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An extra vote, also you wanted DT to clarify his stance but he hasn't commented yet.What do we currently need to do here?![]()
You got a better option?Hmm. The One Piece planet should logically be far larger than Jupiter, so that seems like a massive understatement.![]()
Do you have prove or reason for this thinking because difference between earth and jupiter is very significance. I always thought one piece planet was slightly bigger than earth but not the levels of Jupiter.Hmm. The One Piece planet should logically be far larger than Jupiter, so that seems like a massive understatement.![]()
One Piece planet is accepted as being 1.4 Million km in diameterDo you have prove or reason for this thinking because difference between earth and jupiter is very significance. I always thought one piece planet was slightly bigger than earth but not the levels of Jupiter.
so about same diameter as sunOne Piece planet is accepted as being 1.4 Million km in diameter
It's nearly twice the diameter of the Sun.so about same diameter as sun, ok thanks.
well but nasa website is saying 1.4 million km approximately here.It's nearly twice the diameter of the Sun.
I'm hoping that Oda provides us with a concrete figure one day.
Ah, I was looking at the wrong figure, thanks.well but nasa website is saying 1.4 million km approximately here.
Yeah, the current calc makes assumptions about the Grandline and Alabasta that ultimately makes the planet size a lot smaller than it actually is, I can't wait for Oda to finally confirm a Solar System sized Blue Planet.I'm hoping that Oda provides us with a concrete figure one day.
The situation hasn't changed since this previous reply:What do we currently need to do here?![]()
An extra vote, also you wanted DT to clarify his stance but he hasn't commented yet.
More opinions are needed on how to calculate the feat for a different planet not for that partI’m not a staff member, but why are more opinions needed? Both linked calculations added 6.399 to a value that was already in magnitude form, when it was supposed to be a compensation for a value that isn’t. This led the result to be far higher than it should be.
So it’s likely a typo or a conversion error. As the OP suggested, you just need to fix that.
Yeah that was initially proposed (those e21 and e22 results) but then proposal got changed into "earthquake formula is designed to work specifically on Earth".I’m not a staff member, but why are more opinions needed? Both linked calculations added 6.399 to a value that was already in magnitude form, when it was supposed to be a compensation for a value that isn’t. This led the result to be far higher than it should be.
So it’s likely a typo or a conversion error. As the OP suggested, you just need to fix that.
If we already use 10^4.8 one for different planets then I see no problem with that.@Floxy178 and @KLOL506 I got a question for yall
The earthquake formula for Total Seismic Energy that the calculators and others use is
10^(1.5×Richter Magnitude)×10^9.091
It stops at 10.5 for earth because earth can't take it
If a planet is bigger, can we just use that instead for total seismic energy?
I don't think so. Earthquake magnitude is capped by amount and length of fault lines (limit is around 10) and strength of rocks ( around 9.6). Strongest earthquake is Chilean 1960(magnitude 9.5), and it released 25% of all seismic energy in 20century.If a planet is bigger, can we just use that instead for total seismic energy?
That is wrong. Not the source but the relevance. That 9.6 figure is talking about energy storage ability of rocks from natural tectonic movements. But here we are talking about an external source of energy inserting energy to tectonic plates. A real life example would be if a massive meteorite or some moon/dwarf planet were to collide with the earth, it would cause an earthquake much bigger than 9.6 and destroy most of the earth. External energy is not limited by rock's strength.I don't think so. Earthquake magnitude is capped by amount and length of fault lines (limit is around 10) and strength of rocks ( around 9.6). Strongest earthquake is Chilean 1960(magnitude 9.5), and it released 25% of all seismic energy in 20century.
Larger planets are not beholden to first cap, but still are limited by the second (unless directly stated otherwise some way or another).
Come to think of it, we should prohibit use of this formula for anything higher than 9.5 quake
Isn't total seismic energy only used for natural earthquakes?That is wrong. Not the source but the relevance. That 9.6 figure is talking about energy storage ability of rocks from natural tectonic movements. But here we are talking about an external source of energy inserting energy to tectonic plates. A real life example would be if a massive meteorite or some moon/dwarf planet were to collide with the earth, it would cause an earthquake much bigger than 9.6 and destroy most of the earth. External energy is not limited by rock's strength.
Earthquakes caused by tectonic plate shaking*Isn't total seismic energy only used for natural earthquakes?
Nay, formulas does not differentiate between artificial and natural. One is for just seismic waves, other is total work. And the meteor one is for converting kinetic energy into seismic energy. It does not work here.Isn't total seismic energy only used for natural earthquakes?
Earthquakes caused by meteor impacts and artificial earthquakes use other formulas, the formula for meteors which is "10^((Richter Magnitude + 5.87)/0.67)" for example calculates the kinetic energy of the meteor from the richter magnitude of the earthquake caused by the impact event (View page 11 for the source), while other artificial earthquakes use the formula for radiated waves "10^(1.5*(Richter Magnitude)+4.8)" rather than total seismic energy.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^?????@Floxy178 and @KLOL506 I got a question for yall
The earthquake formula for Total Seismic Energy that the calculators and others use is
10^(1.5×Richter Magnitude)×10^9.091
It stops at 10.5 for earth because earth can't take it
If a planet is bigger, can we just use that instead for total seismic energy?
Why are we using total seismic energy for earthquakes created by punching or emanating energy here?@Floxy178 and @KLOL506 I got a question for yall
The earthquake formula for Total Seismic Energy that the calculators and others use is
10^(1.5×Richter Magnitude)×10^9.091
It stops at 10.5 for earth because earth can't take it
If a planet is bigger, can we just use that instead for total seismic energy?
I mean, the magnitude ultimately doesn't matter. You can't use total seismic energy for artificial quakes, period.That's correct, but if I'm not mistaken what KT asked is more like a general question "is 10.5+ magnitude disqualifier for total seismic when planet is bigger?" rather than a specific feat.
He didn't say anything about earthquake in question being artificial though. Seems more like a general question to me.I mean, the magnitude ultimately doesn't matter. You can't use total seismic energy for artificial quakes, period.