Re: Time/Distance scale for FT (sort of related top safe speed, etal)

> >> To the best of my knowledge, I was first to propose the T = 7.5
> min,
> >> D= 1000km, which also fits 2 to 1 with DS2 and lets you assume 1
> >> thrust point = 1 gee.
> >
> >i'm not entirely sure that's the case.
> >
> >1 mu = 1000 km
> > = 1e6 m
> >
> >1 tu = 7.5 min
> > = 450 s
> >
> >speed 1 = 1 mu / tu
> > = 1e6 m / 450 s
> >
> >dv = a * dt
> >a = dv / dt
> > = (1e6 / 450) / 450
> > =~ 5 m/s^2
> 
> d = 0.5 * at^2
> 1,000,000m = 0.5 * a * 450sec^2
> 9.87654321 m/sec^2 = a
> Close enough to 1 gravity for me.

You're both right, FT's current movement system isn't perfect (although
the vector one is close).  The only way it "fits" is if you assume
instantaneous (or very quick) acceleration at the start of the
turn, while coasting the rest of the turn.

One way of looking at it is from a distance perspective (how much
acceleration does it take to cover X distance (assuming constant
acceleration)), the other is to look at it from a vector perspective
(how much acceleration does it take to change my delta V by Y amount
(assuming constant acceleration)).

In the real world, the thing to keep in mind is:
o given a constant acceleration, the amount your vector changes (dV)
  is twice the distance you traveled (in that time).

In the full thrust world, (assuming constant acceleration), this
isn't true.  They are equal.  Which in turn gives you two different
acceleration, depending on how you calculate it.

But both of you probably already figured that out,
Daryl
-- 
Daryl Lonnon
dlonnon@verinet.com