You know the feat.
So, today I started reading this new seies. It's not bad and since it's in its infant state I thought I can start calcing the feats here.
Now, we are going to calc both the KE of the meteor and the destruction it caused. It's a simple enough math.
Average Floor Height = 3 meter
Number Of Floor = 7
Building Height = 332 pixel = 21 meter
Building Width = 426 pixel = 26.9457 meter
Meteor Width = 30 pixel = 1.8975 meter
Meteor Length = 44 pixel = 2.7831 meter
Assuming the building is rectangular in nature so width in every side is same. Also using the standard 80% of the building is hollow.
Building Volume = 0.2*L*W*H
Building Volume = 0.2*26.9457*26.9457*21 meter^3
Building Volume = 3049.51596 meter^3
The meteor is not perfect sphere. Thus I am going to use the average of length and width and use that as the diameter of the meteor.
Meteor Diameter = (2.7831+1.8975)/2 meter
Meteor Diameter = 2.3403 meter
Meteor Volume = (4/3)*pi*r^3
Meteor Volume = (4/3)*pi*(2.3403/2)^3 meter^3
Meteor Volume = 6.7119 meter^3
Here I am going to use the ordinary chondrites meteors average density which comes out as 3350 kg/m^3
Meteor Mass = p*v
Meteor Mass = 3350*6.7119
Meteor Mass = 22484.9659 kg
So, our meteor is barely 25 tons. As I don't know how much speed it should retain as it touches the ground for now I am skipping it.
Now, the meteor slammed through the building into the ground. It destroyed the building completely and made a crater in the ground.
Now if you look at the scans you will see that the building and the crater was basically pulverized. Pulverization of the ground is 214 joules/cc but I don't remember the concrete pulverization number on the top of my head. So, I will use 8 joules/cc for it.
Building Destruction Energy = 3049.5159*1000000*8 Joules
Building Destruction Energy = 24396127681.81158 Joules
Now, lets calc the volume of the crater.
UnDead = 24 pixel = 1.7 meter
Crater Diameter = 857 pixel = 60.7041 meter
Crater Depth = 153 pixel = 10.8375 meter
Crater Volume = 0.5*(4/3)*pi*r^3
Crater Volume = 0.5*(4/3)*pi*10.8375*(60.7041/2)^2 meter^3
Crater Volume = 20910.51533 meter^3
Crater Destruction Energy = 20910.51533*1000000*214 Joules
Crater Destruction Energy = 4474850281010.02204 Joules
Total Energy = 4474850281010.02204 + 24396127681.81158 Joules
Total Energy = 4499246408691.83363 Joules
Total Energy = 1.0753 Kilotons
Her Kiss Is Like A Meteor = 1.0753 Kilotons
All right, there is no clear way to scale the meteor size here. So lets go at it a bit round about way.
Inside the blue circle we can see the fragments of the meteor. Lets assume the length of that fragment is the entire length of the meteor.
Now in the bottom panel we can see the size of the meteor in comparison to earth. So, we can scale the meteor using earth's size.
Anyway, first lets scale the meteor size.
Earth diameter, d = 532 pixel 12742000 meter
Fragmented piece length, l = 22.8035 pixel = 546169.74686 meter
There is a nice full picture of the whole meteor. We can calc the volume from that.
Meteor length, l = 74.51845 pixel = 546169.74686 meter
Meteor width, w = (30+41.0487+35.5105+32.7566+20.8086)/5 pixel = 32.02488 pixel
Meteor width, w = 234720.6712 meter
Modelling the meteor as a cylinder,
Volume, V = pi*r^2*h = 23633073106815073.3581599 meter^3
Average Density of meteor p = 3350 kg/m^3
Meteor mass, m = v*p = 79170794907830495749.83586 kg
Meteor speed v = 11000 meter/sec
Kinetic Energy = .5*m*v^2 = 4789833091923744992865069751.1275 joules
Kinetic Energy = 1.144797 Exatons of TNT
So, based on the description of the attack it does seems like energy was converted to matter here. But I am going to err on caution and calc it both ways.
Crater size = 853.8413 pixel = 24.4 miles or 39267.99 meters.
