Trigger-verse calc archive

[Stocking Anarchy]

Shiny Chariot blows the Gainax/Trigger star onto the Moon. The Moon is 3474.8km in diameter.

Spoiler

717 pixels = 3474.8km
1 pixel = 3474.8km/717 = 4.84630404km
4.84630404km X 221 = 1071.03319km
4.84630404km X 70 = 339.241283km

A = 1071.03319 X 339.241283/2
= 178546.302km^2
= 181669337000 X 4
= 726677348000m^2

Next the volume of the square middle part as a square.

V = lh
= 339.241283 X 339.241283
= 115084.648km^2
= 115084648000m^2

V = 726677348000 + 115084648000
= 841761996000m^3

Now for the depth of our mark. We can clearly see it's at least as deep as the nearby lunar basins, so we'll scale using the depths of lunar basins (from here).

Spoiler

the-moon - Antoniadi
4km
the-moon - Mare Australe
2.13 km
the-moon - Bailly
4.13 km
the-moon - Compton
3.2 km
the-moon - Coulomb-Sarton Basin
3.9-6 km (average of 4.95km)
the-moon - Mare Crisium
4.57 km
the-moon - Mare Fecunditatis
1.84 km
the-moon - Freundlich-Sharonov Basin
3.6-6.0 km
the-moon - Grimaldi
3.2 km
the-moon - Hertzsprung
4.5-5.3 km (average of 4.9km)
the-moon - Mare Humboldtianum
2.98 km
the-moon - Mare Humorum
2.24 km
the-moon - Mare Imbrium
2.9 km
the-moon - Korolev
4.6-5.4 km (average of 5km)
the-moon - Lorentz
4.45 km
the-moon - Lomonosov-Fleming Basin
2.44 km
the-moon - Mendeleev
4.8-5.0 km (average of 4.9km)
the-moon - Mendel-Rydberg Basin
5.24 or 5.56 km (average of 5.4km)
the-moon - Milne
3.25 km
the-moon - Mare Moscoviense
5.25-5.96 km (average of 5.605km)
the-moon - Mutus-Vlacq Basin
3.0 km
the-moon - Mare Nubium
1.63 km
the-moon - Mare Orientale
6.04 km
the-moon - Planck
4.0 km
the-moon - Schrödinger
4.8 km
https://the-moon.wikispaces.com/Mare+Serenitatis
2.14 km
https://the-moon.wikispaces.com/Mare+Smythii
5.0 km
https://the-moon.wikispaces.com/South+Pole-Aitken+Basin
8.5-12 km (average of 10.25km)

Let's find the average

D = 1.63 + 1.84 + 2.13 + 2.14 + 2.24 + 2.44 + 2.9 + 2.98 + 3.0 + 3.2 + 3.2 + 3.25 + 4 + 4.0 + 4.13 + 4.45 + 4.57 + 4.8 + 4.8 + 4.9 + 4.95 + 4.8 + 5 + 5.0 + 5.4 + 5.605 + 6.04 + 10.25
=113.645/28
= 4.05875km
= 4058.75m

Now for the volume of the point. There are four of them, so we'll times it by four.

V = 841761996000 X 4058.75
= 3.4165015e15m^3

Next we find the energy, using pulverization (you see dust everywhere surrounding the Moon afterwards, as shown below).

E = 3.4165015e15 X 200000000
= 6.833003e23 joules

We'll hold onto that for now. We can see the debris surrounding the Moon afterwards too, so we can find KE.

Spoiler

Timeframe is 4 seconds. Also note beforehand, we see no debris.

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Now onto scaling.

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Spoiler

161 pixels = 3474.8km
1 pixel = 3474.8km/161 = 21.5826087km
21.5826087km X 309 = 6669.02609km

T = 6669.02609km/4s
= 1667256.52m/s

Average mass of the Moon is 3344kg per m^3.

M = 3.4165015e15 X 3344
= 1.1424781e19kg

Now for the kinetic energy.

E = (0.5) X 1.1424781e19 X 1667256.52^2
= 1.5878985e31 joules

Let's add them together for our total energy.

E = 6.833003e23 + 1.5878985e31
= 1.58789857e31 joules
= 3.79516866634799225 yottatons

We can also get the speed for this. Distance to the Moon from Earth is 384,400km.

Spoiler

3 seconds.

T = 384,400km/3s
= 128133333/299792458 X 100
= 42.7406793% C

Final Results
Shining Chariot makes a star on the Moon = 3.795 yottatons
Shining Chariot's Moon arrow speed = 42.741% C

 

[Stocking Anarchy]

Spoiler

The time loop that Akko and her friends get caught in causes multiple stars to glow brighter than usual. If it was just luminosity alone, that'd be easy, but we're talking about an increase in pre-exisiting luminous energy. I'll be looking at Taccocat's Percy Jackson Artemis calc for reference. In order to find the increase in luminosity of the stars, we're going to need to find the bolometric magnitude, for which, we'll first need the apparent magnitude. To find that, we'll be using the below formula;

M = Absolute magnitude
m = Apparent magnitude
D = Distance in parsecs

To make things much easier for us, all of the stars that increased in luminosity are directly named. Pollux is 34 lightyears (10.4244 parsecs) away, has an apparent magnitude of 1.14, is a spectral type of K0 III and a luminosity of 32.7 L.

M = 1.14 - 5((log(10.4244) - 1)
= 1.04974466

Next we find the bolometric magnitude, given the bolometric correction, using the below chart...

Spoiler

Pollux has a spectral class of K0 III, so it's bolometric correction is -0.37.

Mbol = Bolometric magnitude
Mv = Absolute visual magnitude
BC = Bolometric correction

Mbol = 1.04974466 + -0.37
= 0.67974466

Finally, to figuring out the luminous energy of the star.

Lstar = Luminousity of the star
L = Luminousity of the Sun
MbolSun = Bolometric magnitude of the Sun
Mbolstar = Bolometric magnitude of the star

The Sun usually has a bolometric magnitude of 4.75 and a luminousity of 3.846 X 10^26 watts, or 3.84600e26 watts (note also that watts are equal to joules, though watts are in joules per second).

Lstar/3.84600e26 = 10^((4.75 - 0.67974466)/2.5)
Lstar/3.84600e26 = 42.4719436
Lstar = 3.84600e26 X 42.4719436
Lstar = 1.63347095e28 W

We're not done quite yet however. The luminousity of Pollux is 32.7 L, which means it's 32.7 times brighter than the Sun.

L = 32.7 X 3.84600e26
= 1.257642e28 W

Finally, we subtract the energy to find the increase in luminousity.

L = 1.63347095e28 - 1.257642e28
= 3.7582895e27 W

With all that finally out of the way? We just have to do it for all the other stars too. We'll refer to the above chart as we go along.

Gienah is 154 lightyears (47.2166 parsecs) away, has an apparent magnitude of 2.585, is a spectral type of B8 III and has a luminosity of 331 L.

M = 2.585 - 5((log(47.2166) - 1)
= -0.785473555

There's nothing for a spectral type of B8 III in the above chart, so I assume that means a BC 0.

Mbol = -0.785473555 + 0
= 0.785473555

Lstar/3.84600e26 = 10^((4.75 - -0.785473555)/2.5)
Lstar/3.84600e26 = 163.753059
Lstar = 3.84600e26 X 163.753059
Lstar = 6.29794265e28 W

L = 3.84600e26 X 331
= 1.273026e29 - 6.29794265e28
= 6.43231735e28 W

L = 6.43231735e28 - 6.29794265e28
= 1.343747e27 L

Ankaa is 85 lightyears (26.0611 parsecs) away, has an apparent magnitude of 2.377, is a spectral type of K0.5 IIIb and has a luminosity of 80 L.

M = 2.377 - 5((log(26.0611) - 1)
= 0.297036287

The closest I can find to K0.5 IIIb is K0 III, which would be a BC of -0.19

Mbol = 0.297036287 + -0.19
= 0.107036287

Lstar/3.84600e26 = 10^((4.75 - 0.107036287)/2.5)
Lstar/3.84600e26 = 71.9756317
Lstar = 3.84600e26 X 71.9756317
Lstar = 2.7681828e28 W

Ankaa is 80 times more luminous than the Sun.

