Warrior Ancient and Medieval Rules

[Mark Stone]

Jon asked me to repost these questions which he hasn't gotten around to
answering yet, so here they are:

----- Forwarded message from Mark Stone <mark@...> -----

Later Swiss are, on their own and other lists, 1/2 - 3/4 pike for each unit, and
the rest 2HCT. I just want to make sure I understand the implications of this.
There's going to be a strong temptation to play the Swiss in 20 figure units,
with 4 elements of pike and 1 element of 2HCT, and say "well, 80% is pretty
close to 75%, so that's close enough." My understanding of the lists is that
this construction is _not_ allowed. If you want 4 elements of pike in a unit,
you _must_ take at least 2 elements of 2HCT. Can I have that explicitly
confirmed? It'll save many arguments down the road.

I also want to make sure I understand the mechanics of the list rules in the
following situation:
On one side we have a 32 figure Seleucid pike unit, Reg C MI, 2 elements wide
andn 4 deep. On the other side we have a 24 figure Swiss unit, in column,
configure thus: front rank LHI P, next 3 ranks LMI P, last two ranks LHI 2HCT.
To simplify, we'll assume in the following combat that all die rolls are even.

Bound 1: The Seleucids and Swiss charge each other. The Seleucids do 12@2 (P vs.
LHI) +1 (charging) = 12@3 = 30. The Swiss do 12@3 (P vs. MI) +1 (charging) =
12@4 = 36. Result: neither pushes the other back, but the Seleucids, as losers,
become disordered. Now, under the list rules, since the Seleucids did not
"recoil, break or break off", the Swiss can swap a rear element of 2HCT to the
front.

Bound 2: The Seleucids only have two ranks fighting, as do the Swiss, but the
Seleucids get the overlap. The Seleucids do 16@2 (P vs. LHI) -2 (disordered) -1
(facing 2HCT) = 16@-1 = 12. The Swiss do 4@5 (2HCT vs. MI) = 16 + 4@3 (P vs. MI)
= 10 for a total of 26. The Seleucids have taken 1 CPF (they count as a 24
figure unit in this formation) and twice as many, so must take a waver test for
second cause of disorder, and recoil. The Swiss may take the second 2HCT element
and use it to expand, under the list rules.

Bound 3: Assume the Seleucids pass their waver. Where they have been
continuously in contact with the Swiss, they do 8@2 (P vs. LHI) -2 disordered)
-1 (facing 2HCT) +2 (facing shieldless LHI) = 8@1 = 12. Where they are facing
the new 2HCT element that just expanded, this is first contact against that
element so they don't count shiedless. They're two factors less for 8@-1 = 6,
for a total of 18. The Swiss no longer get their second rank of P, but get 8@5
(2HCT vs. MI) +1 (following up) = 8@6 = 40. The Seleucids again take 1 CPF and
twice as many, taking another cause of disorder, and thus another waver test,
and recoiling yet again.

The only part in all of this where I'm unclear is if I have correctly described
when the 2HCT do and do not count shielded. Jon?

----- End forwarded message -----

[joncleaves]

In a message dated 8/25/2003 11:13:59 Central Daylight Time,
mark@... writes:
Later Swiss are, on their own and other lists, 1/2 - 3/4 pike for each unit,
and
the rest 2HCT. I just want to make sure I understand the implications of this.
There's going to be a strong temptation to play the Swiss in 20 figure units,
with 4 elements of pike and 1 element of 2HCT, and say "well, 80% is pretty
close to 75%, so that's close enough." My understanding of the lists is that
this construction is _not_ allowed. If you want 4 elements of pike in a unit,
you _must_ take at least 2 elements of 2HCT. Can I have that explicitly
confirmed? It'll save many arguments down the road.
That one is for Scott to answer


I also want to make sure I understand the mechanics of the list rules in the
following situation:
On one side we have a 32 figure Seleucid pike unit, Reg C MI, 2 elements wide
andn 4 deep. On the other side we have a 24 figure Swiss unit, in column,
configure thus: front rank LHI P, next 3 ranks LMI P, last two ranks LHI 2HCT.
To simplify, we'll assume in the following combat that all die rolls are
even.

Bound 1: The Seleucids and Swiss charge each other. The Seleucids do 12@2 (P
vs.
LHI) +1 (charging) = 12@3 = 30. The Swiss do 12@3 (P vs. MI) +1 (charging) =
12@4 = 36. Result: neither pushes the other back, but the Seleucids, as
losers,
become disordered. Now, under the list rules, since the Seleucids did not
"recoil, break or break off", the Swiss can swap a rear element of 2HCT to the
front.
Right.


Bound 2: The Seleucids only have two ranks fighting, as do the Swiss, but the
Seleucids get the overlap. The Seleucids do 16@2 (P vs. LHI) -2 (disordered)
-1
(facing 2HCT) = 16@-1 = 12. The Swiss do 4@5 (2HCT vs. MI) = 16 + 4@3 (P vs.
MI)
= 10 for a total of 26. The Seleucids have taken 1 CPF (they count as a 24
figure unit in this formation) and twice as many, so must take a waver test
for
second cause of disorder, and recoil. The Swiss may take the second 2HCT
element
and use it to expand, under the list rules.
Good.


Bound 3: Assume the Seleucids pass their waver. Where they have been
continuously in contact with the Swiss, they do 8@2 (P vs. LHI) -2 disordered)
-1 (facing 2HCT) +2 (facing shieldless LHI) = 8@1 = 12. Where they are facing
the new 2HCT element that just expanded, this is first contact against that
element so they don't count shiedless. They're two factors less for 8@-1 = 6,
for a total of 18. The Swiss no longer get their second rank of P, but get 8@5
(2HCT vs. MI) +1 (following up) = 8@6 = 40. The Seleucids again take 1 CPF and
twice as many, taking another cause of disorder, and thus another waver test,
and recoiling yet again.

The only part in all of this where I'm unclear is if I have correctly
described
when the 2HCT do and do not count shielded. Jon?
The last part is wrong. 7.1 says: "Steady foot using pike, LTS, 2HCT or 2HCW
cannot count shieldless (even if they do not possess shields) to frontal
opponents hand-to-hand weapons when charging or being charged." That does not
apply in this case, so the 2HCT are shieldless in bound 3.

Otherwise good.
J


[Non-text portions of this message have been removed]

[scott holder]

Later Swiss are, on their own and other lists, 1/2 - 3/4 pike for each unit, and
the rest 2HCT. I just want to make sure I understand the implications of this.
There's going to be a strong temptation to play the Swiss in 20 figure units,
with 4 elements of pike and 1 element of 2HCT, and say "well, 80% is pretty
close to 75%, so that's close enough." My understanding of the lists is that
this construction is _not_ allowed. If you want 4 elements of pike in a unit,
you _must_ take at least 2 elements of 2HCT. Can I have that explicitly
confirmed? It'll save many arguments down the road.

That one is for Scott to answer

>Consider it explicitly confirmed. There is no "close enough" in Warrior lists.


[Non-text portions of this message have been removed]