Mountain Height = 590.09321 pixel = 27138.26828 meter
Mountain Width = (357.6870+284.0176+200.4120+68.5419)/4 = 227.664625 pixel
Mountain Width = 227.664625 pixel = 10470.25040 meters
Modeling it as cylinder.
Volume, v = pi*r^2*h = 2336609559012.737132 meter^3
Density of Granite, p = 2691 kg/m^3
Mass = p*v = 6287816323303275.62333069 kg
Now, assuming the mountain was created by raising ground,
PE = m*g*(h/2) = 836138186758120797376.789375 Joules or 199.8418 Gigatons
Again, assuming the mountain was created by turning energy into matter,
E = mc^2 = 565149182651151345155987460233976.7688 Joules or 135.07389 Zettatons
Lumen creates a mountain [Low End] = 199.8418 Gigatons
Lumen creates a mountain [High End] = 135.07389 Zettatons
Ryokugyu instead of Shiryuu
A lot of people were bitching about Shiryuu taking Jozu's fruit and beating Mihawk theory. But now that Oda has shit on that the possibility of Shiryuu beating Mihawk has become dim.
Many were concerned about who Zoro was going to fight during Final war. Some were having trouble with timeline regarding when he would fight Mihawk. Some even suggested after Luffy becoming PK, fighting in Final War Oda should make an additional arc solely for Zoro fighting Mihawk.
To me Mihawk was never going to be anything more than the hype train. Whether the PK throne fight or Final war I could never place him in either one of them. Aside from being Zoro's goal he has no purpose in the story.
So him getting beat by someone wouldn't surprise me that much. I hoped it would be Shiryuu but I would be fine with it being Ryo.
Fujitora has come up with a solution for you. And that is abolishing the Shichibukai system.
No marine could touch any Shichibukai because WG provided protection for them. But if Fujitora succeeds then Mihawk will be a fair game.
I predict Oda has reserved Green Bull for this exact purpose. GB will first appear when he goes to fight Mihawk & arrest him. This way, Zoro's opponent for the final war will be set. He will take revenge for his Sensei and become WSS in the same arc SHP + RA will take out the WG.
Tier List Placement
I actually put Shanks above Mihawk. Not by much. IMO if they fight 10 times then Shanks would win 7 times & Mihawk 3 times. That's just the way I view it.
Regarding Ryo > Shanks/Mihawk. I currently place Kizaru on the same level as Shanks (50/50 either way). I do think Ryo is a bit weaker but by fighting Mihawk he will improve & rise to Mihawks level. So more like,
Pre - Fight, Shanks >~ Mihawk > Ryo
Post - Fight, Shanks >~ Mihawk ~ Ryo
I am rooting for him to have a Zoan fruit.
Honestly speaking instead of a paramecia or logia swordsman I prefer Zoan swordsman cause Zoan DF improve physical stats & won't have much effect on the characters primary fighting style whether he is a swordsman or Martial artist.
The War that took out ROX
In Chapter 907 we learned a few things,
Before Roger a pirate captain or organization named Rocks was the top group.
Big Mom & Kaido possibly worked under this organization/captain.
We thought Garp became a Marine Hero because of his fight with Roger, but now learned that it happened before that. Probably when he took down "Rocks" organization or their captain.
Now lets factor in some other points,
Zephyr become Admiral ad the age of 38. He died at the age of 78. So that happened 40 years before the current timeline.
Rocks captain/organization was defeated 40+ years prior to current timeline.
40 years ago Garp, Sengoku & Zephyr were the most prominent vice-admirals of marine. They led an Armada against Rocks pirates to defeat them once & for all.
They also had other strong marines with them like Tsuru, Bogart & others as well who were Rear Admiral at that time.
Meme wasn't directly under Rocks & was just allied captain (like Squardo & Sai etc.). Kaidowas the cabin boy like Shanks.
The whole plan was crafted by Sengoku & Tsuru. They timed it just when Meme, & other allied captains were not present. As a result they only fought the core crew of Rocks pirates.
Garp beat their captain & became "the hero of marine". Sengoku & Zephyr beat their top commanders (current Yonko FM level) & became Admiral.
