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https://vsbattles.fandom.com/wiki/User_blog:Spinosaurus75DinosaurFan/Thanos_crushes_the_Tesseract
This calculation of Thanos crushing the Tesseract needs to be discussed.
Some dude on Naruto Forums calculated:
Minor thing but, from the first Avengers:
Steve: Does Loki need any particular kind of power source?
BANNER: He's got to heat the cube to a hundred and twenty million Kelvin just to break through the Coulomb barrier.
Implying the Tesseract could theorectically survive just under that, yet Thanos just kinda breaks it.
Taking it at face value:
http://www.endmemo.com/physics/radenergy.php
And using a surface area of like 0.02 m^2 as an eyeballed lowball, we get 2.3515E+23 Joules or 56.2 Teratons of tnt per second, so consistency I guess
But then Kepekley23 revised it:
Ring finger diameter = 28px = 19.8mm (average ring size for men is 10) Ôù¥ Edge size = 164px = 0.115971428571m Ôù¥ Surface area = 6*0.115971428571^2 = 0.08069623346m┬▓
Assuming an emissivity of 0.03, aka that of copper; Ôù¥ rE = 0.03*5.6703e-8*120000000^4*0.08069623346 = 2.8463123820422e+22 watts/second, or 6.8 teratons of TNT (Small Country level+)
Could be higher, but yeah, that surface area is nowhere near lowballed anyway.
Matthew Schroeder says the minimum emissivity is 0.92 after a certain temperature, I also vaguely recall DontTalkDT saying that, but the DontTalkDT showed up and said:
Hmmm.... did I say that? If so I perhaps overgeneralized.
For stars and star like objects this should be true (as they are approximately black bodies apparently), but it shouldn't hold in general. As I learnt recently the emissivity of the plasma in a fusion reactor, for example, is on purpose made small through some means or another.
Then Kep said:
For black bodies, emissivity is not necessary, I believe, since it's ~1 by default. We can calculate the radiation with the temperature, area and the Boltzmann constant alone.
I am not sure what is the conclusion.
Note: Former calc group members, calc group members and Kaltias only
This calculation of Thanos crushing the Tesseract needs to be discussed.
Some dude on Naruto Forums calculated:
Minor thing but, from the first Avengers:
Steve: Does Loki need any particular kind of power source?
BANNER: He's got to heat the cube to a hundred and twenty million Kelvin just to break through the Coulomb barrier.
Implying the Tesseract could theorectically survive just under that, yet Thanos just kinda breaks it.
Taking it at face value:
http://www.endmemo.com/physics/radenergy.php
And using a surface area of like 0.02 m^2 as an eyeballed lowball, we get 2.3515E+23 Joules or 56.2 Teratons of tnt per second, so consistency I guess
But then Kepekley23 revised it:
Ring finger diameter = 28px = 19.8mm (average ring size for men is 10) Ôù¥ Edge size = 164px = 0.115971428571m Ôù¥ Surface area = 6*0.115971428571^2 = 0.08069623346m┬▓
Assuming an emissivity of 0.03, aka that of copper; Ôù¥ rE = 0.03*5.6703e-8*120000000^4*0.08069623346 = 2.8463123820422e+22 watts/second, or 6.8 teratons of TNT (Small Country level+)
Could be higher, but yeah, that surface area is nowhere near lowballed anyway.
Matthew Schroeder says the minimum emissivity is 0.92 after a certain temperature, I also vaguely recall DontTalkDT saying that, but the DontTalkDT showed up and said:
Hmmm.... did I say that? If so I perhaps overgeneralized.
For stars and star like objects this should be true (as they are approximately black bodies apparently), but it shouldn't hold in general. As I learnt recently the emissivity of the plasma in a fusion reactor, for example, is on purpose made small through some means or another.
Then Kep said:
For black bodies, emissivity is not necessary, I believe, since it's ~1 by default. We can calculate the radiation with the temperature, area and the Boltzmann constant alone.
I am not sure what is the conclusion.
Note: Former calc group members, calc group members and Kaltias only
