Reference for Common Feats

Throughout fiction and real life, there have been numerous feats demonstrating a certain character's or objects destructive power. This page's purpose is to consolidate calculations for those feats for better convenience in determining an object's or character's Attack Potency or Durability. Please note, these calculations are typically low-ends or averages and may not be a one-size-fits-all due to outliers. I.E. freezing of an average human will most likely not apply to one that is exceedingly large or incredibly diminutive.

Impact Feats

If not slammed into a wall

When being hit by a car, the linear momentum of the car+person system needs to remain the same. Linear momentum is m*v

The values vary based on the vehicle and the speed of course.

For example, assuming the human is 70 kg, the car is 1500 kg and that the car's speed is 11.176 m/s:

FinalSpeed = (MassCar*InitialSpeed):(MassPerson+MassCar)

Using the values above this is 10.677707006369426751592356687898 m/s.

KE of the person is 3990.4699419854760842224836707371 Joules

Peak Human level

Getting Hit by a Car

25 mph or 11.176 m/s (Average suburb speed): ((1500*11.176)/(70+1500))^2*70*0.5 = 3990.47 J or 3.99047 kilojoule (Peak Human level)

45 mph or 20.1168 m/s (Daily City travel speed): ((1500*20.1168)/(70+1500))^2*70*0.5 = 12,929.12 J or 12.929 kilojoules (Peak Human level+)

60 mph or 26.8224 m/s (Traditional interstate travel speed): ((1500*26.8224)/(70+1500))^2*70*0.5 = 22,985.1069 J or 22.985 kilojoules (Wall level)

70 mph or 31.2928 m/s (Highway speed limit): ((1500*31.2928 m/s)/(70+1500))^2*70*0.5 = 31,285.284 J or 31.285 kilojoules (Wall level)

Getting hit by a Pickup Truck

The average pickup trucks can weigh over 4082.3 kg.

25 mph or 11.176 m/s (Average suburb speed) = ((4082.3*11.176)/(70+4082.3))^2*70*0.5 = 4225.45244 joules, or 4.225 Kilojoules - Peak Human level

45 mph or 20.1168 m/s (Daily City travel speed) = ((4082.3*20.1168)/(70+4082.3))^2*70*0.5 = 13690.4659045 joules, or 13.69 Kilojoules - Peak Human level+

60 mph or 26.8224 m/s (Traditional interstate travel speed) = ((4082.3*26.8224)/(70+4082.3))^2*70*0.5 = 24338.6060524 joules, or 24.33 kilojoules - Wall level

70 mph or 31.2928 m/s (Highway speed limit) = ((4082.3*31.2928)/(70+4082.3))^2*70*0.5 = 33127.5471269 joules, or 31.127 kilojoules - Wall level

Getting Hit by a Bus

The average "traditional-sized" school bus weighs in at 10,659.421 kg.

25 mph or 11.176 m/s (Average suburb speed) = ((10659.421*11.176)/(70+10659.421))^2*70*0.5 = 4314.74851771 J or 4.314 kilojoules (Peak Human level)

45 mph or 20.1168 m/s (Daily City travel speed) = ((10659.421*20.1168)/(70+10659.421))^2*70*0.5 = 13979.7851974 J or 13.98 kilojoules (Peak Human level+)

60 mph or 26.8224 m/s (Traditional interstate travel speed) = ((10659.421*26.8224)/(70+10659.421))^2*70*0.5 = 24852.951462 J or 24.852 kilojoules (Wall level)

70 mph or 31.2928 m/s (Highway speed limit) = ((10659.421*31.2928)/(70+10659.421))^2*70*0.5 = 33827.63 J or 33.827 kilojoules (Wall level)

Getting hit by a Semi Truck

The average semi-truck can weigh in excess of 36,287 kg.

25 mph or 11.176 m/s (Average suburb speed) = ((36287*11.176)/(70+36287))^2*70*0.5 = 4354.787 joules, or 4.354 kilojoules - Peak Human level

45 mph or 20.1168 m/s (Daily City travel speed) = ((36287*20.1168)/(70+1500))^2*70*0.5 = 14109.50864 joules, or 14.109 kilojoules - Wall level

60 mph or 26.8224 m/s (Traditional interstate travel speed) = ((36287*26.8224)/(70+36287))^2*70*0.5 = 25083.5709154 joules, or 25.083 kilojoules - Wall level

70 mph or 31.2928 m/s (Highway speed limit) = ((36287*31.2928)/(70+36287))^2*70*0.5 = 34141.5270794 joules, or 34.141 kilojoules - Wall level

If slammed into a wall

However, it should be noted that the above calculations assume that the person is sent flying by the car. In some odd cases in fiction, the car stops and the character tanks the attack. Or in some cases, a character is slammed into a wall by a car. In these cases, the entire KE of the car scales to the character's durability.

KE = 1/2*mass*velocity^2 (Where mass is in kilograms and velocity is in meters per second)

Getting Hit by a Car

0.5*1500*11.176^2 = 9.3677232e4 Joules - Wall level

This value assumes that this is an average-sized car weighing in at 1500 kg and travelling at 25 mph/11.176 m/s.

45 mph or 20.1168 m/s (Daily City travel speed) = 0.5(1500) * 20.1168^2 = 303,514.23168 joules, or 303.5 Kilojoules - Wall level

60 mph or 26.8224 m/s (Traditional interstate travel speed) = 0.5(1500) * 26.8224^2 = 539,580.85632 joules, or 539.5 Kilojoules - Wall level+

70 mph or 31.2928 m/s (Highway speed limit) = 0.5(1500) * 31.2928^2 = 734,429.49888 joules, or 734 Kilojoules - Wall level+

Here are some values for other vehicle types and the like.

Getting hit by a Pickup Truck

The average pickup trucks can weigh over 4082.3 kg.

25 mph or 11.176 m/s (Average suburb speed) = 0.5(4,082.3) * 11.176^2 = 254,945.709462 joules, or 255 Kilojoules - Wall level

45 mph or 20.1168 m/s (Daily City travel speed) = 0.5(4,082.3) * 20.1168^2 = 826,024.098658 joules, or 826 Kilojoules - Wall level+

60 mph or 26.8224 m/s (Traditional interstate travel speed) = 0.5(4,082.3) * 26.8224^2 = 1,468,487.2865 joules, or 1.5 Megajoules - Room level

70 mph or 31.2928 m/s (Highway speed limit) = 0.5(4,082.3) * 31.2928^2 = 1,998,774.362185216 joules, or 2 Megajoules - Room level

Getting Hit by a Bus

The average "traditional-sized" school bus weighs in at 10,659.421 kg.

25 mph or 11.176 m/s (Average suburb speed) = 0.5(10,659.421) * 11.176^2 = 665,696.702668 joules, or 666 Kilojoules - Wall level+

45 mph or 20.1168 m/s (Daily City travel speed) = 0.5(10,659.421) * 20.1168^2 = 2,156,857.31665 joules, or 2.15 Megajoules - Room level

60 mph or 26.8224 m/s (Traditional interstate travel speed) = 0.5(10,659.421) * 26.8224^2 = 3,834,413.00737 joules, or 4 Megajoules - Room level

70 mph or 31.2928 m/s (Highway speed limit) = 0.5(10,659.421) * 31.2928^2 = 5,219,062.14892063232 joules, or 5.22 Megajoules - Room level

Getting hit by a Semi Truck

The average semi-truck can weigh in excess of 36,287 kg.

25 mph or 11.176 m/s (Average suburb speed) = 0.5(36,287) * 11.176^2 = 2,266,177.145056 joules, or 2.27. Megajoules - Room level

45 mph or 20.1168 m/s (Daily City travel speed) = 0.5(36,287) * 20.1168^2 = 7,342,413.94998144 joules, or 7.34 Megajoules - Room level

60 mph or 26.8224 m/s (Traditional interstate travel speed) = 0.5(36,287) * 26.8224^2 = 13,055,127.03695416 joules, or 13 Megajoules - Room level+

70 mph or 31.2928 m/s (Highway speed limit) = 0.5(36,287) * 31.2928^2 = 17,766,828.81723904 joules, or 17.77 Megajoules - Room level+

Falling from Great Heights

The energy of a falling object can be calculated by gravitational potential energy, or PE = mgh.

However, in most cases in fiction, in order to make the character's durability impressive, the height is so great that it reaches terminal velocity (more details about that).

The terminal velocity of a human being is around 53 m/s.

Assuming the person is 70 kg:

KE = 0.5*70*53^2 = 9.8315e4 Joules

Wall level

Approximating that border without air resistance: 53 m/s / 9.8 m/s^2 = 5.4081632653061224s drop time.

r = (1/2)*a*t^2 gives the distance covered by such a long fall.

(1/2)*9.8*5.4081632653061224^2 = 143.316326530612242302 m

Therefore, one would have to drop 143.3 m before this calculation applies.

A Human-Shaped Hole

A common gag in fiction is that someone gets slammed towards a wall so hard that a human-sized hole is left.

The average human body has a surface area of 1.9 m^2. Divide that in half and you get 0.95m^2, or 9,500 cm^2.

Assuming that the average human head's length (meaning, front to back) is 7/8ths of the average human head's height (23.9 cm). That will be used for the depth of the crater.

7/8ths of 23.9 is 20.9125.

20.9125*9500 = 1.9866875e5 cm^3.

For fragmentation (8 j/cm^3):

198,668.75 * 8 = 1.589350e6 joules, Room level

For violent fragmentation (69 j/cm^3):

198,668.75 * 69 = 1.370814375e7 joules, Room level+

For pulverization (214 j/cm^3)

198.668.75 * 214 = 4.25151125e+7 joules, Room level+

If the wall is made out of steel:

Fragmentation (208 j/cm^3):

198,668.75 * 208 = 4.1323100e7 joules, or 0.009 Tons of TNT, Room level+

Violent fragmentation (568.5 j/cm^3):

198,668.75 * 568.5 = 1.12943184e8 joules, or 0.027 Tons of TNT, Room level+

Pulverization (300-1000 J/cm^3)

198,668.75 * 300 or 1000 = 5.9600625e+7 to 1.9866875e+8 joules, Room level+

Getting hit by cannonballs

Using the standardized values, a cannonball weights 32 lb (14.514 kg) and has a speed in between 1250 feet per second (381 m/s), 1450 ft/s (441.96 m/s) and 1700 ft/s (518.16 m/s).

