Studying the rigid body inertia ellipsoid using free-access
online stereo 3D simulation
Authors: Svetoslav Zabunov and Maya Gaydarova
Abstract - The current paper aims at presenting the capabilities and benefits of a free-access
online stereo 3D simulation developed by Svetoslav Zabunov. The simulation is used in
teaching the inertia ellipsoid and the moment of inertia tensor of rigid bodies to students in
universities. The main features of the described simulation are presenting the inertia ellipsoid
of rigid bodies in a stereo 3D view mode; translation of the ellipsoid and its moment of inertia
tensor to an arbitrary point in the body reference frame; display of principal axes of inertia;
calculation of the diagonalized translated moment of inertia tensor and the diagonalizing
rotation matrix. The simulation, subject to the current material, may be observed at the
following web address: http://ialms.net/sim/.
Keywords: Inertia ellipsoid in stereo 3D simulation, Rigid body inertia ellipsoid, Translation
of the moment of inertia tensor in simulation.
1. Introduction
This article discloses a method of teaching the inertia ellipsoid and its properties using
a visual e-learning method. A simulation that has free unrestricted access on the Internet
demonstrates the inertia tensor of various rigid bodies to students in universities around the
world. The major features of the described simulation are the presentation in stereo 3D
environment (Zabunov, 2012) of the inertia ellipsoid of rigid bodies along with translation of
the ellipsoid and its moment of inertia tensor to an arbitrary point in the body reference frame.
The principal axes of inertia are also simultaneously displayed along with the diagonalized
translated moment of inertia tensor and the diagonalizing rotation matrix (Figure 1). The
simulation, subject to the current material, may be observed at the following web address:
http://ialms.net/sim/.
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Figure 1. Displacement of the moment of inertia tensor to a vertex of a rigid body (along all
three axes) and visualization in 3D scene of the corresponding inertia ellipsoid.
2. Prerequisites
In order to present the simulation and its benefits, a short introduction to the inertial
capabilities of rigid bodies is exposed.
Moment of inertia of a rigid body about an axis characterizes quantitatively its inertia
capabilities during a rotational motion about that given axis. Moment of inertia of a rigid body
with volume V about axis OO is calculated as follows (Arnold, 1989):
I OO   r 2 dm   r 2 dV ,
V
V
where  represents the density of the body at the point where the elementary mass dm  dV
is located, and r is the distance from the elementary mass to the axis of rotation OO . About
different axes the body has different moments of inertia. There are countless axes of rotation
and moments of inertia and each of them could be calculated using the formula above. But
this is a rather hard task. Another way of defining the moment of inertia is the relation that it
establishes between two vector variables, characterizing the rotational motion of the rigid
body:
 
L  I ,


where L is the angular momentum, and  is the angular velocity of the rotational motion.
This equation is useful, but it is applicable only if the rotation is realized about a principal


axis of inertia – only in that case vectors L and  would be parallel to each other and the
relation binding them could be presented by the means of a product with a scalar value, as it is
seen in the example above. Such a case is a rotation of a homogenous and axially symmetric
rigid body about its axis of symmetry.
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

In the general case vectors L and  are not parallel. Nevertheless, a relation between
them still exists and it is a tensor of second rang called moment of inertia tensor (Beer
& Johnston, 1988; Goldstein et al., 2001):
 I xx

I O   I xy
 I xz

I xy
I yy
I yz
I xz 

I yz 
I zz 
This variable is in respect to a point in space O , through which passes the axis of rotation. To


describe the relation between vectors L and  , they should be presented in a matrix form
(one-row-matrixes):
L  ωI O

The components of vector L then would be:
Lx   x I xx   y I xy   z I xz
L y   x I xy   y I yy   z I yz
Lz   x I xz   y I yz   z I zz

Obviously, in this general case, the momentary axis of rotation is defined by vector  . The
inertia about the axis of rotation creates an angular momentum about this same axis that is the


projection of the angular momentum vector L along vector  , or in matrix form:
Lω 
Lω ~
( ω ~ is the transposed matrix of ω )

Substituting the angular momentum with the product of angular velocity and the moment of
inertia tensor yields:
Lω 
ωIω~

