Seiken Tsukai no World Break, freezing a country
So since we apparently don’t know any details about this country we can only calculate what was shown in the scene.
The angular size of a single house at the furthest distance in the frame is 0.594 degrees and the height should be about 20 feet (https://www.pipsisland.com/how-tall-is-a-2-story-house/#:~:text=A%20standard%202%2Dstory%20building,then%209%20to%2010.5%20meters.), so the distance (or radius of the circular sector) is 588 metres. Height of the atmosphere is 10,000,000 metres (https://education.nationalgeographic.org/resource/atmosphere/) so the total volume of air being affected is 0.5x588^2x7/18pix10,000,000, which is 2.11x10^12 m^3. Since the guy who cast the spell said not a single blade of grass would grow again we can assume…
Seiken Tsukai no World Break, Edward blitzes his surroundings
So despite the frame being kind of paused from the camera’s perspective Edward still seems too fast for it to keep up. We’ll calculate that.
Edward is about 1.77m tall and his angular size in that frame is 63 degrees, so he should be 1.44 m away using the angsizing calculator.
Apparent speed= 1.44/0.01= 144 m/s
Actual speed= 144 x (5/0.001)= 720000 m/s
Dude cuts through a cloud.
Guy from a thing Creature likes made a cut in the sky.
The cut has an angular size of 65.3 degrees (or more but whatever). Let’s assume that the clouds are nimbostratus, so the distance should be 1828.8m. The length of the cut is therefore 2.34km and the width would be 59.5 m. Since he was portrayed as moving very quickly, we can assume the time period to be 1 second. Density should be 2.29 kg/m^3.
Instead of the cloud being displaced along its width and thickness, as displacing the clouds that way would lead to the cut being surrounded by the cloud (therefore not allowing it to be visible), it instead must have been displaced along its width only. In which case, the equation on the calculations page is not valid for this feat and we must de…
Fire Punch Space Tree Calculation
First I’ll find the mass of the tree by considering how it manages to use its gravitational attraction to counter tidal forces. Since it manages to stay completely undeformed I will consider both the centrifugal tidal forces and the attractive tidal forces together. Since the mass of a hypothetical object on the tree or of the tree would be in both sides of the equation it doesn't require consideration.
L will be the length of the tree, R will be the earth’s radius+ half the length of the tree
Radius of earth= 6371000
Length of tree= 6317000x 233/ 96.5= 15300000
Mass of earth= 5.97 x10^24 kg
The result is 2.01 x10^24 kg
Lol.
So from this point you would usually substitute the mass into the GPE equation to find how much energy was needed to bring…