| Topic: |
Science > Physics |
| User: |
"Gregory L. Hansen" |
| Date: |
19 Oct 2003 07:13:15 PM |
| Object: |
Frame Dragging and Flywheels |
I'm dinking around with an idea for testing frame dragging in general
relativity with neutrons. It seems a ball bearing and a regular
interferometer can't do it, the signal would be a billion times too small.
But now I'm thinking of a neutron storage ring with a much larger flywheel
spinning inside it. A ring had already been built to do a measurement of
the neutron lifetime, diameter is about a meter. In principle another
ring could be made larger to incorporate a larger flyweel or to hold
neutrons of higher energy, but the size I'm thinking of is on the order of
a meter.
What would be a typical mass and highest angular momentum of a flywheel
that size? I think there were some people around here, like Meron, that
know a bit about high performance flywheels.
--
"Let us learn to dream, gentlemen, then perhaps we shall find the
truth... But let us beware of publishing our dreams before they have been
put to the proof by the waking understanding." -- Friedrich August Kekulé
.
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| User: "Sam Wormley" |
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| Title: Re: Frame Dragging and Flywheels |
19 Oct 2003 07:56:46 PM |
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"Gregory L. Hansen" wrote:
I'm dinking around with an idea for testing frame dragging in general
relativity with neutrons. It seems a ball bearing and a regular
interferometer can't do it, the signal would be a billion times too small.
But now I'm thinking of a neutron storage ring with a much larger flywheel
spinning inside it. A ring had already been built to do a measurement of
the neutron lifetime, diameter is about a meter. In principle another
ring could be made larger to incorporate a larger flyweel or to hold
neutrons of higher energy, but the size I'm thinking of is on the order of
a meter.
What would be a typical mass and highest angular momentum of a flywheel
that size? I think there were some people around here, like Meron, that
know a bit about high performance flywheels.
--
"Let us learn to dream, gentlemen, then perhaps we shall find the
truth... But let us beware of publishing our dreams before they have been
put to the proof by the waking understanding." -- Friedrich August Kekulé
The mass of the Earth has about 0.698 billionth total voting power as the
rest of the universe on our local spacetime.
The Earth's mass is about 5.97 x 10^24 kg
How fast (angular velocity) can you spin a lot of mass with radius
of 50 cm or so?
.
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| User: "" |
|
| Title: Re: Frame Dragging and Flywheels |
19 Oct 2003 11:17:35 PM |
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In article <bmv9ar$t1s$3@hood.uits.indiana.edu>, (Gregory L. Hansen) writes:
I'm dinking around with an idea for testing frame dragging in general
relativity with neutrons. It seems a ball bearing and a regular
interferometer can't do it, the signal would be a billion times too small.
But now I'm thinking of a neutron storage ring with a much larger flywheel
spinning inside it. A ring had already been built to do a measurement of
the neutron lifetime, diameter is about a meter.
How do you keep the neutrons stored?
In principle another
ring could be made larger to incorporate a larger flyweel or to hold
neutrons of higher energy, but the size I'm thinking of is on the order of
a meter.
What would be a typical mass and highest angular momentum of a flywheel
that size? I think there were some people around here, like Meron, that
know a bit about high performance flywheels.
Well, the magical number is "critical speed", given by
vc = sqrt(yield_strength/density)
The highest speed a point on the perifery may have is of the order of
vc (up to a geometry dependent factor of the order of one). For the
toughest metallic alloys I can recall this comes to around 5-600 m/s.
Some specialty glasses and the like may get higher but not that much
higher.
Now, the angular momentum will be Iw (where w = v/r), so, since I is
of the order of mr^2 you'll get angular momentum of the the order of
L = m*vc*r
Assuming (simplistically), a flywheel with average thickness t, you'll
get
L = pi*rho*t*r^3*vc
Note that vc*rho = sqrt(rho*Yield_strength) so, unlike in the case
where you go for highest speed here you want to maximize the product
of density and yield_strength. Can't say offhand what'll give the
best value. Highest quality steels will give you
sqrt(rho*Y) = 4-5*10^6 (kg/(m^2*s))
Tungsten will probably do better. So, assume highest value of 10^7.
