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From: mexicodoug <
mexicodoug_at_meteoritecentral.com>
Date: Sat, 29 Dec 2007 17:01:42 -0600 Message-ID: <009401c84a6e$c7879a70$4001a8c0_at_MICASA> Sterling, Hey, the 100% "conversion" was just a footnote on Darren's upper limit for how much 14-C would be produced in his column. You seem to have missed out that I credited 7,000 times more 14-C in your favor :-). What's this, not even a thanks? Since 13-C has a market price of a $1,000 a gram nowadays, your windfall in 14-C rarities of rarities must be worth zillions and zillions! You weren't sent to after school remedial hall or given homework, or even asked to check your own work. I hadn't even the slightest concern how nitrophilic your blissful neutron compatriots are ... But, since you mention that your posts were otherwise perfect (except for that one error that you found and mentioned), maybe I'll reply to my favorite enjoyably entertaining piece you kindly sent me yesterday... no homework intended .... I get the idea that you are terrible with homework, and would solve world hunger before settling down to do what the professor assigned... Best wishes, Doug ----- Original Message ----- From: "Sterling K. Webb" <sterling_k_webb at sbcglobal.net> To: "mexicodoug" <mexicodoug at aol.com>; <meteorite-list at meteoritecentral.com> Sent: Saturday, December 29, 2007 2:00 PM Subject: Re: [meteorite-list] Neutron and Protonproductioninhyper-velocityimpacts > Hi, Doug and Crowd,
>
>> ...100% conversion [?]
>
> As far as the efficiency of the reaction, since the
> atmosphere is roughly 80% nitrogen, well, the odds of
> finding a nitrogen atom are pretty good if you're a
> lonesome thermal neutron. But it isn't content that
> determines the conversion efficiency; it's the cross
> section to the reaction. Nitrogen's is very high at ~1.9
> barns (the best unit name of all). Nitrogen is just
> a sucker for a thermal neutron come-on.
>
> Here's the periodic table arranged by the thermal
> neutron cross section of each element. Note please that
> the other elemental constituents of the atmosphere have
> much, much smaller thermal neutron cross sections than nitrogen:
> http://environmentalchemistry.com/yogi/periodic/crosssection.html
>
> Only hydrogen could weakly complete for those thermal
> neutrons, but there's not much hydrogen available except
> in water vapor. Nitrogen's the thermal neutron hog.
>
> Geesh! Make us go back and check everything?
>
> I made one mistake! I said there was 5000 tons of C14
> in the atmosphere, but 5000 tons is the C14 content of
> the entire planet; the C14 content of the atmosphere is
> 70 tons. (Everybody makes one mistake, right? It's a
> gimme.) Not much for all this fuss, is it? However, it
> strengthens my point about how little C14 it takes to
> affect the resulting isotopic record.
>
> Can we talk about something else now? I'm tired of
> homework.
>
>
> Sterling K. Webb
> --------------------------------------------------------------------
> ----- Original Message -----
> From: "mexicodoug" <mexicodoug at aol.com>
> To: <meteorite-list at meteoritecentral.com>
> Sent: Saturday, December 29, 2007 9:15 AM
> Subject: Re: [meteorite-list] Neutron and
> Protonproductioninhyper-velocityimpacts
>
>
> Hi Darren and friends,
>
> I think your geometry is mostly fine though there is an easier way to
> check
> the nuclear reaction asserted, left in the wake of the meteoroid. It's
> more
> accurate and you don't even need to truncate the atmosphere:
>
> Try this:
>
> Km^2 "nuclear reactor" meteoroid=Pi*.5^2
> =0.785
>
> Km^2 Earth shell=4*Pi*6383^2
> =512,000,000
>
> ratio= 1 to 652,000,000
>
> So we agree (you got 660,000,000) whether you make long cylinders and
> subtract spheres and cut the atmosphere, or just look at the area as
> above.
