Follow

    Marvel Cinematic Universe

    Concept » Marvel Cinematic Universe appears in 146 issues.

    Marvel's superhero movie continuity that is shared between several major character franchises.

    Thor Nidavellir Calculation

    • 99 results
    • 1
    • 2
    • 3
    Avatar image for nwname
    nwname

    10064

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    #1  Edited By nwname  Moderator

    First let me say this: I have seen multiple calculations of the nidavellir feat and none were accurate, they all ignored that the star was an old star that had gone out and had to be reignited. In this calc i will use the actual temperature of the star that is accurate to the visuals of the movie and not the temperature of a young real life neutron star.

    No Caption Provided

    Here i scaled the size difference between the star and the beams diameter. The size i assumed (2000 meters) is close to the one you get via pixel scaling but it doesn't matter as the result of this calc solely on the size radio of the beam and the star, the surface area of Thor that is exposed to the beam and the temperature of the star.

    Surface area of the star = 4 x 3,1415 x 1000^2 = 12.556.000 m^2.

    No Caption Provided
    No Caption Provided

    Going by color temperature of the star it is around 5000 Kelvin.

    P = εAσT4

    Going by this formula of black body radiation the energy output of the star is: 1 (emissivity of the star) x 12 556 000 (surface area) x σ=5.67 × 10-8 W/m2•K^4 (Stefan-Boltzmann constant) x 5000^4 = 4.45307625E+14 watts.

    Surface area of the beam = ~396 m^2 going by 3,1415 x r^2.

    Heat intensity of the beam = 4.45307625E+14 / 396 = 1.124.514.204.545 watts/m^2.

    Thor had to hold it open for a few minutes. Assuming he couldn't do a small part of the process thus no handle, 2 minutes is a safe bet.

    60 x 2 (timeframe in seconds) x 1.124.514.204.545 = 134.941.704.545.448 Joules per m^2. So approximately 135 TJ or ~32 Kilotons of TNT since exposed part perpendicular to the beam including the armor should be ~1 m^2.

    No Caption Provided

    Correction edit: star itself appears to be fompletely white so 6600 K is more fitting. So 1,35200000E+15 / 396 x 120 = 409.696.969.696.969 joules or ~410 TJ or 98 Kilotons of TNT.

    The Updated final version of the calculation is in the Comment #7

    Avatar image for deactivated-5d01cd4d1eb4b
    deactivated-5d01cd4d1eb4b

    2650

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    cool <3

    Avatar image for helloman
    helloman

    30115

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    This is wrong.

    Avatar image for nwname
    nwname

    10064

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    #4 nwname  Moderator

    @helloman said:

    This is wrong.

    I know this is old but why?

    Avatar image for erkan12
    Erkan12

    10904

    Forum Posts

    1017

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 1

    @nwname said:
    @helloman said:

    This is wrong.

    I know this is old but why?

    Probably because Nidavellir is a magical place and you can't use scientific calculations?

    LOL.

    Avatar image for nwname
    nwname

    10064

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    #6 nwname  Moderator

    @erkan12: Its not a magical place, its a super high tech structure (dyson sphere) and the structure doesn't matter for this calc, unless you think the star is magic as well.

    Avatar image for nwname
    nwname

    10064

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    #7  Edited By nwname  Moderator

    Updated calculation:

    • With pixel scaling the star is around 850 pixels and the beam is 8 pixels wide. That means the surface area ratio is 45 156. So the heat is focused that much more compared to the surface of the star.
    • The surface temperature should be more like 6400-6500K instead of 6600K.
    • The dyson sphere had small holes in it and some light was definitely escaping. Im assuming 3% of the heat escapes (doesn't matter much since its a tiny portion)

    So with all this considered the updated calc is: 95126814 W/m^2 star radiation, multiplied by 45 156 is 4 295 546 412 984‬ joules per second or 4 295 Gigawatts (roughly 4.3 TW) accounting for the 3% energy escape that is a total of 500 TJ total heat or equal to the total energy output of 119.5 kilotons of TNT.

