Is the World Engine feat more impressive than Thor's Nidavellir Star tanking feat?

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kell_saloks

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Poll: Is the World Engine feat more impressive than Thor's Nidavellir Star tanking feat? (53 votes)

Yes 26%
No 74%

Image result for superman world engine gif

vs

Image result for thor star feat gif

Explain why for your reason

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vjbthe3

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They're different

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kell_saloks

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@vjbthe3: Which one is more impressive, I know they're different.

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KryptonianKing88

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Honestly not sure, I've heard WE was affecting the core of the Earth but "Full force of a star" is quite the statement

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RBT

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One is a strength feat and other is a durability feat.

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incursion2

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Thors nidavellir feat is the most impressive high tier feat in recent CBM's imo

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Amcu

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Different types of feats.

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kell_saloks

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Thors nidavellir feat is the most impressive high tier feat in recent CBM's imo

I think Superman doesn't get enough credit, Thor's feat was probably higher since Star>>>Planet but Superman did his feat with out getting injured where as Thor was on the verge of death.

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Wot_m8

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Thor's feat is at best, an outlier, at worst, the star only has town-city level power since that is Thor's consistent level.

Superman's feat on the other hand is ok, I guess. Hard to quantify a terraforming machine.

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Ready_4_Madness

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Superman was weakened

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Odimm

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MrTrey

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#11  Edited By MrTrey

The one that's quantifiable and the best striking feat in the DCEU and MCU from their respective high tiers.

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eri123

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Different types of feats.

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ErickAgl17

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They are both different

The star feat is like a burning resistance feat.since stars are basically Heat i think.. or "power projecting resistance feat"

While the world engine seems more like its putting pressure on superman,

basically the star burns you...while the engine crushes you. They are different and cannot be compared.

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buildhare

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You'd need to be blind, deaf and dumb to think they're comparable in terms of overall impressiveness (Star forge feat is among the best durability feats put to screen, the World Engine isn't even DCEU Superman's best).

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DarkDementor101

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Two unquantifiable wanked feats? Tough choice since neither have anything to allow us to calc off of

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ComicGirl21

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WE. It's quantifiable to a degree because we know Earth's properties and can roughly estimate how powerful the gravity beam would have to be to influence the planet whatsoever (at least 1/1000 of earths mass), let alone reshape it's core. Even if we dont know exactly what it was doing, the force mustve been massive, quintillions of tons of force at least.

Thors star feat is completely abstract. Star had no notable gravity (Wasn't pulling Rocket, Thor etc) so it's mass mustve been tiny. Stars this small and so light do not exist. In other words its a star with completely fictional properties, impossible to recalculate into our physics. We have no idea how hot was the beam. Was it hundreds of degrees? Thousands? Millions? How should we know? Abstract feat.

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Nucleon

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Every feat by Superman get brought to ridiculous proportions.

Melting uru >>>>> crushing buildings.

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OnlyOneEmpereor

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Honestly not sure, I've heard WE was affecting the core of the Earth but "Full force of a star" is quite the statement

Thor isn't star level anyways. Comic thor isn't even star level so that is a complete outlier.

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NWName

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@darkdementor101: You can calc the star one going by the stars color, duration and Thor's size compared to the beam. It comes out at 98 kilotons.

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HERMES1220

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They’re different

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CocaColaMan

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The WE feat wasn’t that impressive for a durability feat. It’s GRAVITY, Superman was just getting his weight increased. Being weakened does help him, but it doesn’t even compare to a beam that could deform Uru.

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Crunch5481

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The WE feat wasn’t that impressive for a durability feat. It’s GRAVITY, Superman was just getting his weight increased. Being weakened does help him, but it doesn’t even compare to a beam that could deform Uru.

No that's not what happened. Increased gravity wouldn't push that much water out of the way, the beam was clearly exerting force itself. I lowballed it and calculated it to be about a 10,000 ton feat. The beam displaced a HUGE amount of water, more than I ever realized until recently. Then take into account that it is a beam where the force is concentrated in the center yet the water was pushed back further than the beam diameter, meaning that the center of the beam would be 10,000+ tons of force, then add that Superman was weakened by the atmosphere.

