WARGAMING: HOW TO ELIMINATE FIGURES
FROM THE BATTLEFIELD
One of the main problem in wargaming at 1:1 ratio (but also one of most unique things with
this kind of wargame) is the way to take off the losses from the battlefield.
It is quite simple but at the same time very interesting, allowing to recreate the right “alea”
in fighting. Usually an enemy – according to my rules – can provoke a percentage of losses on
the enemy’s troops (how to calculate this percentage is described in another chapter). Let’s
say that this percentage is 7,7%, i.e. every 100 papersoldiers on the battlefield.
Let’s also say that a unit is fighting with 96 men all included (as for example a wing of a
battalion in a regiment of the Scanian War). This wing will provoke 7,39 losses, i.e. 7,4 losses
(you have to round up ad with the centimes of Euro).
It is then necessary to eliminate this amount of papersoldiers; but from which point to start?
You have to consider the decimal number …,4; so you start to count (having the enemy
in front of you) from right to left, eliminating (in green) 1 papersoldier every 4 papersoldiers.
And so 1,2,3, 4 – eliminated, 5, 6, 7, 8 – eliminated (2 in total), 9, 10, 11, 12 eliminated
(3 in total) ; 13, 14, 15, 16 – eliminated (4 in total) ; 17, 18, 19, 20 – eliminated (5 in total);
21, 22, 23, 24 – eliminated (6 in total), 25, 26, 27, 28 - eliminated (7 in total).
It is possible to count in this way till the third row, than you have to start again
from the first one.
this kind of wargame) is the way to take off the losses from the battlefield.
It is quite simple but at the same time very interesting, allowing to recreate the right “alea”
in fighting. Usually an enemy – according to my rules – can provoke a percentage of losses on
the enemy’s troops (how to calculate this percentage is described in another chapter). Let’s
say that this percentage is 7,7%, i.e. every 100 papersoldiers on the battlefield.
Let’s also say that a unit is fighting with 96 men all included (as for example a wing of a
battalion in a regiment of the Scanian War). This wing will provoke 7,39 losses, i.e. 7,4 losses
(you have to round up ad with the centimes of Euro).
It is then necessary to eliminate this amount of papersoldiers; but from which point to start?
You have to consider the decimal number …,4; so you start to count (having the enemy
in front of you) from right to left, eliminating (in green) 1 papersoldier every 4 papersoldiers.
And so 1,2,3, 4 – eliminated, 5, 6, 7, 8 – eliminated (2 in total), 9, 10, 11, 12 eliminated
(3 in total) ; 13, 14, 15, 16 – eliminated (4 in total) ; 17, 18, 19, 20 – eliminated (5 in total);
21, 22, 23, 24 – eliminated (6 in total), 25, 26, 27, 28 - eliminated (7 in total).
It is possible to count in this way till the third row, than you have to start again
from the first one.
The problem is that the papersoldiers are glued – for Scanian Wars – on stand of
4 papersoldiers each: this means that - in case as above - you have to take off 6 figures, it is possible to take off 1 stand per time (with 4 figures), but the second
stand will suffer just the loss of 3 figures.
In this case it is necessary to leave this
stand on the battlefield, but that unit (or wing) will have a “malus” of 2.
In in the next turn that unit will lose, let’s say, 6,2 papersoldiers; it will be necessary
then to take off 4 other figures (1 stand), with a new additional “malus” of 2 papersoldiers that
added to the previous “malus” of 3 makes a total of 4 figures and so it is possible
to take off in that turn also a second stand plus 1 "malus" to be added the following turn.
At the end of two turns the unit will be visually in this way:
4 papersoldiers each: this means that - in case as above - you have to take off 6 figures, it is possible to take off 1 stand per time (with 4 figures), but the second
stand will suffer just the loss of 3 figures.
In this case it is necessary to leave this
stand on the battlefield, but that unit (or wing) will have a “malus” of 2.
In in the next turn that unit will lose, let’s say, 6,2 papersoldiers; it will be necessary
then to take off 4 other figures (1 stand), with a new additional “malus” of 2 papersoldiers that
added to the previous “malus” of 3 makes a total of 4 figures and so it is possible
to take off in that turn also a second stand plus 1 "malus" to be added the following turn.
At the end of two turns the unit will be visually in this way:
Then the unit will fix itself shifting forwards figures from the back (or simply taking them off directly from the back of the unit to be faster) in this way:
