I am going to run a DS2 game at the GZG-ECC. It is based on the Moon.
I need some help:
1. First, I need some help with DFFG's in a vacuum. At first, I was of the
opinion that they would have a longer range, since it would not diffuse into
the (nonexistent) atmosphere. In this, I pictured that it was held in magnetic
containment for the full distance to the target. After a closer reading, the
plasma is not contained once it leaves the barrel of the gun. Having never
taken a physics class, my question is: Would the plasma remain as a cohesive
ball of gas, maintain the same dispersal pattern or be even more quickly
dispersed?
2. Second, the Moon is about 1/6th the mass of the Earth. However, it is
less dense (and contains fewer heavy metals). But I have been unable to find
what the circumference of the moon. I am trying to figure out the distance to
the horizon on the Moon. If anyone can provide information or insight on this
matter, it will be appreciated.
> 1. First, I need some help with DFFG's in a vacuum. At first, I was of
I'd think it would tend to disperse more rapidly, no air pressure to help
contain it, but I have an English and history degree, so what would I know?
> 2. Second, the Moon is about 1/6th the mass of the Earth. However, it
Mass and density have nothing to do with horizon; and Luna shouldn't have
1/6 MASS, but rather 1/6 GRAVITY. I don't happen to recall the formula
for a horizon, but you can find it in the book by Steven Gillette on
worldbuilding, from the series edited by Ben Bova.
Hi there Brian!
Still working on your Moonbase Xi scenario?
I still think that the DFFG will have increased range bands due to the lack of
atmospheric scattering of the beam...but thats just a guess.
As you say, the Moon will have a greatly reduced horizon so a target may be
only engageable out to a DFFG 4's (new) Mid range band.
BTW, those hostile environment suits I mentioned on the list should work well
for your scenario. I pushed some figures around the other day and 2 chits to
destroy an element seems about right. They are fragile, but it IS the moon
after all!
Cheers,
> I am going to run a DS2 game at the GZG-ECC. It is based on the Moon.
I'd think it would be a wash. Quicker expansion because there's no atmospheric
pressure, but slower expansion because of the lack of atmospheric drag and
turbulence. So leave it as it is (it's simpler that way, anyway). Lasers (as
if it matters!) would have a greater effective range. Slug throwers (HVC, MDC)
would have a greater effective range due to the absence of atmospheric drag,
but the range to the horizon might become an issue.
- Sam
Samuel spake thusly upon matters weighty:
> >I am going to run a DS2 game at the GZG-ECC. It is based on the Moon.
Or is it with slug throwers that the simple mechanical component of the system
precludes an increase in the effective range? The range limitation may not be
'projectile dynamics' based, but rather a result of not being able to aim
reliably enough at distant targets with projectile weapons. If that were the
case, then maximum range at which a hit might be scored might increase, but
effective range would not change. I'd tend to offer a suggestion of doubling
the width of the outer rangeband without affecting the others.
Just an idea.
/************************************************
> At 22:16 12/13/98 -0500, you wrote:
Calculating the distance to the horizon is not that tricky if you assume that
the radius of the surface is much greater than the hieght of the observer.
This is most definitely the case for combatants on a planetoid. You can simply
use this approximate formula:
DISTANCE_TO_HORIZON = (2*RADIUS_OF_SURFACE*HEIGHT_OF_OBSERVER)^1/2
i.e. the distance to horizon is proportional to the square root of the height
of the observer. Note that this assumes a flat, spherical surface free of
terrain that blocks LOS.
Hence, for the moon which has a radius of 1738 km, assuming a 2m high
observer, we get:
Dh = (2*1738000m*2m)^1/2 = 2637m
> Brian Bell
ps. It's not really that hard to calculate this distance for really tall
observers (or really small radii of curvature), but I'm feeling lazy and I
know for a human size observer the difference is negligible. If you feel that
the exact calculation is the only way for you, then I could sketch out the
exact solution, to be left as an exercise to the reader;)
> At 22:16 13/12/98 -0500, you wrote:
My first idea was that once atmosphere had been gotten rid of the range should
increase provided you don't have a magnetic field about to stuff this up. If
it required magnetic containment in the barrel then it would probably need it
outside (charged particles and all). I think the plasma would remain cohesive
longer on the moon than Earth (the moon has no magnetic field to speak of).
Double range would probably be fair. Don't ask me about the size of the moon.
Tony. twilko@ozemail.com.au
> 2. Second, the Moon is about 1/6th the mass of the Earth. However, it
> At 22:16 12/13/98 -0500, I wrote:
One thing to be aware of when using this formula is that this give the
distance to the horizon. You could actually see objects past this distance as
long as they have a height greater than zero. In order to determine the
maximum range at which a target is visible, you would need to use:
MAX_VISIBLE_RANGE( Ho, Ht ) = ((2*R)^0.5)*(Ho^0.5 + Ht^0.5)
Ho = observer height Ht = target height R = radius of curvature of surface
In other words, the sum of the distances to the horizon for both the observer
and the target. Still, I would only use the distance to the observer's horizon
when firing, or at
least give the target hull/turret down status from the observer's
horizon to the MAX_VISIBLE_RANGE.
This is the same effect that led to the realization that the
earth was round - seeing only the masts of ships over the
horizon.
You can see that infantry could use the limited horizon more effectively than
tanks by going to ground, etc. This would also apply to short people. Towers
would be a necessity for all static defences (that is, if you want to see the
enemy coming).
Hope this helps!
> Tony Christney <acc@questercorp.com> wrote:
Would this be right? By the way, anyone know the radius of Earth?
> Hence, for the moon which has a radius of 1738 km, assuming a
Would
> this be right?
Earth stats:
Mass (g) = 5.97x10(**27)
Density (g/cm**3) = 5.5
Diameter = 12,756 km
Mk
> Andrew Martin wrote:
Would
> this be right?
This is what I get also, so I guess your calculator is working OK;)
> On Sun, 13 Dec 1998, Brian Bell wrote:
the trouble with the DFFG, and plasma weapons in general, is that they are
pure PSB. devastatingly effective, but utter fiction. physics thus has
very little to say on the question of range - go with whatever feels
right (i would say more range).
Tom