Then taking r of .5 m and thickness of, say, .1m, and folding
numerical factors neglected in the above you'll get angular momentum
of the order of
L = 1-2*10^5 kg*m^2/s
How much you need?
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
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| User: "Gregory L. Hansen" |
|
| Title: Re: Frame Dragging and Flywheels |
21 Oct 2003 10:19:49 AM |
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In article <zlJkb.10$_4.4439@news.uchicago.edu>,
<mmeron@cars3.uchicago.edu> wrote:
In article <bmv9ar$t1s$3@hood.uits.indiana.edu>,
glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) writes:
I'm dinking around with an idea for testing frame dragging in general
relativity with neutrons. It seems a ball bearing and a regular
interferometer can't do it, the signal would be a billion times too small.
But now I'm thinking of a neutron storage ring with a much larger flywheel
spinning inside it. A ring had already been built to do a measurement of
the neutron lifetime, diameter is about a meter.
How do you keep the neutrons stored?
In principle another
ring could be made larger to incorporate a larger flyweel or to hold
neutrons of higher energy, but the size I'm thinking of is on the order of
a meter.
What would be a typical mass and highest angular momentum of a flywheel
that size? I think there were some people around here, like Meron, that
know a bit about high performance flywheels.
Well, the magical number is "critical speed", given by
vc = sqrt(yield_strength/density)
The highest speed a point on the perifery may have is of the order of
vc (up to a geometry dependent factor of the order of one). For the
toughest metallic alloys I can recall this comes to around 5-600 m/s.
Some specialty glasses and the like may get higher but not that much
higher.
Now, the angular momentum will be Iw (where w = v/r), so, since I is
of the order of mr^2 you'll get angular momentum of the the order of
L = m*vc*r
Assuming (simplistically), a flywheel with average thickness t, you'll
get
L = pi*rho*t*r^3*vc
Note that vc*rho = sqrt(rho*Yield_strength) so, unlike in the case
where you go for highest speed here you want to maximize the product
of density and yield_strength. Can't say offhand what'll give the
best value. Highest quality steels will give you
sqrt(rho*Y) = 4-5*10^6 (kg/(m^2*s))
Tungsten will probably do better. So, assume highest value of 10^7.
Then taking r of .5 m and thickness of, say, .1m, and folding
numerical factors neglected in the above you'll get angular momentum
of the order of
L = 1-2*10^5 kg*m^2/s
How much you need?
I worked out my flywheel assuming hoop stresses, and spun it up to the
material's yield point. Interestingly, mass dropped out. That makes
sense in retrospect, since the angular momentum is proportional to mass,
and the centripetal force on a volume element is proportional to its mass.
If I did my math right, for neutrons going 20 m/s grazing a meter tall
cylindrical flywheel at one meter diameter, yield strength 600 MPa, for
half an hour before the beams are recombined, that would result in a phase
shift of 1e-22 radians.
Dang. It seemed like such a cool idea.
--
"Let us learn to dream, gentlemen, then perhaps we shall find the
truth... But let us beware of publishing our dreams before they have been
put to the proof by the waking understanding." -- Friedrich August Kekulé
.
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| User: "" |
|
| Title: Re: Frame Dragging and Flywheels |
21 Oct 2003 03:25:22 PM |
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In article <bn3iql$e3u$1@hood.uits.indiana.edu>, (Gregory L. Hansen) writes:
In article <zlJkb.10$_4.4439@news.uchicago.edu>,
<mmeron@cars3.uchicago.edu> wrote:
In article <bmv9ar$t1s$3@hood.uits.indiana.edu>,
(Gregory L. Hansen) writes:
I'm dinking around with an idea for testing frame dragging in general
relativity with neutrons. It seems a ball bearing and a regular
interferometer can't do it, the signal would be a billion times too small.
