Science > Physics > Gyroscopes - Usenet Physics FAQ - Reference frames
| Topic: |
Science > Physics |
| User: |
"JM Albuquerque" |
| Date: |
26 Jul 2004 01:04:56 PM |
| Object: |
Gyroscopes - Usenet Physics FAQ - Reference frames |
Around these newsgroups there are current discussions about
inertia, mass, centrifugal force, equivalence principle, gravity, etc.
I've looked at the Physics FAQ:
http://math.ucr.edu/home/baez/physics/index.html
And I've notice the following:
1 - Gyroscopes don't belong to Physics.
2 - Centrifugal force doesn't even exist.
3 - Gravity also is a fictitious force like the centrifugal force.
http://math.ucr.edu/home/baez/physics/General/Centrifugal/centri.html
(See at bottom: Could gravity be a fictitious force too? YES.)
Why gyroscopes don't belong to the actual Physics is the point.
It looks like that the actual Physics don't like the reference frame of
gyroscopes, their 3 rotating motions, 3 force/torque at right angles,
and 3 non inertial reference frames (a triple problem with mass inertia).
Physics goes for inertial reference frames (those non rotating) and simply
discard the rotating frames of reference.
Looking at the universe, where are those non rotating frames (inertial)?
Since all features that one could see in the Universe is rotating and
working like a gyroscope (all of them are flat, like a disk, and all them
most likely precess if an external force is applied)?
In fact there are no true inertial frame of reference, since everything is
rotating at large scale.
If so why inertial reference frames and the equivalence principle were
both chosen?
Experts, please explain why a bicycle wheel don't fall under gravity, if a
precession motion exists?
Please explain this:
http://physics.nad.ru/Physics/English/gyro_tmp.htm
http://physics.nad.ru/Physics/English/gyro_txt.htm
And with 4Mb size the bicycle wheel:
http://www.rci.rutgers.edu/~williebo/zzzgyrovideo.MPG
(The usual bicycle wheel precessing around a vertical axis which doesn't
fall and turn around if allowed to do so).
What happened to the space-time curvature, near the bicycle wheel, in order
not to fall to the ground, like every other non rotating mass will do?
Is there any reasonable explanation why the bicycle wheel doesn't fall?
At such a small speed, how does GRT explain that the bicycle wheel doesn't
fall? Gyroscopes show that locally the space-time curvature clear depends on
the angular speed of the solid body.
No rotating speed and it will fall.
Just spin the body a little and it will not fall. Why?
Once upon a time Physics came to a problem about which frame of
reference is the best:
1 - Inertial reference frames;
2 - Rotating reference frames.
Physics took the first and forget the second.
And Physics failed to explain most of the basic working mechanisms
(those that everybody knows).
I'm not confortable with the fact that Physics doesn't explain the
mechanism by which gyroscopic motion opposes gravity.
It is a clear violation of the equivalence principle, but nobody cares.
But I do care!
Without a good explanation I propose that Usenet Physics FAQ clear say that
gyroscopic motion of matter cannot be explained by any actual theory, at:
http://math.ucr.edu/home/baez/physics/General/open_questions.html
Gyroscopic motion is an open question in Physics.
Or else please explain it.
-------------------------------
I've been studying and analysing gyroscopes.
Also the mechanical to electromagnetic energy conversion (the 3-phase
synchronous generators that power the World, also the homopolar
generator, and so on).
The conclusion I've found is that electromagnetic fields forces and
mechanical gyroscopic forces are very similar. Both apply the right hand
rule and the force field equations are the same in a local reference frame
(at any laboratory here on Earth).
The equations of circular motion, spiral motion, electromagnetism,
gyroscopic torque, etc. etc. could be of the form:
x(t) = K t (cos(wt))^2
y (t) = K t sin(wt)cos(wt)
z(t) = K t sin(wt)cos(wt)
Being:
t = time
K = constant
w = angular speed
The equations:
x(t) = K (cos(wt))^2
y (t) = K sin(wt)cos(wt)
Physically meant a solid disk rotating simultaneously around two
perpendicular axis, being the x,y the positions of a point of the disk
surface projected in a plane.
A circle is:
x(t) = K cos(wt)
y (t) = K sin(wt)
multiplying both by cos(wt) produces another rotation going in and out of
the paper plane in the perpendicular direction.
The first equation x(t) produces all the work (double frequency, always
positive and looks like power in electricity, could be gravity and is the
actual gyroscopic force applied).
The other two equations don't produce any work since the average is zero.
Plotting those x(t) and y(t) equation in a computer software enables one to
see many lovely things, like all the Physics passing in front of your eyes.
(the first "t" refers to linear motion and could be made constant at the
speed of light in one axis direction. The "t" close to "w" is always time
from zero to infinite.)
Them try:
(cos(t))^2 - sin(t)cos(t) for the Doppler shift of light.
Combinations of those circular terms:
cos(t)^2
cos(t)sin(t)
produce many evidence that all the Universe could be a large spinning
device and spin is all that matters. Linear motions are against mother
nature and that's why all the know forces come to day light when linear
motion exists.
A "rotating Mach like" principle perhaps could be the right approach.
But I don't have any theory, nor I want to have.
The above is just a curiosity.
I've been inspired here:
http://www.rci.rutgers.edu/~williebo/
And the best is here:
http://www.rci.rutgers.edu/~williebo/g7Gyroscopicnalysis.pdf
and predictions:
http://www.rci.rutgers.edu/~williebo/za27Predictions.pdf
Some strange things are:
http://www.keelynet.com/gravity/gyroag.htm
There are many other like the one above
True of False is the question ???
I guess that circular motion cannot be discarded form Physics.
Orbits and circular motion is not free fall, it is circular motion.
Remember Pioneer 10 going out of the solar system?
There is so many evidence that Physics must have taken the wrong
road some where that it almost stinks.
.
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| User: "Bilge" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
26 Jul 2004 02:49:12 PM |
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JM Albuquerque:
Around these newsgroups there are current discussions about
inertia, mass, centrifugal force, equivalence principle, gravity, etc.
I've looked at the Physics FAQ:
http://math.ucr.edu/home/baez/physics/index.html
And I've notice the following:
1 - Gyroscopes don't belong to Physics.
2 - Centrifugal force doesn't even exist.
3 - Gravity also is a fictitious force like the centrifugal force.
http://math.ucr.edu/home/baez/physics/General/Centrifugal/centri.html
(See at bottom: Could gravity be a fictitious force too? YES.)
Why gyroscopes don't belong to the actual Physics is the point.
It looks like that the actual Physics don't like the reference frame of
gyroscopes, their 3 rotating motions, 3 force/torque at right angles,
and 3 non inertial reference frames (a triple problem with mass inertia).
Physics goes for inertial reference frames (those non rotating) and simply
discard the rotating frames of reference.
Look in any classical mechanics book under ``space fixed axes'' and
``body fixed axes''.
.
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| User: "Meenken" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
01 Aug 2004 04:03:23 AM |
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"JM Albuquerque" <jm.aREMOV.E@sapo.pt> schreef in bericht
news:2mkva3FogcrpU1@uni-berlin.de...
Around these newsgroups there are current discussions about
inertia, mass, centrifugal force, equivalence principle, gravity, etc.
I've looked at the Physics FAQ:
http://math.ucr.edu/home/baez/physics/index.html
And I've notice the following:
1 - Gyroscopes don't belong to Physics.
2 - Centrifugal force doesn't even exist.
3 - Gravity also is a fictitious force like the centrifugal force.
http://math.ucr.edu/home/baez/physics/General/Centrifugal/centri.html
(See at bottom: Could gravity be a fictitious force too? YES.)
Why gyroscopes don't belong to the actual Physics is the point.
It looks like that the actual Physics don't like the reference frame of
gyroscopes, their 3 rotating motions, 3 force/torque at right angles,
and 3 non inertial reference frames (a triple problem with mass inertia).
Physics goes for inertial reference frames (those non rotating) and simply
discard the rotating frames of reference.
Looking at the universe, where are those non rotating frames (inertial)?
