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Re: Questions about your letter to SD

Hello Steve,

>I am fond of using what relativity texts call "geometrized
>units".  In geometrized units time is also measured in units of
>distance by multiplying conventional time by c (the speed of
>light), and velocity is therefore dimensionless.  In more
>advanced uses even mass is measured in units of distance (using
>the conversion factor G/c^2)!
>The expression in conventional units, off the top of my head,
>seems like it should be:
>v / c = p * c / E
>v = p * c^2 / E

I had seen you use these geometrized units and understood most of them. It's
a bit tricky for me since I always used the conventional notation.
Unfortunately you are one of the very few that use this notation.
Where I made the mistake was that I had not translated the E to E/c^2.
Thank you for the explanation.

> > >about a 2:1 Earth time/spacecraft time ratio.  Accelerating to
> > >0.866c at 1 g will take about 1.14 years of spacecraft time.
> > 
> > I think it is 1.27 years, but it does not make much difference in the
> > calculated values.
>As far as I know the formula I use is correct:
>v = tanh(a' * t')
>v = velocity (unitless, as a fraction of c)
>a' = spacecraft acceleration in its local frame (units of 1/distance)
>t' = spacecraft elapsed time (units of distance)
>Again, this uses geometrized units; for conventional units you'd
>use the formula:
>v = c * tanh(a' * t' / c)
>Perhaps you calculated the global frame time rather than the
>spacecraft local time?

No, I used the same formula you did and still get 1.277 years (a=9.8 c=3E8,

>I believe that I discussed why I considered both fuel and exhaust
>velocity, by citing the example of burning the fuel and using the
>energy to accelerate a lower quantity of reaction mass to a
>higher velocity.  I still believe that such a case results in
>much poorer behavior; at least by the analysis I've done it's
>better to burn the fuel and eject all the waste products at a
>lower velocity than to burn the fuel and leave most or all of the
>waste products on the ship.

I'm sorry, I must have missed that part when I read your letter.
Dumping the waste products is obviously better than dragging them with the

>Now that you've induced me to think about it further, I'd have to
>say that the efficiency of the mass-to-energy conversion from
>burning the fuel is really what's more critical than either the
>fuel type or the reaction exhaust velocity.  Using hydrogen for
>fuel and burning it with fusion means that you can't convert more
>than 1/300 of the fuel mass to energy (using your figure), and
>hence no more than 1/300 of the fueled spacecraft mass converted
>to energy can be applied to accelerating the payload.  A
>self-fueled ship simply cannot reach high relativistic speeds
>with a reasonable fuel-to-payload ratio without being able to
>convert a large fraction of the fuel to energy.

Indeed, the only feasible option for self-fueled ships would be a matter &
anti-matter mixture.

Do you know how efficiently energy can be transferred into anti-matter these
days? Or to put the question in an other way, what are the input energies of
these supercolliders and how many anti-particles can be isolated after a
It's hard to get accurate data, so I haven't a clue of the efficiency to
create anti-matter.

The energy needed for the anti-matter creation probably has to come from
solar power and fusion.

The reason that I followed your derivation quite thorough is that about a
week ago I had finished calculations which should give the same results. My
approach is completely different from yours. My main goal was to calculate
the energy needed for such a trip. I had planned to send it to SD soon. I'd
appreciate it very much if you would look at it before I send it to SD. It's
mainly formulas and about 10Kbyte long. If you are interested I will send it
to you.

Greetings Timothy