But now I'm thinking of a neutron storage ring with a much larger flywheel
spinning inside it. A ring had already been built to do a measurement of
the neutron lifetime, diameter is about a meter.
How do you keep the neutrons stored?
In principle another
ring could be made larger to incorporate a larger flyweel or to hold
neutrons of higher energy, but the size I'm thinking of is on the order of
a meter.
What would be a typical mass and highest angular momentum of a flywheel
that size? I think there were some people around here, like Meron, that
know a bit about high performance flywheels.
Well, the magical number is "critical speed", given by
vc = sqrt(yield_strength/density)
The highest speed a point on the perifery may have is of the order of
vc (up to a geometry dependent factor of the order of one). For the
toughest metallic alloys I can recall this comes to around 5-600 m/s.
Some specialty glasses and the like may get higher but not that much
higher.
Now, the angular momentum will be Iw (where w = v/r), so, since I is
of the order of mr^2 you'll get angular momentum of the the order of
L = m*vc*r
Assuming (simplistically), a flywheel with average thickness t, you'll
get
L = pi*rho*t*r^3*vc
Note that vc*rho = sqrt(rho*Yield_strength) so, unlike in the case
where you go for highest speed here you want to maximize the product
of density and yield_strength. Can't say offhand what'll give the
best value. Highest quality steels will give you
sqrt(rho*Y) = 4-5*10^6 (kg/(m^2*s))
Tungsten will probably do better. So, assume highest value of 10^7.
Then taking r of .5 m and thickness of, say, .1m, and folding
numerical factors neglected in the above you'll get angular momentum
of the order of
L = 1-2*10^5 kg*m^2/s
How much you need?
I worked out my flywheel assuming hoop stresses, and spun it up to the
material's yield point. Interestingly, mass dropped out. That makes
sense in retrospect, since the angular momentum is proportional to mass,
and the centripetal force on a volume element is proportional to its mass.
If I did my math right, for neutrons going 20 m/s grazing a meter tall
cylindrical flywheel at one meter diameter, yield strength 600 MPa, for
half an hour before the beams are recombined, that would result in a phase
shift of 1e-22 radians.
Uggh:-( kinda tough to measure. And way too many orders of magnitude
for any increase in material strength to make a difference. You need
scrith.
Dang. It seemed like such a cool idea.
It was a cool idea. Unfortunately, Mother Nature turns out a *****,
again.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
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| User: "Gregory L. Hansen" |
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| Title: Re: Frame Dragging and Flywheels |
20 Oct 2003 09:04:20 AM |
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|
In article <zlJkb.10$_4.4439@news.uchicago.edu>,
<mmeron@cars3.uchicago.edu> wrote:
In article <bmv9ar$t1s$3@hood.uits.indiana.edu>,
glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) writes:
I'm dinking around with an idea for testing frame dragging in general
relativity with neutrons. It seems a ball bearing and a regular
interferometer can't do it, the signal would be a billion times too small.
But now I'm thinking of a neutron storage ring with a much larger flywheel
spinning inside it. A ring had already been built to do a measurement of
the neutron lifetime, diameter is about a meter.
How do you keep the neutrons stored?
Sextupoles. See NIM 228, 240 (1985). They got storage times of 45
minutes for 2e-6 eV neutrons.
In principle another
ring could be made larger to incorporate a larger flyweel or to hold
neutrons of higher energy, but the size I'm thinking of is on the order of
a meter.
What would be a typical mass and highest angular momentum of a flywheel
that size? I think there were some people around here, like Meron, that
know a bit about high performance flywheels.
Well, the magical number is "critical speed", given by
vc = sqrt(yield_strength/density)
The highest speed a point on the perifery may have is of the order of
vc (up to a geometry dependent factor of the order of one). For the
toughest metallic alloys I can recall this comes to around 5-600 m/s.