The formula for kinetic energy is as follows

KE = 0.5 * m * v^2, where mass = kg and v = m/s

Putting the values into this KE calculator, we get the following:

6 lbs (2.72155 kg)

Low end (381 m/s) = 197.531 kilojoule, 9-B, (Wall level+)

Mid end (441.96 m/s) = 217.7 kilojoule, 9-B (Wall level+)

High end (518.16 m/s) = 265.8 kilojoule, 9-B (Wall level+)

12 lbs (5.44311 kg)

Low-end (381 m/s) = 395 kilojoule, 9-B (Wall level+)

Mid-end (441.96 m/s) = 531.6 kilojoule, 9-B (Wall level+)

High-end (518.16 m/s) = 730.71 kilojoule, 9-B (Wall level+)

18 lbs (8.164663 kg)

Low-end (381 m/s) = 592.6 kilojoule, 9-B (Wall level+)

Mid-end (441.96 m/s) = 797.4 kilojoule, 9-B (Wall level+)

High-end (518.16 m/s) = 1.09606 megajoule, 9-A (Room level)

24 lbs (10.88622 kg)

Low-end (381 m/s) = 790 kilojoule, 9-B (Wall level+)

Mid-end (441.96 m/s) = 1.0632 megajoule, 9-A (Room level)

High-end (518.16 m/s) = 1.46 megajoule, 9-A (Room level)

32 lbs (14.515 kg)

Low-end (381 m/s) = 1.05 megajoules, 9-A (Room level)

Mid-end (441.96 m/s) = 1.41 megajoules, 9-A (Room level)

High-end (518.16 m/s) = 1.94 megajoules, 9-A (Room level)

42 lbs (19.0509 kg)

Low-end (381 m/s) = 1.38 megajoule, 9-A (Room level)

Mid-end (441.96 m/s) = 1.86 megajoule, 9-A (Room level)

High-end (518.16 m/s) = 2.56 megajoule, 9-A (Room level)

Surviving a Fall from Low-Earth Orbit

So we want to calculate how much durability one would need to survive a fall from Low Earth orbit.

Low Earth orbit starts at 160 km.

We will assume that a human like creature falls and that it starts at rest.

For the weight of the creature I will assume 60 kg.

High End

The whole energy of the fall comes from the gravitational potential energy. So we know that in total the kinetic energy on impact can not be higher than the initial gravitational potential energy.

The potential energy is given by the formula GMm/r_1 - GMm/r_2, where M is the mass of earth, m is the mass of the object falling, r_1 is the initial distance from the center of the earth and r_2 is the final distance from the center of the earth. G is the gravitational constant.

Radius of earth is 6371000 m = r_2

r_2 + 160000m = r_1

G = 6.67408*10^-11

M = 5.972*10^24 kg

m = 60 kg

So setting in we get:

(6.67408*(10^-11) * 5.972*(10^24) * 60)/6371000 - (6.67408*(10^-11) * 5.972*(10^24) * 60) / (6371000 + 160000) = 9.1959e7 J

Room level+

Low End

The terminal velocity for a human is 53 m/s, near the ground.

So while someone falling from great heights might initially have a higher speed when going towards the ground the speed will drop towards that value.

0.5*60*53^2 = 8.427e4 J

So at terminal velocity this would only be low end Wall level.

What is Realistic?

The actual value would likely lie somewhere inbetween those two.

One could try to do a more accurate method using the drag equation and the barometric formula, even though I am not quite sure whether that would work (at some point of the fall we would likely talk about supersonic stuff which it usually is hard to get the needed values for).

For now we would stay with Wall level for such a feat.

Also let us mention that this is only for low earth orbit falling. For higher alitudes the potential energy value would go closer to the kinetic energy when falling with escape velocity, while for lower it would mostly just stay the same (the realistic value would go towards to terminal velocity value) except for short falls where not even that much speed if attained.

Bone Breaking Feats

Breaking all the Bones of a Man's Body

On average, the weight of a man's bones is 15% of their body mass, which in of itself is 88.768027 Kilograms. 15% of that is 13.31520405 Kilograms.

The density of bone is 3.88 g/cm^3, which would mean that the total volume would be 13.31520405 divided by 0.00388, which equals 3431.75362113402 cm^3 for our volume.

To get the fragmentation values, we need to use the compressive strength of bones. To quote Wikipedia, "bone has a high compressive strength of about 170 MPa (1800 kgf/cm²), poor tensile strength of 104–121 MPa, and a very low shear stress strength (51.6 MPa)"

So, low end is 51.6, mid is 104, high is 170. Plugging those all into our volume gets us....

Low End: 1.77078.486850515432e5 Joules, Wall level+

Mid End: 3.56902376598e5 Joules, Wall level+

High End: 5.83398115592783e5 Joules, Wall level+

Breaking a Human neck

Volume of a Vertebra

The vertebrae that make up the neck are the cervical vertebrae and are 7 vertebrae in total. However, due to finding info only for vertebra 3 through 7, the smallest one will be calced.

C3 pedicle: The pedicle is roughly a rectangular prism and there are two of them. 5.27 mm x 5.14 mm x 7.08 mm = 0.527 cm * 0.514 cm * 0.708 cm = 0.191781624 cm^3. 0.191781624 cm^3 * 2 = 0.383563248 cm^3

C3 vertebral body: The vertebral body is a cylinder. The mean height is 15.1 mm and the radius 7.34 mm = 2.55575 cc.

Energy to Fragment the C3 Vertebra

The shear strength of bones is 51.6 MPa or J/cc

(0.383563248 + 2.55575) x 51.6 = 151.6685635968 J

Athlete level

Keep in mind, this is just fragmenting most of the C3 vertebra. This does not take into account the lamina.

Breaking a Bone

The durability of a bone depends on the angle of attack.

A bone of a deceased 52-year old woman only required 375 Joules of energy when the force was applied within five degrees of the orientation of the collagen fibres. But the force increased exponentially when they applied it at anything over 50 degrees away from that orientation, up to 9920 Joules when they applied a nearly perpendicular force.

So breaking a bone would require 375-9920 Joules, depending on the angle of attack. That's Peak Human level to Peak Human level+.

Vaporization Feats

Vaporizing a Human

https://m.slashdot.org/story/191575

"Have you ever wondered how much energy is needed to power a phaser set to kill? A trio of researchers at the University of Leicester did, so they ran some tests and found out it would take roughly 2.99 GJ to vaporize an average-sized adult human body. Quoting:First, consider the true vaporization – the complete separation of all atoms within a molecule – of water. With a simple molecular structure containing an oxygen atom bonded to two hydrogen atoms, it takes serious energy to break these bonds. In fact, it takes 460 kilojoules of energy to break just one mole of oxygen-hydrogen bonds — around the same energy that a 2,000-pound car going 70 miles per hour on the highway has in potential. And that's just 18 grams of water! So as you can see, it would take a gargantuan amount of energy to separate all the atoms in even a small glass of water — especially if that glass of water is your analog for a person. The human body is a bit more complicated than a glass of water, but it still vaporizes like one. And thanks to our spies spread across scientific organizations, we now have the energy required to turn a human into an atomic soup, to break all the atomic bonds in a body. According to the captured study, it takes around three gigajoules of death-ray to entirely vaporize a person — enough to completely melt 5,000 pounds of steel or simulate a lightning bolt."

Conclusion

0.71 Tons of TNT (Building level)

Vaporizing an average Building

Wikipedia states that the average size of an owned residence in Japan is 121.7 m^2. When you think "average house/ building" you have in mind a two-storey building, which is approx 6.6 m.

This site gives us a volume of 859.1 m^3. Density of concrete is 2400 kg per cubic meter. So 2,061,840 kg = 2,061,840,000 g.

Using the formula Q = m*c*ΔΤ

Using 3 ends of 80%, 85% and 90% hollowness for the mass.

Concrete's specific heat capacity is 880 J/kg°C.

Change in temperature is from room temperature to concrete's melting point:

Concrete's melting point = 1500 °C.

Room temperature = 20 °C (average of 15 °C and 25 °C) High end: 80% hollowness: 2,061,840,000g * 0.2 * 880J/kg°C * (1500°C - 20°C) = 5.370680832e14 J = 124 kilotons. (Town level)

Mid end: 85% hollowness: 2,061,840,000g * 0.15 * 880J/kg°C * (1500°C - 20°C) = 4.028010624e14 J = 93 kilotons. (Town level)

Low end: 90% hollowness: 2,061,840,000g * 0.1 * 880J/kg°C * (1500°C - 20°C) = 2.685340416e14 J = 62 kilotons. (Town level)

Melting/Heat Feats

Surviving the Heat of the Sun

Surface 1. Radiation: For radiation we need to know the emissivity, surface area and temperature.

The temperature of the sun is about 5500 °C per Wikipedia.

For the surface area we take the surface of the average human body, since we assume that the person is submerged in the sun. The average body surface area is about 1.73 m^2 per this article.

The emissivity is about 1.2 at this temperature per this article.

Now we input this values into this calculator and get 130756044.60407 J/s.

2. Conduction: For conduction we need to know surface area, thickness of the material that the heat is transmitted through, the thermal conductivity of the material and the heat of the sun and the object.

Surface area and temperature of the sun can be taken from the radiation part.

Now for the material were the heat is transmitted through we will take human skin.

