Significance of the solar wind

In article <i7w_c.2421$j62.2045@trnddc04>, jesus_x.TakeThisOut@mozillanews.org (jesus
X) wrote:

 > On 9/4/2004 1:17 PM Stephen Forbes cranked up the brainbox and said:
  > > I have calculated that each second 220,000Kg of air is slowed down by
  > > the solar wind by about 1 to 3% of its tangential velocity, say 10
  > > miles/sec or 36,000 miles/hour.
 >
 > Show your work and source numbers, i did.

OK.
Using some of your numbers.
6,000,000 protons/m^3
450km/s speed of solar wind ( you said 300 to 600 )
So 2.7 x 10^12 protons pass through each square metre each second.
Area of ringworld = 1.55 x 10^21 m^2
So a total of 4.2 x 10^33 protons/sec hit the upper atmosphere.
Factor in mass of proton and relative mass of oxygen molecule gives 2.2 x
10^5 Kg directly affected by solar wind each second.

As a proton is 1/32 the mass of an oxygen molecule, the oxygen will have a
velocity loss of about 1/32 ( momentum calculation ). So 770/32 = 24
miles/sec. I said only 10 to allow for not hitting head on in most cases,
although thinking about it now perhaps I shouldn't have because you will
still need to accelerate the proton by that much even if it takes several
collisions to do so.

 >
 > Second, I'll pretend with you for a moment. In one second 220,000 kg of
 > air is
 > slowed by 2%. That's 220,000kg over an area of 300,000,000,000,000
^^^^^^^^^^^^^^^^^^^
600,000,000,000,000 surely! Thats what it says in my copies of Engineers
and Throne.

 > square miles,
 > or 482,803,200,000,000,000 square meters. That's one kilogram of air per
^^^^^^^^^^^^^^^^^^^^^
I don't think so, its not even close. See above for the correct figure.
You appear to have applied the linear conversion factor when you should
have applied the area conversion factor.

 > 2,194,560,000,000 square meters. And that's only in two dimensions,
 > spread that
 > kilogram of air out in a one molecule thick layer, we're ignoring any
 > measurement of volume. That's an insanely tiny amount of air being
 > acted upon,
 > and will be easily absorbed by the rest of the atmosphere, each and
 > every
 > second. It'll eventually translate into a microscopic heat increase,
 > and play an
 > equally small role in weather on the Ringworld, as opposed to actually
 > creating
 > any actual air displacement, just like on Earth.
 >
  > > This is what happens when an air molecule bounces
  > > off a proton from the solar wind, the loss of tangential velocity is
  > > a fixed quantity that cannot be averaged out over the whole
  > > atmosphere.
 >
 > It can and IS via fluid interactions. That one molecule hits other
 > molecules,
 > and energy is dispersed by each impact.

Agreed, but dispersing the velocity change amongst other molecules only
increases proportionately the mass of air affected. If you can drop the
overall speed to say 1 mph, then 36,000 x 220,000 = 7.92 x 10^9 Kg of air
is affected every second.

FYI I estimate the total mass of air on the Ringworld to be 1.62 x 10^25Kg
or slightly over 2 Earth masses.
So, to slow down say 0.1% of the total atmosphere by 1mph would take about
3,000 years. Why 0.1%? its an estimate for the upper atmosphere which is
directly affected by the solar wind. Personally, I think I should have
used a smaller figure but it still looks like my original 100 years is
wrong. However, it doesn't look like the air is stable enough to allow
evolution to progress very far before problems start.

My scenario is still valid, it just takes a bit longer to happen than I
originally thought but still much less than the supposed lifetime of the
Ringworld.

Stephen

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On 9/5/2004 5:47 PM Stephen Forbes cranked up the brainbox and said:
 > 600,000,000,000,000 surely! Thats what it says in my copies of Engineers
 > and Throne.

Yes, I was off by half.

  >> square miles,
  >> or 482,803,200,000,000,000 square meters. That's one kilogram of air per
 > ^^^^^^^^^^^^^^^^^^^^^
 > I don't think so, its not even close. See above for the correct figure.
 > You appear to have applied the linear conversion factor when you should
 > have applied the area conversion factor.

