Assorted calculation request processing page 2020-04-30

Ongoing Request Welcome

Executive Summary

Assorted calculation request processing page since 2020-04-30

Requests picked by me and put into a blog for approval.

To mods: Please mention which feat calc to approve or I will assume you accept everything.

DC Comics - Some New Gods Smashed Two Planets

ManofSteel1985 wrote:

Can somebody calculate this feat ?

ManofSteel1985 wrote:

I was planning to do such a post a long time ago, but i couldn't (because i'm lazy lol). Anyway, new calculations are required for DC SS tiers, so i made a calculation for this.

Orion fought against an amped Kalibak and Metron stated that if their full might was unleashed they could lay waste to half of Apokolips.

According to this calc diameter of the Apokolips is 778,698,148,700 m.

V = (4/3) * pi * R^3 = 2.47232778+35 m^3

I'll assume that density of the Apokolips is 5500 kg/m^3 (density of the Earth).

So, mass of the Apokolips is 1.35978028e+39 kg

GBE = (3G * M^2) / 5R = 1.90053300e+56 J.

4.75133250E+55 J (475.13 GigaFoe) to each of them. (The two New Gods)

DC Comics - Superman Tanked a Smash of Two Planets

ManofSteel1985 wrote:

If it doesn't mean literally busting half the planet, then we can use this feat .

Qawsedf234 wrote:

Spinosaurus75DinosaurFan wrote:
This only scales to the Flash characters I believe?

No, this scales to the vast majority of DC characters at the moment.

As for the Apok stuff

• Lay waste =/= bust. It just means to lay waste. APok is a massive world city, destroying the building would be enough to lay waste to it

• Superman did withstand Apok and New Gensis ramming into him. But there's an important issue to consider, Superman is human sized and those things are planet sized. He's exposed to such a small fraction of a fraction of a fraction in terms of energy none of the results would be good.

As an example of the second point

• Assuming that in the 4th world the planets are the same ratio as a human to Earth. 2 meters^2 (Superman) / 255,000,000,000,000^2 * 3.801066e+56 = 2.981e+42 joules

• Assuming they are not the same ratio: 2 meters^2 / 1.23616389e+35^2 * 3.801066e+56 = 6.149e+21 joules 

ManofSteel1985 wrote:

Lay waste =/= bust. It just means to lay waste. APok is a massive world city, destroying the building would be enough to lay waste to it.

Okay.

Superman did withstand Apok and New Gensis ramming into him. But there's an important issue to consider, Superman is human sized and those things are planet sized. He's exposed to such a small fraction of a fraction of a fraction in terms of energy none of the results would be good.

Okay, i understood.

If we're going to use "Larger than the largest star" statement, then GBE of the Apokolips (or New Genesis) is 15.72 TeraFoe.

Qawsedf234 wrote:

• No one destroyed Apokolips.

• Going FTL in DC =/= infinite energy. There's a different set of physics that takes over once a person breaches the light barrier. Mass no longer increases is an example of this.

• Them being sphereical means little when both of those planets would be longer than Astromical Units. He's being exposed to a fraction of the energy

In what year was that statement made?

March 1997. It's from Jack Kirby's Fourth World (1997) #1.

Blahblah9755 wrote:

So if both planets have diameters equivalent to the radius of UY Scuti, their radius is 854 solar radii. 

This means they’d have a volume of 8.8126856*10^35 m^3

The new planet would have a volume of 1.7625371 * 10^36 m^3 and radius of 749347425000m

The maximum area of contact for the two planets would be the area of a circle with that radius, or 1.764072*10^24 m^2

Averaging a few different human surface area calculators Superman should have a surface area of 2.355m^2, half of that is 1.1775 m^2

The fraction of half Superman’s area to the total contact area is 1/(9.9384339*10^23). This would be the minimum fraction of the energy he received in the impact.

For the maximum fraction of the energy he could have received, I’ll assume he took all the energy of the portion of the mass of the planets that was directly aligned with him.