Human Skin is around 3mm thick. (wikipedia:Human skin)

It has a thermal conductivity of about 0.209. (http://users.ece.utexas.edu/~valvano/research/Thermal.pdf)

Normal skin temperature is about 33 °C. (http://hypertextbook.com/facts/2001/AbantyFarzana.shtml)

With that we have everything we need. We use this calculator to get a result.

The result is: 658901.0633333334 watts = 658901.0633333334 J/s.

Now we add both results together to get a final value: 658901.0633333334 J/s + 130756044.60407 J/s = 1.3141494566740333*10^8 J/s. Core Now a similar procedure for the core. The core of the sun is about 15.7 million Kelvin hot. The emissivity of the sun at temperatures such as this isn´t known, but the article that is linked to emissivity states says that the minimum lies at 6900 °C. So we will use the minimum emissivity of 0.92 for this. Now we just need to input all values in the calculators again.

1. Radiation: 5.4829665830548E+21 J/s

2. Conductiont: 1892212356.0633333 J/s

5.4829665830548E+21 J/s + 1892212356.0633333 J/s = 5.4829665830566922E+21 J/s

Note: This is for a human in the sun. If the character is a lot bigger or smaller than an average human, or if the character is made from another material, like for example metal, this numbers change.

Maximum internal energy intake If an object is heated it usually doesn´t get hotter than the source of the heat. If the object is as hot as the heat source the energy itself emits to its surroundings should be equal to the energy it is infused with.

That means there is a maximum amount of thermal energy an object can take in through a certain source of heat.

In order to calculate this energy we will just measure how much energy will be necessary to heat the object to this temperature, from the point that it has no internal energy, which should be 0K.

The specific heat capacity of a human body is 3470 J/kg.oC

Average weight of a grown human is around 62 kg.

Surface: The surface of the sun has a temperature of 5.773.2K.

3470*62*5773.2 = 1.242046248E+9J

That is Building Level.

Core: The core of the sun has an temperature of 15 700 000K.

3470*62*15 700 000 = 3.377698E+12J

That is Multi-City Block Level+.

Melting a Plane

Specific Heat Capacity Titanium Ti-6Al-4V = 526.3 J/kg-°C

Steel = 510 J/kg-°C

Aluminium 2024-T3 = 875 J/kg-°C

Melting Point Titanium = 1604 °C

Steel = 1425 °C

Aluminium = 502 °C

Latent Heat of Fusion Titanium = 419000 J/Kg

Steel = 272000 J/Kg (This is for Iron, but is nearly the same though)

Aluminium = 398000 J/Kg

Total Energy = (((526.3)*(7320.98084)*(1604–25)) + ((7320.98084)*(419000))) + (((510)*(23793.1877)*(1604–25)) + ((23793.1877)*(272000))) + (((875)*(148249.862)*(1604–25)) + ((148249.862)*(398000))) = 2.9861275268025227e11 Joules, or 71.37 Tons, City Block level+

Melting a Tank

The mass of a tank is around 60 tons.

Materials of tanks and especially how much of which is there is mostly classified information. Using this article on composite armour we get 10% ceramics and 90% steel, given that the mechanics and everything will be made out of metal. For the ceramics we will assume Alumina, since that is also mentioned as a material used here.

Specific heat of materials: Per this article:

“c” of alumina = 850 J/(kg*K)

“c” of steel = 481 J/(kg*K)

2.2 Latent heat of fusion:

Steel: 260000 J/kg per this article.

Alumina: 620000 J/kg as per this article.

Melting point:

Alumina: 2072 °C (per wikipedia)

Steel: 1425 °C (per this)

Mass of materials: 6000 kg alumina, 54000 kg Steel

Assuming a tank is on average 20 °C warm.

High end:

850 J/(kg*K)*6000 kg *(2072 °C - 20 °C) + 620000 J/kg * 6000 kg + 481 J/(kg * K) * 54000 kg * (2072 °C - 20 °C) + 260000 J/kg * 56000 kg = 8.2043848e10 Joules, City Block level

Low end: 850 J/(kg*K)*6000 kg *(1425 °C - 20 °C) + 620000 J/kg * 6000 kg + 481 J/(kg * K) * 54000 kg * (1425 °C - 20 °C) + 260000 J/kg * 56000 kg = 6.193897e10 Joules, City Block level

Durability to Tank Lava

Lava can be between 700°C and 1250°C. Given that we likely don´t know the heat of the lava let's work with 700 °C.

Emissivity of Lava is between 0.55 and 0.85. At the given temprature it should be around 0.65.

The average human body surface area is 1.73 m^2.

At last we input all this stats in this calculator. That results in 57182.306177806 J/s.

Now part 2 heat transfer through conduction.

Human Skin is around 3 mm thick. (wikipedia:Human skin)

It has a thermal conductivity of about 0.209 (http://users.ece.utexas.edu/~valvano/research/Thermal.pdf)

Normal skin temperature is about 33 °C (http://hypertextbook.com/facts/2001/AbantyFarzana.shtml)

Now we use this calculator. That gives us 80389.06333333334 J/s.

Now we add that together and get: 1.3757136951113934e5 J/s, Wall level

Weather Feats

Destructive Energy of Winds

Air is 1.225 kg/m^3 at sea level. I am going to find the energy of different winds at diffirent speeds and different sizes.

1 m^3 of air:

1 m/s = 0.6125 J = Below Average level

5 m/s = 15.3125 J = Below Average level

10 m/s = 61.25 J = Human level (A little over Low-End wind speed of a thunderstorm)

20 m/s = 245 J = Athlete level+ (A little over the High-End wind speed of a thunderstorm and Low-end speed of an f0 tornado)

40 m/s = 980 J = Peak Human level (Speeds of an F1 tornado and Category 1 hurricane)

50 m/s = 1531.25 J = Peak Human level (An F2 tornado and Cat. 3 hurricane)

70 m/s = 3001.25 J = Peak Human level (An F3 tornado and Cat. 5 hurricane)

90 m/s = 4961.25 J = Peak Human level (An F4 Tornado)

115 m/s = 8100.31 J = Peak Human level (An F5 tornado)

135 m/s = 11162.8 J = Peak Human level+ (Highest wind speed recorded on Earth)

170 m/s = 17701.3 J = Wall level (Great Red Spot wind speeds)

500 m/s = 153125 J = Wall level (Wind speed of Saturn)

600 m/s = 220500 J = Wall level (Wind speed of Neptune)

2415 m/s = 3572240 J = Room level (Fastest wind speed ever found on a planet)

This is only for 1 cubic meter of air and not taking account higher masses of air. And in terms of wind, unless it is a gust, these wind speeds are continous and would keep on delivering the same amount of joules over and over to whatever object.

Creating a Storm

Storms are calculated with either CAPE, condensation, or KE (if applicable). You can read more about that here. Usually the storm clouds extend all the way to the horizon. The visibility on a normal day is 2000 km.

Storm clouds have a height of 8000 m.

π×8,000×20,000^2 = 10053096491487 m^3.

Multiplying that by 1.003 (density of cloud) gives us 10083255780961 kg.

CAPE

"Weak instability": 1.008325578096e16 Joules, 2.40995597059316 Megatons, City level

"Moderate instability": 2.520813945240e16 Joules, 6.02488992648291 Megatons, City level

"Strong instability": 4.033302312384e16 Joules, 9.63982388237265 Megatons, City level

1999 Oklahoma Tornado Outbreak: 5.939037654986e16 Joules, 14.1946406667937 Megatons, City level

1990 Plainfield Tornado: 8.066604624769e16 Joules, 19.2796477647453 Megatons, City level

Condensation So, a storm is generally 1-3 grams per meter. We'll use 1 gram for this, so, it's 10053096491487 g, 10053096491.487 kg.

Now, for condensation, the value is 2264705 j/kg, so, put that with the above and it's

2.2767297889753066335e16 Joules, 5.44151479200599 Megatons, City level

KE

KE is a bit reliant on a specific timeframe, however in this case, the standard assumption is a minute. However, if it takes less then a minute, then you can make your own calc, assuming the storm qualifies for KE Standards

20000/60 is 333.333333333333 m/s

Now, 0.5×10083255780961×333.333333333333^2 is....

5.601808767200e17 Joules, 133.886442810720 Megatons, Mountain level

Tornado Feats

Entire blog here regarding tornado feats.

Earth Feats

Destroying the Surface of the Earth

Earth's circumference = 40075 km

Explosion radius = 20037.50 km

Y = ((x/0.28)^3)

Y is in kilotons, x is radius in kilometers.

Y = ((20037.50/0.28)^3) = 366485260009765.63 Kilotons of TNT

Only 50% of the total energy of the explosion is actually from the blast, so we need to halve the result. This part can be ignored if the explosion was an actual nuclear explosion.

366485260009765.63/2 = 183242630004882.82 Kilotons of TNT, or 183.24 Petatons of TNT, Multi-Continent level

Shaking Different Celestial Bodies

Shaking the Earth

Now, we need to use the Impact Calculator. The circumference of the Earth is 40070 km; plugging in 2000km due to the fact that's the maximum, and playing with other values, we find that an impact that is IV on the Mercalli scale and 3.0 in the Richter magnitude releases an impact energy of around 2.76e+13 joules. We need the seismic energy here, however; and, to get that, we need to divide this value by 10,000.

2.76e+13 J/ 10,000 = 2760000000j

The radius of the Earth is 6563 km or 6563000 meters.

Seismic energy * area = E

4m*pi*(6563000m^2)= 82,473,90.3420392520961213 m^2

82473090.3420392520961213m^2 * 2760000000j = 2.2762572934402834e17 J or Mountain level.

Another method would be this: To find how strong of an impact it truly was, you use this equation:

(Magnitude at distance) + 6.399 + 1.66×log((r/110)×((2×π)/360)) = Richter Magnitude of Earthquake, with r representing the distance away from it.

