Acheiving .5c with Bombtrack Propulsion?
- From: "IsaacKuo" <mechdan@xxxxxxxxx>
- Date: 7 Nov 2005 08:20:08 -0800
The conventional wisdom is that with foreseeable technology,
fast interstellar travel requires either phenomenally efficient
fusion technology, mega-scale beams, antimatter, or some other
exotic technology.
I reject the conventional wisdom, and believe high speed
interstellar travel is possible with relatively modest
future technology. Here is my proposed starship design:
Front view Side view
o---------o o--..__
/ \_______/ \ | | | |""--.._
/ / \ \ | | | | | | | |
/ / \ \ | | | | | | | |
/ / \ \ | | | | | | | |
o--< >--o o-+-+-+-+-+-+-|
\ \ / / | | | | | | | |
\ \ / / | | | | | | | |
\ \_______/ / | | | | | | |_|
\ / \ / | | |__..--""
o---------o o--""
This starship is huge magsail shaped like a basketball net,
with the main payload strung around the rearmost magloop.
This payload is dominated by stored nuclear bombbots which
will be formed into a deceleration bombtrack during the
journey.
The starship is accelerated by traveling along a long
pre-prepared track of nuclear bombbots. As the fast
protium products of the explosions expand outward from
the bombtrack, they hit the conical inner face of the
magsail and propel the sail forward. Assuming a final
fusion stage of D+He3 --> p and He4, the protium
leaves at a speed of .1774c. With a conical taper ratio
of, say, 1:4, an ultimate starship velocity of maybe
..5c is plausible.
So how big/heavy does the starship need to be? This
depends on the mass of the final payload, of course,
and the total mass ratio of the bombtrack to starship.
For example, if the deceleration bombtrack must mass
100 times as much as the final payload, then the
acceleration bombtrack must mass 10,000 times as much.
If the deceleration bombtrack must mass 1000 times as
much as the final payload, then the acceleration
bombtrack must mass 1,000,000 times as much.
Let's assume that the overhead for the bombbot electronics
and stationkeeping thrusters/propellant is relatively
small (say, 5-10%). Then the bombbot is dominated by
the actual nuclear device, and this is where my knowledge
is almost zero. How much overhead is in the driving
stages? Is a D+He3 reaction the best final stage?
Could there be a better final stage which produces more
energetic protium particles? How much of the fast
moving protium products is "lost" to collisions before
reaching the magsail?
My gut feeling is that there should be some relatively
unexotic fission-fusion bomb design which can provide a
bombtrack/payload mass ratio on the order of 1000:1 or
better. But that's just a gut feeling. Without some
more concrete numbers for plausible bomb designs, it's
just a guess.
If it turns out that the mass ratio for using protium
products this way is more like 10,000:1 or worse, then
I have to rethink using protium fusion products. It may
make more sense to use slower heavier particles, and
be satisfied with a much slower cruise speed. For
example, it may be that the optimal products to use are
He4 alphas moving at .0677c. In that case, a cruise
velocity of only .2c would be acheived by this starship
design. That's still rather respectable, for a starship
design which doesn't require exotic technology.
Any help filling in the knowledge gap is appreciated!
Thanks!
Isaac Kuo
.
- Follow-Ups:
- Re: Acheiving .5c with Bombtrack Propulsion?
- From: BDH
- Re: Acheiving .5c with Bombtrack Propulsion?
- From: Allen Thomson
- Re: Acheiving .5c with Bombtrack Propulsion?
- Prev by Date: Re: Tag-team orbits and tidal locking
- Next by Date: Re: Self-contained interstellar vehicle
- Previous by thread: Re: Tag-team orbits and tidal locking
- Next by thread: Re: Acheiving .5c with Bombtrack Propulsion?
- Index(es):
Relevant Pages
|