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Crunch5481

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@nwname said:

@darkdementor101: You can calc the star one going by the stars color, duration and Thor's size compared to the beam. It comes out at 98 kilotons.

Except you're applying real world star knowledge to an impossible star. That star is significantly smaller than is possible in real life and that matters.

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WordWarrior

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The WE feat isn't even impressive tbh.

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NWName

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@nwname said:

@darkdementor101: You can calc the star one going by the stars color, duration and Thor's size compared to the beam. It comes out at 98 kilotons.

Except you're applying real world star knowledge to an impossible star. That star is significantly smaller than is possible in real life and that matters.

You should not accuse without knowing how i did it (but its understandable since all others i saw calced it by ignoring literally everything we saw onscreen). I used a method that is not size dependant (it doesn't matter if its 20km like a real neutron star or 2 km like its on-screen). I also used the on-screen color temperature of 6600 K which gives half a billion times lower result compared to incorrectly using the temperature for a young real life neutron star (1 000 000K).

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DarkDementor101

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@nwname said:
@crunch5481 said:
@nwname said:

@darkdementor101: You can calc the star one going by the stars color, duration and Thor's size compared to the beam. It comes out at 98 kilotons.

Except you're applying real world star knowledge to an impossible star. That star is significantly smaller than is possible in real life and that matters.

You should not accuse without knowing how i did it (but its understandable since all others i saw calced it by ignoring literally everything we saw onscreen). I used a method that is not size dependant (it doesn't matter if its 20km like a real neutron star or 2 km like its on-screen). I also used the on-screen color temperature of 6600 K which gives half a billion times lower result compared to incorrectly using the temperature for a young real life neutron star (1 000 000K).

Could you please link the calculation. Just want to check it out for myself!

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NWName

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TheHercules

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MCU Thor Energy Durability > 616 Thor

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DarkDementor101

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So I might as well mention that I am one of those with the opinion that Nidavellir is a High-Tech Dyson sphere, which allows for many of the physics that takes place in there to be ignored [Super high gravity, Powerful magnetic fields, etc]. So while although I do think that it is possible to create a scientific calc to figure out how much energy was being output by the star, I do not want to go off the limb and say that your calculation is right. I think there are many more variables that go into calculating things such as these, and it is possible that some affect one another in ways you might not have considered in your calc.

That being said, I can contact a few of my professors and ask them a few questions about neutron stars and how these variables relate to one another. If I get anything I'll just post it on your thread

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o0Deadman0o

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"Thor withstanding the neutron star for 47 seconds is equivalent to surviving the largest ever man-made nuclear explosion every 0.012 seconds for 47 seconds, 3,959 50 megaton explosions in total."

Took this from a spacebattle forum post.

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Noone1996

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#31 Noone1996  Online

Nope.

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NWName

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#32  Edited By NWName

@darkdementor101: Okay thanks. Main assumption for my calc was counting it as a basic blackbody like normal stars instead of more like real neutron stars was that it neither displayed the magnetic proverties of a neutron star nor was implied to have extreme spinning speed like real ones. It was more like a normal star but just tiny.

However i would appreciate it if you ask what kinda output a white neutron star can have.

I will add 2 possible reasons why the calc can be wrong to the thread in the meantime.

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nfactor1995

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#33 nfactor1995  Online

WE. It's quantifiable to a degree because we know Earth's properties and can roughly estimate how powerful the gravity beam would have to be to influence the planet whatsoever (at least 1/1000 of earths mass), let alone reshape it's core. Even if we dont know exactly what it was doing, the force mustve been massive, quintillions of tons of force at least.

Thors star feat is completely abstract. Star had no notable gravity (Wasn't pulling Rocket, Thor etc) so it's mass mustve been tiny. Stars this small and so light do not exist. In other words its a star with completely fictional properties, impossible to recalculate into our physics. We have no idea how hot was the beam. Was it hundreds of degrees? Thousands? Millions? How should we know? Abstract feat.

This.

The star was miniscule (like a fraction of the size of Earth's moon type of miniscule) and its only feats were melting metal quickly and failing to burn away Thor's clothes or hair lol.

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ourmanuel

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@rbt said:

One is a strength feat and other is a durability feat.