L = 3.84600e26 X 80
= 3.0768e28 W

L = 3.0768e28 - 2.7681828e28
= 3.086172e27 W

Shaula is 570 lightyears (174.763 parsecs) away, has an apparent magnitude of 1.62, is a spectral class of B2IV+DA7 and a luminosity of 36,300 L.

M = 1.62 - 5((log(174.763) - 1)
= -4.59224746

There's no DA or D class on the above chart, so we'll just take it as B2IV. As IV isn't listed in the above chart, we'll go for an average between III and V. And given III is 0, we'll go with V, that being a BV of -2.2.

Mbol = -4.59224746 + -2.2
= -6.79224746

Lstar/3.84600e26 = 10^((4.75 - -6.79224746)/2.5)
Lstar/3.84600e26 = 41390.3391
Lstar = 3.84600e26 X 41390.3391
Lstar = 1.59187244e31 W

L = 3.84600e26 X 36,300
= 1.396098e31 W

L = 1.59187244e31 - 1.396098e31
= 1.9577444e30 W

Algol is 92.8 lightyears (28.45261 parsecs) away, has an apparent magnitude of 2.12, is a spectral class of B8V, K0IV and A7m (it's a trinary star) and a luminosity of 182 L.

M = 2.12 - 5((log(28.45261) - 1)
= -0.150610555

Given we have multiple stars with varying spectral classes, we'll just go with B8V and a BC of -0.55, as it's the only one there I could find in the above chart.

Mbol = -0.150610555 + -0.55
= -0.700610555

Lstar/3.84600e26 = 10^((4.75 - -0.700610555)/2.5)
Lstar/3.84600e26 = 151.441263
Lstar = 3.84600e26 X 151.441263
Lstar = 5.82443097e28 W

L = 3.84600e26 X 182
= 6.99972e28 W

L = 6.99972e28 - 5.82443097e28
= 1.17528903e28 W

Mimosa is 280 lightyears (85.8484 parsecs) away, has an apparent magnitude of 1.25 is a spectral class of B0.5 III and has a luminosity of 34,000 L.

M = 1.25 - 5((log(85.8484) - 1)
= -3.41866103

The BC of a B0 III star (closest to B0.5 III) is -2.95.

Mbol = -3.41866103 + -2.95
= -6.36866103

Lstar/3.84600e26 = 10^((4.75 - -6.36866103)/2.5)
Lstar/3.84600e26 = 28019.7601
Lstar = 3.84600e26 X 28019.7601
Lstar = 1.07763997e31 W

L = 3.84600e26 X 34,000
= 1.30764e31 W

L = 1.30764e31 - 1.07763997e31
= 2.3000003e30 W

Canopus is 310 lightyears (95.0464 parsecs) away, has an apparent magnitude of −0.74, is a spectral class of A9 II and a luminosity of 10,700 L.

M = −0.74 - 5((log(95.0464) - 1)
= -5.62967836

I couldn't find anything for the BC of A9 II spectral class stars (even when searching the wider web), so fo now we'll use the closest thing of an A7 II BC, that being 0.09.

Mbol = -5.62967836 + 0.09
= -5.53967836

Lstar/3.84600e26 = 10^((4.75 - -5.53967836)/2.5)
Lstar/3.84600e26 = 13057.84
Lstar = 3.84600e26 X 13057.84
Lstar = 5.02204526e30 W

L = 3.84600e26 X 10,700
= 4.11522e30 W

L = 5.02204526e30 - 4.11522e30
= 9.0682526e29 W

We're halfway there now! Then there are the seven major stars of the Horologium constellation; Alpha Horologii, R Horologii, Beta Horologii, Mu Horologii, Zeta Horologii, Eta Horologii and Iota Horologii. Alpha Horologii is 117 lightyears (35.8724 parsecs) from the Sun, has an apparent magnitude of 3.853, is a spectral class of K2 III and a luminosity of 37.61 L.

M = 3.853 - 5((log(35.8724) - 1)
= -1.07919783

The BC of a K2 III star is -0.42.

Mbol = 1.07919783 + -0.42
= 0.65919783

Lstar/3.84600e26 = 10^((4.75 - 0.65919783)/2.5)
Lstar/3.84600e26 = 43.2833502
Lstar = 3.84600e26 X 43.2833502
Lstar = 1.66467765e28 W

L = 3.84600e26 X 37.61
= 1.4464806e28 W

L = 1.66467765e28 - 1.4464806e28
= 2.1819705e27 W

Beta Horologii is 295 lightyears (90.4474 parsecs) away from the Earth, has an apparent magnitude of 4.979, is a spectral type of A3/5 III and has a luminosity of 63 L.

M = 4.979 - 5((log(90.4474) - 1)
= 0.197019564

Given A3 III is listed first, we'll use the BC of that, that being -0.03.

Mbol = 0.197019564 + -0.03
= 0.167019564

Lstar/3.84600e26 = 10^((4.75 - 0.167019564)/2.5)
Lstar/3.84600e26 = 68.1070664
Lstar = 3.84600e26 X 68.1070664
Lstar = 2.61939777e28 W

L = 3.84600e26 X 63
= 2.42298e28 W

L = 2.61939777e28 - 2.42298e28
= 1.9641777e27 W

Mu Horologii is 141.6 lightyears (43.414757 parsecs) from the Sun, has an apparent magnitude of 5.11, is a spectral type of F0 III/IV and has a luminosity of 13 L.

M = 5.11 - 5((log(43.414757) - 1)
= 1.92181313

Given there's no F0 III listed in the chart above, we'll use the BC of F0 III, that being 0.

Mbol = 1.92181313 + 0
= 1.92181313

Lstar/3.84600e26 = 10^((4.75 - 1.92181313)/2.5)
Lstar/3.84600e26 = 13.529282
Lstar = 3.84600e26 X 13.529282
Lstar = 5.20336186e27 W

L = 3.84600e26 X 13
= 4.9998e27 W

L = 5.20336186e27 - 4.9998e27
= 2.0356186e26 W

Zeta Horologii is 160 lightyears (49.0562 parsecs) away from the Sun, has an apparent magnitude of 5.20, is a spectral class of F6 Vluminosity of 16.7 L.

M = 5.20 - 5((log(49.0562) - 1)
= 1.74653048

The above chart doesn't have the BC for F6 V, but given F5 V is -0.02 and F8 V is -0.02, it bears to reason that F6 V shouldn't be that far off, so we'll go with -0.02.

Mbol = 1.74653048 + -0.02
= 1.72653048

Lstar/3.84600e26 = 10^((4.75 - 1.72653048)/2.5)
Lstar/3.84600e26 = 16.1952557
Lstar = 3.84600e26 X 16.1952557
Lstar = 6.22869534e27 W

L = 3.84600e26 X 16.7
= 6.42282e27 W

L = 6.42282e27 - 6.22869534e27
= 1.9412466e26 W

Eta Horologii is 149 lightyears (45.6836 parsecs) away from the Earth, has an apparent magnitude of 5.31, is a spectral class of A6 V + F0 V and has a luminosity of 12.6 L.

M = 5.31 - 5((log(45.6836) - 1)
= 2.0111984

The BC of an A6 V star isn't listed, but given both A5 V and A7 V are 0.02, it stands that A6 V is likewise.

Mbol = 2.0111984 + 0.02
= 2.0311984

Lstar/3.84600e26 = 10^((4.75 - 2.0311984)/2.5)
Lstar/3.84600e26 = 12.2326525
Lstar = 3.84600e26 X 12.2326525
Lstar = 4.70467815e27 W

L = 3.84600e26 X 12.6
= 4.84596e27 W

L = 4.84596e27 - 4.70467815e27
= 1.4128185e26 W

Iota Horologii is 56 lightyears (17.169699999528 parsecs) away from the Solar System, has an apparent magnitude of 5.40, is a spectral type of G0Vp and has a luminosity of 1.64 L.

