References for Common Feats

## **Introduction**

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

## **Impact Feats**

Feats that pertain to events where either objects have an impact on characters or characters have an impact on objects.

### **Getting hit by 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 38646.0699 kg 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:

 Pickup Truck
 Bus
 Semi-Truck

**Results:** The values for getting slammed into a wall / not moving are:

 Pickup Truck
 Bus
 Semi-Truck

The subway train ends are:

* **8 cars & average speed:** 9351396.1820257 Joules, **Wall level**
* **10 cards & average speed:** 11689245.235094 Joules, **Wall level+**
* **8 cars & top speed:** 94480113.721716 Joules/0.02258128 Tons of TNT, **Small Building level**
* **10 cards & top speed:** 118100142.22854 Joules/0.0282266 Tons of TNT, **Small Building level**

#### **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

{\displaystyle m\times 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:

**Final Speed** =

(

M
a
s
s
C
a
r
×
I
n
i
t
i
a
l
S
p
e
e
d

M
a
s
s
P
e
r
s
o
n
M
a
s
s
c
a
r

)

{\displaystyle \left ( \frac{MassCar\times Initial Speed}{MassPerson + Masscar} \right )}

* 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
	=
	14
	,
	975.56957

	{\displaystyle \left ( \frac{1500 \times 11.176}{70 + 1500} \right )^ 2 \times 70 \times 0.5 = 14,975.56957}

	 = 3990.47 Joules, **Street level**
* **45 mph or 20.1168 m/s** (Daily City travel speed)**:** ((1500\*20.1168)/(70+1500))^2\*70\*0.5 = 12929.12 Joules, **Street level+**
* **60 mph or 26.8224 m/s** (Traditional interstate travel speed)**:** ((1500\*26.8224)/(70+1500))^2\*70\*0.5 = 22985.1069 Joules, **Wall level**
* **70 mph or 31.2928 m/s** (Highway speed limit)**:** ((1500\*31.2928 m/s)/(70+1500))^2\*70\*0.5 = 31285.284 J, **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, 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, 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, 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, - 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 Joules, **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 Joules, **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 Joules, **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 Joules, **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**, **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**, **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**, **Wall level**
* **70 mph or 31.2928 m/s (Highway speed limit)** = ((36287\*31.2928)/(70+36287))^2\*70\*0.5 = **34141.5270794 Joules**, **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 = **93677.232 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 = **303514.23168 Joules**, **Wall level**
* **60 mph or 26.8224 m/s (Traditional interstate travel speed)** = 0.5(1500) \* 26.8224^2 = **539580.85632 Joules**, - **Wall level**
* **70 mph or 31.2928 m/s (Highway speed limit)** = 0.5(1500) \* 31.2928^2 = **734429.49888 Joules**, **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 = **254945.709462 Joules**, **Wall level**
* **45 mph or 20.1168 m/s (Daily City travel speed)** = 0.5(4,082.3) \* 20.1168^2 = **826024.098658 Joules**, **Wall level**
* **60 mph or 26.8224 m/s (Traditional interstate travel speed)** = 0.5(4,082.3) \* 26.8224^2 = **1468487.2865 Joules**, **Wall level**
* **70 mph or 31.2928 m/s (Highway speed limit)** = 0.5(4,082.3) \* 31.2928^2 = **1998774.362185216 Joules**, **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 = **665696.702668 Joules**, **Wall level**
* **45 mph or 20.1168 m/s (Daily City travel speed)** = 0.5(10,659.421) \* 20.1168^2 = **2156857.31665 Joules**, **Wall level**
* **60 mph or 26.8224 m/s (Traditional interstate travel speed)** = 0.5(10,659.421) \* 26.8224^2 = **3834413.00737 Joules**, - **Wall level**
* **70 mph or 31.2928 m/s (Highway speed limit)** = 0.5(10,659.421) \* 31.2928^2 = **5219062.14892063232 Joules**, **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 = **2266177.145056 Joules**, **Wall level**
* **45 mph or 20.1168 m/s (Daily City travel speed)** = 0.5(36,287) \* 20.1168^2 = **7342413.94998144 Joules**, **Wall level**
* **60 mph or 26.8224 m/s (Traditional interstate travel speed)** = 0.5(36,287) \* 26.8224^2 = **13055127.03695416 Joules**, **Wall level+**
* **70 mph or 31.2928 m/s (Highway speed limit)** = 0.5(36,287) \* 31.2928^2 = **17766828.81723904 Joules**, **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 (85,200 lb) 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**, **Wall level**

**10 cars:**

38646.0699 x 10 = 386460.699 kg

**11689245.235094 Joules**, **Wall level**

Top speed

**8 cars:**

38646.0699 x 8 = 309168.559 kg

**94480113.721716 Joules**/**0.02258 Tons of TNT**, **Small Building level**

**10 cars:**

38646.0699 x 10 = 386460.699 kg

**118100142.22854 Joules**/**0.0282266 Tons of TNT**, **Small Building level**

### **Stopping Cars**

It is a relatively common task to stop an automobile, so it is reasonable to consider calculating the necessary force to do so. In order to present this calculation in an organized and clear manner, it will be necessary to utilize tables. To begin, we will gather data on the five most commonly encountered types of automobiles:

* cars, SUVs, trucks, buses, and semis (with trailers)

All calculations will be based on the curb weight of the vehicle and the presence of a single driver, with an average weight of 62 kg assumed for the driver.

The following table presents the collected data:

* Car (2016 Toyota Corolla): 2800-2875 lbs or (1270.0576 to 1304.077 kg) (curb weight) / 1332.0576 to 1366.077 kg (with one driver)
* Car (2016 Toyota Camry): 3240 to 3480 lbs (1469.6381 to 1578.5002 kg) (curb weight) / 1531.6381 to 1640.5002 kg (with one driver)
* SUV (Toyota Highlander): 4365 to 4465 lbs (1979.9291 to 2025.2883 kg) (curb weight) / 2041.9291 to 2087.2883 kg (with one driver)
* Pickup Truck (2016 Ram 1500): 4516 to 5663 lbs (2048.4215 to 2568.6915 kg) (curb weight) / 2110.4215 to 2630.6915 kg (with one driver)
* School Bus (Type C; standard school bus): 14910 to 21000 lbs (6763.0567 to 9525.432 kg) (curb weight) / 6825.0567 to 9587.432 kg (with one driver)
* City/Transit Bus (http://onlinepubs.trb.org/onlinepubs/tcrp/docs/TCRPJ-11Task20-FR.pdf ): 20000 to 33000 lbs (9071.84 to 14968.536 kg) (curb weight) / 9133.84 to 15030.536 kg (with one driver)
* Semi (with trailer) : Around 35000 lbs (15875.72 kg) (curb weight) / 15937.72 kg (with one driver) / 80000 lbs (36287.36 kg) (weight limit)

A sudden stop would best be calced using the standard KE formula for mass: KE= 1/2mv²

In order to provide a more comprehensive analysis, I will be comparing data from two countries:

* the United States
* and Japan

Further details will be presented in the following sections:

#### **United States**

The United States provides an ideal starting point for this analysis. While state laws can vary significantly, it is common for fictional works to use Springfield from The Simpsons as a reference point. The creator of The Simpsons has confirmed that this Springfield is located in Oregon, so we will use Oregon's speed limits as a reference. The following table presents these speed limits:

* Freeways: 65-70 mph (29.0576 to 31.2928 m/s)
* Undivided and Divided Rural Roads: 55-70 mph (24.5872 to 31.2928 m/s)
* Freeways (urban): 50-60 mph (22.532 to 26.8224 m/s)
* Residential Roads: 20-25 mph (8.9408 to 11.176 m/s)

**Reference:** https://en.wikipedia.org/wiki/Speed\_limits\_in\_the\_United\_States

Without further ado, assuming one driver and nobody else in the vehicle, here goes:

This is for sudden stops. If you want a more gradual stop, you'll wanna take the velocity and divide by time to get the acceleration. Like for example...
According to descriptions of The Great Frog, a fictional monster from the games Helmet Heroes and Eliatopia, it is capable of abruptly stopping pickup trucks. Assuming a reaction time of 1 second and a scenario in which a pickup truck is traveling on an undivided rural road outside of the fictional city of Alion, the necessary acceleration to stop the vehicle would range from 24.5872 to 31.2928 m/s².

To further illustrate this concept, we can use the lower end of the weight range for pickup trucks and perform the following calculations:

2110.4215\*24.5872=51889.3555 Newtons

Based on Earth's acceleration of gravity (Robby told me Eliatopia was about the same size as Earth, so it works), we got a force equivalent to **5291.2417 kg of force**, or on the very low end of **Class 10**

With the higher acceleration, I got the following.:

2110.4215\*31.2928=66040.9979 Newtons

This would equate to 6734.3076 kg of force, or a bit higher into **Class 10.**

This doesn't include friction coefficient, and I'm sure calcing rounded objects like wheels will be another ass and a half, so what I wrote here is not fully accurate. Now then, let's check out Japan, shall we?

#### **Japan**

It is appropriate to consider Japanese data in this analysis, as Japan is the country of origin for anime and manga. Fortunately, the calculation process will be simpler in this case,

as Japan has two fixed speed limits:

* 100 km/h (27.7777 m/s) for divided national highways
* and 60 km/h (16.6666 m/s) for all other roads, unless otherwise posted.

These speed limits serve as useful reference points for our analysis.

**Reference:** https://en.wikipedia.org/wiki/Speed\_limits\_in\_Japan

That being said, here's the next table:

### **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 70 kg 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 withstand **98315 Joules** 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 many cases in fiction 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 = **98315 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 meters 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. Unusually thin or thick walls may also change the result.

	+ Fragmentation: **1589350 Joules**, **Wall level**
	+ Violent Fragmentation: **13708143.75 Joules**, **Wall level+**
	+ Pulverization: **42515112.5 Joules**/**0.010161 Tons of TNT**, **Small Building level**
* Steel Wall:
	+ Fragmentation: **41323100 Joules**,**0.009876 Tons of TNT**, **Small Building level**
	+ Violent Fragmentation: **112943184 Joules**/**0.0269941 Tons of TNT**, **Small Building level**
	+ Pulverization: **59600625 to 198668750 Joules**, Average of **129134687.5 Joules**/**0.030863 Tons of TNT**, **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**

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 lbs. (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.

### **Pounding a Closed or Locked Door Open**

This feat uses direct measurements on a residential interior door. Actual results may vary.

#### Just Slamming the Door Open

Lateral cross-sectional area of the latch where it touches the gap in the door is 1.333208454 cm²

Brass, a common material used in doors, has a shear strength of 205 to 531 MPa.

**Force**

(1.333208454/10000)\*205000000=27330.7733 newtons

(1.333208454/10000)\*531000000=70793.36889 newtons

**Work/Energy**

Distance needed to cross is 7 mm

27330.7733\*0.007=191.3154131 joules (**Athlete level**)

70793.36889\*0.007=495.5535822 joules (**Street level**)

#### Breaking the Door Down

Latch Volume Touching Hole in Door: 0.315 cm³

Cross-Sectional Area of Hinges: 0.002\*0.0885\*2=3.54\*10-4 m³

Door Pin Volume (within hinges): π\*(0.6/2)²\*8.85=2.502278549 cm³

Diameter of Door Pin: 6 mm

Tensile Strength of Brass: 467.7 to 500.6 MPa

**Low End**

3.54\*10-4\*467700000=165565.8 newtons\*0.006=993.3948 joules\*3=2980.1844 joules (hinges)

2.502278549\*205=512.9671025\*3=1538.901307 joules (door pins)

0.315\*205=64.575 joules (latch)

2980.1844+1538.901307+64.575=4583.660707 joules (total) **(Street Level)**

**High End**

3.54\*10-4\*500600000=177212.4 newtons\*0.006=1063.2744 joules\*3=3189.8232 joules (hinges)

2.502278549\*531=1328.709909\*3=3986.129728 joules (door pins)

0.315\*531=167.265 joules (latch)

3189.8232+3986.129728+167.265=7343.217928 joules (total) **(Street Level)**

## **Cutting Feats**

### **Cutting a sword**

https://www.japanese-sword-katana.jp/katana/1910-1053.htm

Destroying Blades

A knightly (or short) sword blade is typically 5.08 cm wide, and 0.48768 cm thick; a long sword blade is 2.810165975 cm wide, and .414 cm thick; a katana is typically 2.23 cm wide, and 0.7 cm thick

