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stevebugge
I am well aware that the RAW have no function for this, breaking through a barrier is based entirely on it's structure rating. However it seems that in some cases if a large amount of damage is done to a structure in a short period of time then the entire structure should simply fail rather than continue to grow One Square Meter holes. I'm thinking about introducing a houserule for my game that a barrier that takes 4 times it's structure value (or more) in damage from a single attack completely fails, rather than simply developing a 4 square meter hole.

What I'm hoping is to get some wholesale destruction of unarmored windows and internal office framing while still making it difficult to destroy any sort of reinforced structure. I'm also trying to establish a guideline as for just how much explosives would be required to collapse or partially collapse a building, say to cause a catastrophic failure of a load bearing wall. Using this rule it would require 216kg of Commercial Explosives, or 301 KG to hit Heavy Structural Material. (yes this is a serious problem. Every now and then my players decide that the best way to solve a problem is by pancaking a building. This is usually shortly followed by the death or incarceration of the involved characters shortly afterward)
ThreeGee
QUOTE
I am well aware that the RAW have no function for this


The RAW states that a 1 meter square hole is created for each structure rating worth of DV done. So a 2m by 2m section of brick wall would be blown down if you could apply 44 DV to it in one go.
Lagomorph
I think this is a ruling best left to the particular situation and GM, if you make a rule, players will use it. If its left up to the GM, it can be as spectacular or as craptacular as the GM wants.
stevebugge
QUOTE (ThreeGee @ Feb 14 2006, 02:40 PM)
QUOTE
I am well aware that the RAW have no function for this


The RAW states that a 1 meter square hole is created for each structure rating worth of DV done. So a 2m by 2m section of brick wall would be blown down if you could apply 44 DV to it in one go.

Yes that is true, but applying this rule ad nauseum you need to do 66 for a 2x3x.1 section or 88 for a 2x4x.1 section. Using the explosives rules that means 484kg of TNT or ANFO or other Commercial Explosive for the 2x3x.1 or 860kg for the 2x4x.1 section. Up this to Concrete (13 Structure) and it goes to 52 & 301kg for a 2x2x.1, 78 & 676kg for a 2x3x.1, and 104x.1 & 1202kg for a 2x4x.1 section. In short because of the use of the Squareroot in the calculation of damage value in explosives the amount of explosives needed starts getting pretty ridiculous. Using these rules applications like strip mining gold out of granite for example would need atsronomical amounts of explosives. To blast away a 30 meter high by 50 meter wide face 1 meter deep would require 4.6 Megatons of explosive. With 4 successes on a Demolitions test you would still need a little over 850 Kilotons.
stevebugge
QUOTE (Lagomorph)
I think this is a ruling best left to the particular situation and GM, if you make a rule, players will use it. If its left up to the GM, it can be as spectacular or as craptacular as the GM wants.

Very good point. I'm thinking if I use it I'll keep it as a guideline, rather than a hard rule.

Also someone should probably doublecheck my math.
Lagomorph
QUOTE (stevebugge @ Feb 14 2006, 11:54 PM)
QUOTE (ThreeGee @ Feb 14 2006, 02:40 PM)
QUOTE
I am well aware that the RAW have no function for this


The RAW states that a 1 meter square hole is created for each structure rating worth of DV done. So a 2m by 2m section of brick wall would be blown down if you could apply 44 DV to it in one go.

Yes that is true, but applying this rule ad nauseum you need to do 66 for a 2x3x.1 section or 88 for a 2x4x.1 section. Using the explosives rules that means 484kg of TNT or ANFO or other Commercial Explosive for the 2x3x.1 or 860kg for the 2x4x.1 section. Up this to Concrete (13 Structure) and it goes to 52 & 301kg for a 2x2x.1, 78 & 676kg for a 2x3x.1, and 104x.1 & 1202kg for a 2x4x.1 section. In short because of the use of the Squareroot in the calculation of damage value in explosives the amount of explosives needed starts getting pretty ridiculous. Using these rules applications like strip mining gold out of granite for example would need atsronomical amounts of explosives. To blast away a 30 meter high by 50 meter wide face 1 meter deep would require 4.6 Megatons of explosive. With 4 successes on a Demolitions test you would still need a little over 850 Kilotons.

Steve, if you want to make holes in stuff, explosives are not the answer, use the ramming table for a high body vehicle. A 18 body truck, doing 60 meters/turn will do like 54 DV

edit: make sure the truck is expendible though, and preferabbly not manned since it will have to soak 1/2 that amount.
stevebugge
Hmmm It sounds like the Vehicle Ramming rules may come under scrutiny next smile.gif

Seriously this mostly came up because of one player's prediliction for car bombs and the fact that explosive power is a parabolic function.
Ryu
Just decide what effect you consider appropiate. There can be no rule for this without discussing architecture and material sciences.

