Contents
- 1 Introduction
- 2 Impact Feats
- 3 Bone Breaking Feats
- 4 Vaporization Feats
- 5 Melting/Heat Feats
- 6 Weather Feats
- 7 Earth Feats
- 8 Freezing Feats
- 9 Crushing Feats
- 10 Potential Energy/Lifting Feats
- 11 Object Destruction Feats
- 11.1 Destroying a Door
- 11.2 Destroying a Car
- 11.3 Destroying a Tree
- 11.4 Destroying a Wrecking Ball
- 11.5 Breaking off a Lock
- 11.6 Destroying Blades
- 11.7 Destroying a Chimney
- 11.8 Destroying a Barrel
- 11.9 Destroying a Skyscraper
- 11.10 Destroying a Plane
- 11.11 Destroying a Table
- 11.12 Shattering a Windshield
- 11.13 Blowing up Cannons
- 11.14 Destroying Gravestones
- 11.15 Destroying Snow-tipped Mountains
- 12 Star feats
- 13 Creating or destroying a pocket realm
- 14 Mass-energy Conversion Feats - Energy Constructs
- 15 Attacking a Person such that The Person Flew across a Distance before falling onto the Ground
- 16 Miscellaneous Feats
- 17 Speed Feats
- 17.1 Changing Clothes Fast
- 17.2 Running around while things are standing still
- 17.3 Traveling Certain Distances
- 17.4 Galaxy Speed Feats
- 17.5 Escape Velocity
- 17.6 Speed Needed to Light Your Clothes on Fire
- 17.7 Speed Needed to Break the Sound Barrier
- 17.8 Speed Needed to Cause a Sonic Boom audible from the ground
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'sAttack 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
Feats concerning things impacting characters or characters impacting things.
Getting hit be a vehicle
Description: These calculations are about the durability a character needs to have in order to tank getting hit by various vehicles. It is differentiated between getting hit and sent flying and getting hit and remaining in place, like when they are getting slammed into a solid wall.
Requirements: The vehicle in question has to be approximately as heavy or heavier than the reference vehicle used. These weights are 1500 kg for the car, 4082.3 kg for the pickup truck, 10,659.421 kg for the (school) bus and 36,287 kg for the semi-truck. For the New York subway train 38646.0699kg per car were used. Average versions of these types of vehicles may be assumed to fulfill these conditions. The subway train must have 8 or 10 cars respectively and have been driving for long enough to reach its average speed or drive with top speed, for the respective ends. Other vehicles must be shown to reasonably fulfill the assumed speed they were going, for example by looking at the typical speed limit for the street they are on. For the 'slammed into a wall'-ends, the character has to tank the vehicle impact without moving from the place they are standing.
Results: The values for getting hit and beings sent flying:
25mph | 45mph | 60mph | 70mph | |
---|---|---|---|---|
Car | 3990.47 J (Street level) | 12929.12 (Street level+) | 22985.1069 J (Wall level) | 31285.284 J (Wall level) |
Pickup Truck | 4225.45244 J (Street level) | 13690.4659045 J (Street level+) | 24338.6060524 J (Wall level) | 33127.5471269 J (Wall level) |
Bus | 4314.74851771 J (Street level) | 13979.7851974 J (Street level+) | 24852.951462 J (Wall level) | 33827.63 J (Wall level) |
Semi-Truck | 4354.787 J (Street level) | 14109.50864 J (Street level+) | 25083.5709154 J (Wall level) | 34141.5270794 J (Wall level) |
The values for getting slammed into a wall / not moving are:
25mph | 45mph | 60mph | 70mph | |
---|---|---|---|---|
Car | 9.3677232e4 J (Wall level) | 303,514.23168 J (Wall level) | 539,580.85632 J (Wall level) | 734,429.49888 J (Wall level) |
Pickup Truck | 254,945.709462 J (Wall level) | 826,024.098658 J (Wall level) | 1,468,487.2865 J (Wall level) | 1,998,774.362185216 J (Wall level) |
Bus | 665,696.702668 J (Wall level) | 2,156,857.31665 J (Wall level) | 3,834,413.00737 J (Wall level) | 5,219,062.14892063232 J (Wall level) |
Semi-Truck | 2,266,177.145056 J (Wall level) | 7,342,413.94998144 J (Wall level) | 13,055,127.03695416 J (Wall level+) | 17,766,828.81723904 J (Wall level+) |
The subway train ends are:
- 8 cars & average speed: 9351396.1820257 J (Wall level)
- 10 cards & average speed: 11689245.235094 J (Wall level)
- 8 cars & top speed: 94480113.721716 J (Small Building level)
- 10 cards & top speed: 118100142.22854 J (Small Building level)
Calculations:
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
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 (Street 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 (Street 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 - Street 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 - Street 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 (Street 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 (Street 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 - Street 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 - Street 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 traveling 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+
Getting hit by a Subway Train
The weight of a subway car depends on the country, based on tracks from Brazil they can be anywhere from 16 to 30 tons. In Mexico they weigh 28.9 tons.
For this, let's use the NYC subway train:
- A typical revenue train consists of 8 to 10 cars
- A single car weighs 38.6 tons or 38646.0699 kilograms (85,200lb) when empty
- The subway has an average speed of 28.0 km/h, and a top speed of 89 km/h
Average speed
8 cars:
38646.0699 x 8 = 309168.559 kg
9351396.1820257 J = 2.2350373284 Kilograms of TNT (Wall level)
10 cars:
38646.0699 x 10 = 386460.699 kg
11689245.235094 J = 2.7937966623074 Kilograms of TNT (Wall level)
Top speed
8 cars:
38646.0699 x 8 = 309168.559 kg
94480113.721716 J = 22.581289130429 Kilograms of TNT (Small Building level)
10 cars:
38646.0699 x 10 = 386460.699 kg
118100142.22854 J = 28.226611431295 Kilograms of TNT (Small Building level)
Falling from Great Heights
Description: This calculation finds the durability a human would need to have in order to survive falling from so high that it reaches terminal velocity.
Requirements: To use this result the character has to be at least 70kg and should be roughly human shaped. The character needs to fall from a height of at least 143.3m and not be slowed down by anything other than the air resistance working on his own body.
Results: The character would tank 9.8315e4 J on impact with the ground and hence have Wall level durability.
Calculation: 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
Description: This calculates the Attack potency necessary to slam a human into a wall so hard that a human-sized hole is left in the wall. Alternatively, it also equates to the durability of tanking the attack.
Requirements: The hole in the wall has to have the area of a front facing human. Since such feats are often gag-feats particular attention has to be given to consistency of the result. The appropriate amount of destruction for destruction values higher than regular fragmentation needs to be proven. Whether the material of the wall is stone or steel has to be considered.
Results:
- Stone Wall:
- Fragmentation: 1.589350e6 J, Wall level
- Violent Fragmentation: 1.370814375e7 J, Wall level+
- Pulverization: 4.25151125e+7 J, Small Building level
- Steel Wall:
- Fragmentation: 4.1323100e7 J, Small Building level
- Violent Fragmentation: 1.12943184e8 J, Small Building level
- Pulverization: 5.9600625e+7 to 1.9866875e+8 J, Small Building level
Calculation:
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+
For pulverization (214 j/cm^3)
198.668.75 * 214 = 4.25151125e+7 joules, Small Building level
If the wall is made out of steel:
Fragmentation (208 j/cm^3):
198,668.75 * 208 = 4.1323100e7 joules, or 0.009 Tons of TNT, Small Building level
Violent fragmentation (568.5 j/cm^3):
198,668.75 * 568.5 = 1.12943184e8 joules, or 0.027 Tons of TNT, Small Building level
Pulverization (300-1000 J/cm^3)
198,668.75 * 300 or 1000 = 5.9600625e+7 to 1.9866875e+8 joules, 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 cannot 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 in between 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 altitudes the potential energy value would go closer to the kinetic energy when falling with escape velocity, while for lower it would mostly just stay the same (the realistic value would go towards to terminal velocity value) except for short falls where not even that much speed if attained.
Bone Breaking Feats
Breaking all the Bones of a Man's Body
On average, the weight of a man's bones is 15% of their body mass, which in of itself is 88.768027 Kilograms. 15% of that is 13.31520405 Kilograms.
The density of bone is 3.88 g/cm^3, which would mean that the total volume would be 13.31520405 divided by 0.00388, which equals 3431.75362113402 cm^3 for our volume.
To get the fragmentation values, we need to use the compressive strength of bones. To quote Wikipedia, "bone has a high compressive strength of about 170 MPa (1800 kgf/cm²), poor tensile strength of 104–121 MPa, and a very low shear stress strength (51.6 MPa)"
So, low end is 51.6, mid is 104, high is 170. Plugging those all into our volume gets us....
Low End: 1.77078.486850515432e5 Joules, Wall level
Mid End: 3.56902376598e5 Joules, Wall level
High End: 5.83398115592783e5 Joules, Wall level
Breaking a Human neck
Volume of a Vertebra
The vertebrae that make up the neck are the cervical vertebrae and are 7 vertebrae in total. However, due to finding info only for vertebra 3 through 7, the smallest one will be calced.
C3 pedicle: The pedicle is roughly a rectangular prism and there are two of them. 5.27 mm x 5.14 mm x 7.08 mm = 0.527 cm * 0.514 cm * 0.708 cm = 0.191781624 cm^3. 0.191781624 cm^3 * 2 = 0.383563248 cm^3
C3 vertebral body: The vertebral body is a cylinder. The mean height is 15.1 mm and the radius 7.34 mm = 2.55575 cc.
