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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.
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 - Wall 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 - 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) = 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 - Wall 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 - Wall 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 - 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) = 0.5(36,287) * 11.176^2 = 2,266,177.145056 joules, or 2.27. Megajoules - Wall 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 - Wall 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 - Wall 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 - 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.
If the wall is made out of steel:
Toughness of Steel: 87.5 J/cm^{3}
Formula is: (Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
((198,668.75) * (70%)) * (87.5) = 1.21684609375e7 joules, or 0.0029 Tons of TNT, Wall 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 (Wall level)
Mid-end (441.96 m/s) = 1.41 megajoules, 9-B (Wall level)
High-end (518.16 m/s) = 1.94 megajoules, 9-B (Wall level)
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 160 km.
We will assume that a human like creature falls and that it starts at rest.
For the weight of the creature the assumption is 60 kg.
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
Wall level.
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.
The toughness value for a bone is 2.85 j/cc.
Formula is: ((Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
((3431.75362113402 cc) * (70%)) * (2.85 j/cc) = 6.8463484741623699e3 Joules, Peak Human level
Breaking a Human neck
Breaking a Neck It is noted that it takes 15 psi to break a human neck.
Using this conversion, it is 2.9285667750000002e3 Joules or Peak Human level.
Snapping a Neck It is noted to break snap a human neck takes 1,000 to 1,250 foot-pounds of torque.
Using 1,125 foot-pounds of torque as a middle ground.
This conversion site gives me 1525.295 joules or Peak Human level.
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 fibers. 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.
For a middle ground, the value will be accepted as 5,147.5 Joules, 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 (Small 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. The 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 kg * 0.2 * 880J/kg°C * (1500°C - 20°C) = 5.370680832e11 J = 128.362352581262 Tons of TNT. (City District level)
Mid end: 85% hollowness: 2,061,840 kg * 0.15 * 880J/kg°C * (1500°C - 20°C) = 4.028010624e11 J = 96.2717644359465 Tons of TNT. (City Block level)
Low end: 90% hollowness: 2,061,840 kg * 0.1 * 880J/kg°C * (1500°C - 20°C) = 2.685340416e11 J = 64.181176290631 Tons of TNT. (City Block 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 City District 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 = Above Average Human 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 = 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.
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 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 Metropolis 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
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
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.
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
Skulls have been easily destroyed before by large caliber rounds varying from 12-gauge shotgun slugs (At least 2363 ft-lbs or 3204 J), .500 S&W Magnum hollow-point rounds (3000-3900 J) and .308 Winchester/7.62x51mm NATO rounds (Ranging from 3500 to 3700 J), all of which have muzzle energies at around 3000-3900 joules (Peak Human level).
Such damage is even possible with several types of elephant gun rounds (The examples used including the .375 H&H Magnum, .416 Rigby, .458 Lott, .460 Weatherby Magnum, .500 Jeffery, .470 Nitro Express, .500 Nitro Express, .600 Nitro Express and .700 Nitro Express).
Potential Energy/Lifting Feats
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
Toughness value for wood is 0.67 j/cc
Toughness value for steel is 87.5 j/cc
Wood Door Formula is: ((Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
((61947.75 cc) * (70%)) * (0.67 j/cc) = 2.905349475e4 Joules, Wall level
Steel Door
((61947.75 cc) * (70%)) * (87.5 j/cc) = 3.7942996875e6, Wall 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.