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ourmanuel

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"Thor withstanding the neutron star for 47 seconds is equivalent to surviving the largest ever man-made nuclear explosion every 0.012 seconds for 47 seconds, 3,959 50 megaton explosions in total."

Took this from a spacebattle forum post.

heat durability =/= blunt/explosiion durability

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kell_saloks

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@ourmanuel: Spacebattles is 95% full of wankers who use outliers and highends as the average for characters.

They're worst than vs wiki

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Aka_aka_aka_ak

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We have no way of quantifying either

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Arthur_Morgan

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cant compare and no, thors durability is not star level.

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Nucleon

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@comicgirl21 said:

WE. It's quantifiable to a degree because we know Earth's properties and can roughly estimate how powerful the gravity beam would have to be to influence the planet whatsoever (at least 1/1000 of earths mass), let alone reshape it's core. Even if we dont know exactly what it was doing, the force mustve been massive, quintillions of tons of force at least.

Thors star feat is completely abstract. Star had no notable gravity (Wasn't pulling Rocket, Thor etc) so it's mass mustve been tiny. Stars this small and so light do not exist. In other words its a star with completely fictional properties, impossible to recalculate into our physics. We have no idea how hot was the beam. Was it hundreds of degrees? Thousands? Millions? How should we know? Abstract feat.

This.

The star was miniscule (like a fraction of the size of Earth's moon type of miniscule) and its only feats were melting metal quickly and failing to burn away Thor's clothes or hair lol.

And yet it melted uru. I don't care much for "continent-level" or "star-level" denominations, they sound terribly nerdish and manga to my ears, but I do know that Thor took what it takes to melt uru.

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nfactor1995

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#40 nfactor1995  Online

@nucleon: Do we know what it takes to melt uru?

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Chazzer

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#41 Chazzer  Online

@nfactor1995: failing to burn away Thor's clothes or hair lol.

=========

Then you would agree the nuke that Supes got hit with was not that powerful because his suit and hair were fine, right?

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nfactor1995

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#42 nfactor1995  Online

@chazzer: Touche. Point still stands that everything indicates that the star is exponentially weaker, smaller, less intense/hot, etc etc than any star in the real world, and thus we can't use any real world data to analyze the feat unless it is scaled way, way down. Also the issue of being pretty much inherently unquantifiable. Doesn't mean it isn't impressive, but it's been hilariously wanked beyond all reason to the point where people actually believe Thor is anywhere remotely close to "star level," whatever that's supposed to mean.

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Crunch5481

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#43  Edited By Crunch5481

@nwname said:
@crunch5481 said:
@nwname said:

@darkdementor101: You can calc the star one going by the stars color, duration and Thor's size compared to the beam. It comes out at 98 kilotons.

Except you're applying real world star knowledge to an impossible star. That star is significantly smaller than is possible in real life and that matters.

You should not accuse without knowing how i did it (but its understandable since all others i saw calced it by ignoring literally everything we saw onscreen). I used a method that is not size dependant (it doesn't matter if its 20km like a real neutron star or 2 km like its on-screen). I also used the on-screen color temperature of 6600 K which gives half a billion times lower result compared to incorrectly using the temperature for a young real life neutron star (1 000 000K).

No Caption Provided
No Caption Provided

I checked your calc, and I have serious problems.

First I would strongly disagree with your decision to call that star "white". Ignore the arrows, text (its not 24ft I know), lines, etc. for now, but this is directly from the movie and it clearly has a yellow/orange tint to it.

Second, you calculated the total power output of the star and then you made an obviously incorrect assumption that all of that power was focused through the beam. By dividing the total power output by the beam cross-sectional area you assume the entire star output is going through that beam which is blatantly false if you simply look at the image I have here. Clearly you can see that the star is outputting energy in all directions despite being enclosed. It is very inaccurate to make this false assumption. The more accurate way to calculate this would be to take the equation you used:

P = εAσT4

and divide A out from both sides so that you have P/A = epsilon*sigma*T^4 . With this new equation you would be calculating the watts per square meter of the whole star. Then you would multiple (P/A) by the area of the beam cross section and that would give a much more accurate answer to the power of the beam. I went and did that calculation (and I'm using a beam temperature of 5500 since that has a yellow tint) and I got 51,884,043.75 W/m^2 for the power output of the star. Onto problem 3.

