23,473 Pages

## Introduction

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

### Getting Hit by a Car

#### 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

Street level

#### If slammed into a wall

However, it should be noted that the above calculation assumes 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.

When travelling at 45 mph: 0.5(1500) * 20.1168^2 = 303,514.23168 joules, or 303.5 Kilojoules - Wall level

When travelling at 60 mph: 0.5(1500) * 26.8224^2 = 539,580.85632 joules, or 539.5 Kilojoules - Wall level

When travelling at 70 mph: 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 Bus

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

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

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

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

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

##### Getting hit by a Pickup Truck

The average pickup trucks can weigh over 4082.3 kg.

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

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

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

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

##### Getting hit by a Semi Truck

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

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

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

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

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

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

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

If the wall is made out of steel:

Fragmentation:

198,668.75 * 208 = 4.1323100e7 joules, or 0.009 Tons of TNT, Small Building level

Violent fragmentation:

198,668.75 * 568.5 = 1.12943184e8 joules, or 0.027 Tons of TNT, Small Building 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-B (Wall 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-B (Wall level)

High-end (518.16 m/s)= 1.46 megajoule, 9-B (Wall level)

#### 32 lbs (14.515 kg)

Low-end (381 m/s)= 1.05 megajoules, 9-B

Mid-end (441.96 m/s)= 1.41 megajoules, 9-B

High-end (518.16 m/s)= 1.94 megajoules, 9-B

#### 42 lbs (19.0509 kg)

Low-end (381 m/s)= 1.38 megajoule, 9-B (Wall level)

Mid-end (441.96 m/s)= 1.86 megajoule, 9-B (Wall level)

High-end (518.16 m/s)= 2.56 megajoule, 9-B (Wall 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 160km.

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

Small Building 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 inof 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

(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

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 Street level to Street level+.

## Vaporization Feats

### Vaporizing a Human

Conditions

Okay, First off. To vaporize a human thoroughly at once, let’s assume the temperature change is 1800°F or 982.2°C https://www.cremationresource.org/cremation/how-is-a-body-cremated.html

Average body temperature is 97.7°F or 37.5°C

So the temperature change is by 944.72°C

The average human is 62 kilograms

STEP I

60% of human mass is water, or 37.2 kilograms.

The heat capacity of water is 4.1813 kilojoules per kilogram

Plugging the values into this calculator

Specific Heat energy is 146,945,868 joules

Plugging in the mass of water gives us 84,247,026 joules

Adding these two values together we get 231,189,783 J

STEP II

Average amount for body fat is 2.348 kilojoules per kilogram

Fat seems to be 17% of body mass, or 10.54 kilograms going by the numbers shown

Plugging it into the specific heat energy calculator, we get 23,379,855 joules

STEP III

Protein makes up 16% of body mass, which means it makes up 9.92 kilograms of the body

Muscle has a heat capacity of 3.421 kilojoules per kilogram

Plugging it into the specific heat energy calculator, we get 32,060,320 J.

STEP IV

For minerals, it makes up 6% of body mass, or 3.72 kilograms.

We will bone for this, specifically cortical bone, which is 1.313 kilojoules per kilogram.

We get 4,614,353 J

STEP V

Carbohydrates make up merely 1% of human weight, or 0.62 kilograms

Heat energy of sugar (carbohydrate) is 1.255 kilojoules per kilogram.

We get 735,087 joules

Conclusion

Adding them together, we get 291,982,509 J

Small Building level

As noted, we took values that were simplest and closest analogs, plus we did not include the latent heat from anything other than water.

### Vaporizing an average Building

• Turning an average building to ashes
• High end: 80% hollowness: 5.189e14 J = 124 kilotons. (Large Town level)
• Mid end: 85% hollowness: 3.892e14 J = 93 kilotons. (Town level+)
• Low end: 90% hollowness: 2.595e14 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.

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. (https://en.wikipedia.org/wiki/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.

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.

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.

“c” of alumina = 850 J/(kg*K)

“c” of steel = 481 J/(kg*K)

2.2 Latent heat of fusion:

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.

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. (https://en.wikipedia.org/wiki/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 = Street level (Speeds of an F1 tornado and Category 1 hurricane )

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

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

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

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

135 m/s = 11162.8 J = Street 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 = Wall 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, Small City level

"Moderate instability": 2.520813945240e16 Joules, 6.02488992648291 Megatons, Small 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, Small 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

## Earth Feats

### Destroying the Surface of the Earth

Earth's circumference = 40075 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 the Earth

This method assumes that all they're doing is causing the Earth to quake via sheer brute Force. This is what is usually used for the standard Earthquake feat, but, if there's sufficient evidence they're also moving the plates via magic or sheer rule of cool, you can move to the next section.

