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The Calc Request Thread

I'm wondering, can the 20% gun devil's speed be called from CSM chapter 75-76?

Here's the abridged version of events.




The entire 'fight' takes 12 seconds with Gun devil appearing at the coast and after 2 seconds reaches 500 kilometers from Makima's position, keeps moving forward 2 more seconds before stopping, and shooting, the bullets reach Makima's position in less than a single second (pg 4 you can see the bullets hitting the ocean behind makima.) Makima prepares her ability before one of the bullets pierce her brain and in 2 seconds attacks.

Is it possible to calc the Gun devil's speed and bullet speed in these pages? Only problem is I forgot where they are currently. This should be the Tokyo office Iirc.
 
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I'm wondering, can the 20% gun devil's speed be called from CSM chapter 75-76?

Here's the abridged version of events.




The entire 'fight' takes 12 seconds with Gun devil appearing at the coast and after 2 seconds reaches 500 kilometers from Makima's position, keeps moving forward 2 more seconds before stopping, and shooting, the bullets reach Makima's position in less than a single second (pg 4 you can see the bullets hitting the ocean behind makima.) Makima prepares her ability before one of the bullets pierce her brain and in 2 seconds attacks.

Is it possible to calc the Gun devil's speed and bullet speed in these pages? Only problem is I forgot where they are currently. This should be the Tokyo office Iirc.

Well, we have a couple of data, first the total distance is 500km, in 2 seconds he moved an unknown distance and his shots took 1 second to reach Makima.

It is impossible to get the individual speed of the movement, nor the individual speed of the shots, but if it is assumed that his movement speed is similar to that of his shots, we could say that he covered 500 km in 3 seconds.

that would be like mach ~486
 
A question about doing calculations.

Can we use the ImageMeter app? This app allows measuring the size within images easier than doing it with pixels and also has a property to count objects within the image.

 
So I'm wondering if these two speedfeats from JJK can be calced.

First is three-finger Sukuna blitzing Megumi during the cursed womb arc.

0008-022.png


Sukuna here moves so fast that Megumi couldn't see him exit and get behind him despite looking straight at where Sukuna exits.

Second is fifteen-finger Sukuna traveling from where he fights Jogo to Mahoraga.

Scans



Megumi summons Mahoraga at around 11:05. Megumi gets oneshot before 11:07 when Sukuna senses Mahoraga and travels fast enough to save Haruta (blonde ponytail hair dude) from Mahoraga.

Here's a map Gege made detailing the positions of all the fights taking place.

EudVE33XAAUOXnD



Here's a google map of the same area for reference.

B3mZwlh.png



Here is me very badly trying to scale Gege's map to google's.

d3FEWoz.png


shits off center and very much not in scale but its the best I can do rn. Should give a somewhat close visual to where Sukuna fought Mahoraga. hopefully someone else could do better.
 
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Did we ever have or had any members here who ever calced the original Inuyasha manga?




 
just wanted to try my hand at calcing this, not sure what it could be used for though.

Thanks @ChaosTheory123 for giving me some numbers to work with.

G14P8UF.jpg



So, from the perception and reaction time of normal ass people? 0.2 seconds of Unlimited Void is sufficient to cram about 6 months/15.768 million seconds of information into their heads.

That's about the time frame of a normal ass person's average reaction speed at the upper bound of it.

Would basically mean that so long as you can process information and react at a level beyond human, there is a threshold in time during which the power and its debilitating junk information isn't going to dump enough information into your skull to stall your ability to take action. The closer you are to sub-relativistic or relativistic, the less of a threat the power is to you in general.

So that's neat

So based on this

6 months is equal to 15,780,000 Seconds

The human brain can process 11,000,000 bits of info every second. .

So 15,780,000 * 11,000,000 = 1.7358e+14 or 21.6975 terabytes of information in total.

And this is done in 0.2 seconds so...

21.6975/2= 10.84875 terabytes every 0.1 seconds.

and

108.4875 terabytes every second

btw apparently the max memory of the brain is 2.5 petabytes https://www.medanta.org/patient-education-blog/what-is-the-memory-capacity-of-a-human-brain#:~:text=As mentioned in an article,2.5 million gigabytes of memory.

and 1 petabyte is equal to = 1024 terabytes.

so 2,500,000/108.4875= 23044.129508 seconds to overload the human brain's memory.

or around 6.4 hours.


Completely forgot to convert

2500/108.4875= 23.044129508

so around 23 seconds to overload the brain.

someone should probably check my math cause I think I did something wrong but idk where, math is my worst subject.