M = 5.40 - 5((log(17.169699999528) - 1)
= 4.22618647

A spectral type star of G0V would have a BC or -0.05, which is more or less G0Vp.

Mbol = 4.22618647 + -0.05
= 4.17618647

Lstar/3.84600e26 = 10^((4.75 - 4.17618647)/2.5)
Lstar/3.84600e26 = 1.69638886
Lstar = 3.84600e26 X 1.69638886
Lstar = 6.52431156e26 W

L = 3.84600e26 X 1.64
= 6.30744e26 W

L = 6.52431156e26 - 6.30744e26
= 2.1687156e25 W

All that leaves is R Horologii, with a distance of 1000 lightyears (306.601 parsecs) away, has an apparent magnitude of 7.22 and is a spectral class of M5-7e. Unfortunately, it's luminosity isn't listed, and even a search doesn't reveal a direct number, so we'll have to leave it out, for now at least. However, it's finally time to add all of them together!

E = 3.7582895e27 + 1.343747e27 + 3.086172e27 + 1.9577444e30 + 1.17528903e28 + 2.3000003e30 + 9.0682526e29 + 2.1819705e27 + 1.9641777e27 + 2.0356186e26 + 1.9412466e26 + 1.4128185e26 + 2.1687156e25
= 5189217862526000000000000000000 W (5189217862526000000000000000000 joules)
= 1.2402528352117589971 zettatons

Given watts is in joules per second, that is the amount of energy the stars output per second during the timeloop.

Final results
Timeloop causes stars to increase in luminousity (per second) = 1.240 zettatons

The Time Demon broke this time loop, and the party defeated it (although the first time around it resurrected and the second time they needed the Talisman of Time) so at the very least this scales to everyone with Horologium Chamber gear (one of which, the Calamity Wand, is very clearly noted to be able to destroy the world), possibly in base.

 

[Stocking Anarchy]

During the broom race, Amanda uses the Shooting Star to try and win, but Akko boards it and it flies around the world, including emerging from the water with enough force to create a colossal pillar of water that reaches far above cloud level. We also see that the broom race around Luna Nova continues that same day, with no waiting around for Akko to return. First let's get a high end. The Earth is 40,075km in circumference, so what's the timeframe?

Spoiler

29 seconds.

T = 40,075km/29s
= 1381896.55/340.29
= Mach 4060.93788

We can also find the kinetic energy for Akko's durability (Akko ends this covered in scratches, so it would be more than reasonable, as well as the colossal explosion of water). Akko weighs 46kg.

Spoiler

KE = (0.5)mv^2
= (0.5) X 46 X 1381896.55^2
= 4.39216757e13 joules

That's impressive enough by itself, but wait! At one point, Akko flies through the ocean and emerges a few seconds later in a place where the sky is a different colour.


Thing is, this is travelling through water, which is far more viscous that air. The viscosity of water is 8.90e-4 Pa*S (or kg per m^2), while the viscosity of air is while the viscosity of air is 1.81e-5 kg/m^2.

8.90e-4/1.81e-5 = 49.1712707

Ergo, water is 49.1712707 times more viscous than air. Given this is also and it would take 49.1712707 times more energy to move through it.

E = 4.39216757e13 X 49.1712707
= 2.15968461e15 joules
= 516.177010038240951 kilotons

Let's also go with a low end, of the time it took Akko to fly from England to Finland. We can tell it's Finland because it has the same Yeti. Luna Nova is located in England on the St. Michael Alignment, which is based off the Leylines from British folklore.

Spoiler

Now to lay some lines of our own!

Spoiler

139 pixels = 500 miles (804.67200km)
1 pixel = 804.67200km/139 = 5.78900719km
5.78900719km X 321 = 1858.27131km

How long did it take?

Spoiler

Timeframe is 8 seconds.

T = 1858.27131km/8s
= 232283.914/340.29
= Mach 682.60576

KE = (0.5)mv^2
= (0.5) X 46 X 232283.914^2
= 1.24098378e12 joules

E = 1.24098378e12 X 49.1712707^2
= 6.10207494e13 joules
= 14.58430913001912 kilotons

Now let's also find the one for the underwater part. We see her enter at mid-day, then bursts out at night, so that's around 6 time zones. The circumference of the Earth is 40,075km.

L = 40,075km/24
= 1669.79167km X 6
= 10018.75km

Next for the timeframe (we also see the Sun is at it's zenit).

Spoiler

6 seconds.

T = 10018.75km/6s
= 1669791.67/340.29

That's the highest result yet! But what about the kinetic energy?

KE = (0.5)mv^2
= (0.5) X 46 X 1669791.67^2
= 6.41286971e13 joules

E = 6.41286971e13 X 49.1712707
= 3.15328952e15 joules
= 753.6542829827916421 kilotons

Final Results
Shooting Star speed (low end) = Mach 682.606
Shooting Star speed (mid end) = Mach 4060.938
Shooting Star speed (high end) = Mach 4906.967
Akko's kinetic energy on the Shooting Star (low end) = 14.584 kilotons
Akko's kinetic energy on the Shooting Star (mid end) = 516.177 kilotons
Akko's kinetic energy on the Shooting Star (high end) = 753.654 kilotons

 

[Stocking Anarchy]

From episode 6.

Spoiler

The Fountain of Polaris bestows the power of the stars to a witch with hidden radiance. When Akko goes there for the first time, she sees how Chariot went there in the past like she did.

16:57, they enter the Fountain of Polaris, and at 19:26, we have our feat.

Spoiler

The stars embue Chariot with their power. Those stars increase in brightness.

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You can see this with any of the stars, but highlighting one helps demonstrate this. The stars prior to this (and after) aren't shining as much).

Spoiler

Looking at Ampchu's Touhou luminousity calc and Endless Mikes luminousity revision calcs for help. We'll get our luminousity from the average brightness of a star increase to the brightest star. Given that this is the Fountain of Polaris, I'll be going with the values for Polaris, which has an apparent magnitude of 1.99 and is 430 lightyears (4.068e+18m). Now let's get the luminousity...

1.99 = -26.73 - 2.5log((L/3.846*10^26)(146000000000/(4.068e+18))^2)

Going through this step by step...

(146000000000/(4.068e+18))^2) = 1.28808292e-15
1.99 - -26.73 = 1.28808292e-15/((L/3.846*10^26)
28.72 = -2.5Log(1.28808292e-15/((L/3.846*10^26))
28.72/-2.5 = (1.28808292e-15/((L/3.846*10^26))
10^(-11.48800) = (1.28808292e-15/((L/3.846*10^26))
3.25087297e-12 = (1.28808292e-15/((L/3.846*10^26))
3.25087297e-12 X (3.864*10e26) = 1.28808292e-15
1.25613732e16/1.28808292e-15 = 9.75199112e30 joules

Polaris is 2500 times brighter than the Sun.

L = 3.84600e26 X 2500
= 9.61500e29 W

L = 9.75199112e30 - 9.61500e29
= 8.79049112e30

But that's not it! There are approximately 4548 visible in a hemsiphere of the Earth.

E = 8.79049112e30 X 4548
= 3.99791536e34 joules
= 9.5552470363288719568 yottatons

Let's also get the speed of which the magic of the stars crosses to Shiny Chariot.

Spoiler

5 seconds.

T = 4.068e+18m/5s
= 8.13600e17/299792458
= 2713877480 C

For a low end, we'll go with alpha centuri (4.367 light years away, or 41315009973760352m.

T = 41315009973760352m/5s
= 8.26300199e15/299792458
= 27562407.8 C

(146000000000/(8.26300199e15))^2) = 1.28808292e-15
1.99 - -26.73 = 3.12197914e-10/((L/3.846*10^26)
28.72 = -2.5Log(3.12197914e-10/((L/3.846*10^26))
28.72/-2.5 = (3.12197914e-10/((L/3.846*10^26))
10^(-11.48800) = (3.12197914e-10/((L/3.846*10^26))
3.25087297e-12 = (3.12197914e-10/((L/3.846*10^26))
3.25087297e-12 X (3.864*10e26) = 1.28808292e-15
1.25613732e16/3.12197914e-10 = 4.02352887e25 joules

Alpha Centari has a luminousity of 0.0017 L (L being the luminousity of the Sun).

