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starship-design: Deceleration of sail pushed by constant-power beam
Hi all
In my note of 9/11, "Motion of sail driven by constant-power
beam," I wrote--
>The deceleration phase (here assumed without justification to
>be a mirror image of the acceleration phase) needs to be
>addressed in a separate discussion. Timothy has already put a
>lot of thought into it.
The deceleration phase is not a mirror image of the acceleration
phase. In the acceleration phase, the sail/ship is driven by
the beam's radiation pressure alone; in the deceleration phase,
the radiation pressure of the beam continues to provide a for-
ward thrust, but the power of the beam is collected and convert-
ed into a retro-thrust "jet" of, say, protons ejected forward to
slow down the sail/ship.
Timothy, in his 3/29 note, solved the problem of bringing the
sail/ship to a halt with the "jet's" retro-thrust overcoming the
beam's radiation pressure to give a constant deceleration level,
allowing the beam power to take on any variable value required
for the constant deceleration. He found the optimum exhaust
velocity that gives a minimum energy expenditure, but did not
consider the requirement to bring the sail/ship to a halt in a
given distance.
This analysis considers a beam from a constant-output emitter
with the same power that accelerated the sail/ship to a peak
velocity half way to the destination star (per my 9/11 note).
The analysis selects the exhaust velocity that brings the
sail/ship to a halt in the remaining half of the distance.
ANALYSIS
Following the notation of my 9/11 note, the power radiated from
the emitter is Pe, and the power received by the sail, ignoring
inverse-square effects, is Pr. The thrust exerted by the radia-
tion pressure is Pr/c. At the start of the acceleration phase
when Pr = Pe, the mass of the sail/ship is Mo, the initial
acceleration is ao, and the required emitted power is
Pe = T * c
= Mo * ao * c .
When the sail/ship has been accelerated to an apparent velocity,
beta lt-yr/yr, having the velocity parameter, theta, from the
relation: beta = tanh(theta), the received power is reduced by
the Doppler shift according to
Pr = Pe * exp(-theta) ,
and the thrust, Tb, of the received beam's radiation pressure is
Tb = Pr/c
= (Pe/c) * exp(-theta)
= Mo * ao * exp(-theta) .
The beam power is collected and converted with an efficiency,
eta, to the power, Pex, of a retro-thrust exhaust "jet," , or
Pex = eta * Pr .
The exhaust-jet power is the rate of ejection of kinetic energy,
or
Pex = (dM/dt') * (gexh - 1) * c^2 ,
where (dM/dt') is the rate of ejection of propellant mass
(made up of protons, say), and gexh is the jet's energy factor,
(gamma) = 1/sqrt(1 - Vexh^2), derived from the jet's exhaust
velocity, Vexh (lt-yr/yr).
The thrust, Tex, of the exhaust jet is given by
Tex = (dM/dt') * gexh * Vexh * c .
Substituting the dM/dt' derived from the equation for Pex above
gives
Tex = [Pex * gexh * Vexh * c]/[(gexh - 1) * c^2]
= eta * (Pr/c) * [factor] ,
where
[factor] = gexh * Vexh/(gexh - 1) .
Note that the exhaust-jet retro-thrust exceeds the radiation-
pressure thrust, making deceleration of the sail/ship possible,
when eta * [factor] is greater than 1. For Vexh = 0.9, for
example, the value of [factor] is 1.5954, and deceleration is
possible only if the efficiency, eta, of conversion of received
power to exhaust power is greater than 1/1.5954 = 0.6268.
The acceleration, a, of the sail/ship (hopefully, negative) is
given by
a = (Tb - Tex)/M
where M is the mass of the sail/ship at the ship time, t'.
The rate of change of the mass of the sail/ship as propellant is
ejected is obtained from the Tex equation above--
dM/dt' = Tex/(gexh * Vexh * c) .
The rate of change of the velocity parameter is given by the
velocity-parameter equation of motion--
d(theta)/dt' = a/c
= (Tb - Tex)/(M * c) .
