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starship-design: Re: Re: The Size of the Problem
At 09:58 8/5/96, Kelly Starks wrote:
>At 3:07 PM 8/4/96, DotarSojat wrote:
>>SUBJECT: The Size of the Problem
>>The minimum energy requirement to create the antimatter to
>>deliver 1 kg to a distance of 8 lt-yr at 1-g continuous
>>acceleration/deceleration becomes
>>
>> 54.3 kg am / 51.47 kg am/USE = 1.055 USE .
>>IMPLICATIONS
>>
>>....
>>Orders of magnitude reduction in payload and increase in
>>transit time are required to reduce the energy problem to a
>>"manageable" size, to only a few USEs, say.
>>
>>....
>>A few plausible ways around the problem...come to mind:
>>
>>....
>>
>>The group might think of others.
>A few come to mind. ;)
>
>1) Don't use antimatter.
And use at least 250 times the mass of energy fuel? :(
>2) Don't try for a constant 1 g acceleration for the full
>durration of the flight.
>....
While I believe Kelly had a coast phase in mind, his comment
suggested to me trying values of constant acceleration/
deceleration less than 1 g. With some improvements in the cal-
culation from my 4/4/96 memo to the Group, I get the following
reductions in energy requirement, in USE per kilogram of burnout
mass, as the accel/decel value is decreased from 1 g (included
in the table are the peak proper velocity Uend in lt-yr/yr, the
trip time in yr and the mass ratio for the acceleration phase
alone):
Accel/decel Uend Energy/Mbo trip time mass ratio(accel)
(g) (lt-yr/yr) (USE/kg) (yr)
Distance = 8 (actually 7.941) lt-yr
1.0 5.000 1.0510* 4.480 15.11
Distance = 4.35 lt-yr (alpha Centauri)
1.0 3.088 0.4067 3.576 10.58
0.9 2.851 0.3476 3.810 10.03
0.8 2.611 0.2925 4.087 9.47
0.7 2.369 0.2415 4.421 8.91
0.6 2.124 0.1946 4.835 8.35
0.5 1.872 0.1518 5.366 7.79
0.4 1.613 0.1132 6.083 7.22
0.3633 1.516 0.1000 6.417 7.01
0.3 1.342 0.0786 7.128 6.65
0.2 1.049 0.0483 8.867 6.08
0.1 0.707 0.0220 12.751 5.50
0.0476 0.475 0.0100 18.653 5.19
-----------
*was 1.055 in my 8/4/96 memo.
A plot of energy vs trip time shows an elbow at about 0.5 or
0.4 g, where additional expense in energy starts to reach
diminishing returns in reduction in trip time. The point at
0.3633 g for the alpha Centauri trip represents a reduction in
USE/kg by about an order of magnitude from the example trip in
my 8/4/96 memo. The point at 0.0476 g (for alpha Centauri)
represents two orders of magnitude reduction, but at a signif-
icant increase in trip time (to 18.653 yr).
Rex
P.S. For the record, I include below the Fortran program that
did the calculations (incidentally, I regard a Fortran program
as essentially not much more than an ASCII rendering of the
equations):
PROGRAM TRIP !8/11/96
101 FORMAT(2X, 21H Acceleration (gs) = )
102 FORMAT(2X, F8.4, F8.3)
103 FORMAT(2X, F5.3, 4F8.4, F8.3, F8.4, F8.4, F7.2)
DIST = 4.35
XACC = 0.5 * DIST
ETA = 1.
2 CONTINUE
WRITE(*,101)
READ(*,*) AG
IF(AG .EQ. 0.) GO TO 99
ACC = 1.0324 * AG
UEND = SQRT(ACC * XACC * (ACC * XACC + 2.))
GAMEND = SQRT(1. + UEND*UEND)
VEND = UEND/GAMEND
THETA = LOG(UEND + GAMEND) !asinh(Uend)
TACC = THETA/ACC
X1 = 0.05
X2 = X1
1 CONTINUE
X3 = X2
X2 = X1
X1 = X1 + 0.01
GAMEX = 1./SQRT(1. - X1*X1)
R = 1.01
IF(X1 .GT. .05) R = EXP(THETA*(GAMEX-1.+ETA)/(ETA*GAMEX*X1))
Y3 = Y2
Y2 = Y1
Y1 = (GAMEND - 1.) * (GAMEX - 1. + ETA)/((R - 1.) * ETA *
& (GAMEX - 1.))
IF(Y1 .LT. Y2 .AND. X1 .GT. 0.1) THEN
A = ((Y1-Y2)*(X2-X3)-(Y2-Y3)*(X1-X2))/
& ((X1*X1-X2*X2)*(X2-X3)-(X2*X2-X3*X3)*(X1-X2))
B = ((Y1-Y2) - A*(X1*X1-X2*X2))/(X1-X2)
XOPT = -B/(2.*A)
GO TO 3
END IF
GO TO 1
3 CONTINUE
OPTVEXH = XOPT
OPTUEXH = OPTVEXH/SQRT(1. - OPTVEXH*OPTVEXH)
GEXHOPT = SQRT(1. + OPTUEXH*OPTUEXH)
ROPT = EXP(THETA * (GEXHOPT - 1. + ETA)/(ETA * GEXHOPT *
& OPTVEXH))
EFFMAX = (GAMEND - 1.)*(GEXHOPT - 1. + ETA)/((ROPT - 1.) *
& ETA * (GEXHOPT - 1.))
URATIO = UEND/OPTUEXH
AMMBO = (GAMEND - 1.)/(2.* EFFMAX * ETA)
AMMI = (GAMEND - 1.)/(2.* EFFMAX * ETA * ROPT)
WRITE(*,103) UEND, VEND, OPTVEXH, OPTUEXH, EFFMAX, URATIO,
& AMMBO, AMMI, ROPT
ROVER = ROPT * ROPT
FRAM = AMMBO/(ROPT - 1.)
OVAMMBO = FRAM * (ROVER - 1.)
USE = OVAMMBO/51.47
TTRIP = 2. * TACC
WRITE(*,102) USE, TTRIP
GO TO 2
99 STOP
END