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I don't even know where to start with these formulas. Even instability is going to get a sizeable boost. So AGK is likely to go into outlier territory.
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The Problem with Storm/Clouds Calculations
- Thread starter Ugarik
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@Versus
From what I understood, you just have to find distance to the horizon from an observer at cloud height, add it to the distance to the horizon that was originally used and use the new value as radius of storm.
From what I understood, you just have to find distance to the horizon from an observer at cloud height, add it to the distance to the horizon that was originally used and use the new value as radius of storm.
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- #165
Basicly the method remains the same as before. The only thing that is diffirient is that now we need to combine distance to the horizon from observer's PoV and distance to the horizon from clouds' altitude. (Its been decided to use volume of a cylinder just as before)DontTalkDT said:
From what I can tell Mr. Bambu doesn't seem to understand this method. Kaltias has no problem with using this. The rest of the calc group ignored this thread.
The only thing left to do is to define what clouds altitude should we use (Average vs Lowest)
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Given that Mr. Bambu considers this method ill-defined and hard to understand, and that it would likely lead to inflated results, I would prefer considerable evaluations and confirmations from DontTalkDT and the calc group members before we change our standards.
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I do not understand it, that is true. I will point out that... you seem to have shifted the method in question.
In the OP, we have this: S = Sqrt((R+h)^2 - R^2)
Later, you use 14700+sqrt((6371000+460)^2-6371000^2)
So. What changed? 14700 was the given distance to the horizon, why in the world are we adding it to the equation for Distance to the Horizon? Just an example of a thing I do not understand about what is being discussed. I don't claim to be the brightest crayon in the box, don't get me wrong, which is why I generally assume those around me are more capable. But what changed between post one and post two?
In the OP, we have this: S = Sqrt((R+h)^2 - R^2)
Later, you use 14700+sqrt((6371000+460)^2-6371000^2)
So. What changed? 14700 was the given distance to the horizon, why in the world are we adding it to the equation for Distance to the Horizon? Just an example of a thing I do not understand about what is being discussed. I don't claim to be the brightest crayon in the box, don't get me wrong, which is why I generally assume those around me are more capable. But what changed between post one and post two?
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Yes, this only affects horizon calced storms.
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S = Sqrt((R+h)^2 - R^2) is the hormula to determine distance to the horizon. You can still use the old distance to the horizon calculator. Basicly you need to add distance to the horizon from observer height to distance to the horizon from clouds altitude.Mr. Bambu said:
Distance to the horizon was calculated to be 14700 m. Now we need to add to that value distance to the horizon from clouds altitude (I assumed 460 m, which is the lowest crombonimbus altitude) using either formula above or this calc. I've explained why we need distance to the horizon from clouds altitude.
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t makes 90% of them unusable due to leading to outliers, so I doubt it would have been good if it did that.
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Your first equation begins with S =, where S is distance to the horizon. I understand that bit.
But you then later calculate it, with 14700 as the distance to the horizon, but you're adding it. It isn't S = anymore, it has shifted to being S (AKA 14700) + the rest of the calculation. I was seeking elaboration on why this was done.
But you then later calculate it, with 14700 as the distance to the horizon, but you're adding it. It isn't S = anymore, it has shifted to being S (AKA 14700) + the rest of the calculation. I was seeking elaboration on why this was done.
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Okay, finaly I got your question
So it was done because horizon doesn't hide clouds below itself since they are high up in the sky.
For example if clouds altitude is 460 meters and storm raduis = horizon distance = 5 km. Using angular size we can drow how the horizon would look like. atan(460/5000) = 5.26 degrees
5.26 degrees is a lot. Use can fit almost 11 suns beteeen cluds layer and the horizon. That's why we can not use just distance to the horizon alone. Even as a lowball
So it was done because horizon doesn't hide clouds below itself since they are high up in the sky.