Some specialty glasses and the like may get higher but not that much
higher.
Now, the angular momentum will be Iw (where w = v/r), so, since I is
of the order of mr^2 you'll get angular momentum of the the order of
L = m*vc*r
Assuming (simplistically), a flywheel with average thickness t, you'll
get
L = pi*rho*t*r^3*vc
Note that vc*rho = sqrt(rho*Yield_strength) so, unlike in the case
where you go for highest speed here you want to maximize the product
of density and yield_strength. Can't say offhand what'll give the
best value. Highest quality steels will give you
sqrt(rho*Y) = 4-5*10^6 (kg/(m^2*s))
Tungsten will probably do better. So, assume highest value of 10^7.
Then taking r of .5 m and thickness of, say, .1m, and folding
numerical factors neglected in the above you'll get angular momentum
of the order of
L = 1-2*10^5 kg*m^2/s
How much you need?
Don't know how much I need, I haven't started running numbers yet. But I
know there are flywheels made of modern fibers that can store more energy
before flying apart than metal flywheels could, and I thought you were
going on about them once in a discussion about batteries versus flywheels
for energy storage.
I wonder if a staged concept would be worth the effort, with a tungsten
core since the centrifugal forces would be relatively light there, going
to kevlar wrappings or something for strength at the rim. But I think I
can assume there's a lot about flywheel engineering that I don't suspect.
There's a carbon nanotube material that's been made recently that's
significantly stronger than kevlar, although only available in
experimental quantities. If you know the guy, maybe he'll give you a few
meters of it when he's finished with it...
--
"Let us learn to dream, gentlemen, then perhaps we shall find the
truth... But let us beware of publishing our dreams before they have been
put to the proof by the waking understanding." -- Friedrich August Kekulé
.
|
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| User: "" |
|
| Title: Re: Frame Dragging and Flywheels |
20 Oct 2003 03:23:04 PM |
|
|
In article <bn0q14$e8k$2@hood.uits.indiana.edu>, (Gregory L. Hansen) writes:
In article <zlJkb.10$_4.4439@news.uchicago.edu>,
<mmeron@cars3.uchicago.edu> wrote:
In article <bmv9ar$t1s$3@hood.uits.indiana.edu>,
(Gregory L. Hansen) writes:
I'm dinking around with an idea for testing frame dragging in general
relativity with neutrons. It seems a ball bearing and a regular
interferometer can't do it, the signal would be a billion times too small.
But now I'm thinking of a neutron storage ring with a much larger flywheel
spinning inside it. A ring had already been built to do a measurement of
the neutron lifetime, diameter is about a meter.
How do you keep the neutrons stored?
Sextupoles. See NIM 228, 240 (1985). They got storage times of 45
minutes for 2e-6 eV neutrons.
Not bad.
In principle another
ring could be made larger to incorporate a larger flyweel or to hold
neutrons of higher energy, but the size I'm thinking of is on the order of
a meter.
What would be a typical mass and highest angular momentum of a flywheel
that size? I think there were some people around here, like Meron, that
know a bit about high performance flywheels.
Well, the magical number is "critical speed", given by
vc = sqrt(yield_strength/density)
The highest speed a point on the perifery may have is of the order of
vc (up to a geometry dependent factor of the order of one). For the
toughest metallic alloys I can recall this comes to around 5-600 m/s.
Some specialty glasses and the like may get higher but not that much
higher.
Now, the angular momentum will be Iw (where w = v/r), so, since I is
of the order of mr^2 you'll get angular momentum of the the order of
L = m*vc*r
Assuming (simplistically), a flywheel with average thickness t, you'll
get
L = pi*rho*t*r^3*vc
Note that vc*rho = sqrt(rho*Yield_strength) so, unlike in the case
where you go for highest speed here you want to maximize the product
of density and yield_strength. Can't say offhand what'll give the
best value. Highest quality steels will give you
sqrt(rho*Y) = 4-5*10^6 (kg/(m^2*s))
Tungsten will probably do better. So, assume highest value of 10^7.