EDIT: Completely forgot to convert from gigabytes to petabytes.
 
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there’s a feat where a character created so much ice, the sheer mass of it all tipped the entire Earth off it’s axis to the degree where the north and South Pole shifted to the equator

No timeframe was given and it was only explained in narration. Is there a way to calc that? Like how much force or weight would be needed to shift the entire planet?
 
Same episode, Vince kicked the ball beyond Escape Velocity

Space Shuttle is 8.7 meters in diameter with a height of 56.1 meters

the time it took for Vince’s ball to reach it from earth is actually the same as Ashley Q’s ball with China—4 seconds

You can see the Earth and the Space Shuttle breifly be in the same shot.

View attachment 1953
View attachment 1954

View attachment 1955

I got some values here but I don’t really know enough about Angsizing to know how far the Space Shuttle is from The Earth to gauge the distance to know how impressive this feat is

So… outsourcing this to people who can :doge
 
How fast would you have to be to travel from the nearest galaxy of the Milky Way to the earth in 2 minutes~ish?

At approximately 2.5 million light-years away

D= 2.3651826e+22 mts

T= 120 seg

Speed: 2.3651826e+22/120= 1.9709855e+20 m/s or 657,449,994,956.17736816 c
 
I see, that's pretty fast. Would apply to the X-Men's ship from the TAS show. Tldr there's a scene in the final episode where Xavier calls out the Shi'ar and their spaceship takes about 1:40 min to reach the earth.
 
For practical purposes, I will assume that this planet is the size of Earth, and the same amount of air, gravity, water, etc.

In addition to the fact that I am not good at measuring the size of planets without some reference, there are also no measurements to know how much water there is in that planet.

First calc the global earthquake and tsunamis:

Low End:

Generally, a magnitude 10 earthquake should already be felt on the planet, so this would be the minimum interpretation
https://www.wolframalpha.com/input?i=Richter+scale+magnitude+calculator&assumption={"FS"}+->+{{"RichterScaleMagnitudeToEnergy",+"EJ"},+{"RichterScaleMagnitudeToEnergy",+"R"}}&assumption={"F",+"RichterScaleMagnitudeToEnergy",+"R"}+->"10"&assumption="FSelect"+->+{{"RichterScaleMagnitudeToEnergy"},+"dflt"}&lang=es

Earthquake energy (low end): 6.3e19 J

High End:


Magnitud is 4.5 or more

The circumference of a planet like The Earth is 40,000,000 mts so the distance of epicenter is the half

D= 20,000,000 mts or 12,454 miles

Source

First it is relevant to define the term epicenter as it is a word that will be prevalent throughout this blog.

Epicenter: The point on the earth's surface vertically above the focus of an earthquake.

The epicenter is on the surface of the Earth and as Whitebeard's quakes are on the surface also it should eliminate any queries on the validity of the formulas we will be using here concerning his quakes.

...

Most earthquakes occur at the shallow depth 0-70km. For comparison the Japanese quake occured at a depth of 30km I think the above source is reliable.

...

Between 4-4.9 is where it is noticeable to humans outside. We'll of course go for the low end of 70km and use a nomogram to work out the amplitude. You simply draw a line through the known distance to the epicenter from the hypocenter (70km) and the known magnitude and you'll get your amplitude.

h8Dg2eY.png

The red and purple line show you the upper and lower limits of the range. I'm going bang in the middle since the low effects described aren't sufficient and the high effects are too much to the knowledge we have. The blue line is our amplitude and that is 50mm. Which isn't too bad.

...

In actual fact, the amplitude will need to be higher because the epicenter and hypocenter were technically one and the same (as the quake was created on the surface) this is called a "shallow quake" from 0-70km. So we start on the nomograph at 0 and draw across.

KPSVEgi.jpeg

I drew through 2.5 as that is the most we can deduce. However, with a nomogram going up a single unit in magnitude increases the amplitude by 10x.

Why? It's logarithmic.

5000mm for magnitude 4.5.

NirujYy.gif
Where A(mm) is the height of the largest seismic wave (s wave) in millimeters (mm) and delta(t) is the S-P time difference.

Look to GM's version for what I'll use as the s wave height (5000 mm)

The S-P time difference is derived by dividing the Distance to Remote Island in miles by 5.7.