Since all features that one could see in the Universe is rotating and
working like a gyroscope (all of them are flat, like a disk, and all them
most likely precess if an external force is applied)?
In fact there are no true inertial frame of reference, since everything is
rotating at large scale.
If so why inertial reference frames and the equivalence principle were
both chosen?
Experts, please explain why a bicycle wheel don't fall under gravity, if a
precession motion exists?
Please explain this:
http://physics.nad.ru/Physics/English/gyro_tmp.htm
http://physics.nad.ru/Physics/English/gyro_txt.htm
And with 4Mb size the bicycle wheel:
http://www.rci.rutgers.edu/~williebo/zzzgyrovideo.MPG
(The usual bicycle wheel precessing around a vertical axis which doesn't
fall and turn around if allowed to do so).
What happened to the space-time curvature, near the bicycle wheel, in
order
not to fall to the ground, like every other non rotating mass will do?
Is there any reasonable explanation why the bicycle wheel doesn't fall?
At such a small speed, how does GRT explain that the bicycle wheel doesn't
fall? Gyroscopes show that locally the space-time curvature clear depends
on
the angular speed of the solid body.
No rotating speed and it will fall.
Just spin the body a little and it will not fall. Why?
I have read this whole thread,
when I have a bicycle weel and I hold it at both ends of the axle and spin
the weel,and then i let it loose it will all spinning fall to the ground
,what you probably mean is it won't fall over,yes?,
and when you try to bring it from vertical to f.i 45 degrees it resist
you,therefore,if you ride a bicycly you won't fall over to the ground,but as
soon as you stop,you will fall over,why is that?that is,what we are talking
about here,yes?
well,I have been thinking a long time about this problem,but lets assume we
drive in a car and go around a corner,if you take that corner very slow,you
dont need much force to let it take the corner,but now you go very fast,you
have to pull your steering very hard,(therefore we have power steering,we
don't need that for going slow around the corner,do we?),
so in fact ,you have to apply a force to let the car make that
corner,because you have to make an accelleration in a different
direction,(that force for the acceleration comes out of your motor,by
turning your steering wheel),
now,when you go slow,you need a small force,because you spread the force
over a long time to make the corner,
but when you go fast,you have to apply a lot more force to do that same work
i.e.the force needed for the requiered acceleration,to bring that car around
the corner)in a lot shorter time,
well then,for all he particles in your bicycle wheel,holds the same story
,they call that anoloog?
change of direction is acceleration in that other changed direction,and that
takes a force,which over time cost energy,it just depends how quick you want
to accomplish that,
so actually it is a matter of time and f=ma ,
the faster the wheel spins the more force you need to bring all the
particles in that wheel in an other direction,just the same as a car around
the corner(also more force on your steering wheel is requiered?
then I have a question,as gravity and acceleration is equivalent,when you
have a balloon (with light gas in it,i.e. the balloon would go in the air
when not being in the car)
in your car and you go around the corner,we people are pushed the other way
than the corner goes,but what does the balloon do?he should go the other way
as we do? same as the ballon would do outside the car(going against the
gravity),
it should do ,but I am not quite sure
marten
Once upon a time Physics came to a problem about which frame of
reference is the best:
1 - Inertial reference frames;
2 - Rotating reference frames.
Physics took the first and forget the second.
And Physics failed to explain most of the basic working mechanisms
(those that everybody knows).
I'm not confortable with the fact that Physics doesn't explain the
mechanism by which gyroscopic motion opposes gravity.
It is a clear violation of the equivalence principle, but nobody cares.
But I do care!
Without a good explanation I propose that Usenet Physics FAQ clear say
that
gyroscopic motion of matter cannot be explained by any actual theory, at:
http://math.ucr.edu/home/baez/physics/General/open_questions.html
Gyroscopic motion is an open question in Physics.
Or else please explain it.
-------------------------------
I've been studying and analysing gyroscopes.
Also the mechanical to electromagnetic energy conversion (the 3-phase
synchronous generators that power the World, also the homopolar
generator, and so on).
The conclusion I've found is that electromagnetic fields forces and
mechanical gyroscopic forces are very similar. Both apply the right hand
rule and the force field equations are the same in a local reference frame
(at any laboratory here on Earth).
The equations of circular motion, spiral motion, electromagnetism,
gyroscopic torque, etc. etc. could be of the form:
x(t) = K t (cos(wt))^2
y (t) = K t sin(wt)cos(wt)
z(t) = K t sin(wt)cos(wt)
Being:
t = time
K = constant
w = angular speed
The equations:
x(t) = K (cos(wt))^2
y (t) = K sin(wt)cos(wt)
Physically meant a solid disk rotating simultaneously around two
perpendicular axis, being the x,y the positions of a point of the disk
surface projected in a plane.
A circle is:
x(t) = K cos(wt)
y (t) = K sin(wt)
multiplying both by cos(wt) produces another rotation going in and out of
the paper plane in the perpendicular direction.
The first equation x(t) produces all the work (double frequency, always
positive and looks like power in electricity, could be gravity and is the
actual gyroscopic force applied).
The other two equations don't produce any work since the average is zero.
Plotting those x(t) and y(t) equation in a computer software enables one
to
see many lovely things, like all the Physics passing in front of your
eyes.
(the first "t" refers to linear motion and could be made constant at the
speed of light in one axis direction. The "t" close to "w" is always time
from zero to infinite.)
Them try:
(cos(t))^2 - sin(t)cos(t) for the Doppler shift of light.
Combinations of those circular terms:
cos(t)^2
cos(t)sin(t)
produce many evidence that all the Universe could be a large spinning
device and spin is all that matters. Linear motions are against mother
nature and that's why all the know forces come to day light when linear
motion exists.
A "rotating Mach like" principle perhaps could be the right approach.
But I don't have any theory, nor I want to have.
The above is just a curiosity.
I've been inspired here:
http://www.rci.rutgers.edu/~williebo/
And the best is here:
http://www.rci.rutgers.edu/~williebo/g7Gyroscopicnalysis.pdf
and predictions:
http://www.rci.rutgers.edu/~williebo/za27Predictions.pdf
Some strange things are:
http://www.keelynet.com/gravity/gyroag.htm
There are many other like the one above
True of False is the question ???
I guess that circular motion cannot be discarded form Physics.
Orbits and circular motion is not free fall, it is circular motion.
Remember Pioneer 10 going out of the solar system?
There is so many evidence that Physics must have taken the wrong
road some where that it almost stinks.
.
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| User: "JM Albuquerque" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
02 Aug 2004 05:09:38 AM |
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"Meenken" <marten@meenken.fol.nl> wrote:
then I have a question,as gravity and acceleration is equivalent,when you
have a balloon (with light gas in it,i.e. the balloon would go in the air
when not being in the car)
in your car and you go around the corner,we people are pushed the other
way than the corner goes,but what does the balloon do?he should go the
other way as we do? same as the ballon would do outside the car
(going against the gravity), it should do ,but I am not quite sure
marten
The balloon is another story, I guess.
You cannot spin a balloon, like a rock and string does.
The spinning air, that goes along with the balloon, will have more inertia
then the balloon it self, so the balloon is pushed inside and the air
outside.
Again this is a centrifugal problem, not a gyroscopic problem.
I would like very much to discuss the gyroscope... but the gyroscope looks
to be beyond any possible Physical explanation. So far I got about 3
objective and different explanations for the gyroscopic antigravity like
behaviour (none here in this thread). All of the said explanations are very
different and all of then have some new physical concepts involved.
There's only one thing clear to me. Gyroscopes cannot be explained by the
actual Physics based on a space-time geometry. So far I never seen and
explanation based on pure space-time geometry, which clear shows that
space-time geometry is a big balloon that will blow up one day.
.
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| User: "Tom Roberts" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
02 Aug 2004 05:05:41 AM |
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Meenken wrote:
when I have a bicycle weel and I hold it at both ends of the axle and spin
the weel,and then i let it loose it will all spinning fall to the ground
,what you probably mean is it won't fall over,yes?,
and when you try to bring it from vertical to f.i 45 degrees it resist
you,therefore,if you ride a bicycly you won't fall over to the ground,but as
soon as you stop,you will fall over,why is that?that is,what we are talking
about here,yes?
