问题很简单:我仍然有大气压力坐标的问题:

给定图片中的压力依赖性，一个空气包裹移动的速度坐标是什么$\vec v= (u, v=0, w=0)$在<强>压力坐标?

我理解对了吗，在压强坐标中没有w而是ω ?< / p >

I would say we have then the tuple $$(u, v=0, ω)$$ with $$\omega = Dp/Dt = (\partial p/\partial x)_z \cdot Dx/Dt= (\partial p/\partial x)_z \cdot u$$

But I'm not sure, because it seems a bit weird that in the first system we have zero w and in the pressure system we have negative vertical "speed".

Although we normally consider only horizontal components u, v of wind, due to pressure coordinates we have a non zero "vertical" component (because pressure decreases as the parcel moves to the right in the example). How are pressure coordinates than easier to handle when there is an additional component $\omega$ and we have three velocity components to consider instead of just two (of course I know that other equations become more simple as density vanishes...)?

Do I have a mistake in my considerations?

积雨云的顶部通常约为40,000英尺，可以达到超过60,000英尺的高度，分别在245英里或300英里的距离上可见。你能看到多远的雷暴?

我的疯狂表弟说，在，比如说，300英里，说他不能看到这个云的形成-也许只是它的顶部-因为地球的曲率(距离:300英里-曲线在视线以下11.3618英里(也59990.14英尺)。我肯定他错了。但这是合乎逻辑的，地球曲率的计算是合理的。

Upon reviewing the three main methods of estimating vertical motion in prior papers (kinematic, thermodynamic, and QG-omega), I've decided that implementing the kinematic method might be the most straightforward method to understand and also seems to be best suited to my needs. I initially tried solving for the traditional QG-Omega equation, but I am really struggling with "inverting" the laplacian on the left-hand side of the equation, and so have now opted to approach this problem using the kinematic method. However, I'm running into some trouble working out the logic of using this method in my head.

Its derivation seems easy enough. Starting with the continuity equation in pressure coordinates: $$ \frac{dw}{dp} + \frac{du}{dx} + \frac{dv}{dy} = 0 $$

we can rearrange and integrate to yield:

$$ w_{p} = w_{p+\Delta p} + \int_{p}^{p+\Delta p} ( \frac{du}{dx} + \frac{dv}{dy})dp $$

where w is vertical velocity, p is some reference pressure, u is zonal wind, and v is meridional wind.

Now the problem I'm running into is that much of the literature recommends against using the kinematic method for acquiring a realistic estimate of vertical motion. This seems mainly due to the fact that the geostrophic wind is non-divergent, and errors in the ageostrophic wind are large enough to significantly affect horizontal divergence. This makes sense to me for instances where one is using observations, since the goal is to match reality as closely as possible, and observational readings are prone to error both due to instrumentation and error that likely arises from the interpolation process from discrete observations onto a grid.

However, for model data, would this drawback still apply? There are no "errors" since the fields are continuous and the model is simulating everything based on its own assumptions. In that sense, I'm not looking to match reality, but rather match whatever the model's reality is. It seems to me like this is how the WRF model already calculates omega (indicated within the comments of the calc_ww_cp subroutine in the WRF code, although my Fortran is a bit too rusty to actually make sense of it). Basically, I'm looking for the most efficient way of obtaining a realistic quantity representing vertical motion within the model output.

In that sense, is the integration of the continuity equation the best way of accomplishing this for numerical model output? Or is it worthwhile to pursue a different method?

Thanks in advance, and I look forward to any fruitful discussions that can come out of this!

My question is: if weather balloons always pop when they reach a certain altitude, why don’t they just make sure the balloon is carrying a payload that is too heavy for the balloon to reach that altitude (or put less hydrogen). If these are used by meteorologists and the like, wouldn’t it be cheaper to just leave the balloons in the sky instead of constantly putting new ones up?

我似乎找不到任何“一般解释”。(从概念上讲)MJO如何影响TC、季风和ENSO循环

For example, during summer in the northern hemisphere, the Hadley cell in the Northern hemisphere is significantly weaker than the one in the Southern Hemisphere.

I've been told that the southern hemisphere Hadley cell plays a row to carry energy from the SH to the NH in the lower troposphere (elevation) whereas in the upper elevation, energy travels from the summer hemisphere to the winter hemisphere.

What are these differences and what is the concept behind it?

我正在阅读一些关于木星的文章，当我偶然发现这篇文章，这篇文章说大红斑使上层大气非常非常热。就像“比熔岩热”一样。这显然是由于大的声波和大气重力波(不是引力波)，当大的空气袋碰撞时产生，产生巨大的湍流，将原子送入高层大气并将其加热:

巨大的风暴发出大的声波和大气重力波-当大的空气袋碰撞时产生-具有巨大的动能。使原子四处飞舞，提高了现场上空的大气温度。< / p >

However, I didn't come here to ask about Jupiter's Great Red Spot or anything astronomical. This information is not going to be a actual part of the question, as the information above merely suggests where I got this idea from. The article simply poked my curiosity for atmospheric physics. However let's move onto the real question.

