From: rlbell@babbage.waterloo.edu (Richard Bell)
Subject: Re: Star Wars vs. Star Trek, tech & tactics, etc.
Organization: University of Waterloo
Date: Thu, 30 Jul 1992 15:48:52 GMT

In article <1992Jul28.165118.26481@cco.caltech.edu> travis@cco.caltech.edu (Leopold E. Travis) writes:
>
>rlbell@babbage.waterloo.edu (Richard Bell) writes:
>>travis@cco.caltech.edu (Leopold E. Travis) writes:
>>>[Much of my speculation on the Feds vs. the Death Star deleted.]
>>
>>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.
>
>I'll address the issue of quantum black holes later in this post. 
>First, let's examine the conflict assuming no new weapons technologies,
>i.e. both sides are restricted to their standard armaments in the TV
>shows and movies.  
>
>Let's look at the vulnerability of the Death Star's gun.  As I've stated
>before, it must have required about 10^17 megatons to destroy Alderaan. 
>If this feat was accomplished in 10 seconds, this puts the Death Star's
>energy output at 10^16 megatons per second (about 4 x 10^31 Watts, or
>100,000 times the output of the sun (!)).  So if the mechanism of energy
>release is matter-antimatter annihilation, the Death Star must consume
>about 2.3 x 10^14 kilograms of antimatter (and an equal quantity of
>matter, of course) per second when firing its gun.  Clearly this must
>require an incredibly powerful containment field.  The machinery
>involved must take up a good fraction of the Death Star's bulk, and
>mostly be situated deep within it.  So surface photon torpedo hits are
>not going to take it out.  Since the Death Star's radius is 60
>kilometers and a photon torpedo only digs about 0.3 kilometers into it
>(this was a conclusion of my previous post, assuming a yield of 100
>megatons for a photon torpedo), it's going to take dozens of hits upon
>the same spot to get at the bulk of the gun machinery. 
>
The firing port of the planet buster MUST be on the SURFACE of the Death
Star, or it will remove all of the surface above it when it fires.  The feds
do not have to damage the bulk of the machinery, all they have to do is
damage one of the beam focussing elements at one of the weapon ports, and
the Death Star fries itself.  Also, the firing ports will work against the
Death Star by channelling energy back towards the Death Star's innards.
Alternatively, the explosions of the photon torpedoes could be used to
pinch shut all of the firing ports.    

>Furthermore, suppose the Feds do manage to damage the gun machinery. 
>Surely this must be the most heavily backed-up and fail-safed system on
>the Death Star, so it'll take a great deal of damage to cause it to fail
>completely.  And even if the Death Star can only crank out a millionth
>of its usual output, this is still 100,000,000,000 megatons per
>second--easily enough to destroy all life on the target planet (and
>vaporize the oceans and probably the crust as well; recall that the
>arsenals of the United States and the Soviet Union during the cold war
>were on the order of a "mere" 10,000 megatons each, and this would
>probably have been sufficient to wipe out civilization as we know it).

If the planet buster becomes only 99.999999999999999% efficient, it will
suffer the equivalent of a photon hit every time it fires at full power,
but optical components are VERY finicky things, and a photon hit near a
firing port will make it only 99% efficient (it is much more likely for
the losses to be much worse, the only way that I can imagine the planet 
buster to work at all is in a vacuum with all components cooled to absolute
zero, so they are absolutely not going to move and shift alignment;everything
is also superconducting), 99% efficiency means that the DS suffers one percent
of the flow through that port, that part of the weapon will be shut down.
Any typical fed ship has many more photons than the DS has firing ports.

>
>Finally, suppose the Feds do take out the containment field, causing the
>Death Star to explode.  Assuming it carries enough antimatter for 100
>seconds of output from its gun (enough to destroy ten planets), this
>means the blast is going to be about 10^18 megatons--which will still
>wipe out the target planet and the attacking Federation ships.  (Indeed,
>even if the Death Star is 500,000 kilometers from the planet when it
>explodes, the planet's going to receive over 300,000 megatons of energy
>per square kilometer that's exposed.)  Of course, the Empire is then
>without a Death Star.  But if the planet it was attacking happened to be
>the earth, the Federation's not going to be too happy either...
>
>As for the quantum black holes...  Even if they were available,
>channeling the beam back at the Death Star might not be such a hot idea
>(see the previous paragraph).  A Death-Star-busting phaser would
>similarly be a suicide weapon for the planet involved.  A better strategy
>might be to simply drop a sufficiently-massive quantum black hole _into_
>the Death Star (after all, it's not very maneuverable!).  I think this
>might involve a radiation flux as the last of the Death Star's matter
>was sucked in, but I lack the expertise to calculate it.
>
>But suppose that, by whatever means, the Feds do take out the Death
>Star.  What might be the Empire's next move?  Well, how about this... 
>Instead of building a Death Star, why not simply build a robot bomb?  It
>drops out of hyperspace near the target planet and immediately explodes! 
>Even a blast of a few trillion megatons, at the distance the moon, would
>utterly wipe out all life on one hemisphere of the planet (and those on
>the other side had better get some life-support equipment pretty fast). 
>Robot bombs being much easier to build than Death Stars, the Empire
>could produce them relatively quickly.  So once again it looks like
>curtains for the Federation unless the Q or somebody intervenes.
>

Unless the Empire capture a starchart from the feds that lists all of their
valuable planets, and their locations, the feds have more planets than the
empire can destroy with bombs.  The Federation may be small, but it is large
enough to execute general order twenty-four on a galaxy wide scale if it was
faced with systematic annihilation.


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