In our case, it would be, using half of the Circumference of earth,

(4)+6.399+1.66×log((20037.5÷110)×((2×π)÷360)) = Magnitude 11.2328648415393

Now, we take the magnitude and use the formula for a joulecount from said magnitude listed in Earthquake Calculations

10^(1.5*(11.2328648415393)+4.8) is 4.459613919339E21 Joules, 1.06587330768147 Teratons, Small Country level

Shaking the Solar System

For lowball purposes we'll assume the planets are shaking specifically.

Now, we need to use the Impact Calculator. The circumference of the Earth is 40070 km; plugging in 2000km due to the fact that's the maximum, and playing with other values, we find that an impact that is IV on the Mercalli scale and 3.0 in the Richter magnitude releases an impact energy of around 2.76e+13 joules. We need the seismic energy here, however; and, to get that, we need to divide this value by 10,000.

2.76e+13 J/ 10,000 = 2760000000j =

The radius of the solar system is 4545000000000 meters.

Seismic energy * area = E

4m*pi*(4545000000000m^2) = 259,583,831,940,082,794,687,015,143.7379570922657321 m^2

259583831940082794687015143.7379570922657321m^2 * 2760000000 j = 716,451,376,154,628,513,336,161,796,716,761,574.653420596 J or 171.23598856468177587 Yottatons, aka Large Planet level

Shaking the Galaxy

Same formula but now we take the radius of the galaxy which looks to be 1,000,000,000,000,000,000 km.

4m*pi*(1e+21m)^2 = 12,566,370,614,359,172,953,850,573,533,118,011,536,788,677.5975004232838998 m^2

12,566,370,614,359,172,953,850,573,533,118,011,536,788,677.5975004232838998m^2 * 2760000000 j = 34,683,182,895,631,317,352,627,582,951,405,711,841,536,750,169,101,168.263563448 J, aka Solar System level

Shaking the Universe

The radius of the observable universe is 11500000000000 trillion light years, or 1.08798400434679e+29 meters.

4*pi*(1.08798400434679e+29m)^2 = 1.4874928427840380558e59m^2

E = 2760000000 j *1.4874928427840380558e59 m^2 = 4.105480246083945e68 J, aka Galaxy level

The Earth's Rotational Energy

(Picture) The formula of the rotational energy is K = 1/2* Ι*ω^2

The moment of inertia of a sphere is 2/5mR^2

The Earth's angular velocity is 7.3*10^-5 rad/s

Earth's Mass = 5.97e24 kg

Earth's radius = 6372000 m

Κ = 1/2*Ι*ω^2 = 1/5 * m*R^2 *ω^2 = 2.58e29 Joules, Moon level

Splitting the Earth in half

Diametre of the Earth is 12 742 000 metres. Radius is 6 371 000 metres.

No feat, so I'll assume the Earth is split apart by 1 kilometre, or 1000 metres.

The centre of mass of each individual half is 3R/8 from the centre of the sphere.

U = GMm/r

M = m = mass of half of the Earth = 5.97237e+24/2 = 2.986185e+24 kg

G = Gravitational constant = 6.674×10^(−11) m^3⋅kg^(−1)⋅s^(−2)

r = Earth radius = 6 371 000 m

[1]

Here is a picture of the Earth. The diametre of the Earth is 627 pixels, or 12 742 000 metres.

For the split to be visible I'll assume 10 pixels or so. That's 203 222 metres.

Therefore the GPE of the unsplit Earth is still 1.245520136056038e+32 Joules. The split Earth is 1.194708429578599e+32.

So, the final tally would be 5.0811706477439e+30, or Small Planet level.

Vaporizing Earth

Based on this we're looking at the most prevalent elements in the Earth. All of this comprises the mass of the Earth (5.98e24 kg).

• Iron: 32.1% (1.91958e24 kg), Heat Capacity of 460
• Oxygen: 30.1% (1.79998e24 kg), Heat Capacity of 919
• Silicon: 15.1% (9.0298e23 kg), Heat Capacity of 710
• Magnesium: 13.9% (8.3122e23 kg), Heat Capacity of 1050
• Sulfur: 2.9% (1.7342e23 kg), Heat Capacity of 700
• Nickel: 1.8% (1.0764e23 kg), Heat Capacity of 440
• Calcium: 1.5% (8.97e22 kg), Heat Capacity 630
• Aluminium: 1.4% (8.372e22 kg), Heat Capacity of 870

The last 1.2% is a mixture of tons of lesser elements. For the sake of this calc, we'll be ignoring it. All layers of the Earth are solid save for one, which is a layer of molten iron as hot as the surface of the sun. Considering how miniscule the oceans are in relation to the rest of Earth we'll be ignoring these as the only other quote unquote major source of liquid.

Let's get on with the liquid bit first. The inner core is about 1% of the Earth's total volume at 1.0837e21 m^3 total. 1% of that is 1.0837e19 m^3 for the Inner Core.

Density of liquid iron is 6980 kg/m^3. Mass of the Inner Core is 7.564226e22 kg. This means the iron content of Earth is now divided into the categories "Liquid" and "Solid".

• Solid Iron: 1.84393774e24 kg
• Liquid Iron: 7.564226e22 kg

Now we can actually calc the energy needed to vaporize. For the purpose of this calc we will assume the Earth is heated uniformly to the same heat. Let's look at the heat each element needs to vaporize (AKA Boiling Point).

• Iron = 2862 C
• Oxygen = -183 C (so this is pointless)
• Silicon = 3265 C
• Magnesium = 1091 C
• Sulfur = 444.6 C
• Nickel = 2913 C
• Calcium = 1484 C
• Aluminium = 2470 C

So Silicon's heat will be our assumed heat. As a high-end we'll use the boiling point of Tungsten in order to account for truly all elements on Earth- 3414 C is our high-end.

Let's handle heat change first. The core is assumed to maintain heat similar to the surface of our sun all throughout as a starting point- 5778 C, so not relevant. We'll assume the other stuff is the average of their ambient temperatures.

• Mantle: 4000 C - 200 C (2100 C average) and is 84% of Earth's Volume
• Crust: 400 C - 300 C (350 C average) and is 1% of Earth's Volume

For the purposes of this calculation, we will assume all elements are roughly evenly distributed through the sections of Earth aside from the Inner Core- we even know the Outer Core isn't entirely iron.

Outer Core represents 15% of total Earth volume and is comprised of iron and nickel for the most part. The assumptions have to be hefty in order to make up for this- we'll assume half of the world's nickel is present here (0.9%) and the rest is Iron (14.1%, or about 45.338% of remaining iron). Adjusted values below.

• Iron in Inner Core: 7.564226e22 kg
• Iron in Outer Core: 8.36004493e23 kg
• Iron in Mantle/Crust: 1.00793325e24 kg
• Nickel in Core: 5.382e22 kg
• Nickel in Mantle/Crust: 5.382e22 kg

Anything in the Core isn't relevant for heat change since everything there would vaporize from heat anyways if the pressure wasn't so high. So we're ignoring them aside from just shifting states of matter.

We'll put all the things the Crust is made of here. Everything else is assumed to be in the Mantle. We're looking at Oxygen, Silicon, Aluminium, Iron, Calcium, and Magnesium. Divide this 1% volume between them for the following masses:

• Total Volume = 1.0837e19 m^3
• Volume Each = 1.80616667e18 m^3
• Oxygen = 2.06083617e15 kg
• Silicon = 4.20475601e21 kg
• Aliminium = 4.89471168e21 kg
• Iron = 1.42217564e22 kg
• Calcium = 3.61233334e21 kg
• Magnesium = 3.14273001e21 kg

Subtract this from values established at the beginning to get the following table of values.

Mass of Elements

• Mass of Earth = 5.98e24 kg
• Mass of Crustal Iron = 1.42217564e22 kg
• Mass of Crustal Oxygen = 2.06083617e15 kg
• Mass of Crustal Silicon = 4.20475601e21 kg
• Mass of Crustal Calcium = 3.61233334e21 kg
• Mass of Crustal Magnesium = 3.14273001e21 kg
• Mass of Mantle-Based Iron = 9.93711494e23 kg
• Mass of Mantle-Based Nickel = 5.382e22 kg
• Mass of Mantle-Based Oxygen = 1.79998e24 kg
• Mass of Mantle-Based Silicon = 8.98775244e23 kg
• Mass of Mantle-Based Sulfur = 1.7342e23 kg
• Mass of Mantle-Based Magnesium = 8.2807727e23 kg
• Mass of Mantle-Based Aluminium = 7.88252883e22 kg
• Mass of Mantle-Based Calcium = 8.60876667e22 kg
• Mass of Outer-Core Iron = 8.36004493e23 kg
• Mass of Outer-Core Nickel = 5.382e22 kg
• Mass of Inner-Core Iron = 7.564226e22 kg

Now we need Specific Heat energy since we've spent all that time setting this shit up. We won't do anything for the Core elements since their heat doesn't need to change at all for this event to happen.

Specific Heat Energy

As said earlier, as a low-end we assume 3265 C end temperature, as a high-end we assume 3414 C end temperature based on Tungsten. EDIT: Coming back to it, just using the high-end. Results don't change much and it's the only thing that makes logical sense.

The calculation for this is just mass times specific heat times temperature change (which varies). I'm beaten by this so far so I'm using this.

• Low-End Temp Change for Crust Elements = 2915 C
• High-End Temp Change for Crust Elements = 3064 C
• Low-End Temp Change for Mantle Elements = 1165 C
• High-End Temp Change for Mantle Elements = 1314 C

Let's get to it.

• Crustal Iron = 2.005e28 Joules
• Mantle Iron = 6.006e29 Joules
• Crustal Oxygen = 5.803e21 Joules
• Mantle Oxygen = 2.174e30 Joules
• Crustal Silicon = 9.147e27 Joules
• Mantle Silicon = 8.385e29 Joules
• Crustal Magnesium = 1.011e28 Joules
• Mantle Magnesium = 1.143e30 Joules
• Mantle Sulfur = 1.595e29 Joules
• Mantle Nickel = 3.115e28 Joules
• Mantle Aluminium = 9.011e28 Joules
• Mantle Calcium = 7.127e28 Joules

Total Energy of Heat Change = 5.14743701e30 Joules, Small Planet level. But we're far from done.

