[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

Re: Comparison of Sail vs. RAM (Was Laser Aperture Size)

>To recap the problem:
>if a chemical or fusion rocket is used, (with extremely low Ve) then the 
>ship needs planatary sized fuel tanks
>if the Ve (exhaust Velocity) is increased to .9996 (or higher) of C, then 
>mere Kilograms/Sec of RM is needed.  However, in order to do that, one 
>needs lot's of energy, and the momentum of the maser overcomes the 
>momentum of the engine.
>Somewhere in between is a solution.  that is to say, that at some Ve (I'd 
>guess in the .6 to .8 of C range)  The Energy requirements are low enough 
>to reduce the Maser thrust to a value far enough below the engine thrust 
>to allow a not-too-big RM tank to handle it.  Here is a conceptual graph 
>to illustrate what I mean.
>Note:  This Graph is totally arbitrary and qualitative.  I'm fairly 
>confident that the oo point (where the lines intersect) exists, I just 
>have no clue as to how to find it (aside from trial and error which is 
>what I intend to do this weekend)

Since you probably don't have the right formulas trial and error will not
work either, finding these may indeed by a bit nasty. Since I've done such
calculations before, it will be a bit easier for me.

>Any help will be greatly appreciated..

Well, I suppose you want to find the solution that consumes the least energy
(not necessary the least reaction mass).

So we need to optimize the energy.

Energy needed to decelerate a ship depends on it's empty mass, its start
velocity, the exhaust velocity.


(1)  g =            Vexh
          Sqrt[1 - ------]

             M[t] a
(2)  j[t] = --------
             g Vexh

          c           Vstart 
(4)  T = --- ArcTanh[--------]
          a             c

(5)  M[t] = Mo + | j[t] dt

                    a     /
(6)  M[t] = Mo + -------- | M[t] dt
                  g Vexh  /

Solving this differential equation gives:

                    a (T - t)
(7)  M[t] = Mo Exp[-----------]
                     g Vexh

           2      /
(9)  Ek = c (g-1) | j[t] dt

           2                a T
(10) Ek = c (g-1) Mo Exp [--------]
                           g Vexh

           2                 c              Vstart
(11) Ek = c (g-1) Mo Exp [-------- ArcTanh[--------]]
                           g Vexh              c

Finally... E(Mo,Vstart,Vexh)

Now we only need to find its minimum, that's easy: solve dE/dVexh=0

Well, that's not easy but the minima are:

Vstart  Vexh optimal  Fuel:ship-ratio  Energy per kg of ship (in Joules)
 0.1        0.05           7.4            8E14
 0.2        0.10           7.4            3E15
 0.3        0.15           7.3            8E15
 0.4        0.21           7.2            1E16
 0.5        0.27           7.1            2E16
 0.6        0.34           6.9            4E16
 0.7        0.41           6.8            6E16
 0.8        0.51           6.4            9E16
 0.9        0.64           5.9            2E17
 0.99       0.87           4.5            4E17 
 0.9996     0.957          3.6            8E17


P.S. Rex, I suppose some of these formulas will look familiar