estimation of reaction forces for anelastically-supported rigid...

공학석사 학위논문

Estimation of Reaction Forces for

an Elastically-supported Rigid Body by

Reconstructed Displacements Using

Measurements of Accelerometers and Potentiometers

가속도계 및 포텐셔미터 측정치에 의하여

복원된 변위를 이용한 탄성 지지된 강체의 반력 추정

2018년 8월

서울대학교 대학원

건설환경공학부

전 상 범

iii

Abstract

Estimation of Reaction Forces for

an Elastically-supported Rigid Body by

Reconstructed Displacements Using

Measurements of Accelerometers and Potentiometers

SangBum Jeon

Civil and Environmental Engineering

The Graduate School

Seoul National University

This paper proposes a scheme for estimating the reaction forces of the rigid

body supported by the elastic supports for all frequency bands including both

quasi-static and dynamic components. Knowing the elastic supports reaction forces

is important for the maintenance of stability, serviceability and durability of the

rigid body. However, in order to directly measure the reaction forces of the elastic

support, not only the elastic support is damaged but also the maintenance is

difficult and the cost is expensive. To solve this problem, there is a previous study

in which the acceleration is measured by an accelerometer on the surface of the

rigid body, and then the elastic support reaction forces are estimated by using the

rigid body kinematics and the displacement reconstruction scheme. However, there

is a limitation that only dynamic components can be estimated and quasi-static

iv

components cannot be estimated. So, additional potentiometers are used to

compensate the quasi-static components that was lacking only by the

accelerometers and overcome the limitation, so that the elastic support reaction

forces can be estimated for all frequency bands. Applying the global positioning

system equation and the rigid body kinematics to the measured values of the

potentiometers and applying the rigid body kinematics to the measured values of

the accelerometers, the relative displacements and the relative accelerations at the

arbitrary point in the rigid body are estimated. The reconstructed displacement at

the arbitrary point is restored by the combine displacement reconstruction scheme

using not only accelerations but also displacements. If the arbitrary point is set to

the elastic supports, the reaction force can be estimated by using the displacement-

force relation at the elastic support and if the arbitrary point is set to the center of

gravity, the reaction force at the elastic support can be estimated by using

displacement-force relation at the center of gravity and the pseudo-inverse. In order

to verify the proposed scheme, numerical examples are constructed and applied.

Keywords : potentiometer, global positioning system, rigid body kinematics,

displacement reconstruction scheme, elastic support reaction force

Student Number : 2016-26162

v

Contents

Abstract .....................................................................................................iii

Contents......................................................................................................v

List of figures............................................................................................vii

List of tables ...............................................................................................x

1. Introduction ............................................................................................1

1.1 Motivation ......................................................................................1

1.2 Objectives.......................................................................................2

1.3 Organization ...................................................................................3

2. Rigid body motion ..................................................................................5

2.1 Kinematics of the rigid body...........................................................5

2.2 Kinetics of rigid body .....................................................................9

3. Estimation of 6-axis relative displacement and relative acceleration using measurements of potentiometers and accelerometers.............................13

3.1 Estimation of relative displacement using measurements of potentiometers ..............................................................................13

3.2 Estimation of the 6-axis relative behavior .....................................20

3.2.1 Estimation of the 6-axis relative displacement .....................20

3.2.2 Estimation of the 6-axis relative acceleration.......................22

4. Displacement reconstruction using measured displacements and accelerations .........................................................................................27

4.1 The combine displacement reconstruction scheme ........................27

4.2 Separation of the target frequency.................................................36

5. Estimation of reaction forces of elastic support for an elastically-supported rigid body .............................................................................41

vi

5.1 Estimation of the 6-axis force using displacement-force relation ...41

5.2 The displacement-force relation at the elastic support ...................42

5.3 Displacement-force relation at the center of rigid body .................44

6. Numerical verification...........................................................................49

6.1 Details of numerical simulation study ...........................................49

6.2 Numerical simulation study using elastic support displacement-force relation.................................................................................55

6.2.1 Estimation result of the rigid body rotational behavior .........55

6.2.2 Estimation result of the elastic support relative displacement and relative acceleration ................................56

6.2.3 Determination of major factors of the combine displacement reconstruction scheme ...................................56

6.2.4 Estimation result of the elastic support reaction force ..........57

6.3 Numerical simulation study using center of gravity displacement-force relation ...........................................................57

6.3.1 Estimation result of the rigid body rotational behavior .........58

6.3.2 Estimation result of center of gravity of the rigid body relative translation behavior................................................58

6.3.3 Determination of major factors of the combine displacement reconstruction scheme ...................................58

6.3.4 Estimation result of center of gravity of the rigid body 6-axis resultant force..............................................................59

6.3.5 Estimation result of the elastic support reaction force ..........59

vii

List of figures

Figure 2.1 Concept of rigid body motion.....................................................6

Figure 2.2 Components of rigid body kinematics ........................................7

Figure 2.3 Configuration of support excitation ..........................................10

Figure 3.1 Concept of 2-dimensional GPS equation ..................................15

Figure 3.2 Concept of potentiometers setting used in modified GPS equation.................................................................................................................17

Figure 3.3 Different displacement conditions at the same potentiometer measurements ..........................................................................18

Figure 3.4 Components of rigid body kinematics for the displacement ......21

Figure 3.5 Components of rigid body kinematics for the acceleration........23

Figure 4.1 Transfer functions of the displacement reconstruction ..............32

Figure 4.2 Frequency spectrum with noise in the middle band...................37

Figure 4.3 Independently determined transfer functions ............................37

Figure 4.4 Transfer functions with two different target frequencies ...........40

Figure 5.1 Relative displacement and relative acceleration at elastic support.................................................................................................................43

Figure 5.2 Relative displacement and relative acceleration at center of

gravity .....................................................................................45

Figure 5.3 Distributing the resultant force to reaction force at the elastic support ....................................................................................46

Figure 6.1 Details of the accelerometers and the elastic supports setting....50

Figure 6.2 Details of the potentiometers setting.........................................51

Figure 6.3 Translational acceleration and translation displacement of the base .........................................................................................53

viii

Figure 6.4 Angular displacement of the rigid body (numerical example 1 & 2) ........................................................60

Figure 6.5 Angular acceleration of the rigid body (numerical example 1 & 2).................................................................................................................61

Figure 6.6 Angular acceleration of the base (numerical example 1 & 2) ....62

Figure 6.7 Relative displacement at the elastic support1 (numerical example 1) .............................................................63

Figure 6.8 Relative displacement at the elastic support2 (numerical example 1) .............................................................64

Figure 6.9 Relative displacement at the elastic support3

(numerical example 1) .............................................................65

Figure 6.10 Relative acceleration at the elastic support1 (numerical example 1) ...........................................................66



Figure 6.13 Frequency spectrum and Target frequency at the elastic support1

(numerical example 1) ...........................................................69





Figure 6.16 Reaction force at x-axis of the elastic support


Figure 6.17 Reaction force at the elastic support1 (numerical example 1)..73



ix

Figure 6.20 Relative displacement at the center of gravity (numerical example 2) .............................................................76

Figure 6.21 Relative acceleration at the center of gravity (numerical example 2) ...........................................................77

Figure 6.22 Frequency spectrum and Target frequency of translation behavior at the center of gravity (numerical example 2) .........78

Figure 6.23 Frequency spectrum and Target frequency of rotation behavior

at the center of gravity (numerical example 2) .........................79

Figure 6.24 Pseudo-static force at center of gravity (numerical example 2)80

Figure 6.25 Translation force at center of gravity (numerical example 2)...81

Figure 6.26 Moment at center of gravity (numerical example 2)................82




x

List of tables

Table 6.1 Dimension of the rigid body and base size .................................50

Table 6.2 Position of the accelerometers....................................................50

Table 6.3 Position of the elastic supports...................................................50

Table 6.4 Position of the potentiometers (numerical example) ...................51

Table 6.5 Determine target frequency (numerical example 1) ....................57

Table 6.6 Determine target frequency (numerical example 2) ....................59

1

1. Introduction

1.1 Motivation

Knowing reaction forces generated in elastic supports supporting the rigid

body is very important factor for maintenance such as stability, usability and

durability of the rigid body. The reaction forces of the elastic supports can be

measured using a load-cell. However, when installing the load-cell, the elastic

supports are damaged due to cutting off the parts where the reaction forces are

measured and the installation is complicated and expensive.

To solve this problem, Kim (2016) developed a scheme for estimating the

elastic support reaction forces by installing accelerometers on the surface of the

rigid body and the base which is also a rigid body supporting the elastic supports.

Three or more 3-axis accelerometers are installed on the surface of the rigid body

and the base, and the angular accelerations of each of them is estimated by solving

optimization equation for rigid body kinematics. Estimate the relative accelerations

of both ends of the elastic supports from the measured accelerations and the

estimated angular accelerations using rigid body kinematics. By applying the

displacement reconstruction scheme (Hong 2010) to the relative accelerations, the

relative displacements of both ends of the elastic supports can be restored, and the

force generated in the elastic supports can be estimated by using the displacement-

force relation. This scheme can estimate the dynamic component of the reaction

force generated in the mount with considerable accuracy. However, since the

2

accelerometers cannot detect the quasi-static behavior of the rigid body and the

base, it is impossible to estimate the quasi-static reaction force generated in the

elastic supports.

1.2 Objectives

The main objective of this paper is to estimate not only dynamic components

but also quasi-static components of the reaction force. So, relative distance meters,

potentiometers, are added along with accelerometers to indirectly measure the

elastic support reaction forces both the quasi-static components and the dynamic

components. By measuring three or more relative distances between the rigid body

and the base by attaching one end of the three or more potentiometers to one object

point on the rigid body and the other end to the three or more different reference

point on the base, the relative displacements of the object point can be estimated

using the same principle as the global positioning system (GPS). When there are

three or more potentiometers object points on the rigid body, in the same way as

estimating the relative accelerations of the elastic supports, the relative

displacements of the elastic supports can be estimated after estimating the angular

displacements. So, the quasi-static components that Kim 2016 could not restored

are filled with the displacement measured by the potentiometers, and the combined

displacement including the quasi-static component and the dynamic component can

be restored.

The combine displacement reconstruction scheme (Hong 2013) is used to

3

restore the low-frequency component from the potentiometer and the high-

frequency component from the accelerometer. So, the combine displacement

reconstruction scheme complements the disadvantages of the displacement

reconstruction scheme used in Kim (2016), which only can restore the dynamic

displacements from the accelerations but not the quasi-static displacements. By

adding a new condition to determine the two target frequencies separately in the

combine displacement reconstruction scheme, suppressing noise in the middle band

of frequency spectrum is possible. Therefore, after restoring the displacements with

all frequency components without losing the meaningful frequency components, it

is possible to estimate the reaction forces including both the quasi-static

components and the dynamic components by using the force - displacement

relation.

1.3 Organization

This paper is composed of 7 chapters in total. In Chapter 2, the basic of rigid

body motion is explained. Kinematics of the rigid body must be preceded in order

to understand Chapter 3 and Chapter 4, and kinetics of rigid body is necessary in

the process of constructing the numerical examples. Chapter 3 shows the process of

estimating the 6-axis relative displacements and the accelerations using the

potentiometers and the accelerometers. First, estimate the displacements at the

object point on the surface of the rigid body using the GPS equation with the

potentiometer measurements. Second, the estimated displacements can be used to

4

estimate the angular displacements of the rigid body and the relative displacement

at arbitrary point using the rigid body kinematics. Also, the measured accelerations

on the surface of the rigid body can be used to estimate the angular accelerations of

the rigid body and the relative acceleration at arbitrary point using the rigid body

kinematics. Chapter 4 describes the combine displacement reconstruction scheme

that combines the relative displacements and the relative acceleration estimated in

Section 3. The combine displacement reconstruction scheme has two transfer

functions to restore the displacement from both the displacements and the

acceleration. The transfer functions share different target frequencies to reduce the

noise between quasi-static components and the dynamic components. Chapter 5

introduces two methods for estimating the elastic reaction forces using the restored

displacements and Chapter 6 verifies the above scheme through numerical

examples. Lastly, in Chapter 7, the conclusion is presented.