Energy to Fragment the C3 Vertebra
The shear strength of bones is 51.6 MPa or J/cc
(0.383563248 + 2.55575) x 51.6 = 151.6685635968 J
Athlete level
Keep in mind, this is just fragmenting most of the C3 vertebra. This does not take into account the lamina.
Breaking a Bone
The durability of a bone depends on the angle of attack.
A bone of a deceased 52-year-old woman only required 375 Joules of energy when the force was applied within five degrees of the orientation of the collagen 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 Street level to Street level+.
Vaporization Feats
Vaporizing a Human
Conditions
https://www.thoughtco.com/chemical-composition-of-the-human-body-603995
Okay, first off, to vaporize a human thoroughly at once, let’s assume the temperature change is 1800°F or 982.2°C
The normal human body temperature range is typically stated as 36.5–37.5 °C (97.7–99.5 °F) and we shall use 37°C.
So, the temperature change is by 945.22°C
The average human is 62 kilograms
STEP I
We will start with water.
60% of human mass is water, or 37.2 kilograms.
The heat capacity of water is 4182 joules per kilogram at 20 °C
Plugging the values into this calculator
Heat energy spent to change temperature is 147,023,640 joules
Plugging in the mass of water gives us 84,247,052.04 joules
Adding these two values together we get 231,270,692.04 joules
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 23,392,229 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 32,077,288 joules.
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.
(ditto) we get 4,616,795 joules
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.
(ditto) we get 735,476 joules
Conclusion
Adding them together, we get 292,092,454 joules or 0.069818 tons of TNT (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.
Reducing a Human to char
In the case of reducing humans to just dry bone fragments instead of fully vaporizing them, crematoriums recommend a minimum starting temperature of 1400°F or 760°C. https://www.cremationresource.org/cremation/how-is-a-body-cremated.html
Conditions
https://www.thoughtco.com/chemical-composition-of-the-human-body-603995
Average body temperature being 98.6°F or 37°C,
wikipedia:Human body temperature
The temperature change is now by 723°C
The same heat capacity calculator
The average human is 62 kilograms
STEP I
We will start with water.
60% of human mass is water, or 37.2 kilograms.
The heat capacity of water is 4182 joules per kilogram at 20 °C
Plugging the values into this calculator
Specific Heat energy is 112,477,399.20 joules
Plugging in the mass of water gives us 84,247,052.04 joules
Adding these two values together we get 196,724,451.24 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 17,892,746.16 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 24,535,959.36 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.
(ditto) we get 3,531,392.28 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 562,566.30 joules
Conclusion
Adding them together, we get 243,247,115.34 joules or 0.058137456 tons of TNT (Small Building level)
Again, we took values that were the simplest and closest analogs, plus we did not include the latent heat for anything other than water.
Incinerating an average Building
This calc assumes a relatively standard two-story house. If the building in question is significantly different than this, a more specific calc may be needed.
Destruction Value: 1.70 Kilotons of TNT equivalent, or Small Town level
Melting/Heat Feats
Surviving the Heat of the Sun
Surface 1. Radiation: For radiation we need to know the emissivity, surface area and temperature.
The temperature of the sun is about 5500°C per Wikipedia.
For the surface area we take the surface of the average human body, since we assume that the person is submerged in the sun. The average body surface area is about 1.73 m^2 per this article.
The emissivity is about 1.2 at this temperature per this article.
Now we input these 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 where 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. Conduction: 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 a 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
Aluminum 2024-T3 = 875 J/kg-°C
Melting Point Titanium = 1604 °C
Steel = 1425 °C
Aluminum = 502 °C
Latent Heat of Fusion Titanium = 419000 J/Kg
Steel = 272000 J/Kg (This is for Iron, but is nearly the same though)
Aluminum = 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 armor 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
Melting a Car
Description: The energy necessary to melt a car.
Requirements: The car in question may be at most 20°C warm and has to be at least the size of an average car and similar in composition.
Results: 1334006051 joules or 0.3188350982314 tons of TNT or Building level
Calculation: From the destroying a car calc we see there's:
900 kg of steel
180.076 kg of aluminum
22.226 kg of copper
45.3592 kg of glass
183.0245 kg of plastic
About 4.7 kg of rubber. Yes, I'm combining the two kinds. Sue me.
And 131.77764000000002 kg of cast iron.
There is no latent heat of fusion for glass, but melting glass is 2494 J/cm^3
Glass makes up 2.478313012738732% of a car.
Volume of a mid-size car is 3.1e+6 cm^3
(3.1e+6)*2.478313012738732% = 76827.7033949
76827.7033949*2494 = 191608292.267 joules
Specific Heat:
Steel = 481 J/(kg*K)
Aluminum = 870 J/(kg*K)
Copper = 390 J/(kg*K)
Plastic = 1670 J/(kg*K)
Rubber = 2010 J/(kg*K)
Cast iron = 460 J/(kg*K)
Latent Heat of Fusion:
Steel = 260000 J/kg
Aluminum = 396567.46 J/kg
Copper = 206137 J/kg
Plastic = I couldn't find an explicit one for plastic, but I did find one for propylene which is used in plastic so that's going to have to do. 71400 J/kg
Rubber = (On page 512 of this) 16710 J/kg
Cast iron = 247112.54 J/kg
Melting Point:
Steel = 1425 °C
Aluminum = 502 °C
Copper = 1084.62 °C
Plastic = 100 °C
Rubber = 600 °C
Cast iron = 1538 °C
Assuming a car is on average 20°C warm.
Energy = ((481*900*(1425-20))+(900*260000))+((870*180.076*(502-20))+(180.076*396567.46))+((390*22.226*(1084.62-20)+(22.226*206137))+((1670*45.3592*(100-20))+(45.3592*71400))+((2010*4.7*(600-20))+(4.7*16710))+((460*131.77764000000002*(1538-20))+(131.77764000000002*247112.54)) = 1142397758.73 joules
1142397758.73+191608292.267 = 1334006051 joules or 0.3188350982314 tons of TNT or barely Building 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 temperature it should be around 0.65.
The average human body surface area is 1.73 m^2.
At last, we input all these 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 different 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 continuous and would keep on delivering the same number 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 20 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 than 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
Flooding the Entire Earth
This calculation uses this blog for reference.
- Earth's surface area = 510,072,000km^2
Genesis 7:19-20 states the waters went fifteen cubits above the highest mountains. The cubit is an ancient Egypt Roman unit of size which was, on average, equivalent to 45.72cm. So, fifteen cubits would be seven meters.
- Water volume = 510,072,000*8.8850 = 4516687560km^3, or 4.51668756e18 cubic meters
- Water density = 1025kg/m^3 (For an average value between seawater and freshwater)
- Mass of water = 4.6296047e21kg
In order to truly flood the planet, one would need to vaporize all of this water and condense the vapor into rain clouds
- Average global temperature = 16 degrees Celsius
Vaporization
- q1 = (4.6296047e21kg)*(84)*(4186) = 1.6278e27 joules
- q2 = (4.6296047e21kg)*(2264760) = 1.04849435e28 joules
- Total = 1.2112743e+28 joules
Condensation
At the same temperature of 16 degrees Celsius, we approximate the latent heat of condensation with:
- 2500.8 - 2.36*16 + 0.0016(16)^2 - 0.00006(16)^3 = 2463.20384j/g, or 2463203.84 joules per kilogram.
- q3 = (4.6296047e21)*(2463203.84) = 1.140366e28 joules
- Total flood energy = 1.2112743e28+1.140366e28 = 2.3561e28j, or 5.6205 exatons of TNT Multi-Continent level
Creating a Snowstorm
Description: The energy required to create a snowstorm has to be calculated.
Requirements: The snowstorm has to cover the entire sky on a priorly clean day or needs to be at least 20km in radius. The temperature of the land in that area needs to be cooled, such that a total temperature change of at least 20.5°C happens. For the CAPE end the storm needs to demonstrate a moderate instability.
Results:
- Condensation + Temperature Change = 15.73 Megatons of TNT (City level)
- CAPE + Temperature Change = 15.95 Megatons of TNT (City level)
Calculation: The type of clouds that produce snow are Nimbostratus, which usually are 2000 to 4000 meters high, with 3000 meters being the average.
Assuming the snowstorm was created on a clear day and with a good view to horizon, the radius of this cloud would be 20000 meters.
Volume = pi*20000^2*3000 = 3769911184307.75 m^3
The liquid water content of Nimbostratus clouds is 0.001 kg/m^3.
Cloud Mass (Water) = 3769911184307.75 x 0.001 = 3769911184.31 kg
The density of cloud air is 1.003 kg/m^3.
Cloud Mass (Air) = 3769911184307.75 x 1.003 = 3781220917860.67 kg
Now we just need to apply the different methods.
Condensation
The formula for condensation is 2264705 J/kg.
Energy = 2264705 x 3769911184.31 = 8537736708662778.55 Joules, 2.04 Megatons of TNT
CAPE
In this method we're going to assume a moderate instability of 2500 J/kg for the snowstorm.
Energy = 3781220917860.67 x 2500 = 9453052294651675, 2.26 Megatons of TNT
Temperature Change
For this final method we need to calculate the temperature change made at the surface, below the storm. The average temperature between 0 meters (Sea level) and 2000 meters is 8.50°C (281.65 K), and the average density of air is 1.112 kg/m^3. A snowstorm can get temperatures down to -12°C (261.15 K).