    Additional things to consider:

    • Thor has decent regeneration so him taking 500 TJ over 2 minutes time span does not necessarily mean he can take a 500 TJ attack that lasts 2 seconds
    • Blackbody ratiation formulas might not works correctly in some extreme situations like the neutron star being a magnetar or being millions of degrees hot but neither fits the star in the movie so it should not be a problem to use the formula.

    So what does this equal to? Some comparisons to irl and MCU things:

    • Enough to vaporize 14 000 humans each second.
    • Enough to burn through the energy of 1431 first generation arc reactors each second.
    • Enough to power over 3 400 000 000 average US houses.
    Avatar image for nwname
    nwname

    10064

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    #8  Edited By nwname  Moderator

    Still, there are 2 Possible problems with this:

    1. Since its said to be a neutron star it might have some properties like the real ones where it can generate radiation via other mechanics than just blackbody radiation due to the extreme spinning speed and magnetic fields of a neutron star. I personally think the star in Nidavellir is more like a normal star (a basic blackbody) since it did not display rapid rotation or extreme magnetic properties. However these can be due to the Dyson Sphere somehow negating the effects from affecting the outside of the core machine.
    2. Second possible problem is... it being a magic star that doesn't work anything like other stars making it unquantifiable. Main reason for this is it is too small even for a Neutron Star. For all we know it has a "atmosphere" that has extremely low density and that might be what glows making the output far lower.
    Avatar image for erkan12
    Erkan12

    10904

    Forum Posts

    1017

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 1

    #9  Edited By Erkan12

    @nwname said:

    @erkan12: Its not a magical place, its a super high tech structure (dyson sphere) and the structure doesn't matter for this calc, unless you think the star is magic as well.

    It's obvious that's a magical place since Eitri can use dwarven magic and those dwarves created that Star Forge by using the same magic. You can't call it ''Super high tech'' structure when you don't even know how it works, lmao. One thing we know that they have dwarven magic, and a usual neutron star doesn't have a metal core like this;

    No Caption Provided

    If it wasn't magical, there is no way this metal core could exist for centuries while the neutron star burns.

    Avatar image for nwname
    nwname

    10064

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    #10 nwname  Moderator

    @erkan12 said:
    @nwname said:

    @erkan12: Its not a magical place, its a super high tech structure (dyson sphere) and the structure doesn't matter for this calc, unless you think the star is magic as well.

    It's obvious that's a magical place since Eitri can use dwarven magic and those dwarves created that Star Forge by using the same magic (That assumption is baseless, as well as being contradicted by the vfx team saying its a dyson sphere). You can't call it ''Super high tech'' structure when you don't even know how it works, lmao. One thing we know that they have dwarven magic, and a usual neutron star doesn't have a metal core like this (Thats the dyson sphere itself, the things around it are orbital rings);

    No Caption Provided

    If it wasn't magical, there is no way this metal core could exist for centuries while the neutron star burns (Look up Dyson Spheres, thats what they basically are, structures covering the entire star to take its energy).

    Avatar image for erkan12
    Erkan12

    10904

    Forum Posts

    1017

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 1

    #11  Edited By Erkan12

    @nwname: I am not talking about the rings, I am talking about the metal core which stands there in the middle of the space even when there is no power. And another fact is that Eitri uses dwarven magic to create weapons.

    Avatar image for nwname
    nwname

    10064

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    #12  Edited By nwname  Moderator

    @erkan12: ... Did you even look up what a dyson sphere is? It IS the sphere core, not the rings around it. Rings are just additional tech. Yes, it normally would collapse on top of the star ut with sufficient tech (like gravity negation which mcu ships have already shown) it can stay there.

    Whole theme of Asgard was tech combined with magic not magic alone. The structure = super high tech, The enchantments on the weapons = magic.

    Avatar image for crunch5481
    Crunch5481

    5293

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    @nwname:

    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.

    No Caption Provided
    No Caption Provided

    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 my I used 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 his 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: the W-78 (a common US bomb) 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).

    Avatar image for nwname
    nwname

    10064

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    #14  Edited By nwname  Moderator

    @crunch5481 said:

    @nwname:

    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.

