRizicos(αif −ε)= 0

For the first two conditions the mass center of the system is on the axes of
rotation and for the next two conditions the axes of rotation become one of
the principle axes of the inertia of the rotor.
Now the rotor excitation, which is presented by the right side of Eq. (3.4),

is zero. When there is no excitation, the vibrations of the rotor disappear

q(t,α1f , . . . ,αNf)≡0 (4.6)



386 T. Majewski et al.

For the free elements distribution α1f, . . . ,αft, the system is balanced and
there is no vibration.
If we use, for example, two balls or rollers in each drum then from condi-

tions (4.5)we candefine their final positions α1f, . . . ,α4f. The total unbalance
of the rotor can also be presented as two oblique forces in the planes of drums
in which the free elements are located, Fig.3. The resultant force from com-
pensating elements in the drum compensate one of the oblique forces. If there
are only two balls or rollers (one in each drum) then four conditions (4.5) can
be fulfilled only if the masses of the balls are selected according to the rotor
unbalance, Fig.3. For one roller in each drum their masses should be selected
according to the existing unbalance that is unknown. It is a rather theoretical
problem.

Fig. 3. Forces acting on the rotor

If the system has more than four free elements, then conditions (4.5) can
be satisfied not only for one configuration of the balls but also for some other
positions. In this situation the final positions of the free elements are not
defined exactly. If all free elements are located in one drum, they are not able
to compensate the rotor unbalance. The free elements should generate forces
that are opposite to the unbalance. If the rotor has no unbalance, then the
balls take positions for which they compensate each other. If the drum has
two balls, then they occupy positions opposed to each other on one diameter.
The static unbalance indicates the location of balls to be symmetrical with

respect to the vector
−→
Me. For thedynamicunbalance, the free elements occupy

antisymmetric positions with respect to the vector
−→
Md. The distance between

thedrumscannotbe too small because themoment generatedby free elements
would be too small to compensate the dynamic unbalance Md of the rotor.
Conditions (4.5) are necessary to obtain the balanced state of the system

which consists of the rotor and free elements. It is necessary to prove that the
free elements really tend to these positions. Solutions to differential equations
(3.1) and (3.2) should give an answer as to which situations the free elements
compensate or increase the rotor unbalance.



Dynamic compensation of dynamic forces... 387

5. Numerical simulation

It is not possible to determine the exact solutions q(t), αi(t) to equations
(3.1) and (3.2). First results can be obtained from numerical solutions. A sys-
temwith defined parameters was assumed, a software for simulation prepared
and many simulations for different sets of parameters provided. The exam-
ples presented in the paper are presented for the following parameters: rotor
mass – 4kg, mass moment of inertia – A ≈ 0.0018kgm2, B ≈ 0.014kgm2,
R1 = R2 = 0.03m, m1 = m2 = 0.00166kg, z1 = −z2 = 80mm. The most
important parameter was the ratio of the rotor velocity to its natural frequen-
cies. The rotor velocity was taken from the range 20-100rad/s. The diagrams
of the coordinates describing the positions of the rotor and the free elements
show in which way the free elements move with respect to the rotor. We can
point out that the free elements compensate or increase the unbalance, we
can also define the time that is necessary for the free elements to reach the
final positions, and find put in whichway the rotor vibrations change in time.
The computer simulation also gives the answer inwhichway the free elements
behave when they are close to the final positions.

Figure 4 shows the results for the rotor with only static unbalance and one
ball in each of two drums. Themasses of the balls were selected in such away
that they can compensate the rotor unbalance – the conditions (4.5) can be
fulfilled. Velocity of the rotor is bigger than its natural frequencies. It is seen
that the balls are going to the positions for which the system is dynamically
compensated and the vibrations of the rotor vanish.The balls oscillate around
the positions α1f, α2f and the time of decay of vibrations depends on the
viscous damping in the drum.

In normal conditions, the rotor unbalance is not known and therefore in
each drum there should be at least two free elements with masses that can
compensate the biggest unbalance that can happen in the system. In Fig.5,
we can observe vibration of the rotor and motion of four balls for the same
parameters as in the rotor shown inFig.4.The free elements needabout50/ω -
100/ω seconds to reach their final positions.When the balls are close to theirs
final positions, the vibration of the rotor vanishes.