Longsword = 2.810165975\*.414^2 = 0.4816512074688796 cc \* 208 J/cc = 100.18345115352696 J (**Athlete level**)

Shortsword = 2.810165975\*.414\*2.54\*.192 = 0.5673711614937759 cc \* 208 J/cc = 118.01320159070538 J (**Athlete level**)

Katana = .7\*.414\*2.810165975 = 0.814386099585062 cc \* 208 J/cc = 169.3923087136929 J (**Athlete level**)

Slicing shortsword with:

Longsword = 2.54^2\*(.192\*2)\*.414 = 1.0256495616 cc \* 208 J/cc = 213.3351088128 J (**Athlete level**)

Shortsword = 2.54^3\*(.192^2\*2) = 1.208185454592 cc \* 208 J/cc = 251.30257455513603 J (**Athlete level**)

Katana = 2.54^2\*(.192\*2)\*.7 = 1.73419008 cc \* 208 J/cc = 360.71153664 J (**Street level**)

Slicing Katana with:

Longsword = .7\*.414\*2.23 = 0.646254 cc \* 208 J/cc = 134.420832 J (**Athlete level**)

Shortsword = .7\*2.54\*.192\*2.23 = 0.76126848 cc \* 208 J/cc = 158.343 J (**Athlete level**)

Katana = .7^2\*2.23 = 1.0927 cc \* 208 J/cc = 227.2816 J (**Athlete level**)

Longsword = 2.810165975\*.414^2 = 0.4816512074688796 cc \* 568.5 J/cc = 273.81871144605805 J (**Athlete level+**)

Shortsword = 2.810165975\*.414\*2.54\*.192 = 0.5673711614937759 cc \* 568.5 J/cc = 322.5505053092116 J (**Street level**)

Katana = .7\*.414\*2.810165975 = 0.814386099585062 cc \* 568.5 J/cc = 462.97849761410777 J (**Street level**)

Slicing shortsword with:

Longsword = 2.54^2\*(.192\*2)\*.414 = 1.0256495616 cc \* 568.5 J/cc = 583.0817757696 J (**Street level**)

Shortsword = 2.54^3\*(.192^2\*2) = 1.208185454592 cc \* 568.5 J/cc = 686.8534309355521 J (**Street level**)

Katana = 2.54^2\*(.192\*2)\*.7 = 1.73419008 cc \* 568.5 J/cc = 985.88706048 J (**Street level**)

Slicing Katana with:

Longsword = .7\*.414\*2.23 = 0.646254 cc \* 568.5 J/cc = 367.395399 J (**Street level**)

Shortsword = .7\*2.54\*.192\*2.23 = 0.76126848 cc \* 568.5 J/cc = 432.78113088 J (**Street level**)

Katana = .7^2\*2.23 = 1.0927 cc \* 568.5 J/cc = 621.19995 J (**Street level**)

Longsword = 2.810165975\*.414^2 = 0.4816512074688796 cc \* 1000 J/cc = 481.651207 J (**Street level**)

Shortsword = 2.810165975\*.414\*2.54\*.192 = 0.5673711614937759 cc \* 1000 J/cc = 567.37116 J (**Street level**)

Katana = .7\*.414\*2.810165975 = 0.814386099585062 cc \* 1000 J/cc = 814.386099 J (**Street level**)

Slicing shortsword with:

Longsword = 2.54^2\*(.192\*2)\*.414 = 1.0256495616 cc \* 1000 J/cc = 1025.6495616 J (**Street level**)

Shortsword = 2.54^3\*(.192^2\*2) = 1.208185454592 cc \* 1000 J/cc = 1208.185454592 J (**Street level**)

Katana = 2.54^2\*(.192\*2)\*.7 = 1.73419008 cc \* 1000 J/cc = 1734.19008 J (**Street level**)

Slicing Katana with:

Longsword = .7\*.414\*2.23 = 0.646254 cc \* 1000 J/cc = 646.254 J (**Street level**)

Shortsword = .7\*2.54\*.192\*2.23 = 0.76126848 cc \* 1000 J/cc = 761.26848 J (**Street level**)

Katana = .7^2\*2.23 = 1.0927 cc \* 1000 J/cc = 1092.7 J (**Street level**)

## **Bone Breaking Feats**

### **Breaking all the Bones of a Man's Body**

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

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

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: 1.77078486850515432e5 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 Joules**, **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 an Average 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 982.22°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 as the average, which is also the standard figure used for body temperature.

So, the temperature change is by 945.22°C

A handy calculator

A handy specific heat calculator

A handy latent heat calculator

The average human is 62 kilograms.

**STEP I**

We will start with water.

As found in the human composition reference, 62% of human mass is water, or 38.44 kilograms.

The mean specific heat capacity of water is 4.190 kilojoules per kilogram at any point from 0 to 100 °C

We will use this calculator to find the latent heat of the water, which says water has a latent heat of 2264.7057 kJ/kg.

Plugging in the mass of water, we get the following...

(4.19\*38.44\*945.2222222222)+(2264.7057\*38.44)=**239296.181019 kilojoules**

**STEP II**

Average specific heat capacity for body fat is 2.348 kilojoules per kilogram

The latent heat of fusion for body fat is 138.6 to 187.5 kJ/kg (average 163.05 kJ/kg)

The latent heat of vaporization for fat varies based on a study conducted on 14 different fatty acids. A Google Drive document based on the same study made for easier readability gives and average value of 471.2383267499 kJ/kg

Fat seems to be 16% of body mass, or 9.92 kilograms going by the composition values linked at the top of this section.

Plugging the numbers in, we get the following.:

(2.348\*9.92\*945.2222222222)+(163.05\*9.92)+(471.2383267499\*9.92)=**28308.4074369 kilojoules**

**STEP III**

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

For the low end, the protein lysozyme has a specific heat capacity of 1.26 kilojoules per kilogram

For the high end, muscle has a heat capacity of 3.421 kilojoules per kilogram

With these values, we get the following:

**Low End:** (1.26\*9.92\*945.2222222222)=**11814.5216 kilojoules**

**High End:** (3.421\*9.92\*945.2222222222)=**32077.3638044 kilojoules**

**STEP IV**

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

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

We get the following:

(1.313\*3.72\*945.2222222222)=**4616.80561333 kilojoules**

**STEP V**

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

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

The latent heat of fusion for sucrose, a sugar, is 46.2 kJ/mol or 134.9707052 kJ/kg, which decomposes at 186°C

(ditto) we get the following:

(1.255\*0.62\*945.2222222222)+(134.9707052\*0.62)=**819.159248335 kilojoules**

**Conclusion**

Adding them together, we get:

**Low End:** 28308.4074369+239296.181019+11814.5216+4616.80561333+819.159248335=284855.074918 kilojoules; **0.06808199687 tons of TNT (Small Building level)**

**High End:** 28308.4074369+239296.181019+32077.3638044+4616.80561333+819.159248335=305117.917122 kilojoules; **0.07292493239 tons of TNT (Small Building level)**

The simplest and closest analogs were used when all else failed, ~~plus we did not include the latent heat from anything other than water~~ latent heat values were found for fats and sugar (closest analogue).

### **Reducing an Average 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.

62% of human mass is water, or 38.44 kilograms.

The heat capacity of water is 4190 Joules per kilogram at any point from 0 to 100 °C

We will use this calculator to find the latent heat of the water, which says water has a latent heat of 2264.7057 kJ/kg.

Plugging in the values, we get the following:

(4.19\*38.44\*723)+(2264.7057\*38.44)=**203504.269908 kilojoules**

**STEP II**

Average amount for body fat is 2.348 kilojoules per kilogram

The latent heat of fusion for body fat is 138.6 to 187.5 kJ/kg (average 163.05 kJ/kg)

The latent heat of vaporization for fat varies based on a study conducted on 14 different fatty acids. A Google Drive document based on the same study made for easier readability gives and average value of 471.2383267499 kJ/kg

Fat seems to be 16% of body mass, or 9.92 kilograms going by the numbers shown

Plugging them all in, we get the following:

(2.348\*9.92\*723)+(163.05\*9.92)+(471.2383267499\*9.92)=**23132.3718814 kilojoules**

**STEP III**

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

[1]For the low end, the protein lysozyme has a specific heat capacity of 1.26 kilojoules per kilogram

For the high end, muscle has a heat capacity of 3.421 kilojoules per kilogram

Doing the math, we get the following:

**Low End:** (1.26\*9.92\*723)=**9036.9216 kilojoules**

**High End:** (3.421\*9.92\*723)=**24535.95936 kilojoules**

**STEP IV**

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

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

(ditto) we get the following:

(1.313\*3.72\*723)=**3531.39228 kilojoules**

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

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

The latent heat of fusion for sucrose, a sugar, is 46.2 kJ/mol or 134.9707052 kJ/kg, which decomposes at 186°C

There, we get...

(1.255\*0.62\*723)+(134.9707052\*0.62)=**646.248137224 kilojoules**

**Conclusion**
Adding them together, we get:

**Low End:** 23132.3718814+203504.269908+9036.9216+3531.39228+646.248137224=239851.203807 kilojoules; **0.05732581352 tons of TNT (Small Building level)**

**High End:** 23132.3718814+203504.269908+24535.95936+3531.39228+646.248137224=255350.241567 kilojoules; **0.06103017245 tons of TNT (Small Building level)**

The simplest and closest analogs were used when all else failed, ~~plus we did not include the latent heat for anything other than water~~ latent heat values for fats and sugar are found.

#### **Reducing an average human to char, excluding the bones**

Without bone, it'd be more like this:

**Low End:** 23132.3718814+203504.269908+9036.9216+646.248137224=236319.811527 kilojoules; **0.05648179051 tons of TNT**

**High End:** 23132.3718814+203504.269908+24535.95936+646.248137224=251818.849287 kilojoules; **0.06018614944 tons of TNT**

### **Vaporizing an Obese Human**

**Conditions**

https://www.thoughtco.com/chemical-composition-of-the-human-body-603995

The above link is for the average person. According to Penn Medicine, the minimum amount of fat to be considered obese is 25% for an adult man: https://www.pennmedicine.org/for-patients-and-visitors/find-a-program-or-service/bariatric-surgery/who-is-a-candidate/weight-loss-and-obesity-facts

Same principles apply to an obese person as they would an average person with a temperature change to 1800°F or 982.22°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 as the average, which is also the standard figure used for body temperature.

So, the temperature change is by 945.22°C

A handy calculator

A handy specific heat calculator

A handy latent heat calculator

The average human is 62 kilograms.

**STEP 0**

Let's first see how the fat percentage value works out proportionately for a 62-kg person.:

* Water: 38.44 kg
* Protein: 9.92 kg
* Fat: ??? kg
* Minerals: 3.72 kg
* Carbs: 0.62 kg

38.44+9.92+3.72+0.62=52.7 kg

52.7\*4/3=70.26666666666 kg

An obese person has a minimum BMI of 30. For a person who's 5'9" (average height of a man), this would equate to 203.2 lbs (92.16996958 kg)

Let's see how everything stacks up in the case of an obese person.:

* Water: (38.44/70.26666666666)\*92.16996958=50.42239513 kg
* Protein: (9.92/70.26666666666)\*92.16996958=13.012231001 kg
* Fat: 92.16996958/4=23.042492396 kg
* Minerals: (3.72/70.26666666666)\*92.16996958=4.879586625 kg
* Carbs: (0.62/70.26666666666)\*92.16996958=0.813264438 kg

**STEP I**

We will start with water. For an obese person, the mass of water is 50.42239513 kilograms.

The mean specific heat capacity of water is 4.190 kilojoules per kilogram at any point from 0 to 100 °C

We will use this calculator to find the latent heat of the water, which says water has a latent heat of 2264.7057 kJ/kg.

Plugging in the mass of water, we get the following...

(4.19\*50.42239513\*945.2222222222)+(2264.7057\*50.42239513)=**313888.829148 kilojoules**

**STEP II**

Average specific heat capacity for body fat is 2.348 kilojoules per kilogram

The latent heat of fusion for body fat is 138.6 to 187.5 kJ/kg (average 163.05 kJ/kg)

The latent heat of vaporization for fat varies based on a study conducted on 14 different fatty acids. A Google Drive document based on the same study made for easier readability gives and average value of 471.2383267499 kJ/kg

For an obese person, the mass of fat is 23.042492396 kg.