Excellent catch. That's what I get for posting at midnight after several days
with little sleep. The correct area in square meters is 1.55 * 10^21, or
1,550,000,000,000,000,000,000. Which means that each kilogram of air affected by
the solar wind (using _your_ figure of 220,000kg) is spread over an area of
roughly 7,045,454,545,454,545 square meters, which an even LARGER area than I
originally said. Which means the impact is even SMALLER than previously said,
and using YOUR numbers, not numbers I have bothered to verify.

 > FYI I estimate the total mass of air on the Ringworld to be 1.62 x 10^25Kg
 > or slightly over 2 Earth masses.

And how do you estimate this? As I said earlier, show your work. I did.

 > So, to slow down say 0.1% of the total atmosphere by 1mph would take about
 > 3,000 years.

Again, only using your numbers, not numbers I have verified to be valid, it
still seems that your original theory about the Ringworld being void of air in
one century is horribly wrong. You have affected a minuscule portion of the
total atmosphere by a minuscule amount in THREE MILLENIA, or thirty times longer
than you estimated it to take to empty the Ringworld.

 > Personally, I think I should have
 > used a smaller figure but it still looks like my original 100 years is
 > wrong.

Very very wrong. As we have all said all along.

 > However, it doesn't look like the air is stable enough to allow
 > evolution to progress very far before problems start.

Modern Humans are less than half a million years old (between 400,000 and
250,000 years old, depending on sources). Homo Habilus reigned about 2.5 to 1.5
million years ago. So, there was indeed enough time for the various hominids to
evolve, especially considering the number of open niches in the Ringworld ecology.

 > My scenario is still valid,

No it's not, even by your numbers, you're off by at least five orders of
magnitude, and realistically, much more.

 > it just takes a bit longer to happen than I
 > originally thought but still much less than the supposed lifetime of the
 > Ringworld.

A LITTLE bit? A _LITTLE_ bit? Ba. You're wrong, you admitted it, now you're
trying to back away from it and retcon your argument. This isn't a scifi novel.
You're wrong.

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jesus X wrote:

 > On 9/5/2004 5:47 PM Stephen Forbes cranked up the brainbox and said:
 >
  > > FYI I estimate the total mass of air on the Ringworld to be 1.62 x
  > > 10^25Kg
  > > or slightly over 2 Earth masses.
 >
 > And how do you estimate this? As I said earlier, show your work. I
 > did.

Not surprisingly, it's 1.6 x 10^26 kg, or about 26 Earth masses. He's
off by over an order of magnitude.

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steve DeleteThis @antispam.sforbesa.cix.co.uk (Stephen Forbes) wrote in message news:<memo.20040905224715.25321A DeleteThis @sforbesa.compulink.co.uk>...
 > In article <i7w_c.2421$j62.2045@trnddc04>, jesus_x DeleteThis @mozillanews.org (jesus
 > X) wrote:
....
  > > It can and IS via fluid interactions. That one molecule hits other
  > > molecules,
  > > and energy is dispersed by each impact.
 >
 > Agreed, but dispersing the velocity change amongst other molecules only
 > increases proportionately the mass of air affected. If you can drop the

No, it also causes a corresponding decrease in the velocity of the
affected air. The momentum transferred remains unchanged, so the
velocity is inversely proportional to the mass affected. By the time
collisions have transferred that momemntum to the entire Ringworld
atmosphere, the affect on the velocity of the air becomes quite
negligible.

 > FYI I estimate the total mass of air on the Ringworld to be 1.62 x 10^25Kg
 > or slightly over 2 Earth masses.
 > So, to slow down say 0.1% of the total atmosphere by 1mph would take about
 > 3,000 years.

