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Cloud Calculations
- Thread starter Therefir
- Start date
- Status
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The article getting changed was simply talking about the density of the clouds themselves, it's not like they have influence on the authority of how to physically move clouds. Which still requires rearranging/flipping massive bodies of air around the atmosphere. Otherwise, it's like trying to move a tootsie roll center inside a tootsie pop without moving the outer layer.
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Ngl, I am pretty sure it technically still involves regular clouds as there is nothing that say the regular clouds ain’t unaffected by a feat that involves moving the clouds stuff as well as storm clouds by physical means.
The one that has Saitama’s punch overpowered Boro’s strong attack being a example.
Anyway, I am neutral on the proposal and also not a calc member.
I thought we already technically also including the weight of clouds too since clouds do have weight involved as well.
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Also while I am neutral overall, I do leaning towards agreeing with OP on specific aspects.
More specifically, the fact the density of clouds are still affected by the forced movement of a physical attack like Saitama’s punch.
Understand that I am looking at this from a more scientific standpoint, not the mathematical standpoint.
Edit: Made some corrections here as well as removed some redundant phrases.
More specifically, the fact the density of clouds are still affected by the forced movement of a physical attack like Saitama’s punch.
Understand that I am looking at this from a more scientific standpoint, not the mathematical standpoint.
Edit: Made some corrections here as well as removed some redundant phrases.
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I think that's disanalogous because that's a solid and a solid. In this situation, we're talking about a liquid inside a gas. And not one singular blob of liquid, a dispersed spread of liquid droplets inside a gas. You do not need to move all the air to do this.
There's two important things I'd like to point out about this sort of situation:
Firstly, you don't need to move the air in-between at all. If you have a ring of a liquid (or more intuitively, a solid, a liquid would only be different in that it would not be rigidly held together and would break apart), you can move it forward without needing to move the air in the gap of the ring. We can change this step by step, removing parts of the ring until we've got a series of droplets; those droplets need to be moved, but the gaps between them don't necessarily need to be. If you are using something with a vast AoE (like blowing, using a shockwave, or manipulating winds) you would inadvertently move the air as well, so this only matters in the case of water manipulation being used, but still, the tootsie pop example is disanalogous.
Secondly, the air doesn't have to be moved along with it, it just has to be moved out of the way. If you empty a bucket of water, that liquid is not trapping and pushing an ever-growing volume of air underneath it. If I drop the water such that it has a frontal surface area of 1m^2, and travels a height of 20 meters in ~2 seconds, that does not entail a volume of 20m^3 of air being moved at 10 m/s, or anything close to that. All the air around just has to move to get out of the way of the falling water, and return to the equilibrium it was in before, it does not have to take the entire ride along with it.
This is why I think that most cloud feats should be based on a known/assumed/derived volume of the air that pushes the cloud around (plus the water in the cloud itself), rather than the entirety of the air in the cloud itself. The rest of the air only has to move a negligible distance.
@HammerStrikes219 Were you given permission from a staff member to comment here?
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- Thread starter
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Since they are not derailing or anything, I can give him permission to comment.
Summoning the remaining Calc Group members.
@Psychomaster35 @Dark-Carioca @Armorchompy @Migue79 @Wokistan @Jasonsith
Summoning the remaining Calc Group members.
@Psychomaster35 @Dark-Carioca @Armorchompy @Migue79 @Wokistan @Jasonsith
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Still, the main intention of this thread was to slightly change and stablish an standard cloud air density, since as has been repeated several times, the value 1.003 has no basis.
So I will be making some Sandboxes for the storm and cloud pages, which would mainly concern the CAPE and KE calcs. The results would most likely be 40% lower.
So I will be making some Sandboxes for the storm and cloud pages, which would mainly concern the CAPE and KE calcs. The results would most likely be 40% lower.
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@DontTalkDT Thoughts?
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I just came up with a potential contention, but I'm not sure how relevant it is.
Most cloud feats wouldn't have the sort of resolution to primarily affect the water without moving the air much. There would need to be a pretty fine level of control over the forces involved to only really be pushing the water through. The water is just so dispersed and relatively light. It's like using a hammer to hit some grains of sand suspended in water without moving the water very much. If you want to actually retain all of the sand and move it to another location, you're gonna need to bring a lot of the surrounding water with you regardless. Some feats are done by manipulators with control fine enough to pull this off, but I'd expect that most wouldn't be.
