Two different calcs for Boros' ship

[Cryo123]

Either cumulus, stratcumulus, or Nmbostratus.

 

[Marvel_Champion_07]

Zamasu got Low 5-B let's go

 

[Therefir]

Wow, this is already some kind of responsibility, ahah. Looks like I'll do it anyway.

But I'll just say right up front that this will be a High 6-A baseline, so scaling with Orochi is still a problem.

@NikHelton Don't worry, there's no rush, though I think the linear momentum formula is heavily inflating the calc.

If you're going to do it, I suggest using the standard kinetic formula and only using half of the ship's weight for the movement.

Because getting High 6-A/Low 5-B results for slowly tilting a ship that's only 50 to 100 kilometers long is heavily suspicious.

 

[ByAsura]

Also in this panel we can see that Boros' ship is well above the bottom of the clouds, if that affects anything.

This one should be good for calcs.

 

[XVII]

Zamasu got Low 5-B let's go

if this gets accepted (most likely not) only saitama will scale so it's fine

unless boros kick gets upgraded too

 

[NikHelton]

This one should be good for calcs.

It surprises me how you calculated the height of the cloud in pixels in the calculation for this page. 9 pixels lol. The line has not even been completed there. One might as well measure the smallest edge of the cloud and say that its height is 2 km.

 

[NikHelton]

@NikHelton Don't worry, there's no rush, though I think the linear momentum formula is heavily inflating the calc.

If you're going to do it, I suggest using the standard kinetic formula and only using half of the ship's weight for the movement.

Because getting High 6-A/Low 5-B results for slowly tilting a ship that's only 50 to 100 kilometers long is heavily suspicious.

In fact, using KE will only give a level of 6-B / High 6-B.

As for the overestimated results, I haven’t looked at the whole formula yet, but as I understand it, Saitama’s jump speed is used there, and maybe this is the reason

 

[ByAsura]

It surprises me how you calculated the height of the cloud in pixels in the calculation for this page. 9 pixels lol. The line has not even been completed there. One might as well measure the smallest edge of the cloud and say that its height is 2 km.

I didn't make a calculation based on clouds.

 

[Marvel_Champion_07]

only saitama will scale so it's fine

All of Boros' forms would also scale

 

[NikHelton]

I didn't make a calculation based on clouds.

Oh, don't think bad. I was not talking about you, but about the calculation of Zamasu. You just responded to the picture he used.

 

[Cryo123]

@NikHelton Don't worry, there's no rush, though I think the linear momentum formula is heavily inflating the calc.

If you're going to do it, I suggest using the standard kinetic formula and only using half of the ship's weight for the movement.

Because getting High 6-A/Low 5-B results for slowly tilting a ship that's only 50 to 100 kilometers long is heavily suspicious.

It only appears to be slowly tilting because of its size, but you also have to understand it is actually moving multiple kilometers per second.

 

[Cryo123]

Oh and on top of that, the ship has a gravity core that is actively providing upward force, further slowing the ships tilting, which is more reason to use RKE.

Edit: Also, Zamasu's calc should have measured from the cloud's bottom to its shadow to get a more accurate size for the ship. I'll make my own version of the calc attempting to rectify this problem unless someone else is already working on it.

 

[Cryo123]

Slight problem, I have no idea what C is meant to represent in Zamasu's final equation.

 

[Cryo123]

Or the equation he's using.

 

[Cryo123]

Okay, just realised it was the speed of light. I'm an idiot lol.

 

[Cryo123]

Update: My calc got 214.8709369024856819 Petatons of tnt or High 6A. We can use the area destroyed by the landing to determine the toughness of the ship's hull in joules per cubic centimeter. Then we can use that to find the exact power of MB Boros' punch.

If that's rejected though, (which it should not be), it'll be hard to argue that Boros > Orochi.

 

[XVII]

Update: My calc got 214.8709369024856819 Petatons of tnt or High 6A. We can use the area destroyed by the landing to determine the toughness of the ship's hull in joules per cubic centimeter. Then we can use that to find the exact power of MB Boros' punch.

If that's rejected though, (which it should not be), it'll be hard to argue that Boros > Orochi.

ik this is completely unrelated but what do you guys think about the newest chapter?

there aint no way that garou deforming the planet isn't at least 5-C bruh

 

[Marvel_Champion_07]

ik this is completely unrelated but what do you guys think about the newest chapter?

there aint no way that garou deforming the planet isn't at least 5-C bruh

It's already been calced at Low 5-B to 5-B. Mid-end was accepted

 

[XVII]

It's already been calced at Low 5-B to 5-B. Mid-end was accepted

holy shit that was fast

 

[Cryo123]

Should I put my calculation up for evaluation or should I wait?

 

[Kachon123]

Should I put my calculation up for evaluation or should I wait?

