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TheRustyOne

TheRustyOne

VS Battles Calculation Group
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  • Don't DM me for calc evaluations, you can just leave a message here if the Evaluation Thread isn't working for you.
    Viott
    TheRustyOne
    TheRustyOne
    I don't feel like that's a good way to get the volume of all the buildings in the area.

    Also, the Gun Devil was only in Japan for 26 seconds. Meaning what you measured couldn't have taken place in 26 seconds since the Gun Devil would've still been in Japan.

    It'd would have to be an unknown amount less than 26 seconds. Considering how fast the devil is shown to be, highly doubt it stood in one area like this for long, nor do I think dividing the destruction by timeframe is an accurate measure of how powerful it is. Did it fly through and shoot like crazy at the same time? Can't be use in that case.

    That's also ignoring how the destruction clearly goes off screen for who knows how long. There weren't a straight line of buildings in that area either, there would've been sidewalks, streets, and alleyways. Personally, I think this destruction is uncalculatable.
    Viott
    can you check this?
    vsbattles.com

    The boxer ap upgrade

    Old Calc : Viktor plays minecraft : 54343825 Joules or 0.012988485899 tons of tnt (9-A) Recalc : The boxer : viktor ap (Recalc) : 90892970,6799 Joules or 0,0217239414 Tons of TNT (9-A) Pulverization should be used instead of V.frag
    vsbattles.com vsbattles.com
    Apex_Predator_GX
    Hi bud. I hope you're safe and happy. I want to thank you for evaluating my Dino Crisis calculation. I really appreciate your help. :)
    NikHelton
    View previous comments…
    NikHelton
    NikHelton
    The character on the left literally slides down this crater, which is written on the page itself (I dropped it in the comments).
    I just found the volume of the crater and applied the fragmentation value to it


    Regarding the second feat, this is what I came up with together with Ugarik and this method is used as an alternative to the earthquake method in cases when it does not fit the criteria of such
    TheRustyOne
    TheRustyOne
    I don't see him sliding down the crater, he just tripped from the ground shaking. I kind of don't see what you see there.

    Look, it's best if you just ask another calc group member about this.

    For the second calc, I'm not okay with evaluating a method I don't fully understand myself. I don't know when I'll look into that method, but not right now.
    NikHelton
    NikHelton
    Shouldn't the "slide" label be enough?

    Thanks for the help anyway

    tWmzMThD24Y.jpg
    LaserPrecision
    I know this is random, but I noticed a small error regarding your calculation here.

    For the radius of the explosion, you divided the width of the screen. However, the explosion reached the PoV from where it originated. To get an accurate radius of the explosion, you'd have to angsize the distance the explosion is from the screen. Fortunately most of the values you already have. So I took the liberty of doing the math here to see what you think. Px scaling is here.

    5493.889•1083/[1103•2•tan(70deg/2)] = 3851.91m radius

    3851.91^3*((27136*1.37895+8649)^(1/2)/13568-93/13568)^2 = 4593184.695 Tons | 4.593 Megatons

    Not a huge difference obviously. Though it does increase the result a few times.
    LaserPrecision
    LaserPrecision
    I also noticed that the size of the cave is px-scaled incorrectly. Think they mixed up the numbers.

    MHABE.png


    5493.889•1083/[1013•2•tan(70deg/2)] = 4194.13m radius

    4194.13^3*((27136*1.37895+8649)^(1/2)/13568-93/13568)^2 = 5929405.85728 Tons | 5.93 Megatons
    TheRustyOne
    TheRustyOne
    Px scaling is usually preferred over angular size, which considered to be less reliable.

    The explosion does spread out off screen no matter which way you measure. I don't believe using px scaling is wrong in this case.

    Also, why are you using the explosion for the angular size and not the cave? The px size of the explosion is 1920 and not 1013, making your results much lower.

    5493.889•1080/[1920•2•tan(70deg/2)] = 2206.71186 m

    However, I didn't realize I put down 1103 instead of 1013. That is a mistake and I thank you for informing me of it.
    LaserPrecision
    LaserPrecision
    I get that, but ang sizing measures a completely different axis. The width of the screen is the X-axis, and the camera is the Z-axis. There's a less visible X-axis than there is a Y-axis.

    I used the cave because the explosion started there and propagated towards the screen.

    Either way, glad I could help in some way.
    Emerald
    Hey Rusty, sorry for bothering you
    Do you mind explaining to me why this feat done by Iida is considered an LS feat even though it's a kick?
    vsbattles.fandom.com

    MHA Movies: Lifting Strength

    vsbattles.fandom.com vsbattles.fandom.com
    TheRustyOne
    TheRustyOne
    It does not scale to his LS, I said so on the blog itself.