Not really. A bicycle does NOT stay upright due to gyroscopic action, it
does so because (a) the human rider continuously steers the wheels under
the center of gravity, and (b) the geometry of the front fork is
designed to do that automatically if left alone.
well,I have been thinking a long time about this problem,but lets assume we
drive in a car and go around a corner,if you take that corner very slow,you
dont need much force to let it take the corner,but now you go very fast,you
have to pull your steering very hard,(therefore we have power steering,we
don't need that for going slow around the corner,do we?),
so in fact ,you have to apply a force to let the car make that
corner,because you have to make an accelleration in a different
direction,(that force for the acceleration comes out of your motor,by
turning your steering wheel),
Most of the force comes from the tires' friction on the road. In fact,
that where it all comes from if you put the car in neutral (so the
engine does not pull the car through the corner). Note that the turning
force must be perpendicular to the car's motion, and a rear-wheel-drive
car's engine can only exert a force parallel to the car's motion (and
evn that requires friction of tires on road).
now,when you go slow,you need a small force,because you spread the force
over a long time to make the corner,
No. For a circular turn the centripetal force of the turn is
proportional to your mass, proportional to the square of your speed and
inversely proportional to your turning radius.
but when you go fast,you have to apply a lot more force to do that same work
For a circular turn no work is done -- the force is perpendicular to the
direction of motion.
[...]
then I have a question,as gravity and acceleration is equivalent,when you
have a balloon (with light gas in it,i.e. the balloon would go in the air
when not being in the car)
in your car and you go around the corner,we people are pushed the other way
than the corner goes,but what does the balloon do?he should go the other way
as we do? same as the ballon would do outside the car(going against the
gravity),
it should do ,but I am not quite sure
This is a simple experiment to do, just get a helium-filled balloon and
take it for a ride in your car. As expected, the helium balloon goes to
the front of the car when accelerating, to the rear when braking, and
toward the center of any turn. That's because the air is more dense than
the balloon, and is pushed harder than the baloon (all those forces are
proportional to the mass of the object, so there is more force on the
air in the same volumes as the baloon than on the balloon itself).
Tom Roberts tjroberts@lucent.com
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| User: "crynwulf" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
02 Aug 2004 08:08:07 PM |
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Tom Roberts wrote:
Meenken wrote:
when I have a bicycle weel and I hold it at both ends of the axle and
spin
the weel,and then i let it loose it will all spinning fall to the ground
,what you probably mean is it won't fall over,yes?,
and when you try to bring it from vertical to f.i 45 degrees it resist
you,therefore,if you ride a bicycly you won't fall over to the ground,but
as soon as you stop,you will fall over,why is that?that is,what we are
talking about here,yes?
Not really. A bicycle does NOT stay upright due to gyroscopic action, it
does so because (a) the human rider continuously steers the wheels under
the center of gravity, and (b) the geometry of the front fork is
designed to do that automatically if left alone.
In fact the bicycle does stay upright due to gyroscopic action. When you
lean, due to the geometry of the front fork, a torque is applied to the
gyroscopic fron wheel. That torque is in a vertical plane perpenduclar to
the direction of travel. When you lean right, the torque causes the front
of the wheel to turn to the right. The action isn't great, but it does make
bikes (and Harleys) easier to ride.
well,I have been thinking a long time about this problem,but lets assume
we drive in a car and go around a corner,if you take that corner very
slow,you dont need much force to let it take the corner,but now you go
very fast,you have to pull your steering very hard,(therefore we have
power steering,we don't need that for going slow around the corner,do
we?), so in fact ,you have to apply a force to let the car make that
corner,because you have to make an accelleration in a different
direction,(that force for the acceleration comes out of your motor,by
turning your steering wheel),
Most of the force comes from the tires' friction on the road. In fact,
that where it all comes from if you put the car in neutral (so the
engine does not pull the car through the corner). Note that the turning
force must be perpendicular to the car's motion, and a rear-wheel-drive
car's engine can only exert a force parallel to the car's motion (and
evn that requires friction of tires on road).
now,when you go slow,you need a small force,because you spread the force
over a long time to make the corner,
No. For a circular turn the centripetal force of the turn is
proportional to your mass, proportional to the square of your speed and
inversely proportional to your turning radius.
but when you go fast,you have to apply a lot more force to do that same
work
For a circular turn no work is done -- the force is perpendicular to the
direction of motion.
[...]
then I have a question,as gravity and acceleration is equivalent,when you
have a balloon (with light gas in it,i.e. the balloon would go in the
air when not being in the car)
in your car and you go around the corner,we people are pushed the other
way than the corner goes,but what does the balloon do?he should go the
other way as we do? same as the ballon would do outside the car(going
against the gravity),
it should do ,but I am not quite sure
This is a simple experiment to do, just get a helium-filled balloon and
take it for a ride in your car. As expected, the helium balloon goes to
the front of the car when accelerating, to the rear when braking, and
toward the center of any turn. That's because the air is more dense than
the balloon, and is pushed harder than the baloon (all those forces are
proportional to the mass of the object, so there is more force on the
air in the same volumes as the baloon than on the balloon itself).
Tom Roberts
--
Russ Lyttle
Not Powered by ActiveX
http://home.earthlink.net/~lyttlec/philosophy/logos.html
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| User: "Myxococcus xanthus" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
03 Aug 2004 04:19:12 AM |
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crynwulf <lyttlec@earthlink.net> wrote in message news:<XzBPc.7076$9Y6.2841@newsread1.news.pas.earthlink.net>...
In fact the bicycle does stay upright due to gyroscopic action. When you
lean, due to the geometry of the front fork, a torque is applied to the
gyroscopic fron wheel. That torque is in a vertical plane perpenduclar to
the direction of travel. When you lean right, the torque causes the front
of the wheel to turn to the right. The action isn't great, but it does make
bikes (and Harleys) easier to ride.
Please do a google search for references to
Jones, David E.H., "The Stability of the Bicycle", Physics Today (April 1970): 34-40
The gyroscopic contribution to bicycle stability is relatively minor.
Myxococcus xanthus
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| User: "greywolf42" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
03 Aug 2004 12:01:46 PM |
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"Myxococcus xanthus" <mold-guardian@comcast.net> wrote in message
news:ce5e7813.0408030119.77bb3b25@posting.google.com...
crynwulf <lyttlec@earthlink.net> wrote in message
news:<XzBPc.7076$9Y6.2841@newsread1.news.pas.earthlink.net>...
In fact the bicycle does stay upright due to gyroscopic action. When you
lean, due to the geometry of the front fork, a torque is applied to the
gyroscopic fron wheel. That torque is in a vertical plane perpenduclar
to
the direction of travel. When you lean right, the torque causes the
front
of the wheel to turn to the right. The action isn't great, but it does
make
bikes (and Harleys) easier to ride.
Please do a google search for references to
Jones, David E.H., "The Stability of the Bicycle", Physics Today (April
1970): 34-40
The gyroscopic contribution to bicycle stability is relatively minor.
As a teenager -- purely as a scientific experiment ;) -- I repeatedly
released an old bicycle to roll down a local hill. Except for the times it
didn't make the turn, it stayed rolling for several hundred yards. That's
hardly minor.
--
greywolf42
ubi dubium ibi libertas
{remove planet for e-mail}
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| User: "Myxococcus xanthus" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
04 Aug 2004 05:24:42 AM |
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"greywolf42" <mingstb@marssim-ss.com> wrote in message news:<10gvgrk5m6a1q83@corp.supernews.com>...
"Myxococcus xanthus" <mold-guardian@comcast.net> wrote in message
news:ce5e7813.0408030119.77bb3b25@posting.google.com...
The gyroscopic contribution to bicycle stability is relatively minor.
As a teenager -- purely as a scientific experiment ;) -- I repeatedly
released an old bicycle to roll down a local hill. Except for the times it
didn't make the turn, it stayed rolling for several hundred yards. That's
hardly minor.