Then taking r of .5 m and thickness of, say, .1m, and folding
numerical factors neglected in the above you'll get angular momentum
of the order of
L = 1-2*10^5 kg*m^2/s
How much you need?
Don't know how much I need, I haven't started running numbers yet. But I
know there are flywheels made of modern fibers that can store more energy
before flying apart than metal flywheels could, and I thought you were
going on about them once in a discussion about batteries versus flywheels
for energy storage.
Yes, indeed. Note, however, that "more energy" and "more angular
momentum" is not necessarily the same. In a nutshell, for same
geometry, energy is proportional to rho*vc^2 while angular momentum is
proportional to rho*vc. Given the definition of vc, above, this means
that stored energy is proportional to Y (yield strength) while angular
momentum is proportional to sqrt(rho*Y). Now, said modern fibers have
extremely high yield strength, so they win on energy. But, since they
also have low density, it is not necessarily true that they win on
angular momentum. Would have to check some numbers to find out what
is the case on this.
I wonder if a staged concept would be worth the effort, with a tungsten
core since the centrifugal forces would be relatively light there, going
to kevlar wrappings or something for strength at the rim.
Ah, but the stuff in the core contributes little to angular momentum
(remember, m*r^2*w^2), most of the angular momentum does come from the
rim. So you would like to have decent density there. And, as for
"the centrifugal forces being light near the core", you're dealing
with a (reasonably) rigid body, not a collection of free atoms. The
stuff near the center has to support not only itself but the stuff
closer to the rim which is pulling out. So things are more
complicated. Try, for starters, to find the distribution of stresses
in a revolving cylindrical slice. This is a fully solvable problem
and it gives you a good sense of what's going on. Of course, a real
life flywheel will be more complex, but it is nice to have an
analytically solvable example.
But I think I
can assume there's a lot about flywheel engineering that I don't suspect.
True for all of us and any engineering. Principles may be simple but
the devil is in the details.
There's a carbon nanotube material that's been made recently that's
significantly stronger than kevlar, although only available in
experimental quantities. If you know the guy, maybe he'll give you a few
meters of it when he's finished with it...
Meters? You're going to need many millions of kilometers worth of a
nanotube to make something macroscopic.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
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| User: "Edward Green" |
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| Title: Re: Frame Dragging and Flywheels |
21 Oct 2003 09:12:26 PM |
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wrote in message news:<IuXkb.23$_4.10187@news.uchicago.edu>...
....
as for
"the centrifugal forces being light near the core", you're dealing
with a (reasonably) rigid body, not a collection of free atoms. The
stuff near the center has to support not only itself but the stuff
closer to the rim which is pulling out. So things are more
complicated.
For what it's worth, I recall thinking about this recently. There are
two limiting cases of flywheels: if you simply rotated a collection of
spokes not joined by any circumferential elements, obviously you would
have pure spoke stresses; if you merely rotated a rim without spokes
....
To get the most bang out of your material, I suppose you would want to
fail in each mode simultaneously. A general priciple surfaces: this
is a design criteria for submarine hulls; you want each principal
failure mode to kick in about simultaneously at crush depth ...
otherwise you are wasting strength. You could possibly modify the
stress distribution in a rotated disk to sychronize failure modes by
pre-stressing.
But as you've already noted, none of this will make any difference if
you are 10 orders of magnitude out to start with.
.
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| User: "" |
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| Title: Re: Frame Dragging and Flywheels |
21 Oct 2003 10:26:12 PM |
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In article <2a0cceff.0310211812.1a4b6c69@posting.google.com>, (Edward Green) writes:
mmeron@cars3.uchicago.edu wrote in message news:<IuXkb.23$_4.10187@news.uchicago.edu>...
