# Re: Optimum Interstellar Rockets

```Other than complexity, would adjusting the exaust vel for optimum at ships
current vel (I.E. as the ships speed increases.  Changing the exaust
velocity for the optimum for that speed) buy us anything?

If you look at the total amount of fuel my Explorer class needs (about
100,000,000 tons of 6Li) I'm starting to think we should at least work up
numbers for a mass conversion / anti-matter ship.  Thou frankly the idea of
making and carrying a few thousand tons of antimatter bothers me a lot.
(NOT IN MY STARSYSTEM!!!!!)  Ah, how far from a planet would we need to
keep the ship for safty?

Kelly

At 1:26 PM 4/4/96, DotarSojat@aol.com wrote:
>MEMORANDUM
>TO:      LIT/SSD Discussion Group
>FROM:    Rex Finke
>SUBJECT: Optimum Interstellar Rockets (Minimum Antimatter Fuel)
>
>
>INTRODUCTION
>
>Timothy van der Linden points out in his calc.txt that there is
>an optimum ratio of exhaust velocity to final rocket velocity
>relativistically (as I had calculated earlier for non-relativistic
>velocities --undocumented).  The existence of this optimum indic-
>ates that there is a minimum in the amount of antimatter fuel
>required to accelerate a starship to any given final mission
>velocity.
>
>This memo provides the numbers that show how the ratio of the min-
>imum antimatter mass to initial starship mass varies with the de-
>sired mission velocity at the end of the first, acceleration burn.
>
>
>ANALYSIS
>
>We define the following operative quantities:
>
>V = "apparent" velocity = starmap distance/Earth time, in ltyr/yr
>U = "proper" velocity = starmap distance/starship time, in ltyr/yr
>Vend, Uend are the velocities at the end of the acceleration burn
>     (at "burnout")
>Vexh, Uexh are the exhaust velocities
>g = relativistic energy factor "gamma" = 1/sqrt(1 - V^2)
>U = g V
>V = U/sqrt(1 + U^2)
>gend = gamma for Vend
>gexh = gamma for Vexh
>M = starship mass (= Mi initially; = Mbo at burnout)
>r = starship mass ratio = Mi/Mbo
>Ma = annihilation mass used during acceleration burn for rela-
>     tivistic rocket = twice the mass of antimatter = 2 Mam
>Mp = mass of propellant used during acceleration burn for non-
>     relativistic and relativistic rockets
>The propulsive energy efficiency (let's call it eff) is the ratio
>of the final vehicle kinetic energy to the total exhaust kinetic
>energy.
>
>Non-relativistically-
>
>     final vehicle energy = (1/2) Mbo Vend^2
>     total exhaust energy = (1/2) Mp Vexh^2
>     eff = (Mbo/Mp)(Vend/Vexh)^2
>
>Now from the rocket equation
>     Mi/Mbo [= (Mbo + Mp)/Mbo] = exp(Vend/Vexh)
>we get
>     Mp/Mbo = exp(Vend/Vexh) - 1
>
>If we set x = Vend/Vexh  to simplify, we get for the energy effic-
>iency the expression
>     eff = x^2/(exp(x) - 1)
>
>This has a maximum value 0.648 for x = 1.59.
>
>So, if the burnout velocity of a non-relativistic rocket is 1.59
>times its exhaust velocity, the energy efficiency is a maximum of
>64.8 percent.  I.e., the final vehicle energy can be no greater
>than 64.8 percent of the exhaust energy.  This limitation is not
>an important consideration for a non-relativistic rocket because
>energy is subordinate to mass.
>
>Relativistically-
>
>     final vehicle kinetic energy = Mbo (gend - 1) c^2
>     total exhaust kinetic energy = Mp (gexh - 1) c^2 = Ma c^2
>                                               (no energy losses)
>              which gives Mp = Ma/(gexh - 1)
>     but relativistically Mp = Mi - Mbo - Ma
>               Ma/(gexh - 1) = Mi - Mbo - Ma
>                          Ma = (Mi - Mbo)(gexh - 1)/gexh
>
>so the energy efficiency, which is the ratio of the final vehicle
>kinetic energy to the total exhaust kinetic energy, is
>     eff = Mbo (gend - 1) c^2/(Ma c^2)
>         = Mbo (gend - 1) gexh/[(Mi - Mbo)(gexh - 1)]
>         = (gend - 1) gexh/[(Mi/Mbo - 1)(gexh - 1)]
>         = (gend - 1) gexh/[(r - 1)(gexh - 1)]
>
>The relativistic rocket equation, in its "velocity-parameter"
>form, is
>     theta = Vexh ln r
>
>and the definition of the velocity parameter is
>     tanh(theta) = Vend
>or   sinh(theta) = Uend
>
>Note:  asinh(Uend) = ln [Uend + sqrt(Uend^2 + 1)]
>
>so   r = exp[asinh(Uend)/Vexh]
>
>With this relation we have all of the parameters to calculate
>     eff = (gend - 1) gexh/[(r - 1)(gexh - 1)]
>
>The expression for eff is evaluated using a Fortran computer pro-
>gram, OPTVEXH, a description and a copy of which are given in the
>Appendix.
>
>
>RESULTS
>
>The results of the calculations of the optimum Vexh, the maximum
>energy efficiency and the minimum ratios of antimatter mass to
>burn-out mass and to initial mass are given in the table below
>for ascending values of the mission final proper velocity Uend.
>Included in the table are values of Vend, to illustrate the degree
>of saturation of apparent velocity, and of the optimum Uexh, to
>give a value (not otherwise meaningful) to which to relate the
>Uend, in order to examine the behavior of the ratio.