(structural important vs. unimportant walls): obviously, the former will be harder to break, but more dangerous if it happens. Large buildings have safety margins, especially in earthquake regions.

(Brick walls, ferrocrete, the unthinkable SR-materials: Reaction to force and momentum): some will only be deformed by explosions, but not break. Much damage in one attack goes here. A large explosion generates a larger momentum, possibly pushing some elements of the building over. A multitude of smaller explosions will make many holes, but be more locally confined.

(Just, huh, scientific guesstimates. Glad for corrections I am)
hyzmarca
There are worse things than giant loads of explosives, such as a bullet making a perfectly square hole in a plate glass window without causing any other damage.
nick012000
QUOTE (hyzmarca)
There are worse things than giant loads of explosives, such as a bullet making a perfectly square hole in a plate glass window without causing any other damage.

It's not a perfectly square hole. It just has a praticular area.

So, if the bullet were to make a circular hole, it'll just be or radius of the square root of the area/pi.
Austere Emancipator
QUOTE (stevebugge)
Using these rules applications like strip mining gold out of granite for example would need atsronomical amounts of explosives. To blast away a 30 meter high by 50 meter wide face 1 meter deep would require 4.6 Megatons of explosive. With 4 successes on a Demolitions test you would still need a little over 850 Kilotons.

Those kinds of demolitions are generally performed with a row of holes drilled down along the cliff face with small explosive charges set at certain distances down those holes. Not that that makes it any easier to handle with the canon demolition rules, because you'd still need just as much TNT, no matter if you're making 4 1-meter holes, or 1 2-meter hole.

When you have all the time in the world to set up your explosives, you should get a massive bonus to their rating. For example, to collapse that 30 by 50 by 1 meter section of cliff face without that row of holes or calculated positioning of the charges could take much more than 10 times as much explosives -- if you're trying to make do with just one large explosive charge placed at the center of the cliff face, possibly more than 100 times as much. In fact, to manage a 1-meter deep, 40-meter wide crater in solid rock with a non-shaped charge on the surface could easily require an explosion equivalent to 100 or more tons of TNT.

Anyway, if the rules give you a result like 850kt for the required TNT equivalent weight of the explosive, you might be better off applying common sense to the issue. 100kg TNT in a car driven into a building is enough to flatten a small building. 1000kg TNT in a truck driven through a lobby flattens a large section of a 10-story concrete building. 10000kg TNT inside a building can flatten most things, apart from Arcos.
hobgoblin
the reason for the building being flattend isnt so much about the pure force of the blast, but that it takes out a critical part of the structure.

never forget, the oklahoma building came down largely because the explosives where parked next to a support column.

most buildings are like house of cards, only that the cards are more securely attached. still, take out one of the ones at the bottom and all the rest may come falling.

and i would say that drilling a number of holes and filling them with explosives are basicly what the demolition skill is about.

still, i dont know the numbers for SR4 explosives so i dont know how much things have changed there in comparison to SR3. ie, whats the structual numbers for the diffrent materials, and what are the ratings for diffrent types of explosives...
Austere Emancipator
QUOTE (hobgoblin)
the reason for the building being flattend isnt so much about the pure force of the blast, but that it takes out a critical part of the structure.

True. It's just that 10 tonnes of TNT is usually enough to take out all the critical bits within quite a large radius.
hobgoblin
heh, thats true. and it looks good to smokin.gif
Hasaku
QUOTE (stevebugge)
applying this rule ad nauseum you need to do 66 for a 2x3x.1 section or 88 for a 2x4x.1 section.  Using the explosives rules that means 484kg of TNT or ANFO or other Commercial Explosive for the 2x3x.1 or 860kg for the 2x4x.1 section.  Up this to Concrete (13 Structure) and it goes to 52 & 301kg  for a 2x2x.1, 78 & 676kg for a 2x3x.1, and 104x.1 & 1202kg for a 2x4x.1 section.  In short because of the use of the Squareroot in the calculation of damage value in explosives the amount of explosives needed starts getting pretty ridiculous.  Using these rules applications like strip mining gold out of granite for example would need atsronomical amounts of explosives.  To blast away a 30 meter high by 50 meter wide face 1 meter deep would require 4.6 Megatons of explosive.  With 4 successes on a Demolitions test you would still need a little over 850 Kilotons.

How did you arrive at those figures? I did some math (made an Excel spreadsheet) and got much lower numbers, assuming a decent pool for Demolitions tests. The rules for setting explosives on p. 315 state that base DV is modified rating (base rating + demolition hits) times the square root of the kilograms used.