On Earth, we do have large storms. Well not planet-sized storms, but storms the size of entire countries. Typhoon Tip, for example, reached more than 2,000 km in diameter. Typhoon Tip also, was an incredibly turbulent environment, with air pockets colliding. Also, lightning and immense winds could have produced immense sound waves that could have scattered atoms away, and heated up the upper atmosphere to scorching temperatures. It's not just restricted to Typhoon Tip, other energetic hurricanes, like Katrina, Patricia etc. were extremely energetic hurricanes that could have produced sound waves and enough turbulence to heat up the upper atmosphere.

Aside from hurricanes, there are also a plethora of extremely turbulent phenomenon on Earth, such as supercells, derechos, squall lines and so on. These environments have a lot of wind shear, fast winds, updrafts and downdrafts, which have a knack for producing intense turbulence. Rain and hailstones colliding with each other, lightning forming, should produce enough sound and turbulence to heat up the upper atmosphere.

Yet this phenomenon does not occur on Earth. Earth's upper atmosphere is really, really cold, excluding the thermosphere, which is indeed scorching hot, but is diffuse and evenly heated, unlike on Jupiter, where storms heat up the areas of the upper atmosphere above them.

Why don't terrestrial storms heat up the upper atmosphere?

我在物理SE上问过这个问题，但它仍然没有答案。希望地球科学SE在这江南体育网页版个问题上更有知识。< / p >

Wikipedia gives the following equation to calculate the moist adiabatic lapse rate $\Gamma_w$, assuming that there is only one condensible gas (water vapour) mixed in the "dry air":

$\Gamma_w = g\frac{\left(1+\frac{H_v r}{R_{sd}T} \right)}{\left(c_{pd} + \frac{H_{v}^2r}{R_{sw}T^2} \right)}$

Where:

• $\Gamma_{w}$: moist adiabatic lapse rate [K/m]
• $g$: gravitational acceleration [m/s2]
• $H_{v}$: latent heat of vaporization of water [J/kg]
• $R_{sd}$: specific gas constant of dry air [J/kg·K]
• $R_{sw}$: specific gas constant of water vapour [J/kg·K]
• $r={\frac {\epsilon e}{p-e}}$: mixing ratio of the mass of water vapour to the mass of dry air
• $\epsilon = \frac{R_{sd}}{R_{sw}}$: ratio of the specific gas constant of dry air to the specific gas constant for water vapour = 0.622 [dimensionless]
• $e$: water vapour pressure of the saturated air [Pa]
• $p$: pressure of the saturated air [Pa]
• $T$: temperature of the saturated air [K]
• $c_{pd}$: specific heat of dry air at constant pressure [J/kg·K]

What's the form of this equation when the atmosphere composition differs from Earth's, thus allowing multiple condensible gases or even be entirely composed of only one condensible gas (for example 100% water vapour)?

根据科里奥利力，一个空气包在没有其他力的情况下被加速度偏折(霍尔顿，第四版):

这些方程导致圆周运动。< / p >

This figure is from a Matlab script, provided by Holton:

But how is this consistent with the picture, where an object moves along a straight line in the inertial system and the deflection is only apparently because we are in a rotating frame? If the object moves along a straight line in the inertial system, the curve cannot be circular in the rotating frame.

It's due to the disappearance of vertical motion, I think. But in Holton, the third equation is simply left out without comment, and instead only the above two equations are finally written. It is clear that the above equations come out when the vertical velocity is zero, but this is not justified in the explanations. Is this just an empirical observation?

例如，1991年皮纳图博火山爆发后，根据这篇论文，

将大量硫酸气溶胶引入平流层增加了行星反照率(本质上是地球对太阳辐射的反射率)，因为这些气溶胶粒子是有效的散射体，但在太阳波长上只是弱吸收体。

…< / p >

The changes in the Earth's albedo observed by ERBE resulted in a net cooling of approximately 8 W m-2 between 5°S and 5o N, with a net cooling of 4.3 W m-2 between 40° S and 40° N.

What I'm interested in is how exactly this cooling happens. Which is correct:

1. the atmosphere is experiencing a deficit of input energy, or

2. the surface is experiencing a deficit of input energy, which is then communicated to the atmosphere by a decreases longwave surface emission

I suspect (2) is mostly correct, since about 75% of input shortwave solar radiation to the earth is absorbed by the surface, while only 25% by the atmosphere itself, according to ref [1] (and perhaps even less so by volcanic aerosols).

If this is correct, then once the surface does communicate that energy deficit to the atmosphere, will it happen locally? That is, will a slab of the atmosphere nearest the surface be the first to experience a cooling rate?