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 Formula is: ((Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
Steel = 87.5 j/cc
((113924.0506 cc) * (70%)) * (87.5 j/cc) = 6.97784809925e6 Joules
Cast Iron = 40.7 J/cc
((18051.73151 cc) * (70%)) * (40.7 j/cc) = 5.142938307199e5 Joules
Glass = 0.35 J/cc
((9071.84 cc) * (70%)) * (0.35 j/cc) = 2.2226008e3 Joules
Aluminum = 0.327 J/cc
((66694.81481 cc) * (70%)) * (0.327 j/cc) = 1.5266443110009e4 Joules
Copper = 35 J/cc
((2480.580357 cc) * (70%)) * (35 j/cc) = 6.07742187465e4 Joules
Plastic = 0.29875 J/cc
((81890.1566 cc) * (70%)) * (0.29875 j/cc) = 1.7125278998975e4 Joules
Natural Rubber = 1.85 J/cc
((6902.486957 cc) * (70%)) * (1.85 j/cc) = 8.938720609315e3
Synthetic Rubber = 0.1825 J/cc
((13239.9827 cc) * (70%)) * (0.1825 j/cc) = 1.691407789925e3
Total Energy = 7.598160600024624e6 Joules, Wall 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 or (3.8e+7 cc)
Toughness of Wood is 0/67 j/cc
Formula is: ((Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
((3.8e+7 cc) * (70%)) * (0.67 j/cc) = 1.7822e7, Wall level
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. We will use a middle ground of 2,925 kg.
Steel = density of 7.9 g/cc
2925/7.9 = 370.2531645569620253 cc
Energy to Destroy Wrecking Ball
Steel = 87.5 j/cc
Formula is: ((Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
((370.2531645569620253 cc) * (70%)) * (87.5 j/cc) = 2.267800632911392404962e4, Wall level
Breaking off a Lock
Volume of shackle This is a fairly standard lock. This will be 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
Energy to Destroy Shackles
Steel = 86.5 j/cc
Formula is: ((Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
((5.92 cc) * (70%)) * (86.5 j/cc) = 358.456 Joules, 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, the toughness value is 87.5 j/cc.
Formula is: (Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
Longsword = ((104.71 cc) * (70%)) * 87.5 j/cc = 6.4134875e3 joules, Peak Human level
Shortsword = ((200.58 cc) * (70%)) * 87.5 j/cc = 1.2285525e4 joules, Wall level
Destroying a Chimney
Volume of Chimney
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
Noted that stainless steels is used in chimneys, thus the toughness value of stainless steel will be used, 86.75 J/cc
Formula is: ((Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
((559 603.43 cc) * (70%)) * (86.75 J/cc) = 3.398191828675e7 Joules, Wall 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
Toughness of Wood = 0.67 j/cc
Toughness of Steel = 87.5 j/cc
Energy to Destroy Barrel
Some barrels are destroyed completely or just their wooden parts.
Whole Barrel:
Formula is: ((Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
Steel = ((632.9113924 cc) * (70%)) * (87.5 j/cc) = 3.87658227845e4 joules
Wood = ((58441.55844 cc) * (70%)) * (0.67 j/cc) = 2.740909090836e4 joules
6.617491369286e4 joules
Wall level
Just the Wood:
Wood = ((58441.55844 cc) * (70%)) * (0.67 j/cc) = 2.740909090836e4 joules
Wall level
Destroying a Skyscraper
Explosion End: Skyscraper's are noted to be 100 to 150 meters in height, though using a median, I'm using 125 meters.
W = (125m)^3*((27136*1.37895+8649)^(1/2)/13568-93/13568)^2
W = 156.9696009738101485 Tons of TNT or 6.5676081047442163e11 Joules, City Distrct level.
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
Aluminum = 53327288.5 cm3
Toughness =
Formula is: ((Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
Titanium = 47.75 j/cc
((1652591.61 cc) * (70%)) * (47.75 j/cc) = 5.523787456425e7 Joules
Steel = 87.5 j/cc
((3030979.32 cc) * (70%)) * (87.5 j/cc) = 1.8564748335e8 Joules
Aluminum = 0.327 j/cc
((53327288.5 cc) * (70%)) * (0.327 j/cc) = 1.220661633765e7 Joules
Total Energy = 2.530919742519e8 Joules, Room level
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, giving it a toughness value of 0.67 j/cc
Formula is: ((Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
((32774.128 cc)* (70%)) * (0.67 j/cc) = 1.5371066032e4 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
Formula is: ((Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
((37257.99 cc) * (70%)) * (0.67 j/cc) = 1.747399731e4 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
Formula is: ((Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
((25740.74 cc) * (70%)) * (0.67 j/cc) = 1.207240706e4 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
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
Toughness value is 0.35 J/cc
Formula is: ((Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
((5786.2787 cc) * (70%)) * (0.35 j/cc) = 1.4176382815e3 Joules
For the plastic layer:
(11572.5575)*0.038 = 439.757185 cm^3
Toughness value is 0.29875 J/cc
((439.757185 cc) * (70%)) * (0.29875 j/cc) = 91.964221313125 Joules
Total Energy = 1.509602502813125e3, Peak Human level
Blowing up Cannons
This is about blowing up 16th century cannons.