Time frame is: 12,454/5.7 = 2184.9 s

M = log(base10)5000 + 3(log(base10)[8*2184.9]) - 2.92

M= 13.5


Earthquake energy (high end): 1.1220184543019634355910389e+25 J


Wind speed: 42 m/s

The total mass of Earth’s atmosphere is about 5.5 quadrillion tons, or roughly one millionth of Earth’s mass

Atmosphere mass: 5.5e+18 kg

KE: 4.85e+21 J

Now the clouds mass.


Cloud Thickness (Low End) = 2,000 meters

Cloud Thickness (High End) = 8,000 meters

Cloud radius= 20,000,000 mts


Volume (low end)= 2513274122871834590.77 m³

Volume (high end)= 10053096491487338363.08 m³


Density= 1-3 g/m³

Cloud mass (low end)= 2.5132741228718E+15 kg

Cloud mass (mid end)= 7.5398223686155E+15 kg

Cloud mass (mid high end)= 1.0053096491487E+16 kg

Cloud mass (high End)= 3.0159289474462E+16 kg

I'm going to use the formula that @ChaosTheory123 made a while ago to speed up the calculations
Now to get the Latent Heat for condensation of water in clouds

Average Atmospheric Temperature = 15 C

Specific Latent Heat Water in Cloud = (2500.8 - 2.36(15) + 0.0016(15)^2 - 0.00006(15)^3) * 1,000 = 2,465,557.5 joules/kg

Cloud storm energy (low end): 6.1966219e+21 J

Cloud storm energy (mid end): 1.8589866e+22 J

Cloud storm energy (mid high end): 2.4786487e+22 J

Cloud storm energy (high end): 7.4359462e+22 J


Storm's weather manipulation: 6.3e19 J + 6.1966219e+21 J + 4.85e+21 J = 1.1109622e+22 J or 2.6 Teratons

Storm's weather manipulation: 6.3e19 J + 1.8589866e+22 J + 4.85e+21 J = 2.3502866e+22 J or 5.6 Teratons


Storm's weather manipulation: 6.3e19 J + 2.4786487e+22 J + 4.85e+21 J = 2.9699487e+22 J or 7 Teratons


Storm's weather manipulation: 6.3e19 J + 7.4359462e+22 J + 4.85e+21 J = 7.9272462e+22 J or 18.9 Teratons


Storm's weather manipulation: 1.1220184543019634355910389e+25 J + 6.1966219e+21 J + 4.85e+21 J = 1.123e+25J or 2.6 Petatons



Storm's weather manipulation: 1.1220184543019634355910389e+25 J + 1.8589866e+22 J + 4.85e+21 J = 1.124e+25J or 2.6 Petatons



Storm's weather manipulation: 1.1220184543019634355910389e+25 J + 2.4786487e+22 J + 4.85e+21 J = 1.125e+25J or 2.6 Petatons



Storm's weather manipulation: 1.1220184543019634355910389e+25 J + 7.4359462e+22 J + 4.85e+21 J = 1.129e+25J or 2.6 Petatons


Results

Storm's weather manipulation (low end): 2.6 Teratons of TNT

Storm's weather manipulation (low mid end): 5.6 Teratons of TNT

Storm's weather manipulation (mid end): 7 Teratons of TNT

Storm's weather manipulation (mid high end): 18.9 Teratons of TNT

Storm's weather manipulation (high end): 2.6 Petatons of TNT
 
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For practical purposes, I will assume that this planet is the size of Earth, and the same amount of air, gravity, water, etc.

In addition to the fact that I am not good at measuring the size of planets without some reference, there are also no measurements to know how much water there is in that planet.

First calc the global earthquake and tsunamis:

Low End:

Generally, a magnitude 10 earthquake should already be felt on the planet, so this would be the minimum interpretation
https://www.wolframalpha.com/input?i=Richter+scale+magnitude+calculator&assumption={"FS"}+->+{{"RichterScaleMagnitudeToEnergy",+"EJ"},+{"RichterScaleMagnitudeToEnergy",+"R"}}&assumption={"F",+"RichterScaleMagnitudeToEnergy",+"R"}+->"10"&assumption="FSelect"+->+{{"RichterScaleMagnitudeToEnergy"},+"dflt"}&lang=es

Earthquake energy (low end): 6.3e19 J

High End:


Magnitud is 4.5 or more

The circumference of a planet like The Earth is 40,000,000 mts so the distance of epicenter is the half

D= 20,000,000 mts or 12,454 miles



5000mm for magnitude 4.5.