Put it up

 

[Cryo123]

Here it is: they see me rollin', they calcin'

 

[Cryo123]

Like some people here to see if any clear mistakes were made before I put it up for evaluation.

 

[Therefir]

Here it is: they see me rollin', they calcin'

@Ugarik Leaving the weight and scaling of the ship aside, is the formula used here correct?

 

[Cryo123]

@Ugarik Leaving the weight and scaling of the ship aside, is the formula used here correct?

It's the same formula everyone's been using.

 

[Therefir]

It doesn't really justify the discrepancy between using the kinetic energy formula and this momentum-based formula.

That and Ugarik knows about these kind of things (he helped with Sage Centipede's 6666 Leg spin calculation).

 

[Cryo123]

It doesn't really justify the discrepancy between using the kinetic energy formula and this momentum-based formula.

Using the normal KE formula physically doesn't work given the context of the feat.
I've already responded to this so I'll just crt+c it here;
"Because getting High 6-A/Low 5-B results for slowly tilting a ship that's only 50 to 100 kilometers long is heavily suspicious."
"It only appears to be slowly tilting because of its size, but you also have to understand it is actually moving multiple kilometers per second."

"Oh and on top of that, the ship has a gravity core that is actively providing upward force, further slowing the ships tilting, which is more reason to use RKE."

 

[Therefir]

Doesn't really answer my question, moving the whole ship at those speeds gives far inferior results, and I don't care if the ship has a gravity core, you can't calculate the force of a fictional alien technology.

Actually that just confuses me more, how can you use something that hasn't been calculated and cannot be calculated to support the inflated results of the ship?

 

[ByAsura]

That pink line is funky as hell and definitely inflates the result somewhat. Just use the tip of the shadow, like so. There's less perspective issues and parallax error that way.

 

[NikHelton]

In fact, using the method with clouds gives even less result than ByAsura calculation. In the Crew calculation, the cloud height is defined as 2 km (almost), while the height of cumulus clouds (and the clouds are obviously cumulus) ranges from 500 meters to 1.5-2 km. Even if we take the average end, at 1000 m, it will give us the overall result of the ship 3-4 percent less than that of ByAsura.

 

[ByAsura]

Looking at the shadows, the clouds are definitely closer to the middle of the ship.

I'll make my own estimate later.

 

[ByAsura]

If we look at the shadows, we can see the tip of the cloud we're measuring approaches the part of the ship with the two spheroids (I'll use the smallest because the larger one is more deformed).

Cloud to Ground = 50 px

Smallest Spheroid = 30 px
----
Smallest Spheroid = 36 px

Ship Length = 1334 px

Keep in mind, I just eyeballed the length. It's not exactly accurate.

This puts the ship at 18.5, 55.6, or 74 kilometres with the 0.5, 1.5 and 2 kilometre distances.

I feel the fin next to the spheroids would be more accurate for comparison, but inconsistencies in the art make that impossible to use.

 

[Ugarik]

@Ugarik Leaving the weight and scaling of the ship aside, is the formula used here correct?

I don't see why the center of mass of the ship would be in the middle. Its shape is clearly not symmetrical.

I also don't understand the way moment of inertia is calculated.

 

[Cryo123]

I don't see why the center of mass of the ship would be in the middle. Its shape is clearly not symmetrical.

I also don't understand the way moment of inertia is calculated.

It's just an approximation, because everyone's trying to see what methods work and don't work, so going through and finding the center of mass each time is too slow and tedious. Once a method is chosen with proper pixel scaling, we can use the appropriate center of mass.

 

[Ugarik]

And what about its moment of inertia?

 

[Cryo123]

And what about its moment of inertia?

That's just done through the calculator I think.

 

[Ugarik]

I mean no disrespect but it looks like you don't fully understand what moment of inertia is. The calculator you linked uses moment of inertia of a mass point while the ship is a proper rigid body. Moment of inertia of any rigid body (just like its volume) depends on its size and shape.

 

[Cryo123]

I mean no disrespect but it looks like you don't fully understand what moment of inertia is.

Honestly this is not disrespectful at all. I didn't know wat moment of inertia is, and I always love learning, especially about science. I 100% have holes in my knowledge of Classical Physics because for some reason physics classes are not available until jr year. Ironically, I've taught myself more about Quantum Physics and General Relativity than I have about Classical Physics lmao.

The calculator you linked uses moment of inertia of a mass point while the ship is a proper rigid body. Moment of inertia of any rigid body (just like its volume) depends on its size and shape.

Then we should probably find a way to precisely calculate that, unless that links back to my earlier point about first avoiding the time sink and then going in and finishing the calc up.

 

[Emirp sumitpo]

Bump

 

[Cryo123]

Lmao.