But wait, the endcaps don't need to be warm for humans. In fact, we
would like them to be quite cold at the center, around 40C to 50C colder
than the rim, in order to give the air a decent lapse rate and keep the
atmosphere stabile.
Perhaps we do want the endcaps cold, but
... there is no thermodynamic limit on the efficiency, heat pumped
over work input, of a heat pump that is pumping heat from a higher
T region to a lower T region
I don't understand. I thought that Carnot first discovered that the
limit on thermodynamic efficiency has to do with
On Wed, Jul 23, 2003 at 12:57:11PM +, Robert J. Chassell wrote:
But wait, the endcaps don't need to be warm for humans. In fact,
we would like them to be quite cold at the center, around 40C to
50C colder than the rim, in order to give the air a decent lapse
rate and keep
On Wed, Jul 23, 2003 at 01:12:34PM +, Robert J. Chassell wrote:
... there is no thermodynamic limit on the efficiency, heat pumped
over work input, of a heat pump that is pumping heat from a higher
T region to a lower T region
I don't understand. I thought that Carnot
On Sun, Jul 20, 2003 at 11:14:45PM -0400, Erik Reuter wrote:
If the walls are 1km thick and have an average thermal conductivity of
80 W/m K, and if the asteroid has the same temperature/depth profile
at all points on the surface, then radiation and conduction must be
equal and we get
You have been mixing References: headers between this thread and others
and it makes it harder for me to follow.
I always reply to the digest.
I didn't expect that! The message I was replying to previously was about
the habitat, but it referenced a message
Cool air at the axis should be heavier and would try to displace warmer,
less dense air.
This would cause air currents that would speed up heat transfer in the
atmosphere.
I don't think you could expect thermal equilibrium in the scenarios I'm
reading here, but the
Robert,
You have been mixing References: headers between this thread and others
and it makes it harder for me to follow.
Sorry, I did not mean to. I reply to the digest, then remove the
digest's second address, so you do not receive two copies of the same
message, and I try
On Sun, Jul 20, 2003 at 09:43:36PM +, Robert J. Chassell wrote:
I always reply to the digest. Presumably, all the threads in it are
for that digest. It is all one mail message.
I didn't expect that! The message I was replying to previously was about
the habitat, but it referenced a
What is the heat conductivity of dirt, rock, and nickle-iron? Does
anyone one know?
Dirt and rock are similar, in the range 0.2 - 2 W / m K.
Iron is about 84 W / m K
Nickel is 92 W / m K
air at 300K is 0.026 W / m K
Thank you. Am I right in thinking that for air,
- Original Message -
From: Robert J. Chassell [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Sunday, July 20, 2003 5:12 PM
Subject: Re: Irregulars query: air pressure in spinning habitats
What is the heat conductivity of dirt, rock, and nickle-iron? Does
anyone one know
At 10:12 PM 7/20/2003 +, you wrote:
What is the heat conductivity of dirt, rock, and nickle-iron? Does
anyone one know?
Dirt and rock are similar, in the range 0.2 - 2 W / m K.
Iron is about 84 W / m K
Nickel is 92 W / m K
air at 300K is 0.026 W / m K
Thank you.
On Sun, Jul 20, 2003 at 10:12:11PM +, Robert J. Chassell wrote:
Thank you. Am I right in thinking that for air, this is the heat
conductivity for still air, and not the heat transfer capabilities of
moving air?
It is the thermal conductivity of air, which is almost the same whether
the
David Hobby [EMAIL PROTECTED] asked
Robert--
You started all this. Am I right in thinking that you
WANT the habitat to have vaguely terrestrial weather? Something
like a layer of clouds a kilometer or two up? Sounds cool!
Almost: what I want to find out is whether
On Sat, Jul 19, 2003 at 10:55:49AM -0400, Robert J. Chassell wrote:
Presuming the end caps are mostly low-conductive stone or regolith,
the major temperature determiner for the end caps should be, I think,
the air, although indirect light will have an effect.
I think that would be difficult.
At 11:52 AM 7/19/03 -0400, Erik Reuter wrote:
On Sat, Jul 19, 2003 at 10:55:49AM -0400, Robert J. Chassell wrote:
Presuming the end caps are mostly low-conductive stone or regolith,
the major temperature determiner for the end caps should be, I think,
the air, although indirect light will have
On Sat, Jul 19, 2003 at 11:31:54AM -0500, Ronn!Blankenship wrote:
Actually, it is because the soil is a _poor_ conductor of heat, so
it doesn't warm up rapidly in the summer or cool off rapidly in the
winter, so
Actually, no,
At 01:24 PM 7/19/03 -0400, Erik Reuter wrote:
On Sat, Jul 19, 2003 at 11:31:54AM -0500, Ronn!Blankenship wrote:
Actually, it is because the soil is a _poor_ conductor of heat, so
it doesn't warm up rapidly in the summer or cool off rapidly in the
winter, so
Actually, no, it is not conductivity
On Sat, Jul 19, 2003 at 01:19:12PM -0500, Ronn!Blankenship wrote:
Are you suggesting that the air inside the habitat would be moving
fast enough that it would not have time to heat up or cool down to the
same temperature as the interior wall?
