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MCU Thor Storm Calcs

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I think I found a way to find the thickness.

The thickness seems to be around the same size as the Jotunns.

Bad news is, it might nerf the feat and bring it down anywhere between Low 7-B to 7-B, depending on what height you wanna go with for the Jotunns (I do remember them being quite a bit taller than Thor, whose actor is 1.905 meters tall), assuming Grand Canyon times 100 size. Would nicely support the 200 petawatt laser tho
 
KLOL506 said:
I think I found a way to find the thickness.

The thickness seems to be around the same size as the Jotunns.

Bad news is, it might nerf the feat and bring it down anywhere between Low 7-B to 7-B, depending on what height you wanna go with for the Jotunns (I do remember them being quite a bit taller than Thor, whose actor is 1.905 meters tall), assuming Grand Canyon times 100 size. Would nicely support the 200 petawatt laser tho
Click to expand...

This might help though iirc loki is significantly slower than average frost giant
 
chosen said:
Nvm jsut looked on google "frost giant mcu height" they are about 10'
Click to expand...
Seems about decent, since those guys are around half a body taller than Thor.

That case gives me like, 4.63 megatons. Which just so happens to be 10x lower than scaling to the 200 petawatt laser feat. 'Tis casual but what can you do.

But in case y'all haven't noticed, even with this I still fully support Thor scaling to busting Sokovia, as this was still mostly his own power, the vibranium would just repel it back, and Iron Man was only there to reduce the collateral damage by weird "atomic action doubling back" science bullshit. I honestly don't know how the **** on earth that downgrade went through.
 
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Baseline Grand canyon area: 277*18= 4986 square miles or 1.29137e+14 cm^2.

Grand Canyon Area Times 100: 4986*100= 498600 square miles or 1.29137e+16 cm^2

Grand Canyon Area With length and width being 100x bigger and then the two being multiplied: 27700*1800= 49860000 sq mi or 1.29137e+18 cm^2

Giants are 10 ft tall. 3.048 m

Low-end volume: 1.29137e+14*304.8= 3.9360958e+16 cm^3

Mid-end volume: 1.29137e+16*304.8= 3.9360958e+18 cm^3

High-end volume: 3.9360958e+20 cm^3

Calc:

Frag energy of ice is 0.525 J/cc

Low-end: 3.9360958e+16*0.525= 2.0664503e+16 J or 4.938934751434034709 megatons of TNT (Small City level+)

Mid-end: 3.9360958e+18*0.525= 2.0664503e+18 J or 493.89347514340346379 megatons of TNT (Mountain level)

High-end: 3.9360958e+20*0.525= 2.0664503e+20 J or 49.389347514340343537 gigatons of TNT (Island level) (WTF)

Me personally, I'mma stick with the Mid-end. High-end has an area ten thousand times larger than the base area. Which makes no sense.
 
KLOL506 said:
Baseline Grand canyon area: 277*18= 4986 square miles or 1.29137e+14 cm^2.

Grand Canyon Area Times 100: 4986*100= 498600 square miles or 1.29137e+16 cm^2

Grand Canyon Area With length and width being 100x bigger and then the two being multiplied: 27700*1800= 49860000 sq mi or 1.29137e+18 cm^2

Giants are 10 ft tall. 3.048 m

Low-end volume: 1.29137e+14*304.8= 3.9360958e+16 cm^3

Mid-end volume: 1.29137e+16*304.8= 3.9360958e+18 cm^3

High-end volume: 3.9360958e+20 cm^3

Calc:

Frag energy of ice is 0.525 J/cc

Low-end: 3.9360958e+16*0.525= 2.0664503e+16 J or 4.938934751434034709 megatons of TNT (Small City level+)

Mid-end: 3.9360958e+18*0.525= 2.0664503e+18 J or 493.89347514340346379 megatons of TNT (Mountain level)

High-end: 3.9360958e+20*0.525= 2.0664503e+20 J or 49.389347514340343537 gigatons of TNT (Island level) (WTF)

Me personally, I'mma stick with the Mid-end. High-end has an area ten thousand times larger than the base area. Which makes no sense.
Click to expand...
Updated. Thx
 
Actually, I have one last end to suggest, since the 27700*1800 time area is now bunk.

We know the Grand Canyon Area 100 time size to be 498600 sq miles or 1291368.07 square km.

Which means the Grand Canyon would have to be 10x longer (2770 miles is 4457.883 km) and wider (180 miles is 289.682 km) than the IRL thing to the area value.

I'll assume 2 ends. A 1 m displacement, and a 2 m displacement.

But, even if you assume that the area is a perfect square (1291368.07 squared is 1136.383769 km per side), you'd get the same Richter level results regardless of whether you used 4457.883 and 289.682 km as your length and width or 1136.383769 km as both your width, the results are same.

