Experiment12
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Revised Mori Jin.
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Thank you for your attention.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.
No worries. Once that's done we can come back to this thread.Thank you for your attention.
Thank you Antvasima for helping me.@CrimsonStarFallen @Sir_Ovens @Jvando @RegisNex1232 @azontr @Zencha9 @Dienomite22
Are you willing to help out here afterwards please?
I have created a blog but I will send the error later I will send the proof.Once the calc is blogged and evaluated then we can discuss.
do you have a link to the blog?I have created a blog but I will send the error later I will send the proof.
Why don't we use "Gravitational Binding Energy Galaxy" here even though in my discussion it can be used.I just wanna say, you used the Gravitational Binding energy of the galaxy which i don't think we use that here.
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.Why don't we use "Gravitational Binding Energy Galaxy" here even though in my discussion it can be used.
• 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
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.I've created the Blog
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. |
I think your words and words Unshakeable that's true, I agree with the results of Galaxy Level 3-C what about the others?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
(Hopefully thats correct since I rushed 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.
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...?
It's the Oort Cloud- dunno why they said "Kinton" Cloud (different language?)- so it's like an asteroid beltis the kinton cloud a single massive object or is it like an asteroid belt?
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 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
(Hopefully thats correct since I rushed 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.
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...?
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.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
When would be a good time to use the Galaxies GBE for galaxy dispersion?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
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)When would be a good time to use the Galaxies GBE for galaxy dispersion?
I just used the OPs listed. Looking into it, a nebula is 90% Hydrogen, 9% and 1% heavier elements.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.
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 PressureI 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.
Already did.So uh, couldn't we just measure the size of the explosion and just assume the Sun gets destroyed at the edge?