| <blockquote id="quote"><font size="3" face="book antiqua" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Rich Walker</i>
Now, when the next patch is released I will introduce consolidated rebel batteries as I have already done for my next title. I will also offer at least one variant of the historical Antietam with sections.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Don't do it.  A return to the 4 Gun CSA battery vs 6 Gun USA battery formula is a step backwards from the otherwise fair and equitable Corinth model.
 
 Why?
 
 Because, Mr. Tiller's hardwired fire calculation forumla (i.e., Range + Modifier + #Gun FIRE = Fatigue / #Gun Loss) always resolves all friendly sections / batteries stacked in a hex - even when thought to be combined into a single 'large' attack - as multiple <i>separate</i> - individual - attacks.  Thus one on one, anytime a single Yank #6 gun battery fires, it statistically generates a 50% increased Fire factor over any #4 CSA battery.
 
 This was ultimately determined to be unfair way back when.  And, yet, now here we are . . . talkin' bout going forward into the past? [V]
 
 Now I sympathize with anyone's desire to form single large integer battery formations! I really do, dammit!  But, WE can't do it with John Tiller's game, because he employs a stale, inflexible, fixed, dated Fire formula routine that skewers to favor single large interger (i.e. 6 Gun USA battery) fire attacks vs a smaller (i.e., 4 Gun CSA battery ane/or 2 gun section / 1 piece) fire attacks.
 
 Unless, Mr. Tiller rethinks this baby formula, there is NO alternative that is FAIR to all combatants, other than to breakdown <i>all</i> batteries on both sides of the Mason-Dixon.
 
 Exhibit C
 
 Question:  How many 2 gun / 1 piece CSA artillery sections, firing individually as they must (because Mr. Tiller made it thus) over the course of a single defensive / offensive fire turn, are needed to provide a "1" Gun Loss vs a single 6 Gun USA battery?
 
 Answer:  A plethora! [:(]
 
 Question:  How many 6 gun USA batteries are needed to knock out a 2 gun CSA section in a single defensive / offensive fire phase?
 
 Answer:  2
 
 That is, in one consecutive defensive / offensive fire turn two 6 Gun USA batteries netted the following results:  "Fatigue," "1" Gun Loss, "1" Gun Loss, "Fatigue".  (Range=9, Modifier= -20) - Turn 5, Cedar Mountain.
 
 In the CSA defensive fire / offensive fire turn, three 2 gun sections and 1 piece returned fire on above (2) USA batteries, netting the following results:  "Fatigue," "No Effect," "No Effect," "No Effect," "No Effect," "Fatigue," "No Effect," "Fatigue".
 
 Fair?
 
 Why is this so hard?
 
 Fld. Lt. D. Shoeless, CSA
 Secretary of the Cabinet (Ret)
 1st Tenn Provisional Army
 
 
 <center><i>From a certain point onward there is no turning back.  That is the point that must be reached.</i>  --F. Kafka</center>
 
 
 |