I'm still not quite sure what this implies, since some feats (such as cloud dispersal) wouldn't mind if some bits of water got "left behind", dropping aside along with some of the air. This contention may mostly apply to moving a cloud while keeping it intact.
Most cloud feats wouldn't have the sort of resolution to primarily affect the water without moving the air much. There would need to be a pretty fine level of control over the forces involved to only really be pushing the water through. The water is just so dispersed and relatively light. It's like using a hammer to hit some grains of sand suspended in water without moving the water very much. If you want to actually retain all of the sand and move it to another location, you're gonna need to bring a lot of the surrounding water with you regardless. Some feats are done by manipulators with control fine enough to pull this off, but I'd expect that most wouldn't be.
I'm still not quite sure what this implies, since some feats (such as cloud dispersal) wouldn't mind if some bits of water got "left behind", dropping aside along with some of the air. This contention may mostly apply to moving a cloud while keeping it intact.
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Isn't that just what DontTalkDT said? Sorry if I mistook this for something else.
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It's more nuanced in a way that makes more sense to me. I can see the echoes of what DT said in that, but there's some distinctions in wording which I pointed out as issues when responding to DT earlier.
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Maybe this post is still not the full story?
Since air and water are different substances, they'd get pushed different amounts by an oncoming wave of force, which should let some amount of filtering get done by default. Either the water would get pushed further, letting the front push through without needing to displace the air much to move water that far, or the water would get pushed less far, pushing the air out, having other air fill in from the sides, perhaps letting the water get more concentrated so that less air needs to get inadvertently pushed.
But what would it mean for the front to push through without displacing air much?
I'm confused.
On something perhaps more constructive, I believe the OPM calc is flawed on a volume basis, not (just) a density one.
The surface area of the earth is 510,000,000 km^2, that calc finds that the surface area of the region of the clouds that was parted was 61,790,000 km^2; implying that it parted clouds over an eighth of the planet's surface. I did some quick pixel-scaling based on the circle drawn (without accounting for curvature), and the visible parting in the clouds took up 1/50th of the visible surface of the planet.
There's a few other foundational issues with the calc; it assumes that the clouds are parted around to the opposite end of the planet, and it assumes that all of the volume had to move the distance of the slice's furthest sideways extent.
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This is a calc group discussion thread, but since @Therefir give me permission earlier. That is fine.
Also in regards to this.
https://niwa.co.nz/education-and-training/schools/students/clouds#cl3
If you referring to how cloud moves, they are moved by wind in real life and assuming I not misinterpreted your words here too.
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Oh okay. Well, there isn’t much in regards to cloud displacement other than the clouds getting pushed by air as well as not exactly being displacing air much, but rather getting moved by air since wind are considered air currents after all.
The thing I can find in regards to cloud displacement was this article.
https://www.researchgate.net/public..._sky_images_based_on_phase_correlation_theory
Since the air isn’t being pushed necessarily, but more akin to the air pushing the cloud front to move.
Granted, this only relates to cloud movement and not exactly the figures on how much air is being displaced by clouds although I imagine it may not matter in a less significant way.
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I don't think we can keep a method which as resulted in a calc being as severely inaccurate as the OPM calc mentioned above.
GodlyCharmander
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The 1.003kg/m³ gotta go, that much is clear. The chart presented is objectively superior, not a complete fix to the situation, but still, a superior, less problematic method.
If it's agreed that this is the conclusion, then, unless someone presents a better method, it's what should be applied. I don't quite agree with the average because, most type of clouds stay at a certain height, which we can then look at the chart and use the value there (+0.003kg/m³) to measure the weight of a cloud.
If it's agreed that this is the conclusion, then, unless someone presents a better method, it's what should be applied. I don't quite agree with the average because, most type of clouds stay at a certain height, which we can then look at the chart and use the value there (+0.003kg/m³) to measure the weight of a cloud.
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my bad.
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We do need to get this resolved.
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Was there ever a verdict on this? I have some feats that are Waitig on this thread to be concluded...
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The verdict was on mostly cloud density needing to being changed even though the OP’s proposal was also on storm clouds as well.