We thus have two simultaneous differential equations, with
dM/dt' involving exp(-theta) through the dependence of Tex on
Pr, and with d(theta)/dt' involving 1/M. The coupling therefore
is non-linear, and the method of solution I find in my math book
is for linear simultaneous differential equations. Not being a
"mathochist" (one who enjoys suffering in the solution of higher
math problems), I chose to integrate these equations numerically;
see the appended Fortran program SAILTRIP. The implemented
difference equations are
M(n+1) = M(n) - [Tex(n)/(gexh * Vexh)] * [t'(n+1) - t'(n)]
theta(n+1) = theta(n) + [(Tb(n) - Tex(n))/M(n)] *
[t'(n+1) - t'(n)] .
(The SAILTRIP program also includes the acceleration-phase
calculation outlined in my 9/11 note, allowing the program to
cover the whole trip from start to destination.)
Results become consistent to better than three significant
figures for time steps, [t'(n+1) - t'(n)], smaller than 0.01 yr,
with insignificant computation time (about one second for the
deceleration phase) for a time step of 0.001 yr. The decelera-
tion phase is repeated with trial values of the exhaust velocity
until interpolation yields the desired deceleration distance
within a tolerance of 0.0001 lt-yr.
RESULTS
Calculations with the SAILTRIP program were made for a trip to
tau Ceti, whose distance was taken to be 11.9 lt-yr. It was
found that, with the constant-output emitter, the deceleration
grows to exceed 1 g near the destination, where the thrust
increases as the velocity and therefore the Doppler shift
decrease, and where the mass of the sail/ship decreases as the
propellant is depleted. The deceleration is limited in the
calculation to 1 g by the simple expedient of furling the sail.
The calculated values of theta, distance, proper velocity, accel-
eration, apparent (Earth) time, time of emission of radiation
from Earth and mass ratio (ratio of initial mass to the instan-
taneous mass), as functions of ship time, t', are given in the
following table for the tau Ceti trip, for a conversion effic-
iency of received power to exhaust power of 1.0. Also stated are
the values of the exhaust velocity, in lt-yr/yr, that gives the
desired deceleration distance, with the corresponding kinetic
energy, in MeV, of protons having that velocity, and of the final
relative area of the furled sail.
Tship Theta Dist Prop Vel Accel TEarth Temit Mratio
(yr) (rad) (lt-yr) (lt-yr/yr) (g) (yr) (yr)
0.0000 0.0000 0.0000 0.0000 1.0000 0.0000 0.0000 1.0000
0.5000 0.4162 0.1130 0.4283 0.6595 0.5161 0.4031 1.0000
1.0000 0.7092 0.4146 0.7702 0.4920 1.1016 0.6870 1.0000
1.5000 0.9355 0.8776 1.0781 0.3924 1.7838 0.9062 1.0000
2.0000 1.1200 1.4900 1.3693 0.3263 2.5748 1.0848 1.0000
2.5000 1.2756 2.2453 1.6509 0.2793 3.4809 1.2356 1.0000
3.0000 1.4103 3.1399 1.9266 0.2441 4.5059 1.3660 1.0000
3.5000 1.5290 4.1712 2.1983 0.2168 5.6522 1.4810 1.0000
4.0000 1.6350 5.3377 2.4673 0.1949 6.9215 1.5837 1.0000
4.2418 1.6825 5.9500 2.5967 0.1859 7.5797 1.6297 1.0000
Exhaust Velocity = 0.88301; Proton MeV = 1060.5
4.5000 1.6483 6.6085 2.5029 -0.1337 8.2871 1.6786 1.0467
5.0000 1.5731 7.8121 2.3070 -0.1593 9.5904 1.7783 1.1570
5.5000 1.4818 8.9119 2.0867 -0.1971 10.7986 1.8867 1.3067
6.0000 1.3656 9.8934 1.8314 -0.2584 11.9003 2.0069 1.5254
6.5000 1.2058 10.7345 1.5199 -0.3750 12.8791 2.1446 1.8873
7.0000 0.9482 11.3959 1.0968 -0.6833 13.7090 2.3132 2.6602
7.5000 0.4669 11.7924 0.4840 -1.0011 14.3512 2.5588 5.0527
7.9516 0.0000 11.9000 0.0000 -1.0003 14.8195 2.9195 9.4145
Final sail furl = 0.160
Furling the sail to limit the deceleration to 1 g begins about
0.7 yr before arrival at the destination. Even with furling,
the deceleration time is shorter than the acceleration time for
the same distance. The average deceleration is greater than the
average acceleration because the decreasing mass in the deceler-
ation phase overrides the reduction in thrust that comes from the
competition between radiation-pressure push and exhaust retro-
thrust.
