An atmospheric envelope for ground-based telescopes.
Robert Clark
space towers:
Newsgroups: sci.astro, sci.physics, sci.mech.fluids, sci.engr.mech,
sci.space.policy
From: "Robert Clark" <rgregorycl...@yahoo.com>
Date: 28 Mar 2005 12:52:00 -0800
Subject: "Rockets not carrying fuel" and the space tower.
http://groups-beta.google.com/group/sci.astro/browse_frm/thread/ab0c8a53330a521a
Then this could be used to surround ground based scopes with a shroud
that reaches up at least to the stratosphere, say 100,000 ft. The
density of the air at 100,000 ft. is about 10 mbars and is above
weather variations, which occurs in the lower troposphere. It is also
above most water vapor, important for infrared observations.
In the "Rockets not carrying fuel and the space tower" thread I noted
that atmospheric pressure decreases exponentially with altitude for a
constant cross-sectional column of air. Therefore to have the pressure
remain constant with altitude you could have the tower around the scope
have a cross-section that decreases exponentially with altitude at this
same rate. Then the tower could be open to the atmosphere at the top
since the pressure equilibrium will prevent air from rushing down to
the bottom of the tower.
A problem though is the exponential decrease in pressure with altitude
only holds at constant temperature. It may be necessary to have the
tower be heated with a temperature gradient that changes with altitude
to maintain this. Also, since the density of the air at 100,000 ft. is
only about 10 mbars so it might be possible to have an atomically thin
transparent cover at the top to prevent air flow down the tower. Then
this thin cover would only minimally reduce light collection and cause
only minimal distortion.
Another possibility is that such an atmospheric envelope would allow
arbitrarily large liquid mirrors to be constructed on Earth. A key
problem with liquid mirrors is the wind they kick up due to the
rotation distorts the surface:
A Pristine View of the Universe... from the Moon.
"On Earth, LMTs are limited in size to about 6 meters in diameter
because the self-generated wind that comes from spinning the telescope
disturbs the surface. Additionally, like other Earth-based telescopes,
LMTs are subject to atmospheric absorption and distortion, greatly
reducing the range and sensitivity of infrared observing. But the
atmosphere-free moon, Angel says, provides the perfect location for
this type of telescope while supplying the gravity necessary for the
parabolic mirror to form."
http://www.universetoday.com/am/publish/pristine_view_universe_moon.html?2812005
Noted telescope maker Roger Angel also suggests magnetic bearings be
used to solve another problem of large liquid mirror scopes, supporting
the rotating structure with minimal friction:
A Pristine View of the Universe... from the Moon.
"One of the challenges in developing an LMT on the moon is to create
the bearings to spin the platform smoothly and at a constant speed. Air
bearings are used for LMTs on Earth, but with no air on the moon, that
is impossible. Angel and his team are looking at cryogenic levitation
bearings, similar to what's used for magnetic levitation trains to
get a frictionless motion by using a magnetic field. Angel added, "As
a bonus, with the low temperatures on the moon you can do that without
expending any energy because you can make a superconducting magnet that
allows you to make a levitation bearing that doesn't require a
continuous input of electrical power."
"Angel called the bearings a critical component of the telescope.
"With no air on the moon to create wind, there's no limit to size
or reaching the accuracy that you require as long as the bearing is
alright," Angel said."
Then such scopes could be arbitrarily large, 100's of meters or even
kilometers across, with a cost only 100th that of a solid mirror
telescope of comparable size. Liquid mirror telescopes have the
disadvantage that they must be zenith-pointing so can only view a few
degrees of sky around the zenith. But since they are so low cost and
simple to construct we could build many at differenr latitudes to be
able to view most of the sky.
Bob Clark
Uncle Al
>
> In this post I suggested a low-cost means to have permanently erected
> space towers:
>
> Newsgroups: sci.astro, sci.physics, sci.mech.fluids, sci.engr.mech,
> sci.space.policy
> From: "Robert Clark" <rgregorycl...@yahoo.com>
> Date: 28 Mar 2005 12:52:00 -0800
> Subject: "Rockets not carrying fuel" and the space tower.
> http://groups-beta.google.com/group/sci.astro/browse_frm/thread/ab0c8a53330a521a
>
> Then this could be used to surround ground based scopes with a shroud
> that reaches up at least to the stratosphere, say 100,000 ft.
1) Numerical aperture
2) Steering
3) Convection vs. temp and altitude
3) http://www.mazepath.com/uncleal/sunshine.jpg
Hey Clark, you are obsessed with long rigid thingies pointing up.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
operator jay
Therefore to have the pressure
> remain constant with altitude you could have the tower around the scope
> have a cross-section that decreases exponentially with altitude at this
> same rate. Then the tower could be open to the atmosphere at the top
> since the pressure equilibrium will prevent air from rushing down to
> the bottom of the tower.
That doesn't seem right. If air wanted to rush down, why not let it rush
down, and install a turbine for free energy. And, if air were rushing
around, wouldn't it rush UP, from high pressure to low pressure?
George Dishman
"operator jay" <no...@none.none> wrote in message
news:qFBce.478$fF6....@news1.mts.net...
>
> Therefore to have the pressure
>> remain constant with altitude you could have the tower around the scope
>> have a cross-section that decreases exponentially with altitude at this
>> same rate. Then the tower could be open to the atmosphere at the top
>> since the pressure equilibrium will prevent air from rushing down to
>> the bottom of the tower.
>
> That doesn't seem right.
It is complete nonsense. The air would flow
in at the top and fall to the bottom until
the same pressure profile was established
inside as out. Cross section has no effect.
> If air wanted to rush down, why not let it rush
> down, and install a turbine for free energy.
You could, but it takes more energy to pump
the air out in the first place and you get
no more after it is filled.
> And, if air were rushing
> around, wouldn't it rush UP, from high pressure to low pressure?
It reads as though the idea was that the tube
is only open at the top.
George
rgrego...@yahoo.com
George Dishman wrote:
> "operator jay" <no...@none.none> wrote in message
> news:qFBce.478$fF6....@news1.mts.net...
> >
> > Therefore to have the pressure
> >> remain constant with altitude you could have the tower around the
scope
> >> have a cross-section that decreases exponentially with altitude at
this
> >> same rate. Then the tower could be open to the atmosphere at the
top
> >> since the pressure equilibrium will prevent air from rushing down
to
> >> the bottom of the tower.
> >
> > That doesn't seem right.
>
> It is complete nonsense. The air would flow
> in at the top and fall to the bottom until
> the same pressure profile was established
> inside as out. Cross section has no effect.
>
It very definitely has an effect. See the derivation of the
exponential decrease in pressure with altitude at constant temperature
on this page:
Hydrostatic equilibrium of the atmosphere.
http://farside.ph.utexas.edu/teaching/sm1/lectures/node53.html
The point is to have equilibrium the weight pressing down on any
cross-sectional slice must be balanced by the upward force due to the
pressure in that cross-sectional slice. The magnitude of this force is
dependent on the cross-sectional area.
The situation is quite analogous to the tapering proposed to be used
for space towers to have constant stress along its length.
Bob Clark
rgrego...@yahoo.com
If you had a straight cylindrical tower with no taper then the air
would equilibriate just as it does with the normal atmosphere and you
would have the same approximate exponential decrease of pressure with
height with no air flow between top and bottom.
But you do raise a good point. It would appear that if you suddenly
opened the bottom of a tapered tower, then since the pressure was
uniformly at 10 mbars throughout the tower, the outside air at normal
pressure should rush in and up the tower.