No Caption Provided

Third, your beam diameter is much bigger than what I calculated it to be. It's not 24 feet like that picture suggests but I got a smaller diameter nonetheless. How did you calculate the beam diameter? I used a photo from the movie using Thor's body as a reference. The black bars represent his height, since he is not standing straight (shorter than normal) I estimated him to be 6feet tall instead of actual height. The beam is the size of the ring that the bars almost reach, so add an additional foot on each side to get ~62 feet in diameter. This is compared to your 2447.8 cm which is equal to 80.3 feet. Nevertheless this step is unnecessary actually since you can just multiply the power per square meter by 1 to get the power hitting Thor.

Fourth, I would argue ardently that the amount of time Thor was in the beam was actually closer to ~40 seconds, not 2 minutes. That two minutes is only based on Etri's statement that it would take a couple minutes, and not the observations we can make from the film. The entire Nidavellir forge scene seemed to be very time consistent between cuts based on character reactions, locations and the melting process. Therefore it is more accurate to measure the time he spent in the beam using the movie run time. So from when we see the beam start to stop on-screen it is about ~40 seconds, Not 2 full minutes.

Now for the revised calculation:

We will use 51,884,043.75 W/m^2 for the power output of the star, as previously calculated in this post.

We then multiply by 1 square meter to account for Thor's backside area, to get 51,884,043.75 W.

Then we multiply by 40 seconds to get 2,075,361,750 Joules, or 2075 MegaJoules

This amount of energy is equal to 0.000496 Kilotons of TNT versus your 119.5 Kilotons of TNT.

This new value is only 0.000415% of your calculation. Most of this discrepancy comes for your assumption (whether purposeful or not) that the beam contained all the star's output. Even if we make it the full two minutes the energy is only 3x times greater, which is 0.001245% of 119.5 at 0.001488 Kilotons of TNT. EVEN if we use the 6600 Kelvin Temperature that you claimed (which I disagree that the star is white as apparent by the images), then the total energy is 0.0029 Kilotons of TNT AND that is over the full 2 minutes you claimed! 0.0029 Kilotons is still only 0.00243% of your claimed 119.5 Kilotons of TNT. This is insanely small, and quite frankly makes the feat look pathetic. That's only 3,373 kWh, which is NOT a lot of energy. A typical coal power plant at 600MW will produce 14,400kWh of energy in one day of continuous operation. Another comparison: a W-78 Nuclear Warhead designed for three of them to be fitted on a Minuteman III Missle, has a yield of 335-350 Kilotons of TNT per Warhead meaning that one missile would have a combined payload of just over 1 Megaton of TNT, compared to 0.0029Kt. And remember these real world comparisons are compared to 2 minute exposure at 6600degK, versus what I see as more accurate as 40 seconds at 5500degK. All this is in addition to your problems you stated yourself about the emissivity maybe being less than one, which would decrease the energy, and then that the star is impossibly small.

Conclusion: Thor was hit by far less than even 1Kt of energy and NOT anywhere NEAR 119.5 Kt of energy.

TL;DR: Original calc makes bad assumptions that result in HUGE differences in total energy hitting Thor 0.000496 Kt (new calc) vs 119.5 Kt (old calc).

(This post will be posted underneath your calc as well)

Here's the link the nwname's original calc thread: https://comicvine.gamespot.com/marvel-cinematic-universe/4015-56089/forums/thor-nidavellir-calculation-2010870/

Tagging some people from this thread that may be interested.

@nfactor1995, @darkdementor101, @cocacolaman, @wot_m8, @incursion2, @kryptonianking88,@rbt

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NWName

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#44  Edited By NWName

@crunch5481 said:
@nwname said:
@crunch5481 said:
@nwname said:

@darkdementor101: You can calc the star one going by the stars color, duration and Thor's size compared to the beam. It comes out at 98 kilotons.

Except you're applying real world star knowledge to an impossible star. That star is significantly smaller than is possible in real life and that matters.

