| Topic: |
Science > Physics |
| User: |
"sebire" |
| Date: |
26 Apr 2006 05:17:24 AM |
| Object: |
Degrees of Freedom confusion |
My friends and I all seem to be arguing over how many degrees of
freedom this system has (we don't seem to understand the concept...): A
rigid sphere thrown in the air.
One person says four: along, up, rotating forwards, rotating sideways
I say three: your time parameter, and the two rotations,
Another say two, but she says she's guessing...
Does anyone have the definitive answer? I know it's probably a
ridiculously stupid question, but seeing as no one seems to agree...
Thanks,
Natalie
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| User: "Fredrik Bulow" |
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| Title: Re: Degrees of Freedom confusion |
27 Apr 2006 01:46:36 AM |
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"sebire" <sebire@hotmail.com> writes:
My friends and I all seem to be arguing over how many degrees of
freedom this system has (we don't seem to understand the concept...): A
rigid sphere thrown in the air.
One person says four: along, up, rotating forwards, rotating sideways
I say three: your time parameter, and the two rotations,
Another say two, but she says she's guessing...
Does anyone have the definitive answer? I know it's probably a
ridiculously stupid question, but seeing as no one seems to agree...
Thanks,
Natalie
Hi Natalie
I think this is a very good fundamental question.
The short answer to the question is 6 degrees of freedom (DoF). One
way of choosing them is as follows:
In order to make this reasoning more clear let's imagine that there is
a (massless) yellow spot somewhere on the sphere.
Three coordinates (perhaps labeled x,y,y) to say where the center of
mass is. Two coordinates to tell which direction the yellow spot is
pointing in and finaly one more coordinate to tell how the ball is
rotated about the axis that goes through the centre and the yellow
spot.
/Fredrik
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| User: "sebire" |
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| Title: Re: Degrees of Freedom confusion |
28 Apr 2006 10:30:31 AM |
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Part of the reason I'm confused is because I've been told that "if one
has n degrees of freedom, one may use n generalised coordinates to
describe the system".
However, I would have thought that once the ball has been thrown, you
would only need to know t. x, y, z and the rotations will be functions
of t. So how comes it is 6 rather than 1?
Guessing from what I have been told, does this imply that t is not a
generalised coordinate?
My engineering friend says I've forgotten to take into account humidity
- the (cricket) ball might swing...
Natalie
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| User: "PD" |
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| Title: Re: Degrees of Freedom confusion |
28 Apr 2006 12:18:13 PM |
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sebire wrote:
Part of the reason I'm confused is because I've been told that "if one
has n degrees of freedom, one may use n generalised coordinates to
describe the system".
However, I would have thought that once the ball has been thrown, you
would only need to know t. x, y, z and the rotations will be functions
of t. So how comes it is 6 rather than 1?
Guessing from what I have been told, does this imply that t is not a
generalised coordinate?
My engineering friend says I've forgotten to take into account humidity
- the (cricket) ball might swing...
Natalie
Though you haven't asked it, I think I catch a glimmer of what's
perhaps really bothering you and what your friend is muttering about.
Maybe you are assuming that the rotation speeds of the cricket ball are
*constant*, so that you can predict what the rotation angles are just
from initial rotation positions and initial rotation angles. On the
other hand, there are forces (and hence accelerations) that are working
on the x, y, z positions as functions of t, and so those are more
complicated -- gravity, air drag, low-flying birds, etc.
And your friend perhaps is muttering that there are forces that are
causing rotational accelerations as well, such as air drag (which will
be more on a humid than on a dry day). Is that what you're thinking?
PD
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| User: "sebire" |
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| Title: Re: Degrees of Freedom confusion |
28 Apr 2006 04:04:01 PM |
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I was joking about the humidity. I was just thinking that if you knew
the initial velocity and initial rotations, the only thing determining
the entire position of the ball was t, because we know that it's
following a parabolic path, and there's no torque on the ball. So to
me, that suggests one degree of freedom.
I mean, you will still need to know v_x, v_y, v_z, but I didn't think
they counted as degrees of freedom as they don't change during the
flight of the ball.
Perhaps I do not quite understand the definition of a degree of
freedom.
Natalie
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| User: "Fredrik Bulow" |
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| Title: Re: Degrees of Freedom confusion |
30 Apr 2006 12:04:00 AM |
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"sebire" <sebire@hotmail.com> writes:
Guessing from what I have been told, does this imply that t is not a
generalised coordinate?
That is true for Newtonian mechanics. I strongly believe that you
consider time to be a degree of freedom in relativistic mechanics but
I'm not sure about this. Perhaps someone else could confirm this?
I was joking about the humidity. I was just thinking that if you knew
the initial velocity and initial rotations, the only thing determining
the entire position of the ball was t, because we know that it's
following a parabolic path, and there's no torque on the ball. So to
me, that suggests one degree of freedom.