Half Superman’s area* the diameter of the planets is 2800916235600m^3

That’s 1/(3.1463581*10^23) the total volume, and if the planets have uniform density, that’s the fraction of the total energy of the collision which would impact Superman. Since the section of mass passes through the center, it’s average density would be higher than the whole planet’s average, but I don’t think there’s a way to reasonably estimate by how much. 

Since the two planets merged, the fraction of the energy Superman endured should be closer to the minimum anyways.

More discussions here

Star Wars - Yarael Poof Planetary Feat Redux

• The original verdict for Planet level Legends Star Wars Jedi and Sith lords
• The original redux verdict by Phantom Falcon
• Support for redux by Jcphl9

Scans. Please.

• Backup plan for (Small) Planet level Legends Star Wars characters

(Yarael Poof feat debunked; now "low tier" Legends Jedi and Sith lords scales to Kyp Durron who moved a micro singularity at 13.988 petatons of TNT. This is adopted by Death Battle in Obi-Wan Kenobi vs Kakashi Hatake.)

Pokemon - Team Rocket Trio stops a Dragonite

Original calculation by DragonGamerZ913

A forum move is currently happening at the time that I'm posting this so I'm putting this here instead of as a blog post since I can't make any blog posts at the moment.

Jessie stops a Dragonite with a frying pan with help from James and Meowth. I'll be calculating the attack potency needed to perform this feat, since it is used as AP justification on their profile.

Dragonite's mass is listed in the Pokedex as 210.0 kg and is capable of moving up to 1556 MPH. 1 MPH is equal to 0.44704 m/s, so using this conversion, we find that Dragonite can move up to 695.59424 m/s.

Now, let's plug this into the kinetic energy formula. Since Dragonite is completely stopped by Team Rocket's almighty frying pan, the amount of kinetic energy it uses is equal to the amount of kinetic energy that the trio uses.

KE = 1/2 * mass * velocity^2

KE = 1/2 * 210.0 * 695.59424^2

KE = 50,804,391.41 joules

Divide by 4.184e+9 to convert from joules to tons of TNT.

50,804,391.41 / 4.184e+9 = 0.012 tons of TNT (Small Building level)

This was an effort by 3 characters, however, I'm not sure whether this calc would be used since the entire trio is under one profile, or if you'd still need to divide by 3. I'll divide by 3 anyway and just leave it to the staff members' discretion as to which calc should be used.

0.012 / 3 = 0.004 tons of TNT (Wall level+)

Now, I know that on their profile, it says this could be an outlier, but I believe it's worth using this calc since the feat is used as AP justification in the first place.

Pokemon - Team Rocket Trio stops a Dragonite - Evaluations of the Dragonite Calculation

Spinosaurus75DinosaurFan: How much time did it take to stop it?

DragonGamerZ913: I'm having a lot of trouble finding the scene in question. I only found one video of the movie, and from that one, it looks like Dragonite is stopped almost instantly.

If it's instant it should be fine Spinosaurus75 Dinosaur Fan 09:23, April 24, 2020 (UTC)

DragonGamerZ913: I'll need to get a better clip though, just to make sure I don't mess up the information. Unfortunately, such footage is VERY hard to come by

Beavis and Butthead - Fart Explosion

MistaClean wrote:
Priority: 2

Verse: Beavis and Butt-Head

Feat: https://youtu.be/4_cmvvA7PGE%7C Both shrug off a point-blank explosion at 1:31

Calculation is made by Psychomaster35

Psychomaster35 wrote:
Gladly.

Can of beans: 48 px/0.1016 meters (Average can height is 4 inches)

Fire width: 134 px/0.28363333333 meters

Fire width: 5 px/0.28363333333 meters

Explosion diameter: 520 px/29.4978666663 meters (radius is 14.7489333332 meters)

14.7489333332^3*((27136*1.37895+8649)^(1/2)/13568-93/13568)^2 is 0.25785012896 tons, just barely Building level.

You may need to ask an admin to put this in a blog, as I cant put it right now due to the forum move.

Beavis and Butthead - Fart Explosion - Jasonsith comment

Revision required:

Picture

The scene looks like a flame pillar was created, followed by massive smoke.

Flame width was fine. Problem is on the flame pillar height.