You should not accuse without knowing how i did it (but its understandable since all others i saw calced it by ignoring literally everything we saw onscreen). I used a method that is not size dependant (it doesn't matter if its 20km like a real neutron star or 2 km like its on-screen). I also used the on-screen color temperature of 6600 K which gives half a billion times lower result compared to incorrectly using the temperature for a young real life neutron star (1 000 000K).

No Caption Provided
No Caption Provided

I checked your calc, and I have serious problems.

First I would strongly disagree with your decision to call that star "white". Ignore the arrows, text (its not 24ft I know), lines, etc. for now, but this is directly from the movie and it clearly has a yellow/orange tint to it.

This is not the color of the star tho thats the beam that contains the energy of the star (whatever kinda particles its made out of are heated to around 5500K). You can see the color of the star (its surface exposed from the open iris) in the last pic in the OP.

Second, you calculated the total power output of the star and then you made an obviously incorrect assumption that all of that power was focused through the beam. By dividing the total power output by the beam cross-sectional area you assume the entire star output is going through that beam which is blatantly false if you simply look at the image I have here. Clearly you can see that the star is outputting energy in all directions despite being enclosed. It is very inaccurate to make this false assumption. The more accurate way to calculate this would be to take the equation you used:

Its a dyson sphere. Its entire purpose is to collect nearly 100% of the output and the beam should contain the total output of the star as it is the only part of the machine that gives energy to the forge. What do you think the machine does to the collected heat if not used in the beam? It has no connection to any other structure either. The beam is not the visualisation of the radiation coming out of the exposed part alone, the beam did not immediately come out as Thor slowly opened it as at the beginning there was only some light and some negligable amount of particles coming out. And after the thing is fully opened however, the particle beam started and made a shockwave. How does that picture prove the beam only contains the energy coming from the exposed part directly under the opened wall? And the star outputting energy in all directions? Like 90+% of the surface is completely covered by apaque metal how can it come out from the entire surface of the star?

P = εAσT4

and divide A out from both sides so that you have P/A = epsilon*sigma*T^4 . With this new equation you would be calculating the watts per square meter of the whole star. Then you would multiple (P/A) by the area of the beam cross section and that would give a much more accurate answer to the power of the beam. I went and did that calculation (and I'm using a beam temperature of 5500 since that has a yellow tint) and I got 51,884,043.75 W/m^2 for the power output of the star. Onto problem 3.

No Caption Provided

Third, your beam diameter is much bigger than what I calculated it to be. It's not 24 feet like that picture suggests but I got a smaller diameter nonetheless. How did you calculate the beam diameter? I used a photo from the movie using Thor's body as a reference. The black bars represent his height, since he is not standing straight (shorter than normal) I estimated him to be 6feet tall instead of actual height. The beam is the size of the ring that the bars almost reach, so add an additional foot on each side to get ~62 feet in diameter. This is compared to your 2447.8 cm which is equal to 80.3 feet. Nevertheless this step is unnecessary actually since you can just multiply the power per square meter by 1 to get the power hitting Thor.

I assumed 2km diameter for the star which isi like i said in the OP, inconsequential as the beam intensity comes from the surface area ratio between the star and the beam and not the surafce area.

Fourth, I would argue ardently that the amount of time Thor was in the beam was actually closer to ~40 seconds, not 2 minutes. That two minutes is only based on Etri's statement that it would take a couple minutes, and not the observations we can make from the film. The entire Nidavellir forge scene seemed to be very time consistent between cuts based on character reactions, locations and the melting process. Therefore it is more accurate to measure the time he spent in the beam using the movie run time. So from when we see the beam start to stop on-screen it is about ~40 seconds, Not 2 full minutes.

I usually tend to avoid on-screen timeframe as they are often not even relatively accurate due to cuts. But i dont remember the scene entirely so it might be time consistent.

Now for the revised calculation:

We will use 51,884,043.75 W/m^2 for the power output of the star, as previously calculated in this post.

We then multiply by 1 square meter to account for Thor's backside area, to get 51,884,043.75 W.

Then we multiply by 40 seconds to get 2,075,361,750 Joules, or 2075 MegaJoules

This amount of energy is equal to 0.000496 Kilotons of TNT versus your 119.5 Kilotons of TNT.