Try walking a bicycle slowly. Tilt the bicycle to either side as you
walk. Observe the behavior of the front wheel. Gyroscopic action? No.
Myxococcus xanthus
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| User: "Harry Conover" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
09 Aug 2004 11:32:33 PM |
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(Myxococcus xanthus) wrote in message news:<ce5e7813.0408040224.51be16f4@posting.google.com>...
"greywolf42" <mingstb@marssim-ss.com> wrote in message news:<10gvgrk5m6a1q83@corp.supernews.com>...
"Myxococcus xanthus" < > wrote in message
news:ce5e7813.0408030119.77bb3b25@posting.google.com...
The gyroscopic contribution to bicycle stability is relatively minor.
As a teenager -- purely as a scientific experiment ;) -- I repeatedly
released an old bicycle to roll down a local hill. Except for the times it
didn't make the turn, it stayed rolling for several hundred yards. That's
hardly minor.
Try walking a bicycle slowly. Tilt the bicycle to either side as you
walk. Observe the behavior of the front wheel. Gyroscopic action? No.
Bingo! We have a winner!
The stability of a bicycle stems from the caster (or is it camber?)
present in the steering pivot. It has nothing to do with gyroscopic
effects.
Harry C.
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| User: "Phil Holman" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
10 Aug 2004 03:08:53 AM |
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"Harry Conover" <hhc314@yahoo.com> wrote in message
news:7ce4e226.0408092032.1890b8d9@posting.google.com...
mold-guardian@comcast.net (Myxococcus xanthus) wrote in message
news:<ce5e7813.0408040224.51be16f4@posting.google.com>...
"greywolf42" <mingstb@marssim-ss.com> wrote in message
news:<10gvgrk5m6a1q83@corp.supernews.com>...
"Myxococcus xanthus" <mold-guardian@comcast.net> wrote in message
news:ce5e7813.0408030119.77bb3b25@posting.google.com...
The gyroscopic contribution to bicycle stability is relatively
minor.
As a teenager -- purely as a scientific experiment ;) -- I
repeatedly
released an old bicycle to roll down a local hill. Except for the
times it
didn't make the turn, it stayed rolling for several hundred yards.
That's
hardly minor.
Try walking a bicycle slowly. Tilt the bicycle to either side as you
walk. Observe the behavior of the front wheel. Gyroscopic action?
No.
The wheel will turn with a stationary bicycle so you are correct,
Giyroscopic action, No.
The stability of a bicycle stems from the caster (or is it camber?)
present in the steering pivot. It has nothing to do with gyroscopic
effects.
The major contributor to bicycle stability comes from the rider. There
is no such thing as a bicycle that can be ridden by someone who hasn't
learned how to ride a bike and there isn't a bicycle that can't be
ridden by someone who has practiced riding it. Even reverse steering can
be mastered. Zero or even negative trail is easily compensated.
Castor provides the tendency for the front wheel to point straight ahead
when moving forward. When this resistance to turning is just overcome
by the mass imbalance of the front wheel and handlebars at low speed or
gyroscopic forces at higher speed, it is deemed as "stable". However
these forces are so small when compared to the forces that can be input
by the rider, even an "unstable" bicycle is easily ridden.
Gyroscopic forces help in riding a bicycle no hands and overshadow the
mass imbalance at higher speeds which adds to the perception of
stability. People like to use the bicycle traveling along without a
rider to explain how the mechanism is automatic but try putting a 200 lb
cadaver on the bicycle and see how far it goes.
This reminds me of the joke about the two friends driving along in a
car. The passenger says to the driver, you drive like a maniac. This is
nothing says the driver, you should see me when I'm on my own. Oh
jeepers, I'd hate to be in the car with you when you're on you own.
The point being, the enigma of the bicycle which appears to be stable
when no one is riding it is lost when someone sneaks aboard. Well you're
no longer driving on your own. To conclude therefore that the stability
must be inherent in the bicycle and not the rider is incorrect. Riders
do not rely on a mechanism that has them meandering all over the road.
Stability along a chosen path all comes down to rider input where
steering geometry provides only a minor contribution.
PH
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| User: "John Popelish" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
03 Aug 2004 05:06:34 PM |
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greywolf42 wrote:
"Myxococcus xanthus" <mold-guardian@comcast.net> wrote in message
news:ce5e7813.0408030119.77bb3b25@posting.google.com...
The gyroscopic contribution to bicycle stability is relatively minor.
As a teenager -- purely as a scientific experiment ;) -- I repeatedly
released an old bicycle to roll down a local hill. Except for the times it
didn't make the turn, it stayed rolling for several hundred yards. That's
hardly minor.
Good experiment, incorrect conclusion. Try locking the steering
position to straight ahead and repeating the experiment. This does
nothing to reduce the gyroscopic effects on balance, but eliminates
any effect on steering. You might even try riding the bike and see if
you can keep it from falling over without being able to steer.
--
John Popelish
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| User: "crynwulf" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
03 Aug 2004 06:08:50 PM |
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John Popelish wrote:
greywolf42 wrote:
"Myxococcus xanthus" <mold-guardian@comcast.net> wrote in message
news:ce5e7813.0408030119.77bb3b25@posting.google.com...
The gyroscopic contribution to bicycle stability is relatively minor.
As a teenager -- purely as a scientific experiment ;) -- I repeatedly
released an old bicycle to roll down a local hill. Except for the times
it
didn't make the turn, it stayed rolling for several hundred yards.
That's hardly minor.
Good experiment, incorrect conclusion. Try locking the steering
position to straight ahead and repeating the experiment. This does
nothing to reduce the gyroscopic effects on balance, but eliminates
any effect on steering. You might even try riding the bike and see if
you can keep it from falling over without being able to steer.
--
John Popelish
Try this. Hold the front wheel off the ground and get it spinning as fast as
you can. Twist on the handle bars. Feel the gyroscopic effect?
The reason the greywolfs bike stayed up wasn't because the gyroscopic effect
held it up, but because it caused the front wheel to turn to keep the
center of mass between the axles.
--
Russ Lyttle
Not Powered by ActiveX
http://home.earthlink.net/~lyttlec/philosophy/logos.html
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| User: "Tom Roberts" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
03 Aug 2004 07:18:35 PM |
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greywolf42 wrote:
"Myxococcus xanthus" <mold-guardian@comcast.net> wrote in message
news:ce5e7813.0408030119.77bb3b25@posting.google.com...
Please do a google search for references to
Jones, David E.H., "The Stability of the Bicycle", Physics Today (April
1970): 34-40
The gyroscopic contribution to bicycle stability is relatively minor.
As a teenager -- purely as a scientific experiment ;) -- I repeatedly
released an old bicycle to roll down a local hill. Except for the times it
didn't make the turn, it stayed rolling for several hundred yards. That's
hardly minor.
But, of course, the fact that an unoccupied bicycle can roll on its own
has nothing to do with gyroscopic action of its wheels. It can do that
because the geometry of its front fork provides feedback to make it
steer the wheels under its center of gravity. The couple formed by the
lean of the bike, the camber of the front wheel[#], and the weight of
the bike is MUCH larger than the gyroscopic force. And unlike the
latter, it is independent of speed, and works at low speeds (but not at
zero speed, so it does fall when friction brings it to a halt)
[#] extend the axis of the front fork to the ground, and note
it is in front of the wheel's point on the ground. That
distance is called the camber. The front fork is usually
bent forward to reduce the camber to the desired value for
the inclination of the fork. Compare a racing bike and a
touring bike....
If the effect were gyroscopic action then the bicycle would presumably
be stable going backwards also. Try it -- it isn't stable at all and
falls immediately -- the geometry of the front fork (at the rear) causes
oversteer.
Tom Roberts tjroberts@lucent.com
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| User: "greywolf42" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
04 Aug 2004 02:45:37 PM |
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"Tom Roberts" <tjroberts@lucent.com> wrote in message
news:vXVPc.3$Y94.2@newssvr33.news.prodigy.com...
greywolf42 wrote:
"Myxococcus xanthus" <mold-guardian@comcast.net> wrote in message
news:ce5e7813.0408030119.77bb3b25@posting.google.com...
