Fighter groups other than 6 was Re: Fighters and Hangers

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

But only if you are using the OPTIONAL morale rules.

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

And

> --- Laserlight <laserlight@quixnet.net> wrote:

> --- agoodall@att.net wrote:

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

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

> On 5 Mar 2004 at 11:02, Jared Hilal wrote:

The morale rules were optional when given in MT, but the points costs used in
the Fleet Books assume that they are in use. If you do not use the morale
rules you should increase the pionts costs of all your fighter groups.

> On combat between large groups and small groups:

True, however.

> Thus any of the 6-strong groups that can manage to get into a dogfight

Only so long as the dogfight lasts... Where in the FB2 turn sequence are
dogfights resolved? As Oerjan has pointed out already, this is not at all
clear from the existing, scattered, rules.

On the other hand Jack can just make sure that the first four fighters he
activates can't be intercepted by Bob's groups. Bob now has to dogfight
unactivated fighters who have the choice of not being engaged. So Bob get's a
free shot at one of Jack's fighters. No big deal, we assumed that any single
fighter was toast anyway.

> Thus Mr.Pugh's numbers are an inaccurate representation.

At worst it means that Jack has 16 fighters to shoot at the two groups instead
of 20. So it's an average of 10 kills instead of 13
(again assuming that all of Jack's fighters fail to make any re-
rolls, it's 13 vs 16 is re-rolls are factored in). I'd still rather
be in Jack's shoes than Bob's.