# (Re:)^4 starship-design: The Size of the Problem

```At 3:08 PM 8/4/96, I sent the memo "The Size of the Problem."

At 2:47 PM 8/8/96, I wrote:

>Note that 1 cubic mile of (sea) water contains enough deuterium
>to provide about 5,400 USEs from the fusion reaction(s)--
>
>          D + D + D --> He4 + n + p + 21.6 MeV   .

At 9:23 AM 8/12/96, Kelly Starks wrote:

>The problem you listed was the difficulty in manufacturing (not
>to mention storing) the "fuel".  Given that mining and using more
>conventional fuels (like Lithium, duterium, etc..) don't have the
>heavy power costs and system complexity problems.

Actually it wasn't the "difficulty in manufacturing" the antimat-
ter I listed, it was merely the equivalent energy content in USEs
(units equal to the total production of U.S. Electricity in 1987 =
51.47 kg antimatter).  I assumed the manufacturing method (unspec-
ified) of antimatter was 100 percent efficient.  My 8/8 note gave
the volume of water (1/5400 of a cu mi) that would have to be pro-
cessed to extract the deuterium to give 1 USE via the specified
fusion reaction.

>Hell yes I'ld rather have hundreds of times the fuel weight!
>(Obviously due to the weight of the fuel you'ld need more than
>250 times as much, but it would be a lot easier to carry!)

Using antimatter fuel, the example mission to 8 lt-yr at 1-g con-
tinuous acceleration/deceleration is calculated (see my 8/12 note)
to require a mass ratio of at least 15.11 for the acceleration
phase alone.  Using deuterium fuel in the above fusion reaction
(with a Timothy "f" factor of 261) for the same mission, the cal-
culated required mass ratio for the acceleration phase is at least
3.84E12 kg D for each kg delivered to the peak velocity half way
to the destination.  (Note to Timothy: This mass ratio is for an
"optimum" exhaust velocity of 0.09580 c.  One of the reaction
products is a neutron, which can't be used as reaction mass.)
With 1.44E8 kg D per cubic mile of ocean, this mass ratio says you
have to extract all the deuterium from more than 26,000 cubic
miles of ocean to provide the fusion fuel to accelerate 1 kg at
1 g over 4 lt-yr, half way to the destination.

But there is a dramatic reduction in required mass ratio as the
g-level is reduced (and the trip time is increased).  The follow-
ing table shows that the required mass ratio is reduced by about
9 orders of magnitude while the trip time is increased by about a
factor of five:

accel/decel    Uend     trip time   mass ratio(accel)
(g)     (lt-yr/yr)    (yr)
1.0        5.000      4.480         3.842E12
0.5        2.881      6.897         4.869E09
0.2        1.520     11.692         3.659E06
0.1        0.994     16.991         5.929E04
0.05       0.672     24.401         2.675E03
0.0132     0.333     48.064         6.057E01

For the last line, the mass ratio for the full trip would only
be about 3,600 kg D for each kg of final mass, so one would
only have to process about 2.5E-5 cubic miles of water for each
kg delivered to 8 lt-yr in a continuous-g trip time of about
48 yr (24 yr with 1-g accel/coast/1-g decel; see below).  To de-
liver Kelly's 500,000-ton-dry-weight Explorer-class starship on
this mission would require extracting all the deuterium from about
12,500 cubic miles of water.

This may be a more "down-to-Earth" measure of the size of the
problem.  Now we're addressing the "difficulty in manufacturing."

>Thought.  What is the relative weight of an Anti-matter tank to
>the weight of the anti-matter?  Would its weight added to the
>ship start outweighing the advantages of the lighter fuel?

If the deuterium for the fusion engine is carried in a tank made
of an alloy of lithium and aluminum, the anti-hydrogen could be
carried in a tank made of an alloy of anti-lithium and anti-
aluminum, with a mass fraction similar to that of the deuterium
tank.  ;)   ["mass fraction" = (mass of contents)/(sum of masses
of tank and contents)]

>Also you would need to carry (and store) a full round trip worth
>of anti-matter since you couldn't refuel in the target system.

Kelly beware!  I'm on Zenon's side regarding "kamikaze"
missions.  ;)

>You'ld probably get shorter trip times if you had used the same
>power in higher boosts at the start and end of the trips, with
>a coast phase in the middle.  Same power consumption, but higher
>average speed.

Kelly, go to the blackboard and write 100 times "Power is the
rate of use of energy."  Whap, whap, whap.  ;)
If you replace the word "power" with "energy" you're right on.
The table below shows the comparative trip times (to alpha Cen-
tauri: distance = 4.35 lt-yr) for continuous accel/decel at g-
levels below 1, with trip times for 1-g accel/coast/1-g decel
with the same peak velocity (same energy requirement).  The
values of peak velocity, Energy/Mbo and mass ratio (accel alone)
for each entry are given in the table in my note of 8/12 (and
Kelly's response of 8/12).

Accel/decel    trip time  |     1-g accel/decel w/coast
(g)           (yr)    | trip time (yr)  coast time (yr)
1.0          3.576    |     3.576          0.000
0.9          3.810    |     3.582          0.152
0.8          4.087    |     3.603          0.334
0.7          4.421    |     3.646          0.551
0.6          4.835    |     3.720          0.819
0.5          5.366    |     3.845          1.162
0.4          6.083    |     4.051          1.619
0.3          7.128    |     4.407          2.269
0.2          8.867    |     5.091          3.317
0.1         12.751    |     6.813          5.537

For acceleration periods that become a smaller fraction of the
trip time, the average speed tends toward twice the average speed
for continuous accel/decel, leading to cutting the trip time just
in half.

Note: For a reduction in energy requirement by a factor of 10
from the 1-g-all-the-way trip (an accel/decel g of about 0.17,
from the 8/12 table), the trip time for 1 g with coast for the
same energy is increased by less than a factor of 2.
A factor of 10 reduction in energy requirement (for antimatter
fuel) for less than a factor of 2 increase in time says the coast
phase is worth considering.  On the down side, however, are the
requirements for heavier thruster (and power-system) weight for
1 g vs. reduced g, and increased complexity for artificial-gravity
provisions during the coast time (the last column).

>whap, whap.  ;)

I can't imagine how to avoid them.  Also, my Fortran compiler is
dated May 1988 and doesn't allow ENDDO's, so it may not allow
whatever gets around GOTO's.
BTW, the code I provided in my 8/12 note had to be modified to
cover fusion energy.  But the size of the numbers above indicates
to me that possibly nobody would be interested in using the modi-
fied code (except to show I'm wrong).  ;)

Rex

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