```ReplyTo  : Kelly

>We seem to have talked oiurselves out of all the engine concepts?  We can't
>launch fuel over interstellar distences.  The MARS system seems to need
>impossibly large collector arrays, and eiather system would cook the ship.
>
>bummer.

Cheer up Kelly, we will find a way, we can always make a multi-generation
ship, I guess.

Subject  : Photon energy

OK, I agree with you about the aiming and timing 'problem' I wrote you

>> What matters is the energy:momentum ratio of the beamed power.
>> For the photons in the beam that ratio is:
>>
>>  E:p = c:1   where c=3E8
>>
>> For the Asimov moving at 0.7 c that ratio is:
>>
>>  E:p = 0.4c:1
>>
>> The beam needs to have about 1/0.4=2.5 times less momentum than the Asimov,
>> but that is still a lot, so I can't agree with you that the momentum of the
>> beam is neglectable.

WARNING!
I made a mistake here, I forgot that the momentum of the Asimov increased
while it was receiving photons. So the best thing to do is forget the above.

>no, I am talking about the momentum of the photons as opposed to the
>momentum of the ions the "Asimov" will eject as exhaust.  Those will be
>Hydrogen ions or maybe Xenon moving at .9996 or (.99996, depending on how
>much energy you can invest) C  at these speeds, a small mass flow is
>sufficient to slow us down (or speed us up depending on which phase of
>the mission we are in) at a constant 1 G.

This was clear to me, but rethinking this made me realize that your method
can't work. Because adding momentum to the Asimov will only make it move
faster. Transforming it to reverse momentum would surely break one of the
basic physic laws.

In formulas:

- You shoot some photons at the Asimov, that will give an energy Up=p*c
where p is the momentum of the photon.
- To accelerate a mass M to speed v to decelerate the Asimov, we need an
amount of energy Uk=M*c^2*(gamma-1).
- We use all the energy of the received photons for that acceleration,
so  Up=Uk --> p*c=M*c^2(gamma-1) --> p=M*c*(gamma-1)
- Thus to accellerate a mass M to speed v we need to receive photons which
will add us a momentum of p=M*c*(gamma-1).
- But of course we can subtract some momentum that we created by shooting
that mass M away. That momentum equals to p=gamma*M*v.
- Now the problem arises because by receiving photons we gain more momentum
that we loose by shooting that mass M away.
In physics: gamma*M*v is always less than M*c*(gamma-1)

Subject  : Solar array

>increasing the number of transmitters will:
>reduce the amount of photon thrust that each transmitter is subjected to.
>reduce the amount of solar array that each trnsmitter must have.
>reduce the amount of beam jitter (by averaging the errors, they are reduced)
>reduce the heat load of each transmitter.  (the non-Sol side of the solar
>                                panels will make an excellent heat radiator)
>increase the total cost of the mission (hey, you don't get nothing for free)

But it won't reduce the total solar array, which is really big:

Total Solar Power                      : 4E26 Watts
Area of a globe with radius 1E9 metres : 7.9E17 square metres
-> Solar power per square metre        : 5E8 Watt

Mean amount of power needed by the Asimov : 1E18 Watt
-> Size of solar array : 1E18/5E8 = 2E9 square metres
= disc with radius 2.5E4 metres.

Remember 1E9 metres is quite near Sol. You wouldn't like to be there in your
space suit, because 5E8 Joules would be added to your body temperature every
second.

Subject  : Overheating

The main solution for the energy-leak in your letter was 'radiate it away'.

You say the antenna mesh can radiate it away. You assume the leaking-energy
can be radiated away before it can do any damage. Let us assume that the
leaking-energy is ordinary Ohmic resistance and thus emerges as heat in the
mesh. I think that before it can radiate away the mesh is molten or has
disappeared altogether.
So my problem is wheter one can radiate it away fast enough. And if you seem
to be able to guide the energy so well, why not use it.

The amount of energy leaving the crew quarters should not be that much, It
will probably some kind of thermos flask.

```