Shifts in Matter

Now we get into truly changing the matter from solid/liquid to gas. For this we classify everything by solid or liquid. Everything in the inner core is liquid (a small amount of iron)- everything else is held to be solid. This includes the outer core which shifts between liquid and solid.

For solids, they must undergo a state of fusion and vaporization, so we need to multiply them by their values for that in J/kg. For the liquid, it must only undergo the value for vaporization. Oxygen is already gaseous so it needn't be accounted for.

• Solid Iron = 1.84393774e24 kg
• Liquid Iron = 7.564226e22 kg
• Solid Silicon = 9.0298e23 kg
• Solid Magnesium = 8.3122e23 kg
• Solid Sulfur = 1.7342e23 kg
• Solid Nickel = 1.0764e23 kg
• Solid Aluminium = 8.372e22 kg
• Solid Calcium = 8.97e22 kg

Let's start with the irons.

• Liquid Iron Vaporization = 4.700e29 Joules
• Solid Iron Fusion & Vaporization = 4.557e29 Joules & 1.146e31 Joules
• Solid Silicon Fusion & Vaporization = 1.614e30 Joules & 1.154e31 Joules
• Solid Magnesium Fusion & Vaporization = 3.062e29 Joules & 4.357e30 Joules
• Solid Sulfur Fusion & Vaporization = 9.288e27 Joules & 5.299e28 Joules
• Solid Nickel Fusion & Vaporization = 3.204e28 Joules & 6.793e29 Joules
• Solid Calcium Fusion & Vaporization = 1.911e28 Joules & 3.469e29 Joules
• Solid Aluminium Fusion & Vaporization = 3.320e28 Joules & 9.091e29 Joules

Total Fusion + Vaporization Energy = 3.2284828e31 Joules, Small Planet level

the latent heat of fusion/vaporization for Magnesium, Sulfur, and Nickel were calculated since they aren't present on our Calculations page.

Magnesium has Fusion of 8954 J/mol and 127400 J/mol for Vaporization. Magnesium weighs 24.305 g/mol so energy is...

• Fusion: 368401.563 J/kg
• Vaporization: 5241719.81 J/kg

Sulfur has Fusion of 1717.5 J/mol and 9800 J/mol for Vaporization. Sulfur weighs 32.07 g/mol so energy is...

• Fusion: 53554.724 J/kg
• Vaporization: 305581.54 J/kg

Nickel has Fusion of 17470 J/mol and 370400 J/mol for Vaporization. Nickel weighs 58.69 g/mol so energy is...

• Fusion: 297665.501 J/kg
• Vaporization: 6311126.26 J/kg

Total Energy

We're adding together all values denoted in Vaporizing Earth's topic as well as the GBE of Earth.

• GBE of Earth = 2.24e32 Joules
• Matter Shifting of Earth = 3.2284828e31 Joules
• Heat Change of Earth = 5.14743701e30 Joules

Total Energy = 2.614e32 Joules, Planet level.

The World Gets Vaporized: 62.48 Zettatons of TNT, Planet level

Freezing Feats

Freezing a Human

Average human weight = 62 kg

On average 65% of the human body weight is water.

So water mass = 0.65*62 kg.

So total energy = 62 * 3500 * 38 + 0.65*62*1000*333.55 = 21,688,065 Joules. Room level

Crushing Feats

Crushing a Golf Ball

Materials of Golf Ball

A golf ball is made of a rubber core, usually Polybutadiene, and a Ionomer or latex cover, usually Polyurethane

Energy Density of Materials

I will use compressive strength rather than shear since this is crushing the ball.

Polybutadiene = 2.35 MPa on average or 2.35 J/cc

Polyurethane = 7305.75 PSI = 50.37137309 MPa = 50.37137309 J/cc on average

Volume of Ball

The core of the ball is 3.75 cm in diameter. The ball itself can be no less than 4.267 cm in diameter.

The core would be 27.61 cc. The entire ball would be 40.68 cc. To find the volume of the cover, subtract the core volume from the entire volume to get 13.07 cc for the cover.

Energy to Crush Golf Ball

2.35*27.61 = 64.8835 joules for core

13.07*50.37137309 = 658.3538463 joules for cover

723.2373463 Joules in total, Peak Human level

Crushing a Human Skull

Compressive Strength of Bone - 170 MPa

Weight of the Skull - 997 g

Density of Bone - 1.6 g/cm^3

997/1.6 = 623.125 cm^3

170 MPa*623.125 cm^3 = 105,931 J

For shear strength:

Shear Strength of Bone - 51.6 MPa

56.1 MPa*623.125 cm^3 = 34,960 J

Results

Head Crush (Compressive) - 1.05931e5 Joules, Wall level

Head Crush (Shear) - 3.496e4 Joules, Wall level

NOTE: In case the skull is destroyed in a swift blow, the shear value would apply, and in the case of the head being slowly crushed to pieces, the compressive value would apply.

Potential Energy/Lifting Feats

Leaping onto a Roof

Another common feat in fiction is when a character is leaping high in the air usually to jump on a roof of a nearby building.

Small building (10 m)

PE = 70*10*9.81 = 6.867e3 Joules, Peak Human level

Average building (30 m)

PE = 70*30*9.81 = 2.0601e4 Joules, Wall level

Tall building (70 m)

PE = 70*70*9.81 = 4.8069e4 Joules, Wall level

Skyscrapers (300 m)

PE= 70*300*9.81 = 2.0601e5 Joules, Wall level

Snapping a Human Neck

The amount of force necessary to break a neck is around 1000-1250 lbf.

However, techniques can such as neck cranks can greatly reduce the lifting strength necessary through leverage and bodyweight application.

Object Destruction Feats

Destroying a Door

Standard size for a door is 203.2 cm tall, 91.44 cm wide, and 3.334 cm thick.

Volume = 61947.75 cm^3

Fragmentation values for wood and steel can be found here.

Wood Door Fragmentation = 516644.24 Joules or Wall level+.

V. Frag = 1136121.74 Joules or Room level.

Pulverization = 2907827.38 Joules or Room level.

Steel Door Fragmentation = 1.289e7 Joules or Room level.

V. Frag = 3.522e7 Joules or Room level.

Pulverization = 4.058e7 Joules or Room level.

Destroying a Car

Mass and Weight of Materials

The EPA stated that an average vehicle produced in 2016 weighed, on average, 4,035 lbs. or 1830.245 kg

On average, 900 kg of steel is used in the making of a vehicle. or 49.1737444 % of the car.

as of 2015, The average vehicle uses 397 lbs of aluminum. or 180.076 kg at 9.838901349272913 % of the car.

The highest amount of copper used in an average conventional car is 49 lbs. or 22.226 kgat 1.2143729391420275 % of the car.

The amount of glass in an average vehicle is 100 lbs. or 45.3592 kg at 2.478313012738732% of the car

Plastic makes up 10% of the weight of a car. or 183.0245 kg

Tires are made up of 14% natural rubber and 27% synthetic rubber with an average weight of 25 lbs. or 11.3398 kg. 14% of the tires is 1.5875720000000002 kg. 27% is 3.0617460000000003 kg. Since there are 4 tirse we will time these numbers by 4. The total weight lf natural rubber is 6.350288 kg, or 0.3469638217834225 % of the car. The total weight of synthetic rubber is 12.246984 kg, or 0.6691445134394576% of the car.

The amount of cast iron in an average car is about 7.2%. or 131.77764000000002 kg.

This all accounts for about 80.92144004% of the weight for the car.This isn't at 100% but this is as much percentage of materials that could be found, so consider this a low-ball or a near complete fragmentation of a car.

Density of Materials

Steel = an average of 7.9 g/cm³

Aluminum = 2.7 g/cm³

Copper = 8.96 g/cm³

Glass = an average of 5 g/cm³

Plastic = and average of 2.235 g/cc (http://www.tregaltd.com/img/density%20of%20plastics[1].pdf)

Natural Rubber = 0.92 g/cm³

Synthetic Rubber = We will use polybutadiene since it is mostly used in car tires. 0.925 g/cm^3

Cast Iron = an average density of 7.3 g/cm³

Volume of Materials

Steel = 113,924.0506 cm³

Aluminum = 66,694.81481 cm³

Copper = 2,480.580357 cm³

Glass = 9,071.84 cm³

Plastic = 81,890.1566 cm³

Natural Rubber = 6,902.486957 cm³

Synthetic Rubber = 13239.9827 cm³

Cast Iron = 18,051.73151 cm³

Energy to Fragment Materials

To find shear strength from tensile strength, just times the ultimate tensile strength by 0.60.

Steel = 208 j/cc

Cast Iron = 149 MPa or j/cc

Glass = 3.5 j/cc

Aluminium = 40000 PSI = 275.79 megapascales = 275.79 J/cc

Copper = 25,000 PSI = 172.36893 MPa = 172.36893 J/cc

Plastic = It is insanely difficult for me to find plastic mechanical properties. Polypropylene will be used since it is used for most cars, especially in their bumpers. an average of 38.7 MPa = 38.7 j/cc

Natural Rubber = 0.001 GPa = 1 MPa = 1 J/cc

Synthetic Rubber = 4.285714286 MPa = 4.285714286 J/cc

Total Energy

23,696,202.52 Joules for all the steel

2689707.995 Joules for all the iron

31,751.44 Joules for all the glass

18,393,762.98 Joules for all the aluminum

3169149.06 Joules for all the plastic

427,574.9819 Joules for all the copper

56742.783 joules for Synthetic Rubber

6902.486957 Joules for all the natural rubber

Adding this all up is 48,471,794.25 Joules = Room level+

Destroying a Tree

Volume of Tree

A white oak tree will be used since they are somewhat common and are not overly large.