So you just multiplied the surface emission with time? Thats basically what it would take to stay near a 5500K star without any dyson sphere caused effects.

This new value is only 0.000415% of your calculation. Most of this discrepancy comes for your assumption (whether purposeful or not) that the beam contained all the star's output.

It is very much purposeful as the feat involves a machines with one and only purpose of collecting the stars total output as much as possible and only has a single way it uses the collected energy, powering the forges, with the only connection to it being a particle beam(that is clearly not just pure radiation from the exposed area only with visible movement speed and a instantly created shockwave) directed at the forge.

Even if we make it the full two minutes the energy is only 3x times greater, which is 0.001245% of 119.5 at 0.001488 Kilotons of TNT. EVEN if we use the 6600 Kelvin Temperature that you claimed (which I disagree that the star is white as apparent by the images), then the total energy is 0.0029 Kilotons of TNT AND that is over the full 2 minutes you claimed! 0.0029 Kilotons is still only 0.00243% of your claimed 119.5 Kilotons of TNT. This is insanely small, and quite frankly makes the feat look pathetic. That's only 3,373 kWh, which is NOT a lot of energy. A typical coal power plant at 600MW will produce 14,400kWh of energy in one day of continuous operation. Another comparison: a W-78 Nuclear Warhead designed for three of them to be fitted on a Minuteman III Missle, has a yield of 335-350 Kilotons of TNT per Warhead meaning that one missile would have a combined payload of just over 1 Megaton of TNT, compared to 0.0029Kt. And remember these real world comparisons are compared to 2 minute exposure at 6600degK, versus what I see as more accurate as 40 seconds at 5500degK. All this is in addition to your problems you stated yourself about the emissivity maybe being less than one, which would decrease the energy, and then that the star is impossibly small.

Conclusion: Thor was hit by far less than even 1Kt of energy and NOT anywhere NEAR 119.5 Kt of energy.

TL;DR: Original calc makes bad assumptions that result in HUGE differences in total energy hitting Thor 0.000496 Kt (new calc) vs 119.5 Kt (old calc).

How is the assumption bad? Why even build a machine that has the purpose of collesting the entire output if the force is only going to use the energy intensity avaible at the surface of the star?

TL,DR: The beam has a color temperature of 5500K not the star's surface, and the beam contains near the total output of the star as the multi km^2 structure (the dyson sphere) would be usesless otherwise. 40 seconds might be better tho so 40 kt might be more accurate than 120 kt.

(This post will be posted underneath your calc as well)

Here's the link the nwname's original calc thread: https://comicvine.gamespot.com/marvel-cinematic-universe/4015-56089/forums/thor-nidavellir-calculation-2010870/

Tagging some people from this thread that may be interested.

@nfactor1995, @darkdementor101, @cocacolaman, @wot_m8, @incursion2, @kryptonianking88,@rbt

Edit for @darkdementor101 : We don't know what the beam is made out of and in fiction, including MCU, some matter have impossibly high specific heats. And to contain that much energy in only that amount of matter either the density or the specific heat has to be extreme. This is a quicknote without much thought behind it tho will likely come back to this later.

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DarkDementor101

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@nwname@crunch5481: Mind if I but into the conversation?!

@crunch5481 said:

First I would strongly disagree with your decision to call that star "white". Ignore the arrows, text (its not 24ft I know), lines, etc. for now, but this is directly from the movie and it clearly has a yellow/orange tint to it.

Sorry that I had to delete the images, but the post was going to get too long! @nwname, It seems that @crunch5481 is right about the star's actual temperature. The images that you provided shows a much more white beam , whereas most of the images shown by @crunch5481 shows a orange/yellow shade. I think you should rectify that part of your calculation! Although, this also a good time to mention that using the color of the beam itself does not seem to be the best way to find the temperature of the star. Rather, you should use the temperature of the plasma that is being discharged off of the sphere at the center of the rings (Which does seems to be orange/yellow either way) since the concentration of energy into a beam would affect the heat of the beam, and thus give off a different wavelength leading to different color