5

2. Rigid body motion

In this chapter, kinematics and kinetics of the rigid body are briefly described

(Hibbeler, 2004) as these are already established in Kim 2016 even more details,

such as 3-deminsion rotation quaternion. Described equations in this chapter will

be used in next chapters while developing and verifying the scheme. Here, the way

to estimate the angular displacement and acceleration using kinematics of the rigid

body is derived. Kinetics of the rigid body is also derived by using kinematics of

the rigid body and Euler- Lagrange equation on excitation condition.

2.1 Kinematics of the rigid body

A rigid body is a structure that has no deformation or it is negligible. Also, the

rigid body has a constant distance between points at any points in the structure for

all times when external forces act on the structure. In this study, if the deformation

of the structure supported by the elastic support is negligibly smaller than the

deformation of the elastic support, the structure can be assumed as the rigid body.

If the rigid body performs, rigid body motion, the translational motion and the

rotational motion based on the reference point O as shown in Figure 2.1, the

displacement after the rigid body motion at an arbitrary point i is expressed as

follows.

θrurθuuuu iOOiOOOi~-=´+»+= q (2.1)

6

where iu , Ou , qu , θ , ´, ijr and ijr~ denote displacement of the point i ,

translational displacement of the reference point O , angular displacement due to

rigid body rotation based on the reference point O , angular displacement, outer

product operator, relative position vector of any point i and j in the rigid body

and outer matrix of vector ijr , respectively. Knowing the position of the arbitrary

point and the reference point in the rigid body and knowing the translation and

rotational motions of the 3-axis, the motion of the any point in the rigid body can

be described. Here, the angular displacement should be °£q 10 , which can

approximate the trigonometric function by the first-order Taylor approximation.

Therefore, by using Eq. (2.1), if there are measured displacements on the surface of

the rigid body, the displacements at any point in the rigid body can be estimated.

However, since the angular displacements in Eq. (2.1) are not measured values, the

angular displacements must be calculated in order to use rigid body kinematics.

Figure 2.1 Concept of rigid body motion

iOr

z

x

y

i

i

O O

Ou

qu

iiu

bodyRigid

7

Figure 2.2 Components of rigid body kinematics

When the displacements are measured at two arbitrary measurement points i

and j as shown in Figure 2.2, the relation of the two points is as follows according

to the rigid body kinematics.

θru

rθuu

ijij

ijji

~-=

´+=(2.2)

where iu and iju denote measured displacement at the point i and relative

displacement between iu and ju , respectively. Here the matrix ijr~ consisting

of the relative position vectors of the two measurement points is a singular matrix

that it is impossible to calculate the angular displacement with only two

measurements points where r~ is defined as follows.

úúú

û

ù

êêê

ë

é

-

-

-

=

0

0

0~

xy

xz

yz

rr

rr

rr

r (2.3)

Therefore, to calculate the angular displacement of the rigid body, the optimization

equation based on the rigid body kinematics of Eq. (2.2) is derived as follows.

θ

i

jju

iu

bodyRigid

8

å +=Pji

ijij,

2

2

~2

1Min θru

θ&&(2.4)

where P denotes the objective function of the optimization equation. The

following equation is derived by applying the variance to Eq. (2.4).

0)~~~(

)~(~)(

,'

,

,

=+d=

+d=Pd

åå

å

jiij

T

ijji

ij

T

ijT

jiijij

T

ijT

θrrurθ

θrurθθ

(2.5)

Eq. (2.5) should be established for all θd . So, the final equation for estimating the

angular displacement is as follows.

0~)~~(,

1

,åå --=

jiij

T

ijji

ij

T

ij urrrθ (2.6)

The Eq. (2.4) is a quadratic problem that Eq. (2.6) can be also derived by using

primary requirement 0/ =P θdd . Here, at least three points, each point with 3-

axis translational measured displacements, should be known to solve Eq. (2.6).

Otherwise, the rank becomes insufficient and cannot solve Eq. (2.6). Also, by doing

same procedure with the measured accelerations, the same result will be derived.

0~)~~(,

1

,åå --=

jiij

T

ijji

ij

T

ij urrrθ &&&&(2.7)

where iu&& and θ&& denote measured acceleration at the point i and angular

acceleration, respectively.

9

2.2 Kinetics of rigid body

The rigid-body kinetics is necessary to understand the process of constructing

the numerical examples and to understand the resultant force at the center of

gravity in Section 6.3 of Chapter 6. The rigid body kinetics is derived using

kinematics of the rigid body and the Euler-Lagrange equation. The Euler-Lagrange

equation is differential equations that deal with the optimization of a functional

with a function and its derivatives as a domain as follows.

( ) ( )ò=P dttF ,Min ff,f & (2.8)

where P , F , f and f& denote objective function of the Euler-Lagrange

equation, functional, vector function defined in time domain t and derivatives of

the vector functions, respectively. The Euler-Lagrange equation for calculating

kinetics of the rigid body is the function of the total energy, which is the sum of

potential energy and kinetic energy as follows.

)()( ff &KP EEF += (2.9)

where )(fPE and )(f&KE denote potential energy and kinetic energy,

respectively. Here, the potential energy is defined by the 6-axis displacement and

the kinetic energy is defined by the 6-axis velocity.

When excitation occurs by the rigid body2 as shown in Figure 2.3, the

potential energy is the sum of potential energy at each elastic support as follows.

10

Figure 2.3 Configuration of support excitation

å=

--=noE

ibai

TbaP iiii

E1

)()(2

1uukuu (2.10)

where noE and ik denote total number of elastic support and stiffness of i-th

elastic support, respectively. The displacement of the elastic supports is described

with respect to the center of gravity of the rigid body as follows.

å=

+--+--=noE

iGbGaGGi

TGbGaGGP iii

E1

2121 )~~()~~(2

1212121121θrθruukθrθruu (2.11)

The kinetic energy is equal to the sum of kinetic energy of mass elements.

Therefore, the kinetic energy of an arbitrary element point i inside the rigid body is

integrated as follows.

òò ×+×=21 2

1M jjjM iiiK dmdmE uuuu &&&& (2.12)

where iu& and dm denote velocity of the arbitrary element point i and mass

1θ

1G

2G

11Gar12Gar

21Gbr

22Gbr1a

2a

1b

2b

body1Rigid

body2Rigid

2θ

11

element of the rigid body, respectively. 1M and 2M denote total mass of rigid

body1 and rigid body2 in Figure 2.3, respectively. The velocity at the arbitrary

element point is described with rigid body kinematics respect to the center of

gravity of the rigid body as follows.

222111

2211

222111

22222

11111

2

1

2

1

2

1

2

1

)~()~()~()~(2

1

θIθuMuθIθuMu

θruθruθruθru

&&&&&&&&

&&&&&&&&

GT

GTGG

TG

TG

MjjGGjGG

MiiGGiGGK dmdmE

+++=

-×-+-×-= òò(2.13)

where GI denotes moment of inertia which is defined as follows.

i

iGiGiGiGiGiG

iGiGiGiGiGiG

iGiGiGiGiGiG

G dm

yxzyzx

zyzxyx

zxyxzy

òúúú

û

ù

êêê

ë

é

+--

-+-

--+

=22

22

22

)()(

)()(

)()(

I (2.14)

The final rigid body kinetics is derived by applying the Euler-Lagrange

equations to the total energy.

( ) ( )

ff

f

ff,

f

ff,

¶

+¶+

¶

+¶=

=¶

¶+÷÷

ø

öççè

æ

¶

¶=Pd

)()

)((

0,,

SKSK EEEE

dt

d

tFtF

dt

d

&

&

&

&

(2.15)

Since the potential energy does not exist for the velocity vector, the result of the

partial differentiation with respect to the velocity is 0. Also, the kinetic energy does

not have a variable for the displacement vector that the result of partial

12

differentiation for displacement is 0.

0)( =¶

¶+

¶

¶=Pd

ffPK EE

dt

d&

(2.16)

The final matrix equation of the rigid body kinetics is obtained by substituting the

results of Eq. (2.11) and Eq. (2.13) into Eq. (2.16) as follows.

ext

noE

i G

G

GbiTGbi

TGGai

TGbi

TGb

GbiiGaii

GbiTGai

TGaGai

TGai

TGa

GbiiGaii

G

G

G

G

iiiiii

ii

iiiiii

ii

F

θ

u

θ

u

rkrkrrkrkr

rkkrkk

rkrkrrkrkr

rkkrkk

θ

u

θ

u

I

M

I

M

=

÷÷÷÷÷

ø

ö

ççççç

è

æ

úúúúú

û

ù

êêêêê

ë

é

--

--

--

--

+

÷÷÷÷÷

ø

ö

ççççç

è

æ

úúúúú

û

ù

êêêêê

ë

é

å=1

2

1

2

1

2

1

2

1

222122

21

211111

21

2

1

2

1

~~~~~~

~~

~~~~~~

~~

000

000

000

000

&&

&&

&&

&&

(2.17)

In the support excitation problem, the rigid body 2 is excited to obtain the

displacement, velocity, and acceleration of rigid body 1. Here,2Gu and 2θ are

prescribe values and the variation is zero. Therefore, when rigid body 2 is exciting,

kinetics of the rigid body derive as follows

å

å

=

=

÷÷ø

öççè

æúû

ùêë

é

-

-=

÷÷ø

öççè

æúû

ùêë

é

-

-+÷÷

ø

öççè

æúû

ùêë

é

noE

i

G

GbiTGai

TGa

Gbii

noE

i

G

GaiTGai

TGa

GaiiG

G

iii

i

iii

i

1 2

1 11

1

2

211

2

1

111

11

1

~~~

~

~~~

~

0

0

θ

u

rkrkr

rkk

θ

u

rkrkr

rkk

θ

u

I

M

&&

&&

(2.18)

13

3. Estimation of 6-axis relative displacement and relative acceleration using measurements ofpotentiometers and accelerometers

In this chapter, 6-axis relative behavior at the arbitrary point of the rigid body is

estimated using measured displacements and accelerations. Potentiometers measure

relative distance that GPS equation is used to estimate the displacement. So, GPS

equation needs modification to utilized in the condition of potentiometers. With the

estimated displacement the 6-axis relative displacement can be estimate by using

kinematics of the rigid body. In the case of 6-axis relative acceleration, the

measured acceleration is applied to kinematics of the rigid body for estimation.

Here, estimating the 6-aixe relative acceleration using measurements of

accelerometers is also described in Kim 2016.

3.1 Estimation of relative displacement using measurements of potentiometers

The concept of GPS equation needs to be preceded to estimate displacement

using the potentiometers. The GPS equation is a method of calculating a position of

a measuring object point using measurements, change of distance, between the

several measuring reference points and the measuring object point. Here the

position of the measuring object point is solved by optimization problem that is

defined with relative distance-displacement relation (Blewitt 1997). In GPS, the

measuring reference points are satellites. Potentiometers are similar to GPS since

14

they measure relative distances between the measuring reference points and the

measuring object point where potentiometers are attached. So, the GPS equation

can be used to estimate the position of the measuring object point with

potentiometers.

