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From: slj@nmtvax.UUCP
Newsgroups: net.movies.sw
Subject: Death Star Size
Message-ID: <535@nmtvax.UUCP>
Date: Tue, 22-Nov-83 11:09:45 EST
Article-I.D.: nmtvax.535
Posted: Tue Nov 22 11:09:45 1983
Date-Received: Fri, 25-Nov-83 05:31:00 EST
Organization: New Mexico Tech, Socorro, NM
Lines: 51


One question that I have been wondering for quite a while:  how big is 
the Death Star?  I have come up with a figure of about 100 miles.

From the computer graphics used by the Alliance when they brief the 
audience about it, one might say that the DS is about 1/15th the
diameter of Endor.  This is the only view that makes it look that
small:  ususally it looks even larger.  Judging by the only good view
of it from space, when the entire Alliance comes out of hyperspace, 
the DS is about 1/6th the size of Endor, no doubt due to its being
closer to the camera than Endor.  Other views are from too close, and
all one sees is a large piece of it arcing across the sky.  Noting
generally very Earthlike conditions prevailing on Endor, I guess that
it probably isn't smaller than about 6000 miles in diameter.  [If there
are any physicists out there, you might want to point out any major
inadequacies of this guess.]  Assuming that all the views are exaggerated,
and assigning a lower bound of 1/60th the diameter of Endor, this gives
a DS size of about 100 miles.

On the other hand, you might want to chuck out all views from space and
computer graphics.  In that case, take the view we get from Endor when
Vader comes down to fetch Luke back to the Emperor.

    (PUSH)  I swear, that landing platform is the only governmental
    parking space I have seen recently with no 'handicapped only'
    sign next to one of the spots.  Note also that like a good 
    Imperial, Vader or his pilot shows callous disregard for the
    painted parking spaces and sets the shuttle down between them.
    (POP)

The DS from Endor appears to occupy about 5 or 10 degrees of arc.  
Noting that for the DS to remain over the shield generator, it
must be in Endor-stationary orbit.  I don't feel like calculating
what distance from Endor that is, and I'd probably get it wrong if
I tried, but assume that it's three diameters, like with Earth, but
set the diameter of Endor at 6000 miles again.  [Again physicists
can set me straight if they want to.]  Note that in no shot of the
DS does it appear to be any such distance from Endor.  Nonetheless,
for something 18 Kmiles away to subtend 5 degrees of arc is fairly
impressive: 

    sin(THETA) = opposite (mi) / hypotenuse (mi)
    sin(THETA) * hypotenuse (mi) = opposite (mi)
    sin(pi/36) * 18000 (mi) = 1500 (mi)

So I'll stick with my guess about 100 miles.  According to the book,
the new improved DS is twice the size of the old one, so that puts
the old one at about 50 miles.  If anyone can point out any data
which I've overlooked, I'd appreciate hearing about it.

                                   S. Luke Jones