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Viewing blog entries in category: Fire Emblem

  • TTGL
    12cheeper also did a version of this, but never came back to it, so I'll do another.

    0:59 - 1:28

    Flora, a mere maid, freezes the ocean thick enough to stop a fleet of ships. No Dragon Veins required. Average height of a door is 2m.

    78 pixels = 2m
    1 pixel = 2m/78 = 0.0256410256m
    0.0256410256m X 104 = 2.66666666m
    0.0256410256m X 114 = 2.92307692m

    50 pixels = 2.66666666m
    1 pixel = 2.66666666m/50 = 0.0533333332m
    0.0533333332m X 350 = 18.6666666m
    0.0533333332m X 81 = 4.31999999m
    4.31999999m X 2 = 8.63999998m
    0.0533333332m X 42 = 2.23999999m
    0.0533333332m X 302 = 16.1066666m
    16.1066666m + 8.63999998m = 24.7466666m

    18 pixels = 2.92307692m
    1 pixel = 2.92307692m/18 = 0.162393162m
    0.162393162m X 68 = 11.042735m

    Now we have the height of our ships, we can scale to the furthest one (see the above video at 7:18 for more reference and below, see also the map in the corner).

    Now to scaling.

    = 2*atan(55/(310/tan(70/2)))
    = 0.167733436 rad
    = 9.6104179660515463 degrees

    Put that through the angscaler and we get a distance of 65.681m. Add the width of the ship itself to that and it's 90.4276666m. Now for the thickness of the ice. Average height of a man is 1.77m.

    70 pixels = 1.77m
    1 pixel = 1.77m/70 = 0.0252857143m
    0.0252857143m X 10 = 0.252857143m

    4 pixels = 0.252857143m
    1 pixel = 0.252857143m/4 = 0.0632142858m
    0.0632142858m X 12 = 0.75857143m

    With all that done, we can finally get our volume.

    V = πr^2h
    = π X 90.4276666^2 X 0.75857143
    = 19487.1803m^3

    Weight of ice per m^3 is 917kg.

    M = 19487.1803 X 917
    = 17869744.3kg

    Specific heat of water is 4.188, and the average temperature of sea water is 17C. The highest freezing point of salt water is -2C.

    E = mc(delta)t
    = 17869744.3 X 4.188 X 17
    = 1.27225432e9 joules

    On top of that, we can add the heat of the fusion of ice, which is 335.55joules/g, or 335550 joules/kg.

    E = 17869744.3 X 335550
    = 5.9961927e12 joules

    E = 1.27225432e9 + 5.9961927e12
    = 5.99746495e12 joules
    = 1.43342852533 kilotons

    What's more is the final result is even higher, as beyond the ships the ice stretches away to the distance (Though not quite the horizon).


    Final Results
    Flora freezes the ocean = >1.433 kilotons


    As Flora is a mere maid, this should scale to everyone of note. This is also consistent with Camila evaporating a river.
  • TTGL

    11:04

    Corrin makes a bridge of stone. Like last time, I'll be scaling from the in battle landscape.
    Spoiler:


    63 pixels = 1.7m
    1 pixel = 1.7m/63 = 0.026984127m
    0.026984127m X 513 = 13.8428572m

    I'm not sure how to get the length of the bridge, but it's longer than it is wide, so I feel it's safe to say it's about twice as long as it is wide. As for it's height, I'm not to sure, so let's go with half of it's width until I can find some better reference.

    13.8428572m X 2 = 27.6857144m
    13.8428572m/2 = 6.9214286m

    V = lhw
    = 27.6857144 X 6.9214286 X 13.8428572
    = 2652.6333m^3

    It looks enough like granite, which weighs 2691kg/m^3.

    M = 2652.6333 X 2691kg
    = 7138236.21kg

    Let's try GPE.

    Centre of gravity = 27.6857144m/2 = 13.8428572m
    Earths Gravity Pull = 9.807m/s²
    Mass = 1789136100kg

    E = 13.8428572m X 9.807 X 1789136100kg
    = 2.42887572e11 joules
    = 58.0515229445507 tons of TNT

    Final Results
    Corrin makes bridge = 58.056 tons of TNT
  • TTGL

    8:10
    Spoiler:


    I wasn't sure how to scale at first, but I was give the suggestion of scaling off the in battle scenes, so I'll try that. And we indeed get a good view of the river in combat sequence.

    (This is kind of a mess sorry)
    Spoiler:
    Spoiler:


    First for the water at the back on the right.
    = 2*atan(tan(70/2)*(857/510)
    = 1.34483246 rad
    = 77.053224110387234 degrees

    = 2*atan(72/(857/tan(77.053224110387234/2)))
    = 0.0406183341 rad
    = 2.32725911478809 degrees

    16 pixels = 1.7m
    1 pixel = 1.7m/16 = 0.10625m
    0.10625m X 72 = 7.65m

    Calling upon the angscaler, we get a distance of 41.847m away.

