- Thread starter Brian W
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I6 (the obstacle in front of H6) creates one blind hex since the distance from O5 is 5-9 hexes (B10.23). O5 is exactly one level higher than I6, and will therefore not reduce the number of blind hexes - thus no LOS.

G7 (the obstacle in front of F7) also creates one blind hex, but according to A6.42: take the height differences between O6 and G7 (4-2 = 2), subtract by one (2-1 = 1), and reduce the number of blind hexes by the last number (1-1 = 0), thus no blind hexes.

Similarily for D6, where the obstacle is E6. The number of blind hexes is 2 (distance 10), but height difference of O6 and E6 is 3. 3 - 1 = 2, so reduce the number of blind hexes by 2 - again zero blind hexes.

Therefore all those hexes in the diagram are blind.

Rick

Ah, but the rules also say (NRBH) that you can reduce the blind hexes to zero if the obstacle is a hill crest - especially if its part of the "same hill mass", IIRC.st to note, rule A6.42 does say that the minimum is 1 blind hex, regardless of decreases.

Therefore all those hexes in the diagram are blind.

What I've never seen definined is what is part of the "same hill mass".

Sam "No Rule Book here, and completely confused anyway" Belcher

Now I understand what you mean and how you have achieved the answer.

Great stuff.

Rick