Flame pillar height = 213 px = 12.08278 m

I shall make 4 sets of situations: first pair of comparison is a cone volume of fire versus a cylinder volume of fire. Second pair of comparison is normal 20.95% oxygen composition in air versus 100% oxygen composition in air.

Oxygen density is 1.3311 kg/m^3 at room temperature and pressure.

Fire releases 418 000 Joules of energy per 32 grams of oxygen, or 13,062,500 J/kg of oxygen.

Situation	Cone 20.95% oxygen	Cylinder 20.95% oxygen	Cone 100.00% oxygen	Cylinder 100.00% oxygen
Volume of air in flame (m^3)	0.254477877	0.76343363	0.254477877	0.76343363
Volume of oxygen in flame (m^3)	0.053313115	0.159939346	0.254477877	0.76343363
Oxygen mass in flame	0.070965088	0.212895263	0.338735502	1.016206505
Yield from oxygen in flame (J)	926981.457	2780944.371	4424732.492	13274197.47
Yield from oxygen in flame (ton TNT (tier))	0.000221554 (Wall)	0.000664662 (Wall)	0.001057536 (Wall)	0.003172609 (Wall+)

Lowest end to go.

Just one more caveat: This attack potency feat may not even scale to their durability since the fire does not even catch the gang.

Peppa Pig - Puddle Jumping Competition

This.

So Daddy Pig, Mister Elephant, Peppa Pig and her friends joined the puddle jumping competition.

They jumped high, stalled potential energy, and landed onto a puddle to make a splash as big as possible.

Potential energy stalled (J) = mass of participant (in kg) * Earth gravity * jumping height (in m)
OR = 0.5 * mass of participant * (launching speed in m/s)^2

By mgh = 0.5 m v^2 and h₁ = h₀ + v₀ t + 0.5 g t^2
Initial launch speed = 0.5 * Earth gravity * Time elapsed
Jumping height = launch speed * (time elapsed/2) + 0.5 * -gravity * (time elapsed/2)^2

Assume mass of Daddy Pig = 350 kg (the high-end weight of an adult domestic pig)
Start time = 12:06.587
Finish time = 12:09.309
Time elapsed （from launch to entering the puddle) = 2.722 s
Time elapsed/2 = 1.361 s
Gravity (assume on Earth) = 9.80665
Speed = (0.5 * 9.80665 * 2.722) = 13.34685065 m/s
KE = 0.5 * 350 * 13.34685065^2 = 31174.2239 J (Wall level)
(Launch height = 13.34685065 * (2.722/2) + 0.5 * -9.80665 * (2.722/2)^2 = 9.082531867 m)

Now for Mister Elephant.

Assume mass of Mister Elephant = 5443 kg (According to the World Wildlife Fund, an average African elephant weighs about 12,000 pounds (5,443 kilograms).)
Start time = 12:30.246
Finish time = 12:32.275
Time elapsed （from launch to entering the puddle) = 2.029 s
Time elapsed/2 = 1.0145 s
Gravity (assume on Earth) = 9.80665
Speed = (0.5 * 9.80665 * 2.029) = 9.948846425 m/s
KE = 0.5 * 5443 * 9.948846425^2 = 269372.8322 J (Wall level)
(Launch height = 9.948846425 * (2.029/2) + 0.5 * -9.80665 * (2.029/2)^2 = 5.046552349 m)

Now for Peppa Pig herself.

The problem is: Peppa Pig is much smaller than Daddy Pig.

I shall set up 2 ends for Peppa Pig. Low end

Since Peppa Pig later asked 13 of her friends to join and make a group jump as an attempt to break the record by Mister Elephant, I will include the data from there as well.

Picture

... Given the inconsistency of the size representation of different animals in this series, I shall assume all the friends' masses the same as that of Peppa Pig.