Mass = (pi*(20000)^2*2000)*1.112 = 2794760824633.48 kg
C = 1000 J/kg*K
ΔΤ = T(Initial) - T(Final) = 281.65 - 261.15 = -20.5 K
Q = M*C*ΔΤ
Q = (2794760824633.48)*1000*20.5 = 5.86*10^20 Joules, 13.69 Megatons of TNT (
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 as this is an air-blast explosion, 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
However, in the case of a ground-based explosion where the explosion is generated on the ground, we use the following formula:
W = R^3*((27136*P+8649)^(1/2)/13568-93/13568)^2, where W is the yield in tons of TNT, R is the radius and P is the shockwave pressure in bars, where we generally use 1.37895 bars or 20 psi of pressure. There is no need to halve the explosion yield result in the case of ground-based explosions.
R = 20037.50 km or 20037500 m
P = 1.37895 bars
W = 20037500^3*((27136*1.37895+8649)^(1/2)/13568-93/13568)^2 = 6.46570851e+17 tons of TNT or 646.57085 Petatons (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 noticeable 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 joule count 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
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
Diameter of the Earth is 12 742 000 meters. Radius is 6 371 000 meters.
No feat, so I'll assume the Earth is split apart by 1 kilometer, or 1000 meters.
The center of mass of each individual half is 3R/8 from the center 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 diameter of the Earth is 627 pixels, or 12 742 000 meters.
For the split to be visible I'll assume 10 pixels or so. That's 203 222 meters.
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 minuscule 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, Aluminum, Iron, Calcium, and Magnesium. Divide this 1% volume between them for the following masses:
- Total Volume = 1.0837e19 m^3
- Volume Each = 1.80616667e18 m^3
- Oxygen = 2.06083617e15 kg
- Silicon = 4.20475601e21 kg
- Aliminium = 4.89471168e21 kg
- Iron = 1.42217564e22 kg
- Calcium = 3.61233334e21 kg
- Magnesium = 3.14273001e21 kg
Subtract this from values established at the beginning to get the following table of values.
Mass of Elements
- Mass of Earth = 5.98e24 kg
- Mass of Crustal Iron = 1.42217564e22 kg
- Mass of Crustal Oxygen = 2.06083617e15 kg
- Mass of Crustal Silicon = 4.20475601e21 kg
- Mass of Crustal Calcium = 3.61233334e21 kg
- Mass of Crustal Magnesium = 3.14273001e21 kg
- Mass of Mantle-Based Iron = 9.93711494e23 kg
- Mass of Mantle-Based Nickel = 5.382e22 kg
- Mass of Mantle-Based Oxygen = 1.79998e24 kg
- Mass of Mantle-Based Silicon = 8.98775244e23 kg
- Mass of Mantle-Based Sulfur = 1.7342e23 kg
- Mass of Mantle-Based Magnesium = 8.2807727e23 kg
- Mass of Mantle-Based Aluminium = 7.88252883e22 kg
- Mass of Mantle-Based Calcium = 8.60876667e22 kg
- Mass of Outer-Core Iron = 8.36004493e23 kg
- Mass of Outer-Core Nickel = 5.382e22 kg
- Mass of Inner-Core Iron = 7.564226e22 kg
Now we need Specific Heat energy since we've spent all that time setting this shit up. We won't do anything for the Core elements since their heat doesn't need to change at all for this event to happen.
Specific Heat Energy
As said earlier, as a low-end we assume 3265 C end temperature, as a high-end we assume 3414 C end temperature based on Tungsten. EDIT: Coming back to it, just using the high-end. Results don't change much and it's the only thing that makes logical sense.
The calculation for this is just mass times specific heat times temperature change (which varies). I'm beaten by this so far so I'm using this.
- Low-End Temp Change for Crust Elements = 2915 C
- High-End Temp Change for Crust Elements = 3064 C
- Low-End Temp Change for Mantle Elements = 1165 C
- High-End Temp Change for Mantle Elements = 1314 C
Let's get to it.
- Crustal Iron = 2.005e28 Joules
- Mantle Iron = 6.006e29 Joules
- Crustal Oxygen = 5.803e21 Joules
- Mantle Oxygen = 2.174e30 Joules
- Crustal Silicon = 9.147e27 Joules
- Mantle Silicon = 8.385e29 Joules
- Crustal Magnesium = 1.011e28 Joules
- Mantle Magnesium = 1.143e30 Joules
- Mantle Sulfur = 1.595e29 Joules
- Mantle Nickel = 3.115e28 Joules
- Mantle Aluminium = 9.011e28 Joules
- Mantle Calcium = 7.127e28 Joules
Total Energy of Heat Change = 5.14743701e30 Joules, Small Planet level. But we're far from done.
Shifts in Matter
Now we get into truly changing the matter from solid/liquid to gas. For this we classify everything by solid or liquid. Everything in the inner core is liquid (a small amount of iron)- everything else is held to be solid. This includes the outer core which shifts between liquid and solid.
For solids, they must undergo a state of fusion and vaporization, so we need to multiply them by their values for that in J/kg. For the liquid, it must only undergo the value for vaporization. Oxygen is already gaseous so it needn't be accounted for.
- Solid Iron = 1.84393774e24 kg
- Liquid Iron = 7.564226e22 kg
- Solid Silicon = 9.0298e23 kg
- Solid Magnesium = 8.3122e23 kg
- Solid Sulfur = 1.7342e23 kg
- Solid Nickel = 1.0764e23 kg
- Solid Aluminium = 8.372e22 kg
- Solid Calcium = 8.97e22 kg
Let's start with the irons.
- Liquid Iron Vaporization = 4.700e29 Joules
- Solid Iron Fusion & Vaporization = 4.557e29 Joules & 1.146e31 Joules
- Solid Silicon Fusion & Vaporization = 1.614e30 Joules & 1.154e31 Joules
- Solid Magnesium Fusion & Vaporization = 3.062e29 Joules & 4.357e30 Joules
- Solid Sulfur Fusion & Vaporization = 9.288e27 Joules & 5.299e28 Joules
- Solid Nickel Fusion & Vaporization = 3.204e28 Joules & 6.793e29 Joules
- Solid Calcium Fusion & Vaporization = 1.911e28 Joules & 3.469e29 Joules
- Solid Aluminium Fusion & Vaporization = 3.320e28 Joules & 9.091e29 Joules
Total Fusion + Vaporization Energy = 3.2284828e31 Joules, Small Planet level
the latent heat of fusion/vaporization for Magnesium, Sulfur, and Nickel were calculated since they aren't present on our Calculations page.
Magnesium has Fusion of 8954 J/mol and 127400 J/mol for Vaporization. Magnesium weighs 24.305 g/mol so energy is...
- Fusion: 368401.563 J/kg
- Vaporization: 5241719.81 J/kg
Sulfur has Fusion of 1717.5 J/mol and 9800 J/mol for Vaporization. Sulfur weighs 32.07 g/mol so energy is...
- Fusion: 53554.724 J/kg
- Vaporization: 305581.54 J/kg
Nickel has Fusion of 17470 J/mol and 370400 J/mol for Vaporization. Nickel weighs 58.69 g/mol so energy is...
- Fusion: 297665.501 J/kg
- Vaporization: 6311126.26 J/kg
Total Energy
We're adding together all values denoted in Vaporizing Earth's topic as well as the GBE of Earth.
- GBE of Earth = 2.24e32 Joules
- Matter Shifting of Earth = 3.2284828e31 Joules
- Heat Change of Earth = 5.14743701e30 Joules
Total Energy = 2.614e32 Joules, Planet level.
The World Gets Vaporized: 62.48 Zettatons of TNT, Planet level
Freezing Feats
Freezing a Human
Average human weight = 62kg
On average 65% of the human body weight is water.
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
Freezing Earth
Description: Freezing the Earth. Two ends are given. One is for freezing just its atmosphere, the other is for freezing everything including oceans and the Earth's core.
Requirements: For the atmosphere end, at least all air needs to get frozen, all the way up into space. For the whole Earth end, everything including air, oceans, crust, mantle and core needs to be frozen.
Results:
- Freezing Earth's atmosphere: 563.86 Teratons, High 6-B+
- Freezing the entire Earth (plus its oceans): 2.378 Zettatons, Low 5-B
Calculation: This will be a calc for the energy output needed to freeze Earth, its atmosphere, or both.
Freezing the atmosphere for the first part. Info:
1. Atmosphere has a total mass of 5.15×10^18 kg (Wikipedia)
2. It's 78.09% Nitrogen and 20.95% Oxygen, the rest is negligible. (Wikipedia)
E=m*c*ΔT is the formula.
Nitrogen first. Average temperature of the atmosphere is 22.24°C.
E = [4021635000000000000 * 1040 * (295.3 - 63.15)] + (4021635000000000000 * 25702) + (4021635000000000000 * 199190)
E = 1875401006280000000000000J
Now Oxygen.
E = [1078925000000000000 * 919 * (295.3 - 54.35)] + (1078925000000000000 * 13875) + (1078925000000000000 * 213125)
E = 483825628471250000000000J
Now we add them together for a total of 2.35922663475120 × 10^{24} J. Energy needed to freeze the atmosphere.