The frequency of balls oscillations around their final positions is much
smaller than the rotor speed. The vanishing of balls oscillation depends on
their resistance. At the final positions, the free elements generate forces that
are opposed to the oblique forces in planes of the drums generated by the
unbalance.

Other simulations show that for some speeds of the rotor that are close to
its natural frequencies the free elements compensate only the static or dynamic



388 T. Majewski et al.

Fig. 4. Balancing with two balls

unbalance. Below natural frequencies, the free elements occupy positions close
to the unbalance, and the rotor vibration increases.

6. Vibration forces

The inertial force (Eq. (3.6), Fig.2) acting on the free element depends on the
acceleration of the rotor and the position of the free element. It is defined by
the relation

P i =miRiBiq̈ (6.1)

The free elements move slowly with respect to the rotor (α̇ ≪ ω) and the
vibration of the rotor can be written as

q(t)∼=Accosωt−As sinωt (6.2)



Dynamic compensation of dynamic forces... 389

Fig. 5. Balancing with four balls

Accordingly, the acceleration of the vibration is also defined

q̈(t)∼=−ω
2q(t) (6.3)

Substituting relations (6.3) to (6.1), the inertial force will be defined as a
function of time, rotor unbalance and positions of all free elements. We can
calculate the average magnitude of this force during one period of vibration
T =2π/ω



390 T. Majewski et al.

Pi =
1

T

T
∫

o

P i dt=−
1

2
miRiω

2(BciAc+BsiAs) (6.4)

where

Ac = [Acx,Axy,AcΦ,AcΘ]
⊤

As = [Asx,Asy,AsΦ,AsΘ]
⊤

The amplitudes of vibration of the rotor are defined by a relation obtained
from (3.4)

(K−Mω2)Ac−CωAs =ω
2
Pc

(6.5)

CωAc+(K−Mω
2)As =ω

2
Ps

where

Pc = [Pc1,Pc2,Pc3,Pc4]
⊤ Ps = [Ps1,Ps2,Ps3,Ps4]

⊤

and

Pc1 =−Ps2 =Me+
N
∑

i=1

miRicosαi

Pc2 =Ps1 =
N
∑

i=1

miRi sinαi

Pc3 =−Ps4 =Mdcosε−
N
∑

i=1

miRizi sinαi

Pc4 =Ps3 =−Mdsinε+
N
∑

i=1

miRizicosαi

are component forces acting on the rotor. Each component depends on the
static and dynamic unbalance and positions of all free elements with respect
to the rotor.
The vibration force of one element can be presented as a sumof four forces

generated by four components of the rotor vibration

Pi =Pix+Piy +PiΦ+PiΘ (6.6)

and each component is a result of the rotor unbalances Me, Md, and the
unbalance given by each free element, e.g.

Pix =PixMe(αi)+PixMd(αi)+
N
∑

j=1

Pijx(αi−αj) (6.7)



Dynamic compensation of dynamic forces... 391

where

PixMd =−
1

2
miRiω

2axMe sin(αi+ϕx)

PixMd =−
1

2
miRiω

2axMd sin(αi+ε+ϕx)

Pijx =−
1

2
miRiω

2axj sin(αi−αj +ϕx)

axMe, axMd, axj are amplitudes of the rotor vibration in the x direction gene-
rated by the static and dynamic unbalance as well as the element j.
Other components of the vibration forces are similar.

Fig. 6. The vibration force as a function of one free element position

The vibration force Pi acting on each free element is a function of rotor
vibrations that are a function of the unbalance of the rotor and the positions
of all free elements. Figure 6 shows the force P1x if there is one ball and only
static unbalance – balancing in one plane

P1x =−
1

2
m1R1ω

2[axMe sin(αi+ϕx)+ax1 sin(α1−α1+ϕx)] (6.8)

It is seen that the ball has two positions forwhich the force P1 can assume the
zero magnitude. One position is exactly at α1f = π which fulfills conditions
(4.5) and the system is dynamically balanced. For the other position, the free
element increases the unbalance of the system. Which of these positions the
ball will occupy depends on the velocity of the rotor. Only one of them is
a dynamically stable equilibrium. It can be observed in Fig.6 that for one
position of the ball, the derivative of the force with respect to the angular
displacement is negative and it is in the position of equilibrium.
For balancing in two planes, there should be at least one ball or roller in

eachplane, and the forces P1,P2 are functionsof twovariables α1,α2. Figure7
presents the force P2 changing with the position of both balls. It is seen that



392 T. Majewski et al.

there are many positions in which the force acting on one ball disappears.
They are the possible positions of equilibrium of the ball. But the force acting
on the second ball is not zero. The second ball moves with respect to the rotor
which makes the first ball cannot stay in the previous position. The balls can
occupy only such positions for which all vibration forces P1, . . . ,PN become
simultaneously zero.