    No Caption Provided
    No Caption Provided

    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 my I used 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 his 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: the W-78 (a common US bomb) 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.

    Also 2 GJ is hilarious. Dwarves should have just used gasoline or other cheap fuels right?

    Avatar image for erkan12
    Erkan12

    10904

    Forum Posts

    1017

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 1

    @nwname said:

    @erkan12: ... Did you even look up what a dyson sphere is? It IS the sphere core, not the rings around it. Rings are just additional tech. Yes, it normally would collapse on top of the star ut with sufficient tech (like gravity negation which mcu ships have already shown) it can stay there.

    Whole theme of Asgard was tech combined with magic not magic alone. The structure = super high tech, The enchantments on the weapons = magic.

    I did. They surround a natural star in theory. A natural neutron star.

    This is not a neutron star;

    No Caption Provided

    It's dead. Thor said the star is gone out. There is no power in it. And Eitri said the star needed to be awakened. It only works via magic when they move the rings. It's probably related to the dark energy and works via the same magic with the Bifrost.

    They also said that the Dyson Sphere must be as big as a solar system, while Nidavellir was as big as a moon at best. It's completely different.

    Eitri also said that Thor was taking the full force of a star, while Thor was taking only a very small part of it on screen, that's not possible if magic is not involved. Based on what Eitri said, it's magical and you can't use scientific calculations on it, it's ridiculous.

    Avatar image for rukelnikovftw
    RukelnikovFTW

    7558

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    Loading Video...

    Avatar image for deactivated-5dbe38e11d7b7
    deactivated-5dbe38e11d7b7

    323

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    @nwname What would be different in the initial conditions if we could start from the assumption that the beam is from the core?

    @erkan12

    There's some VFX words on a dyson sphere for Nidavellir, but would you accept them?

    Avatar image for nwname
    nwname

    10064

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    #18 nwname  Moderator

    Loading Video...

    He ignores the context and on screen proof completely. It was an old dead neutron star that had to be reignited, making the 1 million K assumption based on a relatively young neutron star wrong. There is also the color of the star indicating a temperature of 5000-6600 K and then there are statements of nidavellir being a dyson sphere. He got everything wrong.

    @reevjar said:

    @nwname What would be different in the initial conditions if we could start from the assumption that the beam is from the core?

    @erkan12

    There's some VFX words on a dyson sphere for Nidavellir, but would you accept them?

    I dont know where the beam is coming from. Normally Dyson spheres collect energy of the star and don't use matter from it. Still the energy absorbed by the sphere from the surface radiation should be the same. It looks like some negligible amount (sompared to the stars mass of even a tiny fraction of the extremely dense surface) of matter taken out of the star getting heated by the collected energy being redirected into the forge.

    Avatar image for deactivated-5dbe38e11d7b7
    deactivated-5dbe38e11d7b7

    323

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    @nwname said:
    @rukelnikovftw said:
    Loading Video...

    He ignores the context and on screen proof completely. It was an old dead neutron star that had to be reignited, making the 1 million K assumption based on a relatively young neutron star wrong. There is also the color of the star indicating a temperature of 5000-6600 K and then there are statements of nidavellir being a dyson sphere. He got everything wrong.

    @reevjar said:

    @nwname What would be different in the initial conditions if we could start from the assumption that the beam is from the core?

    @erkan12

    There's some VFX words on a dyson sphere for Nidavellir, but would you accept them?

    I dont know where the beam is coming from. Normally Dyson spheres collect energy of the star and don't use matter from it. Still the energy absorbed by the sphere from the surface radiation should be the same. It looks like some negligible amount (sompared to the stars mass of even a tiny fraction of the extremely dense surface) of matter taken out of the star getting heated by the collected energy being redirected into the forge.

    Odin: "Heart of a dying star"

    Thor:

    "Awaken the heart of a dying star." 

    Post-feat the beam seems to have turned into superfluids, which have apparently been seen in the core of neutron stars.

    This either means -

    1. Star core

    2. It's not a dying neutron star, just one enclosed by an inactive sphere, since neutron stars are cores unto themselves, no?