Well, in many situation you can use some fact about the problem, like
symmetry or (like in this case) that you know that the trajectory goes
in one plane to determine the time evolution of a few DOF
trivially. That does however not reduce the actual number of DOF.
If some external forces are at work, like for instance the humidity
dependent air drag or gravitational forces from the moon, then that
will change the trajectory of your ball to something more complex than
a parabola, perhaps to something that does not even move in one
plane. However, six coordinates will still be enough to describe the
exact position and orientation of the ball hence the ball still has
six degrees of freedom.
I mean, you will still need to know v_x, v_y, v_z, but I didn't think
they counted as degrees of freedom as they don't change during the
flight of the ball.
Eh... now you got me a bit confused. First of all, v_x, v_y and v_z
are velocities. If any forces at all act on the ball they will not be
constant.
Secondly, velocities are NOT degrees of freedom, the mark the rate of
change of DOF with respect to time. To every (generalised) coordinate
there is a (generalised) speed.
I hope this was somewhat helpful.
/Fredrik
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| User: "Greg Hansen" |
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| Title: Re: Degrees of Freedom confusion |
29 Apr 2006 09:09:48 AM |
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sebire wrote:
I was joking about the humidity. I was just thinking that if you knew
the initial velocity and initial rotations, the only thing determining
the entire position of the ball was t, because we know that it's
following a parabolic path, and there's no torque on the ball. So to
me, that suggests one degree of freedom.
I mean, you will still need to know v_x, v_y, v_z, but I didn't think
they counted as degrees of freedom as they don't change during the
flight of the ball.
Perhaps I do not quite understand the definition of a degree of
freedom.
Natalie
If you know the initial position, velocities, and rotations, you can
find the resulting trajectory. But you don't know the resulting
trajectory until the equations of motion are solved, and the equations
of motion, when solved, will have three positions and three angles
versus time.
An example of a more constrained problem is bowling. The ball rolls on
the floor, so that eliminates the vertical as a degree of freedom-- no
(reasonable) set of initial conditions will make the ball go up and
down. Carts on an air track, a favorite in freshman physics labs, have
one degree of freedom-- there are no initial conditions that can make
them rotate or go sideways or vertically.
Sometimes conserved quantities can be used to eliminate a degree of
freedom. For instance, a pendulum is constrained radially by the rod
that it's attached to. That leaves two angles. But you know (or
assume) that angular momentum in one direction is conserved, and you can
align your system of coordinates with the plane of rotation. And then,
instead of coupled equations in two angles, you just have a
single-variable differential equation to solve.
The number of degrees of freedom is basically the number of differential
equations you need to solve, and the number of non-trivial terms in the
resulting equations of motion, where by trivial I mean x(t)=constant.
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| User: "PD" |
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| Title: Re: Degrees of Freedom confusion |
28 Apr 2006 12:06:04 PM |
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sebire wrote:
Part of the reason I'm confused is because I've been told that "if one
has n degrees of freedom, one may use n generalised coordinates to
describe the system".
However, I would have thought that once the ball has been thrown, you
would only need to know t. x, y, z and the rotations will be functions
of t. So how comes it is 6 rather than 1?
Guessing from what I have been told, does this imply that t is not a
generalised coordinate?
My engineering friend says I've forgotten to take into account humidity
- the (cricket) ball might swing...
Natalie
x, y, z are funtions of t also. It's a projectile.
But how do you know what those functions are?
You have two choices:
1. Measure them. In this case, you have measured positions of x, y, z
as a function of t.
2. Predict (calculate) them. In this case, you need to know initial
values (at t=0, say) of x, y, z, initial values of v_x, v_y, v_z, and a
list of forces (e.g. gravity) acting on the ball.
Take your choice.
But the point is, whatever you choose for x, y, z, you should also
choose for the 3 rotation angles. You seem to think that the rules are
different for x, y, z, than they are for the three rotation angles.
They're not.
PD
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| User: "tadchem" |
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| Title: Re: Degrees of Freedom confusion |
28 Apr 2006 04:12:45 PM |
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CWatters had it right in his second post.
http://scienceworld.wolfram.com/physics/DegreeofFreedom.html
A *generalized* rigid body has 6. Because of the symmetry of the
sphere 2 of those are lost - a rigid sphere can only rotate around one
axis.
Non-rigid bodies may have more. In chemical physics we work with
molecules. A useful model is one in which a molecule is made up of a
finite number of rigid spheres, each of which contributes 6 degrees of
freedom. A simple molecule like toluene (C7H8) has 15 atoms, so it has
15*6 = 90 degrees of freedom, most of which are involved in internal
vibrations.
