Re: [FT] Sensor Range Question

On Mon, 26 Apr 1999, Oerjan Ohlson wrote:

> Thomas Anderson wrote:
> 
> > > one to ten seconds (common time
> > > intervals for RPGs and wargames).
> > 
> > the consensus is, i think, 1 turn = 15 minutes.
> > 
> > there is an equation.
> > 
> > s	length of a move unit (m)
> > f	fraction of the turn over which you burn drives
> > a	acceleration of 1 thrust point when burning (m/s^2)
> > t	length of a turn (s)
> > 
> > s	=	fat^2
> 
> Ahem. The total time you burn your thrusters is f * t, so you get
> 
> s	=	a * (ft)^2

Ahem. The total time you spend travelling is t. if you burn at a for ft,
then you get v = aft. if you then travel at that speed for some time u,
you travel s = uv = uaft. although u = (1-f)t, for very small f, u = t,
so
s = aft^2.

> 'Course, if you assume f = 1 you need more complicated vector movement
> rules to stay realistic <G>

that's the thing; the mechanics clearly describe an impulse-type drive.
i
think it's fairly simple to fix the rules though; it's just a pain
keeping
track of two velocities every turn. of course, those who want realistic
physics can pay the price. having just played BFG, even vanilla
cinematic
is quite realistic enough for me :-).

it's actually possible to construct an equation which does not assume
very
small f, and i believe it is illuminating to do so.

if you consider the turn to have two phases, boost and coast, then

s1	distance covered in boost
s2	distance covered in coast

s	=	s1 + s2
s1	=	vft/2
s2	=	v(1-f)t
v	=	aft
s1	=	a(ft)^2/2
	=	at^2 * (f^2/2)
s2	=	af(1-f)t^2
	=	at^2 * (f - f^2)
s	=	at^2 * ((f^2/2) + (f - f^2))

define

k	=	(f^2/2) + (f - f^2)

so

f	=	1 - (1 - 2k)^(1/2)

and

s	=	kat^2
k	=	s/at^2
a	=	s/kt^2
t	=	(s/ka)^(1/2)

ie, there is a fiddle factor of ((f^2/2) + (f - f^2)) applied to the
simple at^2.

solving for f, i put in s = 1000000 m, a = 10 m/s^2, t = 1000s, and i
get
k = 0.1, which gives f = 0.1055728, plus some more decimal places.

unless i'm being absurdly boneheaded today, that's right, isn't it?

and now, GZG-L will return to your scheduled programming for tonight.

Tom