Not so fast Sterling :-) The size of the crater, which is rare or even
unique... doesn't make mucked-up analyses a requirement!
Short and simple as I just read your reply to me in which you somehow missed
the central point I asked about when you insisted that the crater contains
nothing but powder...let's take a little more of a scientific approach.
My prior post began, "Sterling what model you have accounts for potato sized
meteorites (and powder) scattered in and around meters from the impact, yet
strictly powder inside, especially for a meteorite that sheds like this one
particularly along its natural 'fault' lines."
Please answer that question clearly for my benefit rather than skipping and
speaking of Canyon Diablo and Barringer. A much better comparison, btw, is
Jilin.
As to the ancillary stuff...
Congratulations on ace Mountaineer Mike Fowler who mentioned that 50% of the
atmosphere is under 3.5 miles elevation - it jives within 100 meters to the
calculation I worked on and gives me the confidence I need for checking this
calculation. When you state that "only" 58% of the atmosphere's mass was in
the path of the Peruvian meteorite, just to keep a sensible argument going,
I would suggest you don't introduce bias via adjectives like "only" into the
interpretation. There is an incorrect implication that in this last 2 miles
of atmosphere, cosmic velocity is typically damped. ---not true.
According to my numbers, your 58% estimate was ok for the back of an
envelope, though a little exaggerated. I calculated it to be 62.1% using a
more accurate model (which agrees to M. Fowler's 3.5 mile figure within 100
m) for the atmosphere than your barometric formula. Rather than dump a
bunch of numbers on the list, let me just share this graph, which I just
generated that is useful from sea level to 25 kilometers altitude, so you
can graphically see how much atmospheric mass is traversed for any bolide
around at the Peruvian crater's around October. Don't forget that the
ablative path for most meteorites stops much, much higher than 3800 meters!
www.diogenite.com/Huanocollo.gif
This graphically gives a great idea of how much % of the atmosphere any
meteorite anywhere on Earth passes through to get to any altitude above sea
level, and if you look at it you can see how much of a fraction of the
atmosphere mass is traversed in any segment of the travel from 25Km on down.
Just compare the blue area to the white and you get the idea of of the
FRACTION of the atmosphere traversed. No arithmetic needed - the ratio of
blue to blue+white is the % of the atmosphere for any geographical elevation
and includes luminous paths too..
Sorry, but I can't accept your dismissing unscientifically the arrival of
any meteoritical material generally to the ground as difficult to on one
hand and then on the other calculate all these asides to things even you
don't want to know to such precision! 62% is 62%, not "only" anything. 62%
of the atmosphere is only where it starts in this case in Peru, but this is
another subject. I.e., if it comes in at around a 45 degree angle instead
of vertical, it passes through the full 100% since it doesn't take the
straight path, and you are back to square one. These meteorite was observed
to enter at an angle. Yes, I understand that "on average" meteorites
reaching sea level will go through more atmosphere, but this is a non-issue
when they are conveniently sized and in free fall for that 3800 meters. The
one effect I will agree that will cause a higher velocity, which has nothing
to do with retaining cosmic velocity, is that FREE FALL VELOCITY is greater
in thinner air. There is plenty to be said about that as you would imagine
such as a potential doubling of the energy of impact making a bigger crater
for something the size of Jilin. I don't think it is likely a huge ball is
at the bottom of the crater. Just that there are plenty of kilos that
weren't pulverized in the mucky crater.
Best health,
Doug
The numbers behind the graph, I could post if you want, along with the
modeled temperature in F and C of the atmosphere over its lat/lon. I used
the trapazoidal rule to estimate the percent of the atmospheric mass with
the midpoints of intervals of 200 meters altitude for 0 - 25 kim above sea
level.), and considered that the atmosphere ended at 100Km above sea level.
----- Original Message -----
From: "Sterling K. Webb" <[EMAIL PROTECTED]>
To: "Meteorite List" <meteorite-list@meteoritecentral.com>
Cc: <[EMAIL PROTECTED]>
Sent: Thursday, October 04, 2007 7:17 PM
Subject: Re: [meteorite-list] Entry Dynamics in Peru
Hi, Paul, List,
I just posted:
The air in it weighs 14.7 pounds if you start
at sea level, or 1200 grams...
Wrong!
Grabbing numbers from a column of numbers,
in a hurry, whoops! 1033 grams per sq. cm. is
sea level atmospheric pressure in metric.
The 1200 gram figure is the weight of a cubic
meter of dry air at sea level, in case you're
wondering...
14.7 pounds is 6.668 kg.
Now, if politicians would correct their mistakes
as fast...
Sterling
--------------------------------------------------
----- Original Message -----
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Thursday, October 04, 2007 5:56 PM
Subject: Re: [meteorite-list] Entry Dynamics in Peru
In a message dated 10/4/2007 6:40:52 PM Eastern Daylight Time,
[EMAIL PROTECTED] writes:
The air in it
weighs 14.7 pounds if you start at sea level, or 1200 grams.
Not to quibble...but I always thought that 14 pounds equaled about
6,000 plus or minus a few grams. Have I missed something?
Best regards,
Paul Martyn
Savannah, GA
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