Second part, freezing the core of Earth (outer core + inner core). Info:
1. The outer core weights 1.87 * 10^24 kg (Quora)
2. The inner core has a mass of 10^23kg. (Wikipedia)
3. Both outer core and inner core have a very similar temperature and composition, that is, 5700°K for the temperature and the composition is 88.8% iron, 5.8% nickel, the rest is negligible. (Wikipedia)
With that, we go to the calc. outer core first.
E = [10^23 * 88.8/100 * 460 * (5700 - 1811)] + (10^23 * 88.8/100 * 247112) + (10^23 * 88.8/100 * 6213627) + [10^23 * 11.2/100 * 440 * (5700 - 1728)] + (10^23 * 11.2/100 * 297000) + (10^23 * 11.2/100 * 6311000)
E = 8.261551112 × 10^29 J
Now for the inner core.
E = [1.87 * 10^24 * 88.8/100 * 460 * (5700 - 1811)] + (1.87 * 10^24 * 88.8/100 * 247112) + (1.87 * 10^24 * 88.8/100 * 6213627) + [1.87 * 10^24 * 11.2/100 * 440 * (5700 - 1728)] + (1.87 * 10^24 * 11.2/100 * 297000) + (1.87 * 10^24 * 11.2/100 * 6311000)
E = 1.544910057944 × 10^30 J
For a total of 2.371065 * 10^30J to freeze our Earth's core.
Final parts, crust, upper mantle and mantle. Info:
1. The mantle has a mass of 2.95 * 10^24kg. (Wikipedia)
2. The upper mantle has a mass of 1.06 * 10^24kg. (this states)
3. Earth's crust has an estimated mass of 2.6 * 10^22kg. (Quora)
4. Earth is 32.1% iron, 30.1% oxygen, 15.1% silicon, 13.9% magnesium, the rest is negligible. (Wikipedia. Let's use this for the mantle and for the upper mantle since I don't know any better and it should be about right anyways).
5. Earth's crust is 46.6% oxygen, 27.7% silicon, 8.1% aluminum, 5% iron. The rest is negligible. (Wikipedia)
6. Earth's crust has a temperature of 200°C to 400°C for an average of 573.15°K. (Wikipedia)
7. Earth's upper mantle has a temperature of 200°C to 900°C for an average of 823.15°K. (Wikipedia)
8. Earth's mantle has a temperature of 900°C to 4000°C for an average of 2723.15°K (Wikipedia)
Oof. Alright i think we can start.
Let's start from the most intern layers to the most extern one, so we start with Earth's mantle.
E = [2.95 * 10^24 * 32.1/100 * 460 * (2723.15 - 1811)] + (2.95 * 10^24 * 32.1/100 * 247112.54) + [2.95 * 10^24 * 30.1/100 * 919 * (2723.15 - 54.15)] + (2.95 * 10^24 * 30.1/100 * 13875) + (2.95 * 10^24 * 30.1/100 * 213125) + [2.95 * 10^24 * 15.1/100 * 710 * (2723.15 - 1687)] + (2.95 * 10^24 * 15.1/100 * 1787113) + [2.95 * 10^24 * 13.9/100 * 1020 * (2723.15 - 923)] + (2.95 * 10^24 * 13.9/100 * 348971) + (2.95 * 10^24 * 13.9/100 * 5267489.71) = 7190587580813500000000000000000J
E = 7.190587 * 10^30 J for freezing the Earth's mantle.
Next step, freezing the upper mantle. The parts that are already solid at the temperature of 823.15°K are left out of the equation.
E = [1.06 * 10^24 * 30.1/100 * 919 * (823.15 - 54.15)] + (1.06 * 10^24 * 30.1/100 * 27750) + (1.06 * 10^24 * 30.1/100 * 426250)
E = 3.7033645166 * 10^29J
Next and last step, freezing Earth's crust. The parts that are already solid at the temperature of 573.15°K are left out of the equation.
E = [2.6 * 10^22 * 46.6/100 * 919 * (573.15 - 54.35)] + (2.6 * 10^22 * 46.6/100 * 27750) + (2.6 * 10^22 * 46.6/100 * 426250)
E = 1.12772965552 * 10^28J
For a precise result for the crust part specifically, it's fair to include this calc that Jasonsith made in here.
E = 6.3375485552 * 10^27J is the final result for this part then.
For a precise result for the crust part specifically, it's fair to include this calc that Jasonsith made in here.
For a total of 1.18382125552 * 10^28 J.
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, 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 body weight 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.
Ripping off heads
Credits to vsauce3 for his video.
According to vsauce3's video, it takes about 5000 to 15000 newtons of force to rip out a human head, with the lower-end being supported by accidental hanging drops generating 1000-1260 lbf or 453.592 kgf to 572 kgf (Class 1) which rarely resulted in decapitations, and with higher-end being supported by another certified medical article here, which states a pulling strength of 12000 newtons being required to rip heads off.
5000 newtons of force= 509.8581 kgf (Class 1 lifting strength) (Note: This is the absolute bare minimum required to rip heads off based on the values derived from hanging drops. It should at the very least also suffice for ripping heads off of weaker-than-average people, as not all human bodies have a standardized value).
12000 newtons of force= 1223.6595 kgf (Class 5 lifting strength) (Note: This is the medically-certified standard for casually ripping heads out, which supports vsauce3's higher-end value of 15000 newtons, and should work for ripping out heads of average human beings)
15000 newtons of force= 1529.5743 kgf (Class 5 lifting strength) (Note: This is the higher-end value calculated by vsauce3)
Note: It should be noted that all these numbers are equally correct and applicable, as not all human beings have the same body type or the same strength level. For example, it would be easier to rip off the head of a child than ripping out the head of a world-champion heavyweight lifter.
Ripping off spines
Credits to vsauce3 for his video.
According to vsauce3, the absolute minimum force required to even try and rip out the spine from the skeleton itself would require forces upwards of 1 million newtons, or 101971.621 kgf (Class K lifting strength), to rip it clean from the back and neck muscles as well would require even higher forces and greater pulling distance.
Object Destruction Feats
Destroying a Door
Standard size for a door is 203.2 cm tall, 91.44 cm wide, and 3.334 cm thick.
Volume = 61947.75 cm^3
Fragmentation values for wood and steel can be found here.
Wood Door Fragmentation = 516644.24 Joules
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
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 kg at 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 tires, we will time these numbers by 4. The total weight of 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 100% but this is as much of the percentage of materials that could be found, so consider this a low-ball or a near complete fragmentation of a car.
Density of Materials
Steel = an average of 7.9 g/cm³
Aluminum = 2.7 g/cm³
Copper = 8.96 g/cm³
Glass = an average of 5 g/cm³
Plastic = and average of 2.235 g/cc (http://www.tregaltd.com/img/density%20of%20plastics[1].pdf)
Natural Rubber = 0.92 g/cm³
Synthetic Rubber = We will use polybutadiene since it is mostly used in car tires. 0.925 g/cm^3
Cast Iron = an average density of 7.3 g/cm³
Volume of Materials
Steel = 113,924.0506 cm³
Aluminum = 66,694.81481 cm³
Copper = 2,480.580357 cm³
Glass = 9,071.84 cm³
Plastic = 81,890.1566 cm³
Natural Rubber = 6,902.486957 cm³
Synthetic Rubber = 13239.9827 cm³
Cast Iron = 18,051.73151 cm³
Energy to Fragment Materials
To find shear strength from tensile strength, just times the ultimate tensile strength by 0.60.
Steel = 208 j/cc
Glass = 3.5 j/cc
Aluminum = 40000 PSI = 275.79 megapascales = 275.79 J/cc
Copper = 25,000 PSI = 172.36893 MPa = 172.36893 J/cc
Plastic = It is insanely difficult for me to find plastic mechanical properties. Polypropylene will be used since it is used for most cars, especially in their bumpers. an average of 38.7 MPa = 38.7 j/cc
Natural Rubber = 0.001 GPa = 1 MPa = 1 J/cc
Synthetic Rubber = 4.285714286 MPa = 4.285714286 J/cc
Total Energy
23,696,202.52 Joules for all the steel
2689707.995 Joules for all the iron
31,751.44 Joules for all the glass
18,393,762.98 Joules for all the aluminum
3169149.06 Joules for all the plastic
427,574.9819 Joules for all the copper
56742.783 joules for Synthetic Rubber
6902.486957 Joules for all the natural rubber
Adding this all up is 48,471,794.25 Joules = 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.
White Oak = 30 m height, 1.27 meter diameter
Plugging this into the formula for volume of a cylinder since tree trunks are cylindrical = 38 m^3
Energy to Destroy Tree
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
The weight of a wrecking ball ranges from 450 kg to 5400 kg and they are made of steel.
Steel = density of 7.9 g/cc
450000/7.9 = 56962.02532 cc
5400000/7.9 = 683544.3038 cc
Energy to Destroy Wrecking Ball
Steel = 208 J/cc
Low-end: 208*6962.02532 = 1.184810127e7 Joules, 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
Plugging in the values of the radius of the shackle with the height gives me 1.78 cc x 2 = 3.56 cc for both sides. But, this doesn't take into account the curved portion. So, to find the volume of that, I'll just use the volume of a torus x 0.5.
Orange = Major radius 30 px or 1.249180328 cm
This gives a volume of 2.36 cc
2.36 + 3.56 = 5.92 cc
Since this is just breaking off the lock, the shackle is not usually fragmented completely, so it would be best to just use 1/4 of the volume = 1.48 cc
Energy to Destroy Shackles
To find shear strength from tensile strength, just times the ultimate tensile strength by 0.60.