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Alright, but did you guys conclude the discussion regarding cloud density in private? Then again, I suppose it wasn’t completely opposed anyway.
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Bump
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Yeah, I'm really not sure what's best as an overall solution with all the air and water stuff, but for now at least, using the density chart shown in the OP instead of 1.003 kg/m^3 seems like a very straightforward fix.
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May not be a CGM but I agree and besides this only applies to shit above 2500m to begin with so keeping the old value and specifying this while including the new chart should be a simple solution
(I'm ready for cloud feats to get evaluated now lol)
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2100 Meters actually and technically it is reasonable for clouds to go higher than 2.1k (k as in thousand) meter in a fictional setting/feat as well.
There is also clouds being able to reach higher as well, but that is a different topic.
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Bump
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Sooooo... I kinda forgot about that thread. Can anyone catch me up on where we stand?
If we go with the air option I can look into integrating over the barometric formula to get the mass or something.
If we go with the air option I can look into integrating over the barometric formula to get the mass or something.
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So far I think the consensus is to include air, in which case we go with the chart Therefir made above for altitudes higher than 2500 meters. Anything below that can still use the old density.
He said he's still making the sandbox because the table charts are hard for him to edit.
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Alright, then in that case I might be able to do a slightly more sophisticated model than the table. Albeit I expect the values to be pretty close to each other.
There is the barometric formula which, under the assumption of a standard atmosphere, delivers an approximation of air density up to 85km into the air.
Gonna use that wikipedia page and this one as references for this.
There are two density formula on wikipedia
this one and
this one.
Which one to use depends on whether the temperature lapse rate is zero.
If it's not zero, which is the case if the index b equals 0, 2, 3, 5 or 6, then the first one is used.
If the rate is zero, that is if b equals 1 or 4, the second one is used.
Which value b takes depends on how high up one is. That is well seen in Table 1 on the second page I linked or at the bottom table on the wikipedia page.
The rest of the constants are:
g0 which is the gravitational acceleration: 9.80665 m/s^2
R* which is the universal gas constant: 8.3144598 N·m/(mol·K)
M which is the molar mass of Earth's air: 0.0289644 kg/mol
Now, since the constants change with the height, if we want to calculate mass over multiple altitude ranges (i.e. using several b values) we would calculate the mass for each range separately and then sum up.
In order to calculate the mass separately we want to integrate the air/cloud volume over the density.
I will assume that we use the common construction, where the cloud volume is given by area * cloud thickness. In that case the altitude variable is independent of the position of the cloud and the integral over the volume is given by: Area * Integral over the height.
Therefore let's integrate the two barometric formula's in regard to the height variable.
I'm gonna use the good old integral calculator for this, 'cause I'm lazy.
Let (h - h_b) x L_b = x (integration by substitution so we get factor 1/L_b after integration), p = p_b, T = T_b, and c = 1+ (g_0 x M / R* x L_b) then the formula simplifies to p [T/(T+x)]^c.
Then the anti-derivative is given by (T^c*p*(x+T)^(1-c))/(1-c)*L_b.
Assuming c ≠ 1 (which can't happen).
Now, same game with the second formula.
This time x = (h - h_b), p = p_b and a = (g_0 M / (R* T_b) ). Then the formula simplifies to p*exp(-a x).
The anti-derivative is -(p*e^(-a*x))/a which, by setting the variables back in, is -(p_b*e^(-(g_0 M / (R* T_b) )(h - h_b)))/(g_0 M / (R T_b) ).
Let's make a test whether the result looks plausible:
A 1m^2 are and 1000m height cloud from 0 to 1000m high. That means b = 0, so we use the first formula.
Then T_b = 288.15, L_b = -0.0065, p_b = 1.2250, h_b = 0. Meaning T = 288.15, p = 1.2250, c = 1+ (9.80665 * 0.0289644 / (8.3144598 * -0.0065) ) = -4.25578774055216986602619136436907736975 and for x we get two values for the different heights x1 = 1000 * -0.0065 = -6.5 and x2 = 0.