~
 n ω Inω
  I ω ,
or the projection of the angular momentum along the axis of rotation is derived from the
magnitude of the angular velocity vector and the scalar quantity I ω . It is called a reduced
moment of inertia from the moment of inertia tensor about a given axis of rotation defined by



its direction unit vector n ω or n   in vector form. Thus the reduced moment of inertia


may be expressed using the components of the unit vector n and the moment of inertia
tensor:
~
I ω  n ω Inω
 nx2 I xx  n 2y I yy  nz2 I zz  2nx n y I xy  2nx nz I xz  2n y nz I yz ,

As vector n has unit length its components could be substituted with the direction cosines
defining the axis of rotation:
I ω  I xx cos 2   I yy cos 2   I zz cos 2   2I xy cos cos   2I xz cos cos   2I yz cos  cos 
On the other hand, the reduced moment of inertia may be expressed through angular
momentum and angular velocity as follows:
Iω 
Lω


Lω ~
2
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Finally, the inertial ellipsoid enables the 3D graphical observation of the moment of
inertia tensor properties. The ellipsoid is defined such that its three axes are the reciprocals of
the square roots of the principal moments of inertia (Figure 1, 2, 3 and 4):
x2 y2 z 2
 2  2  1  I xx x 2  I yy y 2  I zz z 2  1
2
a
b
c
Figure 2. The inertia ellipsoid of the non-displaced moment of inertia tensor.
3. Description of the simulation interface and features
The simulation demonstrates rigid body motion under different conditions and has
broad capabilities, only part of which are utilized in the current presentation and the following
example problems (Zabunov, 2010a). A comprehensive tutorial on how to use the simulation
can be found online at http://ialms.net/sim/3d-rigid-body-simulation-tutorial.
Representation of a static rigid body along the inertia ellipsoid is possible, although
the user may also observe the rigid body in rotational motion, and see a large number of
constructions and trajectories (see the ‘View’ drop-down select box in the lower–left of the
interface). Generally, in motion mode the simplest case is when no external forces are acting.
The motion is free. The second option is to introduce an external force with force arm creating
an external torque. In addition, the observer may initiate internal friction and external friction
to the rigid body motion.
The current material describes only a static observation of the chosen rigid body. A
rigid body type is selected from the drop-down select box in the top-right of the interface and
the body dimensions and mass are specified in the fields below. Then, the view mode is
switched to ‘Inertia ellipsoid’. The ellipsoid is drawn along with the principal axes of inertia.
Below the view mode select box, a group of fields is displayed, concerning the ellipsoid view.
The first three fields define the center of the moment of inertia tensor in the body reference
frame - default (0,0,0) (Figure 2). All fields below are read-only. Once a new center for the
moment of inertia is selected, the inertia ellipsoid is redrawn and the translated moment of
inertia tensor is calculated and printed in the fields ‘Translated moment of inertia tensor’. At
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the same time, the translated tensor is diagonalized and its diagonal values are shown in the
‘Diagonalized tensor (elements of the main diagonal)’ fields. The rotation matrix that
diagonalizes the translated tensor is calculated and its components are printed in the
‘Diagonalizing rotation’ fields.
Using the sliders at the top of the interface, the viewer may control camera latitude and
longitude. At any moment, the student may switch the simulation into 3D-stereoscopic view
mode, where the scene may be observed using 3D-stereoscopic anaglyph glasses (check the ‘
Stereo mode’ checkbox).
4. What does the simulation offer
The simulation was developed in order to facilitate students in engineering specialties
while studying mechanics and specifically the moment of inertial tensor. The simulation helps
the construction of the inertia ellipsoid of different rigid bodies and the observation of the
principal moments of inertia.
Figure 3. Displacement of the moment of inertia tensor along the Ox axis.
Due to the simulation, several complex tasks could be solved easily, and the solutions
could be observed in 3D stereo mode. Thus is reached understanding of the investigated
phenomenon. Each step of a solution has its graphical representation, which clarifies the
notion of the used mathematical formulae and presents the acquired results and solutions in
graphical manner, easy enough for the student to comprehend and draw practical
understanding.
The usage of the simulation shall be demonstrated through a few, among many,
example problems and their full solutions.
Example problem 1
It is given a homogenous rigid body with a shape of a rectangular parallelepiped and a
total mass of 10kg (Figure 2). Body reference frame has origin at the center of mass and axes
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parallel to the body’s edges. The body dimensions are 1.0,0.5,0.2 m along the Ox, Oy, Oz axes
respectively.
Calculate the moment of inertia tensor for the center of mass I c .
Hint: The moment of inertia tensor without similarity transformation (rotation) is
 I xx
diagonal I c   0
 0
0
I yy
0
0
0  and I xx , I yy and I zz correspond to the principle moments of
I zz 
inertia of the center of mass along the coordinate axes of the body reference frame, because
the body is homogenous and symmetrical.
Figure 4. Displacement of the moment of inertia tensor along the Ox and Oy axes.
Solution:
The moment of inertia tensor for a homogenous rectangular parallelepiped at its center
of mass in a reference frame oriented along the body edges is calculated by the following
formula:
b 2  c 2
1 
I c  m 0
12 
0