In statistical mechanics a concept called "phase space" is introduced
wherein Avogadro's number (a little over 6.02*10^23) of particles are
used, each of which has 6 degrees of freedom. This leads to the
definition of entropy used in statistical mechanics:
http://scienceworld.wolfram.com/physics/Entropy.html
(Equation 1)
Tom Davidson
Richmond, VA
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| User: "Fredrik Bulow" |
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| Title: Re: Degrees of Freedom confusion |
30 Apr 2006 01:02:49 AM |
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"tadchem" <tadchem@comcast.net> writes:
CWatters had it right in his second post.
http://scienceworld.wolfram.com/physics/DegreeofFreedom.html
A *generalized* rigid body has 6. Because of the symmetry of the
sphere 2 of those are lost - a rigid sphere can only rotate around one
axis.
No, you are mistaken here. Perhaps you are thinking about electrons
that can only have spin up or down? A ball is a very classical object
compared to that. But by all means, go ahead and explain to all of us
how you can describe the direction of the north pole and the rotation
around this axis (say the angle of Sweden) with less than three
coordinates.
Non-rigid bodies may have more. In chemical physics we work with
molecules. A useful model is one in which a molecule is made up of a
finite number of rigid spheres, each of which contributes 6 degrees of
freedom. A simple molecule like toluene (C7H8) has 15 atoms, so it has
15*6 = 90 degrees of freedom, most of which are involved in internal
vibrations.
This is a completely different discussion! We are considering a
CLASSICAL problem.
In what sort of simulation would a 90 DOF model of C7H8 be useful?
The locations of the nucleus only require 15*3=45 coordinates (39 is
enough if the orientation and location of the whole molecule is of no
importance as is often the case).
Six coordinates per atom implies that you take the spatial coordinates
of each atom and another 3 coordinates to determine the attitude of
the individual atoms. Indeed I believe it is just as wrong to
determine the attitude of an atom with three coordinates as it is
wrong to determine the attitude of a ball with one. Surely you are not
doing this?
In statistical mechanics a concept called "phase space" is introduced
wherein Avogadro's number (a little over 6.02*10^23) of particles are
used, each of which has 6 degrees of freedom. This leads to the
definition of entropy used in statistical mechanics:
http://scienceworld.wolfram.com/physics/Entropy.html
(Equation 1)
The phase space is a N dimensional space where the degrees of freedom
are taken as coordinates. To determine a systems time evolution is to
find the path that the system traces through this space as a function
of time. In the case of the ball, the phase space has six dimensions.
/Fredrik
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| User: "CWatters" |
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| Title: Re: Degrees of Freedom confusion |
01 May 2006 11:32:15 AM |
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CWatters had it right in his second post.
http://scienceworld.wolfram.com/physics/DegreeofFreedom.html
A *generalized* rigid body has 6. Because of the symmetry of the
sphere 2 of those are lost - a rigid sphere can only rotate around one
axis.
Actually I'm going to correct myself again..
The school homework answer is 6 and a sphere could be made to move in all 6
(a rigid aeroplane can).
In my second post I was thinking that once thrown it was difficult to
make/maintain the ball rotating in more than one plane....however Beckham
and Botham might disagree.
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| User: "tj Frazir" |
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| Title: Re: Degrees of Freedom ENGLISH for MORONS |
01 May 2006 12:29:32 PM |
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You morons dont get te joke ?
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| User: "tadchem" |
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| Title: Re: Degrees of Freedom confusion |
30 Apr 2006 04:29:43 AM |
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Fredrik Bulow wrote:
"tadchem" <tadchem@comcast.net> writes:
CWatters had it right in his second post.
http://scienceworld.wolfram.com/physics/DegreeofFreedom.html
A *generalized* rigid body has 6. Because of the symmetry of the
sphere 2 of those are lost - a rigid sphere can only rotate around one
axis.
No, you are mistaken here. Perhaps you are thinking about electrons
that can only have spin up or down? A ball is a very classical object
compared to that. But by all means, go ahead and explain to all of us
how you can describe the direction of the north pole and the rotation
around this axis (say the angle of Sweden) with less than three
coordinates.
Very simple. "North" itself defines a body-centric coordinate system
(the number of degrees of freedom is independent of the choice of
coordinates). The 'sense' of the rotation is included in that
definition (rotation in the opposite direction defines "south").
For a spherically symmetric object this is enough. Rotation of less
than 2*pi radians around any arbitrary axis of a 3-D sphere produces a
body indistinguishable from the unrotated body. The moments of inertia
for rotation are indistinguishable, unlike the case with a book.
What is an "angle of Sweden"? What does it have to do with symmetrical
spheres? Are you perhaps adding another distinguishable body to the
system, changing the problem?
