Yeah. If I understand correctly, it´s how the turn is divided.
(put it this way: you have an Arty and a Light Tank in a stand-off, shooting at each other. We know the Arty shoots twice per turn, and the L. Tank shoots five times. Question is, how many shots does the Tank shoot before the Arty gets its first?
Seems that the turn is divided into "ticks"; think of them as turn fractions. So:
at 1/5th of turn = L Tanks shoots (1st shot).
at 2/5th of turn = L Tank shoots (2nd shot)
at 1/2 turn = Arty shoots (1st shot)
at 3/5th turn = L Tank shoots (3rd shot)
at 4/5th turn = L Tank shoots (4th shot)
at 5/5th turn = 2/2 turn = end of turn = L Tank shoots (5th shot) and Arty shoots (2nd shot); coin toss decides who goes first.
Now, imagine that instead of standing still:
- the L Tank advances four squares
-and gets into the Arty´s firing range when it enters the fourth square.
In the above case, the L Tank will advance one square every 1/4th of turn, thus entering the fourth square at the 3/4 of turn. The Arty will only shoot´im once (at the end of the turn) since it cannot shoot twice in 1/4th of turn.
Or so I understand, anyway!