Official Calculations Discussion Thread

[KLOL506]

Question, how do I begin finding the surface area of... well this?

For links, use "vignette".

For surface area, just upscale from the average sized bloke's surface area of 1.7 m^2 (1.6 for women, 1.8 for men.)

 

[Oliver_de_jesus]

How can I calculate this feat? the door is reinforced and is declared Heavy in the novel version

bump

Also, how i can calculate a feat of tear off an arm made of Titanium?

 

[Oliver_de_jesus]

Wait, forget about the door, I already found out how to do this using this calculation. Although, is possible to get the LS from this too?

I still need arm calc though

 

[KLOL506]

bump

Also, how i can calculate a feat of tear off an arm made of Titanium?

You need to find the surface area of the joints that connected the arm to the robot first.

 

[DastardlyDangerousDracoga]

For links, use "vignette".

Ah, so that's what I had wrong.

Does it show up now?

 

[Oliver_de_jesus]

You need to find the surface area of the joints that connected the arm to the robot first.

Apparently the joint measures 124.558824mm in height and 55mm in width (Using the scale to Berto's nose assuming the average), since there are two, double the force should be applied, right?

 

[KLOL506]

Apparently the joint measures 124.558824mm in height and 55mm in width (Using the scale to Berto's nose assuming the average), since there are two, double the force should be applied, right?

You need to account for hollowness, since that arm isn't a solid block of metal.

And no, two arms being used to pull does not multiply the force applied, it just evenly distributes the total force between the arms. So no multiplication of force should occur.

 

[Oliver_de_jesus]

You need to account for hollowness, since that arm isn't a solid block of metal.

I guess 90% hollowness is ok?

 

[Oliver_de_jesus]

doing calculations the total area would be ~ 622.035 cm2
with a Hollowness of 90% would be 62.2035cm2 or 24.489566929 inches

Titanium has a PSI of 30,000 to 200,000 psi (210-1380 MPa), for context the robots were re-designed to accomplish high-risk missions by an international secret organization, so I'll use both ends.


low end

24.489566929x30,000 lb/in^2 =516,536,086 kg(Class M)


high end

24.489566929x200,000 lb/in^2= 3.44357391e9 kg (Class G)

this is well done right? It is the first time I do it and i feels that i made it wrong

 

[ShionAH]

How can I calculate this?

 

[Jasonsith]

I have just made a calculation for one fictional feat whose episode just released.

When after the episode release date time may I post it?

Spoiler

It is about Titan Luz and Eda evading from (just awakened) Titan Belos' grasp and flying into outside Earth atmosphere.
The Owl House Season 3 Final Episode
Released 9 April 2023 10:45 SGT (?)

 

[Oliver_de_jesus]

I have just made a calculation for one fictional feat whose episode just released.

When after the episode release date time may I post it?

Spoiler

It is about Titan Luz and Eda evading from (just awakened) Titan Belos' grasp and flying into outside Earth atmosphere.
The Owl House Season 3 Final Episode
Released 9 April 2023 10:45 SGT (?)

nothing really stops you, the guys from Kaiju n 8 make and publish the calc of a feat in the same day it comes out

 

[DastardlyDangerousDracoga]

Question, how do I begin finding the surface area of... well this?

Ah, so that's what I had wrong.

Does it show up now?

Bump?

 

[Mr. Bambu]

Don't suppose you have GIMP image editing software installed?

 

[Kachon123]

GIMP >>>> any other editing software tbh

 

[Mr. Bambu]

It's quite good, and has a native method of measuring surface area via pixel scaling, allowing you to do it for anything.

 

[Omegabronic]

In the few calcs i did i used paint to count, it took a while, my head started to physically hurt, but i did it

 

[Mr. Bambu]

Right. I'm asking @DastardlyDangerousDracoga if he has GIMP, because I can explain the process for that software.

 

[azontr]

So, thinking about calcing a feat that I could use help with.

A beam is fired (the beam is SoL) and the cat thing at the end of the string of panels intercepted/dodged that beam before it hit its target.

Is there a way to go about calculating this?

Bump.

 

[Kachon123]

Bump.

Find the distance between the target and the tower.

Find how far along the beam was from hitting the target before the cat moved.

Use that distance and divide it by the speed of light to find the timeframe.