Non-rigid bodies may have more. In chemical physics we work with
molecules. A useful model is one in which a molecule is made up of a
finite number of rigid spheres, each of which contributes 6 degrees of
freedom. A simple molecule like toluene (C7H8) has 15 atoms, so it has
15*6 = 90 degrees of freedom, most of which are involved in internal
vibrations.
This is a completely different discussion! We are considering a
CLASSICAL problem.
What is non-classical about the construction of an object from 15
spheres of two different types with (20) flexible linkages? I made no
reference to non-classical properties such as electronic energy states.
I used the example because the internal vibrations which consume 87
degrees of freedom would be more intuitively obvious. I can see now
that I overestimated the intuitive ability f some readers.
Make it a structure of Construx (r) or Legos (r) connected by
long-floppy (non-rigid) springs, and the argument is unchanged.
In what sort of simulation would a 90 DOF model of C7H8 be useful?
IRL the degrees of freedom of C7H8 are used to analyze molecular
vibrations and compared with the optical spectra to identify the many
resonant frequencies that apear in the IR, UV, and Raman spectra.
Some very sophisticated chemistry is being done now by selectively
activating the vibrational modes of a molecule with lasers to weaken
and break specific bonds, leading to highly specific chemical reactions
with high yields. In principle, one could selectively break *only* the
C-H bond in a partic
The locations of the nucleus only require 15*3=45 coordinates (39 is
enough if the orientation and location of the whole molecule is of no
importance as is often the case).
Where is your nucleus *going*? You will need 3 degrees of freedom to
tell me that.
Degrees of freedom are used to specify the location of a rigid body - 3
each.
Degrees of freedom are *also* used to specify the velocity vector of a
rigid body - 3 each.
Six coordinates per atom implies that you take the spatial coordinates
of each atom and another 3 coordinates to determine the attitude of
the individual atoms.
Not "attitude" - velocity. We are dealing with a classical system here.
Heiseberg doesn't come out to play.
Indeed I believe it is just as wrong to
determine the attitude of an atom with three coordinates as it is
wrong to determine the attitude of a ball with one. Surely you are not
doing this?
I care nothing about the ball's 'attitude'. If by 'attitude' you are
referring to the direction the ball is 'facing' it is irrelevant. The
ball is spherically symmetric so it has no 'face.' I am talking
strictly position and motion here. Balls have no 'attitude' in the
sense of orientation.
The phase space is a N dimensional space where the degrees of freedom
are taken as coordinates. To determine a systems time evolution is to
find the path that the system traces through this space as a function
of time. In the case of the ball, the phase space has six dimensions.
When those six are studied for SYMMETRY, it turns out there are only 4
dimensions needed to describe the system. The trajectory happens to
lie entirely within a couple of planes, removing a couple of degrees of
freedom from the problem.
Consider a simple closed-loop trajectory in 3-space. If, by moving
your point of view around you should find that from a certain angle the
'loop' changes its appearance and becomes a 'line' segment, then you
have discovered where one degree of freedom was lost to symmetry. If
you then move around *within* the plane of the loop and discover a
point where every point on the loop seems to be the same distance from
you, you have discovered another place where symmetry has simplified
the problem - you are at the center of a circle.
Symmetry has important physical consequences as well as computational
ones.
Tom Davidson
Richmond, VA
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| User: "sebire" |
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| Title: Re: Degrees of Freedom confusion |
29 Apr 2006 09:31:52 AM |
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Oh dear, so some people say 6, some people say 4, and if you take into
account every atom in the ball, then you have lots...
The example given in my lectures is that of a pendulum, and it's just
oscillating in the vertical plane. That has one degree of freedom, the
angle it makes with the vertical. However, from what people are saying
is that the ball has 3 translational degrees of freedom, even though
that is just going to be moving in one plane, like the pendulum. Surely
then, it will only require one degree of freedom.
I'm beginning to wonder if there is actually a "right" answer. It just
seems to depend on how you define it!
Natalie
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| User: "tadchem" |
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| Title: Re: Degrees of Freedom confusion |
29 Apr 2006 06:38:04 PM |
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sebire wrote:
Oh dear, so some people say 6, some people say 4, and if you take into
account every atom in the ball, then you have lots...
Are you interested in the relative motions of each atom in the ball
with respect to the average of all of the others?
The example given in my lectures is that of a pendulum, and it's just
oscillating in the vertical plane.
A two-dimensional system will have fewer degrees of freedom - 4 per
rigid body. Mechanical constraints (like a pendulum rod/chain/wire)
can reduce this further.
That has one degree of freedom, the
angle it makes with the vertical.
Position AND velocity both count as specifiable 'degrees of freedom."
If it is movable than the angular velocity is also counted.
However, from what people are saying
is that the ball has 3 translational degrees of freedom, even though
that is just going to be moving in one plane, like the pendulum.