    Avatar image for chimeroid
    Chimeroid

    12142

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 1

    User Lists: 0

    @nwname said:
    @rukelnikovftw said:
    Loading Video...

    He ignores the context and on screen proof completely. It was an old dead neutron star that had to be reignited, making the 1 million K assumption based on a relatively young neutron star wrong. There is also the color of the star indicating a temperature of 5000-6600 K and then there are statements of nidavellir being a dyson sphere. He got everything wrong.

    @reevjar said:

    @nwname What would be different in the initial conditions if we could start from the assumption that the beam is from the core?

    @erkan12

    There's some VFX words on a dyson sphere for Nidavellir, but would you accept them?

    I dont know where the beam is coming from. Normally Dyson spheres collect energy of the star and don't use matter from it. Still the energy absorbed by the sphere from the surface radiation should be the same. It looks like some negligible amount (sompared to the stars mass of even a tiny fraction of the extremely dense surface) of matter taken out of the star getting heated by the collected energy being redirected into the forge.

    One idiotic thing that i see from both Because Science and most other YT channels is that they always assume that Mjolnirr weighs trillions of tons, when in fact it isn't heavy at all, making most, if not all, of their calculations mute.

    Avatar image for kryptonianking88
    KryptonianKing88

    10459

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    This thread is hilarious .00049 kt level to melt Uru? the same Uru that no sold the Sokovia explosion and channels enough energy to light up a continent?

    Also, lol at using the star's color. They obviously wouldn't go with blue since it would seem LESS powerful to anyone unfamiliar with stars

    Avatar image for johndeyvido
    Johndeyvido

    501

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    @kryptonianking88:

    LMAO it's funny how fanboys embarrass themselves trying to debunk what was shown on screen often citing scientific reasons. Fictional characters/movies don't follow RL science 100% of the time. Dceu Clark's suit tanked a nuclear explosion while Clark himself nearly died. People using colour of the beam are the worst, vibranium that's weaker than uru took Thor's lightning, IM repulsor blast and Vision's infinity stone blast for an extended period of time before it started to melt slowly but apparently 6k° F is enough to melt uru.

    Avatar image for chimeroid
    Chimeroid

    12142

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 1

    User Lists: 0

    @kryptonianking88:

    LMAO it's funny how fanboys embarrass themselves trying to debunk what was shown on screen often citing scientific reasons. Fictional characters/movies don't follow RL science 100% of the time. Dceu Clark's suit tanked a nuclear explosion while Clark himself nearly died. People using colour of the beam are the worst, vibranium that's weaker than uru took Thor's lightning, IM repulsor blast and Vision's infinity stone blast for an extended period of time before it started to melt slowly but apparently 6k° F is enough to melt uru.

    Well, here's the thing, there is a huge difference between material and the product.

    Maybe 6k F is enough to melt the material, but you get a stronger prodct, that's the point.

    For example, you melt Iron to make Steel.

    Iron melts at 1500 celsius, steel melts at 2500. It gets stronger by getting melted and worked on.

    Avatar image for rajjarsalt
    rajjarsalt

    28127

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    #24  Edited By rajjarsalt

    Imagine unironically calling Superman's feat over a megaton and then simultaneously scrutinizing Thor's feat as a minute fraction of a kiloton.

    Avatar image for johndeyvido
    Johndeyvido

    501

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    @chimeroid:

    Except for what we see on screen. It wasn't melted to get a new product, it was melted to be poured on a mold to get a desired shape. So the metal that melted is the same metal that solidified in the shape of an axe. 6k F isn't scratching the metal, no chance.

    Avatar image for chimeroid
    Chimeroid

    12142

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 1

    User Lists: 0

    @chimeroid:

    Except for what we see on screen. It wasn't melted to get a new product, it was melted to be poured on a mold to get a desired shape. So the metal that melted is the same metal that solidified in the shape of an axe. 6k F isn't scratching the metal, no chance.

    Fair enough, however, the material is far from being the only thing influencing this. Dwarves work with insanely high tech.

    Avatar image for johndeyvido
    Johndeyvido

    501

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    @chimeroid:

    Yes they have high-tech/magic or whatever but still the process is different from melting iron ore to remove impurities and adding other material to get alloys more durable.

