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Liberty: The American Revolution 1775-83» Forums » Rules

Subject: No maximum for blocks in a Battle Hex? rss

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Barry Miller
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Actually this isn't a question as much I'm asking for consensus that I'm not missing something really basic in the rules. According to my interpretation, it's possible (though hardly likely), that an embattled hex could end up with 47 blocks stacked in it. While implausible, I couldn't find anything in the rules to deny the possibility - hence this post. Here's a hypothetical situation to illustrate:

Let's use the Philadelphia Hex as our target hex for this very hypothetical example.

SETUP
- Attacker: British
- Defender: Americans
- The French have entered the war
- After Wintering of the previous year, Philadelphia contains 4 American land blocks and 3 French warship blocks. America also has all of it's remaining blocks stacked in the hexes in Pennsylvania and Maryland.
- During the previous year and turns 1 - 4 of the current year, the British positioned all of their non-Indian forces (22 blocks) surrounding Philadelphia on five sides (leaving the SW side free for the Americans). Each of the five surrounding hexes has no more than four land units each.
- The Americans sense the build-up early-on and so during turns 1-4 are able to move 8 additional blocks inside Philadelphia. And are also able to maneuver his 10 remaining blocks to stack in Baltimore.
- Thusly at the start of the battle, the British have 22 blocks surrounding Philadelphia. The Americans have 15 blocks in Philadelphia and another 10 in Baltimore.

THE BATTLE
- On turn 5 the British declare the SE side of the Philadelphia hex (from Dover) as the main attack, as that's where all the British Warships are. So the British attack the Americans in Philadelphia with 7 blocks (4 land/3 naval).
- As mentioned above, the Americans already have 15 blocks in Philadelphia at the time of the attack (12 land/3 naval).
- Thusly for Combat Round 1 of the battle, there are 22 blocks in the embattled hex.
- Now let's assume that all blocks survive the first round and no one retreats.
- For Round 2, the British arrive with their remaining 15 blocks from the four other hexsides surrounding Philadelphia. And the Americans reinforce Philadelphia with the 10 blocks from Baltimore.

- Thusly for the 2nd round of combat, there are 47 blocks in Philadelphia (every block in the game except for the Indians).

So, back to the original question... while 99.9% unlikely to ever happen, I don't see anything in the rules that makes such a situation impossible.

Am I missing something?

P.S., When I sat down to write this post, I originally intended the entire post to be only a paragraph at most, being essentially a very simple and casual question. But it wasn't long before I felt compelled to illustrate the question with the hypothetical example above. Thusly there's a lot of "creative license" used in coming up with the example and as such I'm sure a lot of questions can be raised regarding how the setup for the battle is possible. So while I think the setup is possible, it is nonetheless admittedly farfetched and used only to illustrate the point of the question.

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Caleb
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Well, I'd love to see the resulting attrition after something like this, but yeah, I guess what you're suggesting could happen, or something like it.

If I were the Americans, though, I'd LOVE for the Brits to do this - they're welcome to Philadelphia if they want to put all their blocks there; I'll just take all the other locations on the entire map!
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