Given: 1) Geosynch is roughly 22,000 km (if I'm not way off base) 2) Given
geosynch is stationary wrt a point on the earth 3) Given a lower orbit will be
(of necessity) faster? I think
4) Given the lower the orbit, the faster the orbit - so the further
off geosynch, the less time you'll be able to see a given point on the earth
before going out of arc
Can some of the math whizzes calculate an altitude where a ship wishing to
supply ortillery support can see a given point on the earth
(we'll call it the battlefield) for a period of at least an hour -
we'll call this the support-fire window. I figure an hour is a few DS2
turns, and at least 12 SG2 turns. Probably long enough to lay down enough
ortillery to matter.
So what altitude would give us an orbit slow enough to maintain visibility on
a point for an hour before going over the horizon? I realize topography
affects this, so an answer that covers a range of altitude would be
acceptable.
I'm just curious: Are we talking about 22,000 km (geosynch)? 400 km (Hubble)?
etc.
Depending on what this window is, and the ortillery window WOULD have
to coincide with fairly well time-wise with planned attacks or no
support would be available (or at least, it might take a lot longer to
arrive). I think I figured out that a shell travelling mach 10 on average
would take about 110 mins to arrive from geosynch... a little long for a
support request. From 400 miles, the answer is something like 171 seconds.
Now, Mach 10 may not be even a reasonable guestimate
for ballistic entry projectiles - I have not modelled terminal
velocity or acceleration due to gravity to guess how long it would really
take... I just picked Mach 10 out of the ether.
If your ortillery is beam, this might be moot except that you'd want to reduce
the amount of atmosphere you fire through to ameliorate diffusion and
defraction and refraction (bending and scattering essentially). You'd still
want to be in close. But your fire missions would arrive rapidfast after the
administrivia of getting them tasked.
But if you are using missile or beam, travel time is an issue. It can be quite
sizeable.
So.... what altitude would be reasonable and give us at least an hour
as a support-window? If someone had someone is gutsy enough to bash up
two formulas, one for the support window
delta-t(support window)= ? (some f(altitude))
and
t (inbound transit) =? (some f(altitude, probably launch v and acc))
then we could get some ideas for the relationship between ortillery ship
altitude, the time the round takes (and hence how long the
support fire should take to arrive) and the time the ship is "on-call"
(the duration of the support window caused by the orbit altitude).
This might be instructive for all of us - it'll give the FTers a
feeling for how close to the surface you have to get and how long you are
useful at various heights when doing ortillery support. For DS and SG players,
it'd give some idea of the range or response times one can expect (we'll still
have to think about the delays in comms and organizational levels but that's
not hard) from ortillery and the length of time ortillery support could be
available.
It may be we can evolve some simple rules which let players choose a lower
altitude support mission (with a smaller window of availability) as a trade
off for faster response on the fire missions, or vice
versa - trade delay time for a longer ortillery support mission.
So, orbital mechanics brains.... and math whizzes.... do your
worst.... :)
You aren't taking into account a "forced" orbit.
If I can match accelerations with the accel on the ground I can pick a point
over the planets surface, match angular velocity with it, and thrust in such a
direction that my ship apparently hovers over it from whatever height I want
as long as I am outside the atmosphere.
Think of it this way, By cancelling part of the planetary gravity by thrusting
away from the planet I can change the "geosynchronous distance" to whatever I
want it to be. Now, with a 1 G accel over Earth I must thrust with the nose
pointed away from the planet so can't fire weapons at the planet. However,
with a 2 G accel my direction changing thrusters would be enough to cancel
gravity and maintain station.
There is a Niven story where this is reversed. He gets in orbit around the
planet and thrusts directly toward the planet. He thus maintains his speed
while decreasing the circle travelled so his speed with respect to a point on
the ground goes way up.
> On Tue, 30 Nov 1999 kaladorn@fox.nstn.ca wrote:
> Given:
umm... about 40 000 km last time i checked. have they changed the universal
gravitational constant again? Arthur C Clarke, in the essay in which he
proposes the geosync commsat, puts it at 26 000 M, which is 42 000 km. i
believe Arthur. he da *man*.
> 2) Given geosynch is stationary wrt a point on the earth
all true.
> Can some of the math whizzes calculate an altitude where a ship
it's a little more complex than that: when the ship is on the target's
horizon, there is a line of sight, but it's through a lot of atmosphere
and the surface at the target is parallel to the line of fire - not good
shooting conditions. what's actually needed is the length of time the ship is
within a certain angular distance of the target point: ie, the time when the
ship is within a certain distance of the midpoint of the sky at the target
(the zenith?).
> I figure an hour is a few DS2
i'll drink to that.
> So.... what altitude would be reasonable and give us at least an hour
these functions should be easily invertible if at all possible!
> This might be instructive for all of us - it'll give the FTers a
too close! 8)
> It may be we can evolve some simple rules which let players choose a
i have/had an ancient pc game where you play an alien fleet commander
attacking planets; you have teleportable shock troops and landerborne heavy
troops (the shock troops have to beam down, build the landing sites ans
survive long enough for the heavies to arrive), and orbital fire support from
your BB flagship and some cruisers: you could do just this with the orbits,
and it also affected how much planetary defence fire they got. given that you
were generally fighting battles at six sites on the planet at once, whilst
also doing strategic bombardment, it was rather tricky. a fantastic game!
i would add two things...
one, you're not considering the possibility of OTH supporting fire; after all,
classic artillery is indirect, so why not ortillery? okay, so the travel time
goes way up, but ICBMs from ships at 1000 km altitude should be a rather
effective weapon.
two, you're assuming firepower is concentrated in ships. i think the
javelin idea (explained well in NivenPournelle's 'footfall' - BUY YOUR
COPY TODAY!) is that you plaster a low orbit with autonomous javelins, and
then call down the closest ones when you need them. thus, there is always fire
support almost directly overhead. oh, and Peter F Hamilton's 'the reality
dysfunction' (bad sf, good space opera) has 'kinetic harpoons' used to great
effect. haDOOOOKEN!
tom
In a message dated 11/30/99 4:30:08 PM Eastern Standard Time,
> kaladorn@fox.nstn.ca writes:
> Given:
Item number 4 assumes that the vessel is in stable orbit--in other
words, not using its propulsion system (whatever it may be) to stay up. With
the
systems available in the FT universe (the GZG-verse?), that is not
really a
valid assumption--no one seems to worry about reaction mass, so running
your thrusters continuously over a few hours isn't a problem.
Rob
In a message dated 99-11-30 18:41:04 EST, you write:
<<
two, you're assuming firepower is concentrated in ships. i think the
javelin idea (explained well in NivenPournelle's 'footfall' - BUY YOUR
COPY TODAY!) is that you plaster a low orbit with autonomous javelins
> [quoted text omitted]
Was this the basis for B5's Centauri Mass Drivers used against the Narn Home
World? BTW: Has anyone come up with rules/stats for these in FT?
Thanks,
Perry
> Given:
That's 22,500 *miles*, or about 36,000 km.
- Sam