Please do a google search for references to
Jones, David E.H., "The Stability of the Bicycle", Physics Today (April
1970): 34-40
The gyroscopic contribution to bicycle stability is relatively minor.
As a teenager -- purely as a scientific experiment ;) -- I repeatedly
released an old bicycle to roll down a local hill. Except for the times
it didn't make the turn, it stayed rolling for several hundred yards.
That's hardly minor.
But, of course, the fact that an unoccupied bicycle can roll on its own
has nothing to do with gyroscopic action of its wheels. It can do that
because the geometry of its front fork provides feedback to make it
steer the wheels under its center of gravity.
LOL! And the geometry, of course, simply acts on it's own. ;)
The question is *how* the feedback is provided.
The couple formed by the
lean of the bike, the camber of the front wheel[#], and the weight of
the bike is MUCH larger than the gyroscopic force. And unlike the
latter, it is independent of speed,
Then the bike would have only one stable speed. And yet, my bicycle stayed
up only at higher speeds. It fell over immediately, if I did not get it
moving fast enough (a short run). And once going, it was stable from about
5 mph to what I later estimated at 30 mph. (The bike took some damage when
it hit a ditch.)
I presume that it is the gyroscopic effect that gives rise to the 'couple'
to which you are referring.
and works at low speeds (but not at
zero speed, so it does fall when friction brings it to a halt)
How about at 0.1 mph? Nope it falls. Even though friction is overcome, and
the handlebars turn.
[#] extend the axis of the front fork to the ground, and note
it is in front of the wheel's point on the ground. That
distance is called the camber. The front fork is usually
bent forward to reduce the camber to the desired value for
the inclination of the fork. Compare a racing bike and a
touring bike....
To a "chopper.".....
I believe you mean "caster", not "camber." "Camber" is the angle of the
wheel from vertical, not the displacement from the turning axis.
http://www.team3s.com/FAQcastcamb.htm
If the effect were gyroscopic action then the bicycle would presumably
be stable going backwards also. Try it -- it isn't stable at all and
falls immediately -- the geometry of the front fork (at the rear) causes
oversteer.
I've also driven forklifts -- which have their steering wheels in the
rear -- and don't lean. And they are unstable in steering going forwards.
In other words, the instability of steering in the following wheel(s) due to
the dynamics of inertia, and is not just dependent upon caster. Examine the
dynamics of a turn with steering in the lead wheel(s) and with steering in
the following wheel(s):
Lead: The wheels initiate a turn to the right. They force the center of
mass to the right, by moving the front of the vehicle to the right. The
inertia of the vehicle will oppose the turning tendency. (Negative
feedback.)
Following: The wheels initiate a turn to the right, by moving the rear of
the vehicle to the left. They force the center of mass to the left. The
inertia of the vehicle will enhance the turning tendency. (Positive
feedback.)
Now, one can engineer a car or bicycle with 'oversteer' by fiddling with the
caster. But this relates to the force needed -- not the balance points.
--
greywolf42
ubi dubium ibi libertas
{remove planet for e-mail}
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| User: "Paul B. Andersen" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
09 Aug 2004 07:26:47 AM |
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"greywolf42" <mingstb@marssim-ss.com> skrev i melding news:10h2ee4n3gvp8bd@corp.supernews.com...
"Tom Roberts" <tjroberts@lucent.com> wrote in message
news:vXVPc.3$Y94.2@newssvr33.news.prodigy.com...
greywolf42 wrote:
"Myxococcus xanthus" <mold-guardian@comcast.net> wrote in message
news:ce5e7813.0408030119.77bb3b25@posting.google.com...
Please do a google search for references to
Jones, David E.H., "The Stability of the Bicycle", Physics Today (April
1970): 34-40
The gyroscopic contribution to bicycle stability is relatively minor.
As a teenager -- purely as a scientific experiment ;) -- I repeatedly
released an old bicycle to roll down a local hill. Except for the times
it didn't make the turn, it stayed rolling for several hundred yards.
That's hardly minor.
But, of course, the fact that an unoccupied bicycle can roll on its own
has nothing to do with gyroscopic action of its wheels. It can do that
because the geometry of its front fork provides feedback to make it
steer the wheels under its center of gravity.
LOL! And the geometry, of course, simply acts on it's own. ;)
The question is *how* the feedback is provided.
The front wheel touches the ground behind the axis
of rotation of the fork.
When the bike is leaned, the front wheel will will
thus turn in the direction it is leaned.
The couple formed by the
lean of the bike, the camber of the front wheel[#], and the weight of
the bike is MUCH larger than the gyroscopic force. And unlike the
latter, it is independent of speed,
Then the bike would have only one stable speed. And yet, my bicycle stayed
up only at higher speeds. It fell over immediately, if I did not get it
moving fast enough (a short run). And once going, it was stable from about
5 mph to what I later estimated at 30 mph. (The bike took some damage when
it hit a ditch.)
If the speed is too small, the wheel will not turn fast enough to
compensate. Remember that for a given angle of the wheel,
the faster you go, the faster will the wheel move in the direction
which cancels the lean.
Loosely put: you have to move the front wheel inn under
the bike faster than it falls.
So it will be stable only above a certain speed.
I presume that it is the gyroscopic effect that gives rise to the 'couple'
to which you are referring.
No.
Try this:
Lean your bike (by leaning your body in the opposite
direction) while riding along a straight line.
You will feel a static torque on the handlebar.
This torque depend only on the leaning angle,
not on the speed.
There is no gyroscopic effect in this case.
Paul
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| User: "Phil Holman" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
09 Aug 2004 09:28:55 PM |
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"Paul B. Andersen" <paul.b.andersen@hia.no> wrote in message
news:cf7qj2$7ct$1@dolly.uninett.no...
"greywolf42" <mingstb@marssim-ss.com> skrev i melding
news:10h2ee4n3gvp8bd@corp.supernews.com...
"Tom Roberts" <tjroberts@lucent.com> wrote in message
news:vXVPc.3$Y94.2@newssvr33.news.prodigy.com...
greywolf42 wrote:
"Myxococcus xanthus" <mold-guardian@comcast.net> wrote in
message
news:ce5e7813.0408030119.77bb3b25@posting.google.com...
Please do a google search for references to
Jones, David E.H., "The Stability of the Bicycle", Physics Today
(April
1970): 34-40
The gyroscopic contribution to bicycle stability is relatively
minor.
As a teenager -- purely as a scientific experiment ;) -- I
repeatedly
released an old bicycle to roll down a local hill. Except for
the times
it didn't make the turn, it stayed rolling for several hundred
yards.
That's hardly minor.
But, of course, the fact that an unoccupied bicycle can roll on
its own
has nothing to do with gyroscopic action of its wheels. It can do
that
because the geometry of its front fork provides feedback to make
it
steer the wheels under its center of gravity.
LOL! And the geometry, of course, simply acts on it's own. ;)
The question is *how* the feedback is provided.
The front wheel touches the ground behind the axis
of rotation of the fork.
When the bike is leaned, the front wheel will will
thus turn in the direction it is leaned.
This is called trail and is not the reason the wheel turns. In fact the
more trail there is the more tendancy for the wheel to keep its forward
direction due to castor.
The couple formed by the
lean of the bike, the camber of the front wheel[#], and the weight
of
the bike is MUCH larger than the gyroscopic force. And unlike the
latter, it is independent of speed,
Then the bike would have only one stable speed. And yet, my bicycle
stayed
up only at higher speeds. It fell over immediately, if I did not
get it
moving fast enough (a short run). And once going, it was stable
from about
5 mph to what I later estimated at 30 mph. (The bike took some
damage when
it hit a ditch.)
If the speed is too small, the wheel will not turn fast enough to
compensate. Remember that for a given angle of the wheel,
the faster you go, the faster will the wheel move in the direction
which cancels the lean.
Loosely put: you have to move the front wheel inn under
the bike faster than it falls.