    I already explained why you cannot discredit sci-fi movies using RL science. Also using screen visuals to calculate a feat is wrong unless vfx does not exist. People were saying Surtur was 300m tall when Hulk hits him based on visual representation but according to the vfx team he was 800m tall almost 3x taller than CV calculated him to be.

    Thor took a concentrated beam from a neutron star while lifting the structure the size of a small moon according to the vfx team so if anyone wants to calculate it using rl science, go for it but to try to disprove it using rl science is just wasting your time because he's fictional character in a fictional movie.

    Avatar image for chimeroid
    Chimeroid

    12142

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 1

    User Lists: 0

    @chimeroid:

    Yes they have high-tech/magic or whatever but still the process is different from melting iron ore to remove impurities and adding other material to get alloys more durable.

    I already explained why you cannot discredit sci-fi movies using RL science. Also using screen visuals to calculate a feat is wrong unless vfx does not exist. People were saying Surtur was 300m tall when Hulk hits him based on visual representation but according to the vfx team he was 800m tall almost 3x taller than CV calculated him to be.

    Thor took a concentrated beam from a neutron star while lifting the structure the size of a small moon according to the vfx team so if anyone wants to calculate it using rl science, go for it but to try to disprove it using rl science is just wasting your time because he's fictional character in a fictional movie.

    Dude, it's very simple, without science that feat is just Thor taking a sunbath in a moderately high temperature. The calc makes this feat out to be a LOT more impressive that it is at face value.

    Avatar image for chuggachugga170
    chuggachugga170

    1506

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    B

    Avatar image for nwname
    nwname

    10064

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    #30 nwname  Moderator
    No Caption Provided

    Update based on this. Beam seemingly gets compressed but later expands too so its not that unlikely that the beam is as hot as the melting point of Uru (but its certainly possible that it was nit as hot as the compressed fire). This new one is assuming it was 50.000K that would cap the energy Thor took at:

    5.67x10^-8 * 2 * (50.000^4 ) * 1 = 708.750.000.000 watts or ~0.7 TW. Assuming he took the beam for 2 minutes he took 85 TJ/20.3 kilotons.

    Avatar image for gangorca
    GangOrca

    13478

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    @nwname: Question, would the energy withstood by Thor not be higher? Isn't 50,000 K the melting point of Uru and not the temperature of the star, like in your calculation?

    Avatar image for nwname
    nwname

    10064

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    #32 nwname  Moderator

    @gangorca: Well it wpuld actually be lower. The beam later gets compressed too which normally increases the temperature meaning the beam itself was <50.000K. Temperature of the star seems to be 4000-6000K based on its color.

    Avatar image for deactivated-62f3a8e120119
    deactivated-62f3a8e120119

    1031

    Forum Posts

    1

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    Wrong thread typed nvm

    Avatar image for deactivated-6310e05cef78c
    deactivated-6310e05cef78c

    3715

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    Avatar image for nwname
    nwname

    10064

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    #41 nwname  Moderator

    @akz: Umm yeah I have already addressed this before in a thread thinking a neutron star is only 6600 kelvins by looking at the color which itself is not completely visible is pure cope and lowball a neutron star is muuuch hotter than 6600K

    @akz: yeah its really simple in temperature T put T = E +6 for low end and T = E+11 or +12

    the feat will become 10^8 times more on low end and 10 ^32 times more high end from the value calculated in the OP .

    Neutron stars that are 600k-1 million Kelvin are extremely rare. It is not the average temperature of a neutron star, it is the average temperature of one that is hot and bright enough to be detected from Earth. Average neutron stars are cold. 6600 K is a more likely temperature.

    Neutron stars that can be observed are very hot and typically have a surface temperature of around 600000 K.

    There are thought to be around one billion neutron stars in the Milky Way, and at a minimum several hundred million, a figure obtained by estimating the number of stars that have undergone supernova explosions. However, most are old and cold and radiate very little; most neutron stars that have been detected occur only in certain situations in which they do radiate, such as if they are a pulsar or part of a binary system.