[1] Petty, G.W., 2006. A first course in atmospheric radiation, 2nd ed. ed. Madison, Wis: Sundog Pub.

我看到像"a half"或“70%”在各种网页上，但找不到实际来源。我可以找到湿度趋势和年度变化的良好来源，但它们只给出百分比变化，例如这些英国气象局趋势。大气的总含水量在很多地方也给出了12,900立方千米，但是我还没有找到一个相应的综合整个大气的储水量

Still I have some troubles with regard to the mathematical formalism:

What does $\left(\frac{\partial \Phi}{\partial x}\right)_p$ mean mathematically? I would have assumed, that $\Phi$ is a function of $x, y, p$: $\Phi = \Phi(x,y,p)$ and, therefore $\frac{\partial \Phi}{\partial x}$ means to keep y,p as constant and consider only the change of $\Phi$ caused by a change in x (this is the definition of partial derivative). Why do we write $\left(\frac{\partial \Phi}{\partial x}\right)_p$ instead? Once we agree on a special set of coordinates we dont have to specify what other coordinates are held constant for the partial derivative. But if we decide to do it, why is it then not written as $\left(\frac{\partial \Phi}{\partial x}\right)_{y,p}$ ? Looks a bit confusing.

Because of this notational discrepancy, I'm afraid I haven't quite gotten the "magic behind it" yet, maybe missing the most important point of all, and therefore just think I got it.

Can somebody explain, where I have my missing point?

• An Introduction to Dynamic Meteorology James R. Holton

• Fluid Dynamics of the Midlatitude Atmosphere Brian J. Hoskins

• Mid-Latitude Atmospheric Dynamics, A First Course Jonathan E. Martin

The amount of air being greater in this direction, why is the atmosphere near the horizon whiter (paler) instead of red, orange like at sunset and sunrise?

At the $CO_2$ 15um wavelength, there is almost no outgoing radiation. OK, it sounds logical, because $CO_2$ absorbs that range of radiation.

But since $CO_2$ is a good absorber, according to Kirchhoff's law it should be a good emitter at the same wavelengths.

The radiation equation in its simplest form says:

$$I_\nu = I_\nu(0) e^{-\tau\nu} + I_\nu^B [1-e^{-\tau_\nu}]$$

I understand that when the optical thickness is large, the outgoing radiation is dominated by the $I_\nu^B$, the Boltzmann value for black-body radiation.

When the atmosphere is optically thick, we would have this situation and, therefore, according to that picture, there should be outgoing radiation also at those wavelengths. Even taking into account the lower temperature of the atmosphere can, in my opinion, not explain, why outgoing radiation is almost zero.

But why is this the case?

想象空气的压力水平为760 hPa，温度为4摄氏度。如果我填写公式:< / p > < p >第一计算2温度的饱和含湿量:< / p > < p > <跨类= >“math-container qs_1美元< / span > <跨类=“math-container”> T_1 < / span >美元(3°C) = 0.622 * ((6.11 * exp(19.85 - 5423 /(273.15 + 3) * 100)) /(76000) * 1000 = 6.18克/公斤= 0.006180公斤/公斤< / p > < p > <跨类= >“math-container qs_2美元< / span > <跨类= >“math-container T_2 < / span >美元(5°C) = 0.622 * ((6.11 * exp(19.85 - 5423 /(273.15 + 5) * 100)) /(76000) * 1000 = 7.12克/公斤= 0.007120公斤/公斤< / p >

填入公式:

$L_m = \frac{0.0098}{1+\frac{2.5*10^6}{1004}\frac{0.007120-0.006180}{5-3}} = 8.34480 \ K\ km^{-1}$

这看起来相当高。我做错了什么吗?< / p >

背景:

对于我的研究，我想并试图通过分析一系列垃圾填埋场岩心钻孔样品和外推数据来预测温室气体排放量(GHG)模型，该模型预测温室气体当量排放量($GHG_{eq}$[吨/年])如果样品可以代表垃圾填埋场作为一个整体并允许自然降解，将产生多少温室气体。请注意，这是基于最小异质性的假设，我知道这对大多数垃圾填埋场来说不是最好的假设，但我希望得到很多数据(总比没有好)。< / p >

As the most notorious GHG from landfill is methane, I'm going to simplify this question to just predicting the amount of methane which would be produced from the (combustible) material in landfill assuming it would just sit there forever.

Note that I have cross-posted my question from the chemistry stack exchange.

The Data:

I have this data which is the average of 3 landfill drill core samples and shows the type of waste which has been calculated from sorting and weighing different types of material found in landfill (Table 1) and the chemical composition and high heating value (HHV)/calorific value of the material (Table 2) which I've had tested in an independent lab. For Table 1, an assumed "type" of plastic, rubber, wood (etc.) is made with a constant chemical composition to simplify the model (although I do recognise that there are many different types of plastics and woods etc.).

