Re: Lag....

Hi People,

I'm going to see what numbers I come up with.  I hope Daryl doesn't
mind.

Daryl Lonnon wrote:
>>From Jon:
>>would have been, if you see what I mean. Say for a ship that's
approximately
>>a 200m sphere, it would have to change it's acceleration enough to be
>>100m from it's predicted point in the period between the targetting
>>pulse arriving and the energy beam - in that case it's moved 100m
differently
>>in .2 of a second. It's a hard G move (and I can't work out exactly
how
>>hard now as I have the flu and that sort of thing is VERY complicated)
> 
> Since this type of stuff has always interested me ... I figure I'll do
> the math.

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.

-- 
     Jerry Han - jhan@undergrad.math.uwaterloo.ca - I disclaim
everything
Dbl Hon AM/CS, University of Waterloo: http://www.uunet.ca/~jerry -
TBFTGOGGI
"They swept away all the streamers after the Labour Day parade,
	Nothing left for a dreamer now, only one final serenade" - Billy
Joel