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BrubakerII -> (10/18/2002 6:53:53 PM)
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Hi Joe
Well of course it all has to do with movement expenditure, and the cost of the terrain in each hex that the unit is moving into and from.
Consider the picture below, which is yours but with the terrain costs overlayed.
[IMG]http://users.esc.net.au/~ooyeah/Forumpic/numbers.jpg[/IMG]
A unit expends its movement OP's each time it moves to a new hex. This (as a general rule) is the cost of a hex it is leaving plus the cost of the hex it is entering divided by two - in otherwords an average of the two hexes. Therefore if a unit is leaving a hex of value 5 and entering a hex of value 3 it will cost it 4 OP's to make the move.
Okay, now considering the picture above, the move west across the river will cost to move into each hex 7,4,1 for a total of 12 and considering the next hexes are 1's also the unit therefore must have an OP value of 12 (because if it was 13 it could move another hex right?). The move south however will take it 5,3,3 which is only 11, but of course the move across the river on the brdige would cost another 4, which it doesn't have.
Out of interest (this is much like the game minesweeper in windows) one would initially think that the hex below the bridge (east of Wiltz) is a 1 OP hex also as the unit can move there as well, except that if you take the unit a different route (ie. S, S, SW, SW) you will find that it can reach that destination if the hex has a value of 2.
Anyway, so to answer your question the reason is that the hex on the other side of the river containing the bridge is of a relatively high value terrain wise. This could I suppose be justified by the difficulty crossing a narrow bridge and the traffic chaos undoubatably eventuating. What the situation does highlight however, is that there doesn't appear to be any inherent cost of crossing a river per se, or a benefit for travelling down a road, as the terrain in the hex provides the basis for the cost of the move, not the road itself.
Brubaker
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