By the time Meme learned about it it was too late. After Rocks was defeated she went her own way and started gaining power & eventually became Yonko. Marines didn't arrest Kaidocause he was a minor & let him go instead.
"They are growing in strength." -- Hina
My theory is that they started growing in power recently. ROX pirates FM was locked in Level 6. He didn't take part in BB's little game cause he had other plans in mind. After BB ransacked Level 6 he escaped and has been gathering up old members & recruiting new ones.
I also think that Luffy will fight the ROX pirates FM after he beats Kaido along with Kidd & Law.
Luffy's future fight
I am hoping that Luffy will fight two Yonko and two Admirals.
First, Kaido. It will be a team battle against Kaido. Most likely the Auction House Trio will join hands.
Now, I am not totally sure about this but I think post-Wano SHC's main villain is going to be ROX pirates. It's possible that Luffy will either fight ROX Captain/FM or Weevil then.
Second, an Admiral most likely Fujitora as he isn’t allowed to set foot in any marine without bringing down Luffy. He will most likely call a Buster Call to assist him against SHC.
Third, it’s going to be BlackBeard. This will happen after BB takes out Shanks and at Raftel. It will be the fight to determine who will be the PK.
Fourth, Akainu. The main villain is going to be Im. But Luffy's opponent is going to be Akainu while Dragon/Sabo takes care of Im/Kong.
In Episode S14E18 after killing Mary Winchester Jack was flying around aimlessly. Sam to a tracker on Jack so he could track Jacks location.
Now Jack flew from Lima, Peru to France in seconds. Jack's tracking signal vanished from Lima at 5:13 and reappeared on Paris at 5:14 (yes I counted the frame as well). So it took him One seconds.
The distance between two points is 6371 miles.
Flying Speed = 6371/1 = 6371 miles/sec
Speed of sound = 343 meter/sec
Jack Flying Speed = Mach 30130.57
Jack Flying Speed = Mach 30130.57
IMO Shin will be promoted to Qin6 six years after current timeline. After that he will wage war against each state in the ten years it will take Qin to unify China. This is how I think Shin's campaign will go.
Year 17, 230BC
17th year of Zheng’s reign(230BC), governor Teng was tasked to attack the state of Han, crushing the Han army and capturing the king of Han. The lands of Han were absorbed into Qin and was renamed Ying Chuan county.
This year Tou conquered Han. This will be Shin's first campaign as Qin6. He will serve as the deputy of Tou.
Year 18, 229 BC
Shi Ji, The Biographies of Assassins, it was mentioned that Li Xin captured the castles of TaiYuan and YunZhong in the dialogue between the crown prince of Yan and Jing Ke, the assassin he was sending to assassinate Ying Zheng.
Shin will serve as a deputy to OuSen and finally slay Houken in this war.
Year 19, 228BC
This year Shin will fight against Wei. His opponent will be GaiMou.
Year 20, 227BC
This will be Shin's first campaign in Chu. Here he will face a pure instinctual GG like Duke Hyou. It will be someone we haven't seen yet. Someone who has an instinct level either 100 or close. This campaign will test Shin's capability as an instinctual Great General.
Year 21, 226BC
– From Zhan Guo Ce: Ch31, Yan No.3, we know that when Shin was pursuing the king of Yan towards Liaodong(辽东), through the suggestion of the king of Dai, the Yan king killed his own crown prince and gave Qin the crown prince’s head. We can deduce from there that Shin was part of Ousen’s invasion in the line above.
Year 21, 225BC
This is his legendary historical war. Here he will lose. I don't want to discuss this further.
Year 25, 222BC
What we know from manga so far,Shi Ba Shou is the lord of Seika City. He does not wish to engage in battle, except for his city, Seika which is close to Yan's border.
In 229BC, Qin attacked Zhao. Great General RiBoku and and General ShiBaShou were tasked to defend against the attack. Qin bribed a minister who was liked by the King of Zhao, with gold to spread news of RiBoku and ShiBaShou plans to rebel. The king of Zhao sent Zhao Cong and Qi general Yan Ju to replace RiBoku. RiBoku refused but was captured by Zhao men and later executed while ShiBaShou was stripped off his duties. (General) Zhao Cong and the Qi general, Yan Ju, was tasked to replace them. Zhao Cong was defeated and Yan Ju fled.