Fig. 7. The vibration force as a function of two free element position

If we know the vibration forces, thenwe can explain why the free elements
change their positions, inwhatdirection they go, andwhat final positions they
occupy. It will allow us to verify the dynamic stability of these positions and
define the efficiency of the method.

7. Stability

Sometimes the balls compensate the rotor unbalance and sometimes they in-
crease it. To define properties of the system, it is necessary to analyze the
dynamic stability of the balls in their final positions α1f, . . . ,αNf. It can be
done if the forces acting on the balls are known (6.4). The final positions of
the balls are stable if the roots λ of determinant (7.1) have negative real parts

∣

∣

∣

∂Pi
∂αj
−κijλ

∣

∣

∣
=0 i,j =1, . . . ,N (7.1)



Dynamic compensation of dynamic forces... 393

where κij is the Kronecker symbol. The derivative of the vibration force with
respect to the position of the ball can be obtained from (6.4)

∂Pi
∂αj
=−
1

2
miRiω

2
[(∂Bic
∂αj
Ac+

∂Bis
∂αj
As

)

+
(

Bic
∂Ac
∂αj
+Bis

∂As
∂αj

)]

(7.2)

At the final positions of the free elements α1f, . . . ,αNf, when the system is
balanced Ac(α1f, . . . ,αNf)=As(α1f, . . . ,αNf)=0, the above relation takes
the form

∂Pi
∂αj
=−
1

2
miRiω

2
[

Bic
∂Ac
∂αj
+Bis

∂As
∂αj

]

(7.3)

The vibration of the rotor is a composition of vibrations from the rotor unba-
lance and the action of all free elements. Then

∂Ac
∂αj
=
∂Acj
∂αj

∂As
∂αj
=
∂Asj
∂αj

The derivative of the amplitude, with respect to the position of the free ele-
ments, can be defined from relations (6.5)

[K−Mω2]
∂Acj
∂αj
−ωC

∂Asj
∂αj

=
∂Qcj
∂αj

(7.4)

ωC
∂Acj
∂αj
+[K−Mω2]

∂Asj
∂αj

=
∂Qsj
∂αj

Developing determinant (7.1), we obtain the characteristic equation

anλ
n+an−1λ

n−1+ . . .+a1λ+ao =0 (7.5)

The roots of the above equation have negative real parts if all the coefficients
are positive and the principle minors of determinates (7.5) are also positive

∆i > 0 for i=2, . . . ,N −1 (7.6)

The coefficients an, . . . ,ao of determinant (7.5) are a function of the rotor
speed.Dependingon the rotor velocity, the roots λ canbepositive ornegative,
and condition (7.1) can be fulfilled or not. Detailed analysis of condition (7.1)
allows us to define the ranges in which the state of full balance of the rotor
can be obtained.
If we take the simplest situation, e.g. only static unbalance of the rotor,

with one free element located in the plane of unbalance (z1 =0), then relation
(7.1) gives

λ=−
1

2
mRω2

[

B1c
∂A1c
∂α1

+B1s
∂A1s
∂α1

]

(7.7)



394 T. Majewski et al.

The system can be balanced if the free element has the mass m=Me/R and
it should occupy the position opposed to the unbalance α1f = π. For these
parameters the vectors B1c,B1s have the form

Bic = [0,1,0,0] Bis = [1,0,0,0]

To make the problem simpler, we can neglect damping of the rotor, and now
the derivatives of the amplitudes take the form

∂Acx1
∂α1

= cxcosα1f =−cx
∂Asx1
∂α1

= cx sinα1f =0

∂Acy1
∂α1

= cy sinα1f =0
∂Asy1
∂α1
= cy cosα1f =−cy

(7.8)

where

cx =
s2x
1−s2x

m

M
R cy =

s2y
1−s2y

m

M
R sx =

ω

ωox
sy =

ω

ωoy

The coefficient cx is negative when the rotor speed is greater than its natu-
ral frequency ωox in the direction x. We have the same situation with the
coefficient cy which is negative for ω > ωoy, and condition (7.7) takes the
form

