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Changing fragmentation, violent fragmentation, and pulverization values (Which uses shear strength, tensile strength, and compressive strength values that don't relate at all to destroying a physical material via physical attacks) to use toughness strength.
Reference: MPa.m1/2, where m1/2 is 1 (Megapascal per (meter^1))/2. Essentially, we'll divide the MP per meter³ in two which will halve our final result.
Table of Values
Wood
Area under the curve: 16260px 100 kN - 69px 10 mm - 94px
Work of deformation: 16260*(100/69)*(10/94) = 2507J
The specimen is 25x10x15 cm wooden block so its volume is 3750 cm³.
Toughness: 2507/3750 = 0.67 J/cm³
Bone
Toughness: 2.85 J/cm³
Glass
Using 0.7 MPa• m½ as the middle ground
Toughness: 0.35 J/cm³
Plastic
So plastic is pretty much impossible to find so instead Polypropylene will be used since this is the type of plastic used most commercially and considered the "safest", having 0.673, 0.644, 0.544, & 0.529 MPa• m½.
Using 0.5975 MPa• m½ as the middle ground
Toughness: 0.29875 J/cm³
Rocks and Minerals
Limestone
Area under the curve: 166400px 1000 psi - 6.895 MPa - 77px 0.0005 (unitless) - 129px
Limestone's toughness: 166400*(6.895/77)*(0.0005/129) = 0.058 J/cm³.
Sandstone
Area under the curve: 71070px 2000 psi - 13.79 MPa - 79px 0.002 (unitless) - 112px
Sandstone's toughness: 71070*(13.79/79)*(0.002/112) = 0.22 J/cm³.
Igneous Rock
Igneous Rock Stress and Strain
Basalt
The fracture toughness of Icelandic basalt was 2.4 MPa m1/2.
Toughness: 2.4 MPa m1/2 = 1.2 J/cm³
Lunar Rock
Diamond
Diamond Stress and Strain The Fracture Toughness of diamond is considered "good" to "exceptional", which gets it a value of 3.4 MPa• m½.
Toughness: 3.4 MPa• m½ = 1.7 J/cm³
Ice
Ice toughness value of 145.7 kPa per meter³ (You'll need to download this particular pdf to see the full research).
Ice toughness = 0.1457 J/cm³
Concrete
Area under the curve: 18134px 20 MPa - 77px 0.05% - 56px
Toughness: 18134*(20/77)*(0.05/100/56) = 0.042 J/cm³
Official Source for reference.
Concrete on average has a fracture toughness of 0.2 - 1.4 MPa• m½.
Toughness: 0.2 - 1.4 MPa• m½ = 0.1 - 0.7 J/cm³.
Soil
Area 81492px 0.05MPa - 67px 1% - 84px
Toughness: 81492*(0.05/67)*(0.01/84) = 0.0072 J/cm³
Metallic Elements
Iron
Iron toughness value is on average 135 MPa• m½.
Iron toughness value is 67.5 J/cm³.
Cast Iron
Iron toughness value is on average 81.4 MPa.m1/2.
Cast Iron toughness value is 40.7 J/cm³.
Steel
Maraging Steel (200 Grade) has a fracture toughness of 175 MPa• m½.
Toughness: 175 MPa• m½ = 87.5 J/cm³.
Titanium
Titanium has a fracture toughness of 95.5 MPa• m½.
Toughness: 95.5 MPa• m½ = 47.75 J/cm³.
Aluminum
According to this aluminum has multiple values, 1.20, 0.93, 0.53, 0.36, and 0.25 MPA• m½.
Using 0.654 MPA• m½ as a middle ground.
Toughness: 0.327 J/cm³.
Copper
Copper ranges from 40 to 100 MPa• m½.
Using 70 MPa• m½ as a middle ground.
Toughness: 35 J/cm³.
Nonmetallic Elements
Silicon
Silicon toughness value is 0.8 - 1 MPa.m1/2.
Silicon Toughness = 0.4 J/cm³ - 0.5 J/cm³
Different levels to destroying material
This will be brief. Moving forward, we will no longer be using fragmentation, violent fragmentation, etcetera. Instead, we will be taking the percentage of the affected volume as the final result, subtracting the volume left intact.
