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Cumulonimbus cloud elevation

ElJoaki5

ElJoaki5

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The cloud calculations page says that the elevation of the base of a Cumulonimbus cloud ranges from 609.6m to 16000m and that 8304.8m is the average. This is a value I’ve seen used in some calcs.

Looking into it this wouldn’t be accurate as the 16000 m is actually referring to the top of the cloud, not the base, and the 8304.8 m would absolutely not be average.

The elevation of the base would actually be much lower. The page already notes that 4000m is the top of the usual height range. This is something that many sources back up. Cumulonimbus are actually categorized as low-level clouds.
So yeah we should fix this calf tissue and change the height range and average height. The usual height range could be 335.28 m up to 1981.2 m based on this source. The "average" would then be 1158.24 m. Which would be backed up by other sources considering >2000m something uncommon.
 
Clouds ain't my realm homie, should've tagged DontTalkDT.

Always tag DT for Calc Policy Page threads.
 
If the sources are accurate then this seems like something worth fixing in our standards.
 
This is somewhat unrelated (maybe?) but I'd like to ask about this site

I've been recommended it to use for cloud calcs and I've seen many calc members use it. What I wanna know is, why are the results between the classic calcs and this calc different?

before it was as such. Let's say we have a Nimbostratus with a 1000x1000 area and 1000m thickness and the height above ground is the low end of 609.6m.
Going by classics, we get 1000000000m^3 and since the cloud density for that is 1 g/m^3, we get a result of 1000000kg.
Putting the same exact numbers into the calculator we get 1100067569.5 kilograms. How? That is a difference of over a thousand, I'd like to know what makes the results this different from the old style and what the formula is. I may be misunderstanding something but if so then please explain it.
 
Arceus0x said:
.
Click to expand...
The 1 g/m3 is referring specifically to the water content of the cloud. Clouds dont have a density of 1 g/m3, just air already has a density of 1225 g/m3 roughly.
The calculator calculates the mass of the air forming the cloud, accounting for differences in air density as elevation increases.
For more info.
 
ElJoaki5 said:
The 1 g/m3 is referring specifically to the water content of the cloud. Clouds dont have a density of 1 g/m3, just air already has a density of 1225 g/cm3 roughly.
Click to expand...
it's 1.225 kg/m^3 which is like 0.001255 g/cm^3.
I get your point though, we never did account for the air, which is also being affected, probably the most affected.
 
Arceus0x said:
it's 1.225 kg/m^3 which is like 0.001255 g/cm^3.
I get your point though, we never did account for the air, which is also being affected, probably the most affected.
Click to expand...
I meant m3 lol. The water is 1 g/m3 and air is 1225 g/m3 roughly
 
ElJoaki5 said:
Tagging randoms cgms 👉👈
@KLOL506 @M3X_2.0 @Damage3245 @TheRustyOne @Floxy178
I sweaar its simple
Click to expand...
I'll look at it later today.
Arceus0x said:
This is somewhat unrelated (maybe?) but I'd like to ask about this site

I've been recommended it to use for cloud calcs and I've seen many calc members use it. What I wanna know is, why are the results between the classic calcs and this calc different?

before it was as such. Let's say we have a Nimbostratus with a 1000x1000 area and 1000m thickness and the height above ground is the low end of 609.6m.
Going by classics, we get 1000000000m^3 and since the cloud density for that is 1 g/m^3, we get a result of 1000000kg.
Putting the same exact numbers into the calculator we get 1100067569.5 kilograms. How? That is a difference of over a thousand, I'd like to know what makes the results this different from the old style and what the formula is. I may be misunderstanding something but if so then please explain it.
Click to expand...
What the calculator does is basically dividing the atmosphere into 7 zones from sea level up to about 85 km, each with its own temperature and lapse rate. It looks what zones are parts of the cloud on and integrates the air density using the barometric formula.
 
Seems ok at a glance, but as KLOL already said it's better to have DT's confirmation.
 
Thank you for helping out. Do we have a sufficient consensus here then? 🙏
 
Antvasima said:
Thank you for helping out. Do we have a sufficient consensus here then? 🙏
Click to expand...
Well yes this is a pretty clear change and we've had multiple cgms agree to it. Would it be ok to proceed without DT or not?
 
I will ask him again via a private message. 🙏
 
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