(Rating + Hits) * Sqr(kg)

The rules for destroying a barrier state that you create a 1m hole for every (Structure Rating) DV, after the attack roll and damage resistance test. You can use Demolitions again as the attack test (in addition to increasing explosive rating), increasing the final DV by hits on this test. Explosives multiply their base DV (which includes modified rating from your first Demolitions test) by 2 when used against barriers, and the barrier resists with Armor vs. explosives attatched directly. Final DV:

((Rating + Demo Hits) * Sqr(kg) * 2) + Demo Hits - Armor Hits

DV to make a hole of (m) meters:

SR * m

Set them equal to each other:

((Rating + Demo Hits) * Sqr(kg) * 2) + Demo Hits - Armor Hits = SR * m

Solve for kg:

kg = (((SR * m) + Armor Hits - Demo Hits) / (2 * (Rating + Demo Hits))) ^ 2

Let's start filling in to make that a little more readable. We'll assume a Demolitions pool of 12, which gives us the average of 4 hits you used in your example. Commercial Explosives have a base rating of 3. Structural Material has a SR of 11 and Armor of 12, giving it 4 average hits on its damage resistance test. Kilograms of Commercial Explosives to make a 6 m^2 hole:

kg = (((11 * 6) + 4 - 4) / (2 * (3 + 4))) ^ 2

kg = (66 / 14) ^ 2

kg = 22.22

22.22kg is a lot less than the 484kg you listed. If I did something wrong, please point it out. The spreadsheet is available here.

CODE

Structural Material
----------------------
Size      kg
------    ------
6          22.22
8          39.51

Concrete
-----------
Size      kg
------    ------
4          14.33
6          31.84
8          56.25

I wasn't sure how to calculate the rock face example.
I used the stats for Concrete, and a 15,000 m^2 hole
since the rules assume a depth of .1 meter.

15k       194M

That's 213.4 kilotons


These numbers are probably still too high to be realistic, but they're a lot saner than your examples.
hobgoblin
hmm, looks like the SR4 demolition may well be a bit more sane then the SR3 version wink.gif

22 kilos to blow a 6 square meter hole in structural material with demolition skill sounds good to me smile.gif

want to low a door of its hinges, a 1 meter hole should do nicely, the rest should be able to swing out of the way (if you managed to take out the lock area that is).
Lagomorph
Hasaku, I think you only get one roll. Setting the explosives counts as the "attack". So your equasion would be doubling up on the successes. I don't have my book so I can't say with certainty, but I think thats how it works.
Hasaku
The section on explosives (p. 315) says "Characters use Demolitions skill to prepare explosives, measuring how well they target key or weak structural points and focus the blast. <snip> See Destroying Barriers, p. 157, for calculating an explosive's effect on a barrier."

The section on destroying barriers (p. 157) says, "If a character is attacking a barrier with intent to destroy it (or create a hole), resolve the attack normally. <snip> A character may use Demolitions skill as the attack skill if he has the proper materials and time to set charges."

So the explosives section can be read as "Use Demolitions to increase the power, then resolve the attempt like an attack as per Destroying Barriers" or "use Demolitions to increase the power, then use the results part of Destroying Barriers to see how big the hole is." Logically, they both represent knowledge of how best to place the explosives, and the attack roll should only be used for actual attacks. However, the rules aren't very clear on this point and could be read both ways.

I reran the calculations without the attack roll, and the difference was very minor. Demolitions skill provides a much greater effect when used to increase the effective Rating of the explosive rather than simply add a few points of DV.
Ryu
More exact mechanics are still far from reality and offer no advantage for gameplay. Trust your GM.

I suggest setting a required quantity of explosives arbitrarily and judge the quality of success after an demolitions test. Flattening a building by filling the basement with C12 will always work, strategically destroying a specified part of a wall not so.
Rotbart van Dainig
Mechanics of Materials are not trivial, but for abstract cases, not too complicated, either:

In this case, it is mostly about pressure tension in the loadbearing walls, tension being force per area (section) - and every material has an allowed maximum of tension above wich it deforms and eventually collapses.

So it suffices to set an approximate value how much reduction in area a structure can take before collapsing, counting all the greatest longitudinal cuts together.
stevebugge
QUOTE (Hasaku)
QUOTE (stevebugge)
applying this rule ad nauseum you need to do 66 for a 2x3x.1 section or 88 for a 2x4x.1 section.  Using the explosives rules that means 484kg of TNT or ANFO or other Commercial Explosive for the 2x3x.1 or 860kg for the 2x4x.1 section.  Up this to Concrete (13 Structure) and it goes to 52 & 301kg  for a 2x2x.1, 78 & 676kg for a 2x3x.1, and 104x.1 & 1202kg for a 2x4x.1 section.  In short because of the use of the Squareroot in the calculation of damage value in explosives the amount of explosives needed starts getting pretty ridiculous.  Using these rules applications like strip mining gold out of granite for example would need atsronomical amounts of explosives.  To blast away a 30 meter high by 50 meter wide face 1 meter deep would require 4.6 Megatons of explosive.  With 4 successes on a Demolitions test you would still need a little over 850 Kilotons.