L = 3.84600e26 X 0.0017
= 6.5382e23 W

L = 4.02352887e25 - 6.5382e23
= 3.95814687e25 W

E = 3.95814687e25 X 4548
= 1.8001652e29 joules
= 43.024980879541118384 exatons

Final Result
The stars embue Shiny Chariot with power (low end) = 43.025 exatons
The stars embue Shiny Chariot with power (high end) = 9.555 yottatons
Magic of the Stars (low end) = 27562407.8 C
Magic of the Stars (high end) = 2713877480 C

 

[Stocking Anarchy]

Vajarois' breathe sends Akko, Sucy and Lotte fly with such force that it parts the clouds above. Note also the clouds before and after she breathes at them. Before...

Spoiler

...and after.

Spoiler

The clouds were parted considerably. The clouds seem to resemble stratocumulus clouds, which float around 2000m off the ground and are between 200m to 500m thick (or an average of 350m).

A compi- lation of observed stratocumulus thicknesses from field studies (Fig. 5b) is remarkably consistent with the sat- ellite data, with the majority of cases showing thick- nesses from 200–500 m, a median value of 320 m, and a tendency for thicker clouds (median 420 m) in mid- and high latitudes.

Akko is 162cm tall (1.62m).

Spoiler

93 pixels = 1.62m
1 pixel = 1.62m/92 = 0.0176086957m
0.0176086957m X 337 = 5.93413045m

Spoiler

95 pixels = 5.93413045m
1 pixel = 5.93413045m/95 = 0.0624645311m
0.0624645311m X 490 = 30.6076202m

That's how tall Vajarois is, and Akko, Lotte and Sucy are slightly higher than that. Distance to the horizon at 30.6076202m is 19.8km (19800m).

H = 2000m - 30.6076202m
= 1969.39238m

While we're at it, we can also get the speed of her roar.

Spoiler

3 seconds, 3 frames.

1s/25 = 40ms
40ms X 3 = 0.12s + 3s = 3.12m

T = 1969.39238m/3.12s
= 631.215506/340.29
= Mach 1.85493404

Spoiler

object degree size = 2*atan(Object_Size/(Panel_Height/tan(70/2)))

= 2*atan(1094/(721/tan(70/2)))
= 1.24664474 rad
= 71.42748215436103 degrees

That's 1369.7m across with the angscaler, or a radius of 684.85m.

Volume as a cylinder.
V = πr^2h
= π X 19800^2 X 350
= 4.31070494e11m^3

Density of a cloud is 1.003kg/m^3.

M = 4.31070494e11 X 1.003
= 432363705482kg

Let's get out timeframe.

Spoiler

3 frames.

1s/25 = 40ms
40ms X 3 = 0.12s

T = 684.85m/0.12s
= 5707.08333 m / s

E = (0.5) X 432363705482 X 5707.08333^2
= 7.04121592e18 joules
= 1.68289099426 gigatons

Now for a low end of just the part of the cloud seen moved onscreen.

V = πr^2h
= π X 684.85^2 X 350
= 515713900m^3

M = 515713900 X 1.003
= 517261042kg

E = (0.5) X 517261042 X 5707.08333^2
= 8.42380301e15 joules
= 2.01333723948 megatons

Final Results
Vajarois' height = 30.608m
Vajarois' exhales (energy - low end) = 2.013 megatons
Vajarois' exhales (energy - high end) = 1.683 gigatons
Vajarois' exhales (speed) = Mach 1.855

 

[Stocking Anarchy]

(Will replace the gif at a later date)
Jasminka effortlessly lifts and throws a big bag of green gunk bigger than herself. Jasminka is 165cm (1.65m) tall.

Spoiler

587 pixels = 1.65m
1 pixel = 1.65m/587 = 0.0028109029m
0.0028109029m X 712 = 2.00136286m
2.00136286m/2 = 1.00068143m

Spoiler

0.0028109029m X 1124 = 3.15945486m
3.15945486m/2 = 1.57972743m

V = 4/3πabc
= 4/3 X π X 1.00068143 X 1.00068143 X 1.57972743
= 6.6261681m^3

I'm going to go with the mass of water for this, which weighs 1000kg/m^3.

M = 6.6261681 X 1000
= 6626.1681kg
= 6.6261681 tons

Next for the timeframe.

Spoiler

1 frame.

1s/25 = 40ms

How high does she lift it?

Spoiler

0.0028109029m X 677 = 1.90298126m

T = 1.90298126m/40ms
= 47.5745315m/s

KE = (0.5) X 6626.1681 X 47.5745315^2
= 7498622.56 joules

Let's also get the GPE.

E = 6626.1681 X 9.807 X 1.90298126
= 123661.109 joules

E = 7498622.56 + 123661.109
= 7622283.67 joules
= 0.001821769519598 tons of TNT

Final Results
Jasminka throws garbage = 1.82kg of TNT
Garbage bag mass = 6.626 tons

 

[Stocking Anarchy]

The armour attaches itself to Louis at rapid speeds, and Amanda can react to this cursed armour. Amanda is 167cm tall.

Spoiler

682 pixels = 167cm (1.67m)
1 pixel = 1.67m/682 = 0.00244868035m
0.00244868035m X 124 = 0.303636363m

Spoiler

188 pixel = 0.303636363m
1 pixels = 0.303636363m/188 = 0.00161508704m
0.00161508704m X 157 = 0.253568665m

Spoiler

0.07 seconds.

7 pixels = 0.253568665m
1 pixel = 0.253568665m/7 = 0.036224095m
0.036224095m X 172 = 6.23054434m

T = 6.23054434m/0.07s
= 89.0077763m/s

Final Results
Possessed Armour attaches to Louis = 89.008m/s

 

[Stocking Anarchy]

Spoiler

Spoiler

The Volcanic Ray spell melts the ground and causes lava to burst up. Continental crust ranges from 30km to 50km, which averages out at 40km (40,000m). First off, let's find the speed.

Spoiler

2.29 seconds. We'll split that in half for the time for the heat ray to reach below the crust and for the lava to burst up.

T = 2.29s/2
= 40,000m/1.14500s
= 34934.4978/340.29
= Mach 102.660959

Amanda is 167cm tall, or 1.67m.

Spoiler

139 pixels = 1.67m
1 pixel = 1.67m/139 = 0.0120143885m
0.0120143885m X 6 = 0.072086331m

Spoiler

0.0120143885m X 505 = 6.06726619m
6.06726619m/2 = 3.03363309m
3.03363309m - 0.072086331m = 2.96154676m

Volume as a cylinder.

V = πr^2h
= π X 3.03363309^2 X 40,000
= 1156474.26m^3

V = πr^2h
= π X 2.96154676^2 X 40,000
= 1102166.11m^3

V = 1156474.26 - 1102166.11
= 54308.15m^3

There are various densities of magma, those being basalt, andesite and rhyolite. We'll go with the average rhyolite, as that weighs the least. Rhyolite magma weighs between 2180 and 2250kg per m^3, which averages out at 2215kg per m^3.

M = 54308.15 X 2215
= 120292552kg

At last for our energy.

KE = (0.5) X 120292552 X 34934.4978^2
= 7.34036662e16 joules

Let's also get the energy for the heat ray to melt the rock. Melting rock takes 4350000000 joules per m^3.

E = 54308.15 X 4350000000
= 2.36240452e14

E = 7.34036662e16 + 2.36240452e14
= 7.36399067e16 joules
= 17.6003601099 megatons

Final Results
Volcanic Ray (speed) = Mach 102.661
Volcanic Ray (energy) = 17.600 megatons

 

[Stocking Anarchy]

Chariots familiar Alcor is summoned from beyond the galaxy and spacetime.

