# Re: New idea Laser launcher/scoop systems

```At 12:24 PM 3/13/96, Timothy van der Linden wrote:
>To Kelly,
>
>>> >My numbers showed that unless the specific impulse were geting out of the
>>> >Explorer classes engines is over 2,000,000 we don't get to go.
>>
>>> I think the specific impulse can be calculated as follows:
>
>F*t = p -> M*g*t = m*Vexh
>
>Specific impulse = M/m = Vexh/(g*t)
>
>This formula says: The momentum of a mass m with velocity Vexh is equal to
>the acceleration of a mass M during t seconds.
>
>1 kg of fusion fuel would work most efficient if it was accelerated to the
>optimum velocity where I made a table for 2 days ago. This gives us Vexh=0.085c
>
>We need to take g=9.8 m/s/s and t=1 second.
>
>That gives for the specific impulse = 0.085c/(9.8*1) = 2.6E6
>
>So while this may look much, I think it is right. Why do you think it
>is not OK?

Well one problem I have is since I don't know ow to find what speed the
reaction particals will be at after the reaction (I have Mev power levels
but I'm not sure how to use them), The equation doesn't tell me much.  For
example, a pound of fuel will fuse and realest its particals at x Mev.
What speed is that?  How much electric power can the generator convert that
to?  Using that electricity, how much mass can we accelerate to the ideal
velocity?

>>The question was, can you get more thrust out of a pound of fusion fuel by
>>using a reactor to convert it to electricity, and use the electricity to
>>accelerate a reaction mass (ignoring engineering losses).  Or would
>>converting the fuel to high speed plasma get you more thrust.
>
>If you ignore energy losses, it doesn't matter, as long as you can regulate
>the exhaust velocity of the plasma (which should not be too hard).

I woud think the velocity would be fixed. I can't think how to increase it
without adding external power.  Decreasing would probably involve taping
some of its power off, and giving it to a larger volume of reaction mass.

>>>> Given that the energy in the fusion reactors was all
>>>> turned into kinetic energy in the exaust plasma, I had assumed using it
>>>> directly would be most efficent.
>>> Ofcourse that depends on the possible efficiency and
>>> on the exhaust speed of the plasma. Can the latter be
>>> regulated by making the outlets smaller or broader?
>>
>> Efficency?  Efficency of what?
>
>The efficiency of the conversion to electricity and then back to kinetic
>energy.

Conversion to electric was about 99% efficency.  I would think a mag
accelerator would be very high efficency.  Actually in our case it would
have to be or we'ld melt the ship.

>>Don't know about using nozzles to accelerate the plasma.
>
>Or to decelerate (depending on the initial velocity)
>
>Tim

Kelly

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

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