Participant	Peppa Pig (low end)	Peppa Pig (high end)	Peppa and 13 friends (low end)	Peppa and 13 friends (high end)	Each of Peppa and 13 friends (low end)	Each of Peppa and 13 friends (high end)
Mass (kg)	50	350	700	4900	50	350
Start time	12:55.777	12:55.777	13:29.338	13:29.338	13:29.338	13:29.338
Finish time (s)	12:56.212	12:56.212	13:31.925	13:31.925	13:31.925	13:31.925
Time elapsed (s)	0.435	0.435	2.587	2.587	2.587	2.587
Speed (m/s)	2.132946375	2.132946375	12.68490178	12.68490178	12.68490178	12.68490178
KE (J)	113.736506 (Street)	796.1555418 (Street)	56317.35656 (Wall)	394221.496 (Wall)	4022.668326 (Street)	28158.67828 (Wall)
Height (m)	0.231957918	0.231957918	8.203960223	8.203960223	8.203960223	8.203960223

... okay so Peppa & co could not win Mister Elephant if Peppa & co were not weighing ~350 kg each. So Peppa would be street level by herself, higher， possibly wall level with friendship, wall level with the magic of friendship puddle jump.

Kirby - Simirror solves a puzzle

Request is made by Eficiente

Idk how to calc this, I'm leaving here all the information needed for it so that it may be copied & pasted in a blog of someone who may want to go over the feat.

Feat: Simirror reflecting a laser (low quality but usable), Kirby reflecting the same laser (better quality).

• Context: The idea of the room is to use the Mirror ability or Simirror to reflect a laser shot at you, killing the enemy who does that, you specifically do this from the other side of that block the laser collides with because the hitbox of most of your moves used to reflect the laser go over the block.
• Why it's a laser: That enemy is called Laser Ball, his lasers are called "Light Amplification by Stimulated Emission of Radiation", move in a straight line, bounce off reflective surfaces and burn on contact.
• Notes:
  • The laser isn't reflected by a regular mirror, it's magical & uses Attack Reflection on everything.
    • This is also why the laser may be reflected without moving in a straight line, the same happens with everything reflected that way.
  • Technically one can mash the attack button repeatedly and have a reflective shield there, without reacting to the laser. But you can react to it perfectly fine and bat it, even from closer than shown in the links above, it's just that the context makes clear the exact distance as to where & how the laser was hit, which I believe makes this the perfect laser-dodging feat the Low Tiers in Kirby should scale to.

Kirby - Simirror solves a puzzle - Jasonsith fix

Picture

Laser displacement = 75 px
Simirror arm displacement = 48 px
Speed = 0.64 c (Relativistic+)

Star Fox - Grenade AP

Star Fox Fandom site description of the grenade

“	“Grenade: This powerful projectile generates a blast radius of 12 feet. It is advised to remain a safe distance from this weapon when deployed.”	„
	~ —Grenade description on the official site.

Blast radius of 12 ft = 3.6576 m

Non-nuclear blast yield = ((3.6576 m/1000/0.28)^3)*1000/2 = 0.001114511 ton TNT = 4,663,115.927 J (Wall)

Star Fox - What does an 8-meter tall Saurian mean

(picture being requested)

This model exists.

Assuming a 8-m tall saurian is as strong as a scaled up human, this means...

Travel speed: (Speed2 = Speed1 * (Height2/Height1)) = 6.35 * (8/1.684) = 30.16627078 m/s (Superhuman)
Weight: (Weight2 = Weight1 * (Height2/Height1)^3) = 62 * (8/1.684)^3 = 6647.148619 kg
Lifting strength: (LS2 = LS1 * (Height2/Height1)^4) = 65 * (8/1.684)^4 = 33105.86624 kg (Class 50)
Striking Strength (standing): (Striking Strength 2 = Striking Strength 1 * (Height2/Height1)^5) = 70 * (8/1.684)^5 = 169,370.4101 J (Wall)
Attack Potency with running acceleration: (not likely combat applicable) 0.5 * 30.16627078 m/s ^ 2 * 6647.148619 kg = 3,024,465.56 J (Wall)

The Boys - How deep into superhuman is The Deep

(moved here)

General formula:

“	One atmosphere is equal to the weight of the earth's atmosphere at sea level, about 14.6 pounds per square inch. If you are at sea level, each square inch of your surface is subjected to a force of 14.6 pounds. The pressure increases about one atmosphere for every 10 meters of water depth. At a depth of 5,000 meters the pressure will be approximately 500 atmospheres or 500 times greater than the pressure at sea level. That's a lot of pressure.	„
	~ Source

The Boys - How deep into superhuman is The Deep - Version 1

Now:

The Deep said in Season 1 he could survive 2 miles under the ocean.