    I calculated the force because I could, it would at best scale to his engine's thrust power. But that isn't lifting strength at all.
    Emerald
    Emerald
    Aight thanks for answering
    SeijiSetto
    hihi hello
    would you mind taking a look at (1) (2) feats? they basically piggyback off existing ones and just serve to get new values for feats.
    God243
    bro i have a question,if a dude survive to a nuke how does it scale if the nuke is 7B(he is very close to the explosion)?low7C or 7B?
    TheRustyOne
    TheRustyOne
    Based on the information you've given me, there is no way to tell.

    You need to give an exact distance for me to give any detailed answer. Even being 1 m away from the center of the explosive can drastically lower the results.
    Damage3245
    Sorry for the delay Rusty. I've been out pretty much all day, so I won't be able to make my response post till tomorrow morning. Will try to get it up as soon as possible.
    Hagane_no_Saiyajin
    I have a couple of issues with this calc:
    • It a little unclear what's happening exactly happening for the feat, or if the calculation really matches up with what's happening
    • The feat was performed over a time frame of 3 seconds
    • What's to say the centinels were digging through dirt instead of rock? They did kick up a lot of dust from their digging
    • Looking at how close the towers are to the edge, the centinels probably wouldn't have needed to dig that much to make the shield towers fall down
    • This feat happens with Team RWBY is in Atlas, so technically it should only apply to characters' Atlas keys, not Post-Haven keys
    • What's to say that the Centinels instead dig out a torus shape or triagular prism shape or pyramid shape?
    TheRustyOne
    TheRustyOne
    My interest in RWBY on this wiki varies depending on my mood, and I have zero interest right now.

    I have no idea when I'll be interested again.

    Ask someone else.
    Hagane_no_Saiyajin
    Hagane_no_Saiyajin
    Well, it's your calc, your responsibility, I've also posted these issues on the Calc Discussion Thread, and I might as well post them on the Ruby discussion thread too
    TheRustyOne
    TheRustyOne
    Not it isn't, stop trying to force me to do what you want. You don't need my opinion or thoughts if you want remove/change my calculation.

    I have zero interest in RWBY as of now. You just need another calc group member's opinion, maybe even the one who evaluated it.

    I will not be responding to this again, I'm respectfully asking you not to bother me on this topic again.
    Hey there, Rusy! I hope you are well. I was wondering if you could help out with adjusting Monstra's KE in the RWBYverse.

    vsbattles.fandom.com

    RWBY: Attacking Atlas​


    vsbattles.fandom.com
    vsbattles.fandom.com
    vsbattles.fandom.com

    RWBY: Monstra​


    vsbattles.fandom.com
    vsbattles.fandom.com
    vsbattles.fandom.com

    RWBY: Oscar's Shield​


    vsbattles.fandom.com
    vsbattles.fandom.com

    Monstra's accepted mass is 1361481844426.29 kg
    Grimm River accepted mass is 2270869563375.8 kg
    2270869563375.8 + 1361481844426.29 = 3.6323514e+12 kg
    Velocity = 1070.29 m/s
    Kinetic Energy = 3.6323514e+12 x 0.5 x 1070.29^2 = 2.0804668e+18 Joules, 497.24349904397706723 Megatons of TNT

    I = P/A = (2.0804668e+18) / (4π((7.2991578933)^2)) = 3.107459e+15 Joules/m^2

    E = ICA

    E = ICA = 3.107459e+15 x 28.745 = 8.9323909e+16 Joules, 21.348926625239005972 Megatons of TNT, City level


    We are gonna add the Grimm River to Monstra's weight since it housed it in the first place. All we need to do know is to add the weight of the liquid it throws up on Atlas.



    In total, it barfs up around five times. Four on screen and one that is mentioned off screen. I was wondering if you could help find the weight of one of the barfs? That way, we can simply mutiply it by five and add it to Monstra's weight and KE. Thank you.
    TheRustyOne
    TheRustyOne
    My interest in RWBY on this wiki varies depending on my mood.

    I'm really out of it right now. Because of that I guess you all should just copy/paste my sandbox of the Curious Cat and change whatever you want with it.

    I have no idea when I'll be interested again.
    Cire
    Gotcha. Who would you recommend I talk to for help?
    TheRustyOne
    TheRustyOne
    IDK.
    Whycantinamemyself
    View previous comments…
    Whycantinamemyself
    Whycantinamemyself
    How would you suggest we find the size of the hole then? If we can’t use the pyramids then there’s not much else to scale it off of.
    TheRustyOne
    TheRustyOne
    That's not my job, I don't know anything about this verse. Best to ask someone who is more familiar with the verse in question.