λ=
1

2
mRω2(cx+ cy)< 0 (7.9)

It is seen that when the rotor speed is greater than ω>ωox and ω>ωoy, the
above condition is fulfilled (cx < 0, cy < 0) and theposition of the free element
α1f = π is stable. For the rotor speed ω < ωox and ω < ωoy, the position
of the free element opposed to the unbalance is unstable and the free element
cannot compensate the rotor unbalance. Using relations (7.8) and (7.9), it is
also possible to define the rotor speed from the range between the minimum
and themaximum natural frequency of the rotor ωomin −ωomax in which the
position α1f =π is also stable.
If there are more free elements placed in different planes (zi 6=0), the di-

rections of vibrations are coupled, there exist static and dynamic unbalances
of the rotor, then condition (7.1) is more complicated. The exact range of the
speed of the rotor in which the positions of the free elements α1f, . . . ,αNf
are stable or unstable can be defined from numerical simulation of the rela-
tions (7.1)-(7.3). If they can stay in theses positions, it means that they can
compensate the dynamic forces generated by the rotor and the systemwill be
balanced.
The presented system can balance itself automatically if the rotor speed is

greater than the natural frequencies. It is also possible in a small range of the
rotor speed between the minimum andmaximum natural frequencies.



Dynamic compensation of dynamic forces... 395

8. Efficiency of the method

If we do not take into consideration the resistance of the free elements in their
motion with respect to the drum, then they will occupy the final positions
α1f, . . . ,αNf, and the system that consists of the rotor and free elements can
be fully balanced. The viscous resistance does not affect the final positions
because it exists only when the free elements move with respect to the rotor.
The viscous resistance can only reduce velocity of the free elements while they
approach the final positions and suppresses their vibrations about the final
positions. Balls or rollers can be taken as the free elements. They roll with
respect to the rotor. The normal reaction between the ball or roller and the
circular path is a function of the angular velocity of the rotor. This reaction
is also a function of rotor vibrations. But when the free elements are close to
the final positions the vibrations are very small and the normal reaction can
be taken as constant. The rolling resistance depends on the normal reaction
and the coefficient of the rolling resistance. The latter one is a function of the
rigidity of the element and the rigidity of the circular path. As a result of the
resistance, there will be a new position of equilibrium of the compensating
element that differences by ∆i with respect to the ”ideal” position αif. If
the free elements are not in the positions α1f, . . . ,αNf, then they do not
completely compensate the static and dynamic unbalance of the rotor. So,
there is a residual unbalance and residual vibrations of the rotor. They are a
function of the speed of the rotor and the coefficient of the rolling resistance.
If the vibration forces are known, then the errors in the positioning of the free
elements can be defined, and in the next step the residual unbalance of the
system can be calculated.
The position of the i-th element is the position of equilibriumwhen

∣

∣

∣Pi(α1f +∆1, . . . ,αnf +∆n)
∣

∣

∣−
∣

∣

∣Fir
∣

∣

∣¬ 0 i=1, . . . ,N (8.1)

where

Fir =
miRiω

2fi
ri sgn(α̇i)

Pi(α1f, . . . ,αNf)= 0

For a small displacement ∆i of the free element with respect to its theoretical
position αif, it can be taken as

Pi(α1f +∆1, . . . ,αNf +∆N)∼=Pi(α1f, . . . ,αNf)+
N
∑

j=1

∆j
∂Pi
∂αj

∣

∣

∣

αjf
=

(8.2)

=
N
∑

j=1

∆j
∂Pi
∂αj

∣

∣

∣

αjf



396 T. Majewski et al.

The derivatives of the vibration forces with respect to the positions of the free
elements are defined by relation (7.3). Relation (8.1) changes into

∣

∣

∣−
1

2
miRiω

2
N
∑

j=1

(

Bic
∂Acj
∂αj
+Bis

∂Asj
∂αj

)

∆j
∣

∣

∣−
∣

∣

∣Fir
∣

∣

∣¬ 0 (8.3)

It is a system of algebraic equations fromwhich the displacements of the free
elements can be defined (i=1, . . . ,N)

∣

∣

∣−
1

2
miRiω

2
[

sinαif

N
∑

j=1

(∂Acxj
∂∆j

−zi
∂AcΨj
∂∆j

−
∂Asyj
∂∆j

+zi
∂AsΘj
∂∆j

)

∆j +

(8.4)