The Method (?):

I've seen the generic methanogenesis combustion reaction as:

$\ce{CO_2 + 4H_2 -> CH_4 + 2H_2O}$

I've also seen an elemental formula for hydrocarbon combustion along the lines of:

$\ce{C_xH_y + $(x + \frac{y}{4})$O_2 -> xCO_2 = $\frac{y}{2}$H_2O + $Heat + Light$}$

I know how to calculate the conversion rate going from any GHG such as $\ce{CO_2}$ or $\ce{CH_4}$ to $GHG_{eq}$ but it's this first step - the predicting of the GHGs themselves with what formula - which I'm lost at.

The Questions:

1. How can I calculate $\ce{CH_4}$ from landfill samples with the attached data and which reaction should be used?
2. (Bonus) What other considerations should be accounted for with any of the other data? (e.g. the Sulfur going into $\ce{SO_2}$ emissions (etc.)).

I'm imagining something simple like two sealed fish tanks, one filled with air that we breath every day, and the other filled with the same air but with a 50% higher concentration of carbon dioxide. Shine a light into them both and measure the temperature.

I think this would convert a lot of "deniers".

是哪种或哪种气体造成的?< / p >

If some gasses in the atmosphere do not refract blue light, what colors DO they refract?

Is it possible to have a breathable atmosphere with a non-blue sky?

I understand that as the wind is bent, the part/vector of the coriolis effect that acts in direct opposition of the PGF increases. But once the wind is bent so that it blows parallel to the isobars, why would the coriolis effect (have the tendency to) be similar to the PGF at that point?

Is it because 1) the wind picks up speed as long as it moves down the gradient at angle less than 90°, and 2) an increase in wind speed increases the coriolis effect? So that there will always be a moment where the coriolis force has grown equal to the PGF (at which point the wind will stop speeding up, because it moves paralel to the isobars)?

我正在阅读这篇之前的问题关于如果水丢失到太空中是如何的，其中一个答案说(意译)地球失去了足够的水，海洋减少了12厘米，如果地球失去了那么多的水，它能通过从太空接收水来弥补损失吗?我做研究< a href = " https://www.google.com/search?q=how%20did%20earth%20get%20its%20water& rlz = 1 casfjy_enus1020& oq = % 20 % 20地球% 20如何% 20 % 20 water& aq = chrome.0.0i512l2j0i390.6038j0j7& sourceid = chrome& ie = UTF-8&安全= active& ssui =“rel = " noreferrer " >源< / > < / p > < blockquote > < p >远离太阳,气温很低,水结冰的对象(如彗星形成,而接近太阳水与岩石材料发生反应形成水合矿物。人们认为，地球继承水的最可能方式是小行星和彗星撞击地球。

地球还在以这种方式接受水吗?< / p >

In the case of waves, are Rossby waves 1 and Kelvin waves 2 examples of that? Like, imagine the Earth as a single planet with no Sun (so no influence by the Sun's heat), no moon (so no tides) and no planetary internal hot core (so no influence by the heat from Earth's internal core). Then assume that somehow water is still liquid and air in its gas form, then, just by Earth's rotation, would there be any waves or wind (even if they would be very subtle)? Would there still be Rossby waves or Kelvin waves for instance? (I found a comment to a question in Quora that indicates that the answer is basically "yes" 3, but no sources are given, so I would like to see if someone could verify that)

有句谚语听起来像:

晚上太阳红红的，水手的喜悦;清晨红日，水手取警告。< / p >

Is it valid and if it is then what is it about?

由于某种原因，我只能在ppm中找到答案，我不知道如何进行这种转换。

BBC刊登了一篇文章极端天气:它是什么，它与气候变化有什么关系，其中包含了这个图形:

这是准确的吗?大气中的高压是否将温暖的空气向下推，然后空气被压缩，变得更热?从基本知识看来，“大气中的高压”这意味着密度更大的空气(可能以气泡的形式?)会取代密度较小的热空气，让密度较小的热空气“漂浮”起来。在它上面(也许热空气也会以气泡的形式向上移动)。撇开低暖空气/高密度空气界面的细节不谈，到底发生了什么?稠密的空气向下移动而温暖的空气向上移动(就像我相信的那样)还是稠密的空气(在图中称为高压)将温暖的空气向下推，就像图中所描绘的那样?< / p >

https://www.ncei.noaa.gov/access/monitoring/monthly-report/global/202113

What could be the reason for this? My idea is that the N. hemisphere has accumulated significantly more continental mass, while the S. hemisphere is almost continent-free. And the water is warming because the Gulf Stream is currently flushing more warm water northwards. Is my idea correct? Do you have other suggestions or are there suitable papers?