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Ugarik
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  • Do you talk calc requests on your wall?
    Ugarik
    Ugarik
    Shoot. But I don't promise i'll make it
    DaReaperMan
    DaReaperMan
    20240221_215908.jpg

    20240221_215858.jpg

    Statesman lifts and throws a giant Clockwork while depowered

    you can read the full thing here

    Also, Statesman is a little over 8 feet tall, since he stands at the max height you can make a character in City of Heroes, though I'd need to set up and AE mission in order to give you a comparison since uh, he can't be found in-game anymore outside of specific quests.
    Reiner04
    Reiner04
    Bump, my apologies for being late, I updated the calc with right shapes.
    Ugarik
    Ugarik
    I accepted the calc but it most graphic editors you can directly measure area of pretty much any arbitrary shape using selection tool
    Reiner04
    Reiner04
    Thank you. I have not done much of calcs before so I've been not so sure on methods but will keep it in mind.
    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?
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