QUOTE (Arsenal @ pg 93)
Cutting requires far less explosives than breaching, but more preparation time.
Makes sense, but I can't get it to work out in practice.
Okay, lets say we are faced with a security door. That counts as Reinforced Material, so we have Structure Rating of 9, and an Armor Rating of 8.
Let's also say that we have a drone carrying 200 foam explosive canisters, each packed with 1 kilogram of Rating 4 foam explosive.
So if we decide to breach the door, lets figure out how much explosive it's going to take.
Arsenal gives us the following:
DV required = (Radius[m] x Structure Rating) + (Armor Rating ÷ 4)
Explosive weight = (DV ÷ DV Multiplier ÷ Explosives Rating)squared in kg
So to blow a 1 meter diameter hole, we need a DV of
(.5 * 9) + (8 / 4)
(4.5) + (2)
6.5 or 7 DV
Let's say we are just putting the explosives on the ground, so the Multiplier is x1.
The weight in kilograms is:
(7 / 1 / 4) squared kilograms
3 kilograms
So 3 kilos of rating 4 foam explosive will put a 1 meter diameter hole in our security door.
Good 'nuff.
So let's say instead of breaching it, we decide to try cutting it.
We are going to attempt to cut a 1 square meter hole out of the door.
We are still using rating 4 foam explosive, and we will shape it into four 1-meter lines.
The DV formula for cutting is: (sqrt of kilos/meter)*(Rating of explosives)
We also get 1 additional AP per net hit made on the demo extended test (assume no net hits, so no AP)
The door can buy 2 hits from it's armor, effectively giving it 11 damage boxes.
If we want to do 11 boxes of damage, we will need almost 9 KILOGRAMS PER ONE METER LINE!!!
DV = (sqrt 9)*(4)
DV = (3)*(4)
DV = 12
Holy shit. You need a grand total of 36 kilograms of the same explosive to cut a 1meter square hole out of that security door, whereas you only need 3 kilos to blow a 1meter diameter hole. WTF???
Please tell me my math is wrong...
[edit]
My math WAS wrong, so I fixed it. But all the correction did was change the required mass of the breaching charge from 5 kilos to 3, exacerbating the problem with the cutting charge calculation.
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