An atmospheric envelope for ground-based telescopes.

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"
Date: 28 Mar 2005 12:52:00 -0800
Subject: "Rockets not carrying fuel" and the space tower.
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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."
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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
Reply to
Robert Clark
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[snip]
1) Numerical aperture 2) Steering 3) Convection vs. temp and altitude 3)
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Hey Clark, you are obsessed with long rigid thingies pointing up.
Reply to
Uncle Al
Therefore to have the pressure
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?
Reply to
operator jay
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.
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.
It reads as though the idea was that the tube is only open at the top.
George
Reply to
George Dishman
...
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.
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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
Reply to
rgregoryclark
rushing
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
Reply to
rgregoryclark
Notice that the area A cancels out in going from eqn (322) to (323).
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.
Not at all, that simply reflects the fact that a thicker tower can support more weight at the same stress.
Reply to
George Dishman
temperature
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
Reply to
rgregoryclark
Ok, I misread the quote, I thought the taper went the other way.
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.
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
Reply to
George Dishman
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 Date: Mon, 08 Nov 2004 12:03:37 -0500 Local: Mon,Nov 8 2004 9:03 am Subject: Re: Question about Poiseuille's formula.
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In that thread the dividing line was given as about Mach 0.3.
Bob Clark
Reply to
Robert Clark
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.
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
Reply to
George Dishman
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.
snipped-for-privacy@yahoo.com wrote:
Reply to
Bruce Durdle
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.
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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
Reply to
Robert Clark
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.
Reply to
Quantum Mirror
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
Reply to
George Dishman
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 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 wrote:
Reply to
Bruce Durdle
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
Reply to
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.
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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
Reply to
Robert Clark
applies
significant
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
Reply to
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."
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.
Reply to
Quantum Mirror

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