This question is not meaningful since the air in
- Original Message -
From: Erik Reuter [EMAIL PROTECTED]
To: Killer Bs Discussion [EMAIL PROTECTED]
Sent: Saturday, July 19, 2003 12:24 PM
Subject: Re: Irregulars query: air pressure in spinning habitats
On Sat, Jul 19, 2003 at 11:31:54AM -0500, Ronn!Blankenship wrote:
Actually
On Sat, Jul 19, 2003 at 05:03:07PM -0500, Dan Minette wrote:
Ronn's closer to right on this than you are.
Dan, you and Ronn are both straying from the topic that we were
discussing, and neither of you have addressed the question that
was asked. Additionally, Ronn made an incorrect statement
At 06:20 PM 7/19/03 -0400, Erik Reuter wrote:
On Sat, Jul 19, 2003 at 05:03:07PM -0500, Dan Minette wrote:
Ronn's closer to right on this than you are.
Dan, you and Ronn are both straying from the topic that we were
discussing, and neither of you have addressed the question that
was asked.
On Sat, Jul 19, 2003 at 06:02:21PM -0500, Ronn!Blankenship wrote:
At 06:20 PM 7/19/03 -0400, Erik Reuter wrote:
Presumably this is because the ground conducts heat much better than
the air and spreads heat out evenly throughout the ground.
As I said, this is _not_ true: the upper few feet
Robert J. Chassell wrote:
Presuming the end caps are mostly low-conductive stone or regolith,
the major temperature determiner for the end caps should be, I think,
the air, although indirect light will have an effect.
Erik Reuter [EMAIL PROTECTED] responded
I think
On Sat, Jul 19, 2003 at 08:12:06PM -0400, Robert J. Chassell wrote:
Hmmm you are right that the temperature is often more or less
constant 2 meters or so underground. However, in my understanding,
dirt and rock are very good insulators, and that heat flows slowly
through them.
Dirt is a
On Sat, Jul 19, 2003 at 07:40:38PM -0500, Ronn!Blankenship wrote:
What I have been saying is that if I were to perform the experiment by
obtaining a 2-meter or so probe with a sharp end, mount a thermocouple
at that end, and drive it into the ground, then place an identical
thermocouple at on
Robert,
You have been mixing References: headers between this thread and others
and it makes it harder for me to follow. (For example, you referenced
the Seth thread on this message). I didn't mention it before but it
has happened several times now, and it is particularly bothersome on
these
Erik Reuter wrote:
... That way we won't get caught in
a loop with me making math mistakes and you saying the result doesn't
make sense to you but you don't want to work out the math.
But actually doing the math makes it like WORK for me. : )
By the way, how did you know, before
Erik Reuter [EMAIL PROTECTED] wrote
If R=5km, m=4.85e-26, g=9.8, k=1.381e-23, T=300, and we note that
h must be in km, then
P/P0 = exp[ -0.115 h ] exp[ +1.15e-2 h^2 ] , h in km, R=5km, h = R
For Rama, with R=8km,
P/P0 = exp[ -0.115 h ] exp[ +7.17e-3 h^2 ] , h in km,
On Mon, Jul 14, 2003 at 10:23:53AM +, Robert J. Chassell wrote:
The main problem is that the pressures calculated for Earth disagree
with the figures I have for a pilot's standard atmosphere. In areas
without clouds, the earth's actual atmosphere is best represented by
a dry adiabatic
But I still wonder what the standard temperature is? What is the
lapse rate? How much does temperature drop per kilometer of increased
altitude? How much does dew point drop?
Those questions are well beyond my calculation abilities. If you are
interested in doing a
On Mon, Jul 07, 2003 at 06:49:35PM -0400, Robert J. Chassell wrote:
But I still wonder what the standard temperature is? What is
the lapse rate? How much does temperature drop per kilometer of
increased altitude? How much does dew point drop?
Those questions are well
On Sun, Jul 13, 2003 at 04:40:53PM -0400, Erik Reuter wrote:
But is it a phenomenological formula? I would think measurements were
made of the lapse rate, and then a curve was fit to the data.
