FT-Airless bodies

5 posts · Jan 9 2001 to Jan 10 2001

From: Bif Smith <bif@b...>

Date: Tue, 9 Jan 2001 20:12:13 -0000

Subject: FT-Airless bodies

I was watching the luna eclipse, and it raised some points in my mind-

Would beam weapons have any range modifiers firing from a airless body?, as
opposed to firing through a atmosphere with it`s attendent difusion effects?

How much of a modifier would have to be used on SML`s/missiles to
represent the power used to climb out of a gravity well, which would be
smaller on a airless body as opposed to out of a larger gravity well for a
atmosphere planet, and how much would have to be used to overcome the
atmosphere it`self (or would the atmophere be a neglible compair to the
gravity
well?).

What would be the requirements for fighters climbing out of a small gravity
well, on a airless body?

These may seam to be pointless questions, but they are for a senario I`m going
to be running, with a brown dwarf star with only one planetary body orbiting
it, which is going to be a captired airless moon. This system is going to be
placed between the NAC and ESU, and have been turned into a resuply base for
forces that operate in the area. The ESU are giong to have
to take out this base, or it will allow raiding/flank attacks while they
advance. The NAC will have some ships based here, with some static defences,
and a lot of fighters.

The base will be called MIDWAY 2 (the idea was raised from a comment about the
original midway was just a lump of rock that allowed american planes to use as
a unsinkable runway in the middle of nowhere).

From: Nyrath the nearly wise <nyrath@c...>

Date: Tue, 09 Jan 2001 19:13:12 -0500

Subject: Re: FT-Airless bodies

> bif smith wrote:

Short answer: Yes.

Long answer:

An atmosphere does reduce the range of beam weapons, but even in airless space
such weapons are subject to the dreaded Inverse Square law.

This is because a beam *spreads* as it travels.

The law says that if you double the distance, the beam strength drops by one
quarter. Doubling means a factor of two, two squared is four.
        In other words, the beam strength is 1/(d^2)

The original law was derived for searchlights and things like that. One would
think that coherent sources of light like lasers would be immune. Alas, lasers
are subject to something called diffraction, which also spreads the beam. Not
quite as bad as the inverse square law, but bad enough. The legendary Erik Max
Francis said that the practical limit of a visible light laser is a few
hundred thousand kilometer before they start suffering serious range losses.

--------start of quote---------------------
> Is the lightspeed delay always the limiting factor? Could you hit

You also have diffractions -- even lasers will diffract.  The extent of
the diffraction (the halfangle of the beam) depends on how large your aperture
is and what frequency of light you're using.

So no, in the general case, a distant, predictably-moving target will
_not_ always be vulnerable.  At extreme distances, you can bathe them
with bright light, but it probably won't cause much damage.

> But suppose you detected an object near Pluto

At any given time, Earth and Pluto will be somewhere between 38.4 au (5.76 Tm)
and 40.4 au (6.06 Tm) apart. The median is, naturally, the
Earth-Pluto distance, which is 39.4 au (5.91 Tm).

Let's say you're firing a 1 GW beam with wavelength 500 nm
(greenish-blue) -- or a frequency of 600 THz -- from a 1 m diameter
aperture.

Diffraction halfangle theta, wavelength lambda, and aperture diameter l are
related by

    sin theta = 2.44 lambda/l.

Here, lambda = 500 nm and l = 1 m, so theta = 1.22 urad.

The beam radius r starts at l/2 (half the width of the aperture), and
increases as a function of distance R as

    r = l/2 + R tan theta.

At the median distance to Pluto -- 5.91 Tm -- the beam radius has
increased to 7.21 Mm, or a little over one Earth radius (6.37 Mm).

From: KH.Ranitzsch@t... (K.H.Ranitzsch)

Date: 10 Jan 2001 08:06 GMT

Subject: Re: FT-Airless bodies

> ----- Ursprüngliche Nachricht -----
mind-
> Would beam weapons have any range modifiers firing from a airless

See Nyrath's answer about beam range modifiers in general. However, I
think he missed the point that you wanted know what the _difference_
would be between a beam shot through an athmosphere and one shot from an

airless body. To answer that: beams shot from a moon can be treated as
beams shot from a spaceship. If you have a free line-of-sight to the
target, the only difference would be the effect of gravity and that is
negligible. (unless you are shooting from a Neutron star or Black Hole
;-)

> How much of a modifier would have to be used on SML`s/missiles to
neglible compair to the gravity well?).
> What would be the requirements for fighters climbing out of a small >
gravity well, on a airless body?
> [quoted text omitted]

These would depend critically on the size of the planet and the density of the
athmosphere, and, frankly, I don't know the formulas for dealing with
athmospheric resistance.

If you want to keep it simple in game terms, you could argue that missilles
and fighters taking off from a planet or moon are equipped with extra boosters
or drop tanks that provide enough energy to reach free space. Thus, they could
be treated exactly like any other missiles or fighters.

Greetings

From: Oerjan Ohlson <oerjan.ohlson@t...>

Date: Wed, 10 Jan 2001 18:35:46 +0100

Subject: Re: FT-Airless bodies

> Nyrath the nearly wise wrote:

> An atmosphere does reduce the range of beam weapons, but

Um... well. Yes, the beam will spread out with range if the range is long
enough, but it isn't as simple as "twice the range means four times the cross
section".

It was six years since I studied ray optics so my memory is a bit hazy; in
particular I don't remember what effect beam *dispersion* has (different
phenomenon from diffraction IIRC), but from what I remember:

If the beam is perfectly "parallell" (the focal point is infinitely far away),
then it will behave as your quote describes. (I snipped the quote for brevity;
can repost it if someone missed it)

However, if you focus the beam at some point *closer* than infinity, then the
focussing and the diffraction will counteract each other until the focal point
has been reached, so the beam will "spread out" slower
than the single-aperture diffraction formula suggests. If the focal
point is close enough to the projector, then the beam will get
*narrower* until it has reached the focal point - not wider. The
cross-section will always be a finite area though; the diffraction (or
was it the dispersion?) stops it from becoming infinitely small at the focal
point.

Beyond the focal point the beam goes out of focus - ie., the focussing
and the diffraction "cooperates", so the beam diverges faster than the
diffraction formula says.

So, the big question is: where do you put the focal point of your beam? You
*want* it to be exactly at the target's location, but can you manage that...?
<g>

> KH Ranitzsch wrote:

> What would be the requirements for fighters climbing out of a small

On an airless body, the formula for dealing with air resistance is
actually *very* simple ;-)

Regards,

From: John Leary <john_t_leary@y...>

Date: Wed, 10 Jan 2001 12:39:37 -0800 (PST)

Subject: Re: FT-Airless bodies

> --- KH.Ranitzsch@t-online.de wrote:

> If you want to keep it simple in game terms, you

Even more simple:

Anything with an effective thrust of 48Gs+, has no
need to be streamlined (although it helps).

The fighter after leaving the surface takes one turn to reach orbit,
thereafter it is free to act.

Bye for now,