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The Sonic Rainboom: Here We Go Again Edition
Was re-watching the Sonic Rainboom, and noticed how it generated winds at Pinkie Pie's house. This is a tad different from the previous method, this one is more along the lines of kinetic energy.
Spoiler: Assumptions Made
- Because the Rainboom is in the shape of a circle, I will assume the air pushed is in the shape of a cylindrical sector, with a length equal to my original calc of 965.6064 km.
- The Rainboom began moving clouds at 00.19.55 seconds in this video and the sky is at it's brightest at 00.24.48 seconds which means 4.93 seconds.
- The Storm at Pinkie Pie's house is 20 km in radius, same as the horizon on a clear day. Watching the video again shows that the sky was near-completely cleared by Rainboom, yet the Storm returns the next day, covering the horizon, presumably the same size as it's predecessor. I've also asked a few knowledgeable people on the subject, and they say the the entire sky covered with clouds to the horizon won't alter visibility unless there are things like mist or fog obscuring the horizon.
- The height of the air affected is equal to the height of the clouds (I mean, there's no way it couldn't if it was affecting clouds, and the air at ground level). They look pretty thick, so I guess nimbostratus clouds?
- All the air between Pinkie Pie's house, and the epicenter of the Rainboom was pushed (would not make sense for it to affect only the air at Pinkie Pie's house, while somehow not touching the air between the two points.
First things first: I have the height and radius of the cylinder right? Well since this is a cylindrical sector calculator, I need to find the angle of the cylinder in degrees. I don't know the angle, what I do know is google can convert radians to degrees. I've been asking around, and it turns out I can find the rads by dividing the length of the arc of the sector by the radius of the circle. if the radius of the storm is 20 km, and goes out behind Pinkie for that same length, then that gives me an Arc length of 40 kilometers.
40000 / 965606.4 = 0.04142474614 rads
Converted to degrees = 2.373463121228736
When inserting the information into the cylindrical sector calculator, I get a volume of 38624255991559.18 meters cubed
Air has an average density of 1.225 kg/m3 (I have been informed that being 2 kilometers high will not change the density very much).
38624255991559.18 * 1.225 = 4.7314714e+13 kilograms
As noted in the assumptions section, I am assuming that the Rainboom began at 00.19.55 seconds, as that is when it interacts with the clouds. 40000 meters in 4.93 seconds is 8113.59026 m/s.
0.5 * 4.7314714e+13 * 8113.59026^2 = 1.55737202e21 joules or 372.220846 Gigatons
But we aren't done yet. This was only a small portion of the Rainboom. Shiba once did a surface area calc for the Rainboom that went something like this: ((2*pi*R)/7000 I do have a method of my own where I divide the surface area of the sector by the surface area of the entire Rainboom as a cylinder, but I'm not sure how accurate that would be, so I'll give out 2 ends for this based on both methods.
Spoiler: High End((2*pi*965606.4)/7000 = 866.726277857
866.726277857 * 372.220846 GT = 322613.588394 GigatonsSpoiler: Low EndThe cylindrical sector calculator gives me a surface area of 42566681591.542 square meters
Surface Area of a cylinder is 2*pi*r*h+2*pi*r^2 2*π*965606.4*2000+2*π*965606.4^2 = 5.8705493e+12 square meters
5.8705493e+12 / 42566681591.542 = 137.914187353
372.220846 * 137.914187353 = 51334.5354919 Gigatons
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