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Ugarik
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  • Helloo, I was told to ask you to evaluate this calc- I know you gave your opinion previously regarding it generating a centrifugal force and thus contributing more in decrease in the planet's effective gravity - but I just did the calc as per how planets gravity works within the show. But either way can u check 2nd part incase first part is not making sense? it simply pulling back planet parts that drifted away back into planet but with actual stated mass that is 36% of planets.
    I recently came across the thread where you suggested a new system for fragmentation, violent fragmentation, and pulverization feats, where you use toughness. I understand why the wiki chose not to go with that, but for me personally, accuracy is very important. Do you have any suggestions for me if I use your toughness system? And do you still hold the stance that doing so is accurate?
    Looking at your calc here, what verifies the Dragon is swinging its sword at lightning speeds?
    Ugarik
    Ugarik
    The lightning was 40 m long and it crossed that distance in two frames. Asuming the lightning was traveling at 440 km/s we can conclude that each frame is 0.000045 seconds long

    ^There are most likely mistakes, but that should be the kinetic energy of tilting the ship.

    @Ugarik How can I calculate the energy absorbed in the collision? I doubt Saitama's jump even reaches 7-B with his kinetic energy alone.
    Ugarik
    Ugarik
    Is is a very rought caclulation beause that ship is not a parallelepiped but nevertheless you got everything right

    As for the absorbed energy. First of all, you need to calculate angular momentum of the ship (rotational velocity times its moment of inertia). Then we can apply the law of conservation of angular momentum. According to that law we can conclude that Saitama had the same amount of angular momentum as the ship.

    If we divide Saitama's angular momentum (which is, as previously concluded, equal to the angular momentum of the ship) by the horizontal distance from the pivet axis we get his linear momentum. So now than we know Saitama's linear momentum we can also find his KE since his mass is given. Momentum divided by mass gives you speed and we all know how to calculate KE using speed and mass. How KE of Saitama minus KE of the ship gives you the energy absorbed by the collision (by is expect that ship's KE is gonna be so low is conparasing that we can skip that part completely)

    The problem with all that is that you can (all you almost certainly will) get FTL speed doing all that. So in that case you need to use relatevistic KE formula which is
    • rKE = c*sqrt(m^2*c^2+p^2)-mc^2
    Ugarik
    Ugarik
    You can learn more about the topic here
    Got sent here through KLOL and Armorchompy. They've both heard that cracking feats can't be calced, but I remember this calc Bambu did a few years ago which finds the value by treating the cracked region as a cylinder with a depth of 1cm, using the extent of the cracks as the radius, and then applying fragmentation over it. So what's the deal here? Would a thread need to be made to remove some calcs, or are other cracking feats fine to be calced?
    Ugarik
    Ugarik
    As far as I understand cracks don't really propogate that way in real world. They're always as deep as an material's layer itself.

    With that being said I still think bambu's method is acceptable as the formula "shear strength×volume" has nothing to do with real world's material science anyway.
    Hey, if you've got the time, would you mind helping me with how to calculate a feat? It involves bending steel.

    Tangle bending a hole in a thick steel door

    I figured since the steel seems very thick and they move it outwards quite a bit that it'd yield pretty good results, though I'm clueless as to how this type of feat should be calculated. I heard you were knowledgeable on the topic.
    Ugarik
    Ugarik
    This is a typical deformation problem. You can solve it just by multipling compresive strength by inner cross-sectional area
    I know you're retired, but you evaluated this calc, do you have any thoughts about my comment on it arguing that the actual result should be 15x lower?

    If you don't want to be bothered by this, lemme know and I'll just send it to other calc group people.
    Ugarik
    Ugarik
    It's okay I'm not bussy right now. And the result would actually be even lower than that. Somehow I totally overlooked it
    Agnaa
    Agnaa
    Oh damn, thanks for giving it a look.
    Привет. Я изменил расчёт Орочи и получил корректные результаты. Также добавил плашку кинетической энергии. Мог бы ты посмотреть?
    Ugarik
    Ugarik
    Как ты думаешь, почему я вышел из группы расчётов?
    NikHelton
    NikHelton
    Ну, ты комментаруешь время от времени, поэтому я и обращаюсь.
    Ты можешь просто сказать и я не буду тебя трогать
    Привет. Подумал вычислить один из подвигов в OPM, где Генос и Псайкос образуют единый взрыв и рассчитать это по формуле ударной волны. И подумал, а можно ли вместо стандартного давления в менее полторы бары использовать давление внутри ядерного взрыва? Я перевёл МПа в бары и вышло 50к бар
    NikHelton
    Ugarik
    Ugarik
    Нельзя потому что там идёт речь о волне избыточного давления
    NikHelton
    NikHelton
    Хм. Можно ли тогда поиграться с таймингом у этого расчёта? То есть формула подразумевает под собой давление ударной волны со скоростью в 3 маха. Если предположить более высокую скорость, и результаты же будут выше.

