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Agnaa

Agnaa

VS Battles Super Moderator Administrator Calculation Group Translation Helper Human Resources Gold Supporter
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  • Currently in a nadir of interest. Not reading my profile or notifications. If there's anything important, DM me. 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==================================================================================================================================================================== Currently in a nadir of interest. Not reading my profile or notifications. If there's anything important, DM me.
    Mr_Sugabz
    Agnaa
    Agnaa
    What do you need the middle value for?

    But it would be 0.13 + ((0.2-0.13)/2), or 0.165
    Mr_Sugabz
    Mr_Sugabz
    hmm I'm arguing with my friend actually. And, is this different from the average formula in mathematics?
    Example: (0.2+0.3+0.4 ........ 0.11+0.12+0.13) Then divided by how many data there are?
    Agnaa
    Agnaa
    The conventional average formula doesn't really work for infinite sets; there's an infinite amount of values between 0.2 and 0.13

    But there's many ways to reach the answer I described. (0.2+0.13)/2 also reaches the same answer for all endpoints.

    And intuitively, you could imagine it working the same way as the ordinary average formula. If you add each of the numbers on either side of the extreme, the average of that pair would cancel out to the midpoint. That gets done for every single number in the infinite set.
    FantaRin_The_First
    Hey, hey, fanta here. Small question... How do you find the cubic meter (m^3) value to find the volume for humanoid-shaped things? Genuinely curious, and wonder if you know or is just really simple - like I just need to find a meter to cubic meter converter online or smth.
    View previous comments…
    FantaRin_The_First
    FantaRin_The_First
    Hey, hey, it's Fanta again, and I wish to inform you of what I did with the information you gave me, well... It's dumb, but, basically, I used the height of a giant mecha (6.94m) and ^3 times it in order to get a volume of 334.3m - afterwards, I used the fact that the mecha was composed out of tungsten and multiplied the volume value I got by the V.Frag Value of Tungsten [1412.5 J/cc].

    ... To make it more quicker to understand, here's the formula I used.

    334.3 m^3 * 1412.5 J/cc = 472198750000 joules, or 112.86 Tons.

    Is there anything wrong with this formula itself, and if so, could you tell me?
    Agnaa
    Agnaa
    Main concern I could have is that giant mechas wouldn't usually be completely solid, like the human body is, so you'd need to reduce that number based on how hollow the mech seems.

    If it seems to be 20% air, you'd multiply it by 0.8, if it seems to be 50% air, multiply by 0.5, etc.

    Also, could be a bit strange to assume tungsten for literally the entire thing, including cables and all of that. V. Frag may or may not be accurate. Etc.
    FantaRin_The_First
    FantaRin_The_First
    True, true. Hm... There is a cockpit area, so maybe 0.8 to 0.5 might be fine. Idk. Never done something like this before.

    And I know, but tungsten is the only confirmed material, so can't really guess. And I am assuming V.Frag cuz it explicitly shows very small shards of metal flying onwards after they explode, could settle for Frag, tho.
    FantaRin_The_First
    Hey, hey, fanta here!

    Agnaa, could you check if I do the numbers right in this calc? It is the thing that I asked for your help for, y'know, the big hole. I literally just want to know if I did the math right, and didn't screw up somewhere.
    View previous comments…
    Agnaa
    Agnaa
    I ran the numbers for one of the triangles, and you applied it perfectly.
    FantaRin_The_First
    FantaRin_The_First
    Thanks, mate! From here on, won't be bothering ya about this calc. Got everything I wanted answered, answered! And now I wait for other CGM to check out my calc and get their two cents 'bout it.
    Agnaa
    Agnaa
    No problem buddy.
    azontr
    View previous comments…
    azontr
    azontr
    Alright, thanks!
    Drite77
    Drite77
    Heeeey, sorry for bringing this up here since Fandom is a bitch and won't notify us

    The reason why I don't think I should double the result is because they have 2 arms. When one arm punches, the other is in standby, and when they retrieve their arm, the other punches at the same time.

    They weren't punching with one hand or don't have 2 arms
    Agnaa
    Agnaa
    idk if that really matters. Their total movement is 398-400 arm lengths, even if it was done with two limbs.