How did you arrive at those figures? I did some math (made an Excel spreadsheet) and got much lower numbers, assuming a decent pool for Demolitions tests. The rules for setting explosives on p. 315 state that base DV is modified rating (base rating + demolition hits) times the square root of the kilograms used.

(Rating + Hits) * Sqr(kg)

The rules for destroying a barrier state that you create a 1m hole for every (Structure Rating) DV, after the attack roll and damage resistance test. You can use Demolitions again as the attack test (in addition to increasing explosive rating), increasing the final DV by hits on this test. Explosives multiply their base DV (which includes modified rating from your first Demolitions test) by 2 when used against barriers, and the barrier resists with Armor vs. explosives attatched directly. Final DV:

((Rating + Demo Hits) * Sqr(kg) * 2) + Demo Hits - Armor Hits

DV to make a hole of (m) meters:

SR * m

Set them equal to each other:

((Rating + Demo Hits) * Sqr(kg) * 2) + Demo Hits - Armor Hits = SR * m

Solve for kg:

kg = (((SR * m) + Armor Hits - Demo Hits) / (2 * (Rating + Demo Hits))) ^ 2

Let's start filling in to make that a little more readable. We'll assume a Demolitions pool of 12, which gives us the average of 4 hits you used in your example. Commercial Explosives have a base rating of 3. Structural Material has a SR of 11 and Armor of 12, giving it 4 average hits on its damage resistance test. Kilograms of Commercial Explosives to make a 6 m^2 hole:

kg = (((11 * 6) + 4 - 4) / (2 * (3 + 4))) ^ 2

kg = (66 / 14) ^ 2

kg = 22.22

22.22kg is a lot less than the 484kg you listed. If I did something wrong, please point it out. The spreadsheet is available here.

CODE

Structural Material
----------------------
Size      kg
------    ------
6          22.22
8          39.51

Concrete
-----------
Size      kg
------    ------
4          14.33
6          31.84
8          56.25

I wasn't sure how to calculate the rock face example.
I used the stats for Concrete, and a 15,000 m^2 hole
since the rules assume a depth of .1 meter.

15k       194M

That's 213.4 kilotons


These numbers are probably still too high to be realistic, but they're a lot saner than your examples.

For the strip mining example I coverted Kilograms to Pounds, which either makes the difference or is where I made the math error. I also Multiplied the Barrier Rating by 10 to account for 1 full meter of thickness.
stevebugge
QUOTE (Rotbart van Dainig)
Mechanics of Materials are not trivial, but for abstract cases, not too complicated, either:

In this case, it is mostly about pressure tension in the loadbearing walls, tension being force per area (section) - and every material has an allowed maximum of tension above wich it deforms and eventually collapses.

So it suffices to set an approximate value how much reduction in area a structure can take before collapsing, counting all the greatest longitudinal cuts together.

This is the kind of help I was looking for. So do you think that a flat number like my proposed 4 x the structure rating works, or should it be more flexible like any single attack that compromises more than say 50% of the barrier's surface area?
hobgoblin
i would say that when a support structure is at 50% of its original rating, it have a potential for collapse. this can come from a single attack, or over time.

but one should allso look at whatever its supposed to carry. ifs its only about a wall thats a number of stacked bricks, then blowing out 50% from the top down dont help much. but doing the same from the bottom up, can spell trouble...

but if there is something then resting on top of said wall, any reduction in the rating carry a potential risk in the wall loosing its ability to carry weight, and therefor its crumbles under the weight of whatever is stacked on top...
Rotbart van Dainig
QUOTE (stevebugge)
This is the kind of help I was looking for.  So do you think that a flat number like my proposed 4 x the structure rating works, or should it be more flexible like any single attack that compromises more than say 50% of the barrier's surface area?

SR translates explosives and skill into volume reduction (lenght x depth x height), so you can stick with those rules and use the latter:

You don't even have to worry about area (lenght x depth) for an aproximation in the case of walls most of the time, just substract the diameter of the hole(s) from wall length.

If that reduces the total loadbearing wall lenght beneath the safe point, it comes down.

50% seems an easy to handle value for the start.

PS:
QUOTE (SR4 @ p. 157, Structure rating)
The Structure rating is the number of “damage boxes” required to destroy a section 1 meter square and about 10 cm thick (approximately the typical wall thickness for a residential or offi ce building).
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