Steel = 208 j/cc
Hardened 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
Destroying Blades
Volumes of Blades
A knightly (or short) sword blade is typically 31 3/8 inches long, 2 inches wide, and .192 inches thick A long sword blade is at least 90 cm long 4.14 mm thick [1]
Red = length 90 cm or 964 px at 0.09336099585 cm a pixel
Orange = Width 30.1 px or 2.810165975 cm
Longsword = 104.71 cc
Shortsword = 200.58 cc
Energy to Destroy Blades
Assuming they are made of steel.
Longsword = 2.177968e4 Joules, Wall level
Shortsword = 4.172064e4 Joules, Wall level
Note: This is the fragmentation of an entire blade, but not the hilt.
Destroying a Chimney
Volume of Chimney
I will use this calculator to find the volume of a hollow cuboid. [2]
length = Red 243.84 cm or 239 px at 1.020251046 cm a pixel
Outer Edge B and C = 60.96 cm
thickness = Orange 11.4 px or 11.63086192 cm
inner Edge B and C = 60.96 - (2 x 11.63086192) = 37.69827616 cm
V = 559,603.43 cc
Energy to destroy chimney
Brick is, on average, 3.49375 MPa or 3.49375 J/cc
let's assume 50% is brick while the other half is cement.
279801.715 x 3.49375 = 977557.2418 Joules
279801.715 x 8 = 2238413.72 Joules
3.215970962e6 Joules in total, 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
Energy to Destroy Barrel
Some barrels are destroyed completely or just their wooden parts.
Whole Barrel:
White oak has an average shear strength of 1935 PSI or 13.34136 MPa = 13.34136 J/cc
Steel = 208 x 632.9113924 = 131645.5696 joules
Wood = 13.34136 x 58441.55844 = 779689.8701 J
911335.4397 joules
Wall level
Just the Wood:
Wood = 13.34136 x 58441.55844 = 779689.8701 J
Wall level
Destroying a Skyscraper
Mass of a Skyscraper = Around 222500 tons
http://theconstructor.org/practical-guide/rate-analysis-for-reinforced-concrete/6954/
154 % = 28 % Cement
154 % = 42 % Sand (which 85 % of Sand or 35.7% of the RC)
154 % = 84 % Coarse (Granite is a good assumption)
Cement = 40454.55 Tons = 40454550 kg
Silica = 51579.55 Tons = 51579550 kg
Granite = 121363.64 Tons = 121363640 kg
Cement = 40454550/1250 = 32363.64 m^3
Silica = 51579550/2650 = 19463.9811 m^3
Granite = 121363640/2700 = 44949.4963 m^3
Fragmentation:
Low End: Using Reinforced Concrete Shear Strength:
(32363640000+19463981100+44949496300) cm^3* 28 J/cc 2.7097592872e12 Joules, or 647.648013 Tons, Multi-City Block level+
High End: Using Each Material Shear Strength:
Cement = 6*32363640000 = 194181840000 J
Silica = 70*19463981100 = 1362478677000 J
Granite = 103.42*44949496300 = 4.64867691e12 J
Total Energy = 6.20533743e12 Joules, or 1.48311124 Kilotons, 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% Aluminum (2024-T3) = 148249.862 kg
Titanium Ti-6Al-4V = 4430 kg/m3
Steel = 7850 kg/m3
Aluminum 2024-T3 = 2780 kg/m3
Titanium = 1652591.61 cm3
Steel = 3030979.32 cm3
Aluminum = 53327288.5 cm3
Fragmentation=
Titanium = 550 MPa = 550 J/cc
Steel = 208 J/cc
Aluminum = 40000 PSI = 275.79 megapascales = 275.79 J/cc
Total Fragmentation = 1.6246502e10 Joules, or 3.88300717 Tons = Large Building level
Note: Shooting a plane down does not equal fragmentation. Fragmentation would apply if the plane were 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 diameter 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
wikipedia:Laminated glass#Specifications
A typical laminated makeup is 2.5 mm glass, 0.38 mm interlayer, and 2.5 mm glass. This gives a final product that would be referred to as 5.38 laminated glass.
For the glass:
11572.5575*0.5 = 5786.27875 cm^3
For the plastic layer:
11572.5575*0.038 = 439.757185 cm^3
Fragmentation of glass is 3.5 j/cc.
According to this the plastic is PVB. Its tensile strength is 19.6 MPa. Shear strength is 0.577 of tensile strength. 11.3092 MPa, or 11.3092 j/cc.
Fragmentation of the glass: 5786.27875*3.5 = 20 251.975625 Joules
Fragmentation of the plastic: 439.757185*11.3092 = 4973.301956602 Joules
In total that's 25 225.2775816 Joules, Wall level
Blowing up Cannons
This is about blowing up 16th century cannons.
Density of cast iron is = 7.8 g/cm^3
9100000 g / 7.8 g/cm^3 = 1166666.667 cm^3 of iron
Grey cast iron has a frag energy of 400 J/cc, a violent frag. energy of 613.5 J/cc and a pulverization energy of 827 J/cc.
Frag (400 J/cc): 1166666.667*400 = 466,666,666.8 J or 0.111536 tons of TNT (Small Building level)
V. Frag (613.5 J/cc): 1166666.667*613.5 = 715,750,000.2045 J or 0.17106836 tons of TNT (Small Building level+)
Pulverization (827 J/cc): 1166666.667*827 = 964,833,333.609 J or 0.2306 tons of TNT (Small Building level+)
Destroying Gravestones
Description: Destroying a commonly sized gravestone
Requirements: The gravestone in question needs to be of a regular size. The gravestone needs to be made out of stone or concrete depending on the result used.
Results/Calculation: A common gravestone has a volume of 14158.423 cm^3 (converting from inches). We'll assume concrete since that's a common material and presumably they're all at least relatively similar in durability.
Frag: 6 j/cc * v = 84950.538 Joules, Wall level
V. Frag: 17 * v = 240693.191 Joules, Wall level
Pulv: 40 j/cc * v = 566336.920 Joules, Wall level
Ordinary rock was also commonly used before concrete became mainstream, so...
Frag: 8 J/cc * v = 113267.384 J, Wall level
V. Frag: 69 J/cc * v = 976931.187 J, Wall level
Pulv: 214 J/cc * v = 3029902.552 J, Wall level
Destroying Snow-tipped Mountains
Description: We calculate the energy necessary to destroy a snow-tipped mountain in various climates and by various methods.
Requirements: The mountain needs to be destroyed in the fashion the end specifies. To qualify for cold climate ends, the mountain can be in a cold but not in a snow-covered region. To qualify for the mild climate, it should be in a mild, not too cold, climate like the central European one. To qualify for the tropical, the mountain needs to be in a region of hot tropical climate like commonly found near the equator. Note that the height of the mountains should be taken as the measurement from sea or, at most, ground level.
Results:
Cold Climate:
- Fragmentation: 1.4476458947741767e16J (Low 7-B+)
- Violent Fragmentation: 1.2485945842427274e17J (7-B)
- Pulverization: 3.8787862193105597e17J (7-B+)
- Vaporization: 4.6505624369620427e19J (6-C)
Mild Climate:
- Fragmentation: 1.3089969389957472e17J (7-B)
- Violent Fragmentation: 1.129009859883832e18J (7-A)
- Pulverization: 3.5072936734217302e18J (7-A+)
- Vaporization: 4.2051526665238379e20J (High 6-C)
Tropical Climate:
- Fragmentation: 8.1544016674617632e17J (7-A)
- Violent Fragmentation: 7.0331714381857708e18J (High 7-A)
- Pulverization: 2.1848699967755362e19J (6-C)
- Vaporization: 2.6196015356720914e21J (High 6-C+)
Calculation: We will calc the power necessary to bust snow-tipped mountains. Now, a mountain being snow-tipped means nothing if its winter obviously, but the permanent snow line, i.e., the height which a mountain needs to have to always have snow on the tip, can be used.
All that one needs to know for that is the approximate climate for the region. Then one can simply look up the snow line altitude for a comparable climate on Earth and use that.
I will do three ends.
The first is a cold climate. With cold climate I mean regions that are cold, but not arctic. One obviously can't use this method if the place is so cold that it could be snowing year-round. For this end I will use the snowline of Southern Scandinavia at least 1200m height.
Second, I will do a mild climate. For that I will use the northern side of the alps, with the snow line being at least 2500m high.
Third I will do tropical climate for places that are really hot. For that I will use the at least 4600m snowline of the New Guinea Highlands.
Now, we still need to get a width for the mountain. For angle of repose reasons I think most long-standing big mountains won't have an average angle of more than 45°. Hence, I will assume that the radius of the mountain is equal to its height.
With that in mind we can get the volumes.