It would appear that this should continue. For imagine a tapered tower
closed at both the top and bottom. No matter what pressure we put in
this tower initially it will be maintained at constant pressure
throughout its entire length. So if normal atmospheric pressure was at
the bottom of this closed tower it would remain so all the up to its
top at 100,000 ft. Now if we open this tower at the bottom, there
should be no flow of air in or out because the pressure at the bottom
is already at normal pressure. Then by opening it at the top there will
be a pressure difference at the top and you should get air flow out at
the top and therefore air in flow at the bottom.
However it would appear the pressure would not remain at normal
pressure in the tower when you have the air flow because of the
Bernoulli principle that says when the velocity increases the pressure
decreases.
The question is how much would it decrease.
Bob Clark
George Dishman
<rgrego...@yahoo.com> wrote in message
news:1114863251.0...@o13g2000cwo.googlegroups.com...
>
> George Dishman wrote:
>> "operator jay" <no...@none.none> wrote in message
>> news:qFBce.478$fF6....@news1.mts.net...
>> >
>> > Therefore to have the pressure
>> >> remain constant with altitude you could have the tower around the
> scope
>> >> have a cross-section that decreases exponentially with altitude at
> this
>> >> same rate. Then the tower could be open to the atmosphere at the
> top
>> >> since the pressure equilibrium will prevent air from rushing down
> to
>> >> the bottom of the tower.
>> >
>> > That doesn't seem right.
>>
>> It is complete nonsense. The air would flow
>> in at the top and fall to the bottom until
>> the same pressure profile was established
>> inside as out. Cross section has no effect.
>>
> ...
>
> It very definitely has an effect. See the derivation of the
> exponential decrease in pressure with altitude at constant temperature
> on this page:
>
> Hydrostatic equilibrium of the atmosphere.
> http://farside.ph.utexas.edu/teaching/sm1/lectures/node53.html
Notice that the area A cancels out in going
from eqn (322) to (323).
> The point is to have equilibrium the weight pressing down on any
> cross-sectional slice must be balanced by the upward force due to the
> pressure in that cross-sectional slice. The magnitude of this force is
> dependent on the cross-sectional area.
You are forgetting there is also an upward
force from the angled sides of the tube.
The bottom line is that eqn 326 is independent
of any variation of the cross section with
height.
> The situation is quite analogous to the tapering proposed to be used
> for space towers to have constant stress along its length.
Not at all, that simply reflects the fact that
a thicker tower can support more weight at the
same stress.
rgrego...@yahoo.com
For a tower tapered *smaller* as you go upward there won't be an
upward force due to the tapered sides.
In the equations on the "Hydrostatic equilibrium of the atmosphere"
page, the cross-section area A won't cancel out if it is changing with
z.
Bob Clark
George Dishman
<rgrego...@yahoo.com> wrote in message
news:1114869854.7...@l41g2000cwc.googlegroups.com...
> George Dishman wrote:
<snip>
Ok, I misread the quote, I thought the taper
went the other way.
> For a tower tapered *smaller* as you go upward there won't be an
> upward force due to the tapered sides.
Correct, there would be a downward force on
the outside of the tower which is a small
fractio of the crushing force when the inside
is evacuated.
Once the air enters, there will be an equal
upward force on the inside so the structure
no longer carries the extra force (weight)
of the outside air, but that upward force
has an equal-but-opposite downward force on
the air, otherwise the air accelerates in a
direction to reduce the unbalanced force.
> In the equations on the "Hydrostatic equilibrium of the atmosphere"
> page, the cross-section area A won't cancel out if it is changing with
> z.
Split the column into numerous narrow columns
side by side. Each column either has the
downward force of the tube or open space at
the top. Air moves sideways to equalise the
pressure at any level so the columns ending
on the inside of the tube have the same
pressure as a free-standing column, and the
pressure for that is the same whether it is
inside or outside the tube.
Bottom line is that the air pressure inside
and out is the same at any given height once
equilibrium is reached. Otherwise we could
drill a small hole halfway up, put a turbine
in the hole and generate free power forever.
George
Robert Clark
That such power *might* be possible doesn't prove this wouldn't work
any more than the fact that the drawing of power from flowing rivers
would be "free" means that can't work.
You might be able to use the Bernoulli principle to determine if there
is a velocity change between the ends given there is a pressure
difference at the top and bottom.
This is usually applied to incompressible fluids but Professor M.S.
Cramer in this post suggests you can apply this energy principle even
for compressible gases as long as the Mach number (ratio of the speed
to the speed of sound) is low:
Newsgroups: sci.astro, sci.physics, sci.mech.fluids, sci.engr.mech,
sci.space.policy
From: M.S. Cramer <macra...@vt.edu>
Date: Mon, 08 Nov 2004 12:03:37 -0500
Local: Mon,Nov 8 2004 9:03 am
Subject: Re: Question about Poiseuille's formula.
http://groups-beta.google.com/group/sci.astro/msg/a2672f6dc7003aca
In that thread the dividing line was given as about Mach 0.3.
Bob Clark
George Dishman
"Robert Clark" <rgrego...@yahoo.com> wrote in message
news:1114888984.3...@o13g2000cwo.googlegroups.com...
> any more than the fact that the drawing of power from flowing rivers
> would be "free" means that can't work.
A flowing river is not in static equilibrium,
the column of air is. For the case of static
equilibrium, the pressure inside must exactly
equal the pressure outside or you can violate
conservation of energy and produce a perpetual
motion machine. In the dynamic case starting
from an internal vacuum, the air will continue
to flow in until it reaches the conditions for
static equilibrium.
> You might be able to use the Bernoulli principle
No, you can't use Bernoulli for the static case.
Try to stick one topic Bob, you were talking of
the static situation and suggesting that tapering
the cross section could influence the relationship
between pressure and height. It doesn't, which you
can see from the conservation argument. If you
follow that then go back to the web pages you
quoted and see if you can follow what they are
saying so that you learn a bit about the subject.
HTH
George
Robert Clark
Perhaps this is something that can be tested.
At constant temperature for a constant cross-sectional column, the
pressure changes with height by:
p(z)= p(0)e^(-mgz/RT), p the pressure, m the molecular weight, g the
gravitational acceleration, 9.81 m/s^2, z the height in kilometers, R
the universal gas constant, 8.31 J/(K mol), T the temperature in
Kelvin. The quantity h = RT/mg is sometimes called the scale height and
the formula is written:
p(z)= p(0)e^(-z/h).
Then for air at molecular weight 29 and at T=288K, h = 8.4 km.
However, to test this we would want something for which the scale
height is smaller so a tower of reasonable height could be used to see
an observable difference. You want the temperature to be low, and the
molecular weight high so the scale height is small. However, the
temperature can't be too low because the pressure formula is dependent
on the ideal gas law which breaks down when the temperature gets so low
as to make the gas become liquid.
Some possibilities might be xenon and radon. Xenon has molecular
weight 131 and boiling point 165K. Then at 170K the scale height is
1.1 km.
Radon has molecular weight 222 and boiling point 211K. Then at 220K
the scale height is .84 km. However, radon usually has some percentage
of radiactive isotopes so you might not want to use that.
Any other cases of a gas (which need not be an element) with a low
ratio of boiling point to molecular weight at 1 atm?
If you use xenon, you could use a straight closed vertical pipe say
100 meters long supported within or along side some other tall
structure or building. You could use this to confirm the exponential
decrease of pressure with height at constant temperature. The pressure
would be less at the top than the bottom by a factor of e^(-.1/1.1) =
.913.
For a closed tapered pipe containing xenon, you want the
cross-sectional area to decrease at the same rate, so for example it
should have area .913 times smaller at the top (at 100 meters high)
than at the bottom. Then you want to confirm this causes the pressure
to remain the same at the top as at the bottom.
