```Bb accident, I mailed this letter only to Kelly yesterday. So here is it for
the rest of you.

Subject : (Star)Wars

> > One very consistent problem in LIT over the last year has been
> > a very limited interest in the engineering realities of a
> > situation, and to much fondness for endless equation wars.
>
>These are not equation wars; as this message shows, you don't
>always have to use math to talk about physics.  We are talking
>about very real physics concepts, and I'm afraid you are the one
>who has a few important ones wrong.
>
>Before we can do the detailed engineering for a starship, we need
>to understand what physical constraints it will be under.  Even
>the best engineering cannot violate the laws of physics.  If we
>are engaged in "equation wars", it is because we are trying to
>figure out the limits of what is possible before we do detailed
>design work based on faulty assumptions.  This is the most
>fundamental interest in engineering realities you could want.

Steve, I've nothing to add, you're completely right.

Timothy

===============================================================================

>From    : Timothy
Subject : Prelaunching

>>> This method of prelaunching fuel packets will cost the
>>> same amount of energy that is needed if you take those
>>> same packets with you from the start.
>
>True, but since the ship doesn't have to supply the energy, it saves an
>incredable amount of fuel.

The ship does not have to supply the energy! It will leave fully fueled. It
is still accelerated by some kind of photon beam.

>To put it bluntly no fusion powered ship could
>carry enough fuel, to accelerate itself and its fuel up to high reletivistic
>speeds.  Estimates are that a ship would need a 1,000,000 to 1 fuel to ship
>mass ratio to get up to \$10 of light speed.  But if the ship doesn't have to
>carry its fuel, a 200 to 1 ratio could get you most of the way to light speed
>(or was that 1/3rd light speed.  Been a while since GES ran the numbers off
>for me.)

Huh, if it doesn't carry its fuel, how do you calculate a ratio?

!! Please be specific if you mean FUEL or REACTION MASS !!

>>> The problem is that there are no accurate numbers of
>>> the density of interstellar debris. So any number is almost
>>> a guess.
>
>Thats been a constand problem for us.  How do you design a ship to travel
>through something you know next to nothing about?

It would be best if we could find a solution without using the interstellar
particles, but at the same time we should keep in mind that we have to
protect us against it. This may sound a bit contradictory but a general
solution would be best.

===========================================================================

>From    : Timothy
Subject : Plasma mirror

>>> Replenishing the "mirror" will probably take lots of ions
>>> or in other words mass that has to be taken with us.
>
>Agreed, but it don't have a handel on the amount of mass.

Why doesn't it have to do with the amount of mass? Ions are particles too.
They may have small masses, but if you have enough of them you could build a
complete dragon-fly sail.

>>> Also using such a light (non heavy) sail will mean that
>>> the "mirror" is accelerated a lot and lot of energy is lost
>>> due to the Doppler effect.
>
>Given that the mirror is the surface of a plasma, and said plasma is being
>continuously being replenished.  I'm not sure the reflective "surface" is
>actually moving?  Althou obviously the particals in the plasma are moving
>(and accelerating) rapidly.  Are the micro waves reflecting off the
>particals?  Or off the area where the plasma is ionized enough to reflect
>them?

Probably all the way in between. But that is not important here, because a
reflection means transfer of momentum, and with it energy. Since the plasma
will be accelerated a lot, it will retrieve a lot of energy.

>>> Also I have doubts how well the "mirror" reflects, ionized
>>> particles attract or repell each other so, it won't take long
>>>  befor the "mirror" has destroyed itself.
>
>The ions in a plasma of the same material wil repell each other.  You might
>be able to do some magnetic tricks to hold it, but it probably wouldn't be
>worth the trouble.
>
>[ Hum --- I wounder if you could wiggle the exausting plasma to for a laser
>to get back some of the energy? ]

Don't talk about plasma as if it where some easy to control stuff. The
movement of the plasma itself would create magnetic fields. It's like
boiling water but much worse.

>As to reflectivity.  That would depend on the nature of the plasma and a lot
>of other variables.  This is probably not a question we could easily figure
>out for ourselves.

>>> Shock wave? I don't understand, please explain again.
>
>You have a mass of plasma being hit with E18 of energy.  It will be HOT, and
>highly ionized.  It will be explosivly expanding.  The light pressure of the
>beam (or the feed mass) will probably keep it from flowing straight up the
>beam.  But it will be moving rapidly to the sides and forward or the ship.
> We should be able to tap this for thrust.

My guess is that this shock wave will move mostly forward instead of backward.

You seem to want to do the same thing Kevin did: Make a easy thing
complicated and so loose complete control over what you are doing.
Whatever method you can think of, it will take just as much energy that the
dragon-fly-sail will use.

==============================================================================

>From    : Timothy
Subject : Floating <> flying

>>> Yes, in fact is does not take any energy to keep floating a
>>> few metres above Earth's surface.
>
>(???????!!!!!!)  SAY WHAT!!  Unless we're on differnt planets you have to
>generate or disapate as much energy as to generate 9.8 m/s^2 of acceleration.

No, I wasn't in the Acme physics store...

formulas:

F = m a
U = F s = m a s
P = U / t

Where F=force, U=energy, m=mass, a=acceleration, s=distance_moved, t=time_needed

In 4.3 seconds I will evenly pull a mass of 1.3 kg 2.6 metres up in Earth's
g=9.8 m/s^2

That will take a force F=1.3*9.8=12.74 Newton
and an amount of energy U=12.74*2.6=33.124 Joule
and will take a power of P=33.124/4.3=7.703 Watt

So if we are talking about weightlifting, there are several phases.

First the weightlifter has to get the weight from the ground to up his head.
That is where the real energy is needed, since there is a distance of
movement involved.
After he/she has lifted the weight and keeps it above his/her head the real
work is done. All that the weightlifter has to do is keep his muscles in
place. But that energy is not added to the weights.

Often it is more difficult to get the weight up than to keep it there, that
is just because of this reason.

```