Cold Climate: 1/3*pi*1200^2*1200 = 1.8095573684677209054e9 m^3 = 1.8095573684677209e15 cm^3
Mild Climate: 1/3*pi*2500^2*2500 = 1.636246173744683978e10 m^3 = 1.636246173744684e16 cm^3
Tropical Climate: 1/3*pi*4600^2*4600 = 1.0193002084327203822e11 m^3 = 1.0193002084327204e17 cm^3
With that we can throw the usual destruction values on it for results:
Cold Climate:
-Fragmentation: 1.8095573684677209e15 * 8 = 1.4476458947741767e16J (Low 7-B+)
-Violent Fragmentation: 1.8095573684677209e15 * 69 = 1.2485945842427274e17J (7-B)
-Pulverization: 1.8095573684677209e15 * 214.35 = 3.8787862193105597e17J (7-B+)
-Vaporization: 1.8095573684677209e15 * 25700 = 4.6505624369620427e19J (6-C)
Mild Climate:
-Fragmentation: 1.636246173744684e16 * 8 = 1.3089969389957472e17J (7-B)
-Violent Fragmentation: 1.636246173744684e16 * 69 = 1.129009859883832e18J (7-A)
-Pulverization: 1.636246173744684e16 * 214.35 = 3.5072936734217302e18J (7-A+)
-Vaporization: 1.636246173744684e16 * 25700 = 4.2051526665238379e20J (High 6-C)
Tropical Climate:
-Fragmentation: 1.0193002084327204e17 * 8 = 8.1544016674617632e17J (7-A)
-Violent Fragmentation: 1.0193002084327204e17 * 69 = 7.0331714381857708e18J (High 7-A)
-Pulverization: 1.0193002084327204e17 * 214.35 = 2.1848699967755362e19J (6-C)
-Vaporization: 1.0193002084327204e17 * 25700 = 2.6196015356720914e21J (High 6-C+)
Star 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)
Destroying the Sun from Earth
Description: The sun gets destroyed by an omnidirectional explosion starting on Earth,
Requirements: The explosion has to be omnidirectional, and its epicenter needs to be on planet Earth. The explosion has to completely destroy the Sun.
Results: 1.0541206e+47 Joules or Solar System level
Calculation: Using our Earth and our Sun
- Radius of the sun: 695500km
- Frontal area of the sun: 1.5188816e+18m^2
- GBE of the Sun: 5.693e41 J
Distance between sun and Earth: 149600000km
Area of the explosion = 4pi(149600000000)^2 = 2.8123738e+23 m^2
- E = 2.8123738e+23/1.5188816e+18*5.693e41
- E = 1.0541206e+47 Joules or (Solar System level)
Creating or destroying a pocket realm
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 starry sky
- Average star distance that human can see in starry night: (4 to 4000 light years)/2 = 2002 light years = 1.894e19 meters
- 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)
- It yields: 1.688e63 joules (Multi-Solar System level)
Creating a pocket dimension containing a moon
Description: Creating a pocket dimension which contains an earthlike planet and a moon.
Requirements: See Creation Feats
Results: 2.8373406e+34 Joules (5-A)
Calculation: See here
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. So get over here and familiarize yourself about the criteria before applying the below table.
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 | Large Island |
Arrow | 0.018 | 1.61773E+15 | Large 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 | Large Mountain |
All grown man on Earth | 3.85E+11 | 3.46015E+28 | Multi-Continent |
Theoretical mass of all life forms on Earth | 1.01835E+13 | 9.15232E+29 | Moon+ |
Theoretical mass of all life forms in our universe | 3.05505E+35 | 2.7457E+52 | Solar System |
Private car | 1311.363636 | 1.17858E+20 | Island |
M1A2 SEPv2 Abrams | 64600 | 5.80586E+21 | Small Country |
Our Moon | 7.342E+22 | 6.59855E+39 | Dwarf Star |
Our Earth | 5.97237E+24 | 5.36761E+41 | Small 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 | Average human |
0.724105801 | 1.730062379 | 0.005043914 | 100 | 2.39006E-08 | Athletic human |
0.75 | 1.791929829 | 0.005224285 | 107.279928 | 2.56405E-08 | Athletic 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/Street |
1.5 | 3.583859657 | 0.01044857 | 429.119712 | 1.02562E-07 | Peak human/Street |
2 | 4.778479543 | 0.013931427 | 762.8794879 | 1.82333E-07 | Peak human/Street |
2.092715875 | 5 | 0.014577259 | 835.25 | 1.9963E-07 | Peak human/Street |
3.222782448 | 7.7 | 0.02244898 | 1980.8789 | 4.73441E-07 | Peak human/Street |
4.101723116 | 9.8 | 0.028571429 | 3208.6964 | 7.66897E-07 | Peak human/Street |
5.23597512 | 12.51 | 0.036472303 | 5228.668341 | 1.24968E-06 | Peak human/Street |
6 | 14.33543863 | 0.041794282 | 6865.915391 | 1.64099E-06 | Peak human/Street |
6.333339138 | 15.13186576 | 0.044116227 | 7650 | 1.82839E-06 | Peak human+/Street+ |
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 | Small building |
358.9007726 | 857.5 | 2.5 | 24566581.81 | 0.005871554 | Small building |
500 | 1194.619886 | 3.48285681 | 47679968 | 0.011395786 | Small building |
717.8015452 | 1715 | 5 | 98266327.25 | 0.023486216 | Small building |
1000 | 2389.239772 | 6.96571362 | 190719872 | 0.045583143 | Small building |
1435.60309 | 3430 | 10 | 393065309 | 0.093944864 | Small building |
1672.449284 | 3995.882346 | 11.64980276 | 533460000 | 0.1275 | Small building+ |
2341.897425 | 5595.354468 | 16.31298679 | 1046000000 | 0.25 | Building |
3589.007726 | 8575 | 25 | 2456658181 | 0.587155397 | Building |
4967.914649 | 11869.53926 | 34.60507073 | 4707000000 | 1.125 | Building+ |
5000 | 11946.19886 | 34.8285681 | 4767996800 | 1.139578585 | Building+ |
6623.886199 | 15826.05235 | 46.14009431 | 8368000000 | 2 | Large building |
7178.015452 | 17150 | 50 | 9826632725 | 2.348621588 | Large building |
9264.4532 | 22135.00005 | 64.53352784 | 16369504368 | 3.912405442 | Large building |
10000 | 23892.39772 | 69.6571362 | 19071987198 | 4.55831434 | Large building |
11941.38067 | 28530.82162 | 83.18023795 | 27196000000 | 6.5 | Large building+ |
14356.0309 | 34300 | 100 | 39306530900 | 9.394486353 | Large building+ |
14811.45982 | 35388.12887 | 103.1723874 | 41840000000 | 11 | City block |
34893.48575 | 83368.90391 | 243.0580289 | 2.32212E+11 | 55.5 | City block+ |
46837.94849 | 111907.0894 | 326.2597357 | 4.184E+11 | 100 | Multi City Block |
50000 | 119461.9886 | 348.285681 | 4.768E+11 | 113.9578585 | Multi City Block |
100000 | 238923.9772 | 696.571362 | 1.9072E+12 | 455.831434 | Multi City Block |
109844.7259 | 262445.3878 | 765.1469031 | 2.3012E+12 | 550 | Multi City Block+ |
143560.309 | 343000 | 1000 | 3.93065E+12 | 939.4486353 | Multi City Block+ |
148114.5982 | 353881.2887 | 1031.723874 | 4.184E+12 | 1000 | Small town |
273109.8245 | 652524.8547 | 1902.404824 | 1.42256E+13 | 3400 | Small town+ |
356707.1885 | 852259.0015 | 2484.720121 | 2.42672E+13 | 5800 | Town |
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 | Large town |
3473595.227 | 8299251.868 | 24196.06958 | 2.3012E+15 | 550000 | Large town+ |
4683794.849 | 11190708.94 | 32625.97357 | 4.184E+15 | 1000000 | Small city |
6371000 | 15221846.58 | 44378.56147 | 7.74125E+15 | 1850203.426 | Small city |
The table below lists out the energy 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 | Large town |
3473595.227 | 8299251.868 | 24196.06958 | 2.30252E+15 | 550316.3282 | Large town+ |
4683794.849 | 11190708.94 | 32625.97357 | 4.18838E+15 | 1001046.261 | Small city |
6371000 | 15221846.58 | 44378.56147 | 7.75625E+15 | 1853788.582 | Small 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
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.
cc refers to cubic centimeters.
Rock has a frag energy of 8 J/cc, a violent frag energy of 69 J/cc and a pulverization energy of 214.35 J/cc.
Steel has a frag energy of 208 J/cc, a violent frag energy of 568.5 J/cc and a pulverization energy of 1000 J/cc.
Concrete has a frag energy of 6 J/cc, a violent frag energy of 17-20 J/cc and a pulverization energy of 40 J/cc.
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
Pulverization:
5.11e5*214.35 = 1.0953285e+8 joules or 0.02618 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
Pulverization:
5.11e5*1000 = 5.11e+8 joules, or 0.1221319 tons of TNT, Small Building level
If the ground is made out of concrete:
Fragmentation:
5.11e5*6 = 3066000 joules or 0.0007328 tons of TNT, Wall level
Violent Fragmentation:
5.11e5*17-20 = 8.687e+6-1.022e+7 joules or 0.002076-0.00244264 tons of TNT, Wall level
Pulverization:
5.11e5*40 = 2.044e+7 joules or 0.0048853 tons of TNT, Wall 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
General Calc: KE and Humans
Description: Kinetic Energy humans get from moving at a certain speed.
Requirements: The character of human-like weight need to move at the specified speed and fulfill the Kinetic Energy Feats standards. This means that usually they must demonstrate performing a bodycheck or similar at specifically that speed. (Performing a bodycheck or going at that speed separately deosn't suffice)
Results/Calculation: Average Human = 62 kg
With that outta the way, let's go.