To test if there would be gas flow if the tapered pipe was open at
both ends, you could enclose an open tapered pipe within a much larger
closed straight pipe containing xenon.
Bob Clark
George Dishman
"Robert Clark" <rgrego...@yahoo.com> wrote in message
news:1114895529....@f14g2000cwb.googlegroups.com...
It has been tested many times, that is where
the basic physics you find in textbooks comes
from.
Just hold a funnel under water in your bath
and see if it naturally flows up or down. Then
invert the funnel and see if the flow reverses.
I say there will be no flow when everything
reaches equilibrium.
George
rgrego...@yahoo.com
The key difference, and why it must be tested under the regime where
the fluid is a gas, is that liquids are (largely) incompressible, while
gases are compressible.
For instance when you calculate that the pressure at the bottom of a
column of water increases at the rate of 1 atm for very 10 meters of
height, that is based on the (approximation) that water is
incompressible.
In contrast, in calculating the exponential decrease of pressure with
altitude for a gas, that uses the version of the ideal gas law that
shows density changing with pressure.
Bob Clark
Bruce Durdle
derives from Bernouilli's equation with v = 0.
If you have a long enough column of air, whether it is enclosed in a
tube or not, the pressure will vary along its length. In a column of air
of atmospheric dimensions, the temperature will vary with altitude.
George Dishman
<rgrego...@yahoo.com> wrote in message
news:1114907770.3...@f14g2000cwb.googlegroups.com...
> the fluid is a gas, is that liquids are (largely) incompressible, while
> gases are compressible.
Again that is irrelevant since a pressure
difference would cause a flow regardless of
compressibility.
In the static situation we are discussing,
compressibility only means the density will
alter with pressure which is taken into
account in eqn 323 on the page you quoted.
In addition "largely" doesn't mean "completly"
so the simple test I proposed is adequate on
all counts.
However, it remains entirely unnecessary since
your suggestion violates conservation of energy.
George
Robert Clark
Bruce Durdle wrote:
> P = rho x g x h whether or not the fluid involved is compressible -
that
> derives from Bernouilli's equation with v = 0.
>
> If you have a long enough column of air, whether it is enclosed in a
> tube or not, the pressure will vary along its length. In a column of
air
> of atmospheric dimensions, the temperature will vary with altitude.
The key difference is that for a incompressible liquid, rho is staying
constant regardless of pressure. But for the compressible gas case, rho
is varying with pressure.
To see this equation can't hold in the regime where the ideal gas law
holds, note that under the ideal gas law: rho = Pm/RT , m the molecular
weight, R the universal gas constant, T the temperature. See for
example equation (325) here:
Hydrostatic equilibrium of the atmosphere.
http://farside.ph.utexas.edu/teaching/sm1/lectures/node53.html
Then if both equations were valid, P = (Pm/RT)gh so RT = mgh. But in
our case T is being held constant while h is varying.
This means Bernoulli also wouldn't hold when I wanted to apply it in
the tapered case with both ends open to see the speed of the gas.
It may be you could modify Bernoulli to hold in that case by making
the pressure vary exponentially with altitude.
Bob Clark
Robert Clark
where
> > the fluid is a gas, is that liquids are (largely) incompressible,
while
> > gases are compressible.
>
> Again that is irrelevant since a pressure
> difference would cause a flow regardless of
> compressibility.
>
> In the static situation we are discussing,
> compressibility only means the density will
> alter with pressure which is taken into
> account in eqn 323 on the page you quoted.
>
> In addition "largely" doesn't mean "completly"
> so the simple test I proposed is adequate on
> all counts.
>
> However, it remains entirely unnecessary since
> your suggestion violates conservation of energy.
>
> George
It can't hold for a liquid for which the ideal gas law can't apply
because the exponential decrease of pressure with altitude is
explicitly dependent on the ideal gas law.
If the violation of the energy principle you're referring to is the
Bernoulli principle then that itself doesn't hold when the pressure is
changing exponentially with altitude.
Bob Clark
George Dishman
> George Dishman wrote:
>> <rgrego...@yahoo.com> wrote in message
>> > The key difference, and why it must be tested under the regime
> where
>> > the fluid is a gas, is that liquids are (largely) incompressible,
> while
>> > gases are compressible.
>>
>> Again that is irrelevant since a pressure
>> difference would cause a flow regardless of
>> compressibility.
>>
>> In the static situation we are discussing,
>> compressibility only means the density will
>> alter with pressure which is taken into
>> account in eqn 323 on the page you quoted.
>>
>> In addition "largely" doesn't mean "completly"
>> so the simple test I proposed is adequate on
>> all counts.
>>
>> However, it remains entirely unnecessary since
>> your suggestion violates conservation of energy.
>>
>> George
>
> It can't hold for a liquid for which the ideal gas law can't apply
> because the exponential decrease of pressure with altitude is
> explicitly dependent on the ideal gas law.
This is the page you cited:
http://farside.ph.utexas.edu/teaching/sm1/lectures/node53.html
and you talked specifically of "the equation of
hydrostatic equilibrium" which is marked (323).
That holds for any gas or liquid and the text
above explains why.
The gas laws only come in after that when they
derive equation (326). Now if you fill the tube
with helium, then things will be different and
if you are lucky, it might even float away, and
if you fill it with hot air, the same might
happen, but if the gas mix and temperature
profile inside are the same as outside then the
pressure variation with height will also be the
same inside as out regardless of the cross
section. Read the text above eqn 323 again if
this is still eluding you.
> If the violation of the energy principle you're referring to is the
> Bernoulli principle
No, I am referring to the principle of
conservation of energy that says you cannot
create a perpetual motion machine from which
you can extract energy indefinitely without
replacing it. We could do that using a turbine
in the tube wall if you could create a pressure
difference at any height just by changing the
cross section.
> then that itself doesn't hold when the pressure is
> changing exponentially with altitude.
The Bernoulli principle relates to moving gas
so is not relevant to this discussion of a
system in static equilibrium in any way.
George
George
Quantum Mirror
able to point straight up. LOL to look in all directions the tube would
need to reach the zenith of the horizon. That is one big tube.
George Dishman
"Quantum Mirror" <jun...@pgrb.com> wrote in message
news:1114964504....@z14g2000cwz.googlegroups.com...
> The dumb part of this argument is the ground based scopes would only be
> able to point straight up. LOL to look in all directions the tube would
> need to reach the zenith of the horizon. That is one big tube.
Or just stick the scope at the top and
it is above the atmosphere. It is a daft
idea but Bob might learn some basic
physics nonetheless.
George
Bruce Durdle
The temperaturem of an atmospheric column of air is not constant, but
varies with altitude - standard atmospheric temperature is 288 K at sea
level, 218 K at 10 km altitude.
Bernouilli's eqwuation is a statement of energy conservation and applies
to any situation, regardless ofg the configuration, if al significant
energy forms are taken into account. With a gas, the thermal energy term
is more significant than the kinetic energy term, and you cannot ignore
the temperature effects.
Robert Clark
I didn't mention it but the method of erecting the tube by using
vented pressurized gas along its length the gas would allow you to
alter the direction the tube is pointing by changing the direction you
vent.
For a zenith pointing liquid mirror scope, they can be useful because
of their low cost, perhaps 1/100th the cost of a traditional telescope,
and because of their ease of construction.
You could have many of these scopes located at different latitudes to
cover the entire sky.
Bob Clark
Robert Clark
I was thinking of this applied to large ground based telescopes.
The Keck telescope has a weight of about 270 tons.
The planned 100 meter OWL telescope has a planned weight of 9,000
metric tons:
The Ultimate Telescope.
http://www.astrosociety.org/pubs/mercury/31_03/ultimate.html
There would also be the problem of maintaining the scopes at such high
altitudes.