Dai was a short-lived state from 228 BC to 222 BC during the Warring States Period. Prince Zhao Jia, older brother of King Youmiu of Zhao, fled with the remnant forces to Dai Commandery after the Conquest of Zhao and was proclaimed the new king of Zhao.
In 222BC, Shin and OuHon destroyed the rest of Yan, capturing King of Yan. They also destroyed Dai, capturing King Jia of Dai.
ShiBaShou will be the Great General of Dai. By that time I doubt there will be any renowned Great General in China. Shin will be the one to take him out while OuHon deals with the remnants of Yan. This will also be the first war since he will lose his lieutenants in Chu & his first solo campaign against a Zhao 3 Great Heavens. This win will make up for his loss in Chu.
Year 26, 221BC
-From Shiji: Biographies of Bai Qi and Wang Jian, Shin was in fact involved in final invasions of Qi (and Yan).
So here, both Shin and OuHon was involved in this campaign. So far we know about 3 Qi Generals. They are GanMo, KanShu and DenKaku. Among them GanMo was supposed to be the Commander-in-chief of Qi Army in the Coalition Army.
OuHon, Shin and KyouKai each takes out a General.
Why two Chu invasion? First, I want to see Shin face off against a true instinctual type. And Chu is a land of mystery so why not an instinctual general from there?
Second, as Chu is a superstate I want Shin to have some success against them before his colossal failure later.
All right, had some free time on my hand. And this was already scaled and ready so I thought why not post the most impressive feat of this pathetic excuse of an emperor.
First, we need the island size. I already scaled it before. I took the mid end.
Whole Cake Island Diameter [Mid End] = 7183.5134 meter
Now we need the size of one of WCI's blocks that's beside the huge towers.
Whole Cake Castle diameter = 491 pixel
Block's Height = 103 pixel = 1506.92847 meter
Now Meme slash entered from one part and exited from other. For the lack of better method I will use the width as the length that the slash crossed.
Block's Height = 260 pixel = 1506.92847 meter
Block's Width = 117 pixel = 678.11781 meter
Lets now scale the hole it made and how much water it raised,
Block's Height = 45 pixel = 1506.92847 meter
Hole diameter = 8 pixel = 267.8983 meter
Water width = (100+36+69)/3 or 68.3333 pixel = 2288.28762 meter
Water Height = (44+61+54+44+26)/5 or 45.8 pixel= 1533.71831 meter
Assuming the block was 80% solid and cylinder shaped,
Hole's Volume = pi*r^2*h = pi*(267.8983/2)^2*678.11781 = 38223895.7192 meter^3
It looks like Meme's attack pulverized it. But it's also true that everything in WCI is made from food ingredient. I am going to play it safe and use 8 j/cc for pulverization here.
Rock Destruction, Q1 = 305791165753944.95927 joules
Volume of Water = pi*r^2*h = pi*(2288.28762/2)^2*1533.71831
Volume of Water = 6307491961.51961 meter^3
Density of sea water is 1024 kg/m^3
Water Weight = 6458871768596.08110671 kg
PE of Water = mgh = 6458871768596.08110*9.8*(1533.71831/2)
PE of Water, Q2 = 48539840477845673.62884 joules
Total DC = Q1 + Q2 = 48845631643599618.5881132109 joules
Total DC = 11.674386148 Megatons
So that was impressive.
Big Mom's Elbaf Spear = 11.674386148 Megatons
A general normally has at least 10,00 soldiers under him. I am guessing will have that many under him too after he comes one. After this war I say it will go like this,
Hi Shin Unit
Total - 14000 Soldiers.
General (10000 Soldiers)
4000 man Commander
I don't see Kyokai skipping a rank & becoming 5K man commander.
Breakdown of 14000 Soldiers
Since we don't know if Kyokai has other 2k, 1k, 300 man commander in her unit or not this isn't a proper breakdown of the unit.
500 man General unit
This is basically the unit Shin personally leads into battle.
500 man Kyokai unit
This is basically the unit Kyokai personally leads into battle.
1000 man unit
Even though they have no official rank I think they each at least commands that much based on their position as lieutenant of HSU & experienced military leader.