Spoiler

http://little-witch-academia.wikia.com/wiki/Chariot_cards

その星の獣は、
強く輝く者に付き従う。
シャイニィロッドが認めた者呼ばれれば,
銀河も時空も超えて現れる。

The star 's beast will follow the strongly shining one.
If it is called it will appear beyond the galaxy and the spacetime.

Alcor, Chariots familiar, actually does fly in from beyond the galaxy and space-time itself. We even see this at the very start of episode 1.

Spoiler

It takes 10 seconds for Alcor to reach Earth. The Solar System resides 8 kiloparsecs (2.46854207e20m) from the centre of the Milky Way. The Milky Way has a diameter of 100,000 lightyears, and thus 50,000 lightyears (4.7302642e20m) in radius.

D = 4.7302642e20m - 2.46854207e20m
= 2.26172213e20m/10s
= 2.26172213e19/299792458
= 75442929600 C

Final Results
Alcor flies in from beyond the galaxy = 75442929600 C

 

[Stocking Anarchy]

It's job done and magic returned to the world, the Shiny Rod disintegrates and returns to the skies and becomes the Big Dipper once again. The furtherest star in the Big Dipper, Dubhe, is 124 lightyears from the Earth (or 1.173e+15km).

Spoiler

Timeframe is 9 seconds.

T = 1.173e+15km/9s
= 1.30333333e17/299792458
= 434745203 C

Final Results
Shiny Rod becomes the Big Dipper = 434745203 C

 

[Stocking Anarchy]

Always listen to technobabble, you never know when it could pay off big. On a rewatch, I noticed a few interesting pieces of technobabble that I, and everyone else, had overlooked. The first was absorbing the mass energy of the Spiral Death Fields compressed space-time, and the other is this...

The Ashtanga vanishes through the portal, and LordGemone gives this explanation of where they went.


By manipulating multidimensional space, the Ashtanga can manipulate probability and move up and down the time axis. The Schrödinger Warp backs this up, as their actions are unknown until they take them, allowing them to exist in multiple different places at the same time.

Spoiler

Super Galaxy Gurren Lagann responds by sending a barage into all nearby points of space-time.

Spoiler

Note also the number of times the single Ashtanga appears on the charts.

Spoiler

Tetukan comments on a planck time coefficient, and Leeron says not to worry about it, and that all of their shots will be hits thanks to the probability manipulation There's another translation where he says "we can fire randomly and still hit them!").

Spoiler

Simon then finishes off with this;

Spoiler

They fire the beams into the past, present and future and blast the dreadnaught out of it's hiding place in the probability space.


Leeron then points out the ship has built up immense energy, like a galaxy is about to be born.

Spoiler

So to sum up what happened; Super Galaxy Gurren Lagann fired attacks into every planck instant and hit with every one of them. If you need further confirmation of this, the Glossary in Final Drill confirms that the Mega Maelstrom Vortex Cannon fired missiles into every point in space and time in every dimension.

Spoiler

(The relevant part)

Spoiler

[Possible space-time simultaneous firing][Deathlete/Episode 26] A Super Galaxy Gurren Lagann's "Mega-Vortex Cannon" or "Maelstrom Cannon" is an all-gun attack that fires as a random variable bullets. Artor in enormous number of enemies in every time-space and dimensions from the near past to the near future.

The kanji being used for every is "あらゆる" which means that indeed yes, the missiles are being fired into every possible point in time alone the time axis. Needless to say, this is going to be incredible. Even if we go with one joule for each attach, it will be more than comparable to a supernova. We can't get the explosions themselves as each was in space, and the shots were all energy beams and FTL anyway (from Tacocats calcs), but thankfully, we can get the energy of the attacks by looking at the damaged Ashtanga. It's covered in smoldering pockmarks that weren't there before.
We'll need the width of the Ashtanga. I previously calc'd their dimensions here, but to repost said information below briefly...

Spoiler

The diameter of the Moon is 3474.2km.

Spoiler

932 pixels = 3474.2km
1 pixel = 3474.2km/932 = 3.7276824km

Spoiler

3.7276824km X 432 = 1610.3588km

Spoiler

364 pixels = 1610.3588km
1 pixel = 1610.3588km/364 = 4.42406264km
4.42406264km X 878 = 3884.327km
744 + 704 = 1448 pixels
4.42406264km X 1448 = 6406.0427km

Now onto the Ashtanga.

Spoiler

105 pixels = 1610.3588km
1 pixel = 1610.3588km/105 = 15.3367505km
15.3367505km X 1072 = 16440.9965km

Spoiler

300 pixels = 16440.9965km
1 pixel = 16440.9965km/300 = 54.8033217km
54.8033217km X 1388 = 76067.0105km
54.8033217km X 584 = 32005.1399km

Now let's scale the height of the Ashtanga to one of the craters.

Spoiler

1712 pixels = 32005.1399km
1 pixel = 32005.1399km/1712 = 18.6945911km
18.6945911km X 116 = 2168.57257km (2168572.57m)
2168572.57m/2 = 1084286.28m

Volume as a demi-sphere.
V = 4/3πr^3
= 4/3 X π X 1084286.28^3
= 5.33974474e18/2
= 2.66987237e18m^3

Given there's no chunks of rock floating around the craters and from the way the missiles exploded, I think it's fair enough to say the craters were violently fragmented. The violent fragmentation of rock per m^3 is 8000000.

E = 2.66987237e18 X 8000000
= 2.1358979e25 joules

We're aren't done yet though! As stated above, the missiles aim for near past -10, near future +8. As the Ashtanga Dreadnaught hasn't instantly blinked back into existence, they would be refering to a few seconds here (the innumerable number of times it pops up on the radar backs this up). There are 10^43 planck instances in a single second. Now let's work some magic!

E = 8 + 10
= 18 X 10^43
= 1.8e44 X 2.1358979e25
= 3.84461622e69 joules

That's pretty impressive, but we're not only talking every point in time, but every point in space too. We can count a minimum amount of points in space through the portals opened by Super Galaxy Gurren Lagann. This is simultaneously the easiest and hardest part...

Spoiler

There are at least 181 portals.

E = 3.84461622e69 X 181
= 6.95875536e71 joules

And that's not the end of it. Not only did it target every point in space and time on a planck scale, but also every dimension, of which there are at least 11.

Spoiler

E = 6.95875536e71 X 11
= 7.6546309e72/10^44
= 76.546309 NINAFOE

As another value, the manga says that prior to the Ashtanga appearing in realspace to battle Super Galaxy Dai-Gurren (before the Galactic Spiral Abyss), the Spiral Radar picks up 10^68 enemies.

Spoiler

Dayakka: Cybela! What happened?
Cybella: The enemies... Their number...
Dayakka: How many!?
Cybela: 10 to the power of...68.

So let's also use the value of 10^68 for the total number of times attacked.

E = 2.1358979e25 X 10^68
= 2.1358979e93/10^44
= 21.358979 TENAEXAFOE

Final Results
Maelstrom Mega Vortex Cannon (low end) = 76.546 NINAFOE
Maelstrom Mega Vortex Cannon (high end) = 21.359 TENAEXAFOE

 

[Stocking Anarchy]

Something that's been around for a while now, but I feel gets somewhat overlooked. Super Galaxy Gurren Lagann is stated to generate energy comparable to a microcosmos.

Spoiler

Later, Leeron comments that it's as if a galaxy is about to be born inside Super Galaxy Gurren Lagann.

Spoiler

Super Galaxy Gurren Lagann has power comparable to a galaxy being born. If the above isn't enough evidence, Final Drill outright confirms it;

Spoiler

(The relevant part)

Spoiler

[Super Galaxy Gurren Lagann] [Mecha/Episode 26] Super Spiral class gunman. The appearence of the transformation of Super Galaxy Dai-Gurren with enough energy to create a galaxy, as the spiral force was transmitted from Ark Gurren Lagann through the transcendental spiral engine from Gurren Lagann. Equipped with a Mega Vortex Cannon, it is protected by a spiral field that is intact even if hit by a planet.

