Not My Calc
So, a tornado starts to form just before the harbor of Ningbo. And this massive storm cloud starts forming from the ocean. This feat is shown below.
Let's look at a diagram of a supercell.
As for why this storm cloud is a supercell, we can see that from the picture/diagram located above, we can see from the diagram that a tornado is located just below said cloud.
This matches what is shown in the very first picture I posted.
Here, we are shown that the cloud splits in this picture, and those clouds pass just over the bump. Now let's compile all the information. I'll list them here.
- Distance from Ningbo to Bump shown in first picture = 128.8km (according to Nukemap statistics in the 3rd image)
- Apparently the clouds split from center of Ningbo to where that bump is shown in the first pic.
- At the very bottom of a supercell cloud is a mesocyclone.
Actual Calculation[]
- Seeing that there is a tornado forming at the bottom of cloud formation, this cloud is a supercell.
- Thickness of cloud = 12000m to 18000m (from image/diagram above)
- Using the lower end for the thickness of cloud in this case, as the cloud in this scenario is at least that thick.
So from this quote here, starting from the tornado that forms from the middle of the supercell, it would mean that the storm cloud spans up to 32km from the middle? Thus...
- Radius of cloud = 32000m, or 32km
Now for volume:
- Volume of cylinder = (Pi)*(Radius^2)*(Height)
- Volume = Pi*((32000m)^2)*(12000m) = 3.8604e13 m^3
- Density = 1.003 kg/m^3 (Does anyone have the source for this)
- Mass = 3.872e13 kg
Timeframe:
So, I made a sequence of the feat below.
- Using 15 seconds for the cloud split since the majority of said cloud split was shown to completely split in 1 panel. From the first three panels shown above, the cloud was just barely starting to split.
Velocity:
- 128.8km/15 seconds = 8586.67m/s
Since the majority of the cloud split was shown to happen in one panel instead of four, we use 15 seconds for this.
Kinetic Energy[]
- KE = 0.5*(Mass)*(Velocity^2)
- KE = 0.5*(3.872e13 kg)*((8586.67m/s)^2)
- KE = 1.42743e21 Joules, or 341.164 Gigatons TNT. Large Island level
Questions[]
- Did the storm cloud cover the entirety of Ningbo, or just a part of it?
Encore: Raoh Parts a Cloud[]
Credit to CHAOSTHEORY123 for the original calc
Let's do this.
Rocks![]
- Raoh's height = 6'11, or 2.11m = 146.17px
- Rock height = 69.123px = ~1m
So, since the rock is shown to vertically upwards, let's use equations for vertical motion of proectiles. Haven't done this since high school, but let's see how this goes.
- Rock height = Y = 1m
- Vfy (final velocity) = 0m
- acceleration = 9.81m/(s^2)
- Viy (initial velocity) = find this value...
So, we use this equation to find "Viy"
- ((Vfy)^2) = ((Viy)^2) + 2*(accelration)*(Y)
Isolate Viy from above equation.
- Viy = sqrt(0+(2*(9.81m/(s^2))*(1m)) = 4.42945m
Now, we use Viy to find "Time". Using the equation
- Vfy = Viy + ((acceleration)*(Time))
- Time = (Vfy - Viy)/(acceleration)
- Time = ((0 - 4.42945)m/s)/(9.81m/(s^2)) = 0.45153 seconds (I know it shows up as negative but set it as positive since there is no negative time)
Clouds![]
- Hole thickness = 85px = 2000m (Original calc uses nimbostratus clouds for thickness. According to google, they are about 2000m thick.)
- Hole diameter = 217px = 5105.883m
- Hole radius = 5105.883m/2 = ~2553m
- Cloud Radius = 405px = 9529.41m
Volume of cloud[]
- Volume of cylinder = (Pi)*(Cloud Radius^2)*(Height) = (3.1416)*(9529.41^2)*(2000) = 5.706e11 m^3
- Density of cloud = 1.003 kg/m^3
- Mass of cloud = 5.723e11 kg
Velocity[]
Object degree size = 2*atan(tan(70/2)*(Object_Size/Panel_Height))
- Object size = diameter of hole = 217px
- Panel height = 1024px
Object degree size = 2*atan(tan(70/2)*((Object_Size = 217px)/(Panel_Height = 1024px))) = 0.2 rads, or 11.4592 deg
- Angle = 11.4592 degrees
- Object size = Diameter of hole = 5105.883m
- Distance to hole = 25444 m
But then you need to add the thickness of cloud since the blast went all the way through the cloud. This means that...
- Distance Raoh's Blast Traveled = distance to hole + cloud thickness = (25444 + 2000)m = 27444m
Kinetic Energy[]
For Raoh's blast speed, Raoh's blast went all the way through the clouds at the same time the rocks from the ground went up 1m.
- It took about 0.45153 seconds for rock to go up 1m.
- That is the same timeframe for Raoh's blast to go through the clouds. Thus, we do...
- Raoh blast speed = (Distance Raoh's Blast Traveled = 27444m)/(0.45153 seconds) = 60780 m/s; Mach 178.6
Now, for KE of Raoh's cloud, it took the same amount of time for said hole in cloud to form in comparison to the rock moving up 1m.
- It took about 0.45153 seconds for rock to go up 1m.
- That is the same timeframe for hole in cloud to form. Thus, we do...
- Cloud velocity = (Radius of hole = 2553m)/(0.45153 seconds) = 5654.11m/s
- Cloud KE = 0.5*(Mass of cloud = 5.723e11 kg)*((5654.11m/s)^2) = 9.148e18 Joules, or 2.178 Gigatons of TNT.
Final Tally[]
- Kasumi Kenshiro splits cloud = 341.164 Gigatons TNT; Large Island level
- Raoh splits cloud = 2.178 Gigatons of TNT; Small Island level
- Raoh blast speed = Mach 178.6
That is about it.