Find the distance that the cat moved.

Use that distance and divide it by the timeframe to get the speed.

 

[ShionAH]

How the HELL can I calculate this?

 

[ByArrow]

How can I calculate this?

 

[Spider-Aty]

Hi so I did discuss this on discord with bambu but would also like to ask some questions too from other calc guys. So first I want to ask about this is this calc https://vsbattles.fandom.com/wiki/User_blog:Ugarik/Daily_Bugle_calc# accurate even if it was a 46 story building and this calc too https://character-feats.fandom.com/...ileaction=toggle_view_desktop#articleComments could you guys also calc these thing feats compared to Spider-Man’s

I also have some Spider-Man feats that need calcs too

 

[Spider-Aty]

Hi so I did discuss this on discord with bambu but would also like to ask some questions too from other calc guys. So first I want to ask about this is this calc https://vsbattles.fandom.com/wiki/User_blog:Ugarik/Daily_Bugle_calc# accurate even if it was a 46 story building and this calc too https://character-feats.fandom.com/...ileaction=toggle_view_desktop#articleComments could you guys also calc these thing feats compared to Spider-Man’s
I also have some Spider-Man feats that need calcs too

I also have more feats to add to it

 

[ArcaneAbyss]

I am trying to calculate the acceleration due to gravity of an artificially created mass construct whose escape velocity supersedes SOL. I have the radius already

Using the 'Vesc = SQRT(2GM/r)" to get its mass using SOL or 99% SOL(as low end) or 18's blast (at high end) as it's escape velocity and the calculated radius, and then using the g=GM/r^2, to get the acceleration due to gravity, I'll Confirm if this formula is applicable. Again, no assumptions needed, radius can be extrapolated, G is a constant, escape velocity is a lowballed SOL or 99% SOL, OR 18's blast - depending on if ftl values can be used in the formula.

Is the above method acceptable for SOL or FTL escape velocities? Or do i need it lowball to less than light speed to low-ball to less than SOL?

I just need to know if it's only viable at less than SOL, or SOL and above can also he used

 

[Spider-Aty]

I am trying to calculate the acceleration due to gravity of an artificially created mass construct whose escape velocity supersedes SOL. I have the radius already


Is the above method acceptable for SOL or FTL escape velocities? Or do i need it lowball to less than light speed to low-ball to less than SOL?

I just need to know if it's only viable at less than SOL, or SOL and above can also he used

SOL is solar level correct I don’t know hat feat you are talking about

 

[Spider-Aty]

I am trying to calculate the acceleration due to gravity of an artificially created mass construct whose escape velocity supersedes SOL. I have the radius already


Is the above method acceptable for SOL or FTL escape velocities? Or do i need it lowball to less than light speed to low-ball to less than SOL?

I just need to know if it's only viable at less than SOL, or SOL and above can also he used

But I think it is above sol after I read your calculation thank you btw for that

 

[SeijiSetto]

SOL is solar level correct I don’t know hat feat you are talking about

I think he's talking about the Pretty Black Hole from DBS, and he's asking if using the standard escape velocity formula for celestial bodies is valid for when the escape velocity is beyond the speed of light, he isn't talking about AP.

I am trying to calculate the acceleration due to gravity of an artificially created mass construct whose escape velocity supersedes SOL. I have the radius already


Is the above method acceptable for SOL or FTL escape velocities? Or do i need it lowball to less than light speed to low-ball to less than SOL?

I just need to know if it's only viable at less than SOL, or SOL and above can also he used

Not a calc group member but I think having higher than SoL escape velocities should work, since that's literally how black holes function to begin with, no?

 

[KingTempest]

no

 

[SeijiSetto]

no

Even aside from the whole Pretty Heart debate since I know you're in that thread, can you provide a reason why other than just "no"?

On both pages for black holes over on Fandom it explicitly mentions that FTL characters generally have no issue with them.

generally considered to inescapable to any entity that does not have either reality warping, FTL speed, or time travel

and the other one says

faster than light characters would still be capable to escape a black hole past the event horizon even though even for them it gets more and more difficult

From that logic am I wrong for inferring that FTL escape velocities are a thing?