As the infantry commander he should at least command that much. Though I think he will become 2000 man commander.
Part of monster trio from the very beginning.
2000 man unit
300 man unit
Can't account for rest 800 soldiers. But there are many units aside from these. So they can be used to fill those numbers.
Lexcorp Laser DC
Feat -- 05:10
To dig 2 miles in underground Lexcorp used laser to destroy boulders along the way.
Human = 301.8447 pixel = 1.7 meter
Hole diameter = 743 pixel = 4.184 meter
Hole Thickness = 143 pixel = 0.8053 meter
Modeling it as cylinder.
Hole Volume, pi*r^2*h = 11.0764 meter^3
Fragmentation value 8 j/cc
Laser DC = 88611368.07582 Joules = 0.02117 tons
To beat Doomsday Superman holds him & goes straight to space. Then body slams him in the ground. It took him 300 frames to reach that level. The video is 23.96 FPS.
Timeframe = 12.5208 seconds
Since except for space we don’t know how far he went I will use 2 different ways to calc the distance he traveled.
First [Low End] for space we will use Kramline line which is 100 km above see surface.
Superman’s Speed = (100*1000/12.5808) = 7986.6666 meter/second
Superman’s Speed = Mach 23.2847
For the next distance calc I am going to use angsizing. First I am going to calc the distance to earth from our pov & then to superman. Subtracting the distance we will get how far Superman traveled.
Panel Height = 1080 pixel
Superman = 36.6196 pixel = 1.8796 meter
Height H = 175.5733 pixel
Width W = 1967.40488 pixel
So, R = (H/2)+(W^2/8H) = 2843.53097 pixel = 6371 km
So, W = 1967.40488 pixel = 4408.01828 km
First distance from our POV to Superman,
object degree size = 2*atan(36.6196/(1080/tan(70/2)))
object degree size = 2.720117530808
Using angular calculator I get Distance = 39.584 meter
Again for distance from our POV to Earth,
object degree size = 2*atan(1967.40488 /(1080/tan(70/2)))
object degree size = 103.8087520
Using angular calculator I get Distance = 1.7279e+3 km
This is actually a low estimate cause Superman goes farther than that. But because of lack of earth shot from his final distance I am using this one.
Distance from Superman to Earth = 1727.8604 km
Superman’s Speed = (1.727.8604*1000/12.5208) = 137999.20260 m/s
Superman’s Speed = Mach 402.33003673469
I am not really sure about the second one though. I have my doubt about it.
Superman's Body Slam
Now, let’s calc the DC of Superman’s Body Slam. First we need the height of Doomsday.
Superman = 500.8732 pixel = 1.8796 meter
Doomsday = 644.4261 pixel = 2.4183 meter
Doomsday = 11.3137 pixel = 2.4183 meter
Crater Diameter = 1393 pixel = 297.7534 meter
Crater Depth = 397 pixel = 84.8586 meter
Modeling the crater as half ellipsoid,
Crater Volume, .5*(4/3)*pi*(d/2)^2*r = 3939205.1778462 meter^3
Pulverization value 214 j/cc
Superman’s Body slam = 842989899822042.67655 Joules
Superman’s Body slam = 201.479421564 kilotons
Lexcorp Laser DC = 0.02117 tons
Superman's Speed = Mach 23.2847 - Mach 402.3300
Superman’s Body slam = 201.479421564 kilotons
DC Legends of Tomorrow S03E08
Size of Midwestern US,
Total area = (199729+116096+183108+200365+180530+225163+250493+213100+145746+94321+149932+169640) km^2
Total area pi*r^2 = 2128223 km^2
Radius r, = 823.0640 kilometers
Inputting the value in Nuclear Weapon Effects Calculator we get yield. Thanks to @Shiba Miyuki for the results.
Supergirl survived it point blank & was only knocked out for few minutes. There wasn't any visible injury.
Supernova Energy [Low End] = 1.708 Teratons
Supernova Energy [Mid End] = 32.35 Teratons
Supernova Energy [High End] = 158.2 Teratons
All calc related discussion goes here.
This is pretty straight forward calc. Lufasu makes a big crater by punching Aries into the ground.