    If it truly isn't possible to find the size then that means you cannot find its size. The current method being used is wrong and highly inflates the results.
    Whycantinamemyself
    Whycantinamemyself
    Rewatching some of the movie now, I see certain things that actually could very well be used. I’ll probably calc it later today as it’ll be relatively simple.
    KobsterHope07
    View previous comments…
    Spinoirr
    Spinoirr
    It's probably just making it float tbh, just meaning it still has the same wight and speed thus the same ke as the calc
    TheRustyOne
    TheRustyOne
    If it works like Uraraka's Quirk then there would be no feat. Uraraka makes things weightless, as it's been stated multiple times.

    It's why she releases her Quirk before it lands when throwing objects, since it'd only have the energy from her throw and not the weight of the object.

    However, I've stated we don't know how Gravity Dust works so that's just an assumption. As you said, it could just make things float and not make it weightless or weigh less. Obviously Monstra isn't weightless or it wouldn't cause any impact when it landed.

    So I imagine it's up to whatever you guys think is reasonable.
    Spinoirr
    Spinoirr
    Honestly given it's shown that thing does have KE still and the gravity dust train still had KE, I say it definitely won't affect the calc at all
    RandomGuy2345
    Yo! You busy right now?
    TheRustyOne
    TheRustyOne
    Very much so.

    Focused on some calculations and preparing for possibly three CRTs for MHA.

    This doesn't include anything from other verses I'm going to help with, and two new verses I'd like to get added to the wiki in the future.

    Currently also studying up on some things I'm not very knowledgeable about yet, as I wish to change that.
    RandomGuy2345
    RandomGuy2345
    Damn 😭

    My apologies for bothering you.
    Whycantinamemyself
    Can you please evaluate the melting values of the hakone N2 mine? this is needed to help with a CRT. I
    TheRustyOne
    TheRustyOne
    I cannot evaluate a calculation that is off site, we don't accept these kinds of calculations anymore.

    We use to and we didn't remove any old ones if they were harmless, but you need to place that into an actual blog on the wiki.

    However, I'm currently too busy to evaluate your calc right now.
    Whycantinamemyself
    Whycantinamemyself
    Am I allowed to just copy and paste while giving credit to the original page? Or do I just have to calculate it myself.
    TheRustyOne
    TheRustyOne
    You can copy and paste it, but you do need to give credit to the original user while linking the original as well.
    LaserPrecision
    Here about your calc. First I'd like to say I'm not MHA fan, so I don't know the context of the feat. Only commented because I saw an opportunity for the feat to be improved.

    That said, you are entirely wrong about it being inapplicable and calc-stacking. First, all of this occurs in the same feat, and is thus not calc-stacking. That would be like claiming feats of people tanking explosions from a dozen meters away using ISL is calc-stacking. Not to mention, the ISL page even mentions it can be used in instances like this:

    When Should we use Inverse Square Law?​

    Inverse Square Law has a numerous amount of applications, here is a small list of some uses:

    Click to expand...

    Example 1: Finding energy based on something destroyed within an explosion​

    A ground explosion with a radius of 5 meters has exactly enough energy so when it hits a brick with an area of 0.07116953508 square meters and a volume of 0.0010692559 cubic meters, it vaporizes it. How much energy does this explosion hold?
    Click to expand...
    The only difference is you'd be calc'ing the energy required to generate KE from a distance rather than destruction at a distance. It is in no way calc-stacking, and is a pretty blatant case of where it can be used. Not to mention it's quite literally an explosion quirk. Explosions are like one of the things most susceptible to ISL. So to say it isn't applicable here is extremely dishonest. So I'm not sure why I'd need to make a thread for this when it has been accepted and used in calcs before, and is even recommended in the calculation page for instances like this.

    Why exactly would it not be useable here?
    View previous comments…
    LaserPrecision
    LaserPrecision
    No problem. And thank you for asking around. Because I've never got an answer on why the most scientifically used formula for inverse square law isn't used. Any post I made got ignored.

    Yeah, I'm curious where it came about myself considering it has like zero references and I've never seen the formula used anywhere when searching through links and studies.
    TheRustyOne
    TheRustyOne
    I've just been informed by other calc group members that you do need to account for the object's surface area as well. To quote them.

    "Yeah you fundamentally need the surface area of whatever's being hit for the equation, so there's an actual frame of reference for how the energy spreads"

    Meaning just doing Area*Energy is incorrect.

    Explosion Area/Object Area*Energy is correct.
    LaserPrecision
    LaserPrecision
    I'll take your word for it, though I admittedly haven't seen anything like that when looking up inverse square law.
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