    И кстати, мог бы пожалуйста глянуть таки рассветы на которые кидал ссылки?
    Heyo question out of curiosity, do we have a formula for calculations that deal with destruction on a macro quantum level? I know that we do for Sub Atomic destruction but for a feat I wanna tackle the character's are Manipulating electrons which is Macro Quantum rather than Sub Atomic.


    Now would I substitute it for Sub Atomic destruction for the time being? I myself have never seen any calcs with macro quantum destruction so I'm a bit confused on what to do here.
    Ugarik
    Ugarik
    I'm sorry but I actually have no idea. I'm not that knowledgeable in physics in general. While I am good in classical mechanics and material science my knowledge is barely above average when it comes to any other field
    Yo, ive saw ur opm moon jump calc and ive found ways to measure boros’ ship. If you’re interested in re calcing the feat, I can give you the measurements
    TimmyTurnero
    TimmyTurnero
    In the anime, but if the manga (main source) presents a way to get a much larger size, then it’s more accurate to use that. This would allow it to be 2.7x larger than what the anime says. The feat would be much more impressive:vvvv
    NikHelton
    NikHelton
    The size of the bullet is not calculated quite correctly due to the perspective in the picture and the shape of the bullet itself. You can find another frame to calculate its size.
    TimmyTurnero
    TimmyTurnero
    Yea, its actually a lot LARGER than whats shown here. I can find another way if needed (its not needed)
    Привет, можешь подсказать можно ли рассчитывать первую ачивку как ударную волну, а вторую как землетрясение? Всё происходит в версе Кенгана и пользователи сделали расчёты этого, которые дали результат в Low 7-C. Аргументом является громкий "бум" и небольшая нестабильность на фрейме.

    Насчёт второго, было показано как сотрясается вершина горы, но расчёт был сделан с использованием всей горы и её основания. Можно ли это использовать?
    qCsXgNS1bsg.jpg
    cOba5tDiMBw.jpg
    Может ли это быть расценено как землетрясение?
    NikHelton
    NikHelton
    Не совсем понял формулу с помощью которой можно найти частоту.
    Длину умножаем на плотность и делим на модуль?
    Ugarik
    Ugarik
    Скорость звука нужно разделить на две длинны f = v/(2L). А корость звука это корень из модуля Юнга деленного на плотность v = sqrt(E/ρ)
    И эту частоту не нужно умножать на 2 пи. Могу объяснить почему если будет много вопросов
    NikHelton
    NikHelton
    В данной формуле f это что?
    Sorry to bother you Ugarik.

    I'm calcing someone throwing a huge rock (with his arms) and I'm on the part of finding the distance. Calc is here.

    When I do center of mass, should I add the length of the arms + center of mass, or should I just do just the length of the arms?
    Hey Ugarik, I wanted to ask about a calc I'm working on.

    A character (over time) evaporates a large volume of water which eventually turns into clouds.

    I assumed it would be a 2 part process with assuming the volume for the evaporated water and the cloud would be relative/equal/the same.
    The first process was calculating the energy to vaporize the body of water, and the second process was calculating the energy to condense it into clouds.

    Would that work, or should I axe one of the processes, or is there another way for me to calculate it?
    I just noticed your comment from half a year ago on this calc saying that cutting doesn't work that way. I based another calc off that method and have directed multiple other users to that calc as a sample for their own rock-cutting feats.

    Did you ever get around to making that sort of method officially rejected?
    Ugarik
    Ugarik
    Fracruring is different from cutting and nearly impossible to calculate.

    But this this method only works on the structural level. It suggests that cutting something takes no energy as long as your blade is infinetly sharp. This is actually not true because molecules within a material are chemicaly bond so you still have to apply some energy to unbind them no matter how sharp your blade is.
    Agnaa
    Agnaa
    So my calc should be rejected and that end removed from the profile?
    KLOL506
    KLOL506
    Now that I look at it, I don't think this is your typical cracking calc. This'll require a bunch of pixel-scaling lines.
    Question, what is the formula for calculating water resistance???
    Ugarik
    Ugarik
    The force in newtons is calculated as F = 0.5*A*c*p*v^2, where
    A - cross sectional area
    c - drag coefficient
    p - water density
    v - velocity
    This is only the pressure term, it should be used if the speed is much higher than the critical velocity.
    You should use a different formula if the speed is low, but I don't know it
    Hey Ugarik, we're still racking our brains over how to find the speed of Genos's beam and EC in chapter 84 of OPM for the EC KE calc. Obviously both are at least supersonic, narratively far higher since we're talking about a dragon and a near dragon respectively. I still don't know of any way to calc the beam speed other than lowballing it at speed of sound, which is very much suboptimal if we can find any decent real life analogue. But I have had an epiphany and I would like your opinion before I invest time into it.

    Death Gatling uses a super-gatling gun to face Garou right? 30mm munitions fired from a gatling gun like a GAU-Avenger have a 1010 m/s fire rate. Garou was able to deflect hundreds of bullets at this velocity yet struggled to escape Genos's beams, could we scale Genos's beam speed to Death Gatling's gun and find EC's speed off that?
    Hello, Jack has made changes to this calculation. Would mind and evaluate this calc once again
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