> Your number has over 1% less 14-C because you assumed the atmosphere mass
> was equally dense from sea level to the top, but in reality the atmosphere
> is creamier near the center of the cake but only frosting at the edges.
> Considering half of the atmosphere is under about 6 km height or even
> lower
> if you make it colder, 5 km is much more averaged estimate and a whole
> lot
> less thinking and math. Maybe I should have used 4 or 6 km, but between
> 3-8
> km gives the same answer, so that's totally moot.
>
> Oh. But there is a minor error in the balancing these nuclear reactions
> which you're not going to like, I'm surprised Sterling didn't have his
> thinking cap on when he answered you. 12-Carbon isn't being converted to
> 14-C according to this black-box scheme - it is the 14-Nitrogen that is.
> I
> don't memorize the atmospheric ratio of nitrogen to carbon off-hand, but I
> suspect it is greater than 7000 to 1. So, at your 100% conversion, you're
> off by a factor that should have, say, 7000 times more 14-C in your
> reality
> check.
>
> There are so many things wrong with this model at this point, I'm just
> gonna
> sigh rather than say what I really think of hypotheses made in a one
> dimensional world to solve three dimensional problems. And this post has
> nothing to do with the big Mmm-word.
>
> Best wishes,
> Doug
>
>
> ----- Original Message -----
> From: "Darren Garrison" <cynapse at charter.net>
> To: <meteorite-list at meteoritecentral.com>
> Sent: Friday, December 28, 2007 3:35 PM
> Subject: Re: [meteorite-list] Neutron and Proton
> productioninhyper-velocityimpacts
>
>
>> On Fri, 28 Dec 2007 14:09:36 -0600, you wrote:
>>
>>>the Hulk comic character! I am the first to respect a thought
>>>experiment:
>>>But what scientific experiment could you propose (or has one already been
>>>done?) to follow through?
>>
>> Okay, I've done a bit of thought experiment on this too-- and bear with
>> me
>> because though hard math has never been my strong suit, I think I have
>> this
>> right.
>>
>> Okay, let's just for a moment imagine that a hypervelocity bolide DOES
>> generate
>> C14. Let's say that a 100 meter bolide converts 100 percent of all
>> carbon
>> into
>> C14 in a column 1,000 meters in diameter. For the sake of simple math,
>> let's
>> say that the top of the atmosphere is defined as 100 km up
>> http://en.wikipedia.org/wiki/K%C3%A1rm%C3%A1n_line and the bolide came in
>> at a
>> 90 degree angle. That hypothetical meteoroid would convert all the C to
>> C14 in
>> a volume of around 78.5 km3 by the formula for the volume of a cylinder
>> being
>> pi*r2*h.
>>
>> Now, to determine the volume of the atmosphere, I took a figure for the
>> diameter
>> of the Earth as 12,756 km. I then added the 100 km for the atmosphere on
>> either
>> side to get a figure of 12,956 km for the Earth plus the atmosphere.
>> Then
>> plugged both numbers into the formula for the volume of a sphere,
>> 4/3*pi*r3.
>> Subtracted the second from the first to get the volume of the Earth's
>> atmosphere, 51,924,386,581 km3.
>>
>> So, after all that, you take the volume of the Earth's atmosphere,
>> compare
>> it to
>> the volume of C14 converted atmosphere in that hypothetical column of air
>> created by the bolide, and you get around 1.51*10e-9 of the Earth's
>> atmosphere
>> converted, or around 1 part in 660,000,000.
>>
>> Earth diameter = 12,756 km
>> Earth volume = 1,086,783,833,910 km3
>>
>> Earth+a diameter = 12,956 km
>> Earth+a volume = 1,138,708,220,492 km3
>>
>> Atmosphere volume = 51,924,386,581 km3
>>
>> meteoroid path = 78.53975 km3
>>
>> 1.51*10e-9
>>
>> 660,000,000
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Received on Sat 29 Dec 2007 06:01:42 PM PST |
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