Lufasu = 19.2093 pixel = 1.7 m
Aries length = 508 pixel = 44.9572 m
Aries length = 57.1401 pixel = 44.9572 m
Crater diameter = 898 px = 706.5358 m
Crater depth = 397 px = 312.3549 m
Crater volume = .5*(4/3)*pi*abc
Crater volume = .5*(4/3)*pi*312.3549*(706.5358/2)^2 m^3
Crater volume = 81642327.2060 m^3
Fragmentation value (8 j/cc) = 653138617648443.65152 joules
Violent Fragmentation (69 j/cc) = 5633320577217826.49440 joules
Pulverization value (214.34 j/c) = 17499216413345926.533492 joules
Low end = 156.1038 kilotons
Mid end = 1.3463 megatons
High End = 4.18241 megatons
All related discussion goes here.
Big Meme apparently has the power to create a storm. When she learns that Luffy beat Cracker she went into a fit of rage & created a thunderstorm.
From the looks of it I think it's a Single-Cell thunderstorm. Single-cell thunderstorms typically last 20–30 minutes. But here it looks it lasted close to a day. Here we see the rain started at 2 pm. This is all after Luffy met with Sanji & fought the enraged army so probably 30 minutes earlier but lets just go with it as is. The marriage was next day. We learn from here that weeding was supposed to be at 10 am. The rain stopped at least 4 hours and 30 minutes before marriage started (including the time SH took to reach Bege's castle) so at 5:30 am at least.
So, our time frame is 2 pm to 5:30 am or 15 hours & 30 minutes.
First , we need the whole area the storm covered. From the scan below we can see from our POV the storm. Since it was created in Whole cake island lets assume it's ground Zero & distance from our POV to the island is the radius. I am going to calc the distance using angular formula.
Island Diameter = 548 pixel = 7183.5134 meter
Cloud Base height = 200 pixel = 2621.7202 meter
Cloud height = 143 pixel = meter
Panel Height = 505 pixel
object degree size = 2*atan(548/(505/tan(70/2)))
object degree size = 74.457258438
Now, putting object degree size & WC Island size in angular calculator we will find the distance.
Distance, r = 4.7271e+3 meter
So area covered = pi*(4.7271e+3)^2 = 70200378.24743 meter^2
Now, since we don't have a time frame & her creation didn't look like she gathered clouds at all I am not going to calc it using KE instead I am going to try a different method.
We need rainfall rate. Light rain is typically 2- 4 mm/hr & moderate rain is typically 5-9 mm/hr. For the sake of low end lets use 3 mm/hr. (Link)
Now, we need the amount of water fell during the rain.
1 mm of rainfall equals 1 liter of water over an area of one square meter. (Link)
Total number of millimeters that fell = (3*15.5) or 46.5 mm
Rate in terms of liters/meter^2 = 46.5 liter/meter^2
Total Water Fell = (46.5*70200378.24743) = 3264317588.505495 liter or 3264317588.505495 kg
So, now we need dew point temperatures cause that's the temperature water vapor has to cool down to form water droplets.
dewpoint = surface temperature - (4.4*cloud base/1000)
dewpoint = 20 - (4.4*2621.7202/1000)
dewpoint = -1.0257525517644 celcious
Now, we need the temperature to calc the energy to cool down water vapor.
Height from surface to mid point of cloud = 2621.7202+(1874.5299/2) = 3558.9852 meter
So, temperature at 3558.9852 meter cools down (6/10000*3558.98521) or 2.135391 Celsius or temperature in the cloud is (20-2.135391) or 17.864609 Celsius.
Q1 = specific heat capacity*mass*temperature change.
Q1 = 1.996*(3264317588.505495*1000)*(17.864609+1.02575)
Q1 = 123081605749218.61574931918 Joules
I am going to calc the latent energy needed to condense this huge mass.
The latent heat of condensation = 2.5 * 10^6 Joules/kg
Latent Energy, Q2 = (3264317588.505495*2.5 * 10^6) Joules
Latent Energy, Q2 = 8160793971263737.5 Joules
Total Energy = Q1+Q2 = 8283875577012956.11574931918 Joules
Big Meme's Storm = 1.97989378 Megatons
All discussion related to this calc goes here.
