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

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.

> >> To the best of my knowledge, I was first to propose the T = 7.5

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,