Liquid mirror scopes have significantly lower weight so this might be
possible with those.
Bob Clark
Robert Clark
Bruce Durdle wrote:
> I think I'm going to regret getting involved in this, but ...
>
> The temperaturem of an atmospheric column of air is not constant, but
> varies with altitude - standard atmospheric temperature is 288 K at
sea
> level, 218 K at 10 km altitude.
>
> Bernouilli's equation is a statement of energy conservation and
applies
> to any situation, regardless ofg the configuration, if al
significant
> energy forms are taken into account. With a gas, the thermal energy
term
> is more significant than the kinetic energy term, and you cannot
ignore
> the temperature effects.
>
Energy conservation always holds but the Bernoulli equation also
assumes the fluid is incompressible and non-viscous (viscosity and
frictional effects can be ignored.)
For the real atmosphere the temperature changes with altitude. To
maintain the tower at constant temperature you may have to maintain it
at constant temperature with heating elements alongs its length.
Bob Clark
Quantum Mirror
vented pressurized gas along its length the gas would allow you to
alter the direction the tube is pointing by changing the direction you
vent."
To evacuate the tube down to a partial vacuum the tube would need to be
super rigid and strong, I don't think pressurized gas will cut the
mustard. Have you ever looked through a telescope? All telescopes are
clock driven because as the earth turns the stars are stationary. The
idea is worthless, chunk it and look for another. Or better yet with
your inablity to see simple physical problems with your ideas maybe you
should take up a new hobby.
George Dishman
> "I didn't mention it but the method of erecting the tube by using
> vented pressurized gas along its length the gas would allow you to
> alter the direction the tube is pointing by changing the direction you
> vent."
>
> To evacuate the tube down to a partial vacuum the tube would need to be
> super rigid and strong, I don't think pressurized gas will cut the
> mustard.
There was a long thread on this with Bob recently,
venting pressurised gas cannot even lift its own
weight for any reasonable length of pipe.
George
Robert Clark
The original idea was to use the tube to lauch spacecraft. It
therefore was already intended to be highly directional. You direct the
orientation of the tube to follow that of the telescope.
If you use pressurized gas rather than liquid the pressure does not
have to be especially strong. In fact for gases since pressure changes
exponentially with altitude you can choose the gas and temperature so
that the pressure is nearly constant over its length (long scale
length.)
Then inputting say 5000 psi pressure at the bottom will result in
close to that at the top. There are light-weight carbon fiber tanks
that already operate at these pressures.
Bob Clark
Robert Clark
?????
Pressurized gas would almost certainly be better than liquid. This is
because they are lighter so the pressure that would need to be input at
the bottom would not need to be as great in order to get suffient
pressure out at the top.
Since the pressure changes exponentially for a gas, by making the
scale height sufficiently long the pressure could even change just by a
few percent over kilometer lengths by using low molecular weight gas at
high temperature.
Bob Clark
George Dishman
I suggest you calculate the thrust you would
get. My experience suggests it will be much
too low to lift the pipe any useful distance
but I will be delighted if you prove me wrong.
This time though, make sure you use the nozzle
exit pressure and area, not the pipe pressure.
George
Robert Clark
There is not an *upward* force from the sides of this tapered tube but
there is a downward force. Once I include that in my calculation I
conclude that there is no dependence on cross-section as you said.
So tapering the container alone can not change the pressure profile.
So if this idea for a telescope shroud is to work you would need to
cover the top with a thin cover. Since the pressure at this altitude is
so low, it could probably be thin enough so as not to affect seeing.
Bob Clark
Quantum Mirror
because they are lighter so the pressure that would need to be input at
the bottom would not need to be as great in order to get suffient
pressure out at the top."
Have you ever worked with a vacuum? The apparatus must be very strong
or an implosion will destroy it . The amount of vacuum needed to help
with atmospheric impairment for a telescope would need a strong rigid
design. I am not even going to attempt to calculate the pressure needed
in a flexible tube to hold a vacuum. I don't think it is possible
because of material strength.
Bruce Durdle
Robert Clark wrote:
>
>
> Energy conservation always holds but the Bernoulli equation also
> assumes the fluid is incompressible and non-viscous (viscosity and
> frictional effects can be ignored.)
> For the real atmosphere the temperature changes with altitude. To
> maintain the tower at constant temperature you may have to maintain it
> at constant temperature with heating elements alongs its length.
>
>
> Bob Clark
>
Bernouilli DOES NOT assume that the fluid is incompressible - you can
apply it to a gas stream just as well as a liquid one. But you need to
take into account that the density of the fluid is a variable rather
than a constant - and if you are talking about major changes in
elevation, "g" also becomes a variable with altitude.
Rand Simberg
Durdle <bmdu...@clear.net.nz> made the phosphor on my monitor glow in
such a way as to indicate that:
>Bernouilli DOES NOT assume that the fluid is incompressible - you can
>apply it to a gas stream just as well as a liquid one. But you need to
>take into account that the density of the fluid is a variable rather
>than a constant - and if you are talking about major changes in
>elevation, "g" also becomes a variable with altitude.
If they're major enough for that to become a significant factor,
there's likely little air to deal with...
Robert Clark
Here is my problem:
I have a tapered tower as below.
/ \
/________\ <--- Cross-sectional area A(z+dz)
/ \
/------------\ <--- Cross-sectional area A(z)
Vertical distance between these cross-sections dz.
This page calculates the pressure on the slice between these
cross-sections for the *non-tapered* case:
Hydrostatic equilibrium of the atmosphere.
http://farside.ph.utexas.edu/teaching/sm1/lectures/node53.html
It gives the difference between the upward force exerted by the
gas and the downward force of a thin slice of the gas of thickness dz
and cross-sectional area A as: [p(z)-p(z+dz)]A .
At equilibrium this must be balanced by the downward force on the
slice due to gravity: rho*A*dz*g, with rho the density, and g the
gravitational acceleration.
So (1.) [p(z)-p(z+dz)]A = rho*A*dz*g, then p(z)-p(z+dz) = rho*dz*g.
You see A cancels out. There is no dependence on cross-section.
But suppose you have the tapered case. If A is dependent on height z,
then what I wanted to say is the equation (1.) becomes: (2.)
p(z)A(z)-p(z+dz)A(z+dz) = rho*A(z)*dz*g, so the area wouldn't just
cancel out in this case. But there are a couple of problems with this.
Look at the tapered diagram above. You see the larger lower
cross-section A(z) extends beyond that of the upper cross-section
A(z+dz). If you are calculating just the vertical forces involved, you
shouldn't include the forces that arise from the portion of A(z) that
don't act vertically on A(z+dz).
The other problem as Dishman mentioned is that I don't include the
downward force caused by the sides of the tower.
So these are the factors that suggest the extension to equation (2.)
shouldn't work. However, on the other hand we know *gases* expand to
fill available volume and pressure will equalize within a fixed volume.
So that pressure from the periphery of A(z) (i.e.,that not directly
under A(z+dz) ) will add to the total pressure anyway, which will
equalize evenly and the total pressure will wind up being as in
equation (2.) anyway. Then the pressure on the sides of the tower will
just be the same as this equalized pressure in the slice.
Here's another argument that supports this interpretation. Suppose you
have a rectangular box filled with a gas. The gas will expand to fill
the box.
_______________________
| |
| |
|_______________________|
But suppose you put an equal mass of gas into another box only half as
large in cross-sectional area as before but of the same height and at
the same temperature as the first box:
_____________
| |
| |
|_____________|
In both cases gas fills the entire box and pressure equalizes
uniformly within the box. But the volume in the second case is only
half that in the first case. Then by the ideal gas law PV=nRT, since
the temperature is the same in both cases the pressure is twice as
large in the second case as the first case. Then you see the pressure
is dependent on cross-sectional area because it determines the volume.