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 toughness value of 40.7 j/cc.
Formula is: ((Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
((1166666.667 cc) * (70%)) * (40.7 j/cc) = 3.323833334283e7 Joules or Wall 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 × 10^{46 } 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 | 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 | Wall |
129.2042781 | 308.7 | 0.9 | 3183829.003 | 0.000760953 | Wall |
143.560309 | 343 | 1 | 3930653.09 | 0.000939449 | Wall |
157.91634 | 377.3 | 1.1 | 4756090.239 | 0.001136733 | Wall |
234.2736864 | 559.7360091 | 1.631883408 | 10467500 | 0.002501793 | Wall |
331.19431 | 791.3026175 | 2.307004716 | 20920000 | 0.005 | Wall |
358.9007726 | 857.5 | 2.5 | 24566581.81 | 0.005871554 | Wall |
500 | 1194.619886 | 3.48285681 | 47679968 | 0.011395786 | Wall+ |
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 | Room |
1672.449284 | 3995.882346 | 11.64980276 | 533460000 | 0.1275 | Room+ |
2341.897425 | 5595.354468 | 16.31298679 | 1046000000 | 0.25 | Small Building |
3589.007726 | 8575 | 25 | 2456658181 | 0.587155397 | Small Building |
4967.914649 | 11869.53926 | 34.60507073 | 4707000000 | 1.125 | Small Building |
5000 | 11946.19886 | 34.8285681 | 4767996800 | 1.139578585 | Small Building |
6623.886199 | 15826.05235 | 46.14009431 | 8368000000 | 2 | Small Building |
7178.015452 | 17150 | 50 | 9826632725 | 2.348621588 | Small Building |
9264.4532 | 22135.00005 | 64.53352784 | 16369504368 | 3.912405442 | Small Building+ |
10000 | 23892.39772 | 69.6571362 | 19071987198 | 4.55831434 | Small Building+ |
11941.38067 | 28530.82162 | 83.18023795 | 27196000000 | 6.5 | 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 | City District |
50000 | 119461.9886 | 348.285681 | 4.768E+11 | 113.9578585 | City District |
100000 | 238923.9772 | 696.571362 | 1.9072E+12 | 455.831434 | City District |
109844.7259 | 262445.3878 | 765.1469031 | 2.3012E+12 | 550 | City District |
143560.309 | 343000 | 1000 | 3.93065E+12 | 939.4486353 | City District |
148114.5982 | 353881.2887 | 1031.723874 | 4.184E+12 | 1000 | City District |
273109.8245 | 652524.8547 | 1902.404824 | 1.42256E+13 | 3400 | City District |
356707.1885 | 852259.0015 | 2484.720121 | 2.42672E+13 | 5800 | City District+ |
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 |
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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, 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 standard thickness of a door is 1 3/8 inches or 3.4925 centimeters or 42.599965203125. 87.3125 cc. Toughness of wood is 0.67 j/cc.
Wood Door
Formula is: ((Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
((42.599965203125 cc) * (70%)) * (0.67 j/cc) = 19.979383680265625 Joules, Subhuman level
Steel Door
Toughness value of steel is 87.5 j/cc.
((42.599965203125 cc) * (70%)) * (87.5 j/cc) = 2.60924786869140625e3 Joules, Peak Human 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
((6.913292625 cc) * (70%)) * (0.67 j/cc) = 3.242334241125 Joules, Subhuman level
Steel Wall
((6.913292625 cc) * (70%)) * (87.5 j/cc) = 423.43917328125 Joules, Peak Human level