Figure 3.1 helps to understand the concept of GPS in two-dimensional. If there

are no errors, three circles meet at one point, object point, so position of the point

can estimated by solving nonlinear equation. However, if there are errors in the

measurement values or the position of the potentiometers, the three circles do not

meet at the one point, so that the problem cannot solve by equation. So, the GPS

equation is defined as the optimization problem as follows.

å

ååå

--+-+-=

--=-=P=P

iAiiAiAiA

iAiiA

iAiAi

iitotal

dzzyyxx

ddA

2222

22

22

2

))()()((

)||(||)||(||min rrrr

(3.1)

where totalÕ , iÕ , TAAAA zyx ][=r , T

iiii zyx ][=r , Aid denote total

objective function, objective function of the i-th potentiometer, position vector of

measuring object point A(unknown), the position vector of i-th potentiometer, and

relative distance between the point A and i-th potentiometer, respectively.

Theoretically, Eq. (3.1) has one solution when three or more and four or more

potentiometers are used in the 2-dimensional and 3-dimensional cases, respectively.

If two or three potentiometers are used for 2-dimensional and 3-dimensional,

respectively, the optimization problem is likely to converge to local minimum.

However, the solution can calculate by setting an initial value close to the solution.

15

Figure 3.1 Concept of 2-dimensional GPS equation

Since there is an absolute value inside Eq. (3.1), it is a nonlinear optimization

problem. Most of the nonlinear optimization equations are solved using Newton-

Raphson method, though Eq. (3.1) has nonlinear least squared error from that it can

be solved by the Gauss-Newton method, which has a relatively small amount of

computation and a stable convergence when the following condition is satisfied

(Bjorck 1996).

AiAA d<<- 0rr (3.2)

where 0Ar denotes initial value to use in Gauss-Newton method. So, if the error

between the initial value and actual solution is very small compare to the relative

distance, Gauss-Newton method can be used. The potentiometers provide higher

measurement frequency than the rigid body’s vibration frequency that )(tAr and

16

)( ttA D-r have almost the same value. Therefore, the condition for the Gauss-

Newton method is fully satisfied when )(0 ttAA D-= rr . Eq. (3.1) can be solved by

repeating the following equation.

iTiiTiiA

iA ΠJJJrr 11 )( -+ -= (3.3)

where iAr ,

iJ and iΠ denote Ar value of i-th iteration, jacobian matrix of iAr

and iP vector with iAr , respectively.

To estimate displacement on the surface of the rigid body, three or four

potentiometers are need. One end of potentiometers should attach to the base, the

measuring reference points, separately and another end should attach to the

measuring object point of the rigid body. It should be noted that if four

potentiometers are used in one set, the attachment points on the base must be

determined because when the four attachment points are on one plane it gives the

same result as using three points. With the set of potentiometers, the displacement

at measuring object point can be estimated by applying modified GPS equation.

In GPS, the measuring reference points are fixed because it uses geostationary

satellites as the measuring reference points to measure the relative distances.

However, the potentiometers measuring reference points are moving because the

reference points move as the base move since those are attached to the base. So, to

use the potentiometers on GPS equation, the displacements at the reference points

of the potentiometers need to be considered.

17

å=

-+-=Pnop

jijij

pij

pi

pi d

1

22 )||(||)(min ρUuu (3.4)

where piu , p

ijU , ijρ , ijd and nop denote displacement of a i-th set

potentiometer at attachment point of rigid body side (measuring object point),

displacement of j-th potentiometer in i-th set at attachment point of base

(measuring reference point), initial relative position vector of the two attachment

points, relative distance measured by the j-th potentiometer in the i-th set and

number of potentiometers in the i-th set, respectively. The superscript p means the

potentiometer related variance. Variations in lower case and in upper case were

used to indicate the variance associated with the rigid body and the base,

respectively. In Figure 3.2, it shows the symbols used in Eq. (3.4) and can be seen

that the displacements of the potentiometers attached to the base is performing

rigid body motion as the base is assumed to be a rigid body. So, pijU is expressed

with rigid body kinematics for the arbitrary reference point of the base as follows.

Figure 3.2 Concept of potentiometers setting used in modified GPS equation

θ

Base

bodyRigid

OUO

pjU

pjR

ijρ

piu

Θ

Elastic support

Potentiometer

18

å=

-+´+-=Pnop

jijij

pij

Opi

pi d

1

22 )||)((||)(min ρRΘUuu (3.5)

where OU , Θ and pijR denote displacement of the arbitrary reference point of

the base, angular displacement of the base and relative position vector between the

arbitrary reference point of the base and the j-th attachment point of potentiometer

on the base in the i-th set, respectively. In Eq. (3.5), the two relative position

vectors are the values that can be obtained by knowing the position of the

potentiometers and the arbitrary reference point of the base and the relative

distances are measured by the potentiometers. However, OU and Θ are values

that cannot be measured by the potentiometers. So, if two or more of the unknown

piu , OU and Θ are not determined, the optimization equation will have

innumerable solutions. From Figure 3.3 it can be easily explain. When the left

figure translates and rotates to the right figure, piu , OU and Θ have different

values even if 1)( ijd and 2)( ijd are the same. In this case, the displacement due

to the rigid body motion cannot be estimated only with the potentiometers.

Figure 3.3 Different displacement conditions at the same potentiometer measurements

1)(Θ

Base

1)( OU

1)( piu

1)( ijd

bodyRigid

2)(Θ2)( piu

Base

2)( OU

2)( ijd

bodyRigid

19

In order to solve the problem, first, piu and OU are changed to relative

displacement as follows.

å=

-+-=Pnop

jijij

uip

uip d

1

22,, )||(||)(min ρΘΔΔ (3.6)

where Op

iu

ip UuΔ -=, denotes relative displacement between the arbitrary

reference point of base and the measuring object point. Here, the dynamic

components of the angular displacement of the base are not considered because

those will be removed by the combine displacement reconstruction scheme that

will describe later. Also quasi-static components of the angular displacement of the

base can be removed because in case of earthquake, the angular displacement is

negligibly smaller than the translation. So, by assuming Θ is 0, the following

optimization equation is determined as follows.

å=

-+=Pnop

jijij

uip

uip d

1

22,, )||(||)(min ρΔΔ (3.7)

Eq. (3.7) is the same form as Eq. (3.1) that it can be solved by Gauss-Newton

method, unlike Eq. (3.5).

20

3.2 Estimation of the 6-axis relative behavior

3.2.1 Estimation of the 6-axis relative displacement

By estimating the three or more 3-axis relative translational displacements on

the rigid body surface using three or more sets of the potentiometers, the angular

displacement of the rigid body can be estimated. Using the angular displacement

and the 3-axis relative translational displacements on the rigid body surface, the

relative displacement of any point in the rigid body can be estimated. Since the

angular displacement of the rigid body is the same at all points, the 6-axis relative

displacement of the arbitrary point is known.

The displacement of the rigid body can be expressed by applying rigid body

kinematics Eq. (2.2) as follows.

θrurθuu ppij

pj

ppij

pj

pi

~-=´+= (3.8)

where ppijr denotes initial relative position vector between j-th and i-th set of

potentiometers measuring object points, respectively. The left figure on Figure 3.4

gives a better understanding of the symbols used in Eq. (3.8). Since, Eq. (3.7)

estimates the relative displacement based on the arbitrary object point of the base,

Eq. (3.8) can be written as follows for the relative displacement.

0~~)()(~,, =+-=+---=+- θrΔΔθrUuUuθruu pp

iju

jpu

ippp

ijOp

jOp

ipp

ijpj

pi (3.9)

The angular displacement of the rigid body can be estimated by defining the

21

following optimization equation with the angular displacement of the rigid body as

unknown. Here, the relative displacements measured by the potentiometers and the

position vectors of each relative displacement are known

å +-=Pji

ppij

ujp

uip

,

2

2,,~

2

1)(Min θrΔΔθ (3.10)

The following equation is derived by applying the variance to Eq. (3.10) which is

the same procedure as Eq. (2.6).

åå -úû

ùêë

é-=

-

ji

ujp

uip

Tij

ji

ppij

Tppij

,,,

1

,

)(~~)~( ΔΔrrrθ (3.11)

The relative translational displacements at the arbitrary point of the rigid body

can be estimated as follows. Here, the angular displacements and the translational

displacements at the object points are applied to the rigid body kinematics.

Figure 3.4 Components of rigid body kinematics for the displacement

pir

piu

Ou

bodyRigid

θ

BaseOU

θ

pju pp

ijr piu

OU

bodyRigid

Elastic support

Potentiometer

Base

22

åå==

´-=-´-=-=nod

i

pi

uip

nod

i

Opi

pi

OOuO

nodnod 1,

1

)(1

])[(1

rθΔUrθuUuΔ (3.12)

where uOΔ , Ou , p

ir and nod denote relative displacement at the arbitrary point

of the rigid body based on arbitrary reference point of the base, displacement of the

arbitrary point of the rigid body, relative position vector between the arbitrary point

of rigid body and the displacement measuring object point of i-th set and total

number of object points to measure displacement, respectively. The right figure of

Figure 3.4 clearly shows the symbols used in Eq. (3.12). Since the angular

displacement is the same at any points in the rigid body, the six-axis displacement

at any point in the rigid body can be known. Since there are several object points

on the rigid body, it is possible to reduce measured error by averaging the

estimated relative displacements.

3.2.2 Estimation of the 6-axis relative acceleration

The acceleration of the rigid body and the base can be express by the

following rigid body kinematics.

aaij

aj

ai rθaa ´+= && (3.13a)

where aia , θ&& and aa

ijr denote measured 3-axis acceleration with the i-th

accelerometer attached on the rigid body, angular acceleration of the rigid body and

relative position vector of the i-th accelerometer based on the j-th accelerometer

23

attached on the rigid body, respectively.

aaij

aj

ai RΘAA ´+= && (3.13b)

where aiA , Θ&& and aa

ijR denote measured 3-axis acceleration with the i-th

accelerometer attached on the base, angular acceleration of the base and relative

position vector of the i-th accelerometer based on the j-th accelerometer attached

on the base, respectively. The left figure on Figure 3.5 gives a better understanding

of the symbols used in Eq. (3.13a) and (3.13b). The accelerometers measure

absolute acceleration, unlike potentiometers. So, the angular acceleration of the

rigid body and the base can be estimated respectively and the 6-axis acceleration at

the arbitrary point can also be estimated.

The angular acceleration can be estimated by the process similar to that for

estimating the angular displacement. So, following optimization problem can be

defined.

Figure 3.5 Components of rigid body kinematics for the acceleration

θ&&

Θ&&

aaijR

ajA

aiA

aja

aaijr

aia bodyRigid

Base

θ&&

Θ&&

aiR

aiA

aira

ia

Elastic support

Accelerometer

Oa

OA

bodyRigid

Base

24

å +-=Pji

aaij

aj

ai

,

2

2

~2

1)(min θraaθ &&&& (3.14a)

å +-=Pji

aaij

aj

ai

,

2

2

~

2

1)(min ΘRAAΘ &&&& (3.14b)

The following equations are derived by applying the variance to Eq. (3.14a) and Eq.

(3.14b) which is the same procedure as Eq. (2.7).

åå -úû

ùêë

é-=

-

ji

aj

ai

Taaij

ji

aaij

Taaij

,

1

,

)()~(~)~( aarrrθ&& (3.15a)

åå -úû

ùêë

é-=

-

ji

aj

ai

Taaij

ji

aaij

Taaij

,

1

,

)()~

(~

)~

( AARRRΘ&& (3.15b)

Using the angular acceleration estimated by Eq. (3.15a) and (3.15b) and the

measured acceleration, the translational acceleration at any point can be estimated,

which is the same as knowing the 6-axis acceleration.

