A4.52 MFs with DT MMC and non-DT leader

[EJ1]

Hi all,

Would someone please present the math to calculate the net available MFs for a stack consisting of a DT MMC (declared at the start of its MPh) with a non-DT leader in which they share 5 PP of kit? Example two of A4.52 states the stack would have six MFs available to it (same with the A4 MF/PP Chart). I don't see it (edit: i.e., I don’t understand the math).

If the stack DTs and shares 5 PP, then 5 MFs; yes, I see that. Similarly, DT MMC with non-DT Hero, then 4 MFs; yes, I see that, too.

Thanks.

 

[Robin Reeve]

This chart, under the IIFT, among your rulebook dividers, should answer your question:

 

[PresterJohn]

A HS carrying five PP has only two MF; if accompanied by a leader, they would have five MF (4.12, 4.42).
If the HS (only) Double Times with the leader, they have six MF. However, if the leader also Double Times, then they have only five MF.

The mathematics would be HS with 8MF and 2 PP and a leader with 6 MF and 1 PP. The PP is two in excess of HS plus leader PP capacity.
8-2=6

 

[EJ1]

Hi Robin,

Thanks for your help. It is my fault for using the colloquialism “I don’t see it.” I should have been more formal. Yes, with my eyes, I see the charts . . . and I have the charts, too; I meant that I don’t understand the math in this case.

Willing to try again? Thanks and cheers.

 

[EJ1]

Hi Prester,

Thanks, but I still do not understand the math in this case. Thus, in steps:

1. Starting with a CX HS moving with a non-CX leader in one stack.
2. Together they have an initial 6 MFs available to them.
3. The HS’s IPC is reduced by 1 for CX and the leader adds his 1 for a net, combined IPC of 3 IPC (as per column two of the A4 chart)
4. Together, they are carrying 5 PP of kit
5. 5 PP of kit – a net IPC of 3 = 2 PP over capacity
6. A gross 6 MFs – 2 PPs over capacity should equal a net of 4 MFs available.
7. My result in line six does not agree with the rule book.

Thus, I’m stuck.

Wait! I think I know get it. The CX added two MFs to the HS. Then, being with the leader for its entire MPh added an additional 2 MFs for a total available to the HS of 8 MFs. Knock the 8 MFs down by 2 for being 2 PPs over IPC and one now gets 6 MFs. If right, my mistake was in line 2.

Nobel laureates can split an atom, but they can’t split a restaurant bill. Ha

Cheers

 

[PresterJohn]

Line 2. The CX HS + leader is 8 MF. The two considered together may only be able to move with 6 MF but the HS has 8 MF.

 

[Philippe D.]

You count the MF available to each unit; in the end the stack will only be able to spend the minimum of the two results.

The leader has 6MF, no questions asked.

The MMC has 4MF, +2 for Double Time, +2 for the leader bonus, -2 for PP (5PP vs 2 IPC and 1 IPC of the leader helping), so that's 6MF.

 

[Robin Reeve]

Just in case, there was an article in an ASL Journal (I cannot remember which one) that introduced the table.

 

[Eagle4ty]

ASL is an infantry game. Like an infantryman, you don't need to know how it works, ruck up and drive on!

 

[Stewart]

A HS carrying five PP has only two MF; if accompanied by a leader, they would have five MF (4.12, 4.42).
If the HS (only) Double Times with the leader, they have six MF. However, if the leader also Double Times, then they have only five MF.

The mathematics would be HS with 8MF and 2 PP and a leader with 6 MF and 1 PP. The PP is two in excess of HS plus leader PP capacity.
8-2=6

An unwounded leader with nothing portaged himself.

 

[PresterJohn]

 

[Old Noob]

Bill Mauldin cartoon: Willy says to Joe "Ya wouldn't be so tired if ya didn't carry all that junk. Get rid of the joker in your deck of cards."

 

[ScottRomanowski]

Also see this expanded set of tables for MF combinations.
@Robin Reeve the article was "Run For The Money" in ASL Annual '96