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Upgrade Jin Mori To Universe Level Or Tier 3-A

• Topic: Mori Jin & other Jecheondaesong summon Kinton Cloud's to defeat Mujin Park.
• Feats:
()

• Information:

— Galaxy Area (100000 Light Years | 9,46073e+20 m) : (https://exoplanets.nasa.gov/blog/1563/our-milky-way-galaxy-how-big-is-space/)
— Diameter Kinton Cloud (19×10^12 light years | 1,79753878979035e+29 m) :
()
— Gravitational Constant (6,67408x10^-11) :
(https://en.m.wikipedia.org/wiki/Gravitational_constant)
— Density of Hydrogen (0.08375 kg/m^3) :
(https://h2tools.org/hyarc/hydrogen-data/basic-hydrogen-properties)

• Evaluation:

— A. Diameter of Kinton Cloud's:
• D = (9,46073e+20÷532)×1,79753878979035e+29=
• D = 3.196622e+47 m

— B. Radius of Kinton Cloud's:
• R = (3.196622e+47÷2)=
• R = 1.598311e+47 m

— C. Kinton Cloud's Volume:
• V = 4/3 × π × r^3
• V = 4/3 × π × 1.598311e+47^3
• V = 1.7103007e+142

— D. Kinton Cloud's Mass:
• Mass = Volume × Density
• Mass = 1.7103007e+142 × 0.08375
• Mass = 1.432377e+141 kg

— E. Method & Result:
• Gravitational Binding Energy
• U = (3 × G × M^2 ÷ (5r)) of Galaxy:
(https://physics.stackexchange.com/questions/246216/binding-energy-of-a-galaxy)
• U = (3 × 6.67 × 10^(-11) × (1.432377e+141^2) ÷ (5(1.7103007e+142))
• U (Energy expended on a Scale)= 4.80086273e129 Joule | Universe Level
• U (Energy released Individually) = 4.80086273e129 ÷ 3
• U = 1.60028758e129 Joule | Universe Level

*Note: Kinton's cloud which is carried out to perform the effectiveness outside the galaxy.
 
You will need to put this into a blog post first. You can get that done by going to your Vs Battle Wiki's account page (Not here on forums) and under the blog section it will give you an option to make a blog (click create new blog). After that is done you will just need to submit it into the calculation evaluation thread where a calc member will evaluate it.
 
PowerToScale said:
You will need to put this into a blog post first. You can get that done by going to your Vs Battle Wiki's account page (Not here on forums) and under the blog section it will give you an option to make a blog (click create new blog). After that is done you will just need to submit it into the calculation evaluation thread where a calc member will evaluate it.
Click to expand...
Thank you for your attention.
 
I just wanna say, you used the Gravitational Binding energy of the galaxy which i don't think we use that here.
 
Experiment12 said:
Why don't we use "Gravitational Binding Energy Galaxy" here even though in my discussion it can be used.
Click to expand...
Discussion with who? I asked about using it for a character in a different series. It was like 2 years ago though I asked that. I can't find atm bur I'm also busy.

Don'tTalk was the one who explained why I couldn't iirc
 
Maybe we should need a lot of people to discuss this thread and maybe it's also necessary to invite members who have knowledge about calculations to make it clearer.
 
Experiment12 said:
• Topic: Mori Jin & other Jecheondaesong summon Kinton Cloud's to defeat Mujin Park.
• Feats:
()

• Information:

— Galaxy Area (100000 Light Years | 9,46073e+20 m) : (https://exoplanets.nasa.gov/blog/1563/our-milky-way-galaxy-how-big-is-space/)
— Diameter Kinton Cloud (19×10^12 light years | 1,79753878979035e+29 m) :
()
— Gravitational Constant (6,67408x10^-11) :
(https://en.m.wikipedia.org/wiki/Gravitational_constant)
— Density of Hydrogen (0.08375 kg/m^3) :
(https://h2tools.org/hyarc/hydrogen-data/basic-hydrogen-properties)

• Evaluation:

— A. Diameter of Kinton Cloud's:
• D = (9,46073e+20÷532)×1,79753878979035e+29=
• D = 3.196622e+47 m
Click to expand...