    You can ask another calc group member, but I think doubling would be ideal. If they say otherwise, then I'll fold. I am relatively new to this, after all.
    FantaRin_The_First
    Hey, hey, I got a question to ask.

    Would it be okay for me to make a user blog that features 3 different calcs from the same show in one user blog? Kinda wanted to make three different user blogs but figured that would be too troublesome and tiresome for CGMs.
    View previous comments…
    Agnaa
    Agnaa
    I'm not sure what destruction values we give for asphalt.

    Using the shapes I drew, I'd do area of a triangle (green triangle 1) + area of a triangle (green triangle 2) + area of an ellipsis (green circle) - area of an ellipsis (yellow circle), and then multiply all of that by the depth. If the crater has an uneven depth, it would be really difficult to account for because of the weird doughnut + triangular indent it's got going on.

    I'd say just a depth of 5.7m for seeming taller than the robots. I wouldn't bother pixel-scaling.
    FantaRin_The_First
    FantaRin_The_First
    I'll figure out the destruction values for asphalt later... Probably settle for just cement or just rocks to make it simpler.

    Thank you for your help. See you later, I suppose.
    FantaRin_The_First
    FantaRin_The_First
    Hey, hey, I am in the finishing touches of my calc - and am going to actually put it in a user blog soon, but want to get something clear first.

    When you said do the area of the shapes and multipy them by the depth do you mean smth like this - example: 7m (base of a triangle times 4m (height of a triangle) times 5.7m (the depth of 5.7m that you mentioned I should use) to get their cubic centimeter / cubic meter?

    Also, once I get all of the cubic centimeter / cubic meter for all of the shapes, do I add them up altogether and then multiply them by the pulverization value of concrete and/or rock?

    Just making sure since I haven't done a calc like this before.
    Eseseso
    Could you open up the Doctor Doom page for just a sec?

    I just want to remove the justification of Doom "destroying Doombots almost comparable to him", since that statement means nothing in scaling and just seems so weird.
    Agnaa
    Agnaa
    Doctor Doom's getting changed real soon, iirc. You should ask in this thread about it.
    Eseseso
    Eseseso
    The thread already did their change and relocked the page, but I'll re ask
    Omnificence
    @Agnaa Do you know if this is a galaxy or universe, or some shiz like that?
    Agnaa
    Agnaa
    The terrible quality makes it too hard to tell.

    Pinging me does nothing btw. Only staff members give notifications by pinging, and I already get notifications for posts on my wall.
    Omnificence
    Omnificence
    I don't suppose you can calculate it, can you?
    Agnaa
    Agnaa
    Nope.

    If you still want me to eyeball from that video, it looks more like a protostar/proto-solar system. Maybe less because of the relative sizes of the characters/spaceships.

    But I could be missing some details that indicate it's a galaxy/universe in the low quality.
    RoTt35
    Hello, I would appreciate if you could evaluate this calc. Thanks in advance
    Agnaa
    Agnaa
    I will not be evaluating that calc.
    RoTt35
    RoTt35
    Np, thanks anyways
    Rikimarox2
    Ello, simple question if you don't mind, but if we have a statement about a character shaking an entire forest, but the forest size wasn't mentioned, what can I do to calc it? Or is it just unquantifiable?
    Jamesthetaker
    Hi there, may I ask a question about lifting strength?

    If there were characters A and B with comparable AP but B had superior lifting strength, then would B be able to crush A with LS with ease?
    Drite77
    Suuuuuup, I don't know if you didn't get a notification (Because I didn't) of the evaluation here but:
    He gave the scan for the 3/4s of a month and it checks out.
    The ship doesn't really have any description itself. Prob could change the speed to a 1800s ship instead of a modern day one if needed
    View previous comments…
    Agnaa
    Agnaa
    I left the question in the calc's comments, but I'll repost here.