White Oak = 30 m height, 1.27 meter diameter

Plugging this into the formula for volume of a cylinder since tree trunks are cylindrical = 38 m^3

Energy to Destroy Tree

The weakest wood that could be found comes from Ceiba pentandra at 350 PSI = 2.41317 MPa = 2.41317 J/cc. The hardest wood that could be found is Dalbergia nigra at 2360 PSI = 16.27163 MPa = 16.27163 J/cc

Low End: 2.41317*38000000 = 9.1700460e7 Joules, 0.022 Tons of TNT, Room level+

High End: 16.27163 x 38000000 = 6.1832194e8 Joules, 0.148 Tons of TNT Small Building level+

The high end is a low ball since Dalbergia nigra is not the hardest type of wood. The low end could go lower since wood like balsa is weaker than Ceiba pentandra.

This also doesn't take into account branches either.

Destroying a Wrecking Ball

Volume of Ball

The weight of a wrecking ball ranges from 450 kg to 5400 kg and they are made of steel.

Steel = density of 7.9 g/cc

450000/7.9 = 56962.02532 cc

5400000/7.9 = 683544.3038 cc

Energy to Destroy Wrecking Ball

Steel = 208 J/cc

Low-end: 208*6962.02532 = 1.184810127e7 Joules, Room level+

High-end: 208*683544.3038 = 1.421772152e8 Joules, or 0.034 Tons of TNT, Room level+

Breaking off a Lock

Volume of shackle This is a fairly standard lock. [2] There will be no measurement to how much energy it takes to completely fragment a lock since most are just broken off. So, it will be just the measurement of the shackle and not the rest of the lock.

The lock is one inch or 61 px. or 0.04163934426 cm a pixel

Red = Portion that is a cylinder is 44 px or 1.832131147 cm

Pluging in the values of the radius of the shackle with the height gives me 1.78 cc x 2 = 3.56 cc for both sides. But, this doesn't take into account the curved portion. So to find the volume of that, I'll just use the volume of a torus x 0.5.

Orange = Major radius 30 px or 1.249180328 cm

This gives a volume of 2.36 cc

2.36 + 3.56 = 5.92 cc

Since this is just breaking off the lock, the shackle is not usually fragmented completely, so it would be best to just use 1/4 of the volume = 1.48 cc

Energy to Destroy Shackles

To find shear strength from tensile strength, just times the ultimate tensile strength by 0.60.

Lock shackles are typically made of Brass, normal Steel, stainless steel, hardened steel, and boron alloy steel

Brass is 235 MPa or 235 J/cc

Steel = 208 j/cc

Hardned Steel = Tensile strength is at least 1000 MPa. 1000 x 0.60 = shear strength 600 MPa = 600 J/cc

Stainless Steel = Tensile strength is 505 MPa. 505 x 0.60 = 303 MPa = 303 J/cc

Cannot find boron alloyed steel tensile or shear strength.

Steel = 307.84 J

Low-End = Peak Human level

Brass = 347.8 J

Mid-Low End = Peak Human level

Stainless Steel = 448.44 J

Mid-High End = Peak Human level

Hardened Steel = 888 J

High-End = Peak Human level

Destroying Blades

Volumes of Blades

A knightly (or short) sword blade is typically 31 3/8 inches long, 2 inches wide, and .192 inches thick A long sword blade is at least 90 cm long 4.14 mm thick [1]

Red = length 90 cm or 964 px at 0.09336099585 cm a pixel

Orange = Width 30.1 px or 2.810165975 cm

Longsword = 104.71 cc

Shortsword = 200.58 cc

Energy to Destroy Blades

Assuming they are made of steel.

Longsword = 2.177968e4 Joules, Wall level

Shortsword = 4.172064e4 Joules, Wall level

Note: This is the fragmentation of an entire blade, but not the hilt.

Destroying a Chimney

Volume of Chimney

I could not find the average size of a chimney so I'll just use this one for a baseline. It is 8 feet tall and 2 feet wide and long.

I will use this calculator to find the volume of a hollow cuboid. [2]

length = Red 243.84 cm or 239 px at 1.020251046 cm a pixel

Outer Edge B and C = 60.96 cm

thickness = Orange 11.4 px or 11.63086192 cm

inner Edge B and C = 60.96 - (2 x 11.63086192) = 37.69827616 cm

V = 559,603.43 cc

Energy to destroy chimney

Brick is, on average, 3.49375 MPa or 3.49375 J/cc

let's assume 50% is brick while the other half is cement.

279801.715 x 3.49375 = 977557.2418 Joules

279801.715 x 8 = 2238413.72 Joules

3.215970962e6 Joules in total, Room level

Destroying a Barrel

Volume of Barrel

Barrels, when empty, weigh around 50 kg or 50,000 grams

Barrels are typically made of oak and steel hoops. Assuming the barrel is 90% wood and 10% steel. The density of white oak is 0.77 g/cc

Wood = 45000/0.77 = 58441.55844 cc

Steel = 5000/7.9 = 632.9113924 cc

Energy to Destroy Barrel

Some barrels are destroyed completely or just their wooden parts.

Whole Barrel:

White oak has an average shear strength of 1935 PSI or 13.34136 MPa = 13.34136 J/cc

Steel = 208 x 632.9113924 = 131645.5696 joules

Wood = 13.34136 x 58441.55844 = 779689.8701 J

911335.4397 joules

Wall level+

Just the Wood:

Wood = 13.34136 x 58441.55844 = 779689.8701 J

Wall level+

Destroying a Skyscraper

Mass of a Skycraper = Around 222500 tons

http://theconstructor.org/practical-guide/rate-analysis-for-reinforced-concrete/6954/

154% = 28% Cement

154% = 42% Sand (which 85% of Sand or 35.7% of the RC)

154% = 84% Coarse (Granite is a good assumption)

Cement = 40454.55 Tons = 40454550 kg

Silica = 51579.55 Tons = 51579550 kg

Granite = 121363.64 Tons = 121363640 kg

Cement = 40454550/1250 = 32363.64 m^3

Silica = 51579550/2650 = 19463.9811 m^3

Granite = 121363640/2700 = 44949.4963 m^3

Fragmentation:

Low End: Using Reinforced Concrete Shear Strength:

(32363640000+19463981100+44949496300) cm^3* 28 J/cc 2.7097592872e12 Joules, or 647.648013 Tons, Multi-City Block level

High End: Using Each Material Shear Strength:

Cement = 6*32363640000 = 194181840000 J

Silica = 70*19463981100 = 1362478677000 J

Granite = 103.42*44949496300 = 4.64867691e12 J

Total Energy = 6.20533743e12 Joules, or 1.48311124 Kilotons, Multi-City Block level+

Another method:

381×129.2×57 mts = 2805836.4 m^3

90% hollowness = 280583640000 cm^3

Fragmentation: Low End: Using Reinforced Concrete Shear Strength: 280583640000 cm^3×28 J/cm^3 = 7.85634192e12 joules or 1.87771078 Kilotons Multi-City Block level+

High End: Using Each Material Shear Strength:

Percentages of material:

154% = 28% Cement

154% = 42% Sand (which 85% of Sand or 35.7% of the RC)

154% = 84% Coarse (Granite is a good assumption)

Volume:

Cement = (280583640000×28)/154 = 51015207300 cm^3

Silica = (280583640000 cm^3×35.7)/154 = 65044389300 cm^3

Granite = (280583640000 cm^3×84)/154 = 153045622000 cm^3

Frag:

Cement = 6 J/cm^3*51015207300 cm^3 = 306091243800 joules

Silica = 70 J/cm^3*65044389300 cm^3 = 4.55310725e12 joules

Granite = 103.42 J/cm^3*153045622000 cm^3 = 1.58279782e13 joules

Total Energy = 306091243800+4.55310725e12+1.58279782e13 Joules = 4.9443539 Kilotons, Multi-City Block level+

Melting:

Specific Heat Capacity:

Silica = 730 J/kg-°C

Alumina = 880 J/kg-°C

Granite = 790 J/kg-°C

Melting point:

Granite = 1237.5 °C Average

Silica = 1600 °C

Alumina = 2050 °C Average

Latent heat of fusion:

Granite = 335000 J/Kg

Silica = 50210 J/mol

(So: Molar Mass = 60.0843 g/mol = 3099121156065 mol)

Alumina = 620000 J/mol

(So: Molar Mass = 101.96 g/mol = 928067727000 mol)

Total Energy (No Cement) = (((790)*(121363640)*(2050-25)) + ((121363640)*(335000))) + (((730)*(51579550)*(2050-25)) + ((3099121156065)*(50210))) + (((880)*(9102272.72)*(2050-25)) + ((928067727000)*(620000))) = 7.3133614000828819e17 Joules, or 174.793533 Megatons, Mountain level (And that's without Cement)

Destroying a Plane

403500 lbs = 183024.521 Kgs

Percentages:

4% Titanium (Ti-6Al-4V) = 7320.98084 kg

13% Steel = 23793.1877 kg

81% Aluminium (2024-T3) = 148249.862 kg

Titanium Ti-6Al-4V = 4430 kg/m3

Steel = 7850 kg/m3

Aluminium 2024-T3 = 2780 kg/m3

Titanium = 1652591.61 cm3

Steel = 3030979.32 cm3

Aluminium = 53327288.5 cm3

Fragmentation=

Titanium = 550 MPa = 550 J/cc

Steel = 208 J/cc

Aluminium = 40000 PSI = 275.79 megapascales = 275.79 J/cc

Total Fragmentation = 1.6246502e10 Joules, or 3.88300717 Tons = Large Building level

Note: Shooting a plane down does not equal fragmentation. Fragmentation would apply if the plane is torn apart completely.

Destroying a Table

Square table

They are between 36 to 44 inches in length. The average of that is 40 inches, or 101.6 cm.

Thickness of the table top ranges from 3/4 inches to 1 3/4 inches. I'll take the average again, 1.25 inches or 3.175 cm.