If it cannot move perpendicular to the plane or be out of the plane
that costs it 2 of its initial 6 degrees of freedom right there.
Surely
then, it will only require one degree of freedom.
If the ball is attached to somehting that keeps it at a fixed (ui.e.
not free) position relative to the axis of the pendulum then that costs
it two more degrees.
I'm beginning to wonder if there is actually a "right" answer. It just
seems to depend on how you define it!
No. You just have to keep a lear head about *what* you are couting -
variables in the descrption of the position and speed of an object.
Remember that if an object is 'free' to *move* along a line, then it is
also free to *be* at a number of different places on that line. That
is two freedoms for each line.
Even a ball following a 1-dimensional track has 2 degrees of freedom.
If it can be *placed* somewhere on the track but it cannot move, then
it has only one degree of freedom.
Tom Davidson
Richmond, VA
Natalie
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| User: "Fredrik Bulow" |
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| Title: Re: Degrees of Freedom confusion |
30 Apr 2006 11:41:14 AM |
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"tadchem" <tadchem@comcast.net> writes:
sebire wrote:
The example given in my lectures is that of a pendulum, and it's just
oscillating in the vertical plane.
A two-dimensional system will have fewer degrees of freedom - 4 per
rigid body. Mechanical constraints (like a pendulum rod/chain/wire)
can reduce this further.
That has one degree of freedom, the
angle it makes with the vertical.
Position AND velocity both count as specifiable 'degrees of freedom."
If it is movable than the angular velocity is also counted.
If memory serves, when I studied analytical mechanics, the definition
of degrees of freedom was given as the minimal number of generalised
coordinates required to describe the system. With that definition a
rigid pendulum restricted to move in a plane has one degree of
freedom. Apparently Tadchem considers generalised velocities to be
degrees of freedom too. That is, if q_1, q_2 ... q_N are the
generalised coordinates and r_1, r_2 ... r_N are the generalised
velocities (such that dq_1/dt= r_1) then I say that there are N DOF
and Tadchem says there are 2*N DOF. Since the definitions always
differ by a factor 2 they are both usefull. :-)
The book I used for my mechanics course was Goldstein's "Classical
Mechanics" so my definition ought to be the one found there. I will
verify this a.s.a.p.
/Fredrik
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| User: "tadchem" |
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| Title: Re: Degrees of Freedom confusion |
30 Apr 2006 01:57:24 PM |
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Fredrik Bulow wrote:
If memory serves, when I studied analytical mechanics, the definition
of degrees of freedom was given as the minimal number of generalised
coordinates required to describe the system. With that definition a
rigid pendulum restricted to move in a plane has one degree of
freedom. Apparently Tadchem considers generalised velocities to be
degrees of freedom too. That is, if q_1, q_2 ... q_N are the
generalised coordinates and r_1, r_2 ... r_N are the generalised
velocities (such that dq_1/dt= r_1) then I say that there are N DOF
and Tadchem says there are 2*N DOF. Since the definitions always
differ by a factor 2 they are both usefull. :-)
The mechanical system can indeed be described with N DOF if there is no
motion. If there is motion, then it is a problem in dynamics rather
than statics, and the generalized velocity coordinates (or, in a
classical system, the generalized momenta) will be required.
The book I used for my mechanics course was Goldstein's "Classical
Mechanics" so my definition ought to be the one found there. I will
verify this a.s.a.p.
I used Slater and Frank's "Mechanics". In Chapter IV they agree that
"The number of generalized coordinates required to determine the system
completely, in a problem involving constraints, is called the 'number
of degrees of freedom' of the system." The chapter then goes into a
derivation of Lagrange's and Hamilton's Equations, bith of which
require consideration of kinetic as well as potential energy.
If you cannot characterize the kinetic energy of a system you cannot
describe the system, and you cannot describe kinetic energy without the
generalized momenta.
With only the N DOF of position, you *can* draw a blueprint...
Tom Davidson
Richmond, VA
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| User: "Fredrik Bulow" |
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| Title: Re: Degrees of Freedom confusion |
01 May 2006 12:03:24 AM |
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"tadchem" <tadchem@comcast.net> writes:
When checking up wikipedia I found *two* mechanics related definitions
of DOF namely:
1. The one I am using
http://en.wikipedia.org/wiki/Degrees_of_freedom_%28engineering%29
2. The one Tadchem is using
http://en.wikipedia.org/wiki/Degrees_of_freedom_%28physics_and_chemistry%29
Since both definitions are obviously used I suggest that we from now on
use the shorthand FDOF for my definition and TDOF for tadchems
definition of DOF.