The mass then is 1 m^2 * ((T^c*p*(x1+T)^(1-c))/(1-c)L_b - (T^c*p(x2+T)^(1-c))/(1-c)L_b)) = ((288.15^-4.25578*1.2250(-6.5+288.15)^(1+4.25578))/((1+4.25578)(-0.0065)) - (288.15^-4.25578*1.2250(0+288.15)^(1+4.25578))/((1+4.25578)*(-0.0065)))) = 1167.62 kg
Using the Table in the OP the average density should be 1.225 kg/m^3 and 1.112 kg/m^3 and the volume is 1000 m * 1 m^2 = 1000 m^3. So the mass should be between 1112 kg and 1225 kg. That fits the result
So the formula passes that test.
Still, would be happy if someone verified that I didn't make mistakes in that. If the formulae work I will try writing a calculator (let's see if I can manage to make it an online one) so that we don't always have to do all that work.
There is the barometric formula which, under the assumption of a standard atmosphere, delivers an approximation of air density up to 85km into the air.
Gonna use that wikipedia page and this one as references for this.
There are two density formula on wikipedia
Which one to use depends on whether the temperature lapse rate is zero.
If it's not zero, which is the case if the index b equals 0, 2, 3, 5 or 6, then the first one is used.
If the rate is zero, that is if b equals 1 or 4, the second one is used.
Which value b takes depends on how high up one is. That is well seen in Table 1 on the second page I linked or at the bottom table on the wikipedia page.
The rest of the constants are:
g0 which is the gravitational acceleration: 9.80665 m/s^2
R* which is the universal gas constant: 8.3144598 N·m/(mol·K)
M which is the molar mass of Earth's air: 0.0289644 kg/mol
Now, since the constants change with the height, if we want to calculate mass over multiple altitude ranges (i.e. using several b values) we would calculate the mass for each range separately and then sum up.
In order to calculate the mass separately we want to integrate the air/cloud volume over the density.
I will assume that we use the common construction, where the cloud volume is given by area * cloud thickness. In that case the altitude variable is independent of the position of the cloud and the integral over the volume is given by: Area * Integral over the height.
Therefore let's integrate the two barometric formula's in regard to the height variable.
I'm gonna use the good old integral calculator for this, 'cause I'm lazy.
Let (h - h_b) x L_b = x (integration by substitution so we get factor 1/L_b after integration), p = p_b, T = T_b, and c = 1+ (g_0 x M / R* x L_b) then the formula simplifies to p [T/(T+x)]^c.
Then the anti-derivative is given by (T^c*p*(x+T)^(1-c))/(1-c)*L_b.
Assuming c ≠ 1 (which can't happen).
Now, same game with the second formula.
This time x = (h - h_b), p = p_b and a = (g_0 M / (R* T_b) ). Then the formula simplifies to p*exp(-a x).
The anti-derivative is -(p*e^(-a*x))/a which, by setting the variables back in, is -(p_b*e^(-(g_0 M / (R* T_b) )(h - h_b)))/(g_0 M / (R T_b) ).
Let's make a test whether the result looks plausible:
A 1m^2 are and 1000m height cloud from 0 to 1000m high. That means b = 0, so we use the first formula.
Then T_b = 288.15, L_b = -0.0065, p_b = 1.2250, h_b = 0. Meaning T = 288.15, p = 1.2250, c = 1+ (9.80665 * 0.0289644 / (8.3144598 * -0.0065) ) = -4.25578774055216986602619136436907736975 and for x we get two values for the different heights x1 = 1000 * -0.0065 = -6.5 and x2 = 0.
The mass then is 1 m^2 * ((T^c*p*(x1+T)^(1-c))/(1-c)L_b - (T^c*p(x2+T)^(1-c))/(1-c)L_b)) = ((288.15^-4.25578*1.2250(-6.5+288.15)^(1+4.25578))/((1+4.25578)(-0.0065)) - (288.15^-4.25578*1.2250(0+288.15)^(1+4.25578))/((1+4.25578)*(-0.0065)))) = 1167.62 kg
Using the Table in the OP the average density should be 1.225 kg/m^3 and 1.112 kg/m^3 and the volume is 1000 m * 1 m^2 = 1000 m^3. So the mass should be between 1112 kg and 1225 kg. That fits the result
So the formula passes that test.
Still, would be happy if someone verified that I didn't make mistakes in that. If the formulae work I will try writing a calculator (let's see if I can manage to make it an online one) so that we don't always have to do all that work.