So it will be stable only above a certain speed.
I presume that it is the gyroscopic effect that gives rise to the
'couple'
to which you are referring.
No.
Try this:
Lean your bike (by leaning your body in the opposite
direction) while riding along a straight line.
You will feel a static torque on the handlebar.
This torque depend only on the leaning angle,
not on the speed.
There is no gyroscopic effect in this case.
You don't have to be riding along. Hold the seat of a stationary bicycle
and lean it slowly and the wheel will turn in the same direction due to
the mass imbalance about the steering axis. Do this very fast and the
wheel will initially turn in the opposite direction (inertia) before
turning in the same direction as the lean. The technique is used to
steer a bicycle while walking along just holding the seat. Gyroscopic
forces do not become dominant until about 15mph.
PH
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| User: "greywolf42" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
09 Aug 2004 11:07:08 AM |
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"Paul B. Andersen" <paul.b.andersen@hia.no> wrote in message
news:cf7qj2$7ct$1@dolly.uninett.no...
"greywolf42" <mingstb@marssim-ss.com> skrev i melding
news:10h2ee4n3gvp8bd@corp.supernews.com...
{snip higher levels}
The question is *how* the feedback is provided.
The front wheel touches the ground behind the axis
of rotation of the fork.
When the bike is leaned, the front wheel will will
thus turn in the direction it is leaned.
A reasonable theory.
Then the bike would have only one stable speed. And yet, my bicycle
stayed
up only at higher speeds. It fell over immediately, if I did not get it
moving fast enough (a short run). And once going, it was stable from
about
5 mph to what I later estimated at 30 mph. (The bike took some damage
when
it hit a ditch.)
If the speed is too small, the wheel will not turn fast enough to
compensate.
Then it is the turning of the wheel that controls the stability. And the
turning of the wheel is driven by .....?
Remember that for a given angle of the wheel,
the faster you go, the faster will the wheel move in the direction
which cancels the lean.
Yes, that is the gyroscopic effect in action.
Loosely put: you have to move the front wheel inn under
the bike faster than it falls.
So it will be stable only above a certain speed.
I presume that it is the gyroscopic effect that gives rise to the
'couple'
to which you are referring.
No.
Try this:
Lean your bike (by leaning your body in the opposite
direction) while riding along a straight line.
You will feel a static torque on the handlebar.
This torque depend only on the leaning angle,
not on the speed.
Hmmm. Tried it. And the torque appeared to vary with speed. Contrary to
your claim. (Of course, I did not use a calibrated force instrument.)
There is no gyroscopic effect in this case.
Now, how do you explain that a bicycle tire -- sans bicycle -- exhibits the
same tendency? That is, it will stay up as it rolls down the hill. There
is no caster to control it.
--
greywolf42
ubi dubium ibi libertas
{remove planet for e-mail}
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| User: "Myxococcus xanthus" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
09 Aug 2004 04:41:06 PM |
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"greywolf42" <mingstb@marssim-ss.com> wrote in message news:<10hf8eht69qh363@corp.supernews.com>...
Now, how do you explain that a bicycle tire -- sans bicycle -- exhibits the
same tendency? That is, it will stay up as it rolls down the hill. There
is no caster to control it.
No one has ever claimed that there ISN'T any gyroscopic effect at all.
What we have stated, is that the gyroscopic contribution to bicycle
stability is minor compared with other effects. Multiple published
studies that have quantified the various contributions to bicycle
stability have all come to the same conclusion.
Myxococcus xanthus
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| User: "greywolf42" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
10 Aug 2004 11:02:14 AM |
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"Myxococcus xanthus" <mold-guardian@comcast.net> wrote in message
news:ce5e7813.0408091341.39a8febf@posting.google.com...
"greywolf42" <mingstb@marssim-ss.com> wrote in message
news:<10hf8eht69qh363@corp.supernews.com>...
Now, how do you explain that a bicycle tire -- sans bicycle -- exhibits
the same tendency? That is, it will stay up as it rolls down the hill.
There is no caster to control it.
No one has ever claimed that there ISN'T any gyroscopic effect at all.
Actually, that is *precisely* the claim that was repeatedly made. Which is
why I bothered to join in at all.
What we have stated, is that the gyroscopic contribution to bicycle
stability is minor compared with other effects.
Now, can you support that claim with quantification?
Multiple published
studies that have quantified the various contributions to bicycle
stability have all come to the same conclusion.
References to those multiple published studies, please. Or, simply provide
the relationships that quantify the contributions.
And the bicycle wheel without the caster? (Auto tires are more stable than
bicycle tires ... because of the flat tread face.)
My task here is done.....
--
greywolf42
ubi dubium ibi libertas
{remove planet for e-mail}
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| User: "Myxococcus xanthus" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
10 Aug 2004 05:07:05 PM |
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"greywolf42" <mingstb@marssim-ss.com> wrote in message news:<10hhtfbfecrlk2c@corp.supernews.com>...
"Myxococcus xanthus" <mold-guardian@comcast.net> wrote in message
news:ce5e7813.0408091341.39a8febf@posting.google.com...
No one has ever claimed that there ISN'T any gyroscopic effect at all.
Actually, that is *precisely* the claim that was repeatedly made. Which is
why I bothered to join in at all.
I never said it. Tom Roberts never said it. Paul Andersen never said
it. Phil Holman never said it. Brian Hutchings never said it.
The only person saying it seems to have been Harry Conover, and that
was only AFTER I posted my statement "No one has ever claimed..."
What we have stated, is that the gyroscopic contribution to bicycle
stability is minor compared with other effects.
Now, can you support that claim with quantification?
Read
http://socrates.berkeley.edu/~fajans/pub/pdffiles/SteerBikeAJP.PDF
Also, do a google search for references to
Jones, David E.H., "The Stability of the Bicycle", Physics Today
(April 1970): 34-40
Many online sources cite the Jones paper (which I read in the
original, a couple of years back), and a few of them offer fairly
complete and reasonably accurate summaries.
Multiple published
studies that have quantified the various contributions to bicycle
stability have all come to the same conclusion.
References to those multiple published studies, please. Or, simply provide
the relationships that quantify the contributions.
And the bicycle wheel without the caster? (Auto tires are more stable than
bicycle tires ... because of the flat tread face.)
My task here is done.....
What task? Intentionally misreading what people post, just so you can
troll?
Myxococcus xanthus
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| User: "greywolf42" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
11 Aug 2004 11:46:02 AM |
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"Myxococcus xanthus" <mold-guardian@comcast.net> wrote in message
news:ce5e7813.0408101407.20838b10@posting.google.com...
"greywolf42" <mingstb@marssim-ss.com> wrote in message
news:<10hhtfbfecrlk2c@corp.supernews.com>...
"Myxococcus xanthus" <mold-guardian@comcast.net> wrote in message
news:ce5e7813.0408091341.39a8febf@posting.google.com...
No one has ever claimed that there ISN'T any gyroscopic effect at all.
Actually, that is *precisely* the claim that was repeatedly made. Which
is
why I bothered to join in at all.
I never said it. Tom Roberts never said it. Paul Andersen never said
it. Phil Holman never said it. Brian Hutchings never said it.
The only person saying it seems to have been Harry Conover, and that
was only AFTER I posted my statement "No one has ever claimed..."
Sheesh.
Here's Tom's claim: "But, of course, the fact that an unoccupied bicycle can
roll on its own has nothing to do with gyroscopic action of its wheels."
Hint: "Nothing to do with gyroscopic action" is equivalent to stating that
there isn't a gyroscopic effect. Nothing = 0.
What we have stated, is that the gyroscopic contribution to bicycle
stability is minor compared with other effects.
Now, can you support that claim with quantification?
Read
http://socrates.berkeley.edu/~fajans/pub/pdffiles/SteerBikeAJP.PDF
Also, do a google search for references to
Jones, David E.H., "The Stability of the Bicycle", Physics Today
(April 1970): 34-40
Many online sources cite the Jones paper (which I read in the
original, a couple of years back), and a few of them offer fairly
complete and reasonably accurate summaries.
Thanks.
