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

---------------------------------------------------------------