I would say that the definition given by tadchem:
Tadchem said:
I used Slater and Frank's "Mechanics". In Chapter IV they agree that
"The number of generalized coordinates required to determine the system
completely, in a problem involving constraints, is called the 'number
of degrees of freedom' of the system." The chapter then goes into a
derivation of Lagrange's and Hamilton's Equations, bith of which
require consideration of kinetic as well as potential energy.
is equivalent with mine. Perhaps tadchem considers momentum to be a
generalised coordinate? In that case it is equivalent with his
definition.
Having said this, I think it is time to return to the original problem
with the ball that someone throws. This system has 6 FDOF or 12 TDOF.
6 TDOF giving the location and momentum of the ball. 3 TDOF giving the
balls current attitude and another three specifying the rotation axis
and the rotation velocity.
Fredrik said:
But by all means, go ahead and explain to all of us
how you can describe the direction of the north pole and the rotation
around this axis (say the angle of Sweden) with less than three
coordinates.
Tadchem said:
Very simple. "North" itself defines a body-centric coordinate system
(the number of degrees of freedom is independent of the choice of
coordinates). The 'sense' of the rotation is included in that
definition (rotation in the opposite direction defines "south").
For a spherically symmetric object this is enough. Rotation of less
than 2*pi radians around any arbitrary axis of a 3-D sphere produces a
body indistinguishable from the unrotated body. The moments of inertia
for rotation are indistinguishable, unlike the case with a book.
What is an "angle of Sweden"? What does it have to do with symmetrical
spheres? Are you perhaps adding another distinguishable body to the
system, changing the problem?
If the ball is totally symmetrical or not is of no importance to the
number for DOF because it still makes sense to say that it has rotated
a certain angle about a certain axis.
If the ball is a point mass, then you have no DOF for attitude or
rotation, but this is a different case all together.
If the body is effected by a torque its axis of rotation will
change. Since this is something that can be changed independently of
everything else it needs its own TDOF. Since three parameters are
needed to specify a direction in space, three TDOF are required. The
same goes for the initial attitude of the ball that requires 3 FDOF.
Hope this cleard things up a bit!
/Fredrik
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| User: "tadchem" |
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| Title: Re: Degrees of Freedom confusion |
01 May 2006 04:53:14 AM |
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Fredrik Bulow wrote:
"tadchem" <tadchem@comcast.net> writes:
When checking up wikipedia I found *two* mechanics related definitions
of DOF namely:
1. The one I am using
http://en.wikipedia.org/wiki/Degrees_of_freedom_%28engineering%29
"degrees of freedom (DOF) are the set of independent displacements that
specify completely the displaced or deformed *position* of the body or
system" (emphasis mine)
As I stated before this is a problem in statics.
2. The one Tadchem is using
http://en.wikipedia.org/wiki/Degrees_of_freedom_%28physics_and_chemistry%29
Physicists are more concerned withthe *state* of the system, not just
the *position.*
Since both definitions are obviously used I suggest that we from now on
use the shorthand FDOF for my definition and TDOF for tadchems
definition of DOF.
Engineers and scientists sometimes use identical terms in different
ways.
Since this forum is "sci.physics" I am comfortable with using the
physicists definitions of DOF. Your definition would be more
acceptable in an engineering group.
I would say that the definition given by tadchem:
is equivalent with mine. Perhaps tadchem considers momentum to be a
generalised coordinate?
That works in Lagrangian and Hamiltonian analyses.
In that case it is equivalent with his
definition.
If the ball is totally symmetrical or not is of no importance to the
number for DOF because it still makes sense to say that it has rotated
a certain angle about a certain axis.
The identifiable number of *motions* of a body depend upon its
symmetry. The symmetry is therefore important toi the *physical* DOFs.
If the ball is a point mass,
....it is an oxymoron. The word "ball" implies at least a non-zero
radious - otherwise we would have called it a point.
If the body is effected by a torque its axis of rotation will
change. Since this is something that can be changed independently of
everything else it needs its own TDOF.
DOFs are not assigned based upon *contingencies.* We are describing
the system as it is, not what *might* happen to it if aliens come by
and hit it with a tractor beam. Should that happen, it would no longer
be a system in free motion. The source of the torque becomes a part of
the system
Hope this cleard things up a bit!
You could probably use a review in mechanics.
Tom Davidson
Richmond, VA
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| User: "Fredrik Bulow" |
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| Title: Re: Degrees of Freedom confusion |
01 May 2006 09:35:31 PM |
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"tadchem" <tadchem@comcast.net> writes:
I think it is time this for this conversation to be brought to an
end.
I have explain what I mean by degrees of freedom and what I mean
by a generalised coordinate. I've also figured out how tadchem defines
these concepts (he counts two degrees of freedom for every degree I
count because he considers (generalised) velocities and (generalised)
coordinates to be degrees of freedom whereas I only count
(generalised) coordinates). I have no problem with this because as
long as it is always made clear which one of these conventions a
person follows, I can follow his reasoning.
