Re: SPACE: Lunar Warfare

Michael Lorrey (
Tue, 14 Jan 1997 17:09:12 -0500

James Rogers wrote:
> At 03:17 PM 1/13/02 -0500, Michael Lorrey wrote:
> >
> >I found a ref ona NASA page stating that one Megaton was 4.2 x 10^22
> >ergs.
> This is equivalent to the ASTM standard.
> >>
> >
> >Yes it is quite small, and could easily be formed into the same
> >stealthed shapes that Mark is puportin for nukes.
> Stealth would play less a role for big rocks. Since a big rock is just "a
> big rock" with no terminal guidance, and only kinetic energy as its energy
> source, there is little you can do to defend against it. It is falling
> whether you like it or not. The best you could do would be to modify the
> trajectory.

True, but anything you put in an induction gun would probably be formed
ahead of time in a lunar refinery. While a simple cylinder would be
easiest to deal with, a penetrating shape would also be easily
attainable, and putting guidance and maneuvering units onboard would
also be possible, though would probably be the most expensive part
(though nowhere near Marks estimates, as his are based on one off custom
devices for a FEW satellites, not high production rates)

> >>
> >> Note that this is not an exact science, and that there are many factors
> >> involved. I was simply demonstrating that nuclear weapons lose most of
> >> their effectiveness in subterranean lunar environments. Nuclear weapons are
> >> hundreds of times more effective on earth.
> >>
> >
> >Thanks for a more expert opinion James. Based on a reference I found of
> >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
> >brain, I had read it somewhere...
> >
> >Additionally, in terms of damage ability, large nukes have, even in
> >atmosphere, an attentuating effectivness proportionate with their yeild.
> >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.
> The scaling factor for nuclear weapons is (Yield1/Yield2)^(1/3), where
> Yield1 and Yield2 are the yields of the nuclear devices in question. This
> scaling factor is "ideal" for all explosives, nuclear or conventional. The
> ideal case more or less holds up until about the 10Mt range, where the
> exponent starts to decrease slightly due to "real" factors.
> If you think about it, a cube root scaling factor makes perfect sense for
> scaling one dimensional energy values in 3D space.

Yes, I was hoping you'd have an idea on this.

> >For examply, a high altitude attack by 3-4 B-52s carrying conventional
> >2000 lb HE bombs can damage as much area as a 10-20 KT nuke, though
> >without the radiation effects. In fact, because the concussive
> >shockwaves are from dozens of sources, rather than one device, they
> >actually are more effective in their area of effectiveness.
> The obvious reason for this being that a single large explosion is poorer at
> distributing destructive energy than many smaller explosions that have been
> spread out. You maximize the area of optimal destructive energy. Cluster
> bombs are designed with the same principle in mind.
> The Oklahoma bombing was not equivalent to a ton of TNT. Although the truck
> carried an estimated 1-1.5 tons of explosive, my "expert" estimate would be
> a damage equivalence of about 1/4 ton of TNT based on the structural damage.

Thanks James. Given this, what is your guesstimate of what an 11 km/s
one ton ferrous rock could do, with a 15 ton impact force?

Size of crater?
If preshaped for penetration, effective depth?
effective radii of shock wave?
anything else?


Michael Lorrey ------------------------------------------------------------ President Northstar Technologies Agent Inventor of the Lorrey Drive

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