...
as for
"the centrifugal forces being light near the core", you're dealing
with a (reasonably) rigid body, not a collection of free atoms. The
stuff near the center has to support not only itself but the stuff
closer to the rim which is pulling out. So things are more
complicated.
For what it's worth, I recall thinking about this recently. There are
two limiting cases of flywheels: if you simply rotated a collection of
spokes not joined by any circumferential elements, obviously you would
have pure spoke stresses; if you merely rotated a rim without spokes
...
Aye, right on.
To get the most bang out of your material, I suppose you would want to
fail in each mode simultaneously. A general priciple surfaces: this
is a design criteria for submarine hulls; you want each principal
failure mode to kick in about simultaneously at crush depth ...
otherwise you are wasting strength.
Good thinking, this.
You could possibly modify the
stress distribution in a rotated disk to sychronize failure modes by
pre-stressing.
Yes. And, you could deviate from plain disk to some more interesting
thickness profile. Don't recall what's optimal in this case. In the
case of a single spoke rotating around its CM, the optimal thickness
profile is gaussian, as I recall. Still ...
But as you've already noted, none of this will make any difference if
you are 10 orders of magnitude out to start with.
Yes, at this point nothing short of miracle will work.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
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| User: "Gregory L. Hansen" |
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| Title: Re: Frame Dragging and Flywheels |
20 Oct 2003 04:16:27 PM |
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In article <IuXkb.23$_4.10187@news.uchicago.edu>,
<mmeron@cars3.uchicago.edu> wrote:
In article <bn0q14$e8k$2@hood.uits.indiana.edu>,
glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) writes:
In article <zlJkb.10$_4.4439@news.uchicago.edu>,
<mmeron@cars3.uchicago.edu> wrote:
In article <bmv9ar$t1s$3@hood.uits.indiana.edu>,
glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) writes:
Don't know how much I need, I haven't started running numbers yet. But I
know there are flywheels made of modern fibers that can store more energy
before flying apart than metal flywheels could, and I thought you were
going on about them once in a discussion about batteries versus flywheels
for energy storage.
Yes, indeed. Note, however, that "more energy" and "more angular
momentum" is not necessarily the same. In a nutshell, for same
geometry, energy is proportional to rho*vc^2 while angular momentum is
proportional to rho*vc. Given the definition of vc, above, this means
that stored energy is proportional to Y (yield strength) while angular
momentum is proportional to sqrt(rho*Y). Now, said modern fibers have
extremely high yield strength, so they win on energy. But, since they
also have low density, it is not necessarily true that they win on
angular momentum. Would have to check some numbers to find out what
is the case on this.
Oh, right.
I should just run any numbers for generic steel to get some idea of
feasibility. If the signal is still a billion times too small to measure,
modern fibers won't make up for that! In that case I may as well save
myself some work. But if I get in the factor of 100 region, there may be
some hope for tweaking flywheels, diameters, neutron energies, etc.
There's a carbon nanotube material that's been made recently that's
significantly stronger than kevlar, although only available in
experimental quantities. If you know the guy, maybe he'll give you a few
meters of it when he's finished with it...
Meters? You're going to need many millions of kilometers worth of a
nanotube to make something macroscopic.
Well, they weren't one carbon nanotube wide! It was carbon nanotubes in
some kind of matrix that I forget, but struck me as odd because "alcohal"
was in the name, which I hadn't thought was a solid. They've gotten
pretty long fibers, with strength and toughness that compare favorably to
kevlar and spider silk. I saw it in a Science News article, maybe I can
find it again.
--
"Let us learn to dream, gentlemen, then perhaps we shall find the
truth... But let us beware of publishing our dreams before they have been
put to the proof by the waking understanding." -- Friedrich August Kekulé
.
