BFP 104 Flying Turrets_Bombardment Question

Ganjulama

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I'm getting ready to play this next week vs. Nelson.

Nelson contends that the 4 hex bombardment range includes the target hex. I'm assuming this rule derives from KGP and VoTG bombardment rules which state, in part, "The final placement hex is the center hex for the Bombardment, and all hexes within [X] hexes of that hex now undergo a Bombardment".

Who is right?
 

Chas

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It includes the target hex.

I'm getting ready to play this next week vs. Nelson.

Nelson contends that the 4 hex bombardment range includes the target hex. I'm assuming this rule derives from KGP and VoTG bombardment rules which state, in part, "The final placement hex is the center hex for the Bombardment, and all hexes within [X] hexes of that hex now undergo a Bombardment".

Who is right?
 
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Just so I'm clear, the '4 hex radius' count includes the initial target hex. So really this affects target hex and every hex within 3 hexes of it?
Looking at C1.72 Harassing Fire, a two-hex radius extends from the “target” hex within “two” hexes (double the normal range of a FFE), so I would say “and every hex within 4 hexes”. But I’m just a newbie.
 

Ganjulama

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Looking at C1.72 Harassing Fire, a two-hex radius extends from the “target” hex within “two” hexes (double the normal range of a FFE), so I would say “and every hex within 4 hexes”. But I’m just a newbie.
Well, I'm a grognard and I'm having a hard time understanding this ruling so your instincts are true young padawan. It contravenes a number of other ASLRB conventions as well:

From KGP:

When the U.S. player purchases a pre-game Bombardment, he records a secret Pre-Reistered hex for it. To resolve each Bombardment. after all on-map set up is completed he places an AR counter in the Pre-Registered hex and makes a C1.31 error DR (with the white dr halved: FRU) to determine the Bombardment Centre Hex. He then places a FFE:C counter in that Centre Hex. All hexes within seven hexes of that hex now undergo Bombardment as per C1.82-C1.823. There are no "spared hexes". After the Bombardment has been fully resolved, its FFE:C is removed and the Bombardment RG is eliminated. Each Bombardment is fully resolved before conducting another Bombardment (if any).

From VoTG:

When (he German player purchases a pre-game Aerial Bombardment, he records a secret Pre-Registered hex for it. To resolve each Bombardment, after all on-map setup is completed he places an FFE:1 counter in the Pre-Reg hex and makes an accuracy dr per C1.732;. The final placement hex is the center hex for the Bombardment, and all hexes within two hexes of that hex now undergo a Bombardment as per C1.82-C1.823. There are no immune hexes. Each Bombardment is fully resolved before conducting another Bombardment (if any). After a Bombardment is resolved the FFE:1 counter is removed and that Bombaidment RG is Eliminated.

The SSR states from this scenario states:

Prior to all set up the Russian secretly records three pre-registered hexes for Bombardment (C1.8) with a four hex radius. These are automatically accurate and there are no immune hexes.

In light of this ruling I think there should be errata posted for this scenario revising this SSR to state:

5. Prior to all set up the Russian secretly records three pre-registered hexes for Bombardment (C1.8). To resolve each Bombardment. after all on-map set up is completed he places an AR counter in the Pre-Registered hex. He then places a FFE:C counter in that Centre Hex. All hexes within THREE hexes of that hex now undergo Bombardment as per C1.82-C1.823. There are no "spared hexes". After the Bombardment has been fully resolved, its FFE:C is removed and the Bombardment is finished.

This revision would clearly state the intention of the designer @Chas . I know this scenario has gotten play before and I would be interested in how others played this SSR @jrv , @JimWhite . Perhaps @rreinesch can tell us how it was played at Texas Team Tournament???
 

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But isn't this really a matter of 6 of one is half-dozen of another? What's the difference between "He then places a FFE:C counter in that Centre Hex. All hexes within THREE hexes of that hex..." and "...pre-registered hexes for Bombardment (C1.8) with a four hex radius" in terms of describing the area?
 

jrv

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But isn't this really a matter of 6 of one is half-dozen of another? What's the difference between "He then places a FFE:C counter in that Centre Hex. All hexes within THREE hexes of that hex..." and "...pre-registered hexes for Bombardment (C1.8) with a four hex radius" in terms of describing the area?
The center hex is at radius zero by my reckoning. The surrounding six hexes are a radius one "circle", as I would read it. If the surrounding six hexes are radius zero, that means the hex itself is radius minus one. I can't see that.

JR
 

Chas

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For full clarity on the discussion here the target hex and all hexes within 4.
What, 7-8 years later, a bunch of playings, and first I can recall this has even been brought up.
 

Paul M. Weir

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This reminds me of the old fence post and wire problem from school days or clarity in the definition of numerical ranges. Eg you need 4 posts for 3 lengths of wire or 5 posts for 4 lengths of wire. In school problems ranges could include one or both or neither limits (0<X<4, 0<=X<4, 0<X<=4 or 0<=X<=4).

ASL usually has an effective inclusive usage, eg if you say a MMC has a range of 4 then that includes the firer hex AND 4 other hexes effectively the "home" hex is not counted against the 4 (IE it's hex "0"). The hexes that can be covered by such a unit is 1 hex (firer's hex) + 6 (adjacent hexs) + 12 (range 2) + 18 (range 3) + 24 (range 4) for a total of 61 hexes within normal range. If N is the range then hexes within normal range is 1+6*sum(1 to N) = 1+6(N(N+1))/2 = 1+3N(N+1). Another way of viewing is if the range (the radius) is N then it covers a "circle" with a "diameter" of 2N+1.

All the above posts have left me a little confused, but
For full clarity on the discussion here the target hex and all hexes within 4.
makes me believe that the target hex does not count against the 4, that the "diameter" is 9 hexes (4+1+4) and the total number of hexes is 61.

That's roughly the same total number of hexes from A1 to F10 on a normal 8"x22" board (very different shape though).

At least that's my reading of it and I could be wrong.
 
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