Plugging the numbers in, we get the following.:

(2.348\*23.042492396\*945.2222222222)+(163.05\*23.042492396)+(471.2383267499\*23.042492396)=**65755.6716843 kilojoules**

**STEP III**

For an obese person, proteins make up 13.012231001 kg

For the low end, the protein lysozyme has a specific heat capacity of 1.26 kilojoules per kilogram

For the high end, muscle has a heat capacity of 3.421 kilojoules per kilogram

With these values, we get the following:

**Low End:** (1.26\*13.012231001\*945.2222222222)=**15497.3068776 kilojoules**

**High End:** (3.421\*13.012231001\*945.2222222222)=**42076.4181176 kilojoules**

**STEP IV**

For an obese man, minerals make up 4.879586625 kg

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

We get the following:

(1.313\*4.879586625\*945.2222222222)=**6055.94164544 kilojoules**

**STEP V**

For an obese person, carbs make up about 0.813264438 kg

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

The latent heat of fusion for sucrose, a sugar, is 46.2 kJ/mol or 134.9707052 kJ/kg, which decomposes at 186°C

(ditto) we get the following:

(1.255\*0.813264438\*945.2222222222)+(134.9707052\*0.813264438)=**1074.50497698 kilojoules**

**Conclusion**

Adding them together, we get:

**Low End:** 65755.6716843+313888.829148+15497.3068776+6055.94164544+1074.50497698=402272.254332 kilojoules; **0.09614537627 tons of TNT**

**High End:** 65755.6716843+313888.829148+42076.4181176+6055.94164544+1074.50497698=428851.365572 kilojoules; **0.10249793632 tons of TNT**

The simplest and closest analogs were used when all else failed, ~~plus we did not include the latent heat from anything other than water~~ latent heat values were found for fats and sugar (closest analogue).

### **Reducing an Obese 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

The above link is for the average person. According to Penn Medicine, the minimum amount of fat to be considered obese is 25% for an adult man: https://www.pennmedicine.org/for-patients-and-visitors/find-a-program-or-service/bariatric-surgery/who-is-a-candidate/weight-loss-and-obesity-facts

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

**STEP 0**

Let's first see how the fat percentage value works out proportionately for a 62-kg person.:

* Water: 38.44 kg
* Protein: 9.92 kg
* Fat: ??? kg
* Minerals: 3.72 kg
* Carbs: 0.62 kg

38.44+9.92+3.72+0.62=52.7 kg

52.7\*4/3=70.26666666666 kg

An obese person has a minimum BMI of 30. For a person who's 5'9" (average height of a man), this would equate to 203.2 lbs (92.16996958 kg)

Let's see how everything stacks up in the case of an obese person.:

* Water: (38.44/70.26666666666)\*92.16996958=50.42239513 kg
* Protein: (9.92/70.26666666666)\*92.16996958=13.012231001 kg
* Fat: 92.16996958/4=23.042492396 kg
* Minerals: (3.72/70.26666666666)\*92.16996958=4.879586625 kg
* Carbs: (0.62/70.26666666666)\*92.16996958=0.813264438 kg

**STEP I**

We will start with water. For an obese person, the mass of water is 50.42239513 kilograms.

The mean specific heat capacity of water is 4.190 kilojoules per kilogram at any point from 0 to 100 °C

We will use this calculator to find the latent heat of the water, which says water has a latent heat of 2264.7057 kJ/kg.

Plugging in the mass of water, we get the following...

(4.19\*50.42239513\*723)+(2264.7057\*50.42239513)=**266939.976794 kilojoules**

**STEP II**

Average specific heat capacity for body fat is 2.348 kilojoules per kilogram

The latent heat of fusion for body fat is 138.6 to 187.5 kJ/kg (average 163.05 kJ/kg)

The latent heat of vaporization for fat varies based on a study conducted on 14 different fatty acids. A Google Drive document based on the same study made for easier readability gives and average value of 471.2383267499 kJ/kg

For an obese person, the mass of fat is 23.042492396 kg.

Plugging the numbers in, we get the following.:

(2.348\*23.042492396\*723)+(163.05\*23.042492396)+(471.2383267499\*23.042492396)=**53732.6112074 kilojoules**

**STEP III**

For an obese person, proteins make up 13.012231001 kg

For the low end, the protein lysozyme has a specific heat capacity of 1.26 kilojoules per kilogram

For the high end, muscle has a heat capacity of 3.421 kilojoules per kilogram

With these values, we get the following:

**Low End:** (1.26\*13.012231001\*723)=**11853.8821973 kilojoules**

**High End:** (3.421\*13.012231001\*723)=**32184.2309499 kilojoules**

**STEP IV**

For an obese man, minerals make up 4.879586625 kg

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

We get the following:

(1.313\*4.879586625\*723)=**4632.18670353 kilojoules**

**STEP V**

For an obese person, carbs make up about 0.813264438 kg

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

The latent heat of fusion for sucrose, a sugar, is 46.2 kJ/mol or 134.9707052 kJ/kg, which decomposes at 186°C

(ditto) we get the following:

(1.255\*0.813264438\*723)+(134.9707052\*0.813264438)=**847.694561497 kilojoules**

**Conclusion**

Adding them together, we get:

**Low End:** 53732.6112074+266939.976794+11853.8821973+4632.18670353+847.694561497=338006.351464 kilojoules; **0.08078545685 tons of TNT**

**High End:** 53732.6112074+266939.976794+32184.2309499+4632.18670353+847.694561497=358336.700216 kilojoules; **0.08564452682 tons of TNT**

The simplest and closest analogs were used when all else failed, ~~plus we did not include the latent heat from anything other than water~~ latent heat values were found for fats and sugar (closest analogue).

#### Reducing an obese human to char, excluding the bones

Without bone, it'd be more like this:

**Low End:** 53732.6112074+266939.976794+11853.8821973+847.694561497=333374.16476 kilojoules; **0.07967833765 tons of TNT**

**High End:** 53732.6112074+266939.976794+32184.2309499+847.694561497=353704.513513 kilojoules; **0.08453740762 tons of TNT**

### **Incinerating an average Building**

* Incinerating/Liquefication of a 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**

### **Vaporization of Iron**

Taken from here, the density of Iron is 7,874 kg/m^3 or 7.874 g/cc. 6,213,627 / 1000 = 6,213.627 to vaporize one gram of Iron. 6213.627 Joules/g \* 7.874 g/cc = 48926.098998 Joules/cc.

Iron characteristics

* Density: 7.874 g/cm^3
* Boiling point: 2862 °C
* Heat of fusion: 13,810 J/mol
* Heat of vaporization: 340,000 J/mol
* Molar Heat Capacity 25.10 J/(mol\*K)
* Molar Mass: 55.8450 g/mol
* Room Temperature: 20 °C (average)

* Heat of Fusion: 247 j/g
* Heat of Vaporization: 6,088 j/g
* Molar Heat Capacity: 0.4494 J/g\*C

(7.874)(0.4494)(6088-20) = 21,472 J

(247)(7.874) + (6,088)(7.874) = 49,881.79 J

49,881.79 + 21,472 = **71,353.79 J**

It actually takes around **71,353.79 j/cc** to vaporize iron.

### **Vaporizing a Volcano**

https://m.youtube.com/watch?v=csIbRU6KUx4&t=9m21s

Height: 266 px or 609.6 m

Width: 181 px or 414.8 m, radius 207.4 m

pi\*207.4^2\*609.6/3\*1e6 = 2.7459402e13 cm^3

Since it's a volcano I'm going to low-ball this and assume it's 80% hollow.

2.7459402e13\*0.2 = 5.4918804e12 cm^3

5.4918804e12\*25700 = 1.4114133e17 Joules, **City level**

Should be violent fragmentation instead

5.4918804e12\*69 = 3.7893975e14 Joules, **Town level+**

### **Vaporizing Tokyo Tower**

Mass = 4000000 kg

Boiling Point = 1370 C

Heat Change = 1354 C (16 C being standard temp)

Specific Heat of Steel = 466 J(kg x K)

Result = 2.524e12 Joules, or **Multi-City Block level**

That's melting. Vaporization can also be done, since we have mass. We will have to use Iron's value for this.

6213627 x 4000000 = 2.486e13 Joules, 5.94 Kilotons of TNT, **Town level** (So close to Low 7-C).

### **Vaporizing The Oceans**

Mass of all oceans is consistently around 1.4e21 kg from multiple sources.

"The worldwide average temperature of all ocean water, from the surface to the seafloor, is 3.5˚C."

Water has a specific heat capacity of 4184 J/kg.

Water's latent heat of vaporization is 2,260,000 J/kg.

Before vaporization happens you have to bring all the water to boiling temperature first. Sea water boils at about a degree higher than regular water, so we're going to ignore it and use 100˚C anyway since it's irrelevant in the grand scheme of things.

Q1 = mcΔT, where m is mass, c is specific heat capacity of water and T is change in temperature [96.5˚C].

Q1 = 5.652584e26 J.

Now, to actually change state.

Q2 = mL, where m is mass and L is water's latent heat of vaporization.

Q2 = 3.164e27 J.

Total energy required = Q1 + Q2.

Energy Required To Vaporize The Oceans = 3.7292584e27 J or **891.3 Petatons, High 6-A.**

## **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 and here. Usually, the storm clouds extend all the way to the horizon. The visibility on a normal day is 20 km. This will give us an area of 1256637061 m^2 for the storm.

Storm clouds have a height of 8000 m, and they usually sit at 4000 m above sea level.

We put these values on this calculator and it will gives us a storm mass of 5421709348262 kg.

**CAPE**

"Weak instability": 5.421709348262e15 Joules, 1.295819633905 Megatons, **Small City level**

"Moderate instability": 1.355427337065e16 Joules, 3.239549084764 Megatons, **Small City level**

"Strong instability": 2.168683739304e16 Joules, 5.183278535623 Megatons, **Small City level+**

1999 Oklahoma Tornado Outbreak: 3.193386806126e16 Joules, 7.632377643705 Megatons, **City level**

1990 Plainfield Tornado: 4.337367478609e16 Joules, 10.366557071246 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\*5421709348262\*333.333333333333^2 is....

3.012060749034e17 Joules, 71.989979661434 Megatons, **City level+**

### **Flooding the Entire Earth**

* Earth's surface area = 4\*pi\*(6378.1 km)^2 = 511201962.310545 km^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 0.006858 kilometers and the highest mountain would be Mt. Everest at 8.8489 km.

* Water volume = 511201962.310545\*(0.006858+8.8489) = 4527080867.34731 km^3

* Water density = 1025 kg/m^3 or 1.025E+12 kg/km^3 (For an average value between seawater and freshwater)

* Mass of water = 4.64025788903099E+21 kg

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 = (4.64025788903099E+21 kg)\*(84)\*(4186) = 1.63162603997263E+27 Joules

* q2 = (4.64025788903099E+21 kg)\*(2264760) = 1.05088198828358E+29 Joules

* Total = 1.06719824868331E+29 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 = (4.64025788903099E+21)\*(2463203.84) = 1.14299010508514E+25 Joules

* Total flood energy = 1.14299010508514E+25+1.06719824868331E+29 = 1.06731254769382E+29 J, or 2.5509382115053E+19 Tons 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** (**7-B**)

## **Earth Feats**

### **Destroying the Surface of the Earth**

#### **Explosion Method**

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)**

#### Melting the Earth's surface

**Part 1: Boiling the Oceans:**

Specific heat capacity of water = 4186J/kg/C | Latent Heat (Vaporization) = 2264705.7

Latent Heat Energy = 1.367544502E+21kg \* 2264705.7 = 3.097086e27j or 740.9 Petatons;

1,3675445020E+21kg \* 4186 \* 1527,5 = 8,744237e27j ou 2,09 Exatons + 0,74 = 2,83 Exatons;

High-Ball (10000-10,5 = 9989,5):

1,367544502E+21kg \* 4186 \* 9989,5 = 5,72e28j ou 13,678 Exatons + 0,74 = 14,418 Exatons;

**Part 2: Melting the earth's crust**

Sodium: 6.54e20 kg | Magnesium: 6.45e20 kg | Potassium: 5.79e20 kg

Total Melting Energy = 3,377675 Exatons;

Silicon = 7.8114e21kg \* 1787000 = 1.39355376e28j or 3.3307 Exatons;

Aluminum = 2.28e21kg \* 396000 = 9.029e26j or 215.8 Petatons;

Iron = 1.56e21kg \* 247000 = 3.8532e26j or 92.0937 Petatons;

Calcium = 1.15e21kg \* 209416.4 = 2.408284e26j or 57.56 Petatons;

Sodium = 6.54e20kg \* 113000 = 7.39e25j or 17.66 Petatons;

