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Re: starship-design: Deceleration scheme



Steve,

> > Using just as much antimatter as matter is a waste of energy for most
> > velocities!
>
>I meant that the fuel mass (which is matter and antimatter in equal
>parts) is 4-5 times the mass of the payload.

I know.

>In a ship that carries its own fuel you get the lowest fuel-payload
>ratio by having the highest-velocity exhaust.  Photons are optimal; the
>mixture of photons and high-velocity particles you get from
>matter/antimatter reactions is about the best you can do.  If you react
>a quantity of antimatter with a larger quantity of matter then you get
>slower exhaust velocity and a higher fuel-payload ratio, and things get
>more or less exponentially worse with decreasing exhaust velocity.

I'm not sure what you mean with "things" but it simply is not true that
higher exhaust velocities are best.

Each final starship velocity has its own optimum for particle exhaust
velocity. This optimum is determined by finding the minimal energy requirements.
So there is a certain exhaust velocity where the energy requirements are
lowest, given a certain final starship velocity.

This means that having an equal quantity of matter & anti-matter is usually
not ideal. In general having a small portion of anti-matter and a larger
portion of matter is the best. The antimatter combined with an equal amount
of matter will provide the energy to exhaust the rest of the matter.
This indeed means that the total fuel mass becomes more, but it will not use
up more energy, just more mass. The essential is, that mass is free, while
energy is not.

There is a delicate balance between the energy available, and the fuel that
has to be carried with the starship.
I've not found an easy way to explain it. I think the reason that it is so
hard to explain this balance, is simply because we are lousy in doing
differential equations by head.

You can check out my page about it. The calculations have been there for
quite a while now:

http://www1.tip.nl/users/t596675/sd/calc/calc.html

Timothy