Multiple published
studies that have quantified the various contributions to bicycle
stability have all come to the same conclusion.
References to those multiple published studies, please. Or, simply
provide
the relationships that quantify the contributions.
And the bicycle wheel without the caster? (Auto tires are more stable
than
bicycle tires ... because of the flat tread face.)
My task here is done.....
What task? Intentionally misreading what people post, just so you can
troll?
LOL! The task of popping pompous proclamations from perniciously proud
proponents of popular perversions of reality.
--
greywolf42
ubi dubium ibi libertas
{remove planet for e-mail}
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| User: "Myxococcus xanthus" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
11 Aug 2004 05:48:29 PM |
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"greywolf42" <mingstb@marssim-ss.com> wrote in message news:<10hkkfvj8dhe3e9@corp.supernews.com>...
Sheesh.
Here's Tom's claim: "But, of course, the fact that an unoccupied bicycle can
roll on its own has nothing to do with gyroscopic action of its wheels."
Hint: "Nothing to do with gyroscopic action" is equivalent to stating that
there isn't a gyroscopic effect. Nothing = 0.
Sheesh.
Tom wrote:
"The couple formed by the lean of the bike, the camber of the front
wheel[#], and the weight of the bike is MUCH larger than the
gyroscopic force."
Hint: "much larger than the gyroscopic force" is equivalent to stating
that the gyroscopic force is non-zero, but relatively minor in its
overall contribution to bicycle stability. Something != 0.
Bicycles have been built where the gyroscopic force has been
completely canceled out by means of a counter-rotating wheel, and they
roll on their own just fine.
Then again, one should not confuse
1) "factors contributing to the stability of an unoccupied bicycle"
2) "factors contributing to the stability of an occupied bicycle"
3) "factors contributing to the stability of a freely rolling wheel"
These are all different systems.
In the case of (1), the primary factor is bicycle geometry.
In the case of (2), the primary factors are active steering and body
weight shifting by the bicycle rider.
In the case of (3), the primary factors are gyroscopic effect and tire
shape (for example, if it has a squarish cross-section).
Let's not argue apples and oranges, OK?
Myxococcus xanthus
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| User: "JM Albuquerque" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
12 Aug 2004 07:21:28 AM |
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"Myxococcus xanthus" <mold-guardian@comcast.net> wrote:
Bicycles have been built where the gyroscopic force has been
completely canceled out by means of a counter-rotating wheel, and they
roll on their own just fine.
I wonder how such bicycles can be constructed with 4 wheels, being two
of then rotating clockwise and two of them rotating counter-clockwise.
And the more important is:
Without gyroscopic effect, how can that bicycle turn at high speed?
At high speed all bicycles and motorbikes must lean side way when turning
to describe a curve. The center of mass of driver + bicycle is out from the
vertical equilibrium. A torque is created due to gravity and they must fall.
The gyroscopic effect shows that an unbalanced gravitic force (torque) can
be compensated by means of precessing (doing a turn) and that's exactly
what the gyroscopic effect does. The gyroscopic effect allows bicycles
to turn and do curves.
When the bicycle + drives center of mass is perfectly vertical there is no
gyroscopic effect (zero) and the bicycle goes straight ahead. When the
gravitic force is unbalanced (the most common situation) it will trend to
fall. Then the bicycle reacts, due gyroscopic effect, and starts making a
turn, due to precession.
Then again, one should not confuse
1) "factors contributing to the stability of an unoccupied bicycle"
2) "factors contributing to the stability of an occupied bicycle"
3) "factors contributing to the stability of a freely rolling wheel"
These are all different systems.
In the case of (1), the primary factor is bicycle geometry.
Not true.
The primary factor to the stability of an unoccupied bicycle is speed.
Apply no speed and the bicycle falls at once.
The bigger the initial speed is the longer the bicycle will remain stable,
the longer will be the travelled distance and the larger will be the radius
of the circular path travelled.
The bicycle never travels straight ahead in a perfect straight line. An
unoccupied bicycle always does a turn depending on speed.
In the case of (2), the primary factors are active steering and body
weight shifting by the bicycle rider.
Only true at low and very low speeds.
Not true at high speeds due to the inertial "square term" on the
gyroscopic effect.
Notice what happen in a bicycle race about 100 meters from the finish
line. All the competitors are riding at maximum to finish first. They are
producing the maximum torque with their feet's on pedals, balancing the
bodies according to the force they want to produce (left or right), and
creating the maximum gravitic unbalance left - right - left -right...
The bicycle is hobbling.
A missed stroke and you got an accident.
Or else, doing turns at high speed proves that factors contributing to the
stability of an occupied bicycle is not active steering nor body weight
shifting by the bicycle rider, because the steering doesn't move (no hands)
and the body of the driver doesn't move also. The gyroscopic effect does
all the work. The only thing the driver needs to do is apply the right
gravitic unbalanced force (constant for a given amount of turn).
Without the gyroscopic effect the bicycle cannot do a simple turn above
a given speed.
In the case of (3), the primary factors are gyroscopic effect and tire
shape (for example, if it has a squarish cross-section).
Never seen a bicycle tire with squared cross-section.
You need a square cross-section whose half cross section is bigger then
the possible side way deviation of the center of mass.
Let's not argue apples and oranges, OK?
And don't forget the trees. Without trees apples and oranges fall equally.
The bicycle speed is the "tree" that you forget to talk about.
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| User: "crynwulf" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
12 Aug 2004 05:31:48 PM |
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JM Albuquerque wrote:
"Myxococcus xanthus" <mold-guardian@comcast.net> wrote:
Bicycles have been built where the gyroscopic force has been
completely canceled out by means of a counter-rotating wheel, and they
roll on their own just fine.
I wonder how such bicycles can be constructed with 4 wheels, being two
of then rotating clockwise and two of them rotating counter-clockwise.
And the more important is:
Without gyroscopic effect, how can that bicycle turn at high speed?
At high speed all bicycles and motorbikes must lean side way when turning
to describe a curve. The center of mass of driver + bicycle is out from
the vertical equilibrium. A torque is created due to gravity and they must
fall. The gyroscopic effect shows that an unbalanced gravitic force
(torque) can be compensated by means of precessing (doing a turn) and
that's exactly what the gyroscopic effect does. The gyroscopic effect
allows bicycles to turn and do curves.
When the bicycle + drives center of mass is perfectly vertical there is no
gyroscopic effect (zero) and the bicycle goes straight ahead. When the
gravitic force is unbalanced (the most common situation) it will trend to
fall. Then the bicycle reacts, due gyroscopic effect, and starts making a
turn, due to precession.
Then again, one should not confuse
1) "factors contributing to the stability of an unoccupied bicycle"
2) "factors contributing to the stability of an occupied bicycle"
3) "factors contributing to the stability of a freely rolling wheel"
These are all different systems.
In the case of (1), the primary factor is bicycle geometry.
Not true.
The primary factor to the stability of an unoccupied bicycle is speed.
Apply no speed and the bicycle falls at once.
The bigger the initial speed is the longer the bicycle will remain stable,
the longer will be the travelled distance and the larger will be the
radius of the circular path travelled.
The bicycle never travels straight ahead in a perfect straight line. An
unoccupied bicycle always does a turn depending on speed.
In the case of (2), the primary factors are active steering and body
weight shifting by the bicycle rider.
Only true at low and very low speeds.
Not true at high speeds due to the inertial "square term" on the
gyroscopic effect.
Notice what happen in a bicycle race about 100 meters from the finish
line. All the competitors are riding at maximum to finish first. They are
producing the maximum torque with their feet's on pedals, balancing the
bodies according to the force they want to produce (left or right), and
creating the maximum gravitic unbalance left - right - left -right...
The bicycle is hobbling.
A missed stroke and you got an accident.
Or else, doing turns at high speed proves that factors contributing to the
stability of an occupied bicycle is not active steering nor body weight
shifting by the bicycle rider, because the steering doesn't move (no
hands) and the body of the driver doesn't move also. The gyroscopic effect
does
all the work. The only thing the driver needs to do is apply the right
gravitic unbalanced force (constant for a given amount of turn).