Magnesium = 6.45e20kg \* 358000 = 2.309e26j or 55.186 Petatons;

Potassium = 5.79e20kg \* 56000 = 3.2424e25j or 7.75 Petatons;

Heating energy:

Low-Ball

Oxygen = 1.277e22kg \* 1538 \* 920 = 1.807e28j or 4.31 Exatons;

Silicon = 7.8114e21kg \* 1538 \* 710 = 8.53e27j or 2.04 Exatons;

Aluminum = 2.28e21kg \* 1538 \* 900 = 3.156e27j or 752.87 Petatons;

Iron = 1.56e21kg \* 1538 \* 450 = 1.079e27j or 257.89 Petatons;

Calcium = 1.15e21kg \* 1538 \* 532 = 9.41e26j or 224.9 Petatons;

Sodium = 6.54e20kg \* 1538 \* 1260 = 1.2674e27j or 302.9 Petatons;

Magnesium = 6.45e20kg \* 1538 \* 1050 = 1.0416e27j or 248.95 Patatons;

Potassium = 5.69e20kg \* 1538 \* 1000 = 8.75e26j or 209.13 Petatons;

Total Heating Energy (Low-Ball) = 8.34664 Exatons;

High-Ball

Oxygen = 1.277e22kg \* 10000 \* 920 = 1.17484e29j or 28.08 Exatons;

Silicon = 7.8114e21kg \* 10000 \* 710 = 5.5462e28j or 13.2557364 Exatons;

Aluminum = 2.28e21kg \* 10000 \* 660.3 = 1.5055e28j or 3.6 Exatons;

Iron = 1.56e21kg \* 10000 \* 450 = 7.02e27j or 1.678 Exatons;

Calcium = 1.15e21kg \* 10000 \* 532 = 6.118e27j or 1.46224 Exatons;

Sodium = 6.54e20kg \* 10000 \* 1260 = 8.2404e27j or 1.9695 Exatons;

Magnesium = 6.45e20kg \* 10000 \* 1050 = 6.7725e27j or 1.61866635 Exatons;

Potassium = 5.79e20kg \* 10000 \* 1000 = 5.79e27j or 1.384 Exatons;

Total Heating Energy = 53.041343 Exatons;

Low-Ball = 2.83 + 8.34664 + 3.377675 = **14.554315 Exatons** (**High 6-A**)

High-Ball = 14.418 + 53.041343 + 3.377675 = **70.837018 Exatons** (**5-C**)

### **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**

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**

* **Description:** The energy necessary to split the Earth in half and separate the halves by a visible distance.

* **Requirements:** The Earth either has to be split to an extent that is visible from outer space (so far away that one can see the Earth in its entirety) or it has to be known that the halves were separated by at least 203km.

* **Results:** 5.0811706477439e+30K - **Small Planet level**.

* **Calculation:**

Diameter of the Earth is 12 742 000 meters. Radius is 6 371 000 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 = 6.674E-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 12742000 meters.

For the split to be visible I'll assume 10 pixels or so. That's 203222 meters.

GPE of the unsplit Earth: 6.674E-11\*(5.97237E24/2)^2/[2\*(3/8)\*6371000] = 1.24552013605604E+32 J

GPE of the split Earth: 6.674E-11\*(5.97237e+24/2)^2/[2\*(3/8)\*6371000+203222] = 1.1947084295786E+32 J

1.24552013605604E+32 - 1.1947084295786E+32 = 5.08117064774388E+30 Joules, or **Low 5-B** **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
* Aluminum = 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**

A handy latent heat calculator

A handy specific heat calculator

Average human weight = 62kg

The average temperature of a human body is 37 °C. Temperature required to cause frostbite is -4 °C.

Temperature change= 37-(-4) = 41 °C

Specific heat capacity of a human is anywhere from 3.47-3.6 kJ/kg or 3470-3600 J/kg. The average is 3535 J/kg

Specific Heat capacity= temperature change \* mass \* heat capacity in J/kg

Specific Heat capacity= (62 \* 3535 \* 41) = 8,985,970 Joules

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

So, water mass = 0.65 \* 62kg= 40.3 kg of water.

Enthalpy of fusion of water is 333.55 J/g or 333550 J/kg.

Enthalpy of Fusion, Q= m \* L, where Q is the energy in Joules, m is the mass in kg, and L is the specific latent heat.

Enthalpy of Fusion= 40.3 \* 333550= 13,442,065 Joules

So total energy = 8,985,970 Joules + 13,442,065 = 22,428,035 Joules or 0.00536043 tons of TNT, **Small Building level**

### **Freezing a human to Absolute Zero**

The average temperature of a human body is 37 °C. Absolute zero is 0 °K or -273.15 °C.

Temperature change= 37-(-273.15)= 310.15 °C

Average human weight = 62kg

Specific heat capacity of a human is anywhere from 3.47-3.6 kJ/kg or 3470-3600 J/kg. The average is 3535 J/kg

Specific Heat capacity= temperature change \* mass \* heat capacity in J/kg

Specific Heat capacity= (62 \* 3535 \* 310.15)= 67,975,575.5 Joules

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

So, water mass = 0.65 \* 62kg= 40.3 kg of water.

Enthalpy of fusion of water is 333.55 J/g or 333550 J/kg.

Enthalpy of Fusion, Q= m \* L, where Q is the energy in Joules, m is the mass in kg, and L is the specific latent heat.

Enthalpy of Fusion= 40.3 \* 333550= 13,442,065 Joules

Temperature required to cause frostbite is -4 °C.

Temperature change= -4 - (-273.15) = 269.15 °C

Specific heat capacity of ice = 2093 J/kg

Specific heat capacity= 40.3 \* 2093 \* 269.15= 22,702,237.285 Joules

So, total energy= 67,975,575.5 + 13,442,065 + 22,702,237.285= **104,119,877.785** Joules or **0.0249 tons of TNT**, **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 × 1024 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 or **2.8294007063E+18 Tons of TNT** or **2.8294 Exatons** (**High 6-A+**)

### **Freezing an average American City**

**Description:** The energy change necessary to completely freeze an average American city.

**Requirements:** The City must have an area of at least 128.27 km^2 or have statements that prove that it is bigger than an average American city. Note that most cities in the modern world and virtually all cities in medieval times are smaller than an average American city. The city has to be frozen into a block of ice with 192m elevation in which even what was formerly air is now replaced by ice. Additionally, it has to be evaluated which method is most appropriate for the feat.

**Results:**

* Method 1: 1.695382968e15 J - **Large Town level**
* Method 2: 9.1726514e+15 J or **Small City level**
* Method 3: 1.241074958e19 J - **Large Mountain level**

**Calculation:**
This will use an average American city

* Area: 128.27 km2
* Elevation = 192 m

Method 1:

Credits to AlexSoloVaAlFuturo

* Volume = 128270000 m^2 \* 192 m = 2.462784e10 m^3

Using the values of air.

* Mass = 2.462784 e10 m^3 \* 1.225 kg/m^3 = 3.0169104e10 kg

Temperature Change = 3.0169104e10 kg\*1003.5 j/kg\*k\*(16C-(-40C)) = 1.695382968e15 J - **Large Town level**

Method 2:

* Volume: 2.462784e10 m^3

Then the mass is 2.462784e10 m^3\*1.165, which is nitrogen's density.

* Mass: 28691433600kg

According to this nitrogen at ambient temperature has specific heat of 1040 J/kg K

* Ambient temperature = 343.15 Kelvin
* Nitrogen melting point = 63.15 Kelvin

ΔT = 343.15 - 63.15 = 280 Kelvin

Now the temperature change:
Q = m\*c\*ΔT

* Q = 8.3549455e+15 J

Not only did the temperature change, but also the physical state

Latent nitrogen-fusing heat = 28500 J/kg

* E = 28691433600\*28500
* E = 8.1770586e+14 J

Total: 9.1726514e+15 or **Small City level**

Method 3:

Credits to AlexSoloVaAlFuturo

Density of frozen nitrogen: 1026.5 kg/m^3

* M = 2.462784e10 m^3\*1026.5 kg/m^3\*0.8 = 2.022438221e13 kg
* Q1 = 2.022438221 e13 kg \* 1040 j/kg\*(16 C-(-195.8 C)) = 4.454865118 e18 J

Vaporization energy oxygen = 2.022438221e13 kg\*199190 j/kg = 4.028494692e18 J

* Q2 = 2.022438221 e13kg\*1040 j/kg\*(-195.8 C-(-210.1 C)) = 3.007770122e17 J

Fusion energy oxygen = 2.022438221e13 kg\*25702 j/kg = 5.198070716e17 J

* E1 = 4.454865118e18 J + 4.028494692e18 J + 3.007770122e17 J + 5.198070716e17 J = 9.303943894e18 J

Density of solid oxygen: 1426 kg/m^3

* M = 2.462784e10 m^3\*1426 kg/m^3\*0.2 = 7.023859968e12 kg
* Q1 = 7.023859968e12 kg\*919 j/kg\*(16 C -(-182.9 C)) = 1.283885042e18 J

Vaporization energy oxygen = 7.023859968e12 kg\*213125 j/kg = 1.496960156e18 J

* Q2 = 7.023859968e12 kg\*919 j/kg\*(-182.9 C -(-218.3 C)) = 2.285044268e17 J

Fusion energy oxygen = 7.023859968e12 kg\*13875 j/kg = 9.745605706e16 J

* E2 = 1.283885042e18 J + 1.496960156e18 J + 2.285044268e17 J + 9.745605706e16 J = 3.106805682e18 J

Total E = 9.303943894e18 J + 3.106805682e18 J = 1.241074958e19 J - **Large Mountain level**

## **Crushing Feats**

### **Crushing a Golf Ball**

**Materials of Golf Ball**

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

**Energy Density of Materials**

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

Polybutadiene = 2.35 MPa on average or 2.35 J/cc

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

**Volume of Ball**

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

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

**Energy to Crush Golf Ball**

2.35\*27.61 = 64.8835 Joules for core

13.07\*50.37137309 = 658.3538463 Joules for cover

723.2373463 Joules in total, **Street level**

### **Crushing a Human Skull**

Skulls have been easily destroyed before by large caliber rounds varying from 12-gauge shotgun slugs (At least 2363 ft-lbs or 3204 J), .500 S&W Magnum hollow-point rounds (3000-3900 J) and .308 Winchester/7.62x51mm NATO rounds (Ranging from 3500 to 3700 J), all of which have muzzle energies at around 3000-3900 joules (**Street level**), with such damage being even possible with several types of elephant gun rounds (The examples used including the .375 H&H Magnum, .416 Rigby, .458 Lott, .460 Weatherby Magnum, .500 Jeffery, .470 Nitro Express, .500 Nitro Express, .600 Nitro Express and .700 Nitro Express) whereas full-powered punches from MMA fighters exceeding 1100 joules, and even full-powered curbstomps from humans, are not enough to wield anywhere close to a similar result.

If one were to crush a skull with their bare hands, they would need to produce at least 500 kilograms of force, which is **Peak Human** lifting strength.

### **Crushing a Whole Human**

The pressure needed to crush a human is commonly agreed on as the following statement: "The human body can withstand 50 psi (pounds per square inch) and that's if it's a sudden impact. However if it's sustained pressure, the body can withstand up to 400 psi if the weight is gradually increased"

This feat applies when a whole person is crushed, so partial crushing like what elephants do do not apply.

The cross-sectional area of a human is listed as 0.68 m². Let's see how many square inches that is.

6800/2.54²=1054.002108 in²

Now that we got the area needed for the feat, let's determine the force needed to crush a whole human.:

Low End: 1054.002108\*50=52700.1054 lbs of force

High End: 1054.002108\*400=421600.8432 lbs of force

That's 52700.1054 to 421600.8432 lbs of force, or **23904.36571 to 191234.9257 kg of force**, giving anyone who crushes a whole human **Class 25** to **Class K**. Because I'll need to calculate the KE needed, I'll convert this to newtons as well and... It sits at 234421.748 to 1875373.984 newtons. As noted from the human-shaped hole feat, the human head would have a length of 20.9125 cm. Let's do the math here.