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So buff or nerf? And at what altitudes will the higher densities apply?Alright, then in that case I might be able to do a slightly more sophisticated model than the table. Albeit I expect the values to be pretty close to each other.
There is the barometric formula which, under the assumption of a standard atmosphere, delivers an approximation of air density up to 85km into the air.
Gonna use that wikipedia page and this one as references for this.
There are two density formula on wikipedia
this one and
this one.
Which one to use depends on whether the temperature lapse rate is zero.
If it's not zero, which is the case if the index b equals 0, 2, 3, 5 or 6, then the first one is used.
If the rate is zero, that is if b equals 1 or 4, the second one is used.
Which value b takes depends on how high up one is. That is well seen in Table 1 on the second page I linked or at the bottom table on the wikipedia page.
The rest of the constants are:
g0 which is the gravitational acceleration: 9.80665 m/s^2
R* which is the universal gas constant: 8.3144598 N·m/(mol·K)
M which is the molar mass of Earth's air: 0.0289644 kg/mol
Now, since the constants change with the height, if we want to calculate mass over multiple altitude ranges (i.e. using several b values) we would calculate the mass for each range separately and then sum up.
In order to calculate the mass separately we want to integrate the air/cloud volume over the density.
I will assume that we use the common construction, where the cloud volume is given by area * cloud thickness. In that case the altitude variable is independent of the position of the cloud and the integral over the volume is given by: Area * Integral over the height.
Therefore let's integrate the two barometric formula's in regard to the height variable.
I'm gonna use the good old integral calculator for this, 'cause I'm lazy.
Let (h - h_b) x L_b = x (integration by substitution so we get factor 1/L_b after integration), p = p_b, T = T_b, and c = 1+ (g_0 x M / R* x L_b) then the formula simplifies to p [T/(T+x)]^c.
Then the anti-derivative is given by (T^c*p*(x+T)^(1-c))/(1-c)*L_b.
Assuming c ≠ 1 (which can't happen).
Now, same game with the second formula.
This time x = (h - h_b), p = p_b and a = (g_0 M / (R* T_b) ). Then the formula simplifies to p*exp(-a x).
The anti-derivative is -(p*e^(-a*x))/a which, by setting the variables back in, is -(p_b*e^(-(g_0 M / (R* T_b) )(h - h_b)))/(g_0 M / (R T_b) ).
Let's make a test whether the result looks plausible:
A 1m^2 are and 1000m height cloud from 0 to 1000m high. That means b = 0, so we use the first formula.
Then T_b = 288.15, L_b = -0.0065, p_b = 1.2250, h_b = 0. Meaning T = 288.15, p = 1.2250, c = 1+ (9.80665 * 0.0289644 / (8.3144598 * -0.0065) ) = -4.25578774055216986602619136436907736975 and for x we get two values for the different heights x1 = 1000 * -0.0065 = -6.5 and x2 = 0.
The mass then is 1 m^2 * ((T^c*p*(x1+T)^(1-c))/(1-c)L_b - (T^c*p(x2+T)^(1-c))/(1-c)L_b)) = ((288.15^-4.25578*1.2250(-6.5+288.15)^(1+4.25578))/((1+4.25578)(-0.0065)) - (288.15^-4.25578*1.2250(0+288.15)^(1+4.25578))/((1+4.25578)*(-0.0065)))) = 1167.62 kg
Using the Table in the OP the average density should be 1.225 kg/m^3 and 1.112 kg/m^3 and the volume is 1000 m * 1 m^2 = 1000 m^3. So the mass should be between 1112 kg and 1225 kg. That fits the result
So the formula passes that test.
Still, would be happy if someone verified that I didn't make mistakes in that. If the formulae work I will try writing a calculator (let's see if I can manage to make it an online one) so that we don't always have to do all that work.
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For clouds floating lower than 2000m it would be a buff, for clouds floating higher up than 2000m it would be a nerf.
It's basically the same as using Therefir's density table in the OP, just (if done right) more accurate as it considers all in-between density values as well, which the table doesn't show. (and correctly "sums up" all the different density values)
It's basically the same as using Therefir's density table in the OP, just (if done right) more accurate as it considers all in-between density values as well, which the table doesn't show. (and correctly "sums up" all the different density values)
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