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Dear Bill,
For 3D motion, in a general case, the "moment of inertia"
about a rotation axis is not enough to determine the relationship
between Torque and Angular Acceleration (Newton-Eulero's law of angular
motion). Often, you need more data, i.e. the three "moments of inertia"
and the six "products of inertia". This array of 9 numbers, forming a 3x3
matrix, is called "tensor of inertia" (actually, three products of inertia
are equal to the other three, so you need just 3 moments and 3 products,
i.e. 6 numbers):
TENSOR OF INERTIA:
Ixx Ixy Ixz
Iyx Iyy Iyz
Izx Izy Izz
The Ixx, Iyy, and Izz elements of the tensor of inertia are just simple
moments of inertia about x, y and z. For instance, as you know, the moment
of inertia of a point mass m about x is:
Ixx = m * (y*y + z*z) (MOMENT OF INERTIA)
The other elements of the inertia tensor, such as Ixy, are the
"products of inertia" (Ixy = Iyx, etc.). For instance, the Ixy product of
inertia for a point mass is the product of its mass m and its x and y
positions:
Ixy = - m * x * y (PRODUCT OF INERTIA)
Of course, the products of inertia of a solid body can be obtained
numerically by summing the products of all the particles contained in the
body. As you know, the same is true for the moments of inertia. Integrals
can be used when the solid has a defined geometrical shape.
In some simple cases, the products of inertia are null, and the moments
of inertia are all what you need to use Newton-Eulero's formula. In these
cases, the tensor of inertia assumes a diagonal form:
DIAGONAL FORM OF THE TENSOR OF INERTIA
Ixx 0 0
0 Iyy 0
0 0 Izz
This singularity occurs when the three carthesian axes of your reference
frame are parallel to the three "principal axes" of the body.
[NOTE: by the way, fortunately, for any given instant you may use a
different ***inertial*** reference system, parallel to the principal axes of
the body at that instant, perform your computations inside that reference
system using the diagonal tensor of inertia, and eventually just transform
***the results*** so that they are resolved along the axes of your global
reference frame; using this trick, you don't need to compute and use the
products of inertia when you want to apply Newton-Eulero's law of angular
motion].
It would be too long to explain how to compute the directions of the
three principal axes. However, if we stay away from unusual math
representations of space, they are always orthogonal, and for most simple
regular solids they can be easily determined:
e.g. For a regular cone, cone frustum or cylinder, one of the principal axes
is "longitudinal", the other two can have any direction orthogonal to the
first. For a regular parallelepiped, the three axes are parallel to the
sides. For a sphere, the three principal axes can have any direction.
For the human body, the directions of the three principal axes vary
depending on the posture (Rick Hinrichs described how to compute them).
Everybody knows the following simple scalar relationship, valid in 2D:
T_z = I_zz * ALPHA_z (equation 1)
where:
T_z = scalar component along z of torque
(z is orthogonal to the 2D image).
I_zz = moment of inertia about z.
ALPHA_z = scalar component along z of angular acceleration.
However, when we use it, we should be always aware that this is not the
complete version of Newton-Eulero's law of angular motion, and it is valid
only if both the following conditions are met:
(1) the rotation occurs about a fixed axis (that I called z, in this
case);
(2) the rotation axis is parallel to a principal axis of the body
The general version of Newton-Eulero's equation, valid for any inertial
reference frame, not necessarily aligned with the three principal axes of
the body, is this:
T = {I}*alpha + omega x ({I}*omega) (equation 2)
where:
{I} = "tensor of inertia" matrix (see above)
T = torque vector
alpha = angular acceleration
omega = angular velocity
x = "cross" symbol indicating vector product
You can see that this equation is extremely more complex than the
corresponding equation for linear dynamics (Newton's second law):
F = m * a (equation 3)
which becomes, in scalar form and carthesian notation:
F_x = m * a_x (equation 4)
F_y = m * a_y (equation 5)
F_z = m * a_z (equation 6)
whereas equation 2, resolved in carthesian notation, becomes::
T_x = Ixx * alpha_x + Ixy * alpha_y + . ..... (equation 7)
T_y = Iyy * alpha_y + ................. (equation 8)
T_z = Izz * alpha_z + ........... (equation 9)
Comparing equations 4,5,6 with 7,8,9, it becomes evident that force and
linear acceleration always have the same direction while torque and angular
acceleration don't.
In some particular cases, you can even apply a torque on a free body
without producing angular acceleration!!!!