Low End: 234421.748\*.209125=49023.44804 joules (**Wall level**)

High End: 1875373.984\*.209125=392187.5844 joules (**Wall level**)

## **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.

### **Rising From the Grave**

**Note:** This applies to those who do this feat literally (ex. zombies), not metaphorically.

**Grave Depth:** 4 to 6 feet (1.2192 to 1.8288 meters): https://feldmanmortuary.com/blogs/blog-entries/1/Blogs/25/Why-Are-Graves-Dug-6-Feet-Deep.html

**Coffin Dimensions:** 84 inches (length) by 28 inches (width) (2.1336 x 0.7112 meters): https://tindallfuneralhome.com/blogs/blog-entries/1/Our-Blogs/147/Are-All-Caskets-The-Same-Size.html

**Max Force Needed to Pull Nail Out:** 165 lbs (74.84274105 kg) average (smooth-shank nail used): https://www.youtube.com/watch?v=kAxGAIFbqu4

**Volume of Soil Pushed Out:** 1.2192\*2.1336\*0.7112=1.850033977 m³ ; 1.8288\*2.1336\*0.7112=2.775050966 m³

**Density of Soil:** 1.1 to 1.6 g/cm³: https://www.sciencedirect.com/topics/agricultural-and-biological-sciences/soil-density

#### Coffin with Halved Lid

Some modern coffins have lids divided into two. This allows for easier movement of the soil when the character opens the coffin while buried in the ground.

Low End: 1.850033977\*1100/2=1017.518688 kg (**Class 5**)

High End: 2.775050966\*1600/2=2220.040773 kg (**Class 5**)

#### Coffin with Full Lid

Other modern coffins still have full lids much like older coffins. These coffins aren't usually nailed, but they can be nailed shut if requested.

Low End: 1.850033977\*1100=2035.037375 kg (**Class 5**)

High End: 2.775050966\*1600=4440.081546 kg (**Class 5**)

#### Coffin That's Nailed Shut

Modern coffins may optionally be nailed shut, but older coffins are always nailed shut. Typically a coffin is nailed shut with 6 to 10 nails: https://www.northwoodscasket.com/build-your-own-coffin

Low End: (1.850033977\*1100)+(74.84274105\*6)=2484.093281 kg (**Class 5**)

High End: (2.775050966\*1600)+(74.84274105\*10)=5188.508956 kg (**Class 10**)

### **Uprooting a Tree**

The feat itself varies depending on the diameter of the tree in question. In a study conducted on tree stumps, the uprooting force of a tree stump is stated to be no higher than 60 kN (6118.297278 kg) of upwards force, indicated for a stump that's 65 cm in diameter (**Class 10**).

## **Object Destruction Feats**

### **Destroying a Door**

Standard size for a door is 2.032 m 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

### **If knocked down**

**Steel Door**

According to this page, each hinge of a door like that weighs 30 pounds or 13.6078 kg, since there are three of them it would be 40.8234 kg in total. Using the 7850 kg/m³ density of steel, this would give us a volume of 5200.433 cc. Using the 358 J/cc fragmentation of steel, it would give us an AP of:

5200.433\*358 = 1,861,755.014 Joules (**9-B: Wall level**)

### **Destroying a Car**

**Mass and Weight of Materials**

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

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

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

The highest amount of copper used in an average conventional car is 49 lbs. or **22.226 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

Cast Iron = 149 MPa or j/cc

Glass = 3.5 j/cc

Aluminum = 40000 PSI = 275.79 megapascals = 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**

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

Low End: 2.41317\*38000000 = 9.1700460e7 Joules, 0.022 Tons of TNT, **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.

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

Brass is 235 MPa or 235 J/cc

Steel = 208 j/cc

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

Short sword = 200.58 cc

Energy to Destroy Blades

Assuming they are made of steel.

Longsword = 2.177968e4 Joules, **Wall level**

Short sword = 4.172064e4 Joules, **Wall level**

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

### **Destroying a Chimney**

**Volume of Chimney**

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

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

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

Outer Edge B and C = 60.96 cm

thickness = Orange 11.4 px or 11.63086192 cm

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

V = 559,603.43 cc

**Energy to destroy chimney**

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

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

279801.715 x 3.49375 = 977557.2418 Joules

279801.715 x 8 = 2238413.72 Joules

3.215970962e6 Joules in total, **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: 222,500 imperial tons; 201,814,058.957 kg.

Using a ratio of 1:5 steel to concrete, as that's the ratio used in the Empire State Building, a relatively old skyscraper that should be typical of most found in the world. While also having less of a steel content than other skyscrapers, such as the Seers/Willis Tower having over 34% of their mass in steel.

Density of structural steel: 7,850 kg/m3

According to this link, reinforced concrete has a density of roughly 2500 kg/m^3

Leaving masses of...

Mass of steel: 201,814,058.957 x (1/6) = 33,635676.493 kg

Mass of concrete: 201,814,058.957 x (5/6) = 168,178,382.464 kg

Leading to volumes of...

Volume of steel: 4,284,799,553.225 cm3

Volume of concrete: 67,271,352,985.639 cm3

Giving yields of...

Frag (Steel [208 J/cm3]): 891,238,307,070.882 joules

Frag (Reinforced Concrete [10 J/cm3]): 672,713,529,856.387 joules

Total: **1.56395e+12 joules, or 373.793 Tons of TNT (8-A)**

V. Frag (Steel [568.5 J/cm3]): 2.43591e+12 joules

V. Frag (Reinforced Concrete [61.2 J/cm3]): 4.11701e+12 joules

Total: **6.55292e+12 joules, or 1.56619 Kilotons of TNT (Low 7-C)**

Pulv. (Steel [Average of 655 J/c3]): 2.80654e+12 joules

Pulv. (Reinforced Concrete [610 J/cm3]): 4.10355e+13 joules

Total: **4.3842e+13 joules, or 10.4785 Kilotons of TNT (7-C)**

### **Destroying a Plane**

403500 lbs = 183024.521 Kgs

Percentages:

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 megapascals = 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 1.016 m.

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 1.016 m.

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**

### **Destroying a Pool Table**

We're calculating the energy necessary to destroy a standard pool (billiard) table.

#### The Slates

We're starting with the stone slabs that make up the table's playing field.

https://en.wikipedia.org/wiki/Billiard\_table

According to Wikipedia we have two standard sizes for the playing field.

One is 254 cm x 127 cm (9ft table) and the other is 234 cm x 117 cm (8ft table). Or in area: 32258 cm^2 and 27378 cm^2.

The slates must be at least 2.54 cm. That gives us the volumes: 2.54 cm \* 32258 cm^2 = 81935.32 cm^3 and 2.54 cm \* 27378 cm^2 = 69540.12 cm^3

Now, taking the destruction values for slate from our destruction value table:

Low End:

* Fragmentation: 69540.12 cm^3 \* 15 J/cc = 1.0431018e6 J
* Violent Fragmentation: 69540.12 cm^3 \* 103.42 J/cc = 7.1918392104e6 J
* Pulverization: 69540.12 cm^3 \* 172.5 J/cc = 1.19956707e7 J

High End:

* Fragmentation: 81935.32 cm^3 \* 15 J/cc = 1.2290298e6 J
* Violent Fragmentation: 81935.32 cm^3 \* 103.42 J/cc = 8.4737507944e6 J
* Pulverization: 81935.32 cm^3 \* 172.5 J/cc = 1.41338427e7 J

For the next part I also want to figure out the weight of the slate. Density is 2600 kg/m^3.

Our high-end volume (9ft) was 81935.32 cm^3 = 0.08193532 m^3 and the low-end volume (8ft) was 69540.12 cm^3 = 0.06954012 m^3.

Multiplying with density:

* High end (9ft): 0.08193532 m^3 \* 2600 = 213.031832 kg
* Low end (8ft): 0.06954012 m^3 \* 2600 = 180.804312 kg

#### Wood

That leaves the wood to destroy. Now those tables have a lot of parts, so measuring the volume of the wood is bothersome. So instead, I will go by weight.

According to this an 8ft pool table with slate weights 1000 Pounds = 453.592 kg and a 9ft pool table weights 1,300 Pounds = 589.6701kg.

We have calculated the weight of the slates above, so we can subtract those.

453.592 kg - 180.804312 kg = 272.787688 kg of wood

589.6701 kg - 213.031832 kg = 376.638268 kg of wood

Now I will need to make a guess at the type of wood. This producer lists several options. I will simply use the thing on the top of the list: Oak. I will use white oak specifically, as the calculation page suggests that as standard assumption.

According to this that type of wood has a density of 0.77\*10^3 kg/m^3.

That gives us a volume of:

* Low end: 272.787688 / (0.77\*10^3) = 0.3542697246753247 m^3 = 354269.7246753247 cm^3
* High end: 376.638268 / (0.77\*10^3) = 0.4891406077922078 m^3 = 489140.6077922078 cm^3

Now applying the destruction values for that wood from our calculations page:

Low End:

* Fragmentation: 354269.7246753247 \* 7.3774 = 2613589.46681974044178 J
* Violent Fragmentation: 354269.7246753247 \* 13.7895 = 4885202.36841038995065 J
* Pulverization: 354269.7246753247 \* 51.297 = 18172974.0666701311359 J

High End:

* Fragmentation: 489140.6077922078 \* 7.3774 = 3608585.91992623382372 J
* Violent Fragmentation: 489140.6077922078 \* 13.7895 = 6745004.4111506494581 J
* Pulverization: 489140.6077922078 \* 51.297 = 25091445.7579168835166 J

#### Summing Up

Now to sum up slate and wood for the final result Low End:

* Fragmentation: 2613589.46681974044178 J + 1.0431018e6 J = 3.65669126681974044e6 J **Wall level**
* Violent Fragmentation: 4885202.36841038995065 J + 7.1918392104e6 J = 1.207704157881039e7 J **Wall level+**
* Pulverization: 18172974.0666701311359 J + 1.19956707e7 J = 3.0168644766670131e7 **Small Building level**

High End:

* Fragmentation: 3608585.91992623382372 J + 1.2290298e6 J = 4.83761571992623382e6 J **Wall level**
* Violent Fragmentation: 6745004.4111506494581 J + 8.4737507944e6 J = 1.5218755205550649e7 J **Wall level+**
* Pulverization: 25091445.7579168835166 J + 1.41338427e7 J = 3.9225288457916884e7 J **Small Building level**

### **Blowing up Cannons**

This is about blowing up 16th century cannons.

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

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

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

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

* **Frag (400 J/cc):** 1166666.667\*400 = **466,666,666.8 J** or **0.111536 tons of TNT (Small Building level)**
* **V. Frag (613.5 J/cc):** 1166666.667\*613.5 = **715,750,000.2045 J** or **0.17106836 tons of TNT (Small Building level+)**
* **Pulverization (827 J/cc):** 1166666.667\*827 = **964,833,333.609 J** or **0.2306 tons of TNT (Small Building level+)**

### **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.

* 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+)

### **Destroying a Sink/Toilet**

As title, note that this is for porcelain ones, some have steel in them which would be higher, so the feat can be used for that variety as a low-ball. This calculation assumes porcelain's destruction values are similar to this site's values for destroying ceramic, since porcelain is a kind of ceramic.

Wall-mounted sink: Roughly 11 kg or 11000 g. Porcelain is 2.403 g/cm^3

* 11000 / 2.403 = 4577.61 cm^3
* Frag: 4577.61 x 3.4 = **15563.874 J**
* VFrag: 4577.61 x 4.53 = **20736.5733 J**

Toilet

* One Piece Toilet/Whole Two Piece Toilet: 40 kg
* 40000 / 2.403 = 16645.85 cm^3
* Frag: 16645.85 x 3.4 = **56595.89 J**
* VFrag: 16645.85 x 4.53 = **75405.70 J**
* Two Piece Toilet Bowl: 25 kg
* 25000 / 2.403 = 10403.66 cm^3
* Frag: 10403.66 x 3.4 = **35372.44 J**
* VFrag: 10403.66 x 4.53 = **47128.58 J**

All Wall level

### **Destroying New York City**

NYC has a land area of 783.84 km^2, a total area of 1,213.37 km^2 and a metro area of 34,490 km^2

* 783.84 km^2 = 783,840,000 m^2
* 1,213.37 km^2 = 1,213,370,000 m^2
* 34,490 km^2 = 34,490,000,000 m^2

Let's assume NYC is vaguely circular (it's not but ehhh) and derive a radius from it.

