- 31,625
- 28,158
While reviewing MCU profiles, I came across this calulcation for the Power Stone's Attack Speed and its Attack Potency.
Context
The gist of the feat and its calc is this; Eson the Searcher uses the Power Stone to surface-wipe an alien world in about 7.28 seconds.
How the calc is interpreting this is that the mass of the world's crust (2.77*10^22 kg) + the mass of the atmosphere combined (5.15*10^18 kg) for a total of 2.770515e+22 kg is moving omnidirectionally a distance of 40,075,000 m, which is the circumference of the world (assuming Earth-like conditions), in that timeframe.
Moving that distance 40,075,000 m in 7.28 seconds gets 5,504,807.69231 m/s or 1.836% Speed of Light.
So using the Omni-directional KE formula; we get (1/12) * 2.770515e+22 * 5504807.69231^2.
Issues
The primary issue I noticed with this calculation is that the distance being used for the K.E. calculation is the full circumference of the world.
Here is the feat in question shown from about 38 seconds onwards to around 46 seconds.
For the mass in question to travel around the circumference of the planet in that timeframe, that'd mean some of it at say the epicenter would have to circle around the entire planet and end up back where it started by the time the feat is over... But that isn't what we see happening at all.
We see the destructions spreading; giant purple fissure lines, the surface being destroyed, the clouds being pushed around... But where is it being shown here that the entire mass of the crust of the planet travelled all the way around the circumference of the planet? Why would mass in the crust on the opposite side of the world to the explosion be moved around that far too?
Moreover, we can see for ourselves from the movement of the clouds in the span of that timeframe that the atmosphere isn't being pushed around that far; just watch the feat and pay attention to how the shockwave of destruction spreads and affects the movement of the clouds. We can see for ourselves that they don't move that far, so why would we assume in the calculation a distance value they travel in the timeframe of the feat?
Also, while we can see the end results of the feat is that the planet's surface is unquestionably destroyed with the seas and greenery wiped out; why does this mean that the entire crust of the planet was blasted away and moved? The crust goes up to 30 kilometers deep on average; you don't need that much to have been destroyed for the surface of the planet to be unrecognizable. If we look at the first few seconds of the feat, from 38 seconds to 40 seconds, the initial wave of destruction is quite shallow around Eson the Searcher. Debris gets kicked up and moved around; but not vast quantities.
I think that the method used for the calc is flawed and that it should be removed.
@Therefir @Spinosaurus75DinosaurFan @DemonGodMitchAubin @Dark-Carioca @AbaddonTheDisappointment As you all took a look at the original calc, I'm hoping that you could leave an evaluation on this thread when you have time.
Context
The gist of the feat and its calc is this; Eson the Searcher uses the Power Stone to surface-wipe an alien world in about 7.28 seconds.
How the calc is interpreting this is that the mass of the world's crust (2.77*10^22 kg) + the mass of the atmosphere combined (5.15*10^18 kg) for a total of 2.770515e+22 kg is moving omnidirectionally a distance of 40,075,000 m, which is the circumference of the world (assuming Earth-like conditions), in that timeframe.
Moving that distance 40,075,000 m in 7.28 seconds gets 5,504,807.69231 m/s or 1.836% Speed of Light.
So using the Omni-directional KE formula; we get (1/12) * 2.770515e+22 * 5504807.69231^2.
Issues
The primary issue I noticed with this calculation is that the distance being used for the K.E. calculation is the full circumference of the world.
Here is the feat in question shown from about 38 seconds onwards to around 46 seconds.
For the mass in question to travel around the circumference of the planet in that timeframe, that'd mean some of it at say the epicenter would have to circle around the entire planet and end up back where it started by the time the feat is over... But that isn't what we see happening at all.
We see the destructions spreading; giant purple fissure lines, the surface being destroyed, the clouds being pushed around... But where is it being shown here that the entire mass of the crust of the planet travelled all the way around the circumference of the planet? Why would mass in the crust on the opposite side of the world to the explosion be moved around that far too?
Moreover, we can see for ourselves from the movement of the clouds in the span of that timeframe that the atmosphere isn't being pushed around that far; just watch the feat and pay attention to how the shockwave of destruction spreads and affects the movement of the clouds. We can see for ourselves that they don't move that far, so why would we assume in the calculation a distance value they travel in the timeframe of the feat?
Also, while we can see the end results of the feat is that the planet's surface is unquestionably destroyed with the seas and greenery wiped out; why does this mean that the entire crust of the planet was blasted away and moved? The crust goes up to 30 kilometers deep on average; you don't need that much to have been destroyed for the surface of the planet to be unrecognizable. If we look at the first few seconds of the feat, from 38 seconds to 40 seconds, the initial wave of destruction is quite shallow around Eson the Searcher. Debris gets kicked up and moved around; but not vast quantities.
I think that the method used for the calc is flawed and that it should be removed.
@Therefir @Spinosaurus75DinosaurFan @DemonGodMitchAubin @Dark-Carioca @AbaddonTheDisappointment As you all took a look at the original calc, I'm hoping that you could leave an evaluation on this thread when you have time.
