Crosspost from Spacesettlers - Might be of interest.

--- In spacesettlers@... "victoriatangoman"
> First off, the author of that post is referring to a mass driver
for
> an earth launch system. He rightly points out the complexity of
such
> a system by using plausible numbers, such as escape velocity of
> 10,000 m/s, cargo mass of 1,000 kg and 1 meter length.
>
> Now we High Frontier types don't usually advocate mass drivers as
> earth launch systems, rather we think they make a lot of sense for
> lunar material launch systems. Note, most people don't even
advocate
> using a lunar mass driver for human capsule launches.
>
> So, our relevent numbers become a lunar escape velocity of 3,000
> m/s, a cargo mass of a fraction of a kilo to a hundred kilos, and
a
> bucket length (into which our lunar material is placed) of at most
> 50 centimeters (this becomes the limiting factor in the switching
> circuitry.)
>
> So let's run through some numbers. Here's my scenario. A launch
> bucket in the shape of a semi-sphere with a radius of 25 cm. This
> corresponds to a volume of 32,738 c.c. Filled with lunar regolith
to
> a capacity of 86.5%, the volume of material is 28,318.45 c.c. with
a
> mass of 78,8856 Kg. For those who think in English units, this is
1
> cubic foot of material with a mass of 173.91 pounds. You can
> substitute in your own optimum cargo package.
>
> Now at an acceleration of 322 gravities, a track length of 1,425
> meters and a period of acceleration of .95 seconds, an escape
> velocity of 3,000 m/s can be achieved. This allows one launch per
> second with a bit of fudge factor included. Of course, theoretical
> mass driver designs should be able to handle higher accelerations
> with shorter track lengths, but we've got to use some numbers, so
I
> chose these.
>
> If we take our 25 cm radius bucket, and want it to fit into one
> track section with no overlap, let's use a 30 cm track section.
>
> So using the author's reasoning but substituting our lunar example
> in place of his earth launch system, we would need an acceleration
> of 3,157.85 m/s^2, and 4,750 track sections.
>
> Our time pulse would be 100 microseconds (1/10,000th of a second)
> and our power at the last section of track would be 747,042,469
> watts.
>
> The 100 ms time pulse allows us to dissipate 10,000 watts for a
> dollar, just as the author points out in his post.
>
> So, using the last formula in his post, the cost of the silicon
> would be . . . M * 2^(1/4) * A^(5/4) * LS^(3/4) * NT^(5/4) / 1250
>
> so plugging in
>
> 78.8856 Kg * 2^(1/4) * (3,157.85 m/s^2)^(5/4) * .3m^(3/4) * 4,750^
> (5/4) / 1250
>
> we get a cost of $28,398,313.11.
>
> What our the differences in the two proposals. Our moon mass
driver
> accelerates at 3,157.85 m/s^2, compared to the Earth system at 500
> m/s^2, our track section is .3 meters compared to 1 meter, our
> escape velecoity is 3,000 m/s compared to 10,000 m/s, our track
> length is 1,425 meters compared to 100,000 meters, our projectile
> mass is 78.8856 kg compared to 1,000 kg, our mass driver has 4,750
> sections compared to 100,000 sections.
>
> So, the earth mass driver hs to be very robust compared to the
lunar
> mass driver and needs $4,000,000,000 of silicon for switching
> purposes compared to our $28,398,313.
>
> Now, our system on the moon will have different operating
> constraints, i.e. operating in vacuum so no heat diffusion into
the
> atmosphere; lunar day temperatures of 250 C, lunar night
> temperatures of -250 C, (if the mass driver is shaded during the
day
> then the environmental temperature will drop.) How this affects
the
> system, I still don't know.
>
> Also, another interesting fact is that the 750 MW we would need to
> power the system could be produced by a surface area of 2,225,000
> m^2 of solar cells operating at 25% efficiency; 2.225 km^2 of
> surface area. Using solar thermal power should decrease the area
to
> 1.59 km^2 operating at 35% efficiency.
>
> Once that power is available, especially if in the form of a SPS
at
> L2, thus available without interuption, we'd be able to launch
6,815
> metric tonnes per day into our mass catcher. That's just shy of
> 2,500,000 metric tonnes per year. That'll go a long way to
building
> our habitat.
>
> Now one thing that I still need to figure out is why this author's
> calculations show a power requirement of 747,042,469 watts for the
> variables I've input, but my earlier calculations of using KE=1/2
(m
> * v^2) 1/2(78.8856 * 3,000^2) = 354,985,200 watts and:
>
> V/A = 3,000 m/s / 3,157.85m/s^2 = .95 sec; thus
>
> 354,985,200/.95 = 373,668,631 watts
>
> This result is exactly half of the 747 MW the author calculated.
> It's very late for me and I'm not sure why there is this
discrepency
> of power requirements. Anyone care to find out and post an answer.
> Obviously, if we need only half the power so much the better for
our
> infrastructure requirements.
>
> So, in conclusion, it looks to me like his math is OK, it's just
> that his conclusion doesn't apply to a lunar mass driver scenario.
>
> --- In spacesettlers@... "xenophile2002"
> > http://www.islandone.org/LEOBiblio/SPBI1SI.HTM
> >
> > There must be something wrong in all of this, or else why would
us
> > High Frontier types maintain our love affair with the mass
> driver.
> > This sounds like it might be another case of 'proving' that
bumble
> > bees can't fly. My math isn't good enough to show this right or
> > wrong, but there must be *something* wrong with this argument.
$4
> > billion dollars just for the silicon?!?
> >
> > Xenophile (couldn't think of anything to put in these
parentheses)