0
2
a  c2
0
0 

0 ,
a2  b2 

where m  10kg is the body mass, and a  1m , b  0.5m and c  0.2m are the body
dimensions. For this reason the answer is the following:
0
0 
0.24

Ic   0
0.87
0  . ■
 0
0 1.04 
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Example problem 2
Find the translated moment of inertia tensor from Example 1 at the vertex of the body
that has only positive coordinates in the body reference frame.
Solution: The parallel axis theorem (Huygens-Steiner theorem) helps to solve the problem:
(1)


I t  I c  m rr ~ 1  r  r 

 I xx  m ry2  rz2

It  
 mrx ry

 mrx rz


 mrx ry
I yy  m

rx2

rz2
 mr y rz
  0.97 - 1.25 - 0.50 
 

 mr y rz
   - 1.25 3.47 - 0.25 
 m rx2  ry2  - 0.50 - 0.25 4.17 

 mrx rz

I zz


Matrix 1 is the 3  3 identity matrix. Vector r is the translation vector. The body is
homogenous and its center of mass coincides with its geometrical center. Hence, the
translation vector pointing to the vertex of the body with only positive coordinates is equal to
0.5 0.25 0.1 (see Figure 1). Figure 3 shows a translations of the moment of inertia tensor
along the Ox axis. Figure 4 shows a translations of the moment of inertia tensor along both
the Ox and Oy axes. Note that in the first case (Figure 3), the translation along a single
coordinate axis yields a diagonal translated tensor and in the second case (Figure 4) four of
the translated tensor’s non-diagonal components are zeroes. ■
Example problem 3
Diagonalize the translated moment of inertia tensor from Example 2.
Solution: From the theory of symmetric matrixes it is known that a 3  3 symmetric
matrix or a symmetric rang 2 tensor can be diagonalized (Arnold, 1989). The diagonalized
tensor will have only three non-zero (meaningful) values. The other three values (products of
inertia) found in the non-dagonalized tensor will be represented by the transformation used to
diagonalize the tensor. This transformation is a similarity transformation using a rotation
matrix R (Zabunov, 2010b). It is so, because under a certain rotation R , the principle axes of
the moment of inertia tensor coincide with the axes of the body reference frame and the
rotated tensor becomes diagonalized:
(2)
 I Dxx
I D  R IR   0
 0
~
0
I Dyy
0
0 
0 
I Dzz 
Let’s observe the eigenvectors n Di  xDi
diagonalized tensor:
n DiI D  n Dii  n DiI D  n Dii  0
(3)
y Di
z Di  and eigenvalues i of the sought
The above matrix equation represents a homogenous system of three linear equations, which
has a non-trivial solution only if the determinant of its coefficients is zero:
I D  1i  0  I Dxx  i I Dyy  i I Dzz  i   0
(4)
Equation (4) has three real roots: 1  I Dxx , 2  I Dyy and 3  I Dzz , thus the diagonalized
tensor is equal to:
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(5)
1 0
I D   0 2
 0 0
0
0 
3 
Applying the similarity transformation (2) to equation (3) leads to:
n DiI D  n Di R ~ IR  n Dii  n Di R ~ I  n Di R ~ i 
ni I  ni i  ni I  ni i  0
(6)
Here vectors ni  xi
(7)
yi
zi  are the eigenvectors of I .
ni  n Di R  n Di  ni R
~
It is obvious that I D and I have the same eigenvalues. Obtaining the eigenvalues of I and
substituting them in (5) gives the wanted diagonalized tensor I D . To find the eigenvalues of
I it is suitable to analyse equation (6). Similarly to (3), this matrix equation represents a
homogenous system of three linear equations, which has a non-trivial solution only if the
determinant of its coefficients is zero:
I xx  i
I xy
I xz
I  1i  0  I xy
I yy  i
I yx  0
(8)
I xz
I yx
I zz  i
Equation (8) is a qubic equation for i with roots 1  0.37 , 2  3.98 and 3  4.25 .
Substituting these eigenvalues in (5) leads to the solution:
0
0 
0.37

