Napoleonic Wargame Club (NWC)

Quick question (spawned by one of Mark Eason's comments in a recent thread):

In the HPS games, when a non-phasing player's artillery battery automatically fires at the enemy's moving player's units (either as "Defensive Fire" or "Opportunity Fire", if I have the terminology correct), does that expend 1 round of ammo from the overall army's ammo supply for <b>each</b> shot taken? (as we know that some defending units can fire multiple times during the enemy's movement phase... whether we like it or not!) Or does it expend only 1 round regardless of how many times it fired? Or does it not expend any rounds??

I'm designing an HPS 4-day scenario, and am trying to gauge how much ammo should be granted to each army. I understand that a round of ammo is expended all the same regardless of whether a 12 gun or 2 gun battery is firing... obviously making the 12 gun battery the preferred battery to fire.

Thanks in advance for any answers or thoughts.

Lt Tim CAvallin, AdR

Here is where it gets tough on a designer: its a percentage that it will expend a round. Just like whether it will cause Low Ammo.

Here is my formula:

Count the number of batteries and add them up. Ammo expenditure is by battery whether its 4 guns or 12.

Now figure out whether the army should be considered fully stocked for ammo, moderately stocked, or low stock.

Divide the number above by:

1 if fully stocked
2/3 if moderately stocked
1/3 if low stock

The result is X

Now the time factor:

For each turn the factor is 2

Add up the total number of turns = T

X * T is the total number of rounds (R)

For meeting engagements divide the total by 3/4
For set piece battles use there is no reduction.

Examples:
32 turn scenario
20 batteries
Moderately stocked - 2/3
X = 13
T = 32 * 2 = 64

X * T = 832

Each battery will fire from 2-4 times in a turn on the average. Sometimes one battery will fire 10 times and others wont fire for several turns as the are limbered.

32 turns, 20 batteries would see a potential of 640 shots if all batteries fired each Offensive Fire phase. This doesnt count in the Defensive fire which is harder to calculate.

If guns are limbered they dont fire. And if guns get captured then you lose 8 rounds per battery.

Now we know that not all guns are going to fire each turn. Adding in that the Defensive Fire will eat up alot of ammo BUT that we have considered that not all of the guns will fire each turn offensively or defensively then this total - 832 - will model a moderate ammo supply to fulfil the player's needs.

If the ammo supply were full then the number would be 20 * 64 or 1280 or an almost unlimited amount of ammo. Again, remember that batteries will not fire every turn. They are bound to be out of action for some time due to maneuver.

And if this was a meeting engagement then the number above would be 1280 * .75 or 960.

If someone sees a fault in my factoring please let me know.

Again, this is a fudge figure and nothing more. There are other things that will determine the ammo supply usage such as weather and player skill. I went with this calculation scheme as it seems to give the players the right amount of ammo.

Oberst Wilhelm Peters
2nd Kuirassiers, Reserve Korps, Austrian Army
[url="http://www.acwgc.org/acwgc_members/burr/Austrian%20Army/Bill_Peters.htm"]Officer Battle Dossier[/url]

Maybe one day we'll eventually get proper artillery supply wagons - this aspect of the game hasn't seen any improvement since the Battleground engine.


Capt Rich White
4th Cavalry Brigade
Cavalry Corps
Anglo-Allied Army

<blockquote id="quote"><font size="3" face="book antiqua" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Bill Peters</i>
<br />

Now figure out whether the army should be considered fully stocked for ammo, moderately stocked, or low stock.

Divide the number above by:

1 if fully stocked
2/3 if moderately stocked
1/3 if low stock

The result is X

20 batteries
Moderately stocked - 2/3
X = 13
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Bill,

Just to avoid confusion[:I] One need to multiply on the modifier not devide. If one devides 20 batteries by 2/3 he receives 30, not 13.

<center></center>
<center><b>Eyo Imperatorskogo Velichestva Leib-Kirassirskogo polku
General-Mayor Anton Valeryevich Kosyanenko
commander of Little Russian grenadiers regiment</b></center>


Thanks Anton. I have to fine tune this to get it right.

My English that is!!

Er, American! [:D]