* 783.84 km^2 = 783,840,000 m^2 (15,795.696286847 meters in radius)
* 1,213.37 km^2 = 1,213,370,000 m^2 (19,652.675812693 meters in radius)
* 34,490 km^2 = 34,490,000,000 m^2 (104,778.37550983 meters in radius)

In order:

* Land area = 15.79 km
* Total area = 19.65 km
* Metro area = 104.77 km

R = Y^(1/3)\*0.28

R = Radius in km

Y = Yield in Kilotons

15.79 = Y^(1/3)\*0.28

Y = 179,337.989204 kilotons (**179.337 Megatons**)

19.65 = Y^(1/3)\*0.28

Y = 345,631.702123 kilotons (**345.631 Megatons**)

104.77 = Y^(1/3)\*0.28

Y = 52,388,593.355184 kilotons (**52.38 Gigatons**)

### **Destroying the Statue of Liberty**

This says that the statue of liberty is 130 tons of iron, and 88 tons of copper.

Iron has a density of 7,874 KG/M^3, while copper has a density of 8,950 KG/M^3. This would mean the volume would be 16.5100330201 and 9.83240223464 meters cubed of copper, or for our purposes, 16510033.0201 and 9832402.23464 centimeters

The shear strength of copper is 150, so, that combined with the J/CC of 20 for Iron makes for...

1805060995.6 Joules, 0.43141993202 Tons, **Building level**

The Statue of Liberty is pretty hollow so it's not really surprising. Then again, I'm not 100% sold on how dense the copper came out, further research may be required to make absolute sure it's on point.

### **Destroying Britannia**

Destroying Britannia.png
Britannia Diameter = 899.99 Km

Y = ((x/0.28)^3)

Y is in Teratons, x is Radius in Kilometers

Britannia Radius = 449.995 Km

Y = ((449.995/0.28)^3)/1000000000 = 4.2 Teratons (**Small Country level+**)

### **Destroying a Continent**

[Calculation 1]: This calculation is Therefir's suggestion for destroying Australia; credit to Therefir on this calculation:

Diameter of Australia = 4000 km

Explosion radius = 2000 km or 2000000 m

W = R^3\*((27136\*P+8649)^(1/2)/13568-93/13568)^2/1000000

W is yield in megatons of TNT, R is radius in meters, and P is pressure of the shockwave in bars, the standard overpressure is 20 psi or 1.37895 bars.

W = 2000000^3\*((27136\*1.37895+8649)^(1/2)/13568-93/13568)^2/1000000 = 642947485.59 Megatons of TNT, or 642.95 Teratons of TNT (**Large Country level+**)

[Calculation 2]: This calculation is for destroying Continental United States; I found that that Australia isn't largely considered a proper continent on its own worldwide. So, I chose the Continental United States since it would approximate a small continent. This case could also apply for a similar feat in the LN since Leon can sink a continent larger than Australia.

Diameter of United States (contiguous) = 2800 miles or 4506.163 km.

Explosion radius = 2253.0815 km or 2253081.5 m

W = R^3\*((27136\*P+8649)^(1/2)/13568-93/13568)^2/1000000

W is yield in megatons of TNT, R is radius in meters, and P is pressure of the shockwave in bars, the standard overpressure is 20 psi or 1.37895 bars.

W =2253081.5^3\*((27136\*1.37895+8649)^(1/2)/13568-93/13568)^2/1000000 = 9.1921e8 Megatons of TNT, or 919.21 Teratons of TNT (**Continental level**)

New: [Calculation 3]: This method uses a method proposed by Ugarik where Antarctica is used as basis.

Area of Antarctica = 5.483 million mi² = 1.42e+13 m²

R = (Area/pi)^0.5

R = (1.42e13/3.14)^0.5 = 2.12e6 m

Explosion radius = 2.12e6 m

W = R^3\*((27136\*P+8649)^(1/2)/13568-93/13568)^2/1000000

W is yield in megatons of TNT, R is radius in meters, and P is pressure of the shockwave in bars, the standard overpressure is 20 psi or 1.37895 bars.

W = 2.12e6^3\*((27136\*1.37895+8649)^(1/2)/13568-93/13568)^2/1000000 = 7.65e8 Megatons of TNT, or 765 Teratons of TNT (**Continental level**)

### **Destroying wooden crates**

All crates can be found here and here

Normally those crates are made using yellow pine trees, which are also know for Ponderosa Pine trees, the density of those trees tend to be 420 kg/m^3, for the destruction, I'll use the values on this site we have the following destruction values:

* Frag: 3.99896 j/cc
* V. Frag: 7.791076 j/cc
* Pulv: 36.68011 j/cc

To save everyone's time, I'll give the measure in cm^3

I'll start with the standard wood crates, then I'll go the heavy ones

#### Equal measures

##### 12 inches (30.48 cm)

Those crates weight 3.62874 Kg, so that gives us 8639.8571428571 cm^3

* Frag: 34.55044312 Kilojoules, **Wall level**
* V. Frag: 67.31378363 Kilojoules, **Wall level**
* Pulv: 316.91091038 Kilojoules, **Wall level**

##### 18 inches (45.72 cm)

Those crates weight 6.80389 Kg, so that gives us 16199.7380952381 cm^3

* Frag: 64.7821046533 Kilojoules, **Wall level**
* V. Frag: 126.21339068009 Kilojoules, **Wall level**
* Pulv: 594.2081753 Kilojoules, **Wall level**

##### 24 inches (60.96 cm)

Those crates weight 10.8862 Kg, so that gives us 25919.5238095238 cm^3

* Frag: 103.651138933333295248 Kilojoules, **Wall level**
* V. Frag: 201.9409798838094496088 Kilojoules, **Wall level**
* Pulv: 950.730984480952031618 Kilojoules, **Wall level**

##### 30 inches (76.2 cm)

Those crates weight 21.7724 Kg, so that gives us 51839.0476190476 cm^3

* Frag: 207.302277866666590496 Kilojoules, **Wall level**
* V. Frag: 403.8819597676188992176 Kilojoules, **Wall level**
* Pulv: 0.000454460317629518 Tons of TNT, **Wall level**

##### 36 inches (91.44 cm)

Those crates weight 28.5763 Kg, so that gives us 68038.8095238095 cm^3

* Frag: 272.08447773333323812 Kilojoules, **Wall level**
* V. Frag: 530.095535949523624022 Kilojoules, **Wall level**
* Pulv: 0.000596479688719498 Tons of TNT, **Wall level**

##### 40 inches (1.016 m)

Those crates weight 38.1018 Kg, so that gives us 90718.5714285714 cm^3

* Frag: 362.779938399999885744 Kilojoules, **Wall level**
* V. Frag: 706.7952846114283488264 Kilojoules, **Wall level**
* Pulv: 0.000795307643174679 Tons of TNT, **Wall level**

##### 48 inches (1.2192 m)

Those crates weight 48.5344 Kg, so that gives us 115558.0952380952 cm^3

* Frag: 462.112200533333180992 Kilojoules, **Wall level**
* V. Frag: 900.3219024152377984352 Kilojoules, **Wall level**
* Pulv: 0.001013069704761904 Tons of TNT, **Wall level**

##### 60 inches (1.524 m)

Those crates weight 73.0284 Kg, so that gives us 173877.1428571429 cm^3

* Frag: 695.327739200000171384 Kilojoules, **Wall level**
* V. Frag: 0.0003237786889729583 Tons of TNT, **Wall level**
* Pulv: 0.001524338605756624 Tons of TNT, **Wall level**

#### 2 equal measures

##### 2 20 inches (50.8 cm), 1 24 inches (60.96 cm)

Those crates weight 9.07185 Kg, so that gives us 21599.6428571429 cm^3

* Frag: 86.376107800000171384 Kilojoules, **Wall level**
* V. Frag: 168.2844590728574767604 Kilojoules, **Wall level**
* Pulv: 792.277275960715857719 Kilojoules, **Wall level**

##### 2 24 inches (60.96 cm), 1 28 3/4 inches (73.025 cm)

Those crates weight 14.515 Kg, so that gives us 34559.5238095238 cm^3

* Frag: 138.202153333333295248 Kilojoules, **Wall level**
* V. Frag: 269.2558765238094496088 Kilojoules, **Wall level**
* Pulv: 0.000302974936635027 Tons of TNT, **Wall level**

##### 2 24 inches (60.96 cm), 1 32 inches (81.28 cm)

Those crates weight 13.6078 Kg, so that gives us 32399.5238095238 cm^3

* Frag: 129.564399733333295248 Kilojoules, **Wall level**
* V. Frag: 252.4271523638094496088 Kilojoules, **Wall level**
* Pulv: 0.00028403874217996 Tons of TNT, **Wall level**

##### 2 22 inches (55.88 cm), 1 36 inches (91.44 cm)

Those crates weight 14.9685 Kg, so that gives us 35639.2857142857 cm^3

* Frag: 142.520077999999942872 Kilojoules, **Wall level**
* V. Frag: 277.6683835857141744132 Kilojoules, **Wall level**
* Pulv: 0.000312440946539538 Tons of TNT, **Wall level**

##### 2 30 inches (76.2 cm), 1 48 inches (1.2192 m)

Those crates weight 23.1332 Kg, so that gives us 55079.0476190476 cm^3

* Frag: 220.258908266666590496 Kilojoules, **Wall level**
* V. Frag: 429.1250460076188992176 Kilojoules, **Wall level**
* Pulv: 0.000482864609312119 Tons of TNT, **Wall level**

##### 2 48 inches (1.2192 m), 1 28 inches (71.12 cm)

Those crates weight 39.4625 Kg, so that gives us 93958.3333333333 cm^3

* Frag: 375.735616666666533368 Kilojoules, **Wall level**
* V. Frag: 732.0365158333330736308 Kilojoules, **Wall level**
* Pulv: 0.000823709847534257 Tons of TNT, **Wall level**

##### 2 48 inches (1.2192 m), 1 60 inches (1.524 m)

Those crates weight 57.6062 Kg, so that gives us 137157.619047619 cm^3

* Frag: 548.48783226666647624 Kilojoules, **Wall level**
* V. Frag: 0.000255402828388874 Tons of TNT, **Wall level**
* Pulv: 0.00120242747466653 Tons of TNT, **Wall level**

##### 2 60 inches (1.524 m), 1 48 inches (1.2192 m)

Those crates weight 68.946 Kg, so that gives us 164157.1428571429 cm^3

* Frag: 656.457848000000171384 Kilojoules, **Wall level**
* V. Frag: 0.0003056789617454248 Tons of TNT, **Wall level**
* Pulv: 0.001439125730708823 Tons of TNT, **Wall level**

##### 2 30 inches (76.2 cm), 1 67 inches (1.7018 m)

Those crates weight 42.6377 Kg, so that gives us 101518.3333333333 cm^3

* Frag: 405.967754266666533368 Kilojoules, **Wall level**
* V. Frag: 790.9370503933330736308 Kilojoules, **Wall level**
* Pulv: 0.000889986528126991 Tons of TNT, **Wall level**

##### 2 24 inches (60.96 cm), 1 72 inches (1.8288 m)

Those crates weight 39.9161 Kg, so that gives us 95038.3333333333 cm^3

* Frag: 380.054493466666533368 Kilojoules, **Wall level**
* V. Frag: 740.4508779133330736308 Kilojoules, **Wall level**
* Pulv: 0.000833177944761791 Tons of TNT, **Wall level**

##### 2 36 inches (91.44 cm), 1 72 inches (1.8288 m)

Those crates weight 57.1526 Kg, so that gives us 136077.619047619 cm^3

* Frag: 544.16895546666647624 Kilojoules, **Wall level**
* V. Frag: 0.000253391747585814 Tons of TNT, **Wall level**
* Pulv: 0.001192959377439 Tons of TNT, **Wall level**

##### 2 48 inches (1.2192 m), 1 72 inches (1.8288 m)

Those crates weight 66.2245 Kg, so that gives us 157677.380952381 cm^3

* Frag: 630.54553933333352376 Kilojoules, **Wall level**
* V. Frag: 0.000293612920287035 Tons of TNT, **Wall level**
* Pulv: 0.00138231923466664 Tons of TNT, **Wall level**

##### 2 24 inches (60.96 cm), 1 84 inches (2.1336 m)

Those crates weight 45.8128 Kg, so that gives us 109078.0952380952 cm^3

* Frag: 436.198939733333180992 Kilojoules, **Wall level**
* V. Frag: 849.8357299352377984352 Kilojoules, **Wall level**
* Pulv: 0.000956261121396704 Tons of TNT, **Wall level**

##### 2 48 inches (1.2192 m), 1 84 inches (2.1336 m)

Those crates weight 90.7185 Kg, so that gives us 215996.4285714286 cm^3

* Frag: 863.761078000000114256 Kilojoules, **Wall level**
* V. Frag: 0.0004022095102123737 Tons of TNT, **Wall level**
* Pulv: 0.001893588135661363 Tons of TNT, **Wall level**