I am confident that 6 FDOF (my degrees of freedom) (or 12 TDOF
(tadchems degrees of freedom)) are required to describe a freely
moving ball. I've also, more than once, given examples as to how these
generalised coordinates might be chosen. It is time for tadchem to
give an example how he would describe the position of a ball using
fewer coordinates. After that he has done that, we can all either come
up with a configuration of the ball that can't be represented using
his coordinate system and dismiss his whole argument or, if we *fail*
to do so, agree with him.
Either way, the argument will be settled!
/Fredrik
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| User: "sebire" |
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| Title: Re: Degrees of Freedom confusion |
26 Apr 2006 05:25:42 AM |
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Ha, the latest offer from another friend is "one degree of freedom":
Once the ball is thrown its position and rotation is defined by time.
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| User: "CWatters" |
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| Title: Re: Degrees of Freedom confusion |
26 Apr 2006 09:33:56 AM |
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"sebire" <sebire@hotmail.com> wrote in message
news:1146047142.191189.140130@y43g2000cwc.googlegroups.com...
Ha, the latest offer from another friend is "one degree of freedom":
Once the ball is thrown its position and rotation is defined by time.
6 I believe. (7 if you count time).
3 rotational (roll, pitch and yaw - think aircraft)
3 translational (forward, back up and down)
I can easily imagine a cricket ball or a baseball moves in all 6 when
thrown.
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| User: "CWatters" |
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| Title: Re: Degrees of Freedom confusion |
26 Apr 2006 09:37:12 AM |
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"CWatters" <colin.watters@turnersNOSPAMoak.plus.net> wrote in message
news:444f84d4$0$2563$ed2619ec@ptn-nntp-reader02.plus.net...
"sebire" <sebire@hotmail.com> wrote in message
news:1146047142.191189.140130@y43g2000cwc.googlegroups.com...
Ha, the latest offer from another friend is "one degree of freedom":
Once the ball is thrown its position and rotation is defined by time.
6 I believe. (7 if you count time).
3 rotational (roll, pitch and yaw - think aircraft)
3 translational (forward, back up and down)
I can easily imagine a cricket ball or a baseball moves in all 6 when
thrown.
Ah let me correct that...
When thrown it may have components of motion in all six but after it's left
the hand it will only rotate in one plane (so 4 would be a good answer ...
unleast until it bounces).
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| User: "tadchem" |
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| Title: Re: Degrees of Freedom confusion |
30 Apr 2006 04:53:59 AM |
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I can see there are a few people who are not clear on the idea of
exactly what a 'degree of freedom' is.
A concise definition used by statisticians is:
"Statisticians use the terms "degrees of freedom" to describe the
number of values in the final calculation of a statistic that are free
to vary. "
(from http://www.animatedsoftware.com/statglos/sgdegree.htm)
In classical physics (dynamics) we describe the time-dependent motion
of a body using forces. The net force (= mass times acceleration, the
second derivative of the 3-dimensional displacement vector with respect
to time) is simply a sum of the individual forces.
An individual force can be a function of the object's location (the
3-dimensional displacement vector, giving us three 'variables' - three
degrees of freedom) or a function of the object's velocity (another
3-dimensional vector, the first derivative of the, giving us three more
'variables' - three more degrees of freedom) - giving us a total of six
variables needed to describe the forces acting on a body.
Things like mass and charge do not vary with time, so thay are
considered 'constants' fotr the equation and not variables. They may
not be the same in *another* problem, but that's *another* problem.
Sometimes additional constraints on a problem will reduce the number of
degrees of freedom by constrianing (removing freedom) one variable to
depend on another. For example if motion were confined to a flat
plane, we could define that plane as the X-Y plane and then Z would not
vary. We could fix Z = 0 (or any other value we choose) and the
problem of describing the position/motion of the body would be partialy
solved - it would have one less degree of freedom.
In general each additional simple constraint removes a degree of
freedom. If motion is confined to a predetermined line as a railroad
track, then 2 degrees of freedom have been lost.
The next lesson will be over the way symmetry affects degrees of
freedom, but I won't go on with that unless I'm sure you have a clear
understanding of just what a degree of freedom *is*.
Tom Davidson
Richmond, VA
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| User: "CWatters" |
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| Title: Re: Degrees of Freedom confusion |
01 May 2006 11:37:12 AM |
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"tadchem" <tadchem@comcast.net> wrote in message
news:1146390839.918189.313830@i40g2000cwc.googlegroups.com...
In general each additional simple constraint removes a degree of
freedom. If motion is confined to a predetermined line as a railroad
track, then 2 degrees of freedom have been lost.