2 miles = 3218.688 m

Pressure at that depth = (1 + (3218.688 m / 10 m * 1) atm) = 322.8688 atm

1 atm = 0.10132738879319 MPa or J/cc

Pressure tanked can be rewritten as 322.8688 * 0.10132738879319 MPa = 32.71545243 MPa or J/cc

The average human has a mass of about 62 kg, or 62,000 grams. That means the average volume of a human is 62,000 cubic centimeters.

Durability tanked by The Deep just by being underwater = 32.71545243 MPa * 62000 cc = 2,028,358.05 J (Wall)

Holy molly.

The Deep said in Season 2 Episode 1 he had been in the Mariana Trench before.

The maximum known depth is 10,984 metres (36,037 ft) (± 25 metres [82 ft) (6.825 miles) at the southern end of a small slot-shaped valley in its floor known as the Challenger Deep.] So the maximum depth The Deep can go into ~= 10,984 m

Pressure at that depth = (1 + 10,984 m / 10 m * 1) atm = 1099.4 atm = 111.3993312 MPa = 111.3993312 J/cc

Durability tanked by The Deep just by being at the Challenger Deep of Mariana Trench = 111.3993312 MPa * 62000 cc = 6,906,758.537 J (Wall)

This is madness. Madness. This~ is~ Sparta~!

The Boys - How deep into superhuman is The Deep - Version 2

Similar to version 1 but only the skin is tanking the ocean pressure.

It is super difficult to find out the human skin volume and mass because the body statistics vary among different institutes. A lot.

To demonstrate,

Approach 1: (NCBI approach)
The average human body has a surface area of 1.9 m^2 or 19000 cm^2.
NCBI suggests human skin density to be 1.02 g/cm^2
Human skin mass = 19000 / 1000 * 1.02 = 19380 g = 19.38 kg

Approach 2: (BBC approach)
In total, (the human skin) accounts for around 16 percent of your body weight
Average human body weight is 62 kg
Human skin mass calculated = 62 kg * 16% = 9.92 kg

Approach 3: (CUNY approach)
The human skin weighs about 3 kg.

Scientific findings of human skin statistics vary a lot indeed.

For now... Basically I use the CUNY approach because the data is most direct.
CUNY says human skin surface area is 20 sqft or 18580.608 cm^2.
Human skin thickness is 0.2 cm.
Human skin volume = 0.2 cm * 18580.608 cm^2 = 3716.1216 cm^3

This is for normal human.
Our average human height is 1.684 m (average of 1.753 m for men and 1.615 m for women).
Chace Crawford, the actor for The Deep, stands 1.791 m.
The Deep's skin volume = 3716.1216 cm^3 * (1.791/1.684)^3 = 4470.441465 cm^3

Durability of The Deep is

• (S1 for diving into 3218.688 m) 4470.441465 * 32.71545243 = 146252.5151 J (Wall)
• (S2 for diving into 10984 m) 4470.441465 * 111.3993312 = 498004.1896 J (Wall)

Speaking of which, human skin's durability (without clothing) by this standard is (1 atm * 0.101327389 MPa/atm * 3716.1216 cc) = 376.5448982 J (Street)

Metroid - Samus Aran survived a crash landing redux

Redux of this.

The working by ChaosTheory123:

Gravitational constant G = 6.67*10^-11
Mass of Zebes M = 4.8*10^27 kg
Zebes radius = 5850000 m
Height at escape above planet surface (arbitrary) = 1000000 m
Change in speed on falling of Samus v = (2GM (1/r - 1/(r+d))^0.5 = 126408.2196 m/s
KE change = 0.5 * 90 * ^2 = 7.19057E+11 J = 171.8586782 ton TNT (Multi-City Block)

The second model by ChaosTheory123 involves Samus absorbing the space ship's explosion, which I highly doubt the validity of such assumption. (That being a passenger should not be fastened to a safety seat belt in a car should a car collide because the KE change in car collision will be absorbed by the passenger.)

