Fighter groups other than 6 was Re: Fighters and Hangers

--- Steve Pugh <steve@pugh.net> wrote:
> On 3 Mar 2004 at 20:44, Jared Hilal wrote:
> > 
> > So isn't a fighter group that has lost 2 fighters to combat just as
> > "unbalanced" as one that started with 4 fighters? 
> 
> No. The group that has lost fighters needs to pass morale tests, the 
> group that started with four doesn't.

But only if you are using the OPTIONAL morale rules.

Then
--- Steve Pugh <steve@pugh.net> wrote:
> On 3 Mar 2004 at 20:44, Jared Hilal wrote:
> > If the problem is that having 12 fighters in three groups of 4 
> > gives an advantage over 12 fighters in 2 groups of 6, 
> 
> Imagine those two forces coming up against each other. The 3 groups 
> of 4 can engage both of the 2 groups of 6 in dogfights and still have
> 1 group left over to attack the opposing carrier. Sure the groups of 
> 6 will probably win the dogfights but not without casualties (which 
> means they need to pass morale tests before attacking the carrier) 
> and not without wasting time.
> 

And

--- Laserlight <laserlight@quixnet.net> wrote:
> Let's say you bring 24 fighters in four groups of six.  Your
> opponent, Monterey "Jack" Limburger, brings 24 fighters in 24 
> groups of one. Now roll initiative and start alternately moving 
> fighter groups...
> 

>--- agoodall@att.net wrote: 
> Here's another example. Let's make it shorter: 12 fighters in 2
> squadrons of 6 for Bob, and 12 fighters in 12 squadrons of 1 for
> Jack. Jack goes first due to initiative. Jack moves a fighter to
> Cruiser 1. Bob moves a fighter squadron to Cruiser 1 to help defend
> it. Jack moves a fighter to Cruiser 2. Bob moves his second squadron
> to defend Cruiser 2. Bob has finished moving. Jack moves the rest of
> his fighters, all 10 of them, to Cruiser 3. Bob spends an endurance
> to fight off two fighters. Jack, meanwhile, slices into a cruiser.
> The question now is: is it worth losing 2 fighters (and potentially
> doing 2 fighters worth of damage to the other guy) in exchange for a
> cruiser?
> 

I had not considered the initiative advantages.

However, this can be easily dealt with in two ways:
1) As the playtest group is revising fighter rules for FB3 anyways,
change the formula for fighter group from:
PV = F * x
where F = constant value of fighter and x = number of fighters in the 
group (from FB1 & 2)

to:
PV = G + (F * x) where G = constant value per group
so that 24 fighters could be divided in any of several ways:

3 groups of 8  =  3G + 3(8F)  =  3G + 24F
4 groups of 6  =  4G + 4(6F)  =  4G + 24F
6 groups of 4  =  6G + 6(4F)  =  6G + 24F
8 groups of 3  =  8G + 8(3F)  =  8G + 24F
12 groups of 2 = 12G + 12(2F) = 12G + 24F
24 groups of 1 = 24G + 24(1F) = 24G + 24F

thus larger numbers of smaller groups will cost more.  Although G
should be a constant and represent the value of having a group, I have
no suggestion for a numeric value as I have no idea how this would
scale to the rest of the FT/FB point system, rather leaving it to those
who crunched the numbers for the system originally.

2) We use a house rule regulating alternate movement:

Once the first player has moved a fighter group, the second player must
move at least as many fighters, even if that is more than one group. 
The first player must then move another group or groups containing at
least one more fighter than the difference between the second player's
move and the first player's first move.  (works like raising and
calling in poker)

E.g. If Bob moves a group of six and Joe moves a group of 5, then Joe
must still move a second group to match Bob's six.  If Joe move a 5 and
a 4, totaling 9 fighters, then Bob must move at least 4 fighters
(9-6=3, and Bob must move 1 more than Joe).

This was originally developed to take care of situations where a player
would move his combat depleted groups first, then move his full
strength groups last.

On combat between large groups and small groups:

--- Steve Pugh <steve@pugh.net> wrote:

<on the 4x6 vs 24x1 example>

> So you can destroy at most 4 of Jack's fighters per turn (and you 
> most probably will kill 4). Jack can in theory kill all of yours on 
> the first turn and on average will kill at least 13 of them 
> (the average for 24 die rolls is 19.2, but taking into account 
> casualties who don't get to fire and wasted re-rolls towards the end 
> of the turn the actual figure will be lower than that - but even if 
> the 20 who can't be engaged don't make any re-rolls they still 
> average 13 kills). 
> 
> Round 2 - it's now 4 rather battered groups vs 20 groups of 1. You'll
> kill 4 at best (though it's less certain this time) and Jack will on 
> average kill at least 10.
> 
> Round 3 - if you have any survivors they'll be facing Jack's 16.

However, just as "Jack" will try not to let you engage more than one
group at a time in dogfights, FT prohibits firing INTO a dogfight.  
(FT, pg 17, right hand column, 3rd full paragraph)

Thus any of the 6-strong groups that can manage to get into a dogfight
are immune from the long range fire of all of the other single groups,
which then would have to close into a furball to attack, allowing the
large group to split fire as per FB1.  Thus Mr.Pugh's numbers are an
inaccurate representation.

J