Note that this would not be the case with an incompressible liquid. In
that case as long as the height was the same the pressure would stay
the same at the bottom. And the pressure would not be uniform
throughout the box but would change linearly with height.
Now notice, the same idea applies to a gas in tapered box as below:
_______
/ \ The pressure would not be the same as in a
/ \ straight vertical box of the same height,
/ \ with the same mass of gas, because the
/ \ volume would be different.
/ \ In effect, the pressure is dependent
/__________________\ on cross-section.
Then, back to my scenario, in trying to arrange it so the pressure
stays constant with altitude you would expect that the pressure lower
down should be higher than at the top ot the tower because it has to
suppport the weight of the gas above. But the point is you widen the
cross-section at the bottom so that the volume increases in such a way
as to cancel out the pressure increase due to the weight.
Bob Clark
George Dishman
"Robert Clark" <rgrego...@yahoo.com> wrote in message
news:1115144657....@o13g2000cwo.googlegroups.com...
> Robert Clark wrote:
>> George Dishman wrote:
>> > <rgrego...@yahoo.com> wrote in message
>> > news:1114863251.0...@o13g2000cwo.googlegroups.com...
>> > >
>> > >
>> > > http://farside.ph.utexas.edu/teaching/sm1/lectures/node53.html
>> >
>> > Notice that the area A cancels out in going
>> > from eqn (322) to (323).
>> >
>> > > The point is to have equilibrium the weight pressing down on any
>> > > cross-sectional slice must be balanced by the upward force due to the
>> > > pressure in that cross-sectional slice. The magnitude of this force
>> > > is
>> > > dependent on the cross-sectional area.
>> >
>> > You are forgetting there is also an upward
>> > force from the angled sides of the tube.
>> >
>> > The bottom line is that eqn 326 is independent
>> > of any variation of the cross section with
>> > height.
>> >
>>
>> There is not an *upward* force from the sides of this tapered tube but
>> there is a downward force. Once I include that in my calculation I
>> conclude that there is no dependence on cross-section as you said.
>> So tapering the container alone can not change the pressure profile.
>>
>> ...
>
> Here is my problem:
>
> I have a tapered tower as below.
>
> / \
> /------------\ <--- Cross-sectional area A(z)
>
> Vertical distance between these cross-sections dz.
>
> This page calculates the pressure on the slice between these
> cross-sections for the *non-tapered* case:
>
> Hydrostatic equilibrium of the atmosphere.
> http://farside.ph.utexas.edu/teaching/sm1/lectures/node53.html
>
> It gives the difference between the upward force exerted by the
> gas and the downward force of a thin slice of the gas of thickness dz
> and cross-sectional area A as: [p(z)-p(z+dz)]A .
> At equilibrium this must be balanced by the downward force on the
> slice due to gravity: rho*A*dz*g, with rho the density, and g the
> gravitational acceleration.
> So (1.) [p(z)-p(z+dz)]A = rho*A*dz*g, then p(z)-p(z+dz) = rho*dz*g.
> You see A cancels out. There is no dependence on cross-section.
>
> The other problem as Dishman mentioned is that I don't include the
> downward force caused by the sides of the tower.
> Now notice, the same idea applies to a gas in tapered box as below:
>
> ________
> / \ The pressure would not be the same as in a
> / \ straight vertical box of the same height,
> / \ with the same mass of gas, because the
> / \ volume would be different.
> / \ In effect, the pressure is dependent
> /__________________\ on cross-section.
>
>
>
> Then, back to my scenario, in trying to arrange it so the pressure
> stays constant with altitude you would expect that the pressure lower
> down should be higher than at the top ot the tower because it has to
> suppport the weight of the gas above. But the point is you widen the
> cross-section at the bottom so that the volume increases in such a way
> as to cancel out the pressure increase due to the weight.
Bob, you are thinking along the right lines.
Split your section into two regions, a central
cylinder and an outer annulus of tapering
width:
p(z+dz)
________ ___
/| |\ ^
/ | | \ |
/ | | \ dz
/ | | \ |
/ | | \ |
/_____|______|_____\ _v_
p(z)
In the central region, the description on
the web page applies so you can find p(z)
as p(z+dz) plus the force due to gravity
acting on the mass within the central
region, i.e. the weight of the gas acting
over the area. Now consider the pressures
on either side of point 'a':
p(z+dz)
________ ___
/| |\ ^
/ | | \ |
/ | | \ dz
/ | | \ |
/ | | \ |
/_____|______|_____\ _v_
<a> p(z)
For the static situation, the pressures
must be the same or there will be a lateral
movement of gas to equalise them. The
conclusion then is that the same formula
applies to the tapered case as for the
constant cross section.
Finally consider points 'b' and 'c':
p(z+dz)
________ ___
/| |\ ^
/ | | \ |
c | | \ dz
/ | | \ |
/ | | \ |
/__b__|______|_____\ _v_
p(z)
The upward pressure at 'b' must be p(z) of
course. The downward pressure at 'b' is
composed of the weight of the gas between
'b' and 'c'(which is negligible) and the
downward force provided by the material of
the structure at 'c'.
If that were the whole story, there would
be a net upward force on the structure but
there is atmosphere outside too which
presses down on point c with equal force.
Imagine removing the tube and letting
everything reach equilibrium, then use
a StarTrek transporter to replace the
tube without disturbing any air molecules.
Nothing has changed, all the pressures
will remain as they were.
George
Robert Clark
> In this post I suggested a low-cost means to have permanently erected
> space towers:
>
> Newsgroups: sci.astro, sci.physics, sci.mech.fluids, sci.engr.mech,
> sci.space.policy
> From: "Robert Clark" <rgregorycl...@yahoo.com>
> Date: 28 Mar 2005 12:52:00 -0800
> Subject: "Rockets not carrying fuel" and the space tower.
>
http://groups-beta.google.com/group/sci.astro/browse_frm/thread/ab0c8a53330a521a
>
> Then this could be used to surround ground based scopes with a
shroud
> that reaches up at least to the stratosphere, say 100,000 ft. The
> density of the air at 100,000 ft. is about 10 mbars and is above
> weather variations, which occurs in the lower troposphere. It is also
> above most water vapor, important for infrared observations.
> In the "Rockets not carrying fuel and the space tower" thread I
noted
> that atmospheric pressure decreases exponentially with altitude for a
> constant cross-sectional column of air. Therefore to have the
pressure
> remain constant with altitude you could have the tower around the
scope
> have a cross-section that decreases exponentially with altitude at
this
> same rate. Then the tower could be open to the atmosphere at the top
> since the pressure equilibrium will prevent air from rushing down to
> the bottom of the tower.
altitude
> only holds at constant temperature. It may be necessary to have the
> tower be heated with a temperature gradient that changes with
altitude
> to maintain this. Also, since the density of the air at 100,000 ft.
is
> only about 10 mbars so it might be possible to have an atomically
thin
> transparent cover at the top to prevent air flow down the tower. Then
> this thin cover would only minimally reduce light collection and
cause
> only minimal distortion.
> Another possibility is that such an atmospheric envelope would allow
> arbitrarily large liquid mirrors to be constructed on Earth. A key
> problem with liquid mirrors is the wind they kick up due to the
> rotation distorts the surface:
>
> A Pristine View of the Universe... from the Moon.
> "On Earth, LMTs are limited in size to about 6 meters in diameter
> because the self-generated wind that comes from spinning the
telescope
> disturbs the surface. Additionally, like other Earth-based
telescopes,
> LMTs are subject to atmospheric absorption and distortion, greatly
> reducing the range and sensitivity of infrared observing. But the
> atmosphere-free moon, Angel says, provides the perfect location for
> this type of telescope while supplying the gravity necessary for the
> parabolic mirror to form."