This is... wrong from the jump?

Why did you have the diameter of the "Kinton"/Oort Cloud be 19 Trillion Lightyears? It's 19 Trillion Kilometers.

And why is it multiplied by the diameter of the galaxy? The blackhole/explosion should only be measured relative to the size of the galaxy, not the Oort Cloud as well.

Also, the end diameter is nearly a sextillion times the radius of the Observable Universe, which by itself should've given away that this calc is completely and utterly wrong.
 
Experiment12 said:
I've created the Blog
Click to expand...
From I've seen you used the uniform sphere for the GBE. And since I have no idea what this is calcing I am assuming you are trying to find the GBE of the Kinton Cloud based on the methodology. Some things to bring up.

1: You used lightyears for the diameter when the Kinton is in kilometres.

2: why multiple the galaxy's area?

3: you used the volume for "r" in the uniform sphere GBE formula while "r" is radius
______________________
To fix it
Radius
Diameter of cloud is 19 trillion km or 1.9e+16m
Radius is 1.9e+16/2= 9.5e+15m

Volume and Mass
volume of sphere V = (4/3)πr^3
V = (4/3)π(9.5e+15^3)= 3.591364E+48m^3

volume x denisty= mass
3.591364E+48x0.08375= 3.0077674e+47kg

GBE
U = (3 × G × M^2 ÷ (5r))
(3×6.67 × 10^(-11)×3.0077674e+47^2÷(5×9.5E+15))= 3.8110266e+68 Joules (3-C Galaxy)

Divide by 3
3.8110266e+68/3= 1.2703422e+68 joules (3-C Galaxy)

Result will vary if using pixel scaling as it seems that the cloud grew bigger from its initial 19 trillion km, which would make sense.
(Hopefully thats correct since I rushed it)

Again this assuming you're trying to find the GBE of the Kinton cloud since I'm am a little confused on the calc, which might be okay...?
 
PowerToScale said:
From I've seen you used the uniform sphere for the GBE. And since I have no idea what this is calcing I am assuming you are trying to find the GBE of the Kinton Cloud based on the methodology. Some things to bring up.

1: You used lightyears for the diameter when the Kinton is in kilometres.

2: why multiple the galaxy's area?

3: you used the volume for "r" in the uniform sphere GBE formula while "r" is radius
______________________
To fix it
Radius
Diameter of cloud is 19 trillion km or 1.9e+16m
Radius is 1.9e+16/2= 9.5e+15m

Volume and Mass
volume of sphere V = (4/3)πr^3
V = (4/3)π(9.5e+15^3)= 3.591364E+48m^3

volume x denisty= mass
3.591364E+48x0.08375= 3.0077674e+47kg

GBE
U = (3 × G × M^2 ÷ (5r))
(3×6.67 × 10^(-11)×3.0077674e+47^2÷(5×9.5E+15))= 3.8110266e+68 Joules (3-C Galaxy)

Divide by 3
3.8110266e+68/3= 1.2703422e+68 joules (3-C Galaxy)

Result will vary if using pixel scaling as it seems that the cloud grew bigger from its initial 19 trillion km, which would make sense.
(Hopefully thats correct since I rushed it)

Again this assuming you're trying to find the GBE of the Kinton cloud since I'm am a little confused on the calc, which might be okay...?
Click to expand...
I think your words and words Unshakeable that's true, I agree with the results of Galaxy Level 3-C what about the others?
 
Not a calc guy but given how most people addressed their issues with the calc 3-C looks fine as long as the calc is correct of course
 
PowerToScale said:
From I've seen you used the uniform sphere for the GBE. And since I have no idea what this is calcing I am assuming you are trying to find the GBE of the Kinton Cloud based on the methodology. Some things to bring up.

1: You used lightyears for the diameter when the Kinton is in kilometres.

2: why multiple the galaxy's area?