Without the gyroscopic effect the bicycle cannot do a simple turn above
a given speed.
In the case of (3), the primary factors are gyroscopic effect and tire
shape (for example, if it has a squarish cross-section).
Never seen a bicycle tire with squared cross-section.
You need a square cross-section whose half cross section is bigger then
the possible side way deviation of the center of mass.
Let's not argue apples and oranges, OK?
And don't forget the trees. Without trees apples and oranges fall equally.
The bicycle speed is the "tree" that you forget to talk about.
If it has 4 wheels it isn't a bicycle but a tetracycle. Two wheel = bicycle,
three wheels = tricycle, four wheels = tetracycle. Unlike a bicycle, a 4
wheeled vehicle doesn't track well unless it has very good alignment. You
know you've got good alignment when you can take your 73 Charger at 150
down the whiteline on a Texas highway and let go of the steering wheel.
Porsche owners need not comment.
--
Russ Lyttle
Not Powered by ActiveX
http://home.earthlink.net/~lyttlec/philosophy/logos.html
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| User: "JM Albuquerque" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
11 Aug 2004 05:04:09 AM |
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"Myxococcus xanthus" <mold-guardian@comcast.net> wrote:
Read
http://socrates.berkeley.edu/~fajans/pub/pdffiles/SteerBikeAJP.PDF
Tell me why the guy and the motorbike page 5 Fig. 6 doesn't fall?
You can't right a word from your own Electronic Parrot.
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| User: "JM Albuquerque" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
09 Aug 2004 05:19:33 PM |
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"Myxococcus xanthus" <mold-guardian@comcast.net> wrote:
"greywolf42" <mingstb@marssim-ss.com> wrote in message
news:<10hf8eht69qh363@corp.supernews.com>...
Now, how do you explain that a bicycle tire -- sans bicycle -- exhibits
the same tendency? That is, it will stay up as it rolls down the hill.
There is no caster to control it.
No one has ever claimed that there ISN'T any gyroscopic effect at all.
What we have stated, is that the gyroscopic contribution to bicycle
stability is minor compared with other effects. Multiple published
studies that have quantified the various contributions to bicycle
stability have all come to the same conclusion.
It is obvious that it all depends on the speed.
The gyroscopic effect depends on the angular speed squared and that
square term is enough for anyone to see what the gyroscopic effect
means to a cyclist.
Obviously that at small speeds the gyroscopic is not enough.
But soon it becomes fundamental due to the squared term, up to
the point that one can drive the bicycle without hands and the hand
stick doesn't move anymore.
Given the square term, gyroscopic effect beats all the other effects
as soon has the angular speed is enough (and it doesn't take to long
as any experienced cyclist can tell you).
The mountain gives a mouse.
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| User: "Phil Holman" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
09 Aug 2004 09:18:19 PM |
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"JM Albuquerque" <jm.aREMOV.E@sapo.pt> wrote in message
news:2nqbepF3hgldU1@uni-berlin.de...
"Myxococcus xanthus" <mold-guardian@comcast.net> wrote:
"greywolf42" <mingstb@marssim-ss.com> wrote in message
news:<10hf8eht69qh363@corp.supernews.com>...
Now, how do you explain that a bicycle tire -- sans bicycle --
exhibits
the same tendency? That is, it will stay up as it rolls down the
hill.
There is no caster to control it.
No one has ever claimed that there ISN'T any gyroscopic effect at
all.
What we have stated, is that the gyroscopic contribution to bicycle
stability is minor compared with other effects. Multiple published
studies that have quantified the various contributions to bicycle
stability have all come to the same conclusion.
It is obvious that it all depends on the speed.
The gyroscopic effect depends on the angular speed squared and that
square term is enough for anyone to see what the gyroscopic effect
means to a cyclist.
Obviously that at small speeds the gyroscopic is not enough.
But soon it becomes fundamental due to the squared term, up to
the point that one can drive the bicycle without hands and the hand
stick doesn't move anymore.
Given the square term, gyroscopic effect beats all the other effects
as soon has the angular speed is enough (and it doesn't take to long
as any experienced cyclist can tell you).
The mountain gives a mouse.
Very good JM. Below about 15mph there isn't enough giro force to assist
with riding no hands.
It can still be done but the technique is a little trickier. One has to
compensate for the mass imbalance about the steering axis which through
inertia, initially causes the wheel to turn away from the lean but then
as the moment imbalance comes into play causes the wheel to turn into
the lean. This supposes that the greater mass is forward of the steering
axis (handle bars, brake levers, stem, more than half the wheel etc.)
which for most bikes is the case.
PH
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| User: "Myxococcus xanthus" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
09 Aug 2004 09:40:28 PM |
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"JM Albuquerque" <jm.aREMOV.E@sapo.pt> wrote in message news:<2nqbepF3hgldU1@uni-berlin.de>...
"Myxococcus xanthus" <mold-guardian@comcast.net> wrote:
"greywolf42" <mingstb@marssim-ss.com> wrote in message
news:<10hf8eht69qh363@corp.supernews.com>...
Now, how do you explain that a bicycle tire -- sans bicycle -- exhibits
the same tendency? That is, it will stay up as it rolls down the hill.
There is no caster to control it.
No one has ever claimed that there ISN'T any gyroscopic effect at all.
What we have stated, is that the gyroscopic contribution to bicycle
stability is minor compared with other effects. Multiple published
studies that have quantified the various contributions to bicycle
stability have all come to the same conclusion.
It is obvious that it all depends on the speed.
The gyroscopic effect depends on the angular speed squared and that
square term is enough for anyone to see what the gyroscopic effect
means to a cyclist.
Obviously that at small speeds the gyroscopic is not enough.
But soon it becomes fundamental due to the squared term, up to
the point that one can drive the bicycle without hands and the hand
stick doesn't move anymore.
Given the square term, gyroscopic effect beats all the other effects
as soon has the angular speed is enough (and it doesn't take to long
as any experienced cyclist can tell you).
The mountain gives a mouse.
What squared term?
Myxococcus xanthus
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| User: "Phil Holman" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
09 Aug 2004 10:53:12 PM |
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"Myxococcus xanthus" <mold-guardian@comcast.net> wrote in message
news:ce5e7813.0408091840.6a5c28da@posting.google.com...
"JM Albuquerque" <jm.aREMOV.E@sapo.pt> wrote in message
news:<2nqbepF3hgldU1@uni-berlin.de>...
"Myxococcus xanthus" <mold-guardian@comcast.net> wrote:
"greywolf42" <mingstb@marssim-ss.com> wrote in message
news:<10hf8eht69qh363@corp.supernews.com>...
Now, how do you explain that a bicycle tire -- sans bicycle --
exhibits
the same tendency? That is, it will stay up as it rolls down
the hill.
There is no caster to control it.
No one has ever claimed that there ISN'T any gyroscopic effect at
all.
What we have stated, is that the gyroscopic contribution to
bicycle
stability is minor compared with other effects. Multiple published
studies that have quantified the various contributions to bicycle
stability have all come to the same conclusion.
It is obvious that it all depends on the speed.
The gyroscopic effect depends on the angular speed squared and that
square term is enough for anyone to see what the gyroscopic effect
means to a cyclist.
Obviously that at small speeds the gyroscopic is not enough.
But soon it becomes fundamental due to the squared term, up to
the point that one can drive the bicycle without hands and the hand
stick doesn't move anymore.
Given the square term, gyroscopic effect beats all the other effects
as soon has the angular speed is enough (and it doesn't take to long
as any experienced cyclist can tell you).
The mountain gives a mouse.
What squared term?
Exactly, methinks someone is thinking 1/2Iw^2 when
Torque (precession) = dL/dt = m*r*v*dTheta/dt
PH
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| User: "crynwulf" |
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| Title: Re: Gyroscopes - Usenet Physics FAQ - Reference frames |
10 Aug 2004 07:57:51 PM |
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Phil Holman wrote:
"Myxococcus xanthus" <mold-guardian@comcast.net> wrote in message
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