#### 3 different measures

##### 24 (60.96 cm), 20 (50.8 cm) and 25 (63.5 cm) inches

Those crates weight 13.1542 Kg, so that gives us 31319.5238095238 cm^3

* Frag: 125.245522933333295248 Kilojoules, **Wall level**
* V. Frag: 244.0127902838094496088 Kilojoules, **Wall level**
* Pulv: 0.000274570644952426 Tons of TNT, **Wall level**

##### 41 (1.0414 m), 28 3/4 (73.025 cm) and 25 1/2 (64.77 cm) inches

Those crates weight 24.494 Kg, so that gives us 58319.0476190476 cm^3

* Frag: 233.215538666666590496 Kilojoules, **Wall level**
* V. Frag: 454.3681322476188992176 Kilojoules, **Wall level**
* Pulv: 0.000511268900994719 Tons of TNT, **Wall level**

##### 48 (1.2192 m), 30 (76.2 cm) and 35 (88.9 cm) inches

Those crates weight 29.9371 Kg, so that gives us 71278.8095238095 cm^3

* Frag: 285.04110813333323812 Kilojoules, **Wall level**
* V. Frag: 555.338622189523624022 Kilojoules, **Wall level**
* Pulv: 0.000624883980402098 Tons of TNT, **Wall level**

##### 48 (1.2192 m), 30 (76.2 cm) and 40 (1.016 m) inches

Those crates weight 27.6691 Kg, so that gives us 65878.8095238095 cm^3

* Frag: 263.44672413333323812 Kilojoules, **Wall level**
* V. Frag: 513.266811789523624022 Kilojoules, **Wall level**
* Pulv: 0.000577543494264431 Tons of TNT, **Wall level**

##### 48 (1.2192 m), 30 (76.2 cm) and 42 (1.0668 m) inches

Those crates weight 40.3697 Kg, so that gives us 96118.3333333333 cm^3

* Frag: 384.373370266666533368 Kilojoules, **Wall level**
* V. Frag: 748.8652399933330736308 Kilojoules, **Wall level**
* Pulv: 0.000842646041989324 Tons of TNT, **Wall level**

##### 48 (1.2192 m), 35 (88.9 cm) and 45 (1.143 m) inches

Those crates weight 44.4521 Kg, so that gives us 105838.3333333333 cm^3

* Frag: 423.243261466666533368 Kilojoules, **Wall level**
* V. Frag: 824.5944987133330736308 Kilojoules, **Wall level**
* Pulv: 0.000927858917037125 Tons of TNT, **Wall level**

##### 58 (1.4732 m), 42 (1.0668 m) and 46 (1.1684 m) inches

Those crates weight 58.967 Kg, so that gives us 140397.619047619 cm^3

* Frag: 561.44446266666647624 Kilojoules, **Wall level**
* V. Frag: 0.000261436070798051 Tons of TNT, **Wall level**
* Pulv: 0.00123083176634913 Tons of TNT, **Wall level**

#### Heavy Duty Crates

##### 36 inches (91.44 cm)

Those crates weight 56.699 Kg, so that gives us 134997.619047619 cm^3

* Frag: 539.85007866666647624 Kilojoules, **Wall level**
* V. Frag: 0.000251380666782755 Tons of TNT, **Wall level**
* Pulv: 0.00118349128021146 Tons of TNT, **Wall level**

##### 2 48 (1.2192 m) and 1 24 (60.96 cm) inches

Those crates weight 66.2245 Kg, so that gives us 157677.380952381 cm^3

* Frag: 630.54553933333352376 Kilojoules, **Wall level**
* V. Frag: 0.00029361269417224 Tons of TNT, **Wall level**
* Pulv: 0.00138231923466664 Tons of TNT, **Wall level**

##### 48 inches (1.2192 m)

Those crates weight 99.7903 Kg, so that gives us 237595.9523809524 cm^3

* Frag: 950.136709733333409504 Kilojoules, **Wall level**
* V. Frag: 0.0004424302395536284 Tons of TNT, **Wall level**
* Pulv: 0.002082945905565989 Tons of TNT, **Wall level**

##### 2 48 (1.2192 m) and 1 80 (2.032 m) inches

Those crates weight 134.717 Kg, so that gives us 320754.7619047619 cm^3

* Frag: 0.000306569183237731 Tons of TNT, **Wall level**
* V. Frag: 0.0005972812445893654 Tons of TNT, **Wall level**
* Pulv: 0.002811978955470955 Tons of TNT, **Wall level+**

##### 2 48 (1.2192 m) and 1 96 (2.4384 m) inches

Those crates weight 179.169 Kg, so that gives us 426592.8571428571 cm^3

* Frag: 0.000407726522944551 Tons of TNT, **Wall level**
* V. Frag: 0.0007943636164094509 Tons of TNT, **Wall level**
* Pulv: 0.003739835785185058 Tons of TNT, **Wall level+**

##### 96 (2.4384 m), 48 (1.2192 m) and 60 (1.524 m) inches

Those crates weight 183.705 Kg, so that gives us 437392.8571428571 cm^3

* Frag: 0.00041804888623327 Tons of TNT, **Wall level**
* V. Frag: 0.0008144744244400436 Tons of TNT, **Wall level**
* Pulv: 0.003834516757460393 Tons of TNT, **Wall level+**

## **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 × 1046**  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: **1.1184707e+42 J** or **267.320913002 Quettatons of TNT** **(4-C, 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)

## **Mass-energy Conversion Feats - Energy Constructs**

Although 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 used process. So please familiarise yourself regarding the criteria before applying the table below.

### **Mass-energy Conversion - The Tally**

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**)

## **Forest Feats**

### **Destroying/Creating a forest**

The destruction value for a Ceiba pentandra tree is 9.1700460e7 Joules, which will be a low end for destruction.

The destruction value for a Dalbergia nigra tree is 6.1832194e8 Joules, this will be a high end for destruction.

For creating a tree, a white oak tree will be used since they are somewhat common and are not overly large. A White Oak = 30 m height, 1.27 meter diameter. Plugging this into the formula for volume of a cylinder since tree trunks are cylindrical gives a volume of 38 m^3. This website says that the density of oak wood is 704 kg/m^3, multiply that by the volume gives a mass of 26,582 kg per tree.

According to this site, a forest has to be at least 1.24 acres in area.

This site provides information on how many trees are in an acre of land; I will be using reforestation values as those make the most sense.

Now, I will use the 10 x 10 value for reforestation, so that is 435 trees per acre as a minimum.

1.24 acres x 435 trees/acre = 538.4 trees.

**Energy to destroy a small forest**

* **Low-End:** 538.4 x 9.1700460e7 Joules = 4.946e10 Joules (**City Block level**)
* **High-End:** 538.4 x 6.1832194e8 = 3.335e11 Joules (**City Block level+**)

**Energy to create a small forest**

538.4 x 26,752 kg = 1.44e7 kg, which according to the table on the Creation Feats page is (**Small Building level**)

### **Splitting a forest with pulverization**

Using Jasonsith's values for sloping a tree, this is the energy to split every tree in half.

**Low-End:** Small trees: 12,622.28361 Joules

**High-End:** Big trees: 201,956.5378 Joules

**Number of Trees:** 538.4

**Low-End Total:** 6,795,837.495624 Joules (**Wall level**)

**High-End Total:** 108,733,399.95152 Joules (**Small Building level**)

### **Splitting a forest with shearing**

This is the value for splitting a forest via shearing the trees.

**First some data:**

Ultimate Tensile Strength of Oak Wood: 5.50 MPa

Shear Strength = 0.6 x Ultimate Tensile Strength

Using Jasonsith's diameters for the trees: 1.27m, and .3175m respectively.

* Area of Circle: pi\*r2
* Number of trees: 538.4

**Calculation to shear tree:**

* Area of the tree cut: pi\*r2 (diameter / cos (30 degrees) / 2)
* Small Tree: 0.091421... (rounded to 0.09142m2)
* Big Tree: 1.462738... (rounded to 1.46274m2)
* 0.09142m2 x 5,500,000 Pa x 0.6 = 301,686 Newtons
* 1.46274m2 x 5,500,000 Pa x 0.6 = 4,827,042 Newtons

As the entire diameter of the tree was sheared, this is what the force is multiplied by as the entire cut needs to follow through the entire diameter of the tree despite slowing down due to the friction caused by the wood during cutting:

* Small Tree: 301,686N x .3175m = 95,785.305 Joules (**Wall level**)
* Big Tree: 4,827,042N x 1.27m = 6,130,343.24 Joules (**Wall level**)

**Calculation to split a forest via shearing:**

* **Low-End:** Small trees: 51,570,808.21 Joules (**Small Building level**)
* **High-End:** Big trees: 3,300,576,800.416 Joules (**Building level**)

## **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**

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

**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.

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 (0.00275 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 / 0.00275) \* 6.35 = 1016000000 m/s; **3.389c** (**FTL**)

Viewing Light as Standing Still

* Speed of Light = 299792458 m/s
* Speed to View Light as Standing Still = (299792458 / 0.00275) \* 6.35 = 692248039381.81 m/s; **2309.09c** (**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 = **30021 c** (**Massively FTL+**)

1 year

* Speed = **2501 c** (**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 = **253878.18773 c** (**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, *Ve* refers to the speed relative to the central body (for example the sun), whereas *Vte* is the speed (at the visible surface of the smaller body) relative to the smaller body (planet or moon).

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:

* At speeds of **950 to 1,000 km/h (590 to 620 mph)** the air flow around the aircraft reaches the speed of sound, and it is reported that the control surfaces no longer affect the direction of flight

### **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.

### **Speed Required for a Human to Run on Water**

**Description:** Speed needed for a human to run on water.

**Requirements:** The character in question has to be running on water without using an underwater bridge, Earth gravity, Mass = 91.71 kg, foot surface area = 330 cm^2, 0.25 m foot depth, 0.234 s to take a step.

**Results:** 30 m/s (**Superhuman**) travel speed with foot strike speed of 48.5 m/s (**Subsonic**)

**Calculation:**

So, taken from here...

A guy wearing special water-repellent shoes weighs about **200 lbs (91.71 kg)** while *g* is the acceleration of gravity. The surface tension of water is *gamma = 0.07 N/m*, and putting numbers in, you find a contact-line perimeter of **12,000 meters long**.

This means each shoe should be **a mile long** and **a mile wide**.

Jamaican runner Usain Bolt, the current world record holder for the 100-meter sprint, ran **10.4 meters per second**. But J.W. Glasheen and T.A. McMahon, two Harvard biologists who studied how the basilisk runs on water, found that in order to mimic the lizard, a human would need to run at almost **30 meters per second**, "a velocity beyond human ability." A man would also need "an average power output almost **15 times greater than the maximum sustained power output for humans**.

Final result = **30 m/s (Superhuman)**

(The foot strike speed is deduced from the jscalc.io online calculator using the parameters set out on the "Requirements" section)

## **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: 1.75 m.

Width: 61 cm, 30.5 for the radius.

So, the volume is 5.11e5 cubic centimeters.

cc refers to cubic centimeters.

Soil has a frag energy of 0.15 J/cc, a violent frag energy of 0.2 J/cc and a pulverization energy of 0.4-1 J/cc.

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, or 0.00098 Tons of TNT, **Wall level**

Violent fragmentation:

5.11e5\*69 = 3.5259e7 Joules, or 0.00843 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 soil instead of hard rock:

Fragmentation:

5.11e5\*0.15 = 7.6714747357545e+4 Joules, **Wall level**

Violent fragmentation:

5.11e5\*0.2 = 1.0228632981006e+5 Joules, **Wall level**

Pulverization:

5.11e5\*0.4-1 = 2.0457265962012-5.114316490503e+5 Joules **Wall level**

5.11e5\*208 = 1.06288e8 Joules, or 0.0254 Tons of TNT, **Small Building level**

Violent fragmentation:

5.11e5\*568.5 = 2.905035e8 Joules, or 0.06943 Tons of TNT, **Small Building level**

Pulverization:

5.11e5\*1000 = 5.11e+8 Joules, or 0.1221319 tons of TNT, **Small Building level**

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.

### **Hurling somebody through the air before they hit the ground**

**Description:** It involves a person being attacked and forcibly propelled through the air before ultimately landing on 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 energy required to send a person flying at a speed across a distance using the Newtonian energy model.

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

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 humans 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

### **Throwing a Person to the Horizon**

A common gag in fiction is to punch or throw a person so far that they reach the horizon or fly out of sight, often with a comedic twinkle to accompany this.