    Now for the water on the left (broken bridge).
    = 2*atan(tan(70/2)*(857/510)
    = 1.34483246 rad
    = 77.053224110387234 degrees

    = 2*atan(48/(857/tan(77.053224110387234/2)))
    = 0.12172125 rad
    = 6.9741139020740723 degrees

    8 pixels = 1.7m
    1 pixel = 1.7m/8 = 0.2125m
    0.2125m X 48 = 10.2m

    Once again we call upon the angscaler to get 62.771m away.

    Now the river fork (dark green).
    = 2*atan(tan(70/2)*(857/510)
    = 1.34483246 rad
    = 77.053224110387234 degrees

    = 2*atan(293/(857/tan(77.053224110387234/2)))
    = 0.712210464 rad
    = 40.8066537123552777 degrees

    31 pixels = 1.7m
    1 pixel = 1.7m/31 = 0.0548387097m
    0.0548387097m X 293 = 16.0677419m

    Once again using the angscaling calculator, we have a distance of 21.599m.

    Now to take that away from the other two...
    41.847m - 21.599m = 20.24800m
    62.771m - 21.599m = 41.17200m

    Next we do the closest part (light green).
    = 2*atan(tan(70/2)*(857/510)
    = 1.34483246 rad
    = 77.053224110387234 degrees

    = 2*atan(221/(857/tan(77.053224110387234/2)))
    = 0.547053873 rad
    = 31.3438780892634909 degrees

    0.0548387097m X 221 = 12.1193548m

    The angscaler gives us 21.599m (same as above, for some reason).


    Elise says it's too deep to go over on horseback, so I'm assuming it's deeper than a horse is tall. A horse is about 1.4 – 1.8 m tall, so let's go with a depth of 2m. Calcing it as a trapezoidal prism. First the right (going with the flat river bed being half the bed and the slopes to the banks being the other halves)...

    7.65m/2 = 3.825m

    V = LH(A + B)/2
    = 41.847m X 2m (7.65m + 3.825m)/2
    = 480.194325m^3

    ...Then the left...

    10.2m/2 = 5.1m

    V = LH(A + B)/2
    = 62.771m X 2m (10.2m + 5.1m)/2
    = 960.3963m^3

    ...then the front bit.

    12.1193548m/2 = 6.0596774m

    V = LH(A + B)/2
    = 21.599m X 2m (12.1193548m + 6.0596774m)/2
    = 392.648916m^3

    V = 480.194325m^3 + 960.3963m^3 + 392.648916m^3
    = 1833.23954m^3
    Now for the next part of the river. This is the best shot I could find...

    27 pixels = 1.7m
    1 pixel = 1.7m/27 = 0.062962963m
    0.062962963m X 92 = 5.7925926m

    = 2*atan(tan(70/2)*(857/510)
    = 1.34483246 rad
    = 77.053224110387234 degrees

    = 2*atan(92/(857/tan(77.053224110387234/2)))
    = 0.23253404 rad
    = 13.323219085159314 degrees

    The anglescaler returns again to give us a distance of 24.798m.

    5.7925926m/2 = 2.8962963m

    V = LH(A + B)/2
    = 24.798m X 2m (5.7925926m + 2.8962963m)/2
    = 215.467067m^3

    At last we can get the total volume.

    V = 3616.65537m^3+ 215.467067m^3
    = 3832.12244m^3

    M = 3832.12244 X 1000
    = 3832122.44kg

    It takes 2260 joules to evaporate a gram of water.

    E = 3832122.44 X 2260000
    = 8.66059671e12 joules
    = 2.06993229207 kilotons

    Final Results
    Camila evaporates a river = 2.07 kilotons


    I'm not sure if I got the scaling for the river right, and I missed the further parts, but that's a calc I've been meaning to do for a bit, and it looks promising in anycase.
  • TTGL
    Here I will try and find the power of individual fire dragons from Fire Emblem (7). I'm not yet sure how to find energy from things yet, so I'll be going by range. First, let's find the size of a dragon. Assuming that ice dragons and fire dragons are the same size, we can scale them off Ninian.



    The average human height is 1.75m. Hectors probably taller than that, but let's stick with that for now.

    1.75 X 4 = 7

    The dragons seem to be slightly longer than they are tall (not including the tail), so let's add 1m.

    Now, as a dragon can fit entirely on one square, let's try and find the size of each square from that...



    That's the room seen there (the room itself is even bigger). One dragon is able to rock it by it's emergance.



    7 X 8 = 56m^2
    12 X 15 = 180
    180 X 56 = 10,080m^2
    (15 X 8 = 120m)
    (12 X 8 = 96m)



    Ninian closes the gate before the dragon can fully crosss, and it's destroyed.



    9 X 9 = 81
    81 X 56 = 4536^2
    (9 X 8 = 72m)

    The standard size of a New York city block is 80 m X 274 m (21,920m), so this may well be low city block level at least.
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