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| User: "" |
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| Title: Re: Frame Dragging and Flywheels |
20 Oct 2003 04:30:36 PM |
|
|
In article <bn1jbb$np0$2@hood.uits.indiana.edu>, (Gregory L. Hansen) writes:
In article <IuXkb.23$_4.10187@news.uchicago.edu>,
<mmeron@cars3.uchicago.edu> wrote:
In article <bn0q14$e8k$2@hood.uits.indiana.edu>,
(Gregory L. Hansen) writes:
In article <zlJkb.10$_4.4439@news.uchicago.edu>,
<mmeron@cars3.uchicago.edu> wrote:
In article <bmv9ar$t1s$3@hood.uits.indiana.edu>,
(Gregory L. Hansen) writes:
Don't know how much I need, I haven't started running numbers yet. But I
know there are flywheels made of modern fibers that can store more energy
before flying apart than metal flywheels could, and I thought you were
going on about them once in a discussion about batteries versus flywheels
for energy storage.
Yes, indeed. Note, however, that "more energy" and "more angular
momentum" is not necessarily the same. In a nutshell, for same
geometry, energy is proportional to rho*vc^2 while angular momentum is
proportional to rho*vc. Given the definition of vc, above, this means
that stored energy is proportional to Y (yield strength) while angular
momentum is proportional to sqrt(rho*Y). Now, said modern fibers have
extremely high yield strength, so they win on energy. But, since they
also have low density, it is not necessarily true that they win on
angular momentum. Would have to check some numbers to find out what
is the case on this.
Oh, right.
I should just run any numbers for generic steel to get some idea of
feasibility. If the signal is still a billion times too small to measure,
modern fibers won't make up for that! In that case I may as well save
myself some work. But if I get in the factor of 100 region, there may be
some hope for tweaking flywheels, diameters, neutron energies, etc.
Yes, that's the idea. That's why I was asking what sort of angular
momentum is needed.
There's a carbon nanotube material that's been made recently that's
significantly stronger than kevlar, although only available in
experimental quantities. If you know the guy, maybe he'll give you a few
meters of it when he's finished with it...
Meters? You're going to need many millions of kilometers worth of a
nanotube to make something macroscopic.
Well, they weren't one carbon nanotube wide! It was carbon nanotubes in
some kind of matrix that I forget, but struck me as odd because "alcohal"
was in the name, which I hadn't thought was a solid. They've gotten
pretty long fibers, with strength and toughness that compare favorably to
kevlar and spider silk. I saw it in a Science News article, maybe I can
find it again.
I vaguely remember something about it, but it is still small stuff
(compared to the scale you need) at the moment.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "tj Frazir" |
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| Title: Re: Frame Dragging and Flywheels |
19 Oct 2003 08:29:17 PM |
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The colder you get it the faster you can spin it.
A laser at its ege wile supper cold with no resistance from any
friction.
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| User: "MorituriMax" |
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| Title: Re: Frame Dragging and Flywheels |
19 Oct 2003 11:55:30 PM |
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"tj Frazir" <GravityPhysics@webtv.net> wrote in message
news:26909-3F933A6D-1@storefull-2154.public.lawson.webtv.net...
The.. .. ..
BS Filter Activated.
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| User: "Gregory L. Hansen" |
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| Title: Re: Frame Dragging and Flywheels |
20 Oct 2003 08:50:59 AM |
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In article <6VJkb.17355$wj.15665@twister.austin.rr.com>,
MorituriMax <newage@sendarico.net> wrote:
"tj Frazir" <GravityPhysics@webtv.net> wrote in message
news:26909-3F933A6D-1@storefull-2154.public.lawson.webtv.net...
The.. .. ..
BS Filter Activated.
Will you quit it? Strength and hardness of materials do tend to increase
when they get cold.
--
"Let us learn to dream, gentlemen, then perhaps we shall find the
truth... But let us beware of publishing our dreams before they have been
put to the proof by the waking understanding." -- Friedrich August Kekulé
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