Because even the escape height is arbitarty I will make a better assumption on the falling height: (Assume no energy loss on falling)

Using the Karman line of 100,000 m instead of an arbitrary 1,000,000 m out of nowhere. (Karman line is one definition used to determine if a spacecraft has left a planet.)

So change of velocity = 42890.5996 m/s
KE change = 0.5 * 90 * 42890.5996^2 =
82782159017 J
= 19.78541086 tons TNT (City block)

I personally would put her here at Zero Suit.

Still 8-B, still bodies any Fett, but way weaker than a tier 7 or 8-A.

Or if you want a super highball:

Escape velocity of Zebes = (2*G*M*(1/R))^0.5 = 330841.9717 m/s
PE absorbed by Samus should she was dragged to Zebes from an infinitely long distance = 0.5 * 90 * 330841.9717^2
= 4.92554E+12 J
= 1177.231946 tons of TNT (Small town)

Still some distance from 9817 tons of TNT.

Extension of a common feat - Destroying half a car over time

This says the current calculated yield to destroy a complete new car adds up to 44,732,125.43 Joules.

Half of it becomes 22366062.72 Joules.

Dividing by 50 means attack potency per second = 447,321.2543 J (Wall)

This applies to Dong Dong Never Die characters and a super low ball for all Street Fighter characters (including Dan Hibiki).

One Piece - Luffy dodges Pacifista laser

At Return to Sabaody Arc, Pacifista PX-5 identified Luffy and proceeded to fire a laser beam at Luffy's head, which was easily dodged and prompting Luffy to claim it was too slow.

One Piece - Luffy dodges Pacifista laser - VBW version

Picture

Long story short Luffy's height is 1.72 m and in the picture he is 469.0516 px. Making his face diameter 52 px or 0.190683 m, further making his half-face radius 0.095341 m.

His straw hat to compare is 112.5877 px long, or 0.412856 m

At the picture where Luffy actually tilted his head in tandem with the laser, his straw hat diameter is 325.0738 px (or 0.412856 m) while the displacement of laser is just 462.9644 px or 0.587983 m.

So speed of Luffy's head compared to speed of light beam = Luffy head displacement vs light displacement

Speed = 1 c * 0.095341 m / 0.587983 m = 0.16215 c (Relativistic)

Faster than Erza Scarlet slicing Deus Sema, but still slower than Naruto Uzumaki (Part II) gongkiu-ing Madara's light fang.

One Piece - Luffy dodges Pacifista laser - Ultraguy

Source

Luffy Waits Patiently for Light to Reach Him (Luffy VS Pacifista)

https://youtu.be/BziEKjRs6ho?t=72

10 frames between Luffy crouching and standing

Image #1

https://i.imgur.com/rod1BS8.png

Luffy (crouched) height: 73 px

Assuming a crouch is ½ total height

Luffy (standing) height: 174 cm

Luffy (crouched) height: 87 cm

(I’m using this calculator for the equations below)

Using standard FOV of 70

2*atan(object_size/(panel_height/tan(FOV/2))) = degrees

2*atan(73/(1080/tan(70/2))) = 5.419441753122

Plug those numbers into this calculator for the distance!

Click “Distance” and then fill out the degree and the object size.

Luffy’s crouched height is the object size, which is 87 cm.

Distance: 919.1 cm

Image #2

https://i.imgur.com/z60Sp0x.png

Luffy (standing) height: 147 px

Luffy (standing) height: 174 cm

Using standard FOV of 70

2*atan(object_size/(panel_height/tan(FOV/2))) = degrees

2*atan(147/(1080/tan(70/2))) = 10.888378337166

Plug those numbers into this calculator for the distance!

Click “Distance” and then fill out the degree and the object size.

Luffy’s standing height is the object size, which is 174 cm.

Distance: 912.85 cm

Distance light travelled while Luffy stood:

919.1 - 912.85

= 6.25 cm

Distance Luffy travelled while standing: 87 cm

Luffy’s speed: 87/6.25 = 13.9 times FTL (FTL+)

Words by Jasonsith: The problem is that there is nothing suggesting the camera movement is 100% in tandem with the laser movement.