 

[KingTempest]

Oh, I thought you meant that any reference to SOL qualifies it as a read black hole

nvm

 

[Spider-Aty]

I think he's talking about the Pretty Black Hole from DBS, and he's asking if using the standard escape velocity formula for celestial bodies is valid for when the escape velocity is beyond the speed of light, he isn't talking about AP.

Not a calc group member but I think having higher than SoL escape velocities should work, since that's literally how black holes function to begin with, no?

You guys are the expert of calcs I asked someone to compare the feats for Spider-Man to the things strength feats.(the guy who told me this has been doing this apparently before this site was created)he said Spider-Man strength feats were better than the things he only has scaling (even though Spider-Man has similar scaling too) so thankyou again

 

[ArcaneAbyss]

SOL is solar level correct I don’t know hat feat you are talking about

SOL as in Speed of light

I think he's talking about the Pretty Black Hole from DBS, and he's asking if using the standard escape velocity formula for celestial bodies is valid for when the escape velocity is beyond the speed of light, he isn't talking about AP.

Exactly

Not a calc group member but I think having higher than SoL escape velocities should work, since that's literally how black holes function to begin with, no?

I just want to be sure, yeah their escape velocity is speed of light, but I also want to know if mftl speeds can be accepted as escape velocity too in the formula, cant be certain if this site has rules against that


Edit: just confirmed with @Psychomaster35, ftl escape velocity still works. Thanks

 

[Spider-Aty]

Can I ask anyone if they recall this feat or compare the calculations done with the scan ugarik used but instead using this scan to calc the feat here instead or combine the scan he used since you can actually see the the width and length better but use the height shown here

 

[Hagane_no_Saiyajin]

I posted a thread, but I thought to bring my query here too. Kagura's feat considering the original quote only says "an astounding amount of heat is released, causing the elimination of the the entire area", gives varying results depending upon the method of calculation:

Using specific heat equations gets 6-C+ to 6-A results (3.87 Gigaton of TNT to 1.513 Petatons of TNT )

Using nuclear fireball calculator gets 6-C results (8.54 Gigatons of TNT to 8.6 Gigatons of TNT)

Using radiation equations gets 9-C+ to High 6-A results (14277.5075 Joules of TNT to 11.97828 Petatons of TNT)

I just wish to know, which is viable?

 

[DastardlyDangerousDracoga]

Don't suppose you have GIMP image editing software installed?

Right. I'm asking @DastardlyDangerousDracoga if he has GIMP, because I can explain the process for that software.

Oh, I didn't think I would have gotten a response so soon to my dilemma so soon. Odd how my notifications didn't catch that ping. But yes, I do have GIMP installed.

 

[Mr. Bambu]

Oh, I didn't think I would have gotten a response so soon to my dilemma so soon. Odd how my notifications didn't catch that ping. But yes, I do have GIMP installed.

Took you long enough

There's a tool on GIMP that may help. What you need to do is to outline the area you want to calculate the surface of (in this instance, the bits of the mech) via any selection tool- the pathing tool is my go to. After that, you need to use the dockable dialog called "histogram", which will tell you the exact number of pixels that are within the selected area. If you have something to pixel scale to the mech, you can learn the length of one pixel- thus every pixel is (Whatever Length)^2. Multiply by quantity of pixels to discern surface area.

For this specific thing you'll need to do this at least twice, and even more if you want extreme accuracy. Once for the front side, once for the backside. Since the guy has a lot of angular turns, you may wish to perform these steps for each segment of his body as well. Good luck.

 

[Flashlight237]

Took you long enough

There's a tool on GIMP that may help. What you need to do is to outline the area you want to calculate the surface of (in this instance, the bits of the mech) via any selection tool- the pathing tool is my go to. After that, you need to use the dockable dialog called "histogram", which will tell you the exact number of pixels that are within the selected area. If you have something to pixel scale to the mech, you can learn the length of one pixel- thus every pixel is (Whatever Length)^2. Multiply by quantity of pixels to discern surface area.

For this specific thing you'll need to do this at least twice, and even more if you want extreme accuracy. Once for the front side, once for the backside. Since the guy has a lot of angular turns, you may wish to perform these steps for each segment of his body as well. Good luck.

People do that? I just drew over pictures with Adobe Illustrator and used a script to determine an object's area.

 

[Mr. Bambu]

stay losing

 

[SeijiSetto]

stay losing

adobe products are always free if ye be willing to sail the high seas