Overview and Feat
In Chaotic, a character known as Aa'une unleashes an attack that causes the sky above to churn into a storm. So, what is the yield for this? I will be using two methods to get a result for this feat. Feat for reference.
The Calculation: Intial Setup
Step 1 of this is finding the area of that storm. To get that I need the diameter of the beam. To get that I need the height of Aa'une. To get that I need the height of Maxxor; luckily I actually have this and don't have to go any deeper down this rabbit hole... he is 9'4", or 284.48 cm.







Sadly he is contorted from the blast. That is easily fixed, though, as the height of his bubble should be at least his full height. His height in blue is 92 pixels. Given is aforementioned real height, we find each pixel is 3.092 cm. Aa'une is 424 pixels in red, or 1311.008 cm tall. To put that into perspective that means he is slightly more than 37 feet tall. Now that we have his height we can find his beam's diameter.







Here we see that Aa'une is 50 pixels in red, and given his real height we find that each pixel is 26.22 cm. The beam's diameter is 23 pixels in blue, or 603.06 cm.






Now we see that the beam is only 9 pixels in blue. Knowing its real diameter we know that each pixel is 67.007 cm at the center of the storm. The storm isn't perfectly lined up with its center, so we need to find the hypotenuse in order to get a true, accurate radius. Triangle length is 688 pixels in aqua. Triangle height is 83 pixels in green. True radius is 692.99 pixels in that case, or 464.3518 meters.
Method 1: Storm Persistent Energy
One option is to assume that he has made these clouds into a true storm. This value entirely hinges on this assumption! To find energy of this, we know that 665 square kilometers gives off 6x10^14 watts (joules per second). When inputting our radius into a circular area formula, we find that our storm is 6.77x10^5 square meters, or 0.677 square kilometers for our purposes. Now we can find the wattage of a storm of this size.
(0.677/665)x(6x10^14)= 6.202x10^11 joules
Method 2: Rotational Kinetic Energy
The second option is to find the kinetic energy the clouds hold that are being rotated. A couple things I should address before getting into the math.
1. The clouds all move at different speeds, but to compensate for this I need to take the average of the clouds.
2. I can't use half of the radius, however, I need to use the radius that is the true middle, the radius that halves the area. This radius is r= 328.3 meters.
3. These clouds are odd, and I don't know the thickness. I'll use 200 meters as a lowball, as even fog is this thick, and 13000 meters as the highball, as this is the thickness of true storm clouds.
To find rotation speed, we need to use the formula V= (2xpixR)/T, when V is speed in meters per second, R is radius in meters, and T is time of a single rotation in seconds. Well, we don't see a full rotation. In fact, we see 1/8th of a full rotation over the course of 2 seconds. This indicates that 1/16th a rotation occurs each seconds, meaning our T value is 16. Plugging in our values gives us a speed of 129.12 meters per second. This is the mean speed of the clouds.
Now to find mass. Once again I'll use 200m and 13000m (13km) as my thickness values.
200m volume: 1.36x10^8 cubic meters x 1.003 = 1.364x10^8 kg. Inputting for KE gives us 1.137x10^12 joules.
13km volume: 8.81x10^9 cubic meters x 1.003 = 8.84x10^9 kg. Inputting for KE gives us 7.369x10^13 joules.
Final Tally
Persistent True Storm Energy:6.202x10^11 joules, 8A, MultiCity Block level
Rotational KE 200m: 1.137x10^12 joules, 8A, MultiCity Block level
Rotational KE 13km: 7.369x10^13 joules, 7C, Town level