Lord of the Rings - Middle-Earth - Osse raises Elenna-nórë

To fix calculation from here.

Lord of the Rings - Middle-Earth - Osse raises Elenna-nórë - Calc 1

Ossë raises Elenna-nórë from the "depths of the ocean".

Average depth of the ocean is 3,688 meters, Elenna-nórë is 435,017 km^2, or 4.35017e+11 m^2.

Minimum volume of raised land is 4.35017e+11 * 3,688 = 1.6043427e+15 m^3

Continental crust has an average density of 2.83 g/cm^3, or 2830 kg/m^3.

Minimum mass of raised land is 1.6043427e+15 * 2830 = 4.5402898e+18 kg

PE = mgh (h cut in half)

Yield = 4.5402898e+18 * 9.11 * 1844 = 7.6271602e+22 J = 18.22 Teratons of TNT (Country)

Lord of the Rings - Middle-Earth - Osse raises Elenna-nórë - Calc 2

After a chat with User:Wokistan, Wok and Amelia agreed 500m was a safer bet.

Volume = 4.35017e+11 * 500 = 2.175085e+14 m^3

Mass = 2.175085e+14 * 2830 = 6.1554906e+17 kg

Low end yield: (height = 500/2 m = 250 m)
= 6.1554906e+17 * 9.11 * 250 = 1.40191E+21 J = 335 Gigatons TNT (Large Island level)

High end yield: (height = 500 m)
= 6.1554906e+17 * 9.11 * 500 = 2.803826e+21 J = 670 Gigatons TNT (Large Island level+)

Lord of the Rings - Middle-Earth - Osse raises Elenna-nórë - Calc 3

As the Mediterranean Sea has an average depth of 1500 m, it makes no sense a piece of large country raised from the bottom of the ocean to ground level stretches over a mere 500 m.

Except 3,646 m, not 1,500 m, is the average depth of Atlantic ocean.

Volume = 4.35017e+11 * 3646 = 1.58607E+15 m^3

Mass = 1.58607E+15 m^3 * 2830 kg/m^3 = 4.48558E+18 kg (Class P)

And I repeat, earth gravity = 9.80665 J/m/kg

Low end yield: (height = 3646/2 m = 1823 m)
= 4.48558E+18 kg * 9.80665 * 1823 = 8.02448E+22 J = 19.179 teratons TNT (Country level)

High end yield: (height = 3646 m)
= 4.48558E+18 kg * 9.80665 * 3646 = 1.6049E+23 J = 38.357 teratons TNT (Country level) - preferred

Danganronpa - Makoto Naegi fell at great heights

Requested by @StrymULTRA

Feat: Makoto Naegi survived a huge fall. While the height of falling is not specified, but the time is (with this @StrymULTRA found that it lasts from 7:42:29 to 07:52:11, making the time frame of 9.82 seconds).

Danganronpa - Makoto Naegi fell at great heights - by StrymULTRA

So, using this calculator, I found that the height of such a long fall is of 472.839 m, so putting it here (with the weight of 65.4 kg due of 52 kg for Makoto + 5 kg for the chair + 8.4 kg for the desk), I got the force of 314263.91 Joules (Wall level).

Danganronpa - Makoto Naegi fell at great heights - by Jasonsith

Problems by StrymULTRA

1.

Picture

The desk and the table look absurdly loosing away from Naegi once they fell together. The chair weight and the desk weight should not add to the weight that Naegi-kun tanked.

Second, Naegi did not take the full impact force - he fell upon some rubbish bag which acted as a not-so-good cushion. Kyoko Kirigiri was dropped into those bags (likely following the same route) on her way to rescue Naegi-kun. Monokuma bluntly said Naegi-kun was to be wasted away in a garbage-strewn pit.

Third, the cinematic time looks past the time required for reaching a terminal velocity.

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

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

As we knew Naegi-kun weighs 52 kg, KE on impact = 0.5*52*53^2 = 91,854.30 Joules (Wall level)

Approximating that border without air resistance: 53 m/s / 9.80665 m/s^2 = 5.404496 s drop time only is required to achieve terminal speed.

