From: rlbell@babbage.waterloo.edu (Richard Bell)
Subject: Re: Star Wars vs. Star Trek, tech & tactics, etc.
Organization: University of Waterloo
Date: Sun, 26 Jul 1992 17:18:20 GMT

In article <1992Jul25.225146.25099@cco.caltech.edu> travis@cco.caltech.edu (Leopold E. Travis) writes:
>
>Returning to the Star Wars vs. Star Trek idea, consider the following
>scenario:  The Death Star drops out of hyperspace near a Federation
>world, accompanied by fifty to a hundred Star Destroyers, and begins
>powering up its 100-quadrillion-megaton planet-zapping weapon of doom.
>Can the Feds stop it in time?
>
>Clearly if it has a shield like the one provided by the moon-based
>generator, the answer is no.  It wouldn't even need the Star Destroyers;
>it could just laugh at the Feds' attacks, blow up the world, and hop
>back into hyperspace.  
>
>So let's suppose it doesn't (perhaps this new shield technology _was_
>invented after the first Death Star was destroyed, or whatever).  The
>Feds can clearly toast the Star Destroyers--which can't fight at FTL
>speeds--but they may buy the Death Star some time.  So the question
>becomes, how many photon torpedoes would be required to destroy the
>Death Star?
>
>I don't know if there's an official figure for the yield of a photon
>torpedo; if so, I'm sure someone will post it.  Let's assume for the
>moment that it's about 100 megatons.  Now a 100-megaton surface burst
>on dry soil will carve out a crater about 1.2 kilometers in diameter,
>and 0.3 kilometers deep (source: Glasstone and Dolan's _The Effects of
>Nuclear Weapons_).  The Death Star's made of metal, but it's mostly
>hollow, so suppose as a rough estimate that a photon torpedo will
>obliterate a portion of this diameter and depth.  It seems likely that
>an area of at least twice this diameter will be rendered non-functional
>from shock and heat effects, so a photon torpedo effectively destroys
>about 4.5 square kilometers of the Death Star's surface.  The area of
>the Death Star being about 45000 square kilometers, this means about
>10000 photon torpedoes would be required to destroy its entire surface.
>
>Of course, one probably wouldn't have to destroy the entire surface
>area to make it unable to fire its gun.  Even so, it seems likely that
>the Death Star could withstand many hundreds of photon torpedoes without
>flinching.  So it looks like there's no way for the Feds to prevent it
>from destroying the planet and then escaping back into hyperspace, since
>only a few minutes at most are required for it to do this (it took half
>an hour at the moon of Yavin because Yavin itself was in the way and
>the Death Star had to maneuver into position).
>
>The above analysis is by no means rigorous; comments and refinements 
>are welcome.
>

The first weapon strike near the gun will cripple it.  The slightest drop
in energy efficiency will instantly destroy the Death Star the next time it
fires.  The Death Star will be destroyed even sooner if the Feds can warp 
space into a set of gravity lenses that send the beam back to its source.
That might require some quantum black holes, but we can plainly see that
the Feds have artificial gravity fields to help them manipute things.

The feds will lose a few planets before they get it right, but the planets
will be only outlying colonies (they will be found first in a search).
Building a planet based Death Star busting phaser will require less effort
than building a planet busting Death Star based weapons (Death Stars being
such puny things in astronomical terms), so Fed planets will stop being 
easy pickings.  If the Feds start giving them to the rebels, the Empire
evaporates almost as quickly as the imperial fleet.