For example if clouds altitude is 460 meters and storm raduis = horizon distance = 5 km. Using angular size we can drow how the horizon would look like. atan(460/5000) = 5.26 degrees
5.26 degrees is a lot. Use can fit almost 11 suns beteeen cluds layer and the horizon. That's why we can not use just distance to the horizon alone. Even as a lowball
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Majority seems to agree with 2000, and for what it's worth so do I.
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Agreement seems to be average of the lowest and highest point of an specific type of cloud.
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> Island level All Might is back
After recalculating All Might's feat, I realized that the results are almost exactly 1000 times higher than before, holy shit, though I probably ****** up something...
After recalculating All Might's feat, I realized that the results are almost exactly 1000 times higher than before, holy shit, though I probably ****** up something...
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Dude
If the multiplier still existed
And AMs win record is actually even now
If the multiplier still existed
And AMs win record is actually even now
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Isn't it like, a massive outlier?
I get that he is the strongest guy around, but storms are so much inflatated with this that you might use energy to mass conversion for feats.
I get that he is the strongest guy around, but storms are so much inflatated with this that you might use energy to mass conversion for feats.
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I'm not a good calcer but I'm leaning towards this.Antvasima said:
Mr. Bambu
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Storms will be 100% unusable in most cases. There are instances where this won't be the case, but often times this will cause them to be unusable. So. Prepare for that.Ricsi-viragosi said:
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Also I find the notions that the clouds would continue moving in a straight line until beyond the horizon rather than bending along with the curvature of the Earth (Which is what the OP is suggesting) utterly ridiculous.
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If this makes 99% of Storm Calcs unusable we shouldn't go with this. I trust DontTalk. I also find that the OP makes some pretty weird assumptions in his post. Such as with clouds that ignore gravity and move into space on a straight line.
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It making stuff outlier has no bearing on the fact that it's what should be used.
We can't low-ball feats just enough for them to not be outliers.
We can't low-ball feats just enough for them to not be outliers.
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The fact that it invalidates storm calcs doesn't matter much tbh. If we're going for accuracy, that fact shouldn't be taken into consideration- at all. Though I do agree that clouds should logically conform to the circular shape of the Earth.
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- #197
Btw I've already explained why KE should never be used anymore
So it's better to use regular clouds destruction method
Clouds volume = 683824252553440 m^3
Mass of the water inside them: 683824252553440*0.0003 = 205147275766 kg
Heat of vaporization - 2264705 J/kg
Total energy - 205147275766×2264705 = 4.6459806×10^17 J - 110.04 Megatons (Mountain level)
So it's better to use regular clouds destruction method
Clouds volume = 683824252553440 m^3
Mass of the water inside them: 683824252553440*0.0003 = 205147275766 kg
Heat of vaporization - 2264705 J/kg
Total energy - 205147275766×2264705 = 4.6459806×10^17 J - 110.04 Megatons (Mountain level)
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Again Ugarik, the result should not matter, what is logically and mathematically correct should. Moving clouds should be KE.
I also find assuming the destruction assumption to be... out of nowhere?
I also find assuming the destruction assumption to be... out of nowhere?
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That's purely arbitrary.
Also, I am suddenly reminded of a very bad old OBD calc about this.
See here for Endless Mike's explanations of why you shouldn't make silly assumptions like this with storms that go all the way to the horizon.
" Ignoring the first two issues, the storm wouldn't have to stretch to the farthest visible horizon to appear to cover it, as long as it took the form of a dome that covered the entire field of view of someone on the ground. Human eyes aren't good at judging distance of incredibly large things like storms that are far away. A crude illustration.
Also, I am suddenly reminded of a very bad old OBD calc about this.
See here for Endless Mike's explanations of why you shouldn't make silly assumptions like this with storms that go all the way to the horizon.
" Ignoring the first two issues, the storm wouldn't have to stretch to the farthest visible horizon to appear to cover it, as long as it took the form of a dome that covered the entire field of view of someone on the ground. Human eyes aren't good at judging distance of incredibly large things like storms that are far away. A crude illustration.
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