deluge.htm
Garrison L. Hilliard
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The Deluge and Evidences of a Creator.
Questions? [INLINE] email me now.
More info on the flood courtesy of the folks at the Christian
Answers Network.
Visit B.J. Corbins Noahs Ark site and read his yet to be published
and fascinating book, Explorers or Ararat.
Visit the "Center For Scientific Creation." I've browsed their site
and highly recommend it. Their is an incredible account of the
deluge and its effects.
A Dinosaur Stampede!?
A Canyon formed in 10 days?
A fossilised human finger in Cretaceous Rock.
One of only two other dedicated Noahs Ark WEB pages.
The second and only other dedicated Noahs Ark WEB site I've found.
This one is quite extensive and well done and has just added a VRML
tour of Noahs Ark. This site is run by Mark Kneisler.
How seaworthy was the Ark?
Evidence of an asteroid impact. When I first read this, I didn't
get it. Then the thought occured to me. 'The fountains of the great
deep were broken up.' Obviously one of the instruments to break up
the fountains of the deep may have been an asteroid impact. Very
interesting article.
Tim Regan
You didn't include an URLs for the Noah's Ark sites but here's some food
for thought on the science for a flood that could cover the Earth from the
'talk.origins' newsgroup.
I didn't write any of this and claim no credit for it.
--------------------------------------------
>How much water are we talking about? This question has been answered
>this before, but I don't know what the number was. Does anyone have
>a quick estimate of the volume of water needed to cover the earth
>as described in Genesis?
Here's some posts I saved from t.o. last time this came through or
so...
******
From: ch...@eso.mc.xerox.com (Chris Heiny)
In article e...@cwis.isu.edu, earl...@cwis.isu.edu (EARLES_CHRISTOPHER_M) writes:
>In article <1994Jan30.0...@scic.intel.com>,
>Seth J. Bradley <sbra...@scic.intel.com> wrote:
>>There's the small problem (amongst many others) that that much water falling
>>that quickly would result in enough kinetic energy to heat the flood waters to
>>the boiling point. How did the fish (and Noah) keep from getting boiled to
>>death?
>
> Good point, actually. I would have never thought of that as a
>counterargument. But, why did all the water have to fall at once? When
>it rains, the whole cloud doesn't just fall to the ground and that's it,
>does it? If so, inform me, because I've been in rainstorms that lasted a
>couple days, and from my estimations, it takes less than a day for an
>object to get from the top of the atmosphere to the ground. There is no
>argument that I know of that can say that it took 40 days & nights for
>all of the clouds to drop their water. No supportable argument says that
>it wasn't a natural rainstorm. If the fish had trouble keeping from
>being boiled alive, I'll remember to guard my lawn from the next
>rainstorm...
You are required to accept 40 days and 40 nights because, if you insist on
literally interpreting the Bible as describing a global flood, then you
must also accept the Bible's description of that Flood. You are also required
to accept that this Flood covered the highest mountain - about 30,000 feet of
water. So let's do a quick bit of math here:
40 days&nights = 960 hours
30,000 feet = 360,000 inches
got that? Ok, so let's divide...
360,000 inches/960 hours = 375 inches/hour
This means, that in one hour, in all places on the earth, it rained 375 inches.
The greatest average annual rainfall in the world is on Mt Waialeale on
Maui, Hawaii - 460 inches/year. The world record for 42 minutes is 12 inches,
Holt, Missouri, on June 22, 1947. Let's divide a little more.
1 hour = 60 minutes
375 inches/hour * 1 hour/60 minutes = 6.25 inches/minute
The world record for 1 minute of rainfall is 1.23 inches (Unionville,
MD, July 4, 1956).
You still think you have any sort of supportable argument for a natural
rainstorm? Allright, let's do a bit more math...
Each cubic inch of water weighs 16.4 grams. 18 grams of water, in vapor form,
occupies about 22 liters of space (this is pure water vapor, no other
gasses, like air, mixed in with it) or about 1343 cubic inches at STP (20
degrees C, 1 atmosphere of pressure).