Even though the trip takes about 8 years of ship time, or about
15 years of Earth time, the total job of the emitter is over in
less than 3 Earth years (consistent with Steve's original predic-
tion, even including the deceleration phase).
I'm not totally comfortable with the calculated mass-ratio
values; they seem too low. I haven't found any analytical or
computational shortcomings, however. The approach laid out here
should be regarded as exploratory, setting up the framework of
the analysis so that future efforts need be devoted only to
refining the details.
There may be performance gains from changing the turnover point
from the halfway point. For example, as the turnover point is
moved earlier than the halfway point, a higher exhaust velocity
(lower [factor] and lower thrust) is allowed, which, together
with the lower peak velocity, calls for a lower mass ratio.
The performance results tabulated above are for 100 percent
conversion efficiency (eta = 1.0) from received power to exhaust
power. The effects of reduced conversion efficiency (eta less
than 1) on required exhaust velocity, final sail furl and, most
importantly, required mass ratio are given in the table below:
eta exhaust velocity final sail furl mass ratio
1.0 0.883 0.160 9.41
0.9 0.849 0.105 15.44
0.8 0.809 0.060 29.05
0.7 0.760 0.029 67.57
Producing high efficiency of conversion from received power to
exhaust power may be as challenging (and as crucial to the
success of the concept) as constructing the emitter or the sail.
Rex
ADDENDUM
PROGRAM SAILTRIP !9/17/96
101 FORMAT(3X, 12HParameter = )
102 FORMAT(2X, 5HTship, 4X, 5HTheta, 6X, 4HDist, 4X,
& 8HProp Vel, 5X, 5HAccel, 4X, 6HTEarth, 4X, 5HTemit, 2X,
& 6HMratio)
103 FORMAT(1X, F6.4, 3X, F6.4, 3X, F7.4, 4X, F8.4, 3X, F7.4,
& 3X, F7.4, 2X, F7.4, F8.4)
104 FORMAT(2X, 19H Exhaust Velocity =, F8.5,
& 14H; Proton MeV =, F7.1)
105 FORMAT(3X, 17HFinal sail furl =, F8.3)
ETA = 1.
AGO = 1.
AO = 1.0324 * AGO
DSTAR = 11.9
DTA = 0.01
DTS = 0.001
DTP = 0.5
1 CONTINUE
WRITE(*,101)
READ(*,*) PAR
IF(PAR .EQ. 0.) GO TO 99
ETA = PAR
C.....Acceleration Phase
IT = -1
TIM = -DTA
ACC = AGO
WRITE(*,102)
2 CONTINUE
IT = IT + 1
FT = IT
TIMN = TIM
TIM = DTA * FT
ITSN = TIMN/DTP
ITS = TIM/DTP
IF(TIM .EQ. 0. .OR. ITS .NE. ITSN) THEN
FITS = ITS
TIMI = DTP * FITS
ARG = AO * TIMI + 1.
THETI = LOG(ARG)
DISTI = 0.5 * (0.5 * AO * TIMI*TIMI + TIMI - THETI/AO)
UI = 0.5 * (ARG - 1./ARG)
AGI = AGO/ARG
TEI = 0.5 * (0.5 * AO * TIMI*TIMI + TIMI + THETI/AO)
TEMI = TEI - DISTI
FMR = 1.
WRITE(*,103) TIMI, THETI, DISTI, UI, AGI, TEI, TEMI, FMR
END IF
THET = LOG(AO * TIM + 1.)
