Re: Starship design revamping

[email removed] writes:
> Sorry, but GURPS is wrong, and Traveller is much closer to being correct.
> For the drives alone it can easily be shown that the "thousand miles to
> the gallon" is correct.
>
> First of all, there was a paragraph buried in the original rules that
> gave fuel consumption as an option. I worked it out and it gave an Isp
> of 600,000 for the drive. (which incidentally matches the figure given
> for a "fusion drive" on a chart that NASA seems to have put in 90% of the
> secondary school classrooms in the US during the 60's) Isp relates thrust
> and fuel consumption as follows: Isp =thrust(in pounds)/fuel consumption
> (in pounds per second). Astute readers will notice slight problems with
> dimensionality in this calculation. More astute readers will realize
> that this is expected when one uses the same unit for mass and force.
> Anyway, to get 6 million pounds opf thrust with such a drive (same as a
> Saturn V) you need 10 pound per second of fuel. This is enough to move
> a 3000 ton ship at 1g. So much for 200 years of fuel...

I'm not sure that the Isp is usable for reactionless thrusterplates...
This "fusion drive" is probably more similar to the "fusion rocket"
in the COACC.

I looked fusion up in a physics textbook and it states that for standard
deuterium fusion with all its secondary stages. (First deuterium fuse
with deuterium to produce various hydrogen and helium isotopes and free
neutrons and assorted other radiation. Then some of these products fuse
to make some more energy.)
The book gives the energy gained from fusion of one atom of deuterium
(including all byproducts etc) to 13E-13 Joules.
The occurance of deuterium in hydrogene it states as 15 to 100,000.
(Deuterium is a hydrogene isotope).
One gram (1E-3 kilos, 1E-6 tons) of hydrogene contains 6.023E23 hydrogene
atoms, and will thus yield 11E9 Jouls.
One ton of Tech15 plant uses 0.009 tons of H2 per hour.
This translates to about 99E12 MWatts per per kl of plant per hour or, in
other words, 99,000,000 MW per kl of plant...

(I hope I havn't done any errors too big to explain away:-)
- -bertil-
- --
Bertil K K Jonell @ Chalmers University of Technology, Gothenburg
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