+cosαif

N
∑

j=1

(

−
∂Acyj
∂∆j

+zi
∂AcΘj
∂∆j

−
∂Asxj
∂∆j

−zi
∂AsΨj
∂∆j

)

∆j
]
∣

∣

∣−|Fir| ¬ 0

When the positioning deviation ∆i is not a small, then equations (8.1) should
be used.
Equations (8.3) or (8.4) define the range ∆imin-∆imax in which the i-

th free element can occupy its position. Because of the rolling resistance, the
deviation of the free element is randomand itdependson the initial conditions.
For the rotor with the static unbalance only, vibration of the rotor center

in the two directions x and y and one free element with themass mR=Me,
relation (8.5) gives

−
2|Fr|

mRω2|Ax cosϕx+Ay cosϕy|
<∆1 <

2|Fr|

mRω2|Axcosϕx+Ay cosϕy|
(8.5)

The deviation ∆1 is proportional to the resistance Fr. If the resistance of the
free element increases, then the error of the positioning increases and residual
unbalance of the system increases as well. The errors of positioning are a
function of rotor amplitudes of vibrations and the shifts angles ϕx, ϕy that
are a function of the rotor speed.
Figure 8 shows themaximum errors in the positioning that can happen if

two balls were used to balance the system. They are a function of the rotor
speed. It is seen that the errors of positioning are smaller for a speed a little
greater than the natural frequencies of the rotor.
The rolling resistance is much smaller than the sliding friction. To obtain

the smallest errors of the method, we should use the free elements as balls
and rollers that can roll instead of slipping. Using other elements as sand,
shots, or liquid involve the slipping and greater friction. The rotor with a
liquid cannot be completely balanced.Under vibration forces, the liquidmoves
to the compensating position. But when the system decreases its unbalance,



Dynamic compensation of dynamic forces... 397

Fig. 8. Maximum errors in the positioning of free elements

then the vibrational forces also vanish and the liquidmoves back (Gawlak and
Majewski, 1991).

The system with the free elements in positions α1f +∆1, . . . ,αNf +∆N
is not completely balanced. If the deviations ∆1, . . . ,∆N are known, then
the final unbalance of the system can be defined. The residual unbalance is a
function of the deviations of all free elements. The residual unbalance will be
defined in the coordinates x0y fixed with the rotor.

The components of the residual static unbalance have the form

MeRx =Me+
N
∑

i=1

miRicos(αif +∆i)

(8.6)

MeRy =
N
∑

i=1

miRi sin(αif +∆i)

The total static unbalance is

MeR =
√

Me2
Rx
+Me2

Ry
(8.7)

When the deviation ∆i is small then

MeRx ∼=−
N
∑

i=1

miRi∆i sinαif

(8.8)

MeRy ∼=−
N
∑

i=1

miRi∆icosαif



398 T. Majewski et al.

The residual dynamic unbalance

MdRx =Mdcosγ−
N
∑

i=1

miRizi sin(αif +∆i)

(8.9)

MdRy =Mdsinγ+
N
∑

i=1

miRizicos(αif +∆i)

For small errors of the positioning of the free elements, the above relations can
be written as

MdRx ∼=−
N
∑

i=1

miRizi∆icosαif

(8.10)

MdRy ∼=−
N
∑

i=1

miRizi∆i sinαif

The total residual dynamic unbalance is

MdR =
√

Md2
Rx
+Md2

Ry
(8.11)

The greater the deviations ∆1, . . . ,∆N, the bigger the systemunbalance. The
total residual unbalance changes with the rotor speed in the same way as the
errors in the positioning of the free elements. Because of the resistance and
friction, the free elements are shifted with respect to the theoretical positions
α1f, . . . ,αNf, and therefore the system cannot be completely balanced. The
smallest resistance results in the smallest residual unbalance.