Either way, first we'll need to determine the kind of magnitude needed to cause the entire Earth to quake. We'll assume that it feels like a Magnitude 4 across the world, just standard noticable shaking with no real damage.

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

### 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

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

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.

## Freezing Feats

### Freezing a Human

Average human weight = 62kg

So water mass = 0.65*62kg.

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

## Crushing Feats

### Crushing a Golf Ball

Materials of Golf Ball

Energy Density of Materials

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

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, Street 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, Street 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, technique can greatly reduce the lifting strength necessary through leverage and bodyweight application. In addition, many fictional cases of neck snapping are outliers, with the characters never demonstrating similar lifting strength in any other capacity.

For these reasons, only use neck snapping as justification for Class 1 if the character has consistently demonstrated such strength with other feats.

## Object Destruction Feats

### Destroying a Door

Volume = 61947.75 cm^3

Wood Door Fragmentation = 516644.24 Joules

V. Frag = 1136121.74 Joules

Pulverization = 2907827.38 Joules

Steel Door Fragmentation = 1.289e7 Joules

V. Frag = 3.522e7 Joules

Pulverization = 4.058e7 Joules

### Destroying a Car

Mass and Weight of Materials

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.pdf)

Natural Rubber = 0.92 g/cm³

Synthetic Rubber = Wewill 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

Steel = 208 j/cc

Glass = 0.75 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

6803.88 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 48,446,846.69 Joules = Small Building level

### Destroying a Tree

Volume of Tree

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

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

Energy to Destroy Tree

Low End: 2.41317*38000000 = 9.1700460e7 Joules, 0.022 Tons of TNT, Small Building 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

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, Wall level+

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

### Breaking off a Lock

Volume of shackle This is a fairly standard lock.

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

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 = Street level

Brass = 347.8 J

Mid-Low End = Street level

Stainless Steel = 448.44 J

Mid-High End = Street level

Hardened Steel = 888 J

High-End = Street level

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

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

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

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, Wall level

### Destroying a Barrel

Volume of Barrel

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:

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

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, Small Town 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 Small Town 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, Small Town 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

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

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, Wall 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, Wall 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, Wall 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

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

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*0.75 = 4339.7090625 Joules

Fragmentation of the plastic: 439.757185*11.3092 = 4973.301956602 Joules

In total that's 9313.011019102 Joules, Street level+

### Blowing up Cannons

This is about blowing up 16th century cannons.

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

20 j/cc for iron frag, so we get...

23333333.34 Joules, 0.0055768005114723 tons of TNT, Small Building level

## Miscellaneous Feats

### Digging up from the Underground

Sometimes characters (usually monsters) burst out from underground.

Assuming the character's height is the height, and that the character's shoulder width is the width:

Height: 175 cm.

Width: 61 cm, 30.5 for the radius.

So the volume is 5.11e5 cubic centimeters.

Fragmentation:

5.11e5*8 = 4.088e6 Joules, Wall level

Violent fragmentation:

5.11e5*69 = 3.5259e7 Joules, or 0.008 Tons of TNT, Small Building level

If the ground is made out of steel:

Fragmentation:

5.11e5*208 = 1.06288e8 Joules, or 0.025 Tons of TNT, Small Building level

Violent fragmentation:

5.11e5*568.5 = 2.905035e8 Joules, or 0.069 Tons of TNT, Small Building level

• Please be noted that this is only for a quick bursting out, not slow digging.

### 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, Wall 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, Wall 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, Street level

Violent fragmentation: 83.35*18.34 = 1528.639 Joules, Street level

Pulverization: 83.35*46.935 = 3912.03225 Joules, Street 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, Street level

Violent fragmentation: 47.625*18.34 = 873.4425 Joules, Street level

Pulverization: 47.625*46.935 = 2235.279375 Joules, Street level

Steel Wall

Fragmentation: 47.625*208 = 9906 Joules, Street level+

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

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

## Creating or destroying a pocket realm with star(s).

### Creating a pocket dimension containing a star at Astronomical unit distance

• It yields: 8.1445131895776341678369398784*10^44 joules or 8.14 Foe (Large Star level)

### Creating a pocket dimension containing a star in the sky

• It yields: 7.503e57 joules (Multi-Solar System level)
• A multiplier format from the result based on the number of stars that were created or destroyed in the pocket realm may be used.

### Creating a pocket dimension containing a starry sky

• It yields: 1.4093732*10^61 joules, or 140.9373 PetaFoe (Multi-Solar System level)

## Start a Discussion Discussions about References for Common Feats

• #### A references for common feats page continuation

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