Since there has been a lot of debate whether Big Meme's feats are impressive or not I have decided to scale the Whole Cake Island. This is the second version. Lets see this one goes.
First I will scale one of the Germa ship. I will use @God Movement Thousand Sunny calc for it.
Thousand Sunny Width = 89.8943 pixel = 26.2 meter
Germa 66 ship width = 488.0368 pixel = 142.2398 meter.
Now I will use this to scale the diameter of the lower level of Whole Cake Castle. All the Germa ships should be of same size. But I doubt Oda used scale here so I will scale four ships & use the average.
Germa 66 ship width = (78+74+76+74)/4 or 75.5 pixel = 142.2398 meter.
Whole Cake Castle lower level diameter = 652 pixel = 1228.3497 meter
TBH the design & size of WCI is pretty inconsistent. But one thing we are sure of is that Germa 66 was situated just behind Whole Cake Castle. I am going to scale the lower level diameter to scale the whole island. Also I am going to use two different scan & use average.
Whole Cake Castle lower level diameter = 52 pixel = 1228.3497 meter
Whole Cake Castle top level diameter = 17 pixel = 401.5758 meter
Whole Cake Castle diameter = 387 pixel = 9141.7564 meter
Again, with different scan,
Whole Cake Castle lower level diameter = 129 pixel = 1228.3497 meter
Whole Cake Castle top level diameter = 26 pixel = 247.5743 meter
Whole Cake Castle diameter = 491 pixel = 4675.3465 meter
So, almost half from what we got from the first scan.
Let's try again then,
Whole Cake Castle lower level diameter = 79 pixel = 1228.3497 meter
Whole Cake Castle top level diameter = 33 pixel = 513.1081 meter
Whole Cake Castle diameter = 462 pixel = 7183.5134 meter
Now, we are getting somewhere. Lets try again,
Whole Cake Castle lower level diameter = 54 pixel = 1228.3497 meter
Whole Cake Castle diameter = 432 pixel = 9826.7976 meter
Now, some can object to the last scaling due to the castle falling down. But I think it falling down has any effect here since we are scaling via diameter. As such we can scale it with no problem.
Nonetheless , I think I will go with the mid end or the value of third scan for island diameter. For top level diameter lets use the average which is (513.1081+247.5743+401.5758)/3 or 387.4194 meter.
If there is any mistake, kindly point it out & I will try to correct it.
Castle Top Level Diameter [Average] = 387.4194 meter.
Whole Cake Island Diameter [Mid End] = 7183.5134 meter
The calc is pretty simple. Nothing more than few scalling at best.
Earth diameter = 1777 pixel = 12,756 km
Explosion diameter = 1245 pixel = 8937.09622 km
Explosion radius = D/2 = 4468.548114 km
Since there is no visible destruction & the explosion was in space it's hard to gauge the DC of the blast. I am going to use nuke calculator for DC,
It's pretty hard to find accurate radius in the calculator so I am going to use the closest I can find ,
Fireball radius (minimum) = 4472.2 kilometres = 10.888888889 petatons
DC of Macrocosm Canon = 10.888888889 petatons of TNT (PtTNT)
All discussion related to the calc goes here
So lets get to it. Any size we scale is going to be low end so there is nothing much I can do here. For average galaxy I am going to use the size of Milky Way. Based on my little research its also the smallest galaxy to naked eye. Size of Milky way also looks to be average . Anyway,
First I am going to scale the Dragon's eye length from one of the nearest galaxy to it. I see two. One around its teeth & one a bit above its eye. I will their average size.
Average Galaxy size = (12.7279 + 22.80350) / 2 = 17.7657 pixel = 110,000 light-years
Dragon Eye Length = 500.22494 pixel = 3097243.401763 light-years
Now,I am going to use the eye to scale the whole body. Dragons whole body is curved so it may not be the most accurate scaling mind you. Nonetheless , I will try my best.
Dragon Eye Length = 26 pixel = 3097243.401763 light-years
Dragon Length = 1537.778600 = 183187485.57011 light years
The scaling looks like a cluster fuck. I know. Not my best work. Anyone is welcome to rescale to more accurately.
Super Shenron Size = 183.187485 million light years