>
>The extreme Uend of 5 ltyr/yr represents the final velocity reach-
>ed at a continuous acceleration, a, of one g (1.0324 ltyr/yr^2)
>over a distance of 3.97 ltyr.  (The acceleration distance
>s = [sqrt(1 + U^2) - 1]/a  .)  Only for destinations beyond about
>8 ltyr or accelerations greater than one g need one consider Uends
>greater than 5 ltyr/yr.
>
>(Note: these calculations assume no energy losses in converting
>annihilation energy to exhaust kinetic energy.  The correction for
>energy losses would be to divide the minMam values by the conver-
>sion efficiency.)
>
>Uend Vend optVexh optUexh maxeff Uend/optUexh minMam/Mbo minMam/Mi
>----non-relativistic----  0.648     1.59         ---        ---
>0.2  0.196 0.124   0.125  0.647     1.60        0.0153     0.0029
>0.5  0.447 0.291   0.304  0.645     1.64        0.0914     0.0174
>1.0  0.707 0.492   0.566  0.640     1.77        0.323      0.0535
>1.1690.76* 0.541   0.643  0.639     1.82        0.422      0.0666
>2.0  0.894 0.691   0.957  0.630     2.09        0.981      0.1211
>3.0  0.949 0.777   1.235  0.622     2.43        1.739      0.1689
>4.0  0.970 0.823   1.450  0.616     2.76        2.537      0.1972
>5.0  0.981 0.852   1.625  0.611     3.08        3.357      0.2210
>--------
>*Timothy's selection
>
>
>OBSERVATIONS
>
>The minimized amount of antimatter is a small fraction of the
>starship's initial mass, less than 25 percent for mission proper
>velocities as high as 5 light-years/year for 100 percent conver-
>sion efficiency.
>
>The maximum energy efficiency decreases slowly as mission proper
>velocity is increased, but remains over 60 percent up to a mission
>proper velocity of 5 light-years/year.
>
>The ratio of the mission proper velocity to the optimum exhaust
>proper velocity increases fairly slowly at first from 1.59 at low
>velocities to almost double at a mission proper velocity of 5
>light-years/year.
>
>The implications of the values of the optimum exhaust velocity
>need to be examined, in terms of their conversion to MeV for
>exhaust particles.
>
>------------------------------------------------------------------
>APPENDIX. Program OPTVEXH
>
>For an input value of final proper velocity Uend, the program cal-
>culates the Vend, the gend and the theta.  Then for values of Vexh
>increasing from 0.01 in increments of 0.01, the program calculates
>the eff until a maximum is passed.  The optimum Vexh is calculated
>by fitting a second-degree curve to the three points that include
>the maximum.  The value of the maximum is simply taken to be the
>value preceding the drop.  The ratios of initial antimatter mass
>to Mbo and Mi are derived from expressions above-
>
>     Ma/Mbo = (r - 1)(gexh - 1)/gexh
>        eff = (gend - 1) gexh/[(r - 1)(gexh - 1)]
>            = (gend - 1)/(Ma/Mbo)
>     Ma/Mbo = (gend - 1)/eff
>        Mam = (1/2) Ma
>    Mam/Mbo = (gend - 1)/(2 eff)
>     Mam/Mi = (Mam/Mbo)(Mbo/Mi)
>            = (Mam/Mbo)/r
>            = (gend - 1)/(2 eff r)
>
>C     PROGRAM OPTVEXH                                 4/2/96
>  101 FORMAT(2X, 21H Final Proper Vel = ?)
>  102 FORMAT(2X, 15H Opt Exh Vel = , F6.4, 18H Max Energy Eff = ,
>     &  F6.4, 17H Antimatter/Mi = , F6.4)
>  103 FORMAT(2X, 8H VEXH = , F4.2, 7H EFF = , F6.4)
>    2 CONTINUE
>      WRITE(*,101)
>      READ(*,*) UEND               !final proper velocity, ltyr/yr
>      IF(UEND .EQ. 0.) GO TO 99
>      VEND = UEND/SQRT(1. + UEND*UEND)
>      GAMEND = 1./SQRT(1. - VEND*VEND)
>      THETA = LOG(UEND + SQRT(UEND*UEND + 1.))   !asinh
>      VEXH = 0.01
>      VEXHN = VEXH
>    1 CONTINUE
>      VEXHNN = VEXHN
>      VEXHN = VEXH
>      VEXH = VEXH + 0.01
>      RN = R
>      R = 1.01
>      IF(VEXH .GT. .05) R = EXP(THETA/VEXH)
>      GAMEX = 1./SQRT(1. - VEXH*VEXH)
>      EFFNN = EFFN
>      EFFN = EFF
>      EFF = (GAMEND - 1.) * GAMEX/((R - 1.) * (GAMEX - 1.))
>C      WRITE(*,103) VEXH, EFF
>      IF(EFF .LT. EFFN .AND. VEXH .GT. 0.1) THEN
>        Y1 = EFF
>        Y2 = EFFN
>        Y3 = EFFNN
>        X1 = VEXH
>        X2 = VEXHN
>        X3 = VEXHNN
>        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)
>        OPTVEXH = -B/(2.*A)
>        AMRATIO = (GAMEND - 1.)/(2.*EFFN*RN)
>        WRITE(*,102) OPTVEXH, EFFN, AMRATIO
>        GO TO 2
>      END IF
>      GO TO 1
>   99 STOP
>      END

----------------------------------------------------------------------

Kelly Starks                       Internet: kgstar@most.fw.hac.com
Sr. Systems Engineer
Magnavox Electronic Systems Company
(Magnavox URL: http://www.fw.hac.com/external.html)

----------------------------------------------------------------------

```