ID   0
3.98
0  (see Figure 1). ■
 0
0
4.25
Example problem 4
Find the rotation matrix R that diagonalizes the tensor from Example 3.
Solution: Each eigenvalue of I substituted in (3) yields a corresponding solution for
n Di , if n Di is constraint to a unit eigenvector:
1  I Dxx  n D1  1 0 0  i
2  I Dyy  n D 2  0 1 0  j
(9)
3  I Dzz  n D 3  0 0 1  k
From (7) and (9) it follows that the rotation matrix R transforms vectors n i to the orthogonal
basis of the body reference frame i, j, k . However, the rotation matrix is defined by the
direction cosines of one basis rotated to another:
cos  x cos  y cos  z 
(10) R  cos  x cos  y cos  z 
 cos  x cos  y cos  z 


Vectors ni are unit vectors and their components are equal to their direction cosines, so:
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(11)
 n1 
R  n 2 
n 3 
The last step is to find the eigenvectors ni . Equation (6) comes to help along with the unit
length of the eigenvectors:
xi I xx  i   yi I xy  zi I xz  0
xi I xy  yi I yy  i   zi I yz  0
(12)
xi I xz  yi I yz  zi I zz  i   0
xi2  yi2  zi2  1
System (12) enables yi and z i to be expressed through xi , using the first three equations, and
substituting in the last equation:
yi  xi
(13)
I xx  i I yz  I xy I xz
I yy  i I xz  I xy I yz
I xx  i I yz  I xy I xz
I zz  i I xy  I xz I yz
  I   I  I I  2  I   I  I I  2 
xx
i yz
xy xz 
xx
i yz
xy xz  
2
xi 1  

1
  I yy  i I xz  I xy I yz   I zz  i I xy  I xz I yz  

 

zi  xi


From the last equation of system (13) xi is found as follows:
(14)
xi  
1
 I xx  i I yz  I xy I xz   I xx  i I yz  I xy I xz 
 

1 
 I yy  i I xz  I xy I yz   I zz  i I xy  I xz I yz 

 

2

2

After substituting xi in the first two equations of (13) yi and z i are found. Note that the
components xi , yi , zi of eigenvector n i do not have defined sign. There are two possibilities
yielding an eigenvector n i along a defined line, but with two possible opposite directions.
The correct direction will be found later. After finding all three eigenvectors n1 , n 2 and n 3
their components are substituted in (11) to generate the sought rotation matrix R (see Figure
1):
0.15 
 0.91 0.38

R  0.40  0.91  0.14 
0.08 0.19  0.98 
However, if the incorrect directions (signs in (14)) were chosen matrix R would yield an
inverse (improper) rotation. This fact can be verified by calculating the determinant of R ,
which should be  1 :
R 1
This means that matrix R is a proper rotation. In case the determinant is  1 , matrix R
should be corrected to a proper rotation by multiplying it with a negative identity (changing
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direction of all eigenvectors) or by changing the sign of one of its rows (changing direction of
one of the eigenvectors). Visit http://ialms.net/sim/ to simulate other examples. ■
5. Conclusion
The simulation of mechanics phenomena gives the opportunity to engineering students
observe setups that are impossible to be created in laboratory conditions and also present
vectors, parameters and constructions in online stereo 3D graphics while the bodies are in
motion or the point of view is modified.
The authors of the presented material express their gratitude to Assoc. Professor
Vesela Decheva.
6. References
Arnold, V. I. (1989). Mathematical Methods of Classical Mechanics Second Edition.
Springer-Verlag
Beer, P. F. & Johnston, E. R. Jr. (1988). Vector Mechanics for Engineers fifth edition.
McGraw-Hill, Inc. ISBN 0-07-079926-1
Goldstein, H., Pool, C., and Safko, J. (2001). Classical Mechanics third edition. Addison
Wesley, Miami
Zabunov, S. (2012). Stereo 3-D Vision in Teaching Physics”, Phys. Teach., 50, 3, pp. 163
Zabunov, S. (2010a). Rigid body motion in stereo 3D simulation. Eur. J. Phys., 31, 1345
Zabunov, S. (2010b). Mathematical Methods in E-Learning Mechanics for Graduates –
Rotation Matrix. Physics Education Journal, Vol. 27, Issue 4, pp. 221, (2010).
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