>
http://www.universetoday.com/am/publish/pristine_view_universe_moon.html?2812005
>
> Noted telescope maker Roger Angel also suggests magnetic bearings be
> used to solve another problem of large liquid mirror scopes,
supporting
> the rotating structure with minimal friction:
>
> A Pristine View of the Universe... from the Moon.
> "One of the challenges in developing an LMT on the moon is to create
> the bearings to spin the platform smoothly and at a constant speed.
Air
> bearings are used for LMTs on Earth, but with no air on the moon,
that
> is impossible. Angel and his team are looking at cryogenic levitation
> bearings, similar to what's used for magnetic levitation trains to
> get a frictionless motion by using a magnetic field. Angel added, "As
> a bonus, with the low temperatures on the moon you can do that
without
> expending any energy because you can make a superconducting magnet
that
> allows you to make a levitation bearing that doesn't require a
> continuous input of electrical power."
> "Angel called the bearings a critical component of the telescope.
> "With no air on the moon to create wind, there's no limit to size
> or reaching the accuracy that you require as long as the bearing is
> alright," Angel said."
>
> Then such scopes could be arbitrarily large, 100's of meters or even
> kilometers across, with a cost only 100th that of a solid mirror
> telescope of comparable size. Liquid mirror telescopes have the
> disadvantage that they must be zenith-pointing so can only view a few
> degrees of sky around the zenith. But since they are so low cost and
> simple to construct we could build many at different latitudes to be
> able to view most of the sky.
>
>
>
> Bob Clark
Astronomers like to wring every last photon from their observations so
having a physical cover at the top of the shroud may be problematical.
The tapered tower providing constant pressure with altitude idea is
also problematical.
So after a web search I was interested to see this report:
David Swenson's electrostatic "invisible wall".
http://amasci.com/weird/unusual/e-wall.html
It concerns the creation of an "invisible wall" due to electrostatic
charge from a large sheet of polypropelene film moving at high speed at
a 3M plant.
I'd like to see a report of this in a peer-reviewed journal if anyone
knows of any.
A similar phenomenon is illustrated by this experiment on Hero's
Fountain:
Expt 013 -- Hero's Fountain.
http://chemmovies.unl.edu/chemistry/beckerdemos/BD013.html
Here, a electrostatically charged balloon held near a Hero's Fountain
causes the fountain to dissipate. See the video at the link in step #4
in the Procedure section.
I was actually thinking of this in regards to creating the tower in
the first place. It would be great if the fluid did not have to be
carried within a tower, which induces frictional and viscosity effects.
But a stream of water sent kilometers into the air would dissipate at
high altitudes. So the idea would be use electrostatically charged air
to keep the water in a tight stream. To create the charge you could
have lasers ionize the air surrounding the water stream.
Bob Clark
George Dishman
Lasers and ionised air? One of the biggest
problems for astronomers is light pollution!
Another is that distortion from the atmosphere
is due to thermal differences. Do you think
this could be done without heating the air
even slightly?
Get real Bob.
George
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
"George Dishman" <geo...@briar.demon.co.uk> wrote in message
news:d5lb4c$h04$1...@news.freedom2surf.net...
...
> Lasers and ionised air? One of the biggest
> problems for astronomers is light pollution!
> Another is that distortion from the atmosphere
> is due to thermal differences. Do you think
> this could be done without heating the air
> even slightly?
I wonder if you could construct a "lens" with ionized air, just
as they carve 3D graphics into blocks of polycarbonate? Be a
little hard on planes, of course... ;>)
David A. Smith
George Dishman
"N:dlzc D:aol T:com (dlzc)" <N: dlzc1 D:cox T:n...@nospam.com> wrote in
message news:Ypqfe.4237$eU.3178@fed1read07...
The ionosphere is permanently charged.
I have on occasion mused about tapping
into that DC source using an argon laser
to create a conductive path. It's a clean
sustainable energy source, but it's not
much more practical than Bob's ideas
(and off-topic in most of these groups).
George
rgrego...@yahoo.com
There are two separate suggestions. In the original "Rockets not
carrying fuel" thread there were various proposals for getting the
fuel/reaction mass to altitude. One was just to send it as a high
velocity fountain. The problem would be it would spread out too far. So
for this proposal, for space propulsion, you would use ionization to
keep the flow in a tight stream.
The second suggestion is in regard to the telescope shroud. You could
have the "invisible wall" only at the top in which case you would not
need to use high powered lasers, for example, by using the technique
that, apparently, worked at the 3M plant.
Or instead of having a physical shroud, you could have an "invisible
wall" formed by ionized air. In this case you would need lasers or some
other method to ionize the air. But the key feature is that the ionized
air would be used to create and preserve the vacuum around the
telescope and all the way up to altitude. So there would not be heated
air around the telescope since it would be maintained in vacuum.
It might be possible to use a single laser to create the ionized
shroud of air. You could rotate the laser at high speed so it formed a
cone of ionization around the telescope.
However, to create sufficient ionization at altitude the power
requirements may be prohibitive.
Bob Clark
George Dishman
<rgrego...@yahoo.com> wrote in message
news:1115572029....@z14g2000cwz.googlegroups.com...
> George Dishman wrote:
>> "Robert Clark" <rgrego...@yahoo.com> wrote in message
>> news:1115565190....@f14g2000cwb.googlegroups.com...
>> >
>> > Astronomers like to wring every last photon from their observations so
>> > having a physical cover at the top of the shroud may be problematical.
>> > The tapered tower providing constant pressure with altitude idea is
>> > also problematical.
>> > So after a web search I was interested to see this report:
>> >
>> > David Swenson's electrostatic "invisible wall".
>> > http://amasci.com/weird/unusual/e-wall.html
The page you quote is entirely speculative, they
didn't actually find out what caused the effect.
I know of a situation where there was a single
turn of wire on the ground. Someone walking along
put one foot in it and found he couldn't move.
The coil undergoing a short circuit test and was
carrying several thousand amps DC. It wasn't a
direct force effect, the field was numbing the
nerves in the guys legs. There are many
possibilities in the case of the 3M effect and
until they figure it out, reproduce it and find
out how to control it you are wasting your time.
Ionising the air would produce a force, the
balloon is charged, not ionised.
The same would be true of the 3M film, the
rollers carry away the opposing charge like
a Van de Graff generator.
> The second suggestion is in regard to the telescope shroud. You could
> have the "invisible wall" only at the top in which case you would not
> need to use high powered lasers, for example, by using the technique
> that, apparently, worked at the 3M plant.
"apparently". There was the appearance of some
sort of force but from the page you quote it
seems they never really understood what caused
it. All the stuff about ionised air is just
speculation.
Again you still have the problem of charging the
air. Simply ionising it wouuld make it glow like
a neon tube.
> Or instead of having a physical shroud, you could have an "invisible
> wall" formed by ionized air. In this case you would need lasers or some
> other method to ionize the air. But the key feature is that the ionized
> air would be used to create and preserve the vacuum around the
> telescope and all the way up to altitude.
The force they produced was a flat plane. Do the
math and see what charge distribution you need
to produce the correct direction of force.
> So there would not be heated
> air around the telescope since it would be maintained in vacuum.
> It might be possible to use a single laser to create the ionized
> shroud of air. You could rotate the laser at high speed so it formed a
> cone of ionization around the telescope.
> However, to create sufficient ionization at altitude the power
> requirements may be prohibitive.
Lighting up the sky with km of ionised gas,
the world's biggest fluorescent tube, would
somewhat defeat the purpose, wouldn't it?