å=

´-=noa

i

ai

ai

O

noa 1

)(1

rθaa && (3.16a)

where Oa , air and noa denote acceleration at the arbitrary point of the rigid body,

relative position vector of the i-th accelerometer based on the arbitrary point of the

rigid body and number of accelerometers attached on the rigid body.

å=

´-=NOA

i

ai

ai

O

NOA 1

)(1

RΘAA && (3.16b)

25

where OA , a

iR and NOA denote acceleration at the arbitrary point of the base,

relative position vector of the i-th accelerometer based on the arbitrary point of the

base and number of accelerometers attached to the base. The right figure in Figure

3.5 gives a better understanding of the symbols used in Eq. (3.16a) and (3.16b). To

reduce the measured error, the average of estimated accelerations is used. The

relative acceleration at the rigid body arbitrary point can be estimated under the

same condition used in Eq. (3.12) as follows.

OOaO AaΔ -= (3.17)

where aOΔ denotes relative acceleration at the arbitrary point of the rigid body

based on the arbitrary point of the base.

27

4. Displacement reconstruction using measured displacements and accelerations

In the study of Kim (2016), the displacement reconstruction scheme (Lee 2010,

Hong 2010) is used to reconstruct dynamic displacement from measured

acceleration. However, the displacement reconstruction scheme has a disadvantage

that it cannot restore the quasi-static displacement. To overcome the disadvantage,

the combine displacement reconstruction scheme (Hong 2013) was developed. The

combine displacement reconstruction scheme restores accurate displacements in

entire frequency band by combining the advantages of the displacements restored

by measurements of the accelerometer and the potentiometer. Since displacements

restored by the accelerometer and the displacements measurements have high

accuracy in high frequency and low frequency component, respectively, it is

possible to accurately restore the displacement for all frequency bands.

4.1 The combine displacement reconstruction scheme

The combine displacement reconstruction scheme is defined by the following

optimization equation.

òò -b

+-=P2

1

2

1

22

2

2

2

)(2

)(2

1)(Min

T

T

T

Tdtuudta

dt

udu (4.1)

where u , u , a , b , 1T and 2T denote displacement to be reconstructed,

28

measured displacement, measured acceleration, regularization factor, integral start

time and integral end time, respectively. When 0=u , it is the same as the

displacement reconstruction scheme used in Kim(2016). Taking the variance in Eq.

(4.1) and solving the primary requirement satisfies the following equation.

0)()(2

1

2

1

2

2

2

2

2

=-db+-d

=Pd òòT

T

T

Tdtuuudta

dt

ud

dt

ud(4.2)

By doing the integration by parts the above equation twice leads to the following

equation

0)()(

)(

2

1

2

1

2

1

3

3

2

2

2

2

22

4

4

=-d--d

+

b--b+dòT

T

T

T

T

T

dt

ad

dt

udua

dt

ud

dt

ud

dtudt

adu

dt

udu

(4.3)

Since the integral term of Eq. (4.3) must be establish for all ud , the final

governing equation and boundary condition are derived as follows.

212

2

22

4

4

where TtTudt

adu

dt

ud<<b+=b+ (4.4a)

213

3

2

2

,atand TTtdt

ad

dt

uda

dt

ud=== (4.4b)

In Eq. (4.4a) and (4.4b), the reconstructed displacement is consists of the two the

displacements as follows.

29

da uuu += (4.5)

where au and du denote reconstructed displacement restored by the measured

acceleration and displacement, respectively. The reconstructed displacement of the

governing equation is divided into two equations for measured acceleration and

displacement as follows.

2

22

4

4

dt

adu

dt

uda

a =b+

uudt

udd

d 22

4

4

b=b+ (4.6)

Fourier transforming Eq. (4.6) derives the following transfer function.

)()()()(

)())(

()(

24

2

24

4

24

2

224

4

uFHuFHuFuF

uFaF

uF

daaa +=b+w

b+

b+w

w=

b+w

b+

w-

b+w

w=

(4.7)

where w , au and )(·F denote angular frequency, displacement estimated by

integrating measured acceleration twice and Fourier transform, respectively. aH

and dH denote the transfer functions of the acceleration and the displacement,

respectively and are given as follows.

24

4

24

4

)2(

)2(

b+p

p=

b+w

w=

f

fHa

24

2

24

2

)2( b+p

b=

b+w

b=

fHd (4.8)

The transfer functions of the acceleration and the displacement are monotone

30

increasing and decreasing function, respectively. Since the transfer functions have

only one undefined coefficient which is regularization factor, it is possible to

determine regularization factor by determining a frequency and the transfer

function value at the frequency for one transfer function as follows.

T

T

TTa

f

ffH a=

b+p

p=

24

4

)2(

)2()( (4.9)

where Tf and Ta denote target frequency and target accuracy, respectively. The

target frequency is the lowest meaningful frequency in the measured acceleration

signals. The target accuracy is the transfer function value of the signal

corresponding to the target frequency. It is best to set the target accuracy as 1, but

the range of the transfer function must be less than 1. Therefore, it is generally

recommended to use target accuracy as 0.99, 0.97 and 0.95, close to 1. The

regularization factor is derived from Eq. (4.9) as follows.

2)2(1

T

T

T fpa

a-=b (4.10)

Normalization of the transfer function to the target frequency can reveal the

general characteristics of the combine displacement reconstruction scheme. By

dividing Eq. (4.8) by the forth power of the target frequency, the normalized

transfer function is derived as follows.

31

44

4

~

~

)~

(l+

=f

ffHa

44

4

~)~

(l+

l=

ffHd (4.11)

where f~

and l denote normalized frequency and regularization factor for

normalized transfer function, respectively. The two coefficients can be expressed as

follows.

Tf

ff =~

T

T

Tf a

a-=

p

b=l

1

)2( 2

2(4.12)

Figure 4.1 shows Eq. (4.11) for target accuracy of 0.97. The sum of the

acceleration transfer functions and the displacement transfer function is 1 in entire

frequency band regardless of the target accuracy. This is because the low frequency

components, which is lacking in the displacement reconstruction scheme used to

recover the dynamic displacement in the study of Kim (2016), is compensated by

the measured displacement. In other words, the combine displacement

reconstruction scheme completely restores the all frequency range of the restored

displacement by mixing the low frequency components of the measured

displacement and high frequency components of the measured acceleration.

The measured displacement and acceleration consists of finite discretized

information. Therefore, the measured displacement and acceleration can be applied

only by discretizing the combine displacement reconstruction scheme defined in

continuous time. So, Eq. (4.6) is discretized into finite elements after the variance

as follows.

32

Figure 4.1 Transfer functions of the displacement reconstruction

å òå ò= D= D

d=db+

d a

a

a

a

k

e t

eea

k

e t

ea

ea

ea

ea dta

dt

uddtuu

dt

ud

dt

ud 2

12

22

1

2

2

2

2

2

)( (4.13a)

where eau , ak2 and atD denote au of e-th element, number of discretized

elements of the acceleration in a time-window and time interval of the acceleration

measurement, respectively.

å òå ò= D= D

db=db+d d

d

d

d

k

e t

eed

k

e t

ed

ed

ed

ed dtuudtuu

dt

ud

dt

ud 2

1

22

1

2

2

2

2

2

)( (4.13b)

where edu , dk2 and dtD denote du of e-th element, number of discretized

elements of the displacement in a time-window and time interval of the

displacement measurement, respectively. Although the frequencies of the measured

displacement and acceleration are different, the time-window size should be same

0.0

0.2

0.4

0.6

0.8

1.0

1.2

10-2 10-1 100 101 102

Hd (aT = 0.97)

Ha (aT = 0.97)

Hd + Ha

Tra

nsfe

r fu

ncti

ons

Normalized frquency, f

Displ. dominant

Transient Acc. dominant

~

33

as follows.

T

wddaaw

f

Ntktkd =D=D= 22 (4.14)

where wd and wN denote window size and time-window size normalized to the

target frequency, respectively. In general, the sampling rate of the measured

acceleration is an integral times of the sampling rate of the measured displacement.

The reconstructed displacement is interpolated with Hermitian shape function

that can be differentiated twice and measured displacement and acceleration are

interpolated with the linear shape function. So, Eq. (4.13a) and (4.13b) are

approximated by the shape functions as follows.

aQuMK aaaaaa tt 242 )())(( D=Db+

aCaQMKu aaaaaaaa ttt 21422 )())(()( D=Db+D= -(4.15a)

where au denotes reconstructed displacement vector restored by the measured

acceleration vector a .

uQuMK dddddd tt 4242 )())(( Db=Db+

uCuQMKu dddddddd ttt 4214242 )())(()( Db=Db+Db= -(4.15b)

where du denotes reconstructed displacement vector restored by the measured

displacement vector u . The matrices in Eq. (4.15a) and (4.15b) are defined as

follows.

34

åò xxx

=e

HTH

i dd

d

d

d1

0 2

2

2

2 NNK daid

eH

THi ,for

1

0=x=åò NNM

åò xx

=e

L

TH

a dd

d1

0 2

2

NN

Q åò x=e

LTHd d

1

0NNQ

(4.16)

where HN and LN denote Hermitian shape function and linear shape function,

respectively. Using the time-window concept in Eq. (4.15a) and (4.15b), the

following impulse response function (FIR) filter can be derived.

å-=

++ D+D=D=a

a

a

k

kpaakpaaaa tptactttu )()()()()( 1

22 ac (4.17a)

where ac and aic )( denote )12( +ak th row of the aC matrix and i-th

column of the ac , respectively.

å-=

++ D+Db=Db=d

d

d

k

kpddkpdddd tptuctttu )()()()()( 1

4242 uc (4.17b)

where dc and dic )( denote )12( +dk th row of the dC matrix and i-th

column of the dc , respectively. The two reconstructed displacements restored by

Eq. (3.17a) and (3.17b) have different sampling rates. Therefore, in order to make

the two sampling rates the same, the displacement with low sampling rate is

interpolated and makes it equal sampling rate to that of the displacement with high

sampling rate. So, the combine displacement reconstruction scheme can be useful

when the displacement of high frequency is needed since the scheme combines the

35

frequency of measured displacement equal to the frequency of measured

acceleration. The combine reconstructed displacement can be written as follows.

edaHaaadaaa tutututu uN )()()()()( x+=+= (4.18)

where ax denotes natural coordinate corresponding to t in e-th element. Here, the

parameters such as the target frequency, the target accuracy, the time-window,

sampling rate or frequency of the measured displacement and the measured

acceleration should be determined.

As mentioned above, the target frequency is determined as the lowest

meaningful frequency in the measured acceleration signal. The target accuracy is

set to 0.99, 0.97 and 0.95, but if there are a lot of noises just below the target

frequency, use lower target accuracy, otherwise use higher target accuracy. It is

recommended to use a normalized time-window size of 5 or more. The frequency

of measured acceleration is recommended to be at least 10 times the highest

meaningful frequency of measured acceleration. The frequency of measured

displacement should be at least twice as high as the target frequency, and if it is

over 4 times, stable results can be estimated.

36

4.2 Separation of the target frequency

As mentioned, the combine displacement reconstruction scheme has a transfer

function of 1 in all frequency domains. This means the combine displacement

reconstruction scheme preserves all signals, but at the same time there is no noise

suppression. Therefore, if there are a clear noises frequency band between the

dynamic signal and the quasi-static signal in the acceleration and displacement

frequency spectrum, the middle band in frequency spectrum, as in Figure 4.2, the

combine displacement reconstruction scheme passes all of these noises. Since the

two transfer functions in Eq. (4.11) share the regularization factor, the displacement

transfer function is automatically determined when the acceleration transfer

function is determined. So, if the two transfer functions can be defined

independently, it will be possible to suppress noise such as the middle band noise

in Figure 4.2. The regularization factor of the normalized transfer function in Eq.

