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Hmmm... doesn't seem to state precise heights at a glance. So I suppose one could only use it as supportive evidence.
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Cloud Calculations
- Thread starter Therefir
- Start date
- Status
- 10,500
- 11,566
Seems like a case-by-case estimation of what is most appropriate would be the most reasonable thing to do.
Generally one should keep in mind that the higher values or the low ends for calcs.
I think taking the averages should usually be acceptable, although, as said, one can derivate from that case-by-case.
- 4,091
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Yeah, it sucks the chart doesn’t specifically listed specific height, but there is a page for this. I believe this was the one.
What Are the Different Types of Clouds? | NOAA SciJinks – All About Weather
Learn about common cloud types and what they can tell us about the weather!
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We should include averages of these ranges in that case, since I can see a lot of calcs using the averages
- 4,091
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Small minor bump
- 32,764
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I believe the default should be the averages of the accepted height-between-ground-to-cloud-bottom values, unless some visuals can give clues as to what exact height should be used.
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In that case, it would look like this:
Cirrus Clouds: 6858 meters
Stratus Clouds: 1219.2 meters
Cumulus Clouds: 914 meters
Stratocumulus Clouds: 811 meters
Nimbostratus Clouds: 1828.8 meters
Cumulonimbus Clouds: 8304.8 meters
Is this okay?
Cirrus Clouds: 6858 meters
Stratus Clouds: 1219.2 meters
Cumulus Clouds: 914 meters
Stratocumulus Clouds: 811 meters
Nimbostratus Clouds: 1828.8 meters
Cumulonimbus Clouds: 8304.8 meters
Is this okay?
- 4,091
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This is fine, I think.
- 15,989
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Bump
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Bump again and again as hammerstrike has said that's fine for the averages
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BTW, this only takes into account cylindrical clouds, what about different cloud shapes?
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The cloud manipulations page needs to be edited in order to reflect the changes. Just didn't get to that yet.
Strictly speaking not just cylindrical clouds, but also any kind of shape that has a volume calculable via base area * height.
If it's not that then you would have no other choice than to integrate the barometric formula per hand (or per integral calculator after some simplification). Well, either that or approximate the shape via a number of the prior mentioned ones and then add up.
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I assume the protocol would be to split it up into separate cloud volumes and add the calced masses of each one
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What about truncated cones?
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K.
Was more so talking about sphere shaped stuff or cones and truncated cones and the like.
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Yeah, for that stuff you would need to integrate the barometric formula over the new volume for a proper value.
Alternative are various ways of approximation. Like, you could use the density at the center of mass (times the volume) for something probably close to the true value. Or use the density at the top (times the volume) for a low-end.
Alternative are various ways of approximation. Like, you could use the density at the center of mass (times the volume) for something probably close to the true value. Or use the density at the top (times the volume) for a low-end.
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Small bump.
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bumpie
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- 10,500
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Easy way: Just take the air density that would apply at the top of the cloud at use that for the whole thing. That gives a low-end.
Hard way: See here for what variables you need to use.
If you're not in the height areas in which L_b = 0 define c = 1+ (g_0 x M / R* x L_b).
Calculate
where h2 is the upper height of the cloud and h1 the bottom height of the cloud.
A(h) is the area of the horizontal crossection of the cloud at the height h.
The A(h) is the term that depends on the form of the cloud now.
For your truncated cone for example: Let r1 be the bottom radius of the cone and r2 the top radius. Then the horizontal crossection's area is a circle and its radius is given by ((r2-r1)/(h2-h1)) * (h-h1) + r1. (linear function that is r2 in in h2 and r1 in h1)
Therefore we get A(h) = pi* (((r2-r1)/(h2-h1)) * (h-h1) + r1)^2.
With that all that's left is to solve the integral. (at least if I made no calculation mistake)
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Soooo with that settled is this ready to go then since we have most of it figured out? oooor am I missing something
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Hard way: See here for what variables you need to use.
If you're not in the height areas in which L_b = 0 define c = 1+ (g_0 x M / R* x L_b).
Calculate
where h2 is the upper height of the cloud and h1 the bottom height of the cloud.
A(h) is the area of the horizontal crossection of the cloud at the height h.
The A(h) is the term that depends on the form of the cloud now.
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Since I finally have some time I'm gonna start editing the cloud page now. ALthough I won't finish today 'cause it's 3am here lol
But yeah, is ready to go.
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okay cool good to hear if there's anything specific you need help adding give me the word but in the meantime I'm gonna continue working on some calcs and put my new cloud stuff up for evaluation when I'm done
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I did an edit to the page.
When doing the common result, I thought about whether taking the average for cumulonimbus is really a good idea.
While it's just wikipedia, it suggests that 200 to 4000m is more common, with higher values existing but being more extreme cases. This one also suggests a lower value for the base, with the top being mostly what goes up high.
So maybe using 4000m or 5000 feet would be better than the average?
When doing the common result, I thought about whether taking the average for cumulonimbus is really a good idea.
While it's just wikipedia, it suggests that 200 to 4000m is more common, with higher values existing but being more extreme cases. This one also suggests a lower value for the base, with the top being mostly what goes up high.
So maybe using 4000m or 5000 feet would be better than the average?
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Two different units?
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4000m or 1524m in metric units. That's the upper ends of the wikipedia and manchester base ranges respectively.
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Ah. Okay then.
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I'm fine with this
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So should we go with the 4000m or 1524m end? If we go with one of those. I currently went with 4000 as it's closer to the average, but I'm open to opinions.
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I don't really have a strong preference, I guess 4000 meters since it's closer to the average?
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same and it doesn't seem anyone has a particular preference so I'll say 4000 is fine by me
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Alright, I added it as such. In that case we're done here.
Randomlamdom
They/Them- 1,676
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One thing, unless i’m misunderstanding cloud height is the distance between sea level and cloud base right? But some seem like overall distance to the top of the cloud, like 16000m for cumulonimbus, 9144m for cirrus or 1828.8 for stratus
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These heights are taken from base to sea level
Randomlamdom
They/Them- 1,676
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So then these should be removed since it seems to be the height for the top of the cloud
Cirrus 9144m
Status 1828.8
Cumulus 1524m
Statocumulus 1219m
Cumulonimbus 16000m
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No they shouldn't be removed have you considered cloud feats such as dispersal that can take place above those cloud layers and need to use those altitudes. No you have not so no they shouldn't be removed
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What makes you say they seem to be that way?
Randomlamdom
They/Them- 1,676
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After reading through the references
Cirrus [12] “form very high in the atmosphere, usually between 15,000 and 30,000 feet” so the top would be 9144m
Stratus [14] “typically rest at a low altitude, found below 6000 ft” and [13] “cloud base can be up to 4,000ft” so 1828.8m would likely be near the top
Cumulus [15] “usually form at altitudes between 1,000 and 5,000ft” 1524m would be the top
Stratocumulus [16] “often occurs at altitudes between 1,000 and 4,000ft” 1219m would be the top
Cumulonimbus [18] “Typical Altitude: 2,000-45,000 ft” and [19] “Altitude 500-16,000 m”, “Peaks typically reach to as much as 12,000 m (39,000 ft), with extreme instances as high as 21,000 m”
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