Peak Human: 9.8 m/s Energy: 2977.24 Joules, Street level
Superhuman: 12.51 m/s Energy: 4851.5 Joules, Street level
Subsonic: 34.3 m/s Energy: 36471.19 Joules, Wall level
Subsonic+: 171.5 m/s Energy: 9.118e5 Joules, Wall level
Transonic: 308.7 m/s Energy: 2.954e6 Joules, Wall level
Supersonic: 377.3 m/s Energy: 4.413e6 Joules, Wall level
Supersonic+: 857.5 m/s Energy: 2.279e7 Joules, Small Building level
Hypersonic: 1715 m/s Energy: 9.118e7 Joules, Small Building level
Hypersonic+: 3430 m/s Energy: 3.647e8 Joules, Small Building level
High Hypersonic: 8575 m/s Energy: 2.279e9 Joules, Building level
High Hypersonic+: 17150 m/s Energy: 9.118e9 Joules, Large Building level
Massively Hypersonic: 34300 m/s Energy: 3.647e10 Joules, Large Building level+
Massively Hypersonic+: 343000 m/s Energy: 3.647e12 Joules, Multi-City Block level
Sub-Relativistic: 0.01 c Energy: 2.786e14 Joules, Town level+
Sub-Relativistic+: 0.05 c Energy: 6.978e15 Joules, Small City level
Relativistic: 0.1 c Energy: 2.807e16 Joules, City level
Relativistic+: 0.5 c Energy: 8.620e17 Joules, Mountain level
Peak Speed for Kinetic Energy: 0.92 c Energy: 8.646e18 Joules, Large Mountain level
That is all. Reminder that Sub-Rel and above requires usage of Relativistic Kinetic Energy.
Speed Feats
Changing Clothes Fast
Description: Characters taking off their clothes in one or two seconds.
Requirements: This should only be used if the clothes are taken off in a realistic fashion, i.e. not ripped off or taken off in a minor show of toon force.
Results:
- 1 second: 8.4216 m/s
- 2 seconds: 4.21 m/s
Calculation:
We will assume our person to be roughly 6 feet.
Shirt
Assuming they remove this from top to bottom and in the fastest manor. We start with the shirt. Which we remove it by pulling it over our heads. Meaning our arm is going as far up as it can go likely. For a 6-foot person. Their arm can about land between 1.5 to 2 feet from elbow to fingertip. To make it easier. We will assume the better end of 2 feet. Or 24 inches
So, the actions you have to take off the shirt is to pull it over our heads and it's off. This means the arms should travel roughly 24 inches when lifting to take it off. And 24 more inches to get back its normal place. So roughly 48 inches. Maybe a bit more or less depending on the type and how someone may change their shirt. But this is a decent average
Shirt = 48 inches
Pants/Zipper/Shoes/Socks
So first. We have to assume you unzip them as most pants come with a zipper. The average zipper is about as long as your longest finger. And may be a bit longer or shorter. But the zipper should at least hit 4 inches. We save this for later
Now the action of taking off pants. Average length for a leg hits 31.8898 inches. So, your arm and body would have to travel the entire length of your leg to take them off. Now this is where shoes and socks come in. Assuming you take them off here rather than going all the way back up just to go back down. To take off your shoes. You'll have to go the length of your own foot to pull them off assuming they aren't easy slip off shoes. A foot on average is 12 inches. And you have 2 feet. And do this for socks as well. So, 12 times 2 times 2. 48 inches for your shoes and socks. Then you go back up the distance of your legs. But that's under Pants. Since you travel 31.8898 inches twice. That's 63.7796 inches. And then we add 4 to that for the zipper
Pants = 67.7796 inches
Shoes = 48 inches
Socks = 48 inches
67.7796 + 48 + 48
That's a grand total of 163.7796 Inches.
But wait. Since this also includes putting on the clothes too. We double it to 327.5592 inches. And since we would likely have to both unzip and result when putting on pants. We add 4 more to consider the extra zipping process that putting them on would provide. And a grand total of 331.5592 Inches. Or 27.629933333 feet.
Now for the timeframe.
Timeframe
1 second = 18.8385909088636 MPH or 8.421603679898383987 MPS. Athletic Human
2 second = 9.419295454431818 MPH or 4.2108018399491999872 MPS. Below Average Human
Running around while things are standing still
Description: Speed for characters running around while various things are standing still.
Requirements: For lightning the Lightning Feats standards need to be fulfilled. For light the Laser/Light Beam Dodging Feats standards need to be met. Additionally, the thing in question must be reliably stated to stand completely still out of the perspective of the moving character
Results/Calculation: For reference, this uses the speed of a snail (0.013 m/s) and the average human running speed (6.35 m/s). The equation is as follows:
Person's Speed = (Object's Speed / Object's Apparent Speed) * Person's Apparent Speed
Viewing Lightning as Standing Still
Speed of Lightning = 440000 m/s
Speed to View Lightning as Standing Still = (440000 / 0.013) * 6.35 = 214923076.92 m/s; 0.72c (Relativistic+)
Viewing Light as Standing Still
Speed of Light = 299792458 m/s
Speed to View Light as Standing Still = (299792458 / 0.013) * 6.35 = 146437085254 m/s; 488.46c (Massively FTL)
Traveling Certain Distances
Description: The speed required to travel certain distances in certain timeframes.
Requirements: It must be proven that the character traveled the given distance in the specified timeframe.
Results/Calculation:
Traverse the distance between the Moon and the Earth
Distance between the Earth and the Moon: 384,400 km
- 1 second: 384399.23 km/s = 1.28c [FTL]
- 5 seconds: 76879.85 km/s = 0.25c [Relativistic]
- 10 seconds: 38439.92 km/s = 0.12c [Relativistic]
- 1 minute: 6406.65 km/s = 0.02137028411 [Sub-Relativistic]
- 10 minutes: 640665.39 m/s = Mach 1867 [MHS+]
- 1 hour: 106777.56 m/s = Mach 311 [MHS]
Traveling an Astronomical Unit
One Astronomical Unit is equal to the average distance between the Earth and the Sun: 149597870700 m = 149597870.7 km
- 1 second: 149597571.50 km/s = 499c [MFTL]
- 10 seconds: 14959757.15 km/s = 49.9c [FTL+]
- 1 minute: 2493292.86 km/s = 8.3c [FTL]
- 10 minutes: 2493292.86 km/s = 8.3c [FTL]
- 10 minutes: 249329.29 km/s = 0.83c [Relativistic+]
- 1 hour: 41554.88 km/s = 0.13c [Relativistic]
Travel 1 Light Year
1 Light Year: 9460730472580.8 km
- 1 second: 9460711551119.85 km/s = 31,557,536c [MFTL+]
- 10 seconds: 946071155111.99 km/s = 3,155,753c [MFTL+]
- 1 minute: 15767858525852 km/s = 525,958c [MFTL+]
- 10 minutes: 1576785258585,20 km/s = 52,595c [MFTL+]
- 1 hour: 2627975430.87 km/s = 8765c [MFTL+]
Traveling an interstellar distance
Distance from the Sun to Alpha Centauri: 4.37 Light Years = 41343392165178,096 km
- 1 second: 41343309478393.76 km/s = 137,906,436c [MFTL+]
- 10 seconds: 4134330947839.38 km/s = 13,790,643c [MFTL+]
- 1 minute: 68909055157973.23 km/s = 2,298,440c [MFTL+]
- 10 minutes: 68905515797.32 km/s = 229,844c [MFTL+]
- 1 hour: 11484252632.89 km/s = 38,307c [MFTL+]
Traveling an Intergalactic distance
Distance from Milky Way to Andromeda: 2,536,802 Light Years = 2400000000000000000000000000 km
- 1 second: 23999952000000004096 km/s = 80,055,222,736,790c [MFTL+]
- 10 seconds: 23999952000000000000000 km/s = 8,005,522,273,679 [MFTL+]
- 1 minute: 39999999200000000000 km/s = 1,334,253,712,279 [MFTL+]
- 10 minutes: 39999999200000000000000 km/s = 133,425,371,227 [MFTL+]
- 1 hour: 66666533333333333334 km/s = 22,237,561,871 [MFTL+]
Galaxy Speed Feats
Description: Traveling various distance involving galaxies in various times.
Requirements: The character needs to be proven to move the given distance in the given time.