For instance, let's say you are standing and rotating with constant
angular velocity about a vertical axis passing through your CM (like a
dancer executing a "piroette"). If you hold a bag with your hand, and the
bag does not move relative to your arm and body, the bag will have constant
and vertical angular velocity as well. This means that it won't have angular
acceleration (alpha=0). However, surprisingly a net torque needs to be
applied
on it, equal to:
T = omega x ({I}*omega) (equation 10)
(the first term of equation 2 disappears because alpha=0)
This net torque is produced by your hand. It doesn't need to be a
couple. If you are not exerting a couple, then the force exerted by your
hand to hold the bag must needs to be "off center" i.e. must needs to have a
non-null lever arm relative to the CM of the bag. You can't help that,
otherwise the bag will rotate outward relative to your body, i.e. its
angular velocity won't be the same as that of your body.
This net torque is needed to change the direction of the angular
momentum of the bag. In fact, that vector is not vertical (see below), and
it is rotating together with the body, with the same angular velocity.
The angular momentum is:
H = {I}*omega
which means (whith a diagonal tensor of inertia):
H_x = Ixx * omega_x
H_y = Iyy * omega_y
H_z = Izz * omega_z
And, since Ixx, Iyy, and Izz are not equal, THE DIRECTION OF THE ANGULAR
MOMENTUM IS NOT THE SAME AS THAT OF THE ANGULAR VELOCITY, whereas the
direction of linear momentum (m*v) is obviously the same as that of linear
velocity (v).
All what I wrote above should be enough to show that THE INERTIA TENSOR
IS MUCH MORE COMPLEX THAN ITS LINEAR ANALOGUE, THE MASS, for two reasons:
(1) it is not a single value and
(2) it is not constant.
This is essentially the reason why angular dynamics is much more complex
than linear dynamics.
For your second question, the answer is: you need to integrate in both
directions. For the third: I don't know.
Regards,
Paolo de LEVA
University Institute of Motor Sciences
Sport Biomechanics
P. Lauro De Bosis, 6
00194 ROME - ITALY
Telephone: (39) 06.367.33.522
FAX/AM: (39) 06.367.33.517
FAX: (39) 06.36.00.31.99
Home:
Tel./FAX/AM: (39) 06.336.10.218
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For 3D motion, in a general case, the "moment of inertia"
about a rotation axis is not enough to determine the relationship
between Torque and Angular Acceleration (Newton-Eulero's law of angular
motion). Often, you need more data, i.e. the three "moments of inertia"
and the six "products of inertia". This array of 9 numbers, forming a 3x3
matrix, is called "tensor of inertia" (actually, three products of inertia
are equal to the other three, so you need just 3 moments and 3 products,
i.e. 6 numbers):
TENSOR OF INERTIA:
Ixx Ixy Ixz
Iyx Iyy Iyz
Izx Izy Izz
The Ixx, Iyy, and Izz elements of the tensor of inertia are just simple
moments of inertia about x, y and z. For instance, as you know, the moment
of inertia of a point mass m about x is:
Ixx = m * (y*y + z*z) (MOMENT OF INERTIA)
The other elements of the inertia tensor, such as Ixy, are the
"products of inertia" (Ixy = Iyx, etc.). For instance, the Ixy product of
inertia for a point mass is the product of its mass m and its x and y
positions:
Ixy = - m * x * y (PRODUCT OF INERTIA)
Of course, the products of inertia of a solid body can be obtained
numerically by summing the products of all the particles contained in the
body. As you know, the same is true for the moments of inertia. Integrals
can be used when the solid has a defined geometrical shape.
In some simple cases, the products of inertia are null, and the moments
of inertia are all what you need to use Newton-Eulero's formula. In these
cases, the tensor of inertia assumes a diagonal form:
DIAGONAL FORM OF THE TENSOR OF INERTIA
Ixx 0 0
0 Iyy 0
0 0 Izz
This singularity occurs when the three carthesian axes of your reference
frame are parallel to the three "principal axes" of the body.
[NOTE: by the way, fortunately, for any given instant you may use a
different ***inertial*** reference system, parallel to the principal axes of
the body at that instant, perform your computations inside that reference
system using the diagonal tensor of inertia, and eventually just transform
***the results*** so that they are resolved along the axes of your global
reference frame; using this trick, you don't need to compute and use the
products of inertia when you want to apply Newton-Eulero's law of angular
motion].
It would be too long to explain how to compute the directions of the
three principal axes. However, if we stay away from unusual math
representations of space, they are always orthogonal, and for most simple
regular solids they can be easily determined:
e.g. For a regular cone, cone frustum or cylinder, one of the principal axes
is "longitudinal", the other two can have any direction orthogonal to the
first. For a regular parallelepiped, the three axes are parallel to the
sides. For a sphere, the three principal axes can have any direction.
