Corporal Kindel
Member
Hello Listers:
B3.42 says:
"B3.42 The one MP penalty for entering a hex already containing a vehicle/wreck is doubled to two MP per vehicle/wreck if entering that hex across a road hexside while using the road movement rate."
So, a vehicle moving across a road hexside using road movement rate into a wreck/vehicle hex would cost 2MP (wreck/vehicle) + ½ MP (road rate cost) = 2.5MP, and then into the (adjoining) road hex (unoccupied) at ½MP, for a grand total of 3MP for the two-hex move.
Now suppose, given the exact same situation, a vehicle enters the first wreck/vehicle hex at non-road movement rate. In this case cost would be 1MP (wreck/vehicle un-doubled) +1MP open ground cost of hex = 2MP for the first hex, and then moves into the (adjoining) road hex (unoccupied) at ½MP, for a grand total of 2.5MP for the two-hex move.
The example assumes a CE CT tank. I just wanted to be sure these calculations were correct. Intuitively it seems like the total costs should be the same in both cases, but I don't think the rule is worded that way, is this correct?
B3.42 says:
"B3.42 The one MP penalty for entering a hex already containing a vehicle/wreck is doubled to two MP per vehicle/wreck if entering that hex across a road hexside while using the road movement rate."
So, a vehicle moving across a road hexside using road movement rate into a wreck/vehicle hex would cost 2MP (wreck/vehicle) + ½ MP (road rate cost) = 2.5MP, and then into the (adjoining) road hex (unoccupied) at ½MP, for a grand total of 3MP for the two-hex move.
Now suppose, given the exact same situation, a vehicle enters the first wreck/vehicle hex at non-road movement rate. In this case cost would be 1MP (wreck/vehicle un-doubled) +1MP open ground cost of hex = 2MP for the first hex, and then moves into the (adjoining) road hex (unoccupied) at ½MP, for a grand total of 2.5MP for the two-hex move.
The example assumes a CE CT tank. I just wanted to be sure these calculations were correct. Intuitively it seems like the total costs should be the same in both cases, but I don't think the rule is worded that way, is this correct?