9. Experiments

To verify the results from theoretical investigation, a laboratory stand was
built and experiments were carried out – Fig.9. Rotor 1 (M = 4.7kg, B ≈
72·10−4kgm2) is supportedon springs 2.At the end of the rotor, there are two
drums3with twoballs 4 in eachdrum.Eachball hasmassof m=21.7·10−3kg
and radius r=8.75mm.Thedrumhas radius R+r=34.75mm.Thedistance
between the drums is l=240mm. The rotor unbalance could be changed by
changing the number and the positions of bolts 5. The distance between the
planes of the bolts is 105mm.
A series of experiments were carried out with different initial positions of

the balls. The balls were kept in these positions by amechanical or electrome-
chanical blocking system. The rotor was driven by an electric motor and we



Dynamic compensation of dynamic forces... 399

Fig. 9. The laboratory stand for verification of the automatic balancing; 1 – rotor,
2 – springs, 3 – drums, 4 – balls, 5 – bolts

could regulate its speed.When the velocity of the rotor was constant, themo-
tor was disconnected from the rotor, the balls were released and they changed
positions with respect to the rotor. The vibrations of the rotor were being re-
corded andmotion of the balls was observed in stroboscopic light. Oscillations
in the horizontal direction of the rotor bearings in one of the experimentswith
the rotor speed greater than its natural frequencies are shown in Fig.10.

Fig. 10. Vibrations of rotor bearings during automatic balancing

The vibrations vanish when the balls are close to their final theoretical
positions. The vibrations do not decay completely because there is a residu-
al unbalance as a result of the errors in the positioning of the balls. In the



400 T. Majewski et al.

final positions, the balls were blocked again. For the given rotor unbalance,
theoretical final positions of the balls were determined. The initial and final
positions of the balls for three experiments with the same dynamic unbalan-
ce are given in Table 1. For this particular rotor unbalance, the balls should
take the following positions to balance the system; α1f = 104

◦, α2f = 256
◦,

α3f =76
◦, α4f =284

◦.

Table 1

No.
Initial position [◦] Final position [◦]
α1 α2 α3 α4 α1t α2t α3t α4t

1 160 200 20 340 110 240 75 280

2 240 280 250 290 80 232 95 300

3 20 60 60 100 90 230 90 270

In these experiments, the minimum error in the positioning of the balls
was ∆min =4

◦ and the maximum error ∆max =26
◦. When the rotor had no

unbalance, then the balls occupied the opposite positions on one diameter in
the drum and in this way they compensated each other.
The experiments show that the balls try to balance the rotor but they do

notoccupyaconstantfinalposition.Theyalso vibrateabout them.Theauthor
explained the reason for this vibration in Majewski (1988). The vibration is
caused by the eccentricity of the circular path of the ball.

10. Conclusion

It was shown that the balls can organize themselves in such a way that they
compensate the unknown rotor unbalance. It is the greatest advantage of the
method.Anunbalanced systemgenerates vibration, andbecauseof the inertial
forces the free elementsmove in the direction generating dynamic forces oppo-
sed to the rotor unbalance. The changing of the positions of the free elements
takes place as long as the unbalance exists. The free elements occupy the final
positions for which there are no vibrations. The principle of the method can
be presented with the block diagram shown in Fig.11.

Fig. 11. A block diagram of the method



Dynamic compensation of dynamic forces... 401

The behavior of the balls depends on the vibration forces. These forces
were defined as a function of the positions of the balls, the rotor unbalance,
and the rotor speed. If the vibration forces are known, then the setting of the
balls with respect to the rotor can be defined. It is also possible to check if
they are dynamically stable or not in these positions, what errors can happen
in their positioning, and what is the efficiency of this method.
There aremany factors thatmake the efficiency of the self-balancing smal-

ler than 100%. The most important is friction of the balls or rollers. The
resistance should be as small as possible and, therefore, it would be better
when the free elements are supported by amagnetic or pneumatic cushion.
The other reason for smaller efficiency is the eccentricity of the drum in

which the balls are located. This problem was not discussed here. A related
research was given byMajewski (1988).
In this paper, a rigid rotor is analyzed. If the rotor unbalance and com-

pensating elements are in different planes, then the rotor undergoes bending
because of the centrifugal forces from the unbalance and the free elements.
The bending can deform the rotor which is another reason that the system
cannot be balanced completely.
This method can be also used for balancing the system in the case of an

accident, e.g. the loss of a blade in a turbine or compressor.
According to conditions (4.5), it is not possible to completely balance the

rotor that rotates about a fixed point. There are two angular components of
vibration. The vibrations vanish when the resultant moment from the unba-
lance and free elements about the point of rotation are zero. It does notmean
the system is balanced. At the point of the rotor support, there is a dynamic
reaction that rotates with the rotor. So, this kind of the system eliminates
vibrations but does not eliminate the excitation completely. This feature, ho-
wever, can be used for elimination of vibrations in general. The idea of a
synchronous eliminator was presented by the author inMajewski (1987, 1994,
2000a,b). It is a self-organizing system. The system detects the unbalance and
changes the configuration of the free elements to eliminate the vibration.