Which is enormously longer than the amount of time it takes for the
transferred momentum to be distributed throughout 100% of the
atmosphere. So it will actually take 3,000,000 years to slow the
entire Ringworld's atmosphere by 1 mile per hour. Given that it's
starting speed is 770 mi/sec, you've got a very long wait before it's
been slowed down enough to drop into the star.<!-- ~MESSAGE_AFTER~ -->
 >> Stay informed about: Significance of the solar wind 

In article <memo.20040905224715.25321A DeleteThis @sforbesa.compulink.co.uk>,
steve DeleteThis @antispam.sforbesa.cix.co.uk (Stephen Forbes) wrote:

 > OK.
 > Using some of your numbers.
 > 6,000,000 protons/m^3
 > 450km/s speed of solar wind ( you said 300 to 600 )
 > So 2.7 x 10^12 protons pass through each square metre each second.
 > Area of ringworld = 1.55 x 10^21 m^2
 > So a total of 4.2 x 10^33 protons/sec hit the upper atmosphere.
 > Factor in mass of proton and relative mass of oxygen molecule gives 2.2
 > x 10^5 Kg directly affected by solar wind each second.

Major oops here. I just reviewed my calculations and discovered that I
factored the mass of oxygen wrongly, I divided by 32 when I should have
multiplied by 32. This is because the ratio of protons to molecules which
collide in this calculation is 1:1.

This means that the correct figure for the mass of air directly affected
by the solar wind is actually 1024 times greater or 2.25 x 10^8 Kg. This
will have a knock-on affect on my later calculations.

 >
 > As a proton is 1/32 the mass of an oxygen molecule, the oxygen will
 > have a velocity loss of about 1/32 ( momentum calculation ). So 770/32
 > = 24 miles/sec. I said only 10 to allow for not hitting head on in most
 > cases, although thinking about it now perhaps I shouldn't have because
 > you will still need to accelerate the proton by that much even if it
 > takes several collisions to do so.

BTW using nitrogen instead of oxygen would have yielded the same result as
it will slow more than oxygen. Same momentum = same result.

snip.

 > Agreed, but dispersing the velocity change amongst other molecules only
 > increases proportionately the mass of air affected. If you can drop the
 > overall speed to say 1 mph, then 36,000 x 220,000 = 7.92 x 10^9 Kg of
 > air is affected every second.

This now becomes 36,000 x 225,000,000 = 8.1 x 10^12 Kg of air affected
every second.

 >
 > FYI I estimate the total mass of air on the Ringworld to be 1.62 x
 > 10^25Kg or slightly over 2 Earth masses.
 > So, to slow down say 0.1% of the total atmosphere by 1mph would take
 > about 3,000 years.

....and to slow down 0.1% of the total atmosphere by 1mph would now take
only 6.3 years.

So although 100 years is too short to lose all the atmosphere on the
Ringworld, there will definitely be problems with air loss in that length
of time.

 > Why 0.1%? its an estimate for the upper atmosphere
 > which is directly affected by the solar wind. Personally, I think I
 > should have used a smaller figure but it still looks like my original
 > 100 years is wrong. However, it doesn't look like the air is stable
 > enough to allow evolution to progress very far before problems start.

Still true but with more urgency. What do you say now Erik, James and
JesusX?

Stephen

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In article <lKQ_c.2762$x12.907@trnddc05>, jesus_x DeleteThis @mozillanews.org (jesus
X) wrote:

  > > FYI I estimate the total mass of air on the Ringworld to be 1.62 x
  > > 10^25Kg or slightly over 2 Earth masses.
 >
 > And how do you estimate this? As I said earlier, show your work. I did.

Air density reduces with height, so I assumed that if it didn't it would
be 5 miles high. I used that and the density of air at sea level to
calculate total air mass.

Stephen

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In article <413C189F.8F2FAACF.DeleteThis@alcyone.com>, max.DeleteThis@alcyone.com (Erik Max
Francis) wrote:

 > Not surprisingly, it's 1.6 x 10^26 kg, or about 26 Earth masses. He's
 > off by over an order of magnitude.

Where did you find that figure? You would need _constant sea level
pressure_ up to 84 kilometres or 52 miles altitude to justify that. If
that much air spread out to a normal distribution it would overflow the
rim walls with no other encouragement needed.

You're the one off by an order of magnitude, not me.

Stephen

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Stephen Forbes wrote:

 > Where did you find that figure? You would need _constant sea level
 > pressure_ up to 84 kilometres or 52 miles altitude to justify that. If
 > that much air spread out to a normal distribution it would overflow
 > the
 > rim walls with no other encouragement needed.
 >
 > You're the one off by an order of magnitude, not me.