Super Galaxy Gurren Lagann can generate power enough to create a galaxy. Given that a lot of transformations are done through mass-energy (giga drills, regeneration, the Death Spiral Abyss), this should be likewise too. We'll use a low end of the Lesser Magellanic Cloud, a mid end of the Milky Way and a high end of IC 1101. The Lesser Magellanic Cloud has a mass of approximately 3,000,000,000 solar masses, the Milky Way has a mass of roughly 1,500,000,000,000 solar masses and IC 1101 has a mass of more than 100,000,000,000,000 solar masses. One solar mass is 1.989 X 10^30kg (or 1.98900e30kg).

(Low end)
M = 3,000,000,000 X 1.98900e30kg
= 5.96700e39kg

E = mc^2
= 5.96700e39 X 299792458^2
= 5.36287215e56/10^44
= 5.36287215 TERAFOE

(Mid end)
M = 1,500,000,000,000 X 1.98900e30
= 2.9835e42kg

E = mc^2
= 2.9835e42 X 299792458^2
= 2.68143608e59 /10^44
= 2.68143608 PETAFOE

(High end)

M = 100,000,000,000,000 X 1.98900e30kg
= 1.98900e44kg

E = mc^2
= 1.98900e44 X 299792458^2
= 1.78762405e61/10^44
= 1.78762405 EXOFOE

Final Results
Super Galaxy Gurren Lagann's engine power (low end) = 5.363 TERAFOE
Super Galaxy Gurren Lagann's engine power (mid end) = 2.681 PETAFOE
Super Galaxy Gurren Lagann's engine power (high end) = 1.788 EXOFOE

 

[Stocking Anarchy]

Spoiler

(Releveant text)

Spoiler

"Super Spiral Field developed. prepare for the shockwave, a huge one will come."
At the same time Dayakka finished talking, a big explosion occurred.
It's a battleship collapse like a satellite.
After the big shockwave that shakes the isolated universe hit the Super Galaxy Dai-Gurren and the violent wave passed away, the enemy fleet was also wiped out.

(I asked a friend who can read Japanese to have a look at this, and they confirmed this is more or less accurate)

An Ashtanga exploding shakes the entire Anti-Spiral pocket universe, which Super Galaxy Dai-Gurren withstands with it's Spiral Shields. Looking back at my old Captain Harlock Noo artefact calc for help here, calculating the energy to shake the planet then applying inverse-square law. The minimum magnitude for an earthquake to be felt for the average person is magnitude 3 (though some extra sensetive people can feel magnitude 2.5 earthquakes).

Spoiler

As such, we'll once again mess around with the impact calculator until we have a magnitude 2.5 impact for a distance of 20,000km (the highest the calculator will take, which is a bit under the. Sure enough, said impact would result 3.72 X 10^12 joules (3720000000000 joules). Now we need to divide that energy by 10,000.

E = 3720000000000/10,000
= 372000000 joules

Now for the surface area of the area affected, that being the observable Universe (46,600,000,000 lightyears, or 4.40870040022e+26m).

A = 4πr^2
= 4 X π X 46,600,000,000^2
= 2.72886278e22LY^2
= 2.44248012179300965e+54m^3

Finally for our energy, which we'll find using inverse-square law.

I = S/4pa
372000000 = S/(4 X π X 4.40870040022e+26^2)
= 372000000 X (4 X π X 4.40870040022e+26^2)
= 9.08602604e62/10^44
= 9.08602604 EXAFOE

Final Results
Ashtanga explodes = 9.086 EXAFOE

 

[Stocking Anarchy]

As there are a few calcs I mean to make and look over involving them, and the images in the old calcs of their sizes where lost in the great PostImage crash, it's time to recalc the sizes of Super Galaxy Gurren Lagann and the Ashtanga. The second movie has the perfect shot from which to scale directly.

Spoiler

The diameter of the Moon is 3474.2km.

Spoiler

932 pixels = 3474.2km
1 pixel = 3474.2km/932 = 3.7276824km

Spoiler

3.7276824km X 432 = 1610.3588km

Spoiler

364 pixels = 1610.3588km
1 pixel = 1610.3588km/364 = 4.42406264km
4.42406264km X 878 = 3884.327km
744 + 704 = 1448 pixels
4.42406264km X 1448 = 6406.0427km

Now onto the Ashtanga.

Spoiler

105 pixels = 1610.3588km
1 pixel = 1610.3588km/105 = 15.3367505km
15.3367505km X 1072 = 16440.9965km

Spoiler

300 pixels = 16440.9965km
1 pixel = 16440.9965km/300 = 54.8033217km
54.8033217km X 1388 = 76067.0105km
54.8033217km X 584 = 32005.1399km

Final Results
Super Galaxy Gurren Lagann main body height = 3884.327km
Super Galaxy Gurren Lagann total height = 6406.043km
Ashtanga total length = 76067.011km

 

[Stocking Anarchy]

There's a couple of space calcs I've been meaning to recalc, so now's as good a time as any to go over them once more.

1. Arc Gurren Lagann's Giga Drill Break

Spoiler

Arc Gurren Lagann attacks an Ashtanga with a Giga Drill, to absolutely no effect at all. Arc Gurren Lagann is officially stated to be between 5km to 10km tall.

Spoiler

612 + 936 = 1548 pixels
1548 pixels = 5km
1 pixel = 5km/1548 = 0.00322997416km
0.00322997416km X 280 = 0.904392765km (904.392765m)
1548 = 10km
1 pixel = 10km/1548 = 0.00645994832km
0.00645994832km X 280 = 1.80878553km (1808.78553m)

While we're on it, we'll find the volume and mass of the giga drill.

Spoiler

44 pixels = 904.392765m
1 pixel = 904.392765m/44 = 20.554381m
20.554381m X 376 = 7728.44726m
7728.44726m/2 = 3864.22363m
20.554381m X 1190 = 24459.7134m
44 pixels = 1808.78553m
1 pixel = 1808.78553m/44 = 41.108762m
41.108762m X 376 = 15456.8945m
15456.8945m/2 = 7728.44725m
41.108762m X 1190 = 48919.4268m

Volume as a cone.

(Low end)
V = πr^2h/3
= π X 3864.22363^2 X 24459.7134/3
= 3.82476262e11m^3

(High end)
V = πr^2h/3
= π X 7728.44725^2 X 48919.4268/3
= 3.05981008e12m^3

For our mass, we'll go with a low end of titanium and a high end of steel, which respectively weigh 4500kg/m^3 and 7850kg/m^3.

(Low low end)
M = 3.82476262e11 X 4500kg
= 1.72114318e15kg

(Low high end)
M = 3.82476262e11 X 7850kg
= 3.00243866e15kg

(High low end)
M = 3.05981008e12 X 4500kg
= 1.37691454e16kg

(High high end)
M = 3.05981008e12 X 7850kg
= 2.40195091e16kg

Next, to scale the distance from the Ashtanga. A single Ashtanga is 76067.0105km long.

Spoiler

1388 pixels = 76067.0105km
1 pixel = 76067.0105km/1388 = 54.8033217km
54.8033217km X 85 = 4658.28234km

Spoiler

object degree size = 2*atan(Object_Size/(Panel_Height/tan(70/2)))

= 2*atan(509/(1617/tan(70/2)))
= 0.296112429 rad
= 16.9659924431093323 degrees

Enter those values into the angscaler, and we can determine that the Ashtanga is 15616km away. Let's get our timeframe.

Spoiler

Timeframe is 10 seconds and 13 frames.

T = 1s/24
= 41.6666667ms X 13
= 0.541666667s + 10s
= 15616km/10.5416667s
= 1481359.68/340.29
= Mach 4353.22719

At last, we come to our kinetic energy

(Low low end)
KE = (0.5)mv^2
= (0.5) X 1.72114318e15 X 1481359.68^2
= 1.8884611e27 joules
= 451.35303537284897857 petatons

(Low high end)
KE = (0.5)mv^2
= (0.5) X 3.00243866e15 X 1481359.68^2
= 3.29431548e27 joules
= 787.36029636711282365 petatons

(High low end)
KE = (0.5)mv^2
= (0.5) X 1.37691454e16 X 1481359.68^2
= 1.51076888e28 joules
= 3.6108242829827918285 exotons

(High high end)
KE = (0.5)mv^2
= (0.5) X 2.40195091e16 X 1481359.68^2
= 2.63545237e28 joules
= 6.2988823374760995648 exotons

Now the hard parts out of the way, time for the easy bit; getting the mass energy to form said giga drill.

