From: Michael Lorrey (retroman@tpk.net)
Date: Tue Jan 14 1997 - 14:53:28 MST
The Low Willow wrote:
>
> On Jun 5, 9:53am, Michael Lorrey wrote:
>
> } I found a ref ona NASA page stating that one Megaton was 4.2 x 10^22
> } ergs.
>
> Equals 10^15 calories equals 10^6*one gigacalorie equals James number.
He said it was 6 kt of TNT, while my calculation was 15 tons....
>
> } > only has an energy yield of 6*10^10 Joules.
> } Yes it is quite small, and could easily be formed into the same
> } stealthed shapes that Mark is puportin for nukes.
>
> Since no one from the Earth crowd has even suggested an active defense
> of Earth, this is a bit beside the point.
>
> } 4.2 x 10^22 ergs per Megaton, I calculated 15 tons of TNT equiv. for the
> } 1 ton rock, so while I was seriously wrong in my own estimate, Mark was
> } under by almost three times. I don't know why that number stuck in my
>
> 3 times versus 3 orders of magnitude. Poor Mark. What was your point
> in saying that? You're the one James cited for confusing force and
> energy units.
I was admitting my own mistake. See, even the hugely arrogant Mike can
admit it when he is wrong. I was also posting the correct answer, which
was not what Mark had posted.
>
> } A 10 MT device will only have a damage area of between 4-8 times larger
> } than a 300 KT device, which is why the powers have gotten away from
> } using big nukes, as they are not as cost effective as using the same
> } amount of material in four or more devices.
>
> Which makes your defense problem much harder, as you now get to try to
> detect and shoot down many more and smaller missiles. 30 times more
> missiles, using the sizes just above. Admittedly Earth has to have that
> many more rockets delivering the warheads, but I think it just might
> manage that for the sake of defense.
Not at all. I was referring to atmospheric impacts. According to James,
earthgov will still need to build and use meganukes to have as much of
an impact under vacuum conditions as a much smaller nuke would on earth,
so not only are we looking at a differential of energy expenditure and
application of four due to gravity, but the two environments of
atmosphere and vacuum also put earth based forces at a disadvantage.
As you all have read my statements before, that war strategically is a
competition of cost effectiveness in battle. These two differences, in
gravitational potential and environment, between earth and the moon are
significant in giving any lunar David a much more even hand against the
terran Goliath.
>
> } an earth based target. While I do not know how to calculate the actual
> } or estimated equivalence, I think that this is a significant factor to
> } consider.
>
> So is the 15 ton rock impact.
>
>
> Antisocial arrogance is only for those of us who deserve it.
--
TANSTAAFL!!!
Michael Lorrey
------------------------------------------------------------
President retroman@tpk.net
Northstar Technologies Agent Lorrey@ThePentagon.com
Inventor of the Lorrey Drive Silo_1013@ThePentagon.com
Website: http://www.tpk.net/~retroman/
Now Featuring:
Mikey's Animatronic Factory
http://www.tpk.net/~retroman/animations.htm
My Own Nuclear Espionage Agency (MONEA)
MIKEYMAS(tm): The New Internet Holiday
Transhumans of New Hampshire (>HNH)
------------------------------------------------------------
Transhumanist, Inventor, Webmaster, Ski Guide, Entrepreneur,
Artist, Outdoorsman, Libertarian, Arms Exporter-see below.
------------------------------------------------------------
#!/usr/local/bin/perl-0777---export-a-crypto-system-sig-RC4-3-lines-PERL
@k=unpack('C*',pack('H*',shift));for(@t=@s=0..255){$y=($k[$_%@k]+$s[$x=$_
]+$y)%256;&S}$x=$y=0;for(unpack('C*',<>)){$x++;$y=($s[$x%=256]+$y)%256;
&S;print pack(C,$_^=$s[($s[$x]+$s[$y])%256])}sub S{@s[$x,$y]=@s[$y,$x]}
This archive was generated by hypermail 2.1.5 : Fri Nov 01 2002 - 14:44:00 MST