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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