Specialist Arms Forum
Warmaster => [WM] Warmaster Fantasy Discussion => Topic started by: Dave on March 15, 2012, 12:54:05 PM
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In driving back enemies, this sentence:
Once this first stand has been positioned, remaining stands move back along the same path
Does that mean the exact same path? An example:
XXX
YYY
|
|ZZ
V
X shoots at Y and drives it back. The player controlling Y chooses to move the left most stand. Do the other stands count as being driving back through Z (and have to check for confusion) or do they follow the left most stand's path and not have to test for it?
On making way, in Diagram 57.1, is the way Unit 2 moved legal?
If the entire unit lies within the path of the friendly unit then all stands must be moved, in this case the player begins with the stand that must move the shortest distance to get out of the path of its friends. This stand is moved the shortest distance out of the path of its friends without changing its orientation. The remaining stands are then rearranged into formation around the first.
The left most stand wasn't arranged around the first. I'm taking around to mean "touching" here.
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On making way, in Diagram 57.1, is the way Unit 2 moved legal?
If the entire unit lies within the path of the friendly unit then all stands must be moved, in this case the player begins with the stand that must move the shortest distance to get out of the path of its friends. This stand is moved the shortest distance out of the path of its friends without changing its orientation. The remaining stands are then rearranged into formation around the first.
The left most stand wasn't arranged around the first. I'm taking around to mean "touching" here.
Back into a legal formation would have been the better description !
Move one shortest distance possible, arrange the rest in legal formation
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As I understand it you would have 2 choices in your examples, either Z denies to make way and Y is confused or you make way with Z and test for confusion on both. (Y once for pushing Z and Z once for making way)
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That's what I assumed but the "along the same path" bit confused me when I re-read it. I wasn't sure if that meant "exactly the same path" or "along the same vector".
Dave