DISTN = DIST
DIST = 0.5 * (0.5 * AO * TIM*TIM + TIM - THET/AO)
IF(DIST .GT. 0.5*DSTAR) THEN
DRAT = (0.5 * DSTAR - DISTN)/(DIST - DISTN)
TST = TIMN + (TIM - TIMN) * DRAT
ARG = AO * TST + 1.
THETAT = LOG(ARG)
DISTT = 0.5 * (0.5 * AO * TST*TST + TST - THETAT/AO)
UT = 0.5 * (ARG - 1./ARG)
AGT = AGO/ARG
TET = 0.5 * (0.5 * AO * TST*TST + TST + THETAT/AO)
TEMT = TET - DISTT
FMT = 1.
WRITE(*,103) TST, THETAT, DISTT, UT, AGT, TET, TEMT, FMT
GO TO 10
END IF
GO TO 2
C.....Deceleration Phase
10 CONTINUE
IPRN = 0
ND = 0
VEXH = 0.9
11 CONTINUE
ND = ND + 1
IF(ND .EQ. 2) VEXH = 0.85
GEXH = 1./SQRT(1. - VEXH*VEXH)
RELF = (GEXH * VEXH)/(GEXH - 1.)
TS = TST
TE = TET
FM = FMT
THETA = THETAT
X = 0.
ACC = 1.0324 * AGT
FURL = 1.
12 CONTINUE
U = SINH(THETA)
GAM = COSH(THETA)
XN = X
X = XN + U * DTS
PR = FURL * FMT * AO * EXP(-THETA)
TBM = PR
TEX = ETA * PR * RELF
FMN = FM
FM = FMN - TEX * DTS/(GEXH * VEXH)
TSN = TS
TS = TSN + DTS
TEN = TE
TE = TEN + GAM * DTS
ACCN = ACC
ACC = (TBM - TEX)/FMN
AGE = (ACC + (ACC - ACCN))/1.0324
IF(AGE .LT. -1.) FURL = -FURL/AGE
THETAN = THETA
THETA = THETAN + ACC * DTS
ITSN = TSN/DTP
ITS = TS/DTP
IF(ITS .NE. ITSN) THEN
FITS = ITS
TSI = DTP * FITS
TSIR = (TSI - TSN)/(TS - TSN)
THETAI = THETAN + (THETA - THETAN) * TSIR
UI = SINH(THETAI)
XI = XN + (X - XN) * TSIR + DISTT
AGI = (ACCN + (ACC - ACCN) * TSIR)/1.0324
TEI = TEN + (TE - TEN) * TSIR
TEMI = TEI - XI
FMI = FMN + (FM - FMN) * TSIR
FMR = FMT/FMI
IF(IPRN .EQ. 1) WRITE(*,103) TSI, THETAI, XI, UI, AGI,
& TEI, TEMI, FMR
END IF
IF(THETA .LT. 0.) THEN
THETR = THETAN/(THETAN - THETA)
TSF = TSN + (TS - TSN) * THETR
THETAF = 0.
DVEXH = VEXHP - VEXH
VEXHP = VEXH
DISTFP = DISTF
DISTF = XN + (X - XN) * THETR + DISTT
IF(ND .GT. 1 .AND. IPRN .EQ. 0) THEN
VEXH = VEXHP + (DSTAR-DISTF) * DVEXH/(DISTFP-DISTF)
IF(ABS(DISTF-DSTAR) .LT. 0.0001) THEN
IPRN = 1
GEXH = 1./SQRT(1. - VEXH*VEXH)
PMEV = 938. * (GEXH - 1.)
WRITE(*,104) VEXH, PMEV
END IF
GO TO 11
END IF
PVEL = 0.
AGF = (ACCN + (ACC - ACCN) * THETR)/1.0324
TEF = TEN + (TE - TEN) * THETR
TEM = TEF - DISTF
FMF = FMN + (FM - FMN) * THETR
FMRAT = FMT/FMF
IF(IPRN .EQ. 1) THEN
WRITE(*,103) TSF, THETAF, DISTF, PVEL, AGF, TEF, TEM,
& FMRAT
WRITE(*,105) FURL
GO TO 1
END IF
GO TO 11
END IF
GO TO 12
99 STOP
END