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Re: starship-design: Re: Re: The Size of the Problem
At 12:50 AM 8/12/96, DotarSojat@aol.com wrote:
>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? :(
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. 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!)
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? 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.
>>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,---
True. The higher speed and longer burn times both dramatically increase
fuel consumption. (Well beyond what seems technically feasable.) Also...
> -- 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
.... 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.
>
>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):
Thanks for including the code!
?? You still use goto's in your code! Bad Rex, Bad. Whap, whap, whap. ;)
> 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
Kelly
----------------------------------------------------------------------
Kelly Starks Phone: (219) 429-7066 Fax: (219) 429-6859
Sr. Systems Engineer Mail Stop: 10-39
Hughes defense Communications
1010 Production Road, Fort Wayne, IN 46808-4106
Email: kgstar@most.fw.hac.com
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