Philippe D.
Elder Member
Do you have to be able to demonstrate the require skill by actually stacking the counters, with instant "falling rubble" rules?
Correct about the German halftrack. I can fit a combo of 14 effective Heroes/LMG in there.
Well, here comes my final say:
For a Level 2 Building with Ground, 1st, and 2nd Level:
34 Heroes with 17 HMG for 119FP @ -34 DRM
34 Heroes with 17 HMG for 119FP @ -34 DRM
34 Heroes with 17 HMG for 119FP @ -34 DRM
In Bypass a SPW 251/1 halftrtrack with a 3FP AAMG plus loaded with 7x Hero, 7x LMG, using up 14 out of 15 available PP for another 21FP totalling 24FP.
This would yield a grand total of 381FP @ -102.
With Captured Use weaponry:
34 Heroes with 17 captured .50 Cals for 136FP @ -34 DRM
34 Heroes with 17 captured .50 Cals for 136FP @ -34 DRM
34 Heroes with 17 captured .50 Cals for 136FP @ -34 DRM
In Bypass a captured US M3 (HMG) halftrtrack firing through its rear target facing for 20FP plus loaded with 5x Hero, 5x LMG, using up 10 out of 10 available PP for another 15FP totalling 35FP.
This would yield a grand total of 443FP @ -102.
My idea to yet 'up' it was to add a SMC Rider to the halftracks per
"D6.2 RIDERS: Riders are Personnel being transported on the outside of an AFV (or Cavalry or motorcyclists). AFV Riders cannot be used until 1942—and even then only by the Russians. In 1943 and thereafter, all nationalities may use Riders. Riders are limited to tanks (not tankettes), TD, SPA, SPAA, Carriers (limited capacity; 6.81), and Assault Guns. Rider capacity is limited to a maximum of 14 PP (as per 6.1); within those timeframes a vehicle not otherwise granted Rider capability can carry one SMC as a Rider and the two PP he possesses. See A5.5 for equivalents. Rider PP cannot be used to satisfy the ammo PP reduction (C10.13) of any Gun that requires a T# (C10.1) to be transported; therefore, a dm 76-82mm mortar is transported by Riders at its normal five PP cost."
Thus, if the 10-3s would have made a difference, one of them could have been placed as a Rider on the US halftrack laden so far with only 4x Hero & 4x LMG, allowing for a fifth Hero+LMG combo as Passenger to use up the 9th and 10th PP while the 10-3 could have directed everything as a Rider. Placing an extra Hero as a Rider does not change anything because Riders may not use SW and only fire within their normal range.
In case the challenge would have been set at range 4 (and not 6), then the extra Hero Rider with his LMG would have added 1.5FP (halved for being a Rider) and an extra -1 Heroic DRM.
So it seems that the final score is settled at
443FP @ -102.
von Marwitz
Yeah, that's my luck as well.And with my luck, I'd roll a 6,6
Well, the good news is that it wouldn't matter.And with my luck, I'd roll a 6,6
But what if Random Selection is boxcars all the way down?Obviously, he has forseen the possibility of boxcars and taken precautions.
von Marwitz
Doesn't matter. As per A9.7, the attack is still resolved when you roll boxcars.But what if Random Selection is boxcars all the way down?
If limbered, the M51 even has 24 FP. However, together with a crew it would count as 1 squad-equivalent for stacking purposes. 10 Heroes with 5x .50 Cals give you 40FP @ -10 DRM, while you "only" get 24 FP @ 0 DRM for the M51.Did anyone take into account fortified Buildings with something like the M51 Multiple .50-cal MG Carriage? It is small so can go into a building and has 20 attack. Just saying other than the 800mm gorilla in the room, the Holy handgrenade and the occasional backpack nuke this has been an interesting thread.
I'm just trying to find away to make that unit plausible.If limbered, the M51 even has 24 FP. However, together with a crew it would count as 1 squad-equivalent for stacking purposes. 10 Heroes with 5x .50 Cals give you 40FP @ -10 DRM, while you "only" get 24 FP @ 0 DRM for the M51.
The thing just ain't gettin' the lead out...
von Marwitz
Ahem, it needs a 9 KIA to kill a cat.But even if you would Cower from 443FP to the 'next column' of 36FP (call that a drop...), the -102 DRM would do in Major von Schoff with a 7 KIA even if he were a cat.
A B12 captured weapon becomes effectively a B10, not a B11.Besides, Oberst von Marwitz argues, that the chance for all 51 .50 Cals to malfunction despite being Captured Equipment is only (a still surprising) 1,2%.
It depends on whether the hex in question is located on the Anglo-American or German turf.Ahem, it needs a 9 KIA to kill a cat.
REALLY?!? I'd have thought Germans would have engineered a superior cat with at LEAST 10 lives. ?It depends on whether the hex in question is located on the Anglo-American or German turf.
On the former, cats are said to have nine lives, on the latter only seven.
I have wondered more than once what the 'opinion' on this is in other European countries. Not unlikely, it might be one of these efforts of going great lengths to be different: Driving on the wrong side, no metric system, not abolishing the mounted lance before 31. December 1927... Probably, you need two extra lives as a cat to put up with all that.
von Marwitz
Yeah. It had long fur, fangs and claws but it only produces one or occasionally two kittens per litter, is grossly overweight, can't cross soft ground, its legs frequently give way, eats trice as much as an alley cat and only capable of walking to the end of the back yard. Oh and the vet bills!REALLY?!? I'd have thought Germans would have engineered a superior cat with at LEAST 10 lives. ?
But if it hits you, it gorges on your scattered entrails like a 'King Tiger' for a week or so...Yeah. It had long fur, fangs and claws but it only produces one or occasionally two kittens per litter, is grossly overweight, can't cross soft ground, its legs frequently give way, eats trice as much as an alley cat and only capable of walking to the end of the back yard. Oh and the vet bills!
PaulAhem, it needs a 9 KIA to kill a cat.
A B12 captured weapon becomes effectively a B10, not a B11.
For 51 SW with an effective B#, the chances of not all breaking, all breaking
B12: 76.23%, 23.77%
B11: 98.82%, 1.18%
B10: 99.9908, 0.0092%
Ahem, it needs a 9 KIA to kill a cat.
A B12 captured weapon becomes effectively a B10, not a B11.
For 51 SW with an effective B#, the chances of not all breaking, all breaking
B12: 76.23%, 23.77%
B11: 98.82%, 1.18%
B10: 99.9908, 0.0092%
I think he may have mistyped (i.e., reversed) the B10 and B12 results, for exactly the reason you suggest.Paul
I am struggling with this.
How can it be more likely that a B12 weapon will all break than a B10 weapon?
Also a 23.77% chance of all breaking seems unlikely as you would need 51 identical dice rolls on the RS dice.
Maybe I'm just missing something.
Gavin
I don't know about the math, but I know there are no adults in the room. -- jimNow I'll step back and wait for the adults in the room to explain my errors.