Re: [FT] Sensor Range Question

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

> or:
> 
> f	=	s/at^2
> a	=	s/ft^2
> t	=	(s/fa)^(1/2)

Should be:

f	=	sqrt(s/a) / t
a	=	s/(ft)^2
t	=	sqrt(s/a) / f

> if you put in s = 1000 km, t = 1000 s (16m40s), f = 0.1, you get a =
10
> m/s2, or ~1 earth gravity, which is nice.

I get a = 1000000 / (1000 * 0.1)^2 = 1000000 / 10000 = 100 m/s^2 ~ 10
g. If you instead assume f ~ 1, a drops to a comfortable ~ 0.1 g.

s = 1000 m, t = 10 s, f = 1 gives a = 10 m/s^2 ~1g, which makes this
scale fit nicely with the atmosphere interface rules in the scenario on
MT p. 39 (where the planetary radius is "infinite" - ie, very large -
compared to the gaming area). With f = 0.1, you're back at the somewhat
uncomfortable a ~ 10g.

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

Best wishes,

Oerjan Ohlson
oerjan.ohlson@telia.com

"Life is like a sewer.
  What you get out of it, depends on what you put into it."
- Hen3ry