(4.11) is the function of the target accuracy of the measured acceleration transfer

function as shown in Eq. (4.12). If the target accuracy is separately determined for

each transfer function, the sum of the transfer functions can be set to other than 1

as shown in Figure 4.3. If the target accuracy of the measured displacement transfer

function is less than the target accuracy of the measured acceleration transfer

function, then the sum of the two transfer functions is greater than 1 in the transient

domains. On the other hand, if the target accuracy of the measured displacement

transfer function is greater than the target accuracy of the measured acceleration

transfer function, then the sum of the two transfer functions is less than 1 in the

37

transient domains. However, in the domain where the normalized frequency is

greater than 1 or less than 0.1, the sum of the two transfer functions is close to 1, so

meaningful signals are not suppressed. Therefore, be separating the regularization

factor independently that define the two transfer functions in Eq. (4.11), the

combine displacement reconstruction scheme can suppress the middle band noises

and preserves the dynamic and quasi-static signals.

Figure 4.2 Frequency spectrum with noise in the middle band

Figure 4.3 Independently determined transfer functions

0.0

2.0

4.0

0.0

0.4

0.7

10-2

10-1

100

101

102

Acceleration

Displacement

Spe

ctru

m o

f A

ccel

erat

ion

Spectrum

of Displacem

ent

Frquency (Hz)

Middle band noise

0.0

0.2

0.4

0.6

0.8

1.0

1.2

10-2

10-1

100

101

102

Hd (ad

T = 0.97)

Hd (ad

T = 0.95)

Ha (aa

T = 0.97)

Ha (aa

T = 0.95)

Hd (ad

T = 0.97) + Ha (aa

T = 0.95)

Hd (ad

T = 0.95) + Ha (aa

T = 0.97)

Tra

nsfe

r fu

ncti

ons

Normalized frquency, f~

38

The regularization factor that defines the two transfer functions in Eq. (4.11)

can be defined independently as follows.

44

4

~

~

)~

(a

af

ffH

l+=

44

4

~)~

(d

dd

ffH

l+

l= (4.19)

where al and dl denote regularization factor for normalized transfer function

of the measured acceleration and displacement, respectively. The condition that the

sum of the two transfer functions is less than or equal to 1 is as follows.

adda HH l£l®£+ 1 (4.20)

The regularization factor for the measured acceleration is same as Eq. (4.12) but it

should be defined with the separate target accuracy as follows.

aT

aT

aa

a-=l

12 (4.21)

where aTa is target accuracy for measured acceleration. Also, the transfer function

of the measured displacement is defined with the regularization factor, the

frequency and the transfer function value at the frequency for one transfer function.

dT

dd

T

ddTd

ffH a=

l+

l=

44

4

)~

()

~( (4.22)

39

where dTa , a

Td

Td

T fff /~

= and d

Tf denote target accuracy for the measured

displacement, normalized target frequency and target frequency for the measured

displacement, respectively. The regularization factor of the measured displacement

for normalized transfer function is undefined coefficient. So, the regularization

factor that determines the transfer function for the measured displacement can

define with the separate target frequency and the target accuracy as follows.

22 )~

(1

dTd

T

dT

d fa-

a=l (4.23)

The target frequency for the measured displacement is determined as the

highest meaningful frequency of the measured displacement signals, opposed to the

target frequency for measured acceleration. The target accuracy for measured

displacement can be set individually, but is defined as same as the target accuracy

of the measured acceleration. Eq. (4.20) is organized in terms of Eq. (4.21) and

(4.23) as follows.

2)~

()1)(1( d

TdT

aT

dT

aT f³aa

a-a-(4.24)

Therefore, if the target frequency and the target accuracies are determined to

satisfy Eq. (4.24), the sum of the transfer functions becomes smaller than 1 in the

middle band, so that the noise between the dynamic displacement and quasi-static

displacement can be suppressed. However, if the values satisfied the equality in Eq.

40

(4.24), the sum of transfer function becomes 1 in entire frequency band which is

the same as using the combine displacement reconstruction scheme without

separating the transfer functions. As mentioned above, the target accuracy is

generally determined by the noise level. However, it would be easier to use the

combine displacement reconstruction scheme when the two target accuracies are

similar values. So, setting the two target accuracies as a value, Eq. (4.24) can be

simply derived as follows.

dTa

T

aT f

~1³

a

a-(4.25)

Figure 4.4 shows Eq. (4.19) with two different target frequencies defined above.

Figure 4.4 Transfer functions with two different target frequencies

0.0

0.2

0.4

0.6

0.8

1.0

1.2

10-3

10-2

10-1

100

101

102

Hd ( f d

T = 0.46, ad

T = 0.97)

Ha ( f a

T = 7.1, aaT = 0.97)

Hd + Ha

Tra

nsfe

r fu

nct

ion

s

Normalized frquency, f~

41

5. Estimation of reaction forces of elastic support for an elastically-supported rigid body

In this chapter, two methods to estimate reaction forces of elastic support are

described. Here, reconstructed displacement at the arbitrary point, restored in

Chapter 4, is applied to the displacement-force relation to estimate force at the

point. Basic procedure of first method, setting the arbitrary point to the point of

elastic support, is same as the scheme establish by Kim 2016. However, in this

paper, by using the combine displacement reconstruction scheme, the pseudo-static

components are also considered while the reaction forces of elastic support are

estimated. In second method, the resultant force is estimated by setting the arbitrary

point to the center of gravity of rigid body and distributes the resultant force to the

elastic supports.

5.1 Estimation of the 6-axis force using displacement-force relation

Using the combine displacement reconstruction scheme described in the

chapter 4, the displacement at the arbitrary point can restored by combining the

relative displacement and the relative acceleration estimated from Eq. (3.12) and

(3.17). Therefore, the restored 6-axis displacement including all the dynamic and

quasi-static displacement can be restored as follows.

),(ˆ uau D OOO ΔΔΔ b=

),(ˆ θθθ &&b= D (5.1)

42

where uOΔ̂ , θ̂ and bD denote combined translational displacement, combined

angular displacement and combine displacement reconstruction filter based on the

regularization factor b , respectively. If the restored 6-axis displacement is applied

to the displacement-force relation at the arbitrary point, the force at arbitrary point

can be estimated as follows.

),ˆ,ˆ( ΘθΔKF uO

OO = (5.2)

where OK and OF denote displacement-force relation and 6-axis force

corresponding to the displacement-force relation, respectively. Here, the

displacement-force relation can have both linear and nonlinear curves.

5.2 The displacement-force relation at the elastic support

To calculate reaction force by displacement-force relation at the elastic support,

the relative displacement at the elastic support need to be restored by combine

displacement reconstruct scheme. By setting the arbitrary point to the elastic

support, the relative displacement and the relative acceleration at the elastic support

can be estimated from the measurements of potentiometers and accelerometers by

using Eq. (3.12) and (3.17). By using Eq. (5.1), the reconstructed relative

displacement at the elastic support can be restored as follows.

),(ˆ,,,

auuiEiEiE ΔΔΔ b= D (5.3)

43

Figure 5.1 Relative displacement and relative acceleration at elastic support

where uiE ,Δ̂ ,

uiE,Δ and

aiE ,Δ denote combined relative translational displacement,

relative displacement and relative acceleration at the i-th elastic support,

respectively. Figure 5.1 gives a better understanding of the symbols used in Eq.

(5.3).

Also, by using Eq. (5.2) the 6-axis reaction force at the elastic support can be

estimated. Here, assuming that there are only axial-forces in the elastic support and

the moment becomes zero. In addition, it is assumed that each axis of translation

displacements is independent to each other and applied to the displacement-force

relation. By the assumptions, Eq. (5.2) can be derived as follows.

)ˆ( ,u

iEEi

Ei ΔKF = (5.4)

where EiK and E

iF denote displacement-force relation at the i-th elastic support

and 3-axis force corresponding to the displacement-force relation at the i-th elastic

θ

OU

uiE ,Δ

pju

Epijr

piu Ep

iir

bodyRigid

Base

θ&&

Θ&&

aiA Elastic support

Potentiometer

Accelerometer

aiE ,Δ

EaiiR

ajA

Eaiir

aja

EaijR

Eaijr

aia

bodyRigid

Base

44

support, respectively. If all the assumptions are established, the reaction force of

elastic support can be estimated using Eq. (5.4).

5.3 Displacement-force relation at the center of rigid body

When displacement-force relation at the center of gravity is known than the

resultant force at the center of gravity can be estimated by using Eq. (5.1) and (5.2).

Here, there is an assumption that the only external force applied to the rigid body is

the reaction force by the elastic support, and the reaction force can be estimated by

distributing the resultant force at the center of gravity to the elastic support. First,

by setting the arbitrary point to center of gravity of the rigid body, relative

displacement and relative acceleration at the center of rigid body can be estimated

from measurements of potentiometers and accelerometers by using Eq. (3.12) and

(3.17). By using Eq. (5.1), the reconstructed relative displacement at center of rigid

body can be restored as follows.

),(ˆ auuCOGCOGCOG ΔΔΔ b= D (5.5)

where uCOGΔ̂ , u

COGΔ and aCOGΔ denote combined relative translational

displacement, relative displacement and relative acceleration at the center of

gravity of the rigid body, respectively. Figure 5.2 gives a better understanding of

the symbols used in Eq. (5.5).

Also, by using Eq. (5.2) and (2.18), the 6-axis resultant force at center of rigid

body can be calculated as follows.

45

åå

å

==

=

úû

ùêë

é --÷

÷ø

öççè

æúû

ùêë

é

-

-=

=

noE

iE

iiTE

i

Eii

noE

i

COG

Eii

TEii

TEi

Eiii

noE

i

Ei

COG

11

1

~)~(

~ˆ

~)~()~(

~Θ

rkr

rk

θ

Δ

rkrkr

rkk

FF

u(5.6)

where COGF denotes 6-axis resultant force at center of gravity of rigid body. Here,

if only the reaction force of the elastic support is all the external force applied to

the elastic support, the following relation holds.

úúúúúúúúúúúúú

û

ù

êêêêêêêêêêêêê

ë

é

úúúúúúúú

û

ù

êêêêêêêê

ë

é

---

---

---=

úúúúúúúú

û

ù

êêêêêêêê

ë

é

Ez

Ey

Ex

Ez

Ey

Ex

Ez

Ey

Ex

Ex

Ey

Ex

Ey

Ex

Ey

Ex

Ez

Ex

Ez

Ex

Ez

Ey

Ez

Ey

Ez

Ey

Ez

COGz

COGy

COGx

COGz

COGy

COGx

F

F

F

F

F

F

F

F

F

rrrrrr

rrrrrr

rrrrrr

M

M

M

F

F

F

,3

,3

,3

,2

,2

,2

,1

,1

,1

,3,3,2,2,1,1

,3,3,2,2,1,1

,3,3,2,2,1,1

000

000

000

100100100

010010010

001001001

(5.7)

Figure 5.2 Relative displacement and relative acceleration at center of gravity

θ&&

Θ&&

Base

bodyRigid

aiA Elastic support

Potentiometer

Accelerometer

aCOGΔ

OA

air

aja

ajr

aia

aiA

aiR

ajA

ajR

pju

θ

BaseOU

bodyRigid

uCOGΔ

pjr

piu

pir

46

Figure 5.3 Distributing the resultant force to reaction force at the elastic support

Eq. (5.7) can be simplified as follows.