Now, a quick divergence - what volume of water is needed to cover the
earth to the depth of 30,000 feet? This is pretty easy to do - we
calculate the volume of a sphere the size of the earth, calculate the
volume of another sphere 30,000 feet greater in radius, subtract the two,
and the difference will be the miminum amount of water needed to complete
the flood.
radius of earth = 3950 miles = 20,856,000 feet (minimum polar radius)
radius of earth + flood = 3950 miles + 30,000 ft = 20,886,000 ft
volume of a sphere = ((4)(pi)(r**3))/3
volume of earth = 3.799983x10**22 cubic feet
volume of earth+flood = 3.816404x10**22 cubic feet
volume of flood = 1.642x10**20 cubic feet (1.116x10**9 cubic miles)
That's a lot of water. If we convert that water to vapor, at 1728 cubic
inches per cubic foot we get
volume of flood = 2.837376x10**23 cubic inches of water
volume of vapor = 3.810596x10**26 cubic inches of vapor
2.025206x10**23 cubic feet of vapor
1.498122x10**12 cubic miles of vapor
That sure looks like a big volume of vapor. But let's try to put that in
context. Assume the following: the minimum altitude for the base of the
vapor canopy is 30,000 feet (that is, it was no lower than the highest
mountain); that the vapor canopy was pure vapor, with no other gasses mixed
into it; and the vapor canopy was STP throughout. How far out into space
would the vapor canopy then extend? That's an easy calculation - take our
volume for the earth +30,000 feet (from above), add the volume of the vapor
canopy, and solve for the radius of that sphere:
volume of earth+30,000' = 3.816404x10**22 cubic feet
volume of earth+30,000'+ vapor canopy = 2.406846x10**23 cubic feet
= 1.653108x10**12 cubic miles
radius of earth+30,000'+vapor canopy = 7308 miles.
depth of vapor canopy = 3358 miles
Ah, you say, "but I've made assumptions above! They're obviously chosen to
make the vapor canopy look bad." Well, if we were to stick in real world
computations for the assumptions, it actually makes things -worse- for the
vapor canopy: raising the minimum altitude decreases the pressure the canopy
would be at, thus increasing its volume and its radius; mixing other gasses
in increases its volume, thus increasing its radius; assuming nonSTP
conditions means decreasing its pressure, raising its volume and thus
increasing its radius.
Still think this is a natural rainstorm?
Let's do some quick heat computations. Forget about the heat needed to
dissipate the gravitational potential as the rain of the flood falls from
a distance some 3238 miles above the highest cloud tops ever measured. Let's
just think about how much heat would be released as that rain condenses out
of the vapor canopy...
To do this, we need to know how much energy is released when the water turns
from vapor to liquid - this is roughly 620 calories per gram at STP [this is
possibly fuzzy memory from Physical Chemistry lo so many years ago - if I'm
full of shit, would the designated t.o chemopedant please step in to pinch
hit?]. Since each cubic inch of water is 16.4 grams, this means that each
cubic inch of rain in the Flood released
620 calories/gram * 16.4 grams/cubic inch = 10168 calories/cubic inch
since it was falling at the minimum rate of 6.25 inches (as we computed above),
this means you need to dissipate
10168 calories/cubic inch * 6.25 inches/minute = 63550
calories/square inch/minute over the entire surface of the earth. Now the
earth's surface is
surface = (4)(pi)(r**2)
surface of earth = 7.87x10**17 square inches
so the earth as a whole had to radiate about 5x10**27 calories per
minute as the vapor canopy condensed into rain for the flood.
Well, that's another impressive number. How can we put it into context?
Let's try this - that 63550 cal/sq in/min is equal to 9,151,200 calories/sq
ft/min, or (about) 5.49x10**8 calories/sq ft/hour. Since 1 BTU = 252
calories, we get 2,180,000 BTU/square foot/hour. Now, on a clear, sunny,
summer day, the solar input from the sun can reach 340 BTU/sq ft/hour. So
according to my math (and I believe I've worked my dimensions correctly, and
that all my units are correct), over the entire duration of the Flood, the
earth had to radiate heat at 6400 times the rate at which it receives it
from the sun.
Do you still think the Flood had any relation to a natural rainstorm?
Chris
PS Unlike the dear departed dolphin, I don't claim this to be my area of
expertise. Feel free to check these figures and let me know where I'm
as full of excrement as the Ark must have been.
******
From: sc...@mayfield.hp.com (Scott Jorgenson)
If I might raise my voice from the crowd of lurkers for a moment:
It seems that there might finally be a use for all of those interesting
articles I've been archiving off of t.o for the last year. Benjamin Franz
(Snowhare) (snow...@xmission.com) wrote:
>> A "vapor canopy" and "subterranenean heated and pressurized reservoirs"
>> simply could not have done what was claimed for them. In fact - either
>> of them would have rendered the Earth uninhabitable for many *MILLIONS*
>> of years. Not by flooding - by sterilizing the planet with superheated
>> steam and raising the surface temperature until the rocks were glowing.
>> The vapor canopy is in and of itself impossible. You just can't pack that
>> much water into the atmosphere, and "orbital canopies" are impossible
>> for kinematic reasons.