Note that this calculation calculates the feat utilizing realistic projectile physics due to the lack of the timeframe, and the specific results may increase greatly depending on the shown speed of the feat. Additionally, this is calculating using the weight and size of a normal human (1.7 m and 70 kg), under good visibility conditions and using the perfect throwing angle. The result may also vary if those factors change.

Ackerman's studies report that the resolution limit of a normal person's view is 5 x 10^-4 rad, which we will double, using 10^-3 rad, to account for minute imperfections and issues that would lower this result. Additionally, he also reports that people with amazing vision under perfect circumstances have a resolution limit of 2 x 10^-4 rad, which we will double (4 x 10^-4) again to glean peak human vision range under more common circumstances. Finally, 2 x 10^-4 rad will also be used for superhuman vision. To glean how far an object can be seen using this method one must simply divide the size of the object by the radians, and the result will be in the unit used. A 3 meter object observed with a vision resolution of 10^-3 rad can be seen at up to 3 / 10^-3 = 3000 meters of distance.

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

* Normal human vision
	+ 1.7 / 10^-3 = 1700
	+ V = sqrt(1700\*9.81) = 129.14 m/s
	+ KE = 129.14^2 x 0.5 x 70 = 583,690 Joules, Wall level
* Peak human vision
	+ 1.7 / (4 x 10^-4) = 4250
	+ V = sqrt(4250\*9.81) = 204.19 m/s
	+ KE = 204.19^2 x 0.5 x 70 = 1.4592e6 Joules, Wall level
* Superhuman vision
	+ 1.7 / (2 x 10^-4) = 8500
	+ V = sqrt(8500\*9.81) = 288.77 m/s
	+ KE = 288.77^2 x 0.5 x 70 = 2.9184e6 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 doesn'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.

### **Punching through a car roof**

The roof of a car is normally 20 gauge, or 0.09525 centimeters

the surface area of a human fist is 25cm2

25cm2 x 0.09525cm = 2.38125cm3

some of the more commonly used materials for car roofs are steel and aluminum

**Steel roofs:**

the frag value of steel is 208j/cm3 while the v.frag value is 568.5j/cm3 and the pulv value is on average 655j/cm3

(frag) 2.38125cm3 x 208j = 495.3 Joules (**Street level**)

(v.frag) 2.38125cm3 x 568.5j = 1353.74 Joules (**Street level**)

(pulv) 2.38125cm3 x 655j = 1559.71 Joules (**Street level**)

**Aluminum roofs:**

the frag value of aluminum is 48.75j/cm3 while the v.frag value is 234j/cm3 and the pulv value is 280j/cm3

(frag) 2.38125cm3 x 48.75j = 116.085 Joules (**Athlete level**)

(v.frag) 2.38125cm3 x 234j = 557.2125 Joules (**Street level**)

(pulv) 2.38125cm3 x 280j = 666.75 Joules (**Street level**)

### **Punching through a skull**

Gonna calc the energy needed to punch through a skull

Average skull thickness: 6.5 millimeters or 0.65 centimeters

Average surface area of fist: 25cm^2

25cm^2 x 0.65 = 16.25^3

Shear strength of bone: 51.6 MPa

Compressive strength of bone: 170 MPa

(Fragmentation) 16.25 x 51.6 = 838.5 Joules (**Street level**)

(Pulverization) 16.25 x 170 = 2762.5 Joules (**Street level**)

### **Shaking the entire Universe**

Surface area of the universe is 2.4320086e+54m². This calculation assumes that the energy spreads across the universe like an earthquake. Said energy is listed out on the Earthquake Calculations page.

* Magnitude 2: 6.309573e+7 times 2.4320086e+54, **1.5344936e+62 Joules**/**3.66752772e+52 Tons of TNT**/**1.53449357983 Exafoe**, **Multi Solar System level**
* Magnitude 3: 1.995262e+9 times 2.4320086e+54, **4.8524943e+63 Joules**/**1.1597739e+54 Tons of TNT**/**48.5249434325 Exafoe**, **Multi Solar System level**
* Magnitude 4: 6.309573e+10 times 2.4320086e+54, **1.5344936e+65 Joules**/**3.66752772e+55 Tons of TNT**/**1.53449357983 Zettafoe**, **Multi Solar System level**

### **Illuminating the entire Universe**

Utilizing the apparent magnitude of Sirius, the brightest star in the Night-Sky, and this formula, with the distance being 46.5 billion light-years (4.399e26 m).

**2.747966869e+47 Joules**/**6.56779844e+37 Tons of TNT**/**2.75 KiloFOE**, **Solar System level**

### **Shaking the Moon**

**Note:** It is highly encouraged to go over the modified intensity listing based on the Mercalli Scale first before choosing any of these. But if it's clear that the exact intensity cannot really be determined, assume that the scale is either in between 4 or 5 based on typical earthquakes.

### **Mercalli Scale**

**Formula:**

* 0.5\*(Mass)\*(Velocity)^2 = Energy in Joules
	+ The moon has a mass of 7.342×10^22 kg

### **Shaking the Solar System**

https://en.wikipedia.org/wiki/Peak\_ground\_acceleration#Comparison\_of\_instrumental\_and\_felt\_intensity
https://en.wikipedia.org/wiki/Modified\_Mercalli\_intensity\_scale#Modified\_Mercalli\_intensity\_scale

This is step one.

Mercalli Intensity IV.

So 1.41 cm/s or 0.0141 m/s.

Mass of Neptune is 1.024e+26 kg. And a radius of 2.4622e+7 m.

Shaking Neptune like this: 0.5 \* 1.024e+26 \* 0.0141^2 = 1.0179072e+22 J

Sun to Neptune distance (Semi-major axis) is 4.5 billion km or 4.5e+12 m

E = 4 \* U \* (Er/Tr)^2, U is energy in joules, Er is explosive radius and Tr is Target Radius. The two radii must be in the same units, but what units you use doesn't matter.

E = 4 \* 1.0179072e+22 \* (4.5e12 / 2.4622e7)^2 = 1.3600239e+33 J or 325.053513384 zettatons of TNT (**5-B, Planet level**)

### **Typical fighting stats**

#### Typical fighting stats - punching

* Some punching speed - Source 1
* Some punching speed - Source 2

Normal human punching speed (scaled to punching speed of a normal researcher) = 15 miles per hour = 6.7056 m/s (**Average human**)

Athletic human punching speed (scaling to average punching speed of Hatton) = 25 mph = 11.176 m/s (**Peak human**)

Some low-end peak human punching speed (scaling to peak punching speed of Hatton) = 32 mph = 14.30528 m/s (**Superhuman**)

~~A higher peak human punching speed (scaling to peak punching speed of Keith Liddell) = 44 mph = 19.66976 m/s (**Superhuman**)~~

A higher peak human punching speed (scaling to peak punching speed of Keith Liddell) = 45 mph = 20.1168 m/s (**Superhuman**)

#### Typical fighting stats - sword striking

* Some sword striking speed - Source 1
* Some sword striking speed - Source 2
* Some sword striking speed - Source 3

Athletic human sword slashing speed (single strike by engineer Sean Franklin) = 70 km/h ~= 19.4444 m/s (**Superhuman**)

Peak human sword stabbing speed (single strike by ajslim @quarte-riposte.com) = 80 mph ~= 35.7632 m/s (**Subsonic**)

Peak athletic human sword slashing speed (single strike by Isao Machii) =
At least 158.29 km/h ~= 43.96944444 m/s (**Subsonic**)
Possibly 350 km/h ~= 97.22222222 m/s (**Subsonic**)

#### Typical fighting stats - knife throwing

* Some knife throwing speed - Source 1
* Some knife throwing speed - Source 2

Low end athletic human knife throwing speed = 26 mph = 41.842944 kph = 11.62304 m/s (**Peak Human**)

High end athletic human knife throwing speed = 30 mph = 48.28032 kph = 13.4112 m/s (**Superhuman**)

Low end peak human knife throwing speed = 33.88958482 mph = 54.54 kph = 15.15 m/s (**Superhuman**)

Mid end peak human knife throwing speed = 35.79098067 mph = 57.6 kph = 16 m/s (**Superhuman**)

High end peak human knife throwing speed = 37.69237652 mph = 60.66 kph = 16.85 m/s (**Superhuman**)

#### Typical fighting stats - baseball throwing

* Baseball travel speed - Source from Guinness

Peak human baseball throwing speed = 105.1 mph = 169.1420544 kph = 46.983904 m/s (**Subsonic**)

#### Typical fighting stats - baseball batting

* Baseball bat swinging speed - Source from Quora

"Average" professional baseballer bat swinging speed =
(Low tier) 70 mph = 31.2928 m/s (**Superhuman+**)
(Mid tier) 75 mph = 33.528 m/s (**Superhuman+**)
(High tier) 80 mph = 35.7632 m/s (**Subsonic**)

* Baseball bat swinging speed - Source from Guinness

Peak human baseball bat swinging speed = 108.1185874 mph = 174 kph = 48.33333333 m/s (**Subsonic**)

#### Typical fighting stats - gun quick drawing and shooting

This site says:

Average shooter can fire Colt Model 1911 bullets at 70-85 rounds per minute, i.e., 0.857142857 second/round to 0.705882353 second/round. (**"Below average human" perception**)

Average "enthusiastic shooter" can fire Colt Model 1911 bullets at 120-180 rounds per minute, i.e., 0.5 second/round to 0.3333 second/round. (**"Below average human" perception**)

Expert speed-shooters can go 420-430 rounds per minute, i.e., 0.142857143 second/round to 0.139534884 second/round. (**Normal Human+ perception**)

Jerry Miculek himself emptied a five-shot revolver in 0.57 seconds in a group the size of a playing card, making the firing frequency 0.114 second/round (**Athletic human perception**).

For extended shots, he also did fire 27 rounds through a 9mm version of the 1911 in just 3.7 seconds or about 0.137037 second/round (**Normal Human+ perception**). This is for extended pistol trigger pulling.

Bob Munden Fastest Quick Draw (Shooting 2 target 6 feet apart within 1/10 second)

Peak human gun drawing speed = 1/20 second/round (**Superhuman perception**)

He himself claimed that he can shoot at within 2/100 sec/round (**Subsonic perception**) but that is unsupported by scientific recording.

### **Other real life statistics**

#### Other real life statistics - Faster than Eye

An MIT findings says:

In the past, normal humans are perceived to perceive images in 1/10 second. (**Peak Human** perception)

A 2014 finding says peak humans are able to perceive images in 13/1000 second. (**Subsonic** perception)

Another unsupported research claims that

UK Air force pilots were able to recognize an image of a plane that was flashed on screen for as little as 1/220th of a second. (**Subsonic+** perception)

**Note:** It has been decided that none of those values should be used for calculations regarding faster than eye speed feats.

#### Other real life statistics - Walking

Wikipedia says. Typical average walking speed is 1.4 m/s and hastier average walking speed reaches 2.5 m/s. Scales to every character who runs as fast as a human can walk (mostly small-sized characters). Also useful on calculating cinematic time.

#### Other real life statistics - Typing

Typing speed of an average human is 40 wpm or 200 characters per minute, i.e., 3.33 key strikes per second (ks/s) (**Below average human** perception)

Typing speed of professional typists reach 60 to 75 wpm or 300 to 375 characters per minute, i.e., 5 ks/s to 6.25 ks/s (**Average human** perception)

Here says:

The highest typing speed ever recorded was 216 wpm (1080 characters per minute or 18 ks/s, **Superhuman**), set by Stella Pajunas in 1946, using an IBM electric typewriter.

Currently, the fastest English language typist is Barbara Blackburn, who reached a peak typing speed of 212 wpm (1060 characters per minute or 17.66666667 ks/s, **Peak human+**) during a test in 2005, using a Dvorak simplified keyboard.

Here says:

It takes 0.25 N ~ 1.5 N for a keyboard strike.

Assume 1 cm travel distance to make a complete strike,

Energy to press a key/button:

Low end: = 0.25 \* 0.01 = 0.0025 J (**Below average human**)

High end: = 1.5 \* 0.01 = 0.015 J (**Below average human**)

Lifting strength (in kg on earth):

Low end: 0.0025 / 9.81 = 0.000254842 kg = 0.254842 gram (**Below average human**)

High end: 0.001529052 / 9.81 = 0.001529051988 kg = 1.529051988 gram (**Below average human**)

Use for enemies that can be defeated by literally pressing a button. Or, for giant-/mini- sized monster rescaling calculations.