ECOG AFF = (5.8)

From Eq. (5.8), pseudo inverse is used to estimate the reaction force at the elastic

support from the resultant force at center of gravity of rigid body as follows.

ECOG FFA =+ (5.9)

where superscript + means pseudo inverse. Therefore, knowing the displacement-

force relation of the center of gravity can estimate the reaction force of each elastic

support. Figure 5.3 helps to understand how the elastic support reaction forces are

estimated from the resultant force at center of the gravity using Eq. (5.9).

However, pseudo inverse also has a mathematical problem. In Eq. (5.7), the

process of finding the elastic support reaction forces using the resultant force of the

center of gravity is the indeterminate equation because it finds nine unknowns by

six equations. The general solution of the indeterminate equation can be expressed

Base

bodyRigidCOGF

EiF

EjrE

ir

EjF

Elastic support

47

using the singular value decomposition as follows.

ååå===

g+=g+=g9

7

9

7

6

1

~)(

jjj

m

jjji

i i

OTi

jm

svFvv

FzF (5.10)

where is , iv , iz , jg , mF and mF

~denote i-th singular value of A , i-th left

singular vector of A , i-th right singular vector of A , undetermined coefficient

multiplied to the j-th left singular vector, general solution of the elastic support

reaction force determined by the undetermined coefficient and solution of the

elastic support reaction force estimated by pseudo inverse. So, the elastic support

reaction force has infinite number of solutions according to the undetermined

coefficient jg . However, the solution of the elastic support reaction force

estimated by the pseudo inverse can be defined by the definition of pseudo inverse

of Eq. (5.10) as follows.

)0(~ 6

1

mi

i i

OTiCOGm

sFv

FzFAF === å

=

+(5.11)

As can be seen in Eq. (5.11), when the 2-norm is the smallest, 0=g j , Eq. (5.10)

has the general solution of the pseudo inverse. The solution given by the pseudo

inverse in Eq. (5.11) is the one of the randomly selected solution from a number of

general solutions. Therefore, the solution obtained by pseudo inverse cannot be

regarded as correct solutions from the mathematical point of view. However, in the

section 6.3 of Chapter 6, it can be seen that the solution estimated from the pseudo

48

inverse is similar to the exact solution. Since the solution cannot be mathematically

explained, it cannot be guaranteed that the calculation using the pseudo inverse is

correct when the conditions are different. So, studies on the choosing the solution

using the pseudo inverse and mathematical verification of the solution estimated by

pseudo inverse must be carried out.

49

6. Numerical verification

To verify the scheme, estimating reaction force, developed in previous chapters,

two numerical examples are used. In the first numerical example, the scheme that

estimates the reaction force using displacement-force relation at the elastic support

is already developed by Kim 2016 and verified. However, Kim 2016 only

estimated the dynamic components for the reaction force of the rigid body.

Therefore, the reaction force are estimated not only the dynamic components but

also the pseudo-static components and verified the scheme. In the second

numerical example, the scheme that estimates the reaction force using

displacement-force relation at the center of gravity and pseudo inverse is verified.

6.1 Details of numerical simulation study

The configuration of numerical example 1, Section 6.2, and numerical

example 2, Section 6.2, are the same. The dimensions of the rigid body and the

base used in numerical examples are shown in Table 6.1. The positions of the rigid

body accelerometer, the base accelerometer and the elastic support measurement

points are shown in Table 6.2, 6.3 and Figure 6.1. The x, y and z axis are positive in

the coordinate direction as shown in Figure 6.1. The positions of the potentiometer

attachment point are shown in Table 6.4, and the configuration of the

potentiometers attachment points are shown in Figure 6.2.

50

Table 6.1 Dimension of the rigid body and base size

AxisRigid body (mm) Base (mm)

Model Reference point Model Reference point

x 350 175 4100 2050

y 1000 500 1780 890

z 660 330 1660 830

Table 6.2 Position of the accelerometers

AxisRigid body accelerometer (mm) Base accelerometer (mm)

1 2 3 4 1 2 3

x 319 216 19 176 2025 1689 0

y 0 380 596 1000 1780 0 953

z 636 660 0 77 1023 1345 1287

Table 6.3 Position of the elastic supports

AxisElastic support (mm)

1 2 3

x 255 176 331

y 1000 0 591

z 437 578 0

Figure 6.1 Details of the accelerometers and the elastic supports setting

zx

y

1Α

: Position of accelerometer installed on the rigid body (aj): Position of accelerometer installed on the base (Ai): Position of Elastic support (Ei)

2Α

3Α

3a2a

1a 4a1E

2E

3E

Base

bodyRigid

51

Table 6.4 Position of the potentiometers (numerical example)

Axis1st set potentiometer (mm) 2nd set potentiometer (mm)

RB B1 B2 B3 RB B1 B2 B3

x 0 -13 -15 -11 350 361 359 363

y 661 662 684 639 234 229 267 197

z 312 291 328 331 541 508 573 569

Axis3rd set potentiometer (mm) 4th set potentiometer (mm)

RB B1 B2 B3 RB B1 B2 B3

x 276 282 301 243 124 137 112 121

y 0 -13 -14 -12 1000 1014 1015 1013

z 636 603 669 672 243 272 266 209

RB : Potentiometer on the rigid bodyB : Potentiometer on the base

Figure 6.2 Details of the potentiometers setting

The mass matrix, stiffness matrix and natural frequency of the numerical

example are shown in Eq. (6.1). The mass of the rigid body in 100 kg, and the

stiffness of elastic support connecting the rigid body and the base is 200 N/mm,

: Position of potentiometer installed

on the rigid body (pi): Position of potentiometer installed on the base (Pi )

1P

1p

2P

3P

z

xy

bodyRigid

Base

52

which is the same for elastic support 1 x, y, z axis, elastic support 2 x, y, z axis, and

elastic support 3 x, y, z axis.

úúúúúúúú

û

ù

êêêêêêêê

ë

é

×

×

×=

2

2

2

mkg35.900000

0mkg65.40000

00mkg96.11000

000kg10000

0000kg1000

00000kg100

M

úúúúúúúú

û

ù

êêêêêêêê

ë

é

×××-

×××--

×-×-×-

-

-

-

=

mkN8.107mkN1.20mkN5.80kN4.47kN2.18

mkN1.20mkN5.42mkN7.10kN4.470kN5


0kN4.47kN2.18kN/m60000

kN4.470kN50kN/m6000

kN2.18kN5000kN/m600

RBK

úúúúúúúú

û

ù

êêêêêêêê

ë

é

×××-

××-×--

×××-

--

-

=

mkN4.47mkN1.20mkN5.80kN4.47kN2.18



0kN6.717kN2.18kN/m60000

kN6.7170kN50kN/m6000

kN2.18kN5000kN/m600

BK

Hz]18.8717.6014.2712.3311.9010.43[=nf

(6.1)

where M , RBK , BK and nf denote mass matrix of the rigid body, stiffness

matrix of the elastic support for the rigid body related displacement, stiffness

matrix of the elastic support for the base related displacement and natural

frequencies of the rigid body, respectively.

53

(a)

-2

-1

0

1

2

0 20 40 60 80 100

Acc

eler

atio

n (

m/s

ec2 )

Time (sec)

(b)

-0.4

-0.2

0.0

0.2

0.4

0 20 40 60 80 100

Dis

pla

cem

ent

(km

)

Time (sec)

Figure 6.3 Translational acceleration and translation displacement of the base(a) Translational acceleration (b) Translational displacement

The excitation force consists of sinusoidal vibration displacement at the center

of gravity of the base. The excitation force is composed of the translational

component and vibration component. As shown in Figure 6.3, the translation

component consists of acceleration, constant velocity and deceleration for 100

seconds. The excitation component consists of the translational displacement and

54

the rotational displacement as shown in Eq. (6.2) as follows.

úúúúúú

û

ù

êêêêêê

ë

é

p-p+p

p+p-p

p+p-p-

=

úúú

û

ù

êêê

ë

é

))7.25(2cos(161

1))1.15(2sin(

198

1))9.7(2sin(

32

1

))8.25(2sin(148

1))2.15(2cos(

112

1))1.7(2sin(

24

1

))2.25(2cos(156

1))5.15(2sin(

148

1))2.7(2sin(

15

1

250

1

ttt

ttt

ttt

U

U

U

Oz

Oy

Ox

úúúúúú

û

ù

êêêêêê

ë

é

p+p

p-p

p+

=

úúú

û

ù

êêê

ë

é

Q

Q

Q

))6.15(2sin(148

1))2.7(2sin(

33

1

))7.15(2cos(173

1))1.7(2sin(

45

1

))4.15(2sin(151

1))3.7(2sin(

51

1

25

1

tt

tt

tt

z

y

x

(6.2)

where OxU , O

yU and OzU are excitation displacement of x, y, and z axis at the

center of gravity of the base, respectively, and xQ , yQ and zQ are rotational

displacement of x, y, and z axis of the base, respectively. The frequencies of the

excitation are set to values existing between the natural frequencies.

The center of gravity displacement, velocity and acceleration of the rigid body

by the configured excitation force are calculated using the linear Newmark-Beta

method. The sampling rate is 3000 Hz and the total analysis time is 100 seconds.

Finally, down sampling is used for the numerical example. The sampling rate of the

accelerometer measurement is 300 Hz and the sampling rate of the potentiometer

measurement is 30 Hz. The accelerometer and the potentiometer measurements

were constructed with the rigid body kinematics and give a relative error of up to 5%

(30.8 dB).

55

6.2 Numerical simulation study using elastic support displacement-force relation

To estimate the relative displacement at the elastic support, estimated angular

displacement and the measurements of the potentiometer are utilized. Also, relative

acceleration at the elastic support is estimated by utilizing estimated angular

acceleration and the measurements of the accelerometer. After setting the target

frequency at the elastic support, the combine displacement and the reaction force at

the elastic support are estimated.

6.2.1 Estimation result of the rigid body rotational behavior

The results of the angular displacement, angular acceleration of the rigid body

and angular acceleration of the base estimation are shown in Figure 6.4, 6.5 and 6.6

for the specific time domain (22~23 seconds). In the case of angular displacements,

the sampling rate of the potentiometer measurements are reduced by 1/10 of the

accelerometer measurement and estimated by ignoring the angular displacement of

the base that have low accuracy. Also, in the case of the x-axis, a large relative error

is applied due to the translational component that simulates the movement of the

base that has lower accuracy than other axes. Although the angular displacements

are inaccurate, the reaction forces of the elastic support are estimated to be high

accuracy because the low frequency component and high frequency component are

obtained at the angular displacement and angular acceleration, respectively, using

the combine displacement reconstruction scheme. The angular accelerations are

estimated with high accuracy for entire time domain.

56

6.2.2 Estimation result of the elastic support relative displacement and relative acceleration

The results of the elastic support relative displacement estimation are shown in

Figure 6.7~6.9. The elastic support relative acceleration results are shown in Figure

6.10~6.12. The results are shown for the specific time domain. In the case of

relative displacements, the accuracy is low for the same reason as angular

displacements. However, using the combine displacement reconstruction scheme,

the low frequency component is derived from the relative displacements and high

frequency component is derived from the relative acceleration that the reaction

forces at elastic support are estimated with high accuracy. The relative

accelerations of the elastic support are estimated with high accuracy for entire time

domain.

