Number of Dice for Shooting

[Eques]

In the shooting section of the rules who shoots at who is calculated by base, eg. each base has to shoot at a base at least partly in front of it and if there are more than one you have to shoot at the nearest etc.

However, under "How many dice to roll" some troops only get 1 dice per 2 bases. But if each base only shoots at one other base how do you marry the two principles up?

Or have I misunderstood?

Thanks

[gozerius]

In most cases where bases get one die per two bases, the front rank and rear rank are combined into a single die. In those cases where several bases are splitting their fire between two separate enemy battlegroups, bases must shoot at those bases which have target priority. It can occur that odd numbers of bases would then be unable to combine against one or the other enemy target. This is most likely to occur when a BG is in a non-rectangular formation, (4,2) or multiple BGs shooting have suffered base losses. Remember that although shooting is targeted by base, dice are accumulated by the target BG, so every base shooting at a base belonging to a particular target BG would count towards the total dice vs that BG.

Example: A BG of seven longbow (4,3) and and 3 bases (2,1) from an adjacent BG target a BG of knights at effective range. The front rank bases total 6 dice and the second rank bases contribute a total of 2 dice (4 bases = 2 dice) for 8 dice against the knights.

[bbotus]

And don't forget the last bullet on page 93-94. If 2 BGs with different POAs are shooting at 1 enemy BG, and 1 base from each BG combine to give you 1 more die ('dice' in the rules), then that die is rolled at the lower POA.

[Eques]

Thanks folks but my query was of a more basic nature!

Say I have 4 Light Horse Shooting at 6 Cavalry.

The BGs happen to be positioned so that each base of the LH will shoot at a different base of the Cavalry under the priority system laid down in the "Shooting" chapter.

Now the LH get 1 dice per 2 bases. If each LH base is each targeted at one Cv base under the priority system then do they each somehow shoot with half a dice?

[hazelbark]

Thanks folks but my query was of a more basic nature!
Say I have 4 Light Horse Shooting at 6 Cavalry.
The BGs happen to be positioned so that each base of the LH will shoot at a different base of the Cavalry under the priority system laid down in the "Shooting" chapter.
Now the LH get 1 dice per 2 bases. If each LH base is each targeted at one Cv base under the priority system then do they each somehow shoot with half a dice?

1 Remember the LH should be 2 bases deep not 4 wide. As both bases count for shooting in the 2nd rank.
2 you add up all the dice shooting at the same BG and roll them together. So 4 bases at 1 dice per 2 will equal 2 dice.
3 Let's say you only had 3 bases then you have 1.5 dice. In shooting you round down. In Impact and melee you round up. So shooting 1.5 dice is 1 dice.
4 Let's say you had 3 superior LH bases and 1 aveage base. The combined 1.5 dice from superior and .5 dice from average use the lower rating. So you get 1 dice shooting as superior and 1 dice shooting as average.