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

(1/2)*9.80665*5.404496^2 = 143.2191 m

Therefore, any realistic drop at a height beyond 143.2191 m would kick in terminal speed. And it does not look like Naegi-kun was experiencing an accelerated fall beside Earth gravity.

However, the cinematic time gives that is 9.82 s which is way past 5.404496 s.

And this is assuming nothing will even further slow the falling speed.

Anyway, 91,854.30 Joules (Wall level) should be a reasonable value if Makoto (un)luckily landed on rough and tough surface with nothing to increase impact time or to reduce falling speed.

And assume Naegi-kun did drop way past terminal velocity, 9.82 s drop would imply a height of (0.5 * 9.82^2 * 9.80665 =) 472.8394 m height at most, meaning a wanked high end = 52 * 9.80665 * 472.8394 = 241122.5 J (Wall level)

Land Before Time - Sharptooth fell at great heights

During a huge earthquake, Sharptooth was knocked off from the edge of a hill by Littlefoot's mother. He then fell into the chasm below, and that big fall just knocked him out.

That sharptooth is... basically a historic t-rex and should therefore weigh about 11.5 metric tonnes or 11,500 kg.

As there is no sign of stopping the fall from 1:54 to 2:02, it is likely the dinosaur reached terminal velocity when falling.

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

Durability tanked on impact = 0.5 * 11,500 * 53^2 = 16,151,750 J (wall+)

Oh that poor sharptooth also briefly clung on the edge of the crumbling hill until Littlefoot's mother knocked him off, so his lifting strength scales to his own body mass (Class 25)

Genndy Tartakovsky's Primal - Night Feeder Sliced a Triceratops

Night Feeder slices some horned/ceratopsian dinosaurs apart

That thing looked like an Achelousaurus.

Picture

Legend says scale length = 734 px = 5 m

Trunk length = 400 px = 2.724796 m

Trunk height = 212 px = 1.444142 m

Head length = 174 px = 1.185286 m

Head height = 96 px = 0.653951 m

Saw thicknesses are around 1.5 mm to 3.0 mm. I say 0.225 cm (or 0.00225 m) thickness is appropriate.

Cut volume = 0.008698 m^3 = 8697.721 cm^3

Assume general human tissue destruction ends
Fragmentation (@4.4 J/cc) = 38269.97 J (Wall)
Violent fragmentation (@7.533 J/cc) = 65519.93 J (Wall)
Pulverisation (@12.9 J/cc) = 112200.6 J (Wall) - currently here

Image Comics - Omni-Man Flies

Omni-Man flew from earth to another planet with life on it in a week.

First, 1 light year = 299792458 * 60 * 60 * 24 * 365.24 = 9.46047E+15 m ("and is a measurement of distance not time")

Time to travel in 1 week = 60 * 60 * 24 * 7 = 604800 s

Low end: The potentially habitable planet TOI 700 d is only 101.4 light years away from our planet earth.

101.4 light year = 9.59292e17 m
Speed = 9.59292e17 m / 604800 s = 1.58613e12 m/s = 5290.76286 c (Massively FTL+)

High end: A 2015 review concluded that the exoplanets Kepler-62f, Kepler-186f and Kepler-442b were likely the best candidates for being potentially habitable.

Kepler-186f is said to be only 490 light-years away from Earth.

490 light year = 4.63563e18 m
Speed = 4.63563e18 m / 604800 s = 7.66473e12 m/s = 25566.8 c (Massively FTL+)

Akira - Tetsuo Jumps to the Moon

Feat at Akira (#27, pg. 49-54)

Earth radius = 6371000 m

Moon radius = 1737400 m

Earth to moon distance = 384400000 m

Earth surface to moon surface distance = 384400000 m - 6371000 m - 1737400 m = 376291600 m

Assume time = 10 s (some time for the audience to realise before Tetsuo shows up in the moon - a lowball actually)

Speed = 376291600 m / 10 s = 37629160 m/s = 0.12551737 c (Relativistic)

Ad Feat - Kool-Aid man becomes a planet

Picture

Ad Feat - Kool-Aid man spins a moon

Picture