(Low low end)
E = mc^2
= 1.72114318e15 X 299792458^2
= 1.54688635e32 joules
= 36.971471080305931878 zettatons

(Low high end)
E = mc^2
= 3.00243866e15 X 299792458^2
= 2.69845729e32 joules
= 64.494677103250481309 zettatons

(High low end)
E = mc^2
= 1.37691454e16 X 299792458^2
= 1.23750907e33 joules
= 295.7717662523900799 zettatons

(High high end)
E = mc^2
= 2.40195091e16 X 299792458^2
= 2.15876582e33 joules
= 515.9574139579350458 zettatons

2. King Kittan's final flight

Spoiler

Space King Kittan flies fast. Might not look fast, but remember Super Galaxy Dai-Gurren is the size of the Moon. The diameter of the Moon is 3474.2km.

Spoiler

260 pixels = 3474.2km
1 pixel = 3474.2km/260 = 13.3623077km
13.3623077km X 153 = 2044.43308km
13.3623077km X 484 = 6467.35693km
13.3623077km X 51 = 681.477693km

Spoiler

973 pixels = 681.477693km
1 pixel = 681.477693km/973 = 0.700388174km
0.700388174km X 1216 = 851.67202km

All we need next is the timeframe.

Spoiler

Timeframe is 19 frames.

T = 1s/24
= 41.6666667ms X 19
= 0.791666667s
= 851.67202km/0.791666667s
= 1075796.24/340.29
= Mach 3161.41009

3. Super Galaxy Dai-Gurren drags the Moon
Super Galaxy Dai-Gurren drags the Moon back into realspace. Straightforward enough. The Moon weighs 7.353 X 10^22kg (or 7.35300e22kg) and as said above, has a diameter of 3474.2km. From when the Moon is first dragged out to when the shining around the Moon stops.

Spoiler

Timeframe is 10 seconds and 5 frames.

T = 1s/24
= 41.6666667ms X 5
= 0.208333333s + 10s
= 10.2083333s
= 3474.2km/10.2083333s
= 340329.797m/s

KE = (0.5)mv^2
= (0.5) X 7.35300e22 X 340329.797^2
= 4.25828299e33 joules
= 1.0177540607074568868 yottatons

Final Results
Arc Gurren Lagann rushes the Ashtanga = Mach 4353.227
Arc Gurren Lagann's Giga Drill kinetic energy (low low end) = 451.353 petatons
Arc Gurren Lagann's Giga Drill kinetic energy (low high end) = 787.360 petatons
Arc Gurren Lagann's Giga Drill kinetic energy (high low end) = 3.611 exotons
Arc Gurren Lagann's Giga Drill kinetic energy (high high end) = 6.299 exotons
Arc Gurren Lagann's Giga Drill mass energy (low low end) = 36.971 zettatons
Arc Gurren Lagann's Giga Drill mass energy (low high end) = 64.495 zettatons
Arc Gurren Lagann's Giga Drill mass energy (high low end) = 295.772 zettatons
Arc Gurren Lagann's Giga Drill mass energy (high high end) = 515.957 zettatons
Super Galaxy Dai-Gurren length = 6467.357km
Space King Kittan's final flight = Mach 3161.410
Super Galaxy Dai-Gurren tows the Moon = 1.018 yottatons

 

[Stocking Anarchy]

(Episode 24)


Yoko and Darry shoot some giant missiles, and knock them off their paths before they explode. I calced the dimensions of Super Galaxy Dai-Gurren previously in my Spiral Abyss calc, though unfortunately the images are gone after the PostImage crash of '18. Thankfully though it won't be hard to rescale. The Moon is 3472.8km in radius.

Spoiler

207 pixels = 3472.8km
1 pixel = 3472.8km/207 = 16.7768116km
16.7768116km X 56 = 939.50145km

Spoiler

323 pixels = 939.50145km
1 pixel = 939.50145km/323 = 2.90867322km
2.90867322km X 180 = 523.56118km
2.90867322km X 483 = 1404.88917km (1404889.17m)

Looking up different warheads, the one I found that was closest in shape to the Anti-Spiral missile was the V-2, which weighed 12500kg and was 14m long.

M2 = (H2/H1)^3*M1
= (1404889.17/14)^3 X 12500
= 1.26314178e19kg

Now that we have the mass of each missile we can find the speed at which they were knocked. Let's also get the speed for Yoko's bullets. Arc Gurren Lagann is described as being "several tens of kilometers" tall...

Spoiler

(The relevant part)

Spoiler

(Crappy translation)

The total length is about tens of kilometers tall.

...and the Space Grapearls aren't that far behind, it would be a fair assumption that Darry's Space Grapearl is at least 30km tall.

Spoiler

object degree size = 2*atan(Object_Size/(Panel_Height/tan(70/2)))

= 2*atan(47/(596/tan(70/2)))
= 0.0746944197 rad
= 4.279675001999448725 degrees

Enter those values through the angscaler and Yoko & Darry are 401.45km away.

Spoiler

The timeframe between Yoko firing and hitting the missile is 8 frames, and the timeframe for the missile getting knocked to the side is 1 frame. There are 24 frames in a second.

T = 1s/24
= 41.6666667ms X 8
= 401.45km/0.333333334s
= 1204350/340.29
= Mach 3539.18716

172 pixels = 523.56118km
1 pixel = 523.56118km/172 = 3.04396035km
3.04396035km X 154 = 468.769894km

T = 468.769894km/41.6666667ms
= 11250477.4m/s

Now for our energy, for which we enter those values through the reletavistic kinetic energy calculator, which yields a result of 8.002E+32 joules, or 191.25239005736136733 zettatons. Now for Darry shooting the other missile.

Spoiler

Timeframe is 14 frames.

T = 1s/24
= 41.6666667ms X 14
= 0.583333334s

211 pixels = 523.56118km
1 pixel = 523.56118km/211 = 2.48133261km
2.48133261km X 164 = 406.938548km

T = 406.938548km/0.583333334s
= 697608.939m/s

KE = (0.5)mv^2
= (0.5) X 1.26314178e19 X 697608.939^2
= 3.07359173e30 joules
= 734.60605401529639494 exotons

Final Results
Anti-Spiral missile size = 1404.88917km
Yoko shoots missile (speed) = Mach 3539.18716
Yoko shoots missile (energy) = 191.253 zettatons
Darry shoots missile = 734.606 exotons

 

[Stocking Anarchy]

Space King Kittan uses Super Galaxy Dai-Gurren's missile (created from Simon's Spiral Power) to blow open a hole in the Spiral Death Machine's protective field. This is of note because the Death Spiral Machine is constantly absorbing mass and converting it to energy, but a huge amount of Spiral Power would disrupt that field, creating an opening. Meaning an even greater amount of energy is required to open it.

BioComputer LordGenome: The Death Spiral Machine is surrounded by a region of super-ultra-dense space. Even the Super Galaxy Dai-Gurren will be lucky to survive 300 seconds within it. A Spiral conversion field is in place around the machine. This field protects the machine by converting mass into Spiral Power.

Leeron: It negates the pressure and generates energy at the same time. Two birds with one stone...

BioComputer LordGenome: Still, that has a fatal flaw. A massive dose of Spiral Power will disrupt that process, and cause a localized collapse of the field.

We can scale the size of Space King Kittan off Arc Gurren Lagann. Arc Gurren Lagann has a height of 5km (this figure is also in the novelizations and the script).

Spoiler

367 pixels = 5000m
1 pixel = 5000m/367 = 13.6239782m
13.6239782m X 120 = 1634.87738m
13.6239782m X 167 = 2275.20436m (2.27520436km)

Now we have Space King Kittan's height, we can find the minimum distance between it and the Death Spiral Machine's protective field.