Results/Calculation:
Milky Way
Distance from Earth to the edge of Milky Way = 25000 light years
1 second
Speed = Distance / Time
Speed = 25000 light years / 1 second
Speed = 788953737500c or 788 billion c (Massively FTL+)
1 minute
Speed = 25000 light years / 60 s
Speed = 13149228958c or 13 billion c (Massively FTL+)
1 hour
Speed = 25000 light years / 3600 s
Speed = 219153815c or 219 million c (Massively FTL+)
1 day
Speed = 25000 light years / 86400 s
Speed = 9131408c or 9 million c (Massively FTL+)
1 week
Speed = 25000 LY / 604800 s
Speed = 1304486c or 1.3 million c (Massively FTL+)
1 month
Speed = 25000 LY / 2,628e+6 s
Speed = 30021c (Massively FTL+)
1 year
Speed = 2501c (Massively FTL+)
From Earth to nearest Galaxy
Distance from Earth to Andromeda = 2.537.000 light years
1 second
Speed = Distance / Time
Speed = 2537000 LY / 1 s
Speed = 80063025281529c or 80 trillion c (Massively FTL+)
1 minute
Speed = 2537000 LY / 60 s
Speed = 1334383754692c or 1.3 trillion c (Massively FTL+)
1 hour
Speed = 2537000 LY / 3600 s
Speed = 22239729244c or 22 billion c (Massively FTL+)
1 day
Speed = 2537000 LY / 86400 s
Speed = 926655385c or 926 million c (Massively FTL+)
1 week
Speed = 2537000 LY / 604800 s
Speed = 132379340c or 132 million c (Massively FTL+)
1 month
Speed = 2537000 LY / 2,628e+6 s
Speed = 3046538c or 3 million c (Massively FTL+)
1 year
Speed = 253878.18773c (Massively FTL+)
From Earth to farthest known Galaxy
Distance from Earth to Abel 2218 = 13 billion light years
1 second
Speed = Distance / Time
Speed = 13 billion LY / 1 s
Speed = 410255943500152960c or 410 quadrillion c (Massively FTL+)
1 minute
Speed = 13 billion LY / 60 s
Speed = 6837599058335884c or 6 quadrillion c (Massively FTL+)
1 hour
Speed = 13 billion LY / 3600 s
Speed = 113959984305598.05c or 113 trillion c (Massively FTL+)
1 day
Speed = 13 billion LY / 86400 s
Speed = 4748332679399.92c or 4 trillion c (Massively FTL+)
1 week
Speed = 13 billion LY / 604800 s
Speed = 678333239914.2742c or 678 billion c (Massively FTL+)
1 month
Speed = 13 billion LY / 2,628e+6 s
Speed = 15610956754.19152c or 15 billion c (Massively FTL+)
1 year
Speed = 1300913062.84929c or 1.3 billion c (Massively FTL+)
Escape Velocity
Description: Escape Velocities for various celestial objects.
Requirements: The object must be launched from the surface of the celestial object such that it never returns to it on its own (escapes far into space). Does not apply to flying objects.
Results:
Escape velocity of Earth
Using the Escape Velocity equation = V(escape) = sqrt[2*G*Mass/radius]
- Mass of Earth = 5.972e24kg
- Radius of Earth = 6371km, or 6371000m
- G = 6.67408e−11 m^3/(kg*(s^2)
- V(escape) = sqrt[2*[6.67408e−11]*[5.972e24 kg]/[6371000m]] = 11185.8m/s, or ~11.2km/s
Escape velocity of a planet 10 times the size of Earth but with the same density
- Radius of Planet = [6371*10]km, or 63710000m
- Volume of Planet = [4/3]*[Pi]*[[Radius = 63710000]^3] = 1.0832e24m^3
- Density of Planet = 5510kg/m^3
- Mass = Volume x Density = [1.0832e24m^3]*[5510kg/m^3] = 5.9685e27kg
- G = 6.67408e−11 m^3/(kg*(s^2)** V(escape) = sqrt[2*[6.67408e−11]*[5.9685e27kg]/[63710000m]] = 111825m/s, or ~111.2km/s
What you need to know here is that the surface escape velocity increases in direct proportion to said planet's diameter.
- Since the size of the planet increased by a factor of 10, the surface escape velocity of the planet would increase by a factor of 10.
Escape velocity of a planet 10 times denser than Earth but with the same volume
- Radius of Planet = [6371]km, or 6371000m
- Volume of Planet = [4/3]*[Pi]*[[Radius = 6371000]^3] = 1.0832e21m^3
- Density of Planet = 55100kg/m^3
- Mass = Volume x Density = [1.0832e21m^3]*[55100kg/m^3] = 5.96843e25kg
- G = 6.67408e−11 m^3/(kg*(s^2)
- V(escape) = sqrt[2*[6.67408e−11]*[5.96843e25kg]/[6371000m]] = 35362m/s, or ~35.362km/s
What you need to know here is that the surface escape velocity increases at the sqrt [density increase] of said planet.
- Since the density of the planet increased by a factor of 10, the surface escape velocity of the planet would increase by a factor of sqrt[10].
List of escape velocities
In this table, the left-hand half gives the escape velocity from the visible surface (which may be gaseous as with Jupiter for example), relative to the center of the planet or moon (that is, not relative to its moving surface). In the right-hand half, V_{e} refers to the speed relative to the central body (for example the sun), whereas V_{te} is the speed (at the visible surface of the smaller body) relative to the smaller body (planet or moon).
Location | Relative to | V_{e} (km/s) | Location | Relative to | V_{e} (km/s) | System escape, V_{te} (km/s) | |
---|---|---|---|---|---|---|---|
On the Sun | The Sun's gravity | 617.5 | |||||
On Mercury | Mercury's gravity | 4.25 | At Mercury | The Sun's gravity | ~ 67.7 | ~ 20.3 | |
On Venus | Venus's gravity | 10.36 | At Venus | The Sun's gravity | 49.5 | 17.8 | |
On Earth | Earth's gravity | 11.186 | At Earth | The Sun's gravity | 42.1 | 16.6 | |
On the Moon | The Moon's gravity | 2.38 | At the Moon | The Earth's gravity | 1.4 | 2.42 | |
On Mars | Mars' gravity | 5.03 | At Mars | The Sun's gravity | 34.1 | 11.2 | |
On Ceres | Ceres's gravity | 0.51 | At Ceres | The Sun's gravity | 25.3 | 7.4 | |
On Jupiter | Jupiter's gravity | 60.20 | At Jupiter | The Sun's gravity | 18.5 | 60.4 | |
On Io | Io's gravity | 2.558 | At Io | Jupiter's gravity | 24.5 | 7.6 | |
On Europa | Europa's gravity | 2.025 | At Europa | Jupiter's gravity | 19.4 | 6.0 | |
On Ganymede | Ganymede's gravity | 2.741 | At Ganymede | Jupiter's gravity | 15.4 | 5.3 | |
On Callisto | Callisto's gravity | 2.440 | At Callisto | Jupiter's gravity | 11.6 | 4.2 | |
On Saturn | Saturn's gravity | 36.09 | At Saturn | The Sun's gravity | 13.6 | 36.3 | |
On Titan | Titan's gravity | 2.639 | At Titan | Saturn's gravity | 7.8 | 3.5 | |
On Uranus | Uranus' gravity | 21.38 | At Uranus | The Sun's gravity | 9.6 | 21.5 | |
On Neptune | Neptune's gravity | 23.56 | At Neptune | The Sun's gravity | 7.7 | 23.7 | |
On Triton | Triton's gravity | 1.455 | At Triton | Neptune's gravity | 6.2 | 2.33 | |
On Pluto | Pluto's gravity | 1.23 | At Pluto | The Sun's gravity | ~ 6.6 | ~ 2.3 | |
At Solar System galactic radius | The Milky Way's gravity | 492–594 | |||||
On the event horizon | A black hole's gravity | 299,792.458 (speed of light) |
The last two columns will depend precisely where in orbit escape velocity is reached, as the orbits are not exactly circular (particularly Mercury and Pluto).
Speed Needed to Light Your Clothes on Fire
Description: How fast one needs to run to set their clothes on fire.
Requirements: The character has to wear regular clothes and they must start burning.
Results: 2,500 km/h (Supersonic)
Calculation: Taken from this source...
That’s the temperature to completely incinerate your entire body – your clothes will catch fire long before you reach that point. Nylon has an ignition point of about 500°C and wool will catch fire at 230°C. Which means that with the right attire, you could trot along at a leisurely 2,500 km/h and still burst into flames. 2,500 km/h (Supersonic)
Speed Needed to Break the Sound Barrier
Description: The speed needed to break the sound barrier.
Requirements: The 343 m/s value applies to regular small objects. The 1000 km/h value is for the speed of the airflow around aircrafts.
Results/Calculation: In dry air at 20 °C (68 °F), the speed of sound is 343 meters per second (about 767 mph, 1234 km/h or 1,125 ft/s)
From page 13 of the "Me 262 A-1 Pilot's Handbook" issued by Headquarters Air Materiel Command, Wright Field, Dayton, Ohio as Report No. F-SU-1111-ND on January 10, 1946:
Speed Needed to Cause a Sonic Boom audible from the ground
Description: Speed needed for an aircraft to cause a sonic boom that can be heard from the ground.
Requirements: The object in question has to be flying at least 9100m above the ground and cause a sonic boom heard on the ground.
Results: Mach 1.12
Calculation: The speed of sound, a critical speed known as Mach 1, is approximately 1,235 km/h (767 mph) at sea level and 20 °C (68 °F).
Depending on the aircraft's altitude, sonic booms reach the ground 2 to 60 seconds after flyover. However, not all booms are heard at ground level. The speed of sound at any altitude is a function of air temperature. A decrease or increase in temperature results in a corresponding decrease or increase in sound speed. Under standard atmospheric conditions, air temperature decreases with increased altitude. For example, when sea-level temperature is 59 degrees Fahrenheit (15 °C), the temperature at 30,000 feet (9,100 m) drops to minus 49 degrees Fahrenheit (−45 °C). This temperature gradient helps bend the sound waves upward. Therefore, for a boom to reach the ground, the aircraft speed relative to the ground must be greater than the speed of sound at the ground. For example, the speed of sound at 30,000 feet (9,100 m) is about 670 miles per hour (1,080 km/h), but an aircraft must travel at least 750 miles per hour (1,210 km/h) (Mach 1.12) for a boom to be heard on the ground.