Second, you calculated the total power output of the star and then you made an obviously incorrect assumption that all of that power was focused through the beam. By dividing the total power output by the beam cross-sectional area you assume the entire star output is going through that beam which is blatantly false if you simply look at the image I have here. Clearly you can see that the star is outputting energy in all directions despite being enclosed. It is very inaccurate to make this false assumption. The more accurate way to calculate this would be to take the equation you used:

P = εAσT4

and divide A out from both sides so that you have P/A = epsilon*sigma*T^4 . With this new equation you would be calculating the watts per square meter of the whole star. Then you would multiple (P/A) by the area of the beam cross section and that would give a much more accurate answer to the power of the beam. I went and did that calculation (and I'm using a beam temperature of 5500 since that has a yellow tint) and I got 51,884,043.75 W/m^2 for the power output of the star. Onto problem 3

I'll address this part tomorrow. Getting late here!

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KryptonianKing88

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@crunch5481: I think Thor did take the beam for 2-3 min, showing it for that long would make the scene draw out

.0005 kt seems WAY too low, I doubt even the lowest calc of the Sokovia explosion would go that low.

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NWName

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@darkdementor101: I adressed the color and other parts in post 44, posted just seconds before yours.

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#48  Edited By NWName
@kryptonianking88 said:

@crunch5481: I think Thor did take the beam for 2-3 min, showing it for that long would make the scene draw out

Thats what i thought too.

.0005 kt seems WAY too low, I doubt even the lowest calc of the Sokovia explosion would go that low.

This would have extremely weird implications of a feat involving multi kilometer structure and that is describes as something only Thor can accomplish being comparable to just 50 liters of fuel burning and Thor being unable to take the heat lighting for no more that a fraction of a second. Lmao dwarves should have used good old fashioned gasoline right? This is downplay to impossible degree.

I already adressed why i think the post was wrong but this just adds to it.

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Nucleon

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#49  Edited By Nucleon

@nfactor1995 said:

@nucleon: Do we know what it takes to melt uru?

Nothing less than the concentrated heat of a dying star. =)

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@nwname: You can also see the star color in the pic I gave that wasn't the beam. I gave both specifically because I knew you'd say that the beam could be different color than the star, but instead you just ignored the picture showing the whole star where it is orange tinted.

Is it stated to be a dyson sphere? I don't recall that being said. Just because it is surrounded by metal doesn't mean that it is a dyson sphere. The sphere seems to just contain it. Again I will reference this picture:

No Caption Provided

Look at this image and tell me that the beam contains the full power output of the star in it. (It doesn't...obviously) I don't know how to make this more obvious, but if we can see light and energy pouring out in all directions from the star then you cannot say that 100% of the energy is in the beam. This is regardless of whether or not the sphere is meant to be a dyson sphere or not because we can see that it does not get 100% of the energy. As to your question for what the sphere does if not contain the energy, it is simple, the machine turns the star on. When the rings align together the star ignites. So clearly there is some interaction between the rings and the center metal sphere despite there being no physical connection. This picture does not necessarily prove that the beam only contains the energy from that part of the star, but we can visually see how much energy is NOT in that beam and I can say with certainty through visual observation alone that there is significantly less energy in that beam then the total output of the star .

Okay.

I've checked the scene multiple times in the past and to me yes it seems time-consistent.

That's what it would take to basically be in contact with the surface of the star for that amount of time for that given square meter, and that is essentially what Thor did. Instead of him going to the surface the surface came to him in beam form.

Just because there is only way to use the energy they are collecting does not say anything about the amount of energy they collect. Again we can see they don't collect it all from that image. Also based on how the beam looks and seems to work with the focusing rings, it seems to me that is simply pulling plasma from the star into a beam. That is very different from a dyson sphere. Also if it is a dyson sphere (which it isn't)you were also assuming 100% efficiency, which is verifiably false from visual observation.

That assumption is bad because we can see all the energy not being collected, so obviously 100% can't be used. Then by looking at the beam mechanism and how it works, it looks like the star is being drained, not that all the energy from it's surface is being concentrated into a beam, that's not what it looks like at all. The purpose is to contain the star, ignite when the beams align, and then to drain a beam from the star. It is not a dyson sphere, certainly not a typical one, since when do dyson spheres use beams of plasma.