George
Dirk Bruere at Neopax
>>only minimal distortion.
>> Another possibility is that such an atmospheric envelope would allow
>>arbitrarily large liquid mirrors to be constructed on Earth. A key
>>problem with liquid mirrors is the wind they kick up due to the
>>rotation distorts the surface:
Which gives me a chance to ask a question that's mildly bothered me for some
40yrs, since I was a pre-teen!
If the problem with (say) a mercury mirror is its liquid state why not spin it
to get the right shape and then freeze it with LN2? I assume this won't work but
am not quite sure why.
--
Dirk
The Consensus:-
The political party for the new millenium
http://www.theconsensus.org
fkasner
> Robert Clark wrote:
>
>>> only minimal distortion.
>>> Another possibility is that such an atmospheric envelope would allow
>>> arbitrarily large liquid mirrors to be constructed on Earth. A key
>>> problem with liquid mirrors is the wind they kick up due to the
>>> rotation distorts the surface:
>
>
> Which gives me a chance to ask a question that's mildly bothered me for
> some 40yrs, since I was a pre-teen!
>
> If the problem with (say) a mercury mirror is its liquid state why not
> spin it to get the right shape and then freeze it with LN2? I assume
> this won't work but am not quite sure why.
>
You forget that mercury is a metal (even if a liquid at room
temperature) and so has a large thermal coefficient of expansion and
thus will severly distort as it is cooled. I used to use mercury for
creating an artificial horizon when doing sextant readings on dry land
with no clear view of the horizon. It had better be relatively windless
when doing such things with mercury.
FK
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
"Dirk Bruere at Neopax" <di...@neopax.com> wrote in message
news:3e7smcF...@individual.net...
> Robert Clark wrote:
>
>>>only minimal distortion.
>>> Another possibility is that such an atmospheric
>>> envelope would allow arbitrarily large liquid
>>> mirrors to be constructed on Earth. A key
>>>problem with liquid mirrors is the wind they
>>> kick up due to the rotation distorts the surface:
>
> Which gives me a chance to ask a question
> that's mildly bothered me for some 40yrs,
> since I was a pre-teen!
>
> If the problem with (say) a mercury mirror is
> its liquid state why not spin it to get the right
> shape and then freeze it with LN2? I assume
> this won't work but am not quite sure why.
I believe they actually do this for composite mirrors. Locate
the glass blank at a certain radius, heat it until soft, spin it
up, and maintain until "cool". Then all the "roughing" work for
that location is done. The mechanisms for spinning something are
not vibration free...
David A. Smith
Brian Whatcott
<di...@neopax.com> wrote:
>Robert Clark wrote:
>
>>>only minimal distortion.
>>> Another possibility is that such an atmospheric envelope would allow
>>>arbitrarily large liquid mirrors to be constructed on Earth. A key
>>>problem with liquid mirrors is the wind they kick up due to the
>>>rotation distorts the surface:
>
>Which gives me a chance to ask a question that's mildly bothered me for some
>40yrs, since I was a pre-teen!
>
>If the problem with (say) a mercury mirror is its liquid state why not spin it
>to get the right shape and then freeze it with LN2? I assume this won't work but
>am not quite sure why.
Thermal gradients are bad news for seeing.
Brian Whatcott Altus OK
Brian Whatcott
wrote:
>...I used to use mercury for
>with no clear view of the horizon. It had better be relatively windless
>when doing such things with mercury.
>FK
This was a traditional technique (oil or molasses were also used)
A roof was usually provided to hold off the wind effects - this would
be "talcum" i.e. a form of mica - or glass. It was the former that
went with Lewis & Clark
Brian Whatcott Altus OK
Dirk Bruere at Neopax
Cool it slowly.
Plus I assume that mercury is a good conductor of heat.
Dirk Bruere at Neopax
> Dirk Bruere at Neopax wrote:
>
>> Robert Clark wrote:
>>
>>>> only minimal distortion.
>>>> Another possibility is that such an atmospheric envelope would allow
>>>> arbitrarily large liquid mirrors to be constructed on Earth. A key
>>>> problem with liquid mirrors is the wind they kick up due to the
>>>> rotation distorts the surface:
>>
>>
>>
>> Which gives me a chance to ask a question that's mildly bothered me
>> for some 40yrs, since I was a pre-teen!
>>
>> If the problem with (say) a mercury mirror is its liquid state why not
>> spin it to get the right shape and then freeze it with LN2? I assume
>> this won't work but am not quite sure why.
>>
>
> You forget that mercury is a metal (even if a liquid at room
> temperature) and so has a large thermal coefficient of expansion and
> thus will severly distort as it is cooled. I used to use mercury for
Now that sounds more plausible...
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
"Dirk Bruere at Neopax" <di...@neopax.com> wrote in message
news:3e94rcF...@individual.net...
> Brian Whatcott wrote:
...
>>>If the problem with (say) a mercury mirror is
>>>its liquid state why not spin it to get the
>>>right shape and then freeze it with LN2? I
>>>assume this won't work but am not quite
>>>sure why.
>>
>> Thermal gradients are bad news for seeing.
>
> Cool it slowly.
It is during operation, where the air in contact with the cold
mirror (can you say condensation?) is at a different temperature
than ambient. The thermal gradient would be through the air, and
stratification would not help you if your telescope were not
pointed straight up.
> Plus I assume that mercury is a good
> conductor of heat.
Better than glass. But it also oxidizes quite readily. And much
heavier to mount.
David A. Smith
Dirk Bruere at Neopax
> Dear Dirk Bruere at Neopax:
>
> "Dirk Bruere at Neopax" <di...@neopax.com> wrote in message
> news:3e94rcF...@individual.net...
>
>>Brian Whatcott wrote:
>
> ...
>
>>>>If the problem with (say) a mercury mirror is
>>>>its liquid state why not spin it to get the
>>>>right shape and then freeze it with LN2? I
>>>>assume this won't work but am not quite
>>>>sure why.
>>>
>>>Thermal gradients are bad news for seeing.
>>
>>Cool it slowly.
>
>
> It is during operation, where the air in contact with the cold
> mirror (can you say condensation?) is at a different temperature
> than ambient. The thermal gradient would be through the air, and
> stratification would not help you if your telescope were not
> pointed straight up.
Well, the answer to that one is obvious - cover it with flat glass in a box
filled with cooled argon or similar. The thermal gradient would be through the
glass, or multiple layers of glass. Or maybe try siting it in antarctica...
bz
> N:dlzc D:aol T:com (dlzc) wrote:
>
>> Dear Dirk Bruere at Neopax:
>>
>> "Dirk Bruere at Neopax" <di...@neopax.com> wrote in message
>> news:3e94rcF...@individual.net...
>>
>>>Brian Whatcott wrote:
>>
>> ...
>>
....
> Well, the answer to that one is obvious - cover it with flat glass in a
> box filled with cooled argon or similar. The thermal gradient would be
> through the glass, or multiple layers of glass. Or maybe try siting it
> in antarctica...
>
Why not use a metal that has a melting point just above room temperature?
http://www.mcp-group.com/alloys/
--
bz
please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.
bz...@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
Brian Whatcott
<di...@neopax.com> wrote:
>Brian Whatcott wrote:
>
>> On Mon, 09 May 2005 02:28:43 +0100, Dirk Bruere at Neopax
>> <di...@neopax.com> wrote:
>>
>>
>>>Robert Clark wrote:
>>>
>>>
>>>>>only minimal distortion.
