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*To*: starship-design@lists.uoregon.edu*Subject*: starship-design: Deceleration of sail pushed by constant-power beam*From*: DotarSojat@aol.com*Date*: Fri, 20 Sep 1996 02:46:46 -0400*Reply-To*: DotarSojat@aol.com*Sender*: owner-starship-design

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

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