6.2.3 Determination of major factors of the combine displacement reconstruction scheme

To restore the combine displacement of the elastic support, the target

frequency need to be determined. The target frequency is determined by analyzing

the frequency spectrum of the relative displacements and the relative accelerations

at the elastic support. The frequency spectrum and target frequency of the elastic

support relative displacement and relative acceleration are shown in Figure

6.13~6.15. The target frequency of numerical example 1 is determined as shown in

Table 6.5. If there is no low-frequency component in the relative displacement, the

displacement restored only by the relative acceleration. The target accuracy is 0.97

which is generally used and time window size is the standard window.

57

Table 6.5 Determine target frequency (numerical example 1)

Elastic support

Target frequency (Hz)

Displacment Acceleration

x y z x y z

Elastic support 1 0.46 - - 7.10 7.20 7.10

Elastic support 2 0.59 - - 7.10 7.20 7.10

Elastic support 3 0.59 - - 7.10 7.20 7.10

6.2.4 Estimation result of the elastic support reaction force

The results of the estimation of the reaction forces at the elastic support are

shown in Figure 6.16~6.19. The results for x-axis of each elastic support are shown

for the entire time domain to verify the pseudo-static components are estimated

with high accuracy and the rest only shown for the specific time domain. Since, the

elastic support stiffness is constant that the results for the reconstructed

displacement at elastic support are not shown. It can be seen that the elastic support

reaction force is estimatedd with high accuracy for entire time domain.

6.3 Numerical simulation study using center of gravity displacement-force relation

To Estimate the translational behavior at the center of gravity of the rigid body,

estimated angular displacement, angular acceleration, the measurement of the

potentiometer and the accelerometer are utilized. After setting the target frequency

for the 6-axis behavior of the rigid body, 6-axis combine displacement and the 6-

axis resultant force at center of gravity of the rigid body are estimated. The reaction

forces at elastic support are estimated by using pseudo inverse to the resultant force.

58

6.3.1 Estimation result of the rigid body rotational behavior

The results of the angular displacement, angular acceleration of the rigid body

and angular acceleration of the base estimation are same as Section 6.2.

6.3.2 Estimation result of center of gravity of the rigid body relative translationbehavior

The results of the rigid body relative translation displacement and acceleration

estimation are shown in Figure 6.20 and 6.21, showing the specific time domain. In

the case of the translation displacement, the accuracy is low for the same reason as

angular displacement. However, the resultant force of center of gravity of rigid

body is estimated with high accuracy because by the combine displacement

reconstruction scheme, low and high frequency component are derived from the

translation displacement and acceleration, respectively. The relative acceleration of

the rigid body is estimated with high accuracy for entire time domain.

6.3.3 Determination of major factors of the combine displacement reconstruction scheme

To estimate the combine displacement at center of gravity of the rigid body,

the target frequency need to be determined. The target frequency is determined by

analyzing the frequency spectrum of the 6-axis behavior at center of gravity of the

rigid body and shown in Figure 6.22 and 6.23. The target frequency of numerical

example 2 is determined as shown in Table 6.6. If there is no low-frequency

component in the 6-axis displacement, the displacement is restored only by the 6-

axis acceleration. The target accuracy is 0.97 with the standard window.

59

Table 6.6 Determine target frequency (numerical example 2)

6축

Traget frequency (Hz)

Displacement Acceleration

x y z x y z

Translation 0.59 - - 7.10 7.20 7.10

Rotation - 0.28 0.28 7.30 7.10 7.10

6.3.4 Estimation result of center of gravity of the rigid body 6-axis resultant force

The results of the estimation of the 6-axis resultant force at center of gravity of

the rigid body are shown in Figure 6.24~6.26. The results for x-axis of translation

force, y-axis and z-axis of moment are shown for the entire time domain to verify

the pseudo-static components are estimated with high accuracy, but moments have

less pseudo-static components that hardly can see. The rest of results only show the

specific time domain. Since, the stiffness is constant that the results for the

reconstructed 6-axis displacement at center of gravity of the rigid body are not

shown. Here, the angular displacement of the base is measured and the stiffness

matrix for the angular displacement of the base is known. It can be seen that the 6-

axis resultant force is calculated with high accuracy for entire time domain.

However, if the stiffness matrix for the base angular displacement is not known, it

can be confirmed that the accuracy decreases.

6.3.5 Estimation result of the elastic support reaction force

The results of the estimation of the reaction forces at the elastic support are

shown in Figure 6.27~6.29. The results are shown for the specific time domain. It

can be seen that the elastic support reaction force is calculated with high accuracy

for entire time domain.

60

-10

-5

0

5

10

22 22.2 22.4 22.6 22.8 23

ExactEstimated

An

gula

r di

spla

cem

ent (

rad,

x 1

0-3

)

Time(sec)

(a)

-10

-5

0

5

10

22 22.2 22.4 22.6 22.8 23

ExactEstimated

An

gula

r di

spla

cem

ent (

rad,

x 1

0-3

)

Time(sec)

(b)

-10

-5

0

5

10

22 22.2 22.4 22.6 22.8 23

ExactEstimated

An

gula

r di

spla

cem

ent (

rad,

x 1

0-3

)

Time(sec)

(c)Figure 6.4 Angular displacement of the rigid body (numerical example 1 & 2)

(a) x-axis (b) y-axis (c) z-axis

61

-70

-35

0

35

70

22 22.2 22.4 22.6 22.8 23

ExactEstimated

An

gula

r ac

cele

rati

on (

rad/

sec

2)

Time(sec)

(a)

-70

-35

0

35

70

22 22.2 22.4 22.6 22.8 23

ExactEstimated

An

gula

r ac

cele

rati

on (

rad/

sec

2)

Time(sec)

(b)

-70

-35

0

35

70

22 22.2 22.4 22.6 22.8 23

ExactEstimated

An

gula

r ac

cele

rati

on (

rad/

sec

2)

Time(sec)

(c)Figure 6.5 Angular acceleration of the rigid body (numerical example 1 & 2)


62

-10

-5

0

5

10

22 22.2 22.4 22.6 22.8 23

ExactEstimated

An

gula

r ac

cele

rati

on (

rad/

sec

2)

Time(sec)

(a)

-10

-5

0

5

10

22 22.2 22.4 22.6 22.8 23

ExactEstimated

An

gula

r ac

cele

rati

on (

rad/

sec

2)

Time(sec)

(b)

-10

-5

0

5

10

22 22.2 22.4 22.6 22.8 23

ExactEstimated

An

gula

r ac

cele

rati

on (

rad/

sec

2)

Time(sec)

(c)Figure 6.6 Angular acceleration of the base (numerical example 1 & 2)


63

-6

-3

0

3

6

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Dis

plac

emen

t (m

m)

Time(sec)

(a)

-6

-3

0

3

6

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Dis

plac

emen

t (m

m)

Time(sec)

(b)

-6

-3

0

3

6

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Dis

plac

emen

t (m

m)

Time(sec)

(c)Figure 6.7 Relative displacement at the elastic support1 (numerical example 1)


64

-6

-3

0

3

6

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Dis

plac

emen

t (m

m)

Time(sec)

(a)

-6

-3

0

3

6

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Dis

plac

emen

t (m

m)

Time(sec)

(b)

-6

-3

0

3

6

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Dis

plac

emen

t (m

m)

Time(sec)



65

-6

-3

0

3

6

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Dis

plac

emen

t (m

m)

Time(sec)

(a)

-6

-3

0

3

6

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Dis

plac

emen

t (m

m)

Time(sec)

(b)

-6

-3

0

3

6

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Dis

plac

emen

t (m

m)

Time(sec)



66

-40

-20

0

20

40

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Acc

eler

atio

n (m

/sec

2)

Time(sec)

(a)

-40

-20

0

20

40

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Acc

eler

atio

n (m

/sec

2)

Time(sec)

(b)

-40

-20

0

20

40

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Acc

eler

atio

n (m

/sec

2)

Time(sec)

(c)Figure 6.10 Relative acceleration at the elastic support1 (numerical example 1)


67

-40

-20

0

20

40

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Acc

eler

atio

n (m

/sec

2)

Time(sec)

(a)

-40

-20

0

20

40

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Acc

eler

atio

n (m

/sec

2)

Time(sec)

(b)

-40

-20

0

20

40

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Acc

eler

atio

n (m

/sec

2)

Time(sec)



68

-40

-20

0

20

40

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Acc

eler

atio

n (m

/sec

2)

Time(sec)

(a)

-40

-20

0

20

40

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Acc

eler

atio

n (m

/sec

2)

Time(sec)

(b)

-40

-20

0

20

40

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Acc

eler

atio

n (m

/sec

2)

Time(sec)



69

0

4

8

0.0

0.3

0.6

0.01 0.1 1 10 100 1000

acceleration

displacement

Spe

ctru

m o

f A

ccel

erat

ion

Spectru

m o

f Disp

lacement

Frequency (Hz)

fT

a = 7.10 Hz

fT

d = 0.46 Hz

(a)

0

4

8

0.0

0.3

0.6

0.01 0.1 1 10 100 1000

acceleration

displacement

Spe

ctru

m o

f A

ccel

erat

ion

Spectru

m o

f Disp

lacement

Frequency (Hz)

fT

a = 7.20 Hz

(b)

0

4

8

0.0

0.3

0.6

0.01 0.1 1 10 100 1000

acceleration

displacement

Spe

ctru

m o

f A

ccel

erat

ion

Spectru

m o

f Disp

lacement

Frequency (Hz)

fT

a = 7.10 Hz

(c)Figure 6.13 Frequency spectrum and Target frequency at the elastic support1

(numerical example 1)(a) x-axis (b) y-axis (c) z-axis

70

0

4

8

0.0

0.3

0.6

0.01 0.1 1 10 100 1000

acceleration

displacement

Spe

ctru

m o

f A

ccel

erat

ion

Spectru

m o

f Disp

lacement

Frequency (Hz)

fT

a = 7.10 Hz

fT

d = 0.59 Hz

(a)

0

4

8

0.0

0.3

0.6

0.01 0.1 1 10 100 1000

acceleration

displacement

Spe

ctru

m o

f A

ccel

erat

ion

Spectru

m o

f Disp

lacement

Frequency (Hz)

fT

a = 7.20 Hz

(b)

0

4

8

0.0

0.3

0.6

0.01 0.1 1 10 100 1000

acceleration

displacement

Spe

ctru

m o

f A

ccel

erat

ion

Spectru

m o

f Disp

lacement

Frequency (Hz)

fT

a = 7.10 Hz



71

0

4

8

0.0

0.3

0.6

0.01 0.1 1 10 100 1000

acceleration

displacement

Spe

ctru

m o

f A

ccel

erat

ion

Spectru

m o

f Disp

lacement

Frequency (Hz)

fT

a = 7.10 Hz

fT

d = 0.59 Hz

(a)

0

4

8

0.0

0.3

0.6

0.01 0.1 1 10 100 1000

acceleration

displacement

Spe

ctru

m o

f A

ccel

erat

ion

Spectru

m o

f Disp

lacement

Frequency (Hz)

fT

a = 7.20 Hz

(b)

0

4

8

0.0

0.3

0.6

0.01 0.1 1 10 100 1000

acceleration

displacement

Spe

ctru

m o

f A

ccel

erat

ion

Spectru

m o

f Disp

lacement

Frequency (Hz)

fT

a = 7.10 Hz



72

-900

-450

0

450

900

0 20 40 60 80 100

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)

(a)

-900

-450

0

450

900

0 20 40 60 80 100

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)

(b)

-900

-450

0

450

900

0 20 40 60 80 100

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)

(c)Figure 6.16 Reaction force at x-axis of the elastic support (numerical example 1)

(a) elastic support1 (b) elastic support2 (c) elastic support3

73

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)

(a)

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)

(b)

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)

(c)Figure 6.17 Reaction force at the elastic support1 (numerical example 1)


74

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)

(a)