>>>>>Another possibility is that such an atmospheric envelope would allow
>>>>>arbitrarily large liquid mirrors to be constructed on Earth. A key
>>>>>problem with liquid mirrors is the wind they kick up due to the
>>>>>rotation distorts the surface:
>>>
>>>Which gives me a chance to ask a question that's mildly bothered me for some
>>>40yrs, since I was a pre-teen!
>>>
>>>If the problem with (say) a mercury mirror is its liquid state why not spin it
>>>to get the right shape and then freeze it with LN2? I assume this won't work but
>>>am not quite sure why.
>>
>>
>>
>> Thermal gradients are bad news for seeing.
>
>Cool it slowly.
>Plus I assume that mercury is a good conductor of heat.
The thermal gradient of concern for 'seeing' is that of the column of
air over the mirror!
Brian Whatcott Altus, OK
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
"bz" <bz...@ch100-5.chem.lsu.edu> wrote in message
news:Xns965161A427CD6WQ...@130.39.198.139...
> Dirk Bruere at Neopax <di...@neopax.com> wrote in
> news:3e96f2F...@individual.net:
>
>> N:dlzc D:aol T:com (dlzc) wrote:
>>
>>> Dear Dirk Bruere at Neopax:
>>>
>>> "Dirk Bruere at Neopax" <di...@neopax.com> wrote in message
>>> news:3e94rcF...@individual.net...
>>>
>>>>Brian Whatcott wrote:
>>>
>>> ...
>>>
> ....
>> Well, the answer to that one is obvious - cover it
>> with flat glass in a box filled with cooled argon or
>> similar. The thermal gradient would be through
>> the glass, or multiple layers of glass. Or maybe
>> try siting it in antarctica...
>>
> Why not use a metal that has a melting point
> just above room temperature?
> http://www.mcp-group.com/alloys/
Aluminum would be such a candidate, depending on how you define
"just above". But the coefficient of thermal expansion (and
hence the optical chracteristic of the mirror) of metals is much
higher than for glass...
David A. Smith
bz
news:PnUfe.4428$eU.2132@fed1read07:
I would say that 660 C is not JUST above room temperature.
And, I am not sure what you were trying to accomplish, but using frozen
mercury for a telescope mirror would NOT be a good idea. Woods metal or
some other low melting metal alloy would be better.
Even better is to use an active mirror. Pizoelectric material, mirror made
up of hundreds or thousands of small mirrors that can each be tilted.
Computer control used to keep the image focused as the atmospheric density
varies.
Un-twinkle Un-twinkle little star, how I wonder what distance you are.
Robert Clark
That the phenomenon observed was a true effect is suggested by
electrostatic experiments suggest as these:
J4-12: ELECTROSTATIC FORCE - MOVING LUMBER.
PURPOSE: To demonstrate polarization of water molecules.
http://jedlik.phy.bme.hu/~hartlein/physics.umd.edu/deptinfo/facilities/lecdem/j4-12.htm
This is from a page of undergraduate physics experiments. It
demonstates the effect of electrostatic charge on polar molecules such
as water. A piece of lumber is made to rotate over a low friction
support by bringing close a rod charged with static electricity by
rubbing with a wool cloth. The effect arises from the polar molecules
of water in the wood.
This is effect is also illustrated by this experiment:
J4-11: POLAR AND NONPOLAR LIQUIDS
PURPOSE: To demonstrate that non-uniform electric fields produce a
force on polar molecules.
http://jedlik.phy.bme.hu/~hartlein/physics.umd.edu/deptinfo/facilities/lecdem/j4-11.htm
Here a stream of water is deflected by a source of electrostatic
charge while a stream of carbon tetrachloride (CCl4), a non-polar
molecule, is not.
The static charge built up on a rod or a balloon by rubbing with a
cloth amounts to perhaps a few thousand volts. Dave Swenson in the
phenomenon observed at 3M estimated the static electricity was in the
megavolt range. Human beings and all living beings are composed of
mostly liquid water. Then considering the deflections of water at a few
thousand volts in these lab experiments, it is conceivable that
voltages a thousand times higher could have the effect observed at the
3M plant.
If this is the explanation then this "invisible wall" would work
against human beings and other living things but not inanimate objects
that did not contain water.
However, it is known that large Van der Graaf generators can produce
static voltages in the megavolt range:
Construction of the Van de Graaff Generator.
http://www.mos.org/sln/toe/construction.html
Have experiments been done on the effect of such large generators on
large amounts of water?
Note that with this explanation of the 3M plant phenomenon it wouldn't
work to create an invisible wall around our telescope because it
wouldn't effect the non-polar molecules in the air.
The method of ionizing the air with lasers might still work though.
Bob Clark
rgrego...@yahoo.com
I presume there is also a repulsive effect here in this case as well:
Expt 013 -- Hero's Fountain.
http://chemmovies.unl.edu/chemÂistry/beckerdemos/BD013.html
since in the video the fountain is made to stop when the charged
balloon is brought close to the stream.
But this page says a stream of falling water will be *attracted* to a
charged balloon:
Lesson 1: Charge and Charge Interactions
Polarization.
http://www.glenbrook.k12.il.us/gbssci/phys/Class/estatics/u8l1e.html
Also curious is what this page on this experiment says of it:
J4-11: POLAR AND NONPOLAR LIQUIDS.
"NOTE: Although the demonstration apparently works as predicted, this
demonstration does not work according to theory, and should probably
not be done and explained in the traditional manner."
http://www.physics.umd.edu/lecdem/services/demos/demosj4/j4-11.htm
And in this video version of the moving lumber experiment, the lumber
is *attracted* to the charged rod whether it is charged positive or
negative:
Question #131
http://www.physics.umd.edu/lecdem/outreach/QOTW/arch7/q131.htm
Bob Clark
George Dishman
> <rgrego...@yahoo.com> wrote in message
> news:1115990444....@f14g2000cwb.googlegroups.com...
> In the 3M phenomenon, there appears to be a repulsive effect.
Or there is an attraction to the building outside
the "tent" formed by the film due to a build up of
charge on the person. They never investigated so
there is little to learn.
> I presume there is also a repulsive effect here in this case as well:
>
> Expt 013 -- Hero's Fountain.
> http://chemmovies.unl.edu/chemčšstry/beckerdemos/BD013.html
>
> since in the video the fountain is made to stop when the charged
> balloon is brought close to the stream.
If you look closely, it is because the slight deviation
causes the water to fall back on the nozzle blocking
it. It is hard to see which way it is moved but it
depends on whether the water is in droplets which can
then be charged as well as the balloon (the direction
then depeds on whether the charges are the same or
opposite) or whether the stream is continuous hence
earth by the conductivity of the water. In that case
only polarisation would apply which is generally
attractive since it pulls opposite charge to the
nearer side. The next page explains why.
> But this page says a stream of falling water will be *attracted* to a
> charged balloon:
Yes, if you ever try it, that is the most usual result.
> Lesson 1: Charge and Charge Interactions
> Polarization.
> http://www.glenbrook.k12.il.us/gbssci/phys/Class/estatics/u8l1e.html
That's a good page, look at the bit headed:
"How Does Polarization Explain the Balloon
and the Wall Demonstration?"
> Also curious is what this page on this experiment says of it:
>
> J4-11: POLAR AND NONPOLAR LIQUIDS.
> "NOTE: Although the demonstration apparently works as predicted, this
> demonstration does not work according to theory, and should probably
> not be done and explained in the traditional manner."
> http://www.physics.umd.edu/lecdem/services/demos/demosj4/j4-11.htm
The devil is in the details ;-)
> And in this video version of the moving lumber experiment, the lumber
> is *attracted* to the charged rod whether it is charged positive or
> negative:
>
> Question #131
> http://www.physics.umd.edu/lecdem/outreach/QOTW/arch7/q131.htm
Yes, that is the normal result of polarisation.
George