I did some reading and it seems things are both simpler and more complex
than my question implied.
On Thu, Jul 10, 2003 at 11:29:26PM -0400, David Hobby wrote:
Erik Reuter wrote:
but since R_e = 6370km, and h = 1km, (1 + h / R_e) = 1 is an
excellent approximation so the formula becomes
P/P0 = exp[ -( h / R_e )( R_e m g / k / T ) ]
= exp[ -739 ( h / R_e )]
I can't
On Wed, Jul 09, 2003 at 09:11:22PM -0400, David Hobby wrote:
Robert J. Chassell wrote:
We may have interpreted the configuration differently. I interpreted
C as meaning a torus, or donut, or `like the inner tube of a tire'.
Agreed.
The short columns must have the same
I goofed in calculating the air pressures in Rama. I hope these
calculations are correct!
(mapconcat
'(lambda (h)
Calculate air pressures in a spinning space habitat
(format %f \n
(let ((e 2.718181828)
(R 8.0)) ; radius of
Erik Reuter wrote:
...
The same as in case 1.
Yes, I agree.
P/P0 = exp[ - ( h / R )^2 / 3.45 ]
Since h/R = 1/5 = 0.2, P/P0 = 0.988
(Although a pressure of .988 bar seems a bit high--a kilometer of
height makes a much larger pressure difference on Earth.)
As I said
[We agree for A and B.]
c) wheel like, with only the rim having air, the spokes
separate from the rim
In this configuration, the relevant maximum height is, I think,
the ceiling. Perhaps I am wrong -- does someone know?
I'm not sure what you are
Robert J. Chassell wrote:
We may have interpreted the configuration differently. I interpreted
C as meaning a torus, or donut, or `like the inner tube of a tire'.
Agreed.
The short columns must have the same pressure distribution as
the long columns in the spokes,
habitats that are wider than they are long are intrinsically stable
I'm having trouble visualizing any of this. When you say 'longer than they
are wide' do you mean like a cigarette or a can? And you are saying a
habitat that is more like a wheel is more stable, right?
Right.
Robert J. Chassell wrote:
...
For those figuring out the air pressure question, would there be
differences if
a) the structure was disc like, completely open on the inside
(other than support structures)
b) wheel like, with the rim having air and four (or x) spokes
Erik Reuter [EMAIL PROTECTED]
The only way I know of to get a simple equation is to make a
couple approximations:
(1) The temperature of all the air is the same, 300K
(2) The air all rotates with the endcaps, speed proportional to
radial distance from center
The second
On Mon, Jul 07, 2003 at 04:06:48PM +, Robert J. Chassell wrote:
Erik Reuter [EMAIL PROTECTED]
The only way I know of to get a simple equation is to make a
couple approximations:
(1) The temperature of all the air is the same, 300K
(2) The air all rotates with the
I figured out some of the values relating to Rama; the air is thin
and the acceleration figures are not consistent with other claims.
Erik Reuter [EMAIL PROTECTED] wrote:
Rama had a radius of about 8km. They entered near the axis and
began descending in spacesuits. After descending 2km,
By the way, does anyone know why so many science fiction writers
descripe spinning space habitats as being longer than they are wide?
Such habitats are intrinsically unstable. But habitats that are wider
than they are long are intrinsically stable
I know that the habitats are supposed to have
Erik Reuter wrote:
On Mon, Jul 07, 2003 at 12:38:36AM -0400, David Hobby wrote:
Yes, but I gave arguments showing that it did not give
reasonable results.
Not at all convincing arguments, with no numbers or equations.
They may not have convinced you, but they were
On Mon, Jul 07, 2003 at 02:47:03PM -0400, David Hobby wrote:
Oh. I thought that I had read the first one. Sorry.
The cylindrical sea melts in the first one, that was what I was
referring to. But they go in before it melts, and that is when most of
the numbers are given. It melts and
Robert J. Chassell wrote:
By the way, does anyone know why so many science fiction writers
descripe spinning space habitats as being longer than they are wide?
Such habitats are intrinsically unstable. But habitats that are wider
than they are long are intrinsically stable
I know that
In a message dated 7/7/2003 12:08:13 PM US Mountain Standard Time,
[EMAIL PROTECTED] writes:
Robert J. Chassell wrote:
By the way, does anyone know why so many science fiction writers
descripe spinning space habitats as being longer than they are wide?
