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"Possibly" 3-C Saitama

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Milky Way Galaxy end is almost exactly the same as what they scale now, so if it gets accepted i guess no match would need to be removed.
 
Probably just misunderstood how to apply the math rather than anything malicious.
I think we can still apply the rule to non-spherical destruction as long as it is done from a distance and we use the area of the beam that covered the object that it destroyed.
 
Therefir accepted the milky way and Andromeda ends.

I'm guessing that means he's in favor of

4-A (Hole extended out across the Milky Way), possibly 3-C (Hole extended out to another galaxy)

What are y'alls thoughts on that?

Also, I don't think that the value needs to be halved if they both tanked the explosion at the epicenter anyway.
 
What would this mean for blast? Would he also get the possibly 3C rating or no?
 
Well I haven't updated votes but it seems like Qawsed is outvoted on this? Though now this thread has turned into a bit more than just a slight upgrade. There's now 2 versions of 4-A that need to be considered.

I'll have to update the OP once I'm on my PC.
 
It's actually 5-1.

Therefir, Marvel, and KingTempest are okay with this change.
 
Content mods don’t count
Frankly, I think they do 😂

And they are also content mods that are very knowledgeable on the verse, so imo they are worth more than just another discussion mod.
 
Frankly, I think they do 😂

And they are also content mods that are very knowledgeable on the verse, so imo they are worth more than just another discussion mod.
not how that works unfortunately, lol.
 
not how that works unfortunately, lol.
Eh, it's not a site wide CRT or anything. Dismissing the majority mod opinion doesn't do much overall.

turned into a bit more than just a slight upgrade.
No, it's a downgrade. The 3-C rating is like 10,000x lower than the previous proposed one and the visible star end is also 1,000x lower than the current one.
 
What are the options for the ends here exactly?

"4-A, possibly 3-C", or just "4-A"?
 
Eh, it's not a site wide CRT or anything. Dismissing the majority mod opinion doesn't do much overall.


No, it's a downgrade. The 3-C rating is like 10,000x lower than the previous proposed one and the visible star end is also 1,000x lower than the current one.
It's a downgrade and upgrade.

And what I meant is that now we are changing the entire rating since the 4-A is being adjusted as well.

So now we have to decide which 4-A value to use.
 
I say just go with the readjusted current version. Since the assumptions are all the same.
That's a fair take, but what would you say is the main reason why it's more likely that the clash between the two of them only reached as far as the outskirts of the Milky Way and not possibly as far as Andromeda Galaxy for example?
 
We end up with a lower 4-A rating while a possibly 3-C is rejected?
Lower 4-A, with a "possibly 3-C"

"possibly 3-C" has been majority-accepted, so far, at least.

Pretty sure there's a 48 hour thing though.

I say just go with the readjusted current version. Since the assumptions are all the same.
Yeah, fair, as much as I would say Milky Way is more fair, this is the easiest way to get this through.
 
but what would you say is the main reason why it's more likely that the clash between the two of them only reached as far as the outskirts of the Milky Way and not possibly as far as Andromeda Galaxy for example?
Murata is shown to draw galaxies as spirals or in easily visible patterns. None of the lights destroyed by the punch in the time reversal picture fit his typical artistic display.

In addition no one comments on the beam destroying another galaxy or anything like that. The general calc assumption is to take the more resonable end if the information is lacking and I believe that end is just the visible star end.
 
The visible star end only works if we were looking at the destruction from inside the Earth's atmosphere.

Plus, Murata believes that we can see galaxies in space like this. If in the time reversal panel we don't see any galaxies, it's simply because there were no galaxies in that direction.
 
Do we even know for sure if that's the same hole the Squared Punch caused instead of just being a random shot in space?
like everything else related to the feat, we have absolutely no information regarding it and it could very well be absolutely nothing at all.
 
Wait, which calculation is currently being used here? The inverse square law apparently isn't usable if the explosion isn't omni-directional from what I'm being told?

Doesn't that mean Qawsedf234 calculation cannot be used, since it's using the Inverse Square Law?

Inverse Square Law = area of the larger object or explosion/area of the object x the initial value

Explosion Area = 3.2943492e+38 m^2

Sun Frontal Area = 3.0438995e+18 m^2

Initial Value (Sun GBE) = 5.693e+41 Joules

3.2943492e+38/3.0438995e+18*5.693e+41 = 6.1614156e+61 Joules (4-A)

I'm having some trouble understanding this, what is going on here?
 
The old calc is an omnidirectional explosion, mine is sector area. Both use inverse square law but mine is smaller due to it being a cone rather than a sphere.
I was told that directional explosions can't be used for the inverse square law and it has to be omnidirectional. Is this a directional or an omnidirectional explosion?

I felt like I understood what was happening and I understand why the original calculation is wrong.

But now my brain is all jumbled and I have no idea what's going on anymore.

Edit: Also, how are you getting your Sector Area? I'm not getting the same numbers as you. Am I using the calculator incorrectly?
 
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I was told that directional explosions can't be used for the inverse square law and it has to be omnidirectional.
Sector area is the area of a circle for an arc radius. The calc assumes a standard energy density spread over the area which can destroy the sun. But you can use inverse square law for anything, it just has to be within reason since while on planets you can quickly rack up unrealistic numbers.

Edit: Also, how are you getting your Sector Area? I'm not getting the same numbers as you. Am I using the calculator incorrectly?
45 degree for the angle and radius for the radius. I'm on mobile atm so I can't provide the picture.

If the numbers are wrong, my imagination would be that all we would need to do is find the hypotenuse.
 
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