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)

(b)

-850

-425

0

425

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)



75

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)

(a)

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)

(b)

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)



76

-6

-3

0

3

6

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Dis

plac

emen

t (m

m)

Time(sec)

(a)

-6

-3

0

3

6

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Dis

plac

emen

t (m

m)

Time(sec)

(b)

-6

-3

0

3

6

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Dis

plac

emen

t (m

m)

Time(sec)

(c)Figure 6.20 Relative displacement at the center of gravity (numerical example 2)


77

-40

-20

0

20

40

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Acc

eler

atio

n (m

/sec

2)

Time(sec)

(a)

-40

-20

0

20

40

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Acc

eler

atio

n (m

/sec

2)

Time(sec)

(b)

-40

-20

0

20

40

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Acc

eler

atio

n (m

/sec

2)

Time(sec)

(c)Figure 6.21 Relative acceleration at the center of gravity (numerical example 2)


78

0

3

6

0.0

0.2

0.4

0.01 0.1 1 10 100 1000

acceleration

displacement

Spe

ctru

m o

f A

ccel

erat

ion

Spectru

m o

f Disp

lacement

Frequency (Hz)

fT

a = 7.10 Hz

fT

d = 0.59 Hz

(a)

0

3

6

0.0

0.2

0.4

0.01 0.1 1 10 100 1000

acceleration

displacement

Spe

ctru

m o

f A

ccel

erat

ion

Spectru

m o

f Disp

lacement

Frequency (Hz)

fT

a = 7.20 Hz

(b)

0

3

6

0.0

0.2

0.4

0.01 0.1 1 10 100 1000

acceleration

displacement

Spe

ctru

m o

f A

ccel

erat

ion

Spectru

m o

f Disp

lacement

Frequency (Hz)

fT

a = 7.10 Hz

(c)Figure 6.22 Frequency spectrum and Target frequency of translation behavior

at the center of gravity (numerical example 2)(a) x-axis (b) y-axis (c) z-axis

79

0

5

10

0.0

0.5

1.0

0.01 0.1 1 10 100 1000

acceleration

displacement

Spe

ctru

m o

f A

ccel

erat

ion

Spectru

m o

f Disp

lacement (x 1

0-3)

Frequency (Hz)

fT

a = 7.30 Hz

(a)

0

5

10

0.0

0.5

1.0

0.01 0.1 1 10 100 1000

acceleration

displacement

Spe

ctru

m o

f A

ccel

erat

ion

Spectru

m o

f Disp

lacement (x 1

0-3)

Frequency (Hz)

fT

a = 7.10 Hz

fT

d = 0.28 Hz

(b)

0

5

10

0.0

0.5

1.0

0.01 0.1 1 10 100 1000

acceleration

displacement

Spe

ctru

m o

f A

ccel

erat

ion

Spectru

m o

f Disp

lacement (x 1

0-3)

Frequency (Hz)

fT

a = 7.10 Hz

fT

d = 0.28 Hz

(c)Figure 6.23 Frequency spectrum and Target frequency of rotation behavior

at the center of gravity (numerical example 2)(a) x-axis (b) y-axis (c) z-axis

80

-2000

-1000

0

1000

2000

0 20 40 60 80 100

ExactEstimated

Tra

nsla

tion

for

ce (

N)

Time(sec)

(a)

-800

-400

0

400

800

0 20 40 60 80 100

ExactEstimated

Mom

ent

(N m

)

Time(sec)

(b)

-800

-400

0

400

800

0 20 40 60 80 100

ExactEstimated

Mom

ent

(N m

)

Time(sec)

(c)Figure 6.24 Pseudo-static force at center of gravity (numerical example 2)

(a) x-axis translation (b) y-axis rotation (c) z-axis rotation

81

-2000

-1000

0

1000

2000

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Tra

nsla

tion

for

ce (

N)

Time(sec)

(a)

-2000

-1000

0

1000

2000

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Tra

nsla

tion

for

ce (

N)

Time(sec)

(b)

-2000

-1000

0

1000

2000

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Tra

nsla

tion

for

ce (

N)

Time(sec)

(c)Figure 6.25 Translation force at center of gravity (numerical example 2)


82

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Mom

ent

(N m

)

Time(sec)

(a)

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Mom

ent

(N m

)

Time(sec)

(b)

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Mom

ent

(N m

)

Time(sec)

(c)Figure 6.26 Moment at center of gravity (numerical example 2)


83

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)

(a)

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)

(b)

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)



84

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)

(a)

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)

(b)

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)



85

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)

(a)

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)

(b)

-800

-400

0

400

800

22 22.2 22.4 22.6 22.8 23

ExactEstimated

Rea

ctio

n fo

rce

(N)

Time(sec)



87

7. Summary and conclusion

In this study, the scheme for estimating the reaction forces of the rigid body

supported by the elastic supports using the potentiometers and the accelerometers

measurements is proposed. This scheme restores the displacement in entire

frequency including the low frequency and the high frequency components, such as

pseudo-static and dynamic components, using potentiometer measurements and

acceleration measurements. Here, Developed theories like the GPS equation and

the combine displacement reconstructed scheme are utilized to estimate reaction

forces of the elastic supports.

The 6-axis relative displacement and relative acceleration at the arbitrary point

of the rigid body are estimated using the potentiometers and accelerometers

measurements. To estimate the relative displacement at the arbitrary point of the

rigid body, the angular displacements are estimated by estimating the relative

displacement at the object point of the potentiometers using modified GPS

optimization equation. Therefore, the relative displacement at the object point and

the angular displacement are utilized for the rigid body kinematics to estimate the

6-axis relative displacement at the arbitrary point of the rigid body. The 6-axis

relative acceleration at the arbitrary point of the rigid body can also be estimated as

same way as the 6-axis relative displacement, by estimating the angular

acceleration with the measured acceleration and utilizing the angular acceleration

and the measured acceleration.

The combine displacement reconstruction scheme is used to restore the

88

displacement of entire frequency band. The acceleration and the displacement show

strength in the high frequency and low frequency band, respectively. In order to use

only the strength of both measurements, the combine displacement reconstruction

scheme is used. Also, to reduce the noise in the middle band frequency, the transfer

functions are separated with two different target frequencies. Therefore, by using

the combine displacement reconstruction scheme, it is possible to restore the

displacements that combine both the quasi-static behavior and the dynamic

behavior. By applying the reconstructed displacement to the displacement-force

relation, the reaction force in the elastic supports can be estimated. The numerical

examples show that the proposed scheme works well.

89

Reference

Al-Rawashdeh 2014

Al-Rawashdeh Y M, Elshafei M, and Al-Malki M F, "In-Flight Estimation of

Center of Gravity Position Using All-Accelerometers" Sensors, vol. 14, no. 9, pp.

17567-17585, 2014.

Bjorck 1996

Ake Bjorck, "Numerical Methods for Least-Squares Problems", SIAM,

Philadelphia, ISBN 0-89871-360-9, 1996

Blewitt 1997

Geoffrey Blewitt, "Basics of the GPS Technique: Observation Equations",

Department of Geomatics, University of Newcastle Newcastle upon Tyne, NE1

7RU, United Kingdom, 1997

Chopra 2012

Chopra A K, "Dynamics of structures: theory and applications to earthquake

engineering", Prentice Hall, Upper Saddle River, NJ, 2012.

Hamming 1989

Hamming R W, "Digital filters", Third Edition, Prentice-Hall, Englewood Cliffs,

NJ, 1989.

Hibbeler 2004

Hibbeler R C, "Engineering Mechanics: Dynamics", Twelfth Edition, Prentice

Hall, Upper Saddle River, NJ, 2004.

Hjelmstad 1995

Hjelmstad K D, Banan M R, and Banan M R, "Time-domain parameter estimation

algorithm for structures. I: Computational aspects", Journal of engineering me-

chanics, vol. 121, no.3, pp. 424-434, 1995.

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Hong 2010

Hong Y H, Kim H K, and Lee H S, "Reconstruction of dynamic displacement and

velocity from measured accelerations using the variational statement of an inverse

problem", Journal of Sound and Vibration, vol. 329, no. 23, pp. 4980-5003, 2010.

Hong 2013

Yun Hwa Hong, Se Gun Lee, Hae Sung Lee, "Design of the FEM-FIR filter for

displacement reconstruction using accelerations and displacements measured at

different sampling rates", Mechanial Systems and Signal Processing, Vol. 38,

No. 2, pp. 460-481, 2013.07.

Kreyszig 2006

Kreyszig E, "Advanced engineering mathematics", Tenth Edition, John Wiley,

Hoboken, NJ, 2006.

Kim 2016

Kim J R, “Estimation Reaction Force and Position of Center of Gravity for Elastic

Supported Rigid Body Using Measured Acceleration”, M.S. Thesis, Seoul

National University, Seoul, South Korea

Lee 1999

Lee H S, Kim Y H, Park C J, and Park H W, "A new spatial regularization scheme

for the identification of the geometric shape of an inclusion in a finite body",

International Journal for Numerical Methods in Engineering, vol. 46, no. 7, pp.

973-992, 1999.

Lee 2010

Hae Sung Lee, Yun Hwa Hong, Hyun Woo Park,"Design of an FIR Filter for the

Displacement Reconstruction Using Measured Acceleration in Low-frequency

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Van Karsen 2007

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93

초록

가속도계 및 포텐셔미터 측정치에 의하여

복원된 변위를 이용한 탄성 지지된 강체의 반력 추정

전상범

건설환경공학부

서울대학교 대학원

이 논문은 탄성받침으로 지지된 강체의 탄성받침 반력을 동적 성분뿐만

아니라 준정적 성분을 포함한 모든 주파수 영역에 대해서 추정하는

기법을 제안한다. 탄성받침 반력을 아는 것은 강체 구조물의 안정성,

사용성, 내구성 등 유지관리에 중요하다. 그러나 탄성받침의 반력을 직접

측정하기에는 탄성받침에 손상이 생길 뿐만 아니라 비용이 비싸며

유지관리가 어렵다는 단점을 가지고 있다. 이러한 문제점을 해결하기

위해 강체 표면에서 가속도계로 가속도를 측정하여 강체운동식과

변위재구성기법을 사용하여 탄성받침 반력을 추정하는 선행연구가

있지만 동적 성분만 추정이 가능하며 준정적 성분은 추정하지 못한다는

한계점을 가지고 있다. 이 한계점을 극복하기 위해서 포텐셔미터를

추가적으로 사용하여 가속도계만으로 부족하였던 준정적 성분이 보완

94

가능하게 되어 모든 주파수 영역에 대해서 탄성받침 반력을 추정 할 수

있다. 포텐셔미터의 측정치에 GPS공식 및 강체운동식을 적용하고

가속도계 측정치에 강체운동식을 적용하여 각각 강체 내의 임의점에서

상대변위와 상대가속도를 계산한다. 이 때 가속도와 변위를 합성하여

변위을 복원하는 합성변위재구성기법을 사용하여 상대변위에서 저주파

성분과 상대가속도에서 고주파 성분을 각각 복원하여 변위를 재구성한다.

임의점을 탄성받침으로 선정하여 탄성받침에서의 힘-변위관계를

사용하여 반력을 추정 할 수 있으면 임의점을 강체의 무게중심을 선정한

경우 무게중심에서의 힘-변위관계와 의사역행렬을 사용하면 탄성받침의

반력을 추정할 수 있다. 제안된 기법은 수치예제를 구성하여 검증을

한다.

주요어 : 포텐셔미터, GPS, 강체운동식, 변위재구성기법, 탄성받침반력

학 번 : 2016-26162

estimation of reaction forces for anelastically-supported rigid...

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