Such habitats are
Erik Reuter [EMAIL PROTECTED] wrote, regarding the pressure at
the center of a spinning space habitat with an 8 km radius:
Where did you get 0.9? Note that the 3.45 number has a 1/R factor
in it. If R goes from 5km to 8km, then 3.45 goes to 2.16. Then,
exp[- 1/2.16] is 0.63. The
At 06:05 PM 7/7/2003 +, you wrote:
By the way, does anyone know why so many science fiction writers
descripe spinning space habitats as being longer than they are wide?
Such habitats are intrinsically unstable. But habitats that are wider
than they are long are intrinsically stable
I know
Kevin Tarr wrote:
...
A can like structure would have more surface area where the gravity is,
obviously one reason to use it in stories. But how are they unstable? Just
asking because I don't know. Do you mean because they can tumble? If so,
how much inertia would a wheel like structure need
What is the air pressure near the center of a spinning O'Neil type
space habitat when the pressure at the rim is 1 atmosphere?
What is the equation that tells you the pressure?
On Earth, sea level air pressure averages approximately 1 bar
(1013.25 mb or 1 atmosphere) and decreases as the
On Sun, Jul 06, 2003 at 07:22:39PM +, Robert J. Chassell wrote:
What is the air pressure near the center of a spinning O'Neil type
space habitat when the pressure at the rim is 1 atmosphere?
What is the equation that tells you the pressure?
Tricky. The only way I know of to get a simple
At 05:26 PM 7/6/03 -0400, Erik Reuter wrote:
On Sun, Jul 06, 2003 at 07:22:39PM +, Robert J. Chassell wrote:
What is the air pressure near the center of a spinning O'Neil type
space habitat when the pressure at the rim is 1 atmosphere?
What is the equation that tells you the pressure?
At 05:26 PM 7/6/03 -0400, Erik Reuter wrote:
On Sun, Jul 06, 2003 at 07:22:39PM +, Robert J. Chassell wrote:
What is the air pressure near the center of a spinning O'Neil type
space habitat when the pressure at the rim is 1 atmosphere?
What is the equation that tells you the
On Sun, Jul 06, 2003 at 07:36:53PM -0400, David Hobby wrote:
The formula above seems to be assuming that pressure at the axis
is zero, which is unrealistic.
Actually, having a pressure near 1 atmosphere at the center is
unrealistic. There is nothing to provide such a pressure. The
Erik Reuter wrote:
On Sun, Jul 06, 2003 at 07:36:53PM -0400, David Hobby wrote:
The formula above seems to be assuming that pressure at the axis
is zero, which is unrealistic.
Actually, having a pressure near 1 atmosphere at the center is
unrealistic. There is nothing to
On Sun, Jul 06, 2003 at 09:28:52PM -0400, David Hobby wrote:
Sure there is something to provide the pressure! The habitat
won't let air out, and you've pumped it up to the point where there is
1 atm at the rim. Or?
No, that is counteracted by the rotational motion.
But look,
On Sun, Jul 06, 2003 at 09:28:52PM -0400, David Hobby wrote:
Appeal to Authority: Rama had air at the axis, and Clarke was usually
good on physics.
Rama had a radius of about 8km. They entered near the axis and began
descending in spacesuits. After descending 2km, they found the pressure
was
At 09:52 PM 7/6/03 -0400, Erik Reuter wrote:
On Sun, Jul 06, 2003 at 09:28:52PM -0400, David Hobby wrote:
[snip]
FWIW, a Google search for space habitat atmospheric pressure returns
23,900 hits, some of which actually seem to relate to the question at hand
. . .
--Ronn! :)
I always knew
Erik Reuter wrote:
On Sun, Jul 06, 2003 at 09:28:52PM -0400, David Hobby wrote:
Sure there is something to provide the pressure! The habitat
won't let air out, and you've pumped it up to the point where there is
1 atm at the rim. Or?
No, that is counteracted by the rotational
On Sun, Jul 06, 2003 at 11:19:38PM -0400, David Hobby wrote:
As I said, I'm not calculating the distribution just now. It's not
really a fair tactic to say that I must.
Sure it is fair. The question asked for an equation (or specific
numbers). I gave an equation. You say it is wrong, but
Erik Reuter wrote:
On Sun, Jul 06, 2003 at 11:19:38PM -0400, David Hobby wrote:
As I said, I'm not calculating the distribution just now. It's not
really a fair tactic to say that I must.
Sure it is fair. The question asked for an equation (or specific
numbers). I gave an equation.
On Mon, Jul 07, 2003 at 12:38:36AM -0400, David Hobby wrote:
Yes, but I gave arguments showing that it did not give
reasonable results.
Not at all convincing arguments, with no numbers or equations.
I believe I said about constant. I can deal with .75
atmospheres when I go into
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