>>
>> Anyone with a first year understanding of physics can do the math
>> needed to show it can't work. But it *sounds* "reasonable" to the naive.
>> Sounding reasonable is not enough.
To which JOHN ALLEN MARSH (jam...@eos.ncsu.edu) replied:
>
> Since neither side has done the math, I would like to see your "first year"
> analysis.
And Mark Christensen (ma...@eomer.hw.stratus.com) also replied:
>
> I've had first and second year physics. So let's see the math. I'm certainly
> open to mathematical proofs of what you just described are impossible.
Well, I too have had first and second year physics, but probably couldn't
recall enough to demonstrate the natural impossibility of the vapor canopy,
if I had to. Fortunately, Terrance J. Gaetz (ga...@cfa.harvard.edu) did so
earlier this year. He certainly isn't the only one to have done so; but his
is the one I saved and have ready-at-hand. I hope he doesn't mind if I
replay two of his posts on the subject here, for the benefit of (at least)
other curious lurkers such as myself.
Another good source, of course, is Arthur Strahler's _Science_and_Earth_
History_. Strahler talks about physical problems with the "founts of the
deep", as well. I highly recommend this book.
Remember, what follows is credited to Terrance Gaetz, but any mistakes/
omissions in duplicating his comments are mine. And remember, the question
this addresses is: could the vapor canopy and its collapse have been
physically possible, without resort to miracle. I won't get into the
Omphalian theological/epistemological problems with invoking such miracles
(the "deceiver God" and all of that), or I'll never be allowed to lurk again!
Subsiding back into the silent masses,
Scott
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Newsgroups: talk.origins
From: ga...@cfa281.harvard.edu (Terrance J. Gaetz)
Subject: Re: Vapor Canopy (was: Chrispiracy/Kind def)
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Date: Mon, 31 Jan 1994 21:21:02 GMT
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I'm bored reading design review documents, so I thought I'd do some
of Chris Bodenmiller's work for him.
Suppose we have a canopy sufficient to produce a wimpy 1 km deep
worldwide flood (I grew up just over 1 km above sea level).
First, we'll try a canopy floated on the atmosphere (ignoring the
fact that this is Rayleigh-Taylor unstable and wouldn't
last longer than it takes water to fall). A 1 km deep layer
of water covering the entire surface of the earth has a mass of
about (12 * 1e5 * (6.4e8)^2) = 5 x 10^23 grams. For comparison,
the mass of the atmosphere is 5.136 x 10^21 grams (according to
the CRC handbook). That means that the atmosphere would have
been squished to about 1 percent of its current volume and have
had a pressure of about 100 atmospheres. The atmosphere's scale
height is currently about 8.5 km, so, as a ballpark guess, the
water-air interface is at about 85 m, call it 100 m. (That's
probably ok to within a factor of two, anyway. Anyone care to do
a proper calculation?)
I couldn't find a table of water opacities, but Jackson (Classical
Electrodynamics) has a figure showing the absorption coefficient
of water from 100 to 10^22 Hz. Unfortunately this means that the
entire optical band covers about 1.5 mm on the graph which makes it
hard to read. The absorption coefficient in the visible ranges from
about 10^(-2) to 4 x 10^(-4) cm^(-1). Let's take 5 x 10^(-4) cm^(-1)
as an eyeball average; that gives an e-folding length of 20 m.
Yikes! That means that the sunlight is attenuated by a factor
of about 10^22. People, we're going to need all the light focusing
you guys can muster.
I dunno, it doesn't sound much like paradise to me: 100 atmospheres
of pressure, really *really* dark, and a kilometer of water poised
just over your head and waiting to fall. The Ocean of Damocles...
We haven't even got enough to cover the mountains yet.
OK, that doesn't look too promising, so let's put the stuff in
orbit. It can't be a sphere of liquid water for sound mathematical
reasons: all points of a fluid sphere can't be simultaneously
in a stable orbit. Instead, make the water into icecubes and have
them orbiting in an ensemble of nonintersecting orbits which approximate
a sphere. There's still the opacity problem unless you put them up
in a really high orbit: to get the layer as thin as 10 m, say, you'd
have to move them out to 10 times the earth's radius. Once they're that
far out, why wouldn't they stay there? You'll also have to deal with
the evaporation of the ice; comets are active by the time they
get in as far as the earth's orbit.