    Can you post where you got that formula from? I don't recognize it. And using another formula I've seen accepted in other calcs, I got a result about 3.53x lower. So either you're using the formula incorrectly, or one of these calcs needs to be updated.
    Click to expand...
    Drite77
    Drite77
    Agnaa
    Agnaa
    Oh, airburst vs ground, doh. Yeah that's fine.
    Arnoldstone18
    Agnaa
    Agnaa
    I'm glad that I don't have to because I won't :3
    Arnoldstone18
    Arnoldstone18
    41ee9620-7454-49a5-a080-e7467b8efad0.png


    Understandable 🫠
    Arkenis
    Hey, got a calc you could do for me if u can. I'm struggling to figure out what to use and how to calc it really given the type of reaction speed.
    0091-013.png

    0091-014.png


    The projectiles are all moving at mach 3 and the guy reacting is 6ft. and the sword is is moving, he's not swinging it to react to them, that sword just fast.
    Agnaa
    Agnaa
    I don't understand what the hell's going on here. How do know the projectiles are moving at mach 3?
    Arkenis
    Arkenis
    They're stated to in previous chapter.
    Agnaa
    Agnaa
    It would be best to link that statement as well when asking for the calc to be made, but idk how to calc that, sorry.
    FantaRin_The_First
    Hey, hey, would it be possible for you to look this calc? ... The first one, to be exact. It is simply a calc that calc a giant's size and GPE to get its AP - it (and the lesser calc) serves as the foundation for an entire verse, but I really just need somebody to see if I fudge smth up with the giant calc.
    Agnaa
    Agnaa
    I'm putting calc evals off for a minute. I might get to it eventually, but you should prolly ask someone else.
    FantaRin_The_First
    FantaRin_The_First
    Thank you for taking the time to reply, Agn. Will do what you said.
    Propellus
    Hey Agnaa, I'm not sure if you're knowledgable in Pythagoras theorem, but this calc could really use some corrections using that formula.
    Agnaa
    Agnaa
    The length of the side opposite the right angle of a right-angled triangle is equal to the square of the other two sides.
    XitSign
    Hello, and terribly sorry for the both on this, going around to try to get some input.

    I have a fairly important CRT for the verse Puzzle and Dragons, however due to its niche nature it can be really hard to get input on these.

    If you can give your thoughts, it would be greatly apricated, but if not I 100% understand. Either way, thank you so much.
    Agnaa
    Agnaa
    I cannot, sorry m8.
    XitSign
    XitSign
    Totally understandable, thank you for getting back to me!
    Expectro2000xxx
    Hi, did a calc to some friends about the size of a giant bear and his GPE, when you have free time can you take a look? Based on the comments I did two mistakes but corrected them so I would like confirmation if now is fine.
    View previous comments…
    Expectro2000xxx
    Expectro2000xxx
    Sorry to bother again, but someone in the comments had a problem with the pixel scaling so in the end I used another image that he posted to pixelscale, so if you could look again at the calc to see if the pixel scaling is fine? The math itself is the same as previously so that part should be good.
    Agnaa
    Agnaa
    Evaluated.
    Expectro2000xxx
    Expectro2000xxx
    Thank once again.
    Majinere566
    Hey sorry if I'm disturbing but can you check this calc out? I need it for an urgent revision. Thanks in advance

    vsbattles.fandom.com

    Tsuna Clocks Lightspeed Bermuda

    vsbattles.fandom.com vsbattles.fandom.com
    Rikimarox2
    Ello! Could you check this calc please? Honestly had no idea what to do, and if using assumptions like that are not allowed. The feat is an important one, but not sure if the method I used for the calc is allowed.
    View previous comments…
    Rikimarox2
    Rikimarox2
    Ah I'm not sure I understand what to do, do I just use something like this:

    Radius = 50 meters, with the height of 20-30 meters instead of taking the mountain's entire volume? Did you mean I should take like, 0.05% or 0.004% of the volume of the mountain and be done with it?
    Agnaa
    Agnaa
    Naw you'd use either 20m or 50m as the height, derive the radius from that (can be done using the pixel scaling on the mountain/island level requirements page) and use that to get volume.

    Could be another number too if calc group finds it suitable. People might wanna do it the other way and get a certain radius (maybe 100m), derive a height from that through the pixel-scaling, and get volume from that.
    Rikimarox2
    Rikimarox2
    Aha I think I understand. I've redone the calc with the edit, so when you have time, I'd appreciate it if you could check it out. (Let's hope I did it the right way)
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