So would the balls moment of inertia constrain it to revolve in just one
plane _once it had left the hand_? Can you make a ball tumble through the
air?
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| User: "tadchem" |
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| Title: Re: Degrees of Freedom confusion |
01 May 2006 04:07:48 PM |
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CWatters wrote:
"tadchem" <tadchem@comcast.net> wrote in message
news:1146390839.918189.313830@i40g2000cwc.googlegroups.com...
In general each additional simple constraint removes a degree of
freedom. If motion is confined to a predetermined line as a railroad
track, then 2 degrees of freedom have been lost.
So would the balls moment of inertia constrain it to revolve in just one
plane _once it had left the hand_?
Assuming it is symmetrical, yes. Symmetry means that all three moments
of inertia are equal; none is 'preferred'. Spinning around one axis
puts the axes of both other moments of inertia into the equatorial
plane with 0 angular momentum associated with either.
Can you make a ball tumble through the
air?
"Tumbling" requires some asymmetry. A sphere doesn't have it.
Tom Davidson
Richmond, VA
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| User: "CWatters" |
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| Title: Re: Degrees of Freedom confusion |
01 May 2006 04:41:15 PM |
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"tadchem" <tadchem@comcast.net> wrote in message
news:1146517668.072191.283810@i40g2000cwc.googlegroups.com...
CWatters wrote:
"tadchem" <tadchem@comcast.net> wrote in message
news:1146390839.918189.313830@i40g2000cwc.googlegroups.com...
In general each additional simple constraint removes a degree of
freedom. If motion is confined to a predetermined line as a railroad
track, then 2 degrees of freedom have been lost.
So would the balls moment of inertia constrain it to revolve in just one
plane _once it had left the hand_?
Assuming it is symmetrical, yes. Symmetry means that all three moments
of inertia are equal; none is 'preferred'. Spinning around one axis
puts the axes of both other moments of inertia into the equatorial
plane with 0 angular momentum associated with either.
Can you make a ball tumble through the
air?
"Tumbling" requires some asymmetry. A sphere doesn't have it.
I guess we best pick at the seam a bit :-)
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| User: "Greg Hansen" |
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| Title: Re: Degrees of Freedom confusion |
01 May 2006 06:14:37 PM |
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tadchem wrote:
CWatters wrote:
Can you make a ball tumble through the
air?
"Tumbling" requires some asymmetry. A sphere doesn't have it.
My brother showed me a juggling trick. He got two pins and a ball
going, and then announced "The double flip!", and threw the ball a
little higher.
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| User: "tj Frazir" |
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| Title: Re: Degrees of Freedom confusion |
30 Apr 2006 02:41:58 PM |
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It l=A9=A9Ks like English ..
HA HA HA
thats funnie the degrees of freedom ,,is engish .
haa hee
Degrees of Freedom (df) (from Internet Glossary of Statistical Terms)
Address:http://www.animatedsoftware.com/statglos/sgdegree.htm
Changed:10:38 AM on Thursday, May 8, 2003
Evry partical in orbit changed the orbital path to an eleipial orbit .
The eliptcal orbit moves the atom that wount move untill the orbits
change .
The eliptical orbits will move the center of mass from the center of
gravity. F is the distance from the center of the atoms G to the center
of the atoms Mass.
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| User: "T Wake" |
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| Title: Re: Degrees of Freedom confusion |
30 Apr 2006 03:04:17 PM |
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"tj Frazir" <GravityPhysics@webtv.net> wrote in message
news:5769-44551306-222@storefull-3212.bay.webtv.net...
It lİİKs like English ..
You wouldn't know, you illiterate fool.
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| User: "PD" |
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| Title: Re: Degrees of Freedom confusion |
26 Apr 2006 10:34:12 AM |
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sebire wrote:
My friends and I all seem to be arguing over how many degrees of
freedom this system has (we don't seem to understand the concept...): A
rigid sphere thrown in the air.
One person says four: along, up, rotating forwards, rotating sideways
I say three: your time parameter, and the two rotations,
Another say two, but she says she's guessing...
Does anyone have the definitive answer? I know it's probably a
ridiculously stupid question, but seeing as no one seems to agree...
Thanks,
Natalie
The question can be asked two ways:
1) How many degrees of freedom does *this* rigid sphere thrown in the
air have?
2) How many degrees of freedome does *any* rigid sphere thrown in the
air have?
The degrees of freedom basically count how many scalar functions of
time do you need to describe the motion, or at least that's the way
I'll set it up.
I believe the answer to (1) is 3: The ball translates in a plane (2D)
and rotates about a single axis (1D)
I believe the aswer to (2) is 6: Any ball's translation can be
described by three coordinates, and any ball's rotation can be
described by three coordinates.
PD
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