For the human body, the directions of the three principal axes vary
depending on the posture (Rick Hinrichs described how to compute them).
Everybody knows the following simple scalar relationship, valid in 2D:
T_z = I_zz * ALPHA_z (equation 1)
where:
T_z = scalar component along z of torque
(z is orthogonal to the 2D image).
I_zz = moment of inertia about z.
ALPHA_z = scalar component along z of angular acceleration.
However, when we use it, we should be always aware that this is not the
complete version of Newton-Eulero's law of angular motion, and it is valid
only if both the following conditions are met:
(1) the rotation occurs about a fixed axis (that I called z, in this
case);
(2) the rotation axis is parallel to a principal axis of the body
The general version of Newton-Eulero's equation, valid for any inertial
reference frame, not necessarily aligned with the three principal axes of
the body, is this:
T = {I}*alpha + omega x ({I}*omega) (equation 2)
where:
{I} = "tensor of inertia" matrix (see above)
T = torque vector
alpha = angular acceleration
omega = angular velocity
x = "cross" symbol indicating vector product
You can see that this equation is extremely more complex than the
corresponding equation for linear dynamics (Newton's second law):
F = m * a (equation 3)
which becomes, in scalar form and carthesian notation:
F_x = m * a_x (equation 4)
F_y = m * a_y (equation 5)
F_z = m * a_z (equation 6)
whereas equation 2, resolved in carthesian notation, becomes::
T_x = Ixx * alpha_x + Ixy * alpha_y + . ..... (equation 7)
T_y = Iyy * alpha_y + ................. (equation 8)
T_z = Izz * alpha_z + ........... (equation 9)
Comparing equations 4,5,6 with 7,8,9, it becomes evident that force and
linear acceleration always have the same direction while torque and angular
acceleration don't.
In some particular cases, you can even apply a torque on a free body
without producing angular acceleration!!!!
For instance, let's say you are standing and rotating with constant
angular velocity about a vertical axis passing through your CM (like a
dancer executing a "piroette"). If you hold a bag with your hand, and the
bag does not move relative to your arm and body, the bag will have constant
and vertical angular velocity as well. This means that it won't have angular
acceleration (alpha=0). However, surprisingly a net torque needs to be
applied
on it, equal to:
T = omega x ({I}*omega) (equation 10)
(the first term of equation 2 disappears because alpha=0)
This net torque is produced by your hand. It doesn't need to be a
couple. If you are not exerting a couple, then the force exerted by your
hand to hold the bag must needs to be "off center" i.e. must needs to have a
non-null lever arm relative to the CM of the bag. You can't help that,
otherwise the bag will rotate outward relative to your body, i.e. its
angular velocity won't be the same as that of your body.
This net torque is needed to change the direction of the angular
momentum of the bag. In fact, that vector is not vertical (see below), and
it is rotating together with the body, with the same angular velocity.
The angular momentum is:
H = {I}*omega
which means (whith a diagonal tensor of inertia):
H_x = Ixx * omega_x
H_y = Iyy * omega_y
H_z = Izz * omega_z
And, since Ixx, Iyy, and Izz are not equal, THE DIRECTION OF THE ANGULAR
MOMENTUM IS NOT THE SAME AS THAT OF THE ANGULAR VELOCITY, whereas the
direction of linear momentum (m*v) is obviously the same as that of linear
velocity (v).
All what I wrote above should be enough to show that THE INERTIA TENSOR
IS MUCH MORE COMPLEX THAN ITS LINEAR ANALOGUE, THE MASS, for two reasons:
(1) it is not a single value and
(2) it is not constant.
This is essentially the reason why angular dynamics is much more complex
than linear dynamics.
For your second question, the answer is: you need to integrate in both
directions. For the third: I don't know.
Regards,
Paolo de LEVA
University Institute of Motor Sciences
Sport Biomechanics
P. Lauro De Bosis, 6
00194 ROME - ITALY
Telephone: (39) 06.367.33.522
FAX/AM: (39) 06.367.33.517
FAX: (39) 06.36.00.31.99
Home:
Tel./FAX/AM: (39) 06.336.10.218
---------------------------------------------------------------
To unsubscribe send SIGNOFF BIOMCH-L to LISTSERV@nic.surfnet.nl
For information and archives: http://isb.ri.ccf.org/biomch-l
---------------------------------------------------------------