This will change how feats moving forward are handled as now, the more of the volume is left unharmed, the less will be factored into the final equation.
Cloud Related Feats
Turns out that clouds have a huge issue too. Mainly the idea of cloud splitting or affecting a portion of a cloud feats, the core idea behind it is that a cloud has a lot of weight it can hold, the typical cumulus cloud “weighs” about 1 billion 400 million pounds, though the issue is this is not the same kind of weight as let’s say, me lifting a 40 kg box means I’m lifting the whole 40 kg part of it (ignoring friction and work). While a cloud can weigh this much, the mass being spread over such a large volume of space, the density, or weight (mass) for any chosen volume, is very small. Basically meaning that one area of a cloud is not equal to the full weight of the cloud, the same with splitting it, affecting a portion of it, etcetera. Due to this, cloud splitting or portion of cloud feats should be percentage based, the only ones that will keep full mass are cloud feats where the entire cloud is affected, and it must be affected in a literal instant. The reason for this is since the mass is spread over a cloud, if it takes a timeframe to disperse the entirety of it, the entire mass is not being affected, however it’s similar to that of a chain reaction.
Cooling Calcs
This one will be a pretty long explanation due to the very science behind these.
Heating falls under a “Thermodynamic System”.
“A thermodynamic system is a body of matter and/or radiation, confined in space by walls, with defined permeabilities, which separate it from its surroundings. The surroundings may include other thermodynamic systems, or physical systems that are not thermodynamic systems. A wall of a thermodynamic system may be purely notional, when it is described as being 'permeable' to all matter, all radiation, and all forces”
A system is typically differentiated between being open, closed or isolated. In an open system matter and heat can be added and removed. In a closed system only heat can be added or removed. An isolated system can not be changed.
Which leads to cooling.
Yet again we have an object which we use as our closed system. This time however we can’t simply add energy into it. After all, that would simply heat it up. Technically we could try and use energy to force the molecules to stop, but the more we try to stop them, the hotter everything gets and the more they try to move, so this would result in needing infinite energy to cool something down. The more obvious solution is to pass on the energy to the environment or even better, a different system. Such a system could be the air that is being moved by a fan. The air passes by our initial system and the energy moves from one system to the other. However, in this case the energy that is needed to power our second system is not proportional to the energy removed from our initial system.
The way we originally treated this was we assumed that the energy is directly removed from the system, without the need of any other system or the environment. This is far too assumptious however to be used as a solution.
An example of an issue with a cooling feat, let’s say we use someone like Paula who uses her PSI Freeze to create a cold swirl of air that freezes the air itself, essentially turning it into ice. This feat doesn’t work for a major reason regarding the substance, nitrogen is actually very volatile even at very low temperatures, meaning that it changes states of aggregation a lot. If you were to put this “out in the open”, it would immediately turn gaseous again.
Cloud Creation
Cloud creation falls under a similar issue.
To quote NASA:
“Clouds are created when water vapor, an invisible gas, turns into liquid water droplets. These water droplets form on tiny particles, like dust, that are floating in the air.”
Said condensation releases the thermal energy stored within the water vapor due to the change in its state of aggregation, from gaseous to liquid. This is also what is responsible for the energy within a storm.
The way we originally treated this was we assumed that the latent heat of condensation is what determines the amount of energy required to create a cloud. To put this into simpler terms, it is like putting a lid on a pot with boiling hot water and thinking that the energy it takes to put on said lid is equal to the heat that comes from the water condensing on the lid. The issue there should be pretty obvious.
Minor Changes
These are minor changes that will be added overtime.
Pixel Scaling
Pixel Scaling innately should have a picture not grabbed from a video, for lower feats this isn't an issue, but for higher feats where a pixel can be millions of kilometers, this could cause major inflation in the calculation. Please look out for these.
Cloud Volume Calculation Change
This part was originally on the page "Using the two values from above one can approximate the volume of the clouds. A simple acceptable approximation is given by pi*(viewing distance)^2*(cloud thickness)." But I for the life of me cannot find where this formula comes from, looking it up the best formula I could find was volume of clouds getting length * width * height which seem to naturally make more sense. The other two methods that are noted as better on the page should be the primarily used methods too: "There are better, but more complicated, formula model the clouds as the intersection of a spherical shell and a spherical sector."