No, that doesn't imply constant pressure. Do you even know what
pressure is? Take a vertical column of air on the Ringworld (or on the
Earth, for that matter). At the bottom of the column the pressure is
101 kPa. That means that the total weight of the air resting on that
column is 101 kN for every 1 m^2 of area. It doesn't matter what the
pressure or density distribution of the air in that column is; you know
it's there due to gravity, so you know it's all accounted for. You
know, first year physics stuff.

Assuming a constant gravitational field throughout the atmosphere (an
approximation valid and routinely used for Earth, and even more valid
for the Ringworld), then you know the weight of the atmosphere is simply
w = m g, letting you calculate the areic mass of the atmosphere from the
pressure and the gravity.

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steve.RemoveThis@antispam.sforbesa.cix.co.uk (Stephen Forbes) wrote in message news:<memo.20040906185137.7295B.RemoveThis@sforbesa.compulink.co.uk>...
 > In article <lKQ_c.2762$x12.907@trnddc05>, jesus_x.RemoveThis@mozillanews.org (jesus
 > X) wrote:
 >
   > > > FYI I estimate the total mass of air on the Ringworld to be 1.62 x
   > > > 10^25Kg or slightly over 2 Earth masses.
  > >
  > > And how do you estimate this? As I said earlier, show your work. I did.
 >
 > Air density reduces with height, so I assumed that if it didn't it would
 > be 5 miles high. I used that and the density of air at sea level to
 > calculate total air mass.

You just assumed 5 miles? Why do that, when you can calculate it
exactly?

The weight of the atmosphere due to centrifugal forces is equal to the
acceleration due to centrifugal forces * the mass of the atmosphere.
That is the force that the atmosphere is exerting downward on the
Ringworld due to atmospheric pressure. That's approximately one Eath
atmosphere of pressure times the surface area of the Ringworld.

m*g = P*A => m = P*A/g =
(10000 Pascals * 6E14 mi^2 * (1609 m/mi)^2)/(0.992 * 9.82m/s^2)
= 1.6e24 kg

Mass of the Earth: 5.98e24kg

Mass of Ringworld atmosphere: 0.27 Earth masses.

I may be missing something, but that's the answer I got.<!-- ~MESSAGE_AFTER~ -->
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On 9/6/2004 1:51 PM Stephen Forbes cranked up the brainbox and said:
 > Air density reduces with height, so I assumed that if it didn't it would
 > be 5 miles high. I used that and the density of air at sea level to
 > calculate total air mass.

Then there's your problem. You made too many assumptions. Where did you get 5
miles? Did you pull it out of thin air?

While I honestly don't know if there are ready-made formulas for this kind of
calculation I'd wager there is but I don't know it. I'd suggest trying to find
that, rather than guessing. With the volume of the Ringworld, I can't see how 2
earth masses of air would suffice, by any stretch.

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On 9/6/2004 1:11 PM Stephen Forbes cranked up the brainbox and said:
 > ...and to slow down 0.1% of the total atmosphere by 1mph would now take
 > only 6.3 years.

 > Still true but with more urgency. What do you say now Erik, James and
 > JesusX?

Still untrue. Using your latest set of incorrect numbers, it would still take
6000 years to slow the Ringworld's atmosphere by 1 MPH. And, that's ignoring the
friction of the Ringworld's land masses which would help speed the atmosphere
back up. So, you're STILL wrong by orders of magnitude.

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On 9/6/2004 3:58 AM Erik Max Francis cranked up the brainbox and said:
 > Not surprisingly, it's 1.6 x 10^26 kg, or about 26 Earth masses. He's
 > off by over an order of magnitude.

While that seems much more reasonable, can you tell me how you arrived at that
figure?

And, we've finally gotten him to admit a mistake, and according to the latest
message, he admits to being wrong TWICE now, in the same calculation. I think
maybe we should start looking for two-headed, three-legged camels manipulating
things behind the scenes.

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jesus X wrote:

 > While that seems much more reasonable, can you tell me how you arrived
 > at that
 > figure?

The surface pressure will be due to the weight of the atmosphere above
it. That's a force per unit area, and for an approximately constant
gravitational field (valid approximation for Earth, even moreso for
Ringworld), that also tells you the mass of the atmosphere:

  P = m g/A

  m = P A/g

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