    Wikipedia

    10^11 to 10^12 only last for instants after its birth. It falls down to 10^6K in just a few years. Most neutron stars are billions of years old. 10^32 times higher energy than mass-energy equivalent of the star each second. Again intuitively you should be able to tell a star can not radiate far more mass than it has... guess not.

    Not all radiation has to be visible. That doesn't change the fact that black body objects at temperatures above 10000K will look blue. The higher the temperature goes the more dominant blue becomes in our vision.

    No Caption Provided
    @akz said:

    @nwname what say to this?

    Bullshit based on a misunderstanding that neutron stars hot enough to be seen from earth = average temperature. Also entirely ignores on screen proof.

    Avatar image for nwname
    nwname

    10064

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    #43 nwname  Moderator

    @akz said:

    @nwname: In calc, did he take 98 KT of tnt over the whole duration or per sec?

    Whole duration. Uru took considerably more to melt tho as Thor was only blocking a small section of the beam.

    Avatar image for kjp
    Kjp

    633

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    This just prove that both mcu Thor and dceu sup are fodders

    Avatar image for nwname
    nwname

    10064

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 0

    #46 nwname  Moderator

    @nwname said:

    I didn't see this lol let me address it

    Your first misconception is that you think neutron stars cool down immediately after formation no true at all .

    https://astronomy.stackexchange.com/questions/14387/what-happens-over-time-as-a-neutron-star-cools

    after 10000 - 100000 years the star's surface will come to 1 million degrees .

    and after that the star will cool as T = kt^1/2 . ( find out the unit for t)

    LMAO things like mass energy equivalence don't suit a noob like you I will explain the things that I know :-

    1) the eqn E = mc^2 where m is the rest mass is a small part of the real picture the original eqn is frame dependent

    E^2 = (Pc)^2 + m^2c^4 here m is the rest mass and P is the momentum of the body in the observer's frame of reference so one single body can have different energies contained with respect to different observers.

    so you saying a star can't emit more than the energy contained in its mass is completely meaning less since that energy itself is not absolute .

    Also how much energy any body emits in unit time is independent of its mass it only depends on the surface temperature and the exposed non insulated area but considering what all things you say it is not very surprising now.

    about that color part to the naked human eye it would look more towards red than blue due to red shift (don't know much about it so explain it later on )

    I didnt say it comes down to 1000s instantly but yes they do start to cool down the moment they form. At faster rates at first too, the link you gave says this aswell. 10^12-11 K doesnt last long at all. I have no misconception here. Most neutron stars are billions of years old thus cold. The thread you linked yourself says they would get down to Sun temperature in a billion years (shorter than the average multi-billion age) without reheating processes like sucking in matter from the outside or fast rotation (which iirc is applicale to millisecond pulsars). So thanks for extra proof helping my case.

    Again with the unrelated stuff. Are you trying to make it look confusing? What does energy momentum relation got to do with a stationary star? It can have different total relativistic energy compared to different frames of reference but its rest mass doesnt change with different reference frames. Are you actually trying to say a star can release more than what it has (from and to a certain frame of reference)?

    I know that. However its just obvious a star can not radiate away more than it has.

    How is redshift relevant? Running out of ways to bullshit? Only type of redshift that might be relevant here is gravitational redshift i guess but even ignoring the fact that the star in the movie clearly lacked a gravitational field like that and going solely for a real case, UV light produced will be redshifted into blue so a hot star would still seem blue.

    Avatar image for antebellum
    Antebellum

    3144

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 2

    Didn't understand nothing but seems plausible.

    Avatar image for supermanwin1875
    supermanwin1875

    4232

    Forum Posts

    0

    Wiki Points

    0

    Followers

    Reviews: 0

    User Lists: 1

    lmfao

    This edit will also create new pages on Comic Vine for:

    Beware, you are proposing to add brand new pages to the wiki along with your edits. Make sure this is what you intended. This will likely increase the time it takes for your changes to go live.

    Comment and Save

    Until you earn 1000 points all your submissions need to be vetted by other Comic Vine users. This process takes no more than a few hours and we'll send you an email once approved.