Hi People,
I'm going to see what numbers I come up with. I hope Daryl doesn't mind.
> Daryl Lonnon wrote:
Or, you're like me, just trying to achieve maximum pain in the shortest
amount of time. (8-)
Let's see here...
Let's assume as above - 200m sphere target
0 velocity relative to firing ship Laser aimed at centre of vessel
Computational speed is infinite
Range of 100,000 km (1/3 of a light second)
It takes a 1/3 of a second for the firing vessel to detect the
target, the target instantly discovers it has been detected, and has
2/3rds of a second to maneuver 100m. (1/3 of a second for the pulse
to get back to the firing ship, 1/3 of a second for the laser to range
out to the target. Note, no time to compute firing solution)
Total evasion time = 2/3rd of a second
g = 10 m/s^2.
Displacement Equation: x=a*t^2
Solving for a:
100/(4/9) = 100/4 * 9 m/s^2 = 25 * 9 m/s^2 = 225m/s^2 =22.5g
Therefore, the vessel must "instantly" accelerate 22.5g to escape.
However, let's try something a little more realistic...
Assumptions - 200m sphere target
100m/s velocity relative to firing ship
10m/s acceleration relative to firing ship. (1 g)
Laser aimed at centre of vessel
Computational/Laser rotational speed is 2/3 second
(Time to swing your turret, move your ship, focus your magnets, discharge your
capacitor to firing levels, whatever.
I picked 2/3 for easy of mind math - this should properly
be k.)
Computational/Evasive speed is 1/3 second
(Time for the computer on the target to realize it has been targeted, and to
take evasive action)
Range of 100,000 km (1/3 of a light second)
It takes 1/3 of a second for the firing vessel to detect the target.
It takes the target 1/3 of a second to realize it has been pinged. The
engines are on, so assume near instantaneous throttle up. This 1/3 of a
second is taken in getting the range pulse back to the firing ship,
which takes 2/3 of a second to compute a solution, lock on, and fire,
and
a 1/3 of a second for the beam to reach the projected target point.
Total evasion time available to the target: 1 second
Displacement Equation: x= vt + a*t^2
xoriginal = 100 (1) + 10 (1) = 110m
xnew = 100(1) + (10 + aevasive)*1
|xoriginal-xnew| = 100 for a clean miss (i.e. my original displacement
must differ from my "evasive" displacement by 100 meters.)
Assume the computer is dumb, and just opts for a straight head acceleration:
|xoriginal-xnew| = |110 - 100 - 10 - aevasive|
= |- aevasive|
Therefore, aevasive=100 which is 10g's. A lot better.
However, I'm also assuming that acceleration is "instantaneous", that it only
occurs in a linear fashion, and a whole bunch of other things.
I think I'm going to work on this problem at home, and come back with
more stuff. (8-)
Please take this as a first approximation. (8-)
J.
Man you guys are impressing the hell out of me with all this physics and math
stuff. I'm having to reach back 10 years to my college physics class to
roughly follow your equations.
I just noticed Jerry Han's e-address, though, so I don't feel so
bad.
Jerry---jhan@undergrad.math....etc
----
> Alan Brain writes:
@:) The unacceptable error is 100 metres in 100 million metres, ie 1 @:) in 1
million. Assume a 10 metre long weapon, this means a
@:) deflection of 1/100,000 of a millimetre. Not impossible, but not
@:) easy.
Seems like it'd be easier with energy weapons like lasers.
@:) Then add tracking errors. The ship itself is not infinitely rigid, @:) and
will flex somewhat as the thrust changes. Basically, you need @:) a lot of
computation, a lot of sensors within the ship to measure
@:) distortions between sensors and weapons, and a mirror/lens that
@:) reacts at light speed or close to it.
Most of these things are fairly doable. I'm curious as to whether modern
warships use stuff like this? Isn't it possible to measure the
deflection of a fiber-optic cable? I imagine you could run the stuff
from a sensor to a gun mount and figure how the ship is twisting. Actually
ships might flex less in space.
@:) The next problem is scanning using EM sensors. Probability of
@:) counter-detection is approximately proportional to Peak Pulse
@:) Energy. Probability of detection is proportional to average
@:) energy.
I just have to mention again that there is at least a small reason to believe
that sensors could be developed that almost completely eliminate the
possibility of counter detection because they don't emit any radiation that
actually hits the target. Quantum stuff, you know. Kinda creepy but I guess
they've actually done it in the lab.
@:) In space, you don't have sea returns, seagulls etc. But you do @:) have
dust particles, large EM sources (stars), etc etc. Now add
@:) decoys - a light balloon with a small rocket motor, and it becomes
@:) tricky.
And it might be interesting to have a game in which players could place bogey
markers on the table that would be indistinguishable (except with effort) from
real spacecraft. Solar flares can do interesting stuff too and a good
simulation of them might work well in an FT game.
> Actually, another interesting related problem is the degree of error
The unacceptable error is 100 metres in 100 million metres, ie 1 in 1 million.
So if the end of the weapon is deflected by 1 part in 1 million, the shot will
miss. Assume a 10 metre long weapon, this means a
deflection of 1/100,000 of a millimetre.
Not impossible, but not easy. Then add tracking errors. The ship itself is not
infinitely rigid, and will flex somewhat as the thrust changes. Basically, you
need a lot of computation, a lot of sensors within the ship to measure
distortions between sensors and weapons, and a
mirror/lens that reacts at light speed or close to it.
Note that in Naval Combat today, typically a Frigate will flex 0.5-2.0
degrees between the location of a weapons mount, and a sensor mount, in sea
state 3. Sea State 6 means at least double, and the pitching of the ship can
cause a 76mm gun barrel to go at least a degree off true.
The next problem is scanning using EM sensors. Probability of
counter-detection is approximately proportional to Peak Pulse Energy.
Probability of detection is proportional to average energy. So you try to
spread the pulse, to have a low peak, yet high average energy. But this means
dwelling on the target for a long time. Automatic Trackers
often use 2/3 or 3/4 logic to avoid false alarms. (2/3 means if you send