101.6*101.6*3.175 = 32 774.128 cm^3

This is a low-ball since it doesn't account for the table legs. Assuming the table is made out of wood:

Fragmentation: 32774.128*8.34 = 2.7333622752e5 Joules, Wall level+

Violent fragmentation: 32774.128*18.34 = 6.0107750752e5 Joules, Wall level+

Pulverization: 32774.128*46.935 = 1.53825369768e6 Joules, Room level

Rectangular table

36 to 40 inches wide, and 48 inches for a four-people table. I'll take 38 inches as the width.

48 inches is 121.92 cm. 38 inches is 96.25 cm. The thickness is 3.175 cm as said above.

121.92*96.25*3.175 = 37 257.99 cm^3

Fragmentation: 37257.99*8.34 = 3.107316366e5 Joules, Wall level+

Violent fragmentation: 37257.99*18.34 = 6.833115366e5 Joules, Wall level+

Pulverization: 37257.99*46.935 = 1.74870376065e6 Joules, Room level

Round table

According to the same website above, round tables are around the same size as square tables. So let's say a diametre of 101.6 cm.

pi*(101.6/2)^2*3.175 = 25 740.74 cm^3

Fragmentation: 25740.74*8.34 = 2.146777716e5 Joules, Wall level+

Violent fragmentation: 25740.74*18.34 = 4.720851716e5 Joules, Wall level+

Pulverization: 25740.74*46.935 = 1.2081416319e6 Joules, Room level

Shattering a Windshield

Normal glass

Danny Hamilton measured the windshield's dimensions to be 46 inches for the top length, 35 inches for height and 56.5 inches for bottom length. That's 116.84 cm, 88.9 cm and 143.51 cm.

Area of a trapezium is (a+b)/2*h

(116.84+143.51)/2*88.9 = 11 572.5575 cm^2

wikipedia:Laminated glass#Specifications

A typical laminated makeup is 2.5 mm glass, 0.38 mm interlayer, and 2.5 mm glass. This gives a final product that would be referred to as 5.38 laminated glass.

For the glass:

11572.5575*0.5 = 5786.27875 cm^3

For the plastic layer:

11572.5575*0.038 = 439.757185 cm^3

Fragmentation of glass is 3.5 j/cc.

According to this the plastic is PVB. It's tensile strength is 19.6 MPa. Shear strength is 0.577 of tensile strength. 11.3092 MPa, or 11.3092 j/cc.

Fragmentation of the glass: 5786.27875*3.5 = 20 251.975625 Joules

Fragmentation of the plastic: 439.757185*11.3092 = 4973.301956602 Joules

In total that's 25 225.2775816 Joules, Wall level+

Blowing up Cannons

This is about blowing up 16th century cannons.

According to Wikipedia, by the 16th century they could weigh about 9100 kg and were largely cast iron.

Density of cast iron is = 7.8 g/cm^3

9100000 g / 7.8 g/cm^3 = 1166666.667 cm^3 of iron

Grey cast iron has a frag energy of 400 J/cc, a violent frag. energy of 613.5 J/cc and a pulverization energy of 827 J/cc.

Frag (400 J/cc): 1166666.667*400 = 466,666,666.8 J or 0.111536 tons of TNT (Small Building level+)

V. Frag (613.5 J/cc): 1166666.667*613.5 = 715,750,000.2045 J or 0.17106836 tons of TNT (Small Building level+)

Pulverization (827 J/cc): 1166666.667*827 = 964,833,333.609 J or 0.2306 tons of TNT (Small Building level+)

Stars feats

Average Neutron Stars GBE

Gravitational Binding Energy Equation for stars is (3*G*M^2)/(r(5-n))

The average neutron star is 1.4 Solar Masses with a radius of 10 kilometers as stated here and there.

• Solar mass is 1.989 × 10^30 kilograms
• Mass of the average star is (1.4*1.989 × 10^30) kilograms
• Radius is 10000 meters.
• Assuming a n (which can go from 0.5 to 1) is 0.5
• G is a constant of 6.67408x10^-11

Calculation

• (3*6.67408*10^-11*((1.4*1.989 * 10^30))^2)/((5-0.5)*10000) = 3.4 × 10⁴⁶ Joules (Solar System Level)

Creating or destroying a pocket realm with star(s)

Creating a pocket dimension containing a star at Astronomical unit distance

The assumption will be that the radius of the pocket dimension is 1 AU (an Earth-Sun distance).

The planet inside the pocket dimension is Earth.

Formula is E= 4*U*((Er/Br)^2), U is GBE of Earth, Er is the explosion's radius, Br is the Earth's radius, and E is the yield.

Sun at the center and planet at the "edge" of the pocket dimension 1 AU= 1,496*10^11 meters

GBE= 2,24*10^32 J

Earth's radius= 6.373.044,737 m

E= 4*2,24*10^32*((1,496*10^11:(6,373044737*10^6))^2)=

8,96*10^32*((9,534074926552*10^5)^2)=

8,96*10^42*(90,898584705107524194608704)=

8,1445131895776341678369398784*10^44 or 8,14 Foe Solar System level

Sun at the "edge" and planet at the center of the pocket dimension

Creating a pocket dimension containing a starry sky

Using the distance between the average star distance that the human eye can see couple with the average numbers of stars which the human eye can see on a clear sky which is around 2500 along with the average star size:

• Average star distance that human can see in starry night: (4 to 4000 light years)/2 = 2002 light years = 1.894e19 meters.
• I will be using the Gravitational Binding Energy of the sun for the average stars= 5.693e41 joules

The radius of the sun for the average star: 695510000 m

4*5.693e41*(1.894e19/695510000)^2 = 1.688e63 joules, (Multi-Solar System level)

Some references:

• http://apfloat.appspot.com/
• https://www.forbes.com/sites/startswithabang/2016/08/20/ask-ethan-how-many-stars-in-the-night-sky-still-exist/#2736d8f962eb
• https://www.theregister.co.uk/2006/12/01/the_odd_body_naked_eye_vision/
• https://spaceplace.nasa.gov/light-year/en/

Mass-energy Conversion Feats - Energy Constructs

While we know that E = mc^2, matter-energy conversion should only be used for a calculation if it is clearly stated that this is the progress used.

Mass-energy Conversion - The Tally

Object	Mass (kg)	Energy (J)	Tier
Pistol round 28 gr. (1.8 g) SS195LF JHP	0.0018	1.61773E+14	Town
FN Five-seven pistol	0.744	6.68663E+16	City
120mm Main Gun M829A3 ammo	10	8.9874E+17	Mountain
Rheinmetall 120mm Main Gun	4507	4.05062E+20	Island
Arrow	0.018	1.61773E+15	Town
Bow	18.18181818	1.63407E+18	Mountain+
European Longsword	1.4	1.25824E+17	City
Sledgehammer	9.1	8.17854E+17	Mountain
Boxing glove	0.8	7.18992E+16	City
Arm of a grown man	3.534	3.17615E+17	City+
A grown human	62	5.57219E+18	Island
All grown man on Earth	3.85E+11	3.46015E+28	Multi-Continent+
Theoretical mass of all life forms on Earth	1.01835E+13	9.15232E+29	Moon+
Theoretical mass of all life forms in our universe	3.05505E+35	2.7457E+52	Solar System
Private car	1311.363636	1.17858E+20	Island
M1A2 SEPv2 Abrams	64600	5.80586E+21	Small Country
Our Moon	7.342E+22	6.59855E+39	Large Planet
Our Earth	5.97237E+24	5.36761E+41	Star
Our Sun	1.9885E+30	1.78715E+47	Solar System
Our Solar System	1.99125E+30	1.78962E+47	Solar System
Our galaxy - the Milky Way	2.28674E+42	2.05519E+59	Multi-Solar System

Note: Source for mass of all life forms on Earth

I assume there are 100*10^9 planets that has a similar mass of life forms on Earth, and 300*10^9 such galaxies in the universe.

Mass-energy Conversion - Quick application

1. Some novice magician created a longsword as an energy construct and is accepted as a mass-energy conversion feat.

Energy used = 1.25824E+17 J = 30072576.9 tons of TNT (City level)

2. Some crazy doomsday robot attempted to turn all Earth life forms into energy, which the hero and the rival/nemesis stopped.

Energy yield by the doomsday robot = 9.15232E+29 J = 2.18746E+20 tons of TNT (Moon level+)

Energy countered by the hero and the rival/nemesis individually = 1.09373E+20 tons of TNT = 4.57616E+29 J (Moon level)

3. Some crazy cosmic tyrant snapped and decimated half of all life forms away into energy from the universe.

Energy possibly used = 50% * 2.7457E+52 J = 1.37285E+52 J = 3.28119E+42 tons of TNT (Solar System level)

Attacking a Person such that The Person Flew across a Distance before falling onto the Ground

We assume an average 2016 Japanese male at 25-29 is picked.

The target weighs at 66.82 kg and stands at 1.7185 m.

To make a target fall, the center of gravity is likely falling from roughly half his own height to roughly ground floor.

Height to fall = 1.7185/2 = 0.85925 m

By PE to KE formula, mgh = 0.5 m v^2
(9.81)(0.85925) = (0.5) v^2
v = ((2)(9.81)(0.85925))^0.5 = 4.105908547
time to fall to this speed = 4.105908547 / 9.81 = 0.418543175 s

Now, the kinetic energy from the yield of an attack should 1-to-1 scale to the target hit who flies at a distance before hitting the ground - in 0.418543175 s.

AP of an attack = Kinetic energy carried by the target = 0.5 x mass x (velocity)^2

The table below lists out the enrgy required to send a person flying at a speed across a distance using the Newtonian energy model.

