# Re: New idea Laser launcher/scoop systems

```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?

>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).

>>> 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.

>Don't know about using nozzles to accelerate the plasma.

Or to decelerate (depending on the initial velocity)

Tim

```