Storm Calculation Issues

[Flashlight237]

Okay, so I've been all "WAIT A MINUTE!!" for the past hour over this. While this wasn't the start of the issues for me, personally, it is at the start of the page, so I'll start there. So, first of all, we currently have this.:

"First we need the radius of the cloud. That can be measured using pixel scaling, but for the average cumulonimbus cloud we will be using 20 kilometers or 20000 meters, as that's the viewing distance in clear days."

While I can understand the use of 20 km, as that is based on the lack of available knowledge at the time, the keyword here is "average cumulonimbus cloud". The average cumulonimbus cloud (which we equate to thunderstorms) is 15 miles (24.14016 km) in diameter, which equates to a far smaller radius of 12.07008 km. This figure was initially found on Wikipedia (https://en.wikipedia.org/wiki/Cumulonimbus_cloud ) with a citation from the NOAA (https://web.archive.org/web/20090825000832/http://www.nssl.noaa.gov/primer/tstorm/tst_basics.html ), but the figure can be found in this PDF by the NOAA as well: https://www.weather.gov/media/grr/brochures/nwsthunderstorms&lightning.pdf

This would mean our overall figures for thunderstorms 36.4217078% what they are now. There's also the fact that a thunderstorm is in no way in hell an indicator of a clear day (and I'm dying on the hill that it ain't!), that's an entirely different revision that I'm not willing to get into.

The second thing I'll get into is our handling of hurricanes. How the frick is creating a hurricane with a 300-mile (482.8032 kilometers) diameter (https://www.weather.gov/source/zhu/...tuff/hurricane_anatomy/hurricane_anatomy.html ) that lasts for weeks on end weaker than creating a thunderstorm with a 15-mile diameter that lasts for 30-60 minutes depending on the documentation? According to the page itself, it has to do with wattage. We usually just roll with one singular second's worth of wattage, but for a thunderstorm, it seems we basically just took the total energy value and called it a day. That's dissonance if I ever saw it.

As I said before, an average thunderstorm lasts 30 (https://www.weather.gov/media/grr/brochures/nwsthunderstorms&lightning.pdf ) to 60 (https://www.nssl.noaa.gov/education/svrwx101/thunderstorms/types/ ) minutes depending on the documentation. For the highest storm on the wiki (8000 j/kg with a 11.8 km tall cloud), the wattage would be about this much.:

5.23*10^16/1800=2.90555555*10^13 watts
5.23*10^16/3600=1.452777777*10^13 watts

This equates to 3472.222222 to 6944.444444 tons of TNT, which is Low 7-C+ to 7-C. This is before the suggested value nerf, which would make these values Low 7-C to Low 7-C+.

For comparison, Hurricane Katrina let out a total energy output of 6*10^19 calories, or 2.5104*10^20 joules: http://www.atmo.arizona.edu/students/courselinks/spring07/atmo336s3/lectures/sec2/hurricanes4.html

Hurricane Katrina lasted 8 days (with no minute times of start and end shown), which is 691200 seconds: https://en.wikipedia.org/wiki/Hurricane_Katrina

This equates to the following average wattage:

2.5104*10^20/691400=3.631944444*10^14 watts

This equates to 86805.55556 tons of TNT, which is 7-C+.

If we used the same logic we did for thunderstorms, creating a hurricane on the same magnitude as Katrina would've been 6-C+ as the value would've been worth 60 gigatons of TNT.

Even then, using the bare minimum of 10^15 joules (https://en.wikipedia.org/wiki/Thunderstorm ), we'd be getting a wattage of 2.7777 to 5.5555*10^11 watts, or 66.39048226 to 132.7809645 tons of TNT, which is a measly 8-B+ to 8-A. I can't find any figures for how many lightning strikes a thunderstorm produces on average, so I don't really have anything to compare the figures to. Then again, you can say a 360 kJ jolt of electricity can power a 100-watt light bulb for an hour and it'd be rated at 100 joules on this site instead, so I dunno.