Spoiler

object degree size = 2*atan(Object_Size/(Panel_Height/tan(70/2)))

= 2*atan(132/(1617/tan(70/2)))
= 0.0773189632 rad
= 4.43005026769831467 degrees

Enter that into the angscaling calculator, and we have a distance of 29412m. Now, to scale the width of the hole.

Spoiler

object degree size = 2*atan(Object_Size/(Panel_Height/tan(70/2)))

= 2*atan(2324/(1616/tan(70/2)))
= 1.19626928 rad
= 68.54118090532377 degrees

The gap is 40083m wide, or a radius of 20041.5m.

Spoiler

432 pixels = 40083m
1 pixel = 40083m/432 = 92.7847222m
92.7847222m X 652 = 60495.6389m
60495.6389m/2 = 30247.8194 meters

Now we have that, let's have the amount of matter turned into energy per second. We can see several trails leading into the Death Spiral Field.

Spoiler

Timeframe is 2 frames.

T = 1s/24
= 41.6666667ms X 2
= 83.3333334ms

870 pixels = 40083m
1 pixel = 40083m/870 = 46.0724138m
46.0724138m X 4 = 184.289655m
46.0724138m X 9 = 414.651724m
414.651724m - 184.289655m = 230.362069m

T = 230.362069m/83.3333334ms
= 2764.34483m/s

That's the amount of matter absorbed every second. With that, we can find the area of mass that would have been absorbed. Volume as a cylinder.

V = πr^2h
= π X 20041.5^2 X 2764.34483
= 3.4882093e12m^3

As for our masses, we'll go with a low end of water and a high end of uranium. Water has a mass of 1000kg/m^3, while uranium has a mass of 19,050kg.

(Low end)

M = 3.4882093e12 X 1000
= 3.4882093e15kg

E = mc^2
= 3.4882093e15 X 299792458^2
= 3.13504617e32 joules
= 74.929401768642449611 zettatons

(High end)

M = 3.4882093e12 X 19,050
= 6.64503872e16kg

E = mc^2
= 6.64503872e16 X 299792458^2
= 5.97226296e33 joules
= 1.4274051051625238351 yottatons

Now we have that, let's also find the speed with which Space King Kittan throws the missile.

Spoiler

Timeframe is 12 frames.

T = 1s/24
= 41.6666667ms X 12
= 0.5s

T = 29412m/0.5s
= 58824/340.29
= Mach 172.864322

Final Results
Space King Kittan height = 2.275km
Super Galaxy Dai-Gurren's missile opens up Death Spiral Machine's field (low end) = >74.929 zettatons
Super Galaxy Dai-Gurren's missile opens up Death Spiral Machine's field (high end) = >1.427 yottatons
Space King Kittan throws the missile = Mach 172.864

 

[Stocking Anarchy]

The old calc for this vanished for some reason...thankfully, I had it saved to Gdocs, so I can repost it here.

Spoiler

(Relevant text)

Spoiler

A big explosion occurs. A torrent of light that shakes the galaxy!

(I've checked with a friend and this is an accurate translation.)

King Kittan's sacrifice shakes the entire galaxy. Looking back at my old Noo shake the Universe calc for help. First, the energy to shake one planet noticably enough to be felt, or at least magnitude 2.5, though the highest I could get by playing around with the impact calculator was magnitude 2.3, which results in 3.72 X 10^12 Joules. Next we need the area that get's shook. Divide that by 10,000 and you have 372000000 joules The radius of the Milky Way is 52,850 lightyears, or 3720000000000 joules of seismic energy.

I = S/4pa
372000000 = S/(4 X π X 4.99988926e20^2)
S = 372000000 X (4 X π X 4.99988926e20^2)
= 1.1686207e51/10^44
= 11.686207 MEGAFOE

Final Result
Kittan's final sacrifice = 11.686 MEGAFOE

 

[Stocking Anarchy]

General Adiane (in Sayrune) sends a giant tidal wave crashing into Dai-Gurren. Unfortunately, as all the old calcs with Gurren Lagann's size in them have crashed, we're going to need to rescale. Kamina is 184cm tall.

Spoiler

52 pixels = 184cm (1.84m)
1 pixel = 1.84m/52 = 0.0353846154m
0.0353846154m X 544 = 19.2492308m
0.0353846154m X 298 = 10.5446154m

Now scaling Gurren Lagann to Dai-Gunzan/Dai-Gurren...

Spoiler

142 pixels = 19.2492308m
1 pixel = 19.2492308m/142 = 0.135557963m
0.135557963m X 2162 = 293.076316m

At last we can get the height of the wave.

Spoiler

470 pixels = 293.076316m
1 pixel = 293.076316m/470 = 0.62356663m
0.62356663m X 381 = 237.578886m
0.62356663m X 2076 = 1294.52432m

Next we need the distance it travelled and the width of the area it covered.

Spoiler

257 pixels = 237.578886m
1 pixel = 237.578886m/257 = 0.924431463m
0.924431463m X 2880 = 2662.36261m

object degree size = 2*atan(Object_Size/(Panel_Height/tan(70/2)))

= 2*atan(257/(1612/tan(70/2)))
= 0.150793492 rad
= 8.63983066966117619 degrees

When we enter this into the angscaler, we get a distance of 1572.5m of the wave to Dai-Gurren. To add the minimum distance it travelled after passing Dai-Gurren...

L = 1572.5m + 1294.52432m
= 2867.02432m

With length, height and width, we can add all of those together for our volume, and from volume, our mass. Volume as a rectangular prism.

V = lhw
= 237.578886 X 2662.36261 X 2867.02432
= 1.8134535e9m^3

Water weighs 1000kg/m^3.

M = 1.8134535e9 X 1000kg
= 1813453500000kg

So that's all the dimensions required for size, all we need now is the speed. How fast did the wave traverse the screen?

Spoiler

Timeframe is 23 frames.

T = 1s/24
= 41.6666667ms X 23
= 0.958333334s

T = 1294.52432m/0.958333334s
= 1350.80799m/s

With everything we need, let's get our kinetic energy!

KE = (0.5)mv^2
= (0.5) X 1813453500000 X 1350.80799^2
= 1.65448818e18 joules
= 395.43216539196941994 megatons

Final Results
Gurren Lagann height = 19.249m
Dai-Gurren length = 293.076m
Sayrune creates a giant tidal wave = 395.432 megatons

 

[Stocking Anarchy]

Sayrune moves underwater at incredible speeds. Scaling Sayrune from Gurren Lagann. From my previous calc, Kamina is 184cm tall.

Spoiler

52 pixels = 184cm (1.84m)
1 pixel = 1.84m/52 = 0.0353846154m
0.0353846154m X 544 = 19.2492308m
0.0353846154m X 298 = 10.5446154m
0.0353846154m X 39 = 1.38m

Next, we scale Gurren Lagann's head to Sayrune.

Spoiler

87 pixels = 1.38m
1 pixel = 1.38m/87 = 0.015862069m
0.015862069m X 1598 = 25.3475863m
1598 + 112 + 97 + 692 + 244 + 1223 + 225 = 4191 pixels
0.015862069m X 4191 = 66.4779312m

Not sure if the tail would be visible above water or not, so we'll go with a low end of the main body and a high end of the full length.

Spoiler

Timeframe is 4 frames.

T = 1s/24
= 41.6666667ms X 4
= 0.166666667s

(Low end)
28 pixels = 25.3475863m
1 pixel = 25.3475863m/28 = 0.905270939m
0.905270939m X 442 = 400.129755m

T = 400.129755m/0.166666667s
= 2400.77853/340.29
= Mach 7.05509574

(High end)
28 pixels = 66.4779312m
1 pixel = 66.4779312m/28 = 2.37421183m
2.37421183m X 442 = 1049.40163m

T = 1049.40163m/0.166666667s
= 6296.40977/340.29
= Mach 18.5030702

Final Results
Sayrune's total length = 66.478m
Sayrune's underwater speed (low end) = Mach 7.055
Sayrune's underwater speed (high end) = Mach 18.503