shooting 1 dice per two bases at 2 different BGs

[player]

K1K1K1K1/K2K2K2K2

K1K1/K2K2/K3K3/K4K4
X1X1/X2X2/X3X3/X4X4

In the diagram above my 4 BGs of Polish Knights at the bottom supported by rear Xbow get shooting dice of 1 per two bases against the two BGs of knights in single rnak at the top. However, my opponent says that the left two of my BGs, K1X1 and K2X2 only have 1 and half dice against his BG K1 and 2 and half dice against his BG K2 because my K2X2 BG is split opposite his two BGs.
I say that I get 4 shooting dice and that 2 dice should be used against the right hand BG and that K2X2 dice can be used either to the left or the right as my choice, otherwise I completely lose it. The one dice from K1X1 is completly useless on its own against the opponents K1 BG as need 2 hits of course to cause any effect. This seems completly unfair as well as illogical as effectively I would lose 2 dice when I should have 4 dice and no shooting is then possible against K1 BG. Surely I should decide which way the 1 dice from K2X2 is used?
Which is correct? Any opinions as the 'target priority' section is not explicit in this scenario

[philqw78]

They get 1 dice per crossbow base, not 1 dice per 2 of The BG. The knights do not shoot, and The crossbow are first rank of shooters

[berthier]

What Phil said but specifically answered on p. 89, 1st column, last sentence of 1st paragraph. Although the relevant part of the sentence in question is talking about Byzantine cavalry, the application still holds true in your example.

[player]

That is interesting about the Xbow getting 1 dice per base rather than 1 per two. Is that version 3 or the same as vesrion 2?

If in the example above, say it was bow* or back rank bow what would the answer be? What I didn't mention was that also K3X3 is partially in front of the left K1 opposing BG , though less than half, if that makes any difference?

[player]

So in all cases where the back rank are shooters and the first rank are not then the back rank are counted as front rank and therefore get 1 dice per base?

[grahambriggs]

So in all cases where the back rank are shooters and the first rank are not then the back rank are counted as front rank and therefore get 1 dice per base?

Yes, if they are in effective range (and cavalry only have an effective range. The relevant wording at the top of page 89 is "If the front rank of a battle group are not shooters, only the second rank can shoot, but counts as the first shooting rank in determining the number of dice". So your crossbow troops count as front rank shooting, and the table on P88 shows you that they get 1 dice per base.

[grahambriggs]

That is interesting about the Xbow getting 1 dice per base rather than 1 per two. Is that version 3 or the same as vesrion 2?

If in the example above, say it was bow* or back rank bow what would the answer be? What I didn't mention was that also K3X3 is partially in front of the left K1 opposing BG , though less than half, if that makes any difference?

It was in version 2 as well, exactly the same wording. If it was bow* the principle would be the same but bow* only gets 1 dice per two bases whether in the 1st, 2nd or 3rd rank. I don't think there are any troops in the list which are front rank non-shooters, 2nd rank bow*. Bow (as opposed to bow*) second rank behind non shooters are treated just the same; they shoot as if in the front rank so get 1 dice per base.

[grahambriggs]

What I didn't mention was that also K3X3 is partially in front of the left K1 opposing BG , though less than half, if that makes any difference?

X3 will shoot according to target priorities on p79. It's shooting from the position of K3. So the priority target is whichever enemy base is in effective (4MU) range straight ahead of the base and closest. If you are lined up exactly facing each other and are at the "join" between K1 and K2 then both enemy battle groups are equally close, so you may choose which one. If you are at an angle, one target will be nearer to your troops than the other, so you'll have to shoot at the nearer one.

[player]

so basically if say 70% of the BG of 4 bases in two ranks is facing one BG and the other 30% is facing another BG and if it is 1 die per 2, the one die is effectively wasted because you can't have half a die? It seems that the logical position would be that the 1 die should be allocted to one side or another, if you get my understanding, rather than be wasted?

[philqw78]

You do not have 1 dice per 2. You have one dice per first rank of shooters. They shoot at the base most directly to their front. If exactly 50/50 for a base you choose which base to shoot at

However

AABB
_XX_
_XX_

X is a BG of Cavalry, all have bow or crossbow

The first rank shoot 1 to A and 1 to B
The second rank provide no dice as there is only 1 base, not 2

[grahambriggs]

so basically if say 70% of the BG of 4 bases in two ranks is facing one BG and the other 30% is facing another BG and if it is 1 die per 2, the one die is effectively wasted because you can't have half a die? It seems that the logical position would be that the 1 die should be allocted to one side or another, if you get my understanding, rather than be wasted?

The back rank die will be wasted if the front rank bases are shooting at different battle groups. There are ways to avoid this irritation though:

- if you have several battle groups all lined up and shooting at the enemy you tend not to get wastage like that, as the base that would perhaps be wasted often combines it's shooting with the BG next door.
- if you line up exactly opposite the enemy bases you'll get situations where you have two enemy bases that are ahead of a front rank base and both are nearest. So you get the opiton of who to shoot at. This can mean that you get to shoot 3 dice at one enemy BG, as opposed to 1 dice each at two.