Next problem is the heat of reentry. Orbital velocities are about
8 km/s which means about a kinetic energy 3.2 x 10^11 ergs/g of
material, or about 7600 cal/g. Yowza. Let's try to get rid of
some of it. The water outweighs the atmosphere by a factor of 100,
so the atmosphere is of little use as a heat sink. Instead, we'll put
the icecubes close to absolute zero and dump the kinetic energy
into heating the icecubes:
ice, -273 C to 0 C: 0.5 x 273 = 136 cal/g
melt the ice: 80 cal/g
water, 0 C to 100 C: 1 x 100 = 100 cal/g
boil the water: 595 cal/g
A total of about 1000 cal/g used, 6600 to go.
100 atmospheres is less than the critical pressure for water
(217.5 atmospheres) and we have plenty of energy left over, so
we'll have 100 atmospheres of *super*heated steam to deal with.
(And they say Venus has an unpleasant climate...)
I hope Noah has a good wood preservative, especially as he'd
have to deal with even worse conditions if you want to have the
mountains covered.
That should be enough to get the creation science canopy theorists
started. Have fun!
Terry Gaetz -- ga...@cfa.harvard.edu
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Newsgroups: talk.origins
From: ga...@cfa281.harvard.edu (Terrance J. Gaetz)
Subject: Re: On the "Genesis Flood"
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[I'm bored again, so time for another excursion into canopy theory.]
It has been pointed out that the arguments concerning conversion of
the energy in the orbiting canopy left out the possibility of converting
some of it into electromagnetic energy. That is true. Of course,
Canopy Theory is still a young creation science, so various facets
remain to be explored and worked out; there may even be an Ignoble Prize
in it for someone.
Let's assume, for the sake of argument, that instead of heating and
melting the ice, heating and boiling the water, and producing a *dense*
atmosphere of hyperheated steam, the orbital energy of the canopy is
converted directly into electromagnetic radiation. That's not physically
possible of course, but it serves as a useful limiting case.
Suppose the orbiting canopy of ice cubes contains sufficient water
to flood the whole world to a depth of 1 km. (As I noted in my
last posting, this isn't nearly enough to cover mountains.)
This time, instead of looking at the disposition of the reentry energy
on a per-gram basis, I'll look at the total energy involved in the
Canopy Reentry Event (or CRE).
A 1 km layer of water has a mass of 12 x 10^5 cm x (6350 x 10^5 cm)^2
or 5 x 10^23 g. (The "12" is a really 4 pi; I'm being biblical
and setting pi to 3 for ease of computation.) This mass is in
the form of ice cubes orbiting with velocity of about 8 km/sec
so the total kinetic energy of the ensemble of icecubes is about
(m v^2 / 2) = 1.6 x 10^35 ergs, or about 40 seconds worth of
the entire power output of the sun. The earth actually intercepts
only a tiny fraction of the sun's output though: the ratio of the
area of a disk with the earth's radius to the area of a sphere with
the radius of the earth's orbit:
(3 x (6350 x 10^5 cm)^2) / (12 x (1.5 x 10^13 cm)^2)
or 4.5 x 10^(-10). Apply the inverse of this as a correction factor
to obtain the canopy kinetic energy in terms of the amount of solar
power intercepted by the earth: 9 x 10^10 seconds worth. Now a
year is about 3 x 10^7 seconds, so the kinetic energy in the canopy
is equivalent to the total amount of solar energy intercepted by
the Earth for a period of close to 3000 years or 10^6 days.
The CRE is documented as taking 40 days (and nights), so the energy
release is equivalent to having 25000 suns overhead.
Actually, the energy will really be spread over the whole surface
of the Earth rather than just the cross sectional area, and say half
is radiated outwards rather than Earthwards, so it would be more like
having only 3000 suns overhead. The Earth's surface temperature
would be raised to about (280 x (3000)^(1/4) - 273) or of order
1800 C.
As I noted above, it is not physically possible for the kinetic
energy in the canopy to be converted directly and completely into
EM radiation; in practice, the energy in the CRE would go into
flashing the water into steam. Recall that of the 7600 cal/g of
orbital kinetic energy available, about 1000 cal/g is used up in
converting the ice into steam leaving 6600 cal/g. (Hmmm. I
wonder if CDC was a subcontractor...) Now, thanks to the value for the
specific heat for steam posted by Glenn Morton, the final temperature
of the steam (assuming no radiative losses) can be estimated as
100 C + 6600 cal/g x 4 C-g/cal = 26500 C. Bearing in mind that the
photospheric temperature of the sun is about 5700 C, it seems
pretty evident that some electromagnetic radiation will be emitted.
The challenge remaining for our Canopy Theorists is to determine the
actual post-CRE conditions; these conditions should be intermediate
between the 100-atmospheres-of-hyperheated-steam scenario and the
baking-under-3000-suns scenario.
I look forward to seeing the detailed models.
--
Terry Gaetz -- ga...@cfa.harvard.edu