3: you used the volume for "r" in the uniform sphere GBE formula while "r" is radius
______________________
To fix it
Radius
Diameter of cloud is 19 trillion km or 1.9e+16m
Radius is 1.9e+16/2= 9.5e+15m

Volume and Mass
volume of sphere V = (4/3)πr^3
V = (4/3)π(9.5e+15^3)= 3.591364E+48m^3

volume x denisty= mass
3.591364E+48x0.08375= 3.0077674e+47kg

GBE
U = (3 × G × M^2 ÷ (5r))
(3×6.67 × 10^(-11)×3.0077674e+47^2÷(5×9.5E+15))= 3.8110266e+68 Joules (3-C Galaxy)

Divide by 3
3.8110266e+68/3= 1.2703422e+68 joules (3-C Galaxy)

Result will vary if using pixel scaling as it seems that the cloud grew bigger from its initial 19 trillion km, which would make sense.
(Hopefully thats correct since I rushed it)

Again this assuming you're trying to find the GBE of the Kinton cloud since I'm am a little confused on the calc, which might be okay...?
Click to expand...
This is still wrong, as while the math seems good, it relies on the density of Hydrogen Gas in order to get the mass of the affected volume, despite how Hydrogen Gas is over a trillion times denser than the vacuum of space.
 
From my knolwedge i don't think you can use GBE for the Oort cloud. the oort cloud is made up of millions of floating space rocks each with their own gravitonal pull.

I could be wrong though
 
TheUnshakableOne said:
From my knolwedge i don't think you can use GBE for the Oort cloud. the oort cloud is made up of millions of floating space rocks each with their own gravitonal pull.

I could be wrong though
Click to expand...
The Oort Cloud does have a GBE- like how the Milky Way does, but that's for permanent dispersion of its countless orbiting parts. It's much lower than total destruction via Inverse Square Law. And we don't use Galaxy GBE because Galaxy dispersion is much rarer than Galaxy Destruction in fiction, so it's not a good benchmark for 3-C / Galaxy Tier.

In other words, a correct calc on the Oort Cloud's GBE should be much less than the partial Galaxy Destruction calc that's currently used to scale God of Highschool Mori
 
ElajRuengies said:
The Oort Cloud does have a GBE- like how the Milky Way does, but that's for permanent dispersion of its countless orbiting parts. It's much lower than total destruction via Inverse Square Law. And we don't use Galaxy GBE because Galaxy dispersion is much rarer than Galaxy Destruction in fiction, so it's not a good benchmark for 3-C / Galaxy Tier.

In other words, a correct calc on the Oort Cloud's GBE should be much less than the partial Galaxy Destruction calc that's currently used to scale God of Highschool Mori
Click to expand...
When would be a good time to use the Galaxies GBE for galaxy dispersion?
 
TheUnshakableOne said:
When would be a good time to use the Galaxies GBE for galaxy dispersion?
Click to expand...
Like a spiral galaxy being turned into bunch of scattered lights flying away from each other, without any part of the galaxy being voided/blown up/destroyed (assuming the whole thing is in a sped up timeframe otherwise the star systems would be going FTL)

Still, in terms of energy on the AP Chart, Galaxy Dispersion is just Solar System level, which feels like a huge misnomer and so the person/item performing the feat would likely just be listed as Galaxy level anyway
 
ElajRuengies said:
This is still wrong, as while the math seems good, it relies on the density of Hydrogen Gas in order to get the mass of the affected volume, despite how Hydrogen Gas is over a trillion times denser than the vacuum of space.
Click to expand...
I just used the OPs listed. Looking into it, a nebula is 90% Hydrogen, 9% and 1% heavier elements.

Honestly, like I said before I doubt that this method would be viable.
 
PowerToScale said:
I just used the OPs listed. Looking into it, a nebula is 90% Hydrogen, 9% and 1% heavier elements.

Honestly, like I said before I doubt that this method would be viable.
Click to expand...
Yeah- you weren't at fault the math itself was fine, but the OP's set up for the calc is fundamentally wrong. And the Density of Hydrogen Gas in Space =/= Density of Hydrogen Gas under Earth's Atmospheric Pressure
 
So uh, couldn't we just measure the size of the explosion and just assume the Sun gets destroyed at the edge?
 
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