Time frame is: 12,454/5.7 = 2184.9 s

M = log(base10)5000 + 3(log(base10)[8*2184.9]) - 2.92

M= 13.5


Earthquake energy (high end): 1.1220184543019634355910389e+25 J

Edit...

(I'll calculate the clouds and wind later when I have time)
Thank you!
 
For practical purposes, I will assume that this planet is the size of Earth, and the same amount of air, gravity, water, etc.

In addition to the fact that I am not good at measuring the size of planets without some reference, there are also no measurements to know how much water there is in that planet.

First calc the global earthquake and tsunamis:

Low End:

Generally, a magnitude 10 earthquake should already be felt on the planet, so this would be the minimum interpretation
https://www.wolframalpha.com/input?i=Richter+scale+magnitude+calculator&assumption={"FS"}+->+{{"RichterScaleMagnitudeToEnergy",+"EJ"},+{"RichterScaleMagnitudeToEnergy",+"R"}}&assumption={"F",+"RichterScaleMagnitudeToEnergy",+"R"}+->"10"&assumption="FSelect"+->+{{"RichterScaleMagnitudeToEnergy"},+"dflt"}&lang=es

Earthquake energy (low end): 6.3e19 J

High End:


Magnitud is 4.5 or more

The circumference of a planet like The Earth is 40,000,000 mts so the distance of epicenter is the half

D= 20,000,000 mts or 12,454 miles



5000mm for magnitude 4.5.




Time frame is: 12,454/5.7 = 2184.9 s

M = log(base10)5000 + 3(log(base10)[8*2184.9]) - 2.92

M= 13.5


Earthquake energy (high end): 1.1220184543019634355910389e+25 J


Wind speed: 42 m/s



Atmosphere mass: 5.5e+18 kg

KE: 4.85e+21 J

Now the clouds mass.


Cloud Thickness (Low End) = 2,000 meters

Cloud Thickness (High End) = 8,000 meters

Cloud radius= 20,000,000 mts


Volume (low end)= 2513274122871834590.77 m³

Volume (high end)= 10053096491487338363.08 m³


Density= 1-3 g/m³

Cloud mass (low end)= 2.5132741228718E+15 kg

Cloud mass (mid end)= 7.5398223686155E+15 kg

Cloud mass (mid high end)= 1.0053096491487E+16 kg

Cloud mass (high End)= 3.0159289474462E+16 kg

I'm going to use the formula that @ChaosTheory123 made a while ago to speed up the calculations


Cloud storm energy (low end): 6.1966219e+21 J

Cloud storm energy (mid end): 1.8589866e+22 J

Cloud storm energy (mid high end): 2.4786487e+22 J

Cloud storm energy (high end): 7.4359462e+22 J


Storm's weather manipulation: 6.3e19 J + 6.1966219e+21 J + 4.85e+21 J = 1.1109622e+22 J or 2.6 Teratons

Storm's weather manipulation: 6.3e19 J + 1.8589866e+22 J + 4.85e+21 J = 2.3502866e+22 J or 5.6 Teratons


Storm's weather manipulation: 6.3e19 J + 2.4786487e+22 J + 4.85e+21 J = 2.9699487e+22 J or 7 Teratons


Storm's weather manipulation: 6.3e19 J + 7.4359462e+22 J + 4.85e+21 J = 7.9272462e+22 J or 18.9 Teratons


Storm's weather manipulation: 1.1220184543019634355910389e+25 J + 6.1966219e+21 J + 4.85e+21 J = 1.123e+25J or 2.6 Petatons



Storm's weather manipulation: 1.1220184543019634355910389e+25 J + 1.8589866e+22 J + 4.85e+21 J = 1.124e+25J or 2.6 Petatons



Storm's weather manipulation: 1.1220184543019634355910389e+25 J + 2.4786487e+22 J + 4.85e+21 J = 1.125e+25J or 2.6 Petatons



Storm's weather manipulation: 1.1220184543019634355910389e+25 J + 7.4359462e+22 J + 4.85e+21 J = 1.129e+25J or 2.6 Petatons


Results

Storm's weather manipulation (low end): 2.6 Teratons of TNT

Storm's weather manipulation (low mid end): 5.6 Teratons of TNT

Storm's weather manipulation (mid end): 7 Teratons of TNT

Storm's weather manipulation (mid high end): 18.9 Teratons of TNT

Storm's weather manipulation (high end): 2.6 Petatons of TNT
So if i got it right, the earthquakes are either island level via Richter scale or continent via Chaos' formula, while the storms are country to small continental?
 
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