Tidbits of Megahexes

[JimmyPlenty]

How do you handle the onesey-twosey hexes when figuring range for megahexes?

[larsdangly]

I drop them.

[tomc]

You can count hexes and divide by three if MH aren't marked.

[Shostak]

We count every hex.

[larsdangly]

What does it mean to 'count every hex' when the unit of measure is the MH?

[larsdangly]

Originally Posted by tomc

You can count hexes and divide by three if MH aren't marked.

But you will often have a remainder. What do you do with that?

[TippetsTX]

I assume we are talking about missile weapon (or spell) ranges, right? Those are the only things using megahexes for their range increments that I am aware of.

If that's the case, then the remainders don't matter. RAW says that hitting a target within 2 MH has no DX penalty. A target more than 2 MH away, but less than 4 MH is -1 to hit and so on. Or are we talking about calculations when the attacking figure's position is not used as the measurement origin point?

[tomc]

Originally Posted by larsdangly

But you will often have a remainder. What do you do with that?

I'm assuming that each MH is 3 hexes across. So dividing the hexes by three tells you how many MHs away you are from your target. The remainder gets rounded up and counts as another MH.

You can do the geometry and adjust the assumed width of 3 to account for the specific alignment of the MH, but at that point you're better off with a measuring stick and admitting to yourself that you're playing a miniatures game.

If you want to be a stickler, you can note that you get a free hex to the border of the hex your in, as well as next two MH. Then there's a -1 DX penalty for each two MH beyond that.

So the first 7 hexes are free, and then -1 DX for each group of 6 beyond that, and -1 DX for that last partial group of 6 (if it has more than zero hexes).

The DX penalty is ((H-7) DIV 6), unless (((H-7) MOD 6) > 0), in which case it's ((H-7) DIV 6) + 1. But I ignore that last part.

In English the penalty is the number of hexes to the target, minus 7, then divide by 6. And if your GM is out to get you, subtract another from your DX for any hexes left over. :)

Simple, no?

What I really do is count out the 7 free hexes, then subtract 1 DX for every 6 hexes past that. I played for years over a homemade blank hex board with a sheet of clear plastic over it. I kept telling myself I'd go back and mark the MH borders with heavy marker, but never did.

[tomc]

Originally Posted by larsdangly

What does it mean to 'count every hex' when the unit of measure is the MH?

If the unit of measure is MH, then it's easy, The first two MH are no penalty, then -1DX for every two MH beyond.

[JimmyPlenty]

I know the idea of MH is to make things easy right? So what to do without a lot of calc?

I think Larsdangly hit it. But I think I'd only drop them for the heroes and add + 1MH for foes.

This way, the players always feel they aren't getting shorted and it's really not a big deal. + or - 1 either way. (although 1 is a lot more in TFT than the d20 games)