References

1. Alfriend K.T., 1974, Partially filled viscous nutation damper, Journal of
Spacecraft and Rockets, II, 383-395

2. Blekhman I.I., 1971, Synchronization of Dynamic Systems, Moscow, Nauka
(in Russion)

3. Bovik P., Hogfors C., 1986,Autobalancing of rotors, Journal of Sound and
Vibrations, 111, 3



402 T. Majewski et al.

4. Breal-Kjear, 1973, Dynamic balancing,Technical Review, 3

5. Chang C.O., Chou C.S., 1991, Stability analysis of a freely precesing gyro-
scope carrying a mercury ring damper, Journal of Sound and Vibration, 146,
39, 491-506

6. Gawlak G., 1986, Balancing of the grinding wheels, 3rd Conf. on Production
Engineering, Design and Control ”PEDAC”, Alexandria

7. Gawlak G.,Majewski T., 1991,Dynamic of device for automatic balancing
of the rotor systems,Archive ofMechanical Engineering,XXXVIII, 3, 185-199

8. Harris C.M. (edit.), 1988,Shock and Vibration Handbook, McGraw-Hill Bo-
ok Company

9. KrawczenkoW.I., 1983, Investigation of the stability of the ball balancer (in
Russian),Maszinostrojenie, I, 25-27

10. Majewski T., 1976,Automatic balancing of the rotor supported elastically in
one direction,Archive of Mechanical Engineering,XXII, 3, 377-390

11. Majewski T., 1978,Automatic balancing of the rotor elastically supported in
two directions (in Polish), Journal of Theoretical and Applied Mechanics, 16,
1, 26-39

12. MajewskiT., 1980,Automatic balancing in twoplanes,Archive ofMechanical
Engineering, XVII, 192-212

13. Majewski T., 1987, Synchronous vibration eliminator for an object having
one degree of freedom, Journal of Sound and Vibration, 112, 401-413

14. Majewski T., 1988, Position errors occurrence in self balancers used on rigid
rotors of rotatingmachinery,Mechanism and Machine Theory, 23, 1, 71-78

15. Majewski T., 1994, Synchronous elimination of mechanical vibration,Publi-
shing House of the Warsaw University of Technology, Mechanics, 155

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17. MajewskiT., 2000b,Synchronouseliminationofvibration in theplane,Part2:
Method efficiencyand its stability,Journal of Sound andVibration,232, 3, 571-
584

18. Majewski T., Domagalski R., 1997, Selection of system parameters for
achieving appropriate balance class, Conference on Mechatronics’97, Warsaw,
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de Mecanique, 11, 397-402

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Dynamic compensation of dynamic forces... 403

21. Sokolowska R., 1981, Automatic balancing in rotation about a fixed point,
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Automatyczna kompensacja sił dynamicznych w dwóch płaszczyznach dla

sztywnego wirnika

Streszczenie

Wpracy przedstawionometodę automatycznego równoważenia sił dynamicznych
działających na sztywny niewyważony wirnik. W tym celu użyto swobodnych ele-
mentów (kulek lub rolek) umieszczonych w dwóch płaszczyznach. Elementy swobod-
ne wirujące razem z wirnikiem mogą zmieniać swoje położenie i generują siły, które
mogą zrównoważyć niewyważenie wirnika. W pracy przedstawiono model fizyczny
i matematyczny. Ustalono położenia elementów swobodnych, w których równoważą
niewyważenie wirnika. Przeprowadzono symulację komputerową, która wykazała że
elementy swobodne rzeczywiście dążą do tych położeń. Zdefiniowano siły wibracyj-
ne, które wymuszają zmianę położenia elementów swobodnych względem wirnika.
Wykazano, że siły wibracyjne zanikają, gdy elementy znajdują się w położeniach
równoważących niewyważenie. Zbadano stateczność położeń końcowych. Analizowa-
nowpływ oporów ruchu na końcowy stanwyważeniawirnika. Zbudowano stanowisko
laboratoryjne i przeprowadzono eksperymenty, które wykazały, że elementy swobod-
ne rzeczywiście przemieszczają się do położeń, w których równoważą siły dynamiczne
działające na wirnik.

Manuscript received May 8, 2006; accepted for print July 5, 2006