So yeah, it seems we got a lotta things to work on in this regard. It's sitewide, but at the same time, if I put this in Staff Discussion, the first person to type in this thread would be like "This should go in Calc Group Discussion" or some nonsense like that, so I'm putting it here. You have some serious checking to do, peeps.

 

[Arkenis]

Not understanding all of this but from what I can tell this is about creating and sustaining the clouds? I have a feat here where I used the 20km so wondering if this crt would effect it

 

[DemonGodMitchAubin]

Ugh... Here we go again with Cloud Calcs being consistently changed, making them the most unreliable calcs ever

 

[CloverDragon03]

Okay, so I've been all "WAIT A MINUTE!!" for the past hour over this. While this wasn't the start of the issues for me, personally, it is at the start of the page, so I'll start there. So, first of all, we currently have this.:

"First we need the radius of the cloud. That can be measured using pixel scaling, but for the average cumulonimbus cloud we will be using 20 kilometers or 20000 meters, as that's the viewing distance in clear days."

While I can understand the use of 20 km, as that is based on the lack of available knowledge at the time, the keyword here is "average cumulonimbus cloud". The average cumulonimbus cloud (which we equate to thunderstorms) is 15 miles (24.14016 km) in diameter, which equates to a far smaller radius of 12.07008 km. This figure was initially found on Wikipedia (https://en.wikipedia.org/wiki/Cumulonimbus_cloud ) with a citation from the NOAA (https://web.archive.org/web/20090825000832/http://www.nssl.noaa.gov/primer/tstorm/tst_basics.html ), but the figure can be found in this PDF by the NOAA as well: https://www.weather.gov/media/grr/brochures/nwsthunderstorms&lightning.pdf

This would mean our overall figures for thunderstorms 36.4217078% what they are now. There's also the fact that a thunderstorm is in no way in hell an indicator of a clear day (and I'm dying on the hill that it ain't!), that's an entirely different revision that I'm not willing to get into.

This is misunderstanding why we use the 20 km figure. We use that for clouds that cover all the way to the horizon, not because that's the actual average diameter of a cumulonimbus cloud, so this whole "average cumulonimbus cloud" thing is not only kinda meaningless, but creates an unnecessary amount of work for the benefit it provides

 

[Flashlight237]

This is misunderstanding why we use the 20 km figure. We use that for clouds that cover all the way to the horizon, not because that's the actual average diameter of a cumulonimbus cloud, so this whole "average cumulonimbus cloud" thing is not only kinda meaningless, but creates an unnecessary amount of work for the benefit it provides

It's literally in the flipping article. If you don't like the idea, kindly remove that terminology or, like I said, change it to actually fit the definition so the wiki becomes less of a laughingstock. And, no, accuracy is NOT "meaningless" or "unnecessary", and I'm sick of people claiming such because they're too lazy to even acknowledge reliable sources.

 

[CloverDragon03]

And with that attitude you're getting absolutely nowhere so preferably tone it down. Cool? Cool

 

[CloverDragon03]

As for the arguments themselves, we find cloud width through the following:

• If the cloud extends to the horizon, we use horizon distance as the radius
• If not, chances are we can directly measure the cloud via pixel scaling or some other method

Only if neither of the above are true would there be a need for an "average" figure

 

[M3X_2.0]

I can agree with using the 24km figure if the thunderstorm is completely shown, otherwise, on cases where it goes beyond horizon distance, just try to calc it using the pov height or lowballing to 20km.

 

[CloverDragon03]

I can agree with using the 24km figure if the thunderstorm is completely shown, otherwise, on cases where it goes beyond horizon distance, just try to calc it using the pov height or lowballing to 20km.

Yeah, this makes the most sense to me. I don't think it should be a replacement in the instance of clouds that go past the horizon, but rather, something that should be used in the absence of any other potentially reliable figure

 

[Dalesean027]

I can agree with using the 24km figure if the thunderstorm is completely shown, otherwise, on cases where it goes beyond horizon distance, just try to calc it using the pov height or lowballing to 20km.

@Flashlight237 you never gave your thoughts here, is this fine by you?