Range (m)	Speed (m/s)	Speed (Mach)	Energy in Joules	Energy in Tons of TNT	Tier
0.5	1.194619886	0.003482857	47.679968	1.13958E-08	Below Average human
0.724105801	1.730062379	0.005043914	100	2.39006E-08	Average human
0.75	1.791929829	0.005224285	107.279928	2.56405E-08	Average human
1	2.389239772	0.006965714	190.719872	4.55831E-08	Athletic human
1.024040244	2.446677679	0.007133171	200	4.78011E-08	Athletic human+
1.254188037	2.99655594	0.008736315	300	7.17017E-08	Peak human
1.5	3.583859657	0.01044857	429.119712	1.02562E-07	Peak human
2	4.778479543	0.013931427	762.8794879	1.82333E-07	Peak human
2.092715875	5	0.014577259	835.25	1.9963E-07	Peak human
3.222782448	7.7	0.02244898	1980.8789	4.73441E-07	Peak human
4.101723116	9.8	0.028571429	3208.6964	7.66897E-07	Peak human
5.23597512	12.51	0.036472303	5228.668341	1.24968E-06	Peak human
6	14.33543863	0.041794282	6865.915391	1.64099E-06	Peak human
6.333339138	15.13186576	0.044116227	7650	1.82839E-06	Peak human+
8.868448661	21.18885025	0.061775074	15000	3.58509E-06	Wall
10	23.89239772	0.069657136	19071.9872	4.55831E-06	Wall
14.3560309	34.3	0.1	39306.5309	9.39449E-06	Wall
50	119.4619886	0.348285681	476799.68	0.000113958	Wall
71.78015452	171.5	0.5	982663.2725	0.000234862	Wall+
100	238.9239772	0.696571362	1907198.72	0.000455831	Room
129.2042781	308.7	0.9	3183829.003	0.000760953	Room
143.560309	343	1	3930653.09	0.000939449	Room
157.91634	377.3	1.1	4756090.239	0.001136733	Room
234.2736864	559.7360091	1.631883408	10467500	0.002501793	Room+
331.19431	791.3026175	2.307004716	20920000	0.005	Room+
358.9007726	857.5	2.5	24566581.81	0.005871554	Room+
500	1194.619886	3.48285681	47679968	0.011395786	Room+
717.8015452	1715	5	98266327.25	0.023486216	Room+
1000	2389.239772	6.96571362	190719872	0.045583143	Room+
1435.60309	3430	10	393065309	0.093944864	Small building
1672.449284	3995.882346	11.64980276	533460000	0.1275	Small building+
2341.897425	5595.354468	16.31298679	1046000000	0.25	Building
3589.007726	8575	25	2456658181	0.587155397	Building+
4967.914649	11869.53926	34.60507073	4707000000	1.125	Large Building
5000	11946.19886	34.8285681	4767996800	1.139578585	Large Building
6623.886199	15826.05235	46.14009431	8368000000	2	Large building
7178.015452	17150	50	9826632725	2.348621588	Large building
9264.4532	22135.00005	64.53352784	16369504368	3.912405442	Large building
10000	23892.39772	69.6571362	19071987198	4.55831434	Large building
11941.38067	28530.82162	83.18023795	27196000000	6.5	Large building+
14356.0309	34300	100	39306530900	9.394486353	Large building+
14811.45982	35388.12887	103.1723874	41840000000	11	City block
34893.48575	83368.90391	243.0580289	2.32212E+11	55.5	City block+
46837.94849	111907.0894	326.2597357	4.184E+11	100	Multi City Block
50000	119461.9886	348.285681	4.768E+11	113.9578585	Multi City Block
100000	238923.9772	696.571362	1.9072E+12	455.831434	Multi City Block
109844.7259	262445.3878	765.1469031	2.3012E+12	550	Multi City Block+
143560.309	343000	1000	3.93065E+12	939.4486353	Multi City Block+
148114.5982	353881.2887	1031.723874	4.184E+12	1000	Multi City Block+
273109.8245	652524.8547	1902.404824	1.42256E+13	3400	Multi City Block+
356707.1885	852259.0015	2484.720121	2.42672E+13	5800	Multi City Block+
500000	1194619.886	3482.85681	4.768E+13	11395.78585	Town
1000000	2389239.772	6965.71362	1.9072E+14	45583.1434	Town
1077272.815	2573863.055	7503.973922	2.21334E+14	52900	Town+
1255629.525	3000000	8746.355685	3.0069E+14	71866.6348	Town+
1481145.982	3538812.887	10317.23874	4.184E+14	100000	Town+
3473595.227	8299251.868	24196.06958	2.3012E+15	550000	Town+
4683794.849	11190708.94	32625.97357	4.184E+15	1000000	City
6371000	15221846.58	44378.56147	7.74125E+15	1850203.426	City

The table below lists out the enrgy required to send a person flying at a speed across a distance using the relativistic energy model.

Range (m)	Speed (m/s)	Speed (Mach)	Energy in Joules	Energy in Tons of TNT	Tier
1255629.525	3000000	8746.355685	3.00713E+14	71872.03271	Town+
1481145.982	3538812.887	10317.23874	4.18444E+14	100010.4517	Town+
3473595.227	8299251.868	24196.06958	2.30252E+15	550316.3282	Town+
4683794.849	11190708.94	32625.97357	4.18838E+15	1001046.261	City
6371000	15221846.58	44378.56147	7.75625E+15	1853788.582	City

One thing: I include a dataset for a distance of 9264.4532 m as the farthest horizon a human eye can see. Working: Average US human height = (1.753 + 1.615)/2 = 1.684 m Earth mean radius = 6371000 m
For two identical human to see each other at a distance, the farthest distance the one would travel away from the other standing still yet seeing each other can see each other = Arc(G1-M-G2) = 2 times Arc(G1-M)
G1-M = OM * angle(G1-O-M)
cos(angle(G1-O-M))= OM / (H1-G1 + G1-O) = 6371000 / (6371000 + 1.684)
angle(G1-O-M) = 0.00072708 rad
Arc(G1-M-G2) = 4632.2266 * 2 = 9264.4532 m

Picture

Miscellaneous Feats

Throwing a Person to the Horizon

Another common gag in fiction is that a person is punched/thrown so hard they reach the horizon/they fly out of sight.

On a normal day the visibility is usually 20 km.

Since an angle of 45 degrees requires the least force, that will be used as a low-ball.

Range of trajectory formula for 45 degrees angle is R = V^2/g. So now we can extract initial velocity from it: V = sqrt(R*g).

V = sqrt(20000*9.81) = 442.95 m/s

KE = 70*442.95^2*0.5 = 6.8671645875e6 Joules, Room level

Throwing a Person above the Clouds

Cloud height is usually 2000 m.

Formula is (close to earth): initial speed = sqrt(2*9.81*peak height). So in this case sqrt(2*9.81*2000) = 198 m/s

Using 70 kg for the human weight: 0.5*70* 198^2 = 1.37214e6 Joules, Room level

Punching a Hole through Doors

The average surface area of a human fist is 25 cm^2. The average thickness of a door is 3.334 cm thick. 83.35 cm^3. Values taken from here. For pulverization I'll use the average value.

Wood Door

Fragmentation: 83.35*8.34 = 695.139 Joules, Peak Human level

Violent fragmentation: 83.35*18.34 = 1528.639 Joules, Peak Human level

Pulverization: 83.35*46.935 = 3912.03225 Joules, Peak Human level

Steel Door

Fragmentation: 83.35*208 = 1.73368e4 Joules, Wall level

Violent fragmentation: 83.35*568.5 = 4.7384475e4 Joules, Wall level

Pulverization: 83.35*655 = 5.459425e4 Joules, Wall level

Punching through a Wall

Walls are 3/4 inch thick. That's 1.905 cm.

The human fist is 25 cm^2.

25 cm^2*1.905 = 47.625 cm^3

Wood Wall

Fragmentation: 47.625*8.34 = 397.1925 Joules, Peak Human level

Violent fragmentation: 47.625*18.34 = 873.4425 Joules, Peak Human level

Pulverization: 47.625*46.935 = 2235.279375 Joules, Peak Human level

Steel Wall

Fragmentation: 47.625*208 = 9906 Joules, Peak Human level+

Violent fragmentation: 47.625*568.5 = 2.70748125e4 Joules, Wall level

Pulverization: 47.625*655 = 3.1194375e4 Joules, Wall level

Flattening a Human/Humanoid

The body is 16% skin, 15% bone, 15.5% fat (this is the average between athletic men/women) and roughly 45% muscle. That last source actually says 40%, but this source says 40-50% and uses 50% in its pie chart, which shows everything else I've said to be pretty accurate. As such I'll say 45% as a middle ground.

This only adds up to 86.5‬% of the body. Another 7% (generally speaking) is blood, and I'm sure the other 4 is miscellaneous and impossible to calculate.

Now we'll add the compressive strength of all of those factors. Bone is the easiest answer with a compressive strength of 170 MPA. Skin isn't really something you can crush because it's more of an elastic sort of thing, but there is something called the "mean elastic modulus."

A mean elastic modulus (mean of course meaning average) is the measure of rigidity or stiffness of a material. This is as close as we're going to get to actual compressive strength so this seems fine to use. The mean elastic modulus for skin is a surprising 97 MPA.

This source gives muscle a low compressive strength of 1.0467 MPA. Unfortunate for that to be so, but at least this is moving along briskly.

Finally for fat. This has technically already been answered because body fat is just more skin, and so it also has a mean elastic modulus of 97 MPA as sourced above. So now for the j/cc required to pulverize an entire human body flat.

97 + 97 + 1.0467 + 170 = 365.0467 j/cc. That's for the entire body. This result can only be used if the outer body, inner body and bones have all been flattened.

If only the outer and inner body have been flattened, that's 195.0467‬ j/cc.

Flattening the average human

Average shoulder width is 40.64 cm, average head width (front to back) is 18 cm and the average height is of course 176.5 cm as you'll notice if you look over my calcs.

Anyhow, the volume of the average human thus is 129113.28‬ cubic cm.

If only the outer and inner body are flattened, this is 2.5183119e7 J. 9-A, Room level+.

If the entire body (outer and inner body + bones) has been flattened, this is 4.7132376‬e7 J. 9-A, Room level+.

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