Assuming a 1 m displacement I get a Richter Scale magnitude value of 9.01, or 493.6377 megatons of TNT (Mountain level), which is shockingly similar to my mid-end frag value.

Assuming a 2 m displacement, you get a Richter Scale magnitude value of 9.21, which is roughly 984.936 megatons of TNT (Mountain level+)

At least we know the outlier stuff was a bunch of bullshit from the beginning. Good to know. Heh.

Now, might I suggest you kneel before me and my glorious purpose of restoring the heyday of 7-A MCU?
 
KLOL506 said:
Actually, I have one last end to suggest, since the 27700*1800 time area is now bunk.

We know the Grand Canyon Area 100 time size to be 498600 sq miles or 1291368.07 square km.

Which means the Grand Canyon would have to be 10x longer (2770 miles is 4457.883 km) and wider (180 miles is 289.682 km) than the IRL thing to the area value.

I'll assume 2 ends. A 1 m displacement, and a 2 m displacement.

But, even if you assume that the area is a perfect square (1291368.07 squared is 1136.383769 km per side), you'd get the same Richter level results regardless of whether you used 4457.883 and 289.682 km as your length and width or 1136.383769 km as both your width, the results are same.

Assuming a 1 m displacement I get a Richter Scale magnitude value of 9.01, or 493.6377 megatons of TNT (Mountain level), which is shockingly similar to my mid-end frag value.

Assuming a 2 m displacement, you get a Richter Scale magnitude value of 9.21, which is roughly 984.936 megatons of TNT (Mountain level+)

At least we know the outlier stuff was a bunch of bullshit from the beginning. Good to know. Heh.

Now, might I suggest you kneel before me and my glorious purpose of restoring the heyday of 7-A MCU?
Click to expand...
Mcu hulk and Thor vs ultimate hulk and Thor when
 
KLOL506 said:
Actually, I have one last end to suggest, since the 27700*1800 time area is now bunk.

We know the Grand Canyon Area 100 time size to be 498600 sq miles or 1291368.07 square km.

Which means the Grand Canyon would have to be 10x longer (2770 miles is 4457.883 km) and wider (180 miles is 289.682 km) than the IRL thing to the area value.

I'll assume 2 ends. A 1 m displacement, and a 2 m displacement.

But, even if you assume that the area is a perfect square (1291368.07 squared is 1136.383769 km per side), you'd get the same Richter level results regardless of whether you used 4457.883 and 289.682 km as your length and width or 1136.383769 km as both your width, the results are same.

Assuming a 1 m displacement I get a Richter Scale magnitude value of 9.01, or 493.6377 megatons of TNT (Mountain level), which is shockingly similar to my mid-end frag value.

Assuming a 2 m displacement, you get a Richter Scale magnitude value of 9.21, which is roughly 984.936 megatons of TNT (Mountain level+)

At least we know the outlier stuff was a bunch of bullshit from the beginning. Good to know. Heh.

Now, might I suggest you kneel before me and my glorious purpose of restoring the heyday of 7-A MCU?
Click to expand...
I slep. Don't wake me up until we use 4-C Pre Ragnarok Thor
 
OneBleachHurricane said:
8-A IRON MAN VS 8-A OPTIMUS PRIME
9TiWsf.gif
VS
optimusprime.gif


8-A GENERAL GRIEVOUS

general-grievous.gif
Click to expand...
Me an intellectual:
The Shipgirls from Azur Lane
mutsuki-azure-lane.gif


And this guy too
kamen-rider-den-o.gif
 
Last edited:
IG should be at least 4-B since it was claimed that nothing has been seen like that before, then again there's black holes and stuff so...
 
Last edited:
KLOL506 said:
Actually, I have one last end to suggest, since the 27700*1800 time area is now bunk.

We know the Grand Canyon Area 100 time size to be 498600 sq miles or 1291368.07 square km.

Which means the Grand Canyon would have to be 10x longer (2770 miles is 4457.883 km) and wider (180 miles is 289.682 km) than the IRL thing to the area value.

I'll assume 2 ends. A 1 m displacement, and a 2 m displacement.

But, even if you assume that the area is a perfect square (1291368.07 squared is 1136.383769 km per side), you'd get the same Richter level results regardless of whether you used 4457.883 and 289.682 km as your length and width or 1136.383769 km as both your width, the results are same.

Assuming a 1 m displacement I get a Richter Scale magnitude value of 9.01, or 493.6377 megatons of TNT (Mountain level), which is shockingly similar to my mid-end frag value.

Assuming a 2 m displacement, you get a Richter Scale magnitude value of 9.21, which is roughly 984.936 megatons of TNT (Mountain level+)
Click to expand...
Considering how how high that is we could get High 7-A back
 
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