如何计算w-wind吗?- 江南体育网页版- - - - -地球科学堆江南电子竞技平台栈交换 最近30从www.hoelymoley.com 2023 - 07 - 10 - t05:08:37z //www.hoelymoley.com/feeds/question/9493 https://creativecommons.org/licenses/by-sa/4.0/rdf //www.hoelymoley.com/q/9493 7 如何计算w-wind吗? //www.hoelymoley.com/users/3194 2017 - 01 - 20 - t06:56:50z 2017 - 01 - 23 t16:21:11z < p >我的模型(AGCM)不计算w-wind(风速的垂直分量)。然而,我确实有以下可用变量从AGCM输出:< / p > < pre > <代码>除,v-wind位势、欧米茄和一些其他变量(如温度、含湿量)23压力水平。< /代码> < / pre > < p >除和v-wind风速的水平分量,< a href = " https://en.wikipedia.org/wiki/Geopotential " rel = " nofollow noreferrer " >位势< / >是地球引力的潜力,ω是解决< a href = " https://en.wikipedia.org/wiki/Omega_equation " rel = " nofollow noreferrer " > < / >ω方程。< / p > < p >我想计算通过使用除w-wind v-wind位势ω。所建议的< em > JeopardyTempest < / em >下面的回答,我可以使用方程计算w-wind <强> w =−ω/ρg < / >强。这是多么可靠,如果我不使用<强> w =−ωRT / pg > < /强大? < / p > //www.hoelymoley.com/questions/9493/-/9495 # 9495 10 回答如何计算w-wind由JeopardyTempest吗? JeopardyTempest //www.hoelymoley.com/users/6298 2017 - 01 - 20 - t11:07:59z 2017 - 01 - 21 t00:52:39z < p >ω,ω是气象学w密切相关。它可以使用链式走向w (< a href = " http://www.sjsu.edu/faculty/watkins/omega.htm " rel = " noreferrer " > < / >这提醒从沃特金斯SJSU帮助)……< / p > < pre > <代码>ω= dp / dt = (dp / dz) (dz / dt) = (dp / dz) w < /代码> < / pre > < p > <一口> < em > [p是压力,t是时间,z是高度,和w身高垂直运动坐标)< / em > < /一口> < / p > < p >通常一般气象可以估计dp / dz使用< a href = " https://www.e-education.psu.edu/meteo300/node/7 " rel = " noreferrer " > < / >静压近似,大概认为重力和浮力平衡。这使得方程:< / p > < pre > <代码> dp / dz =−ρg < /代码> < / pre > < p > <一口> < em >(ρ是空气密度和g是重力加速度)< / em > < /一口> < / p > < p >所以意味着你可以继续< / p > < pre > <代码>ω= (dp / dz) w =−ρgw < /代码> < / pre > < p >导致<强> w =−ω/ρg < / >强。< / p > < p > <强>请注意,这并不总是有效的。< /强> < a href = " https://www.google.com/url?sa=t& rct = j& q =, esrc = s&源= web& cd = 1, cad = rja& uact = 8, ved = 0 ahukewjjmqzpuddrahujlsykhcarchqqfggdmaa& url = https % % 2 f % 2 3 fwww.amazon.com % 2 fintroduction-dynamic-meteorology-international-geophysics % 2自民党% 2 f0123848660& usg = AFQjCNGdSbK_Pfo2P8HbJIyT5NckdVTUpg& sig2 = 7 ccpxd8f00nu6vubkhtujw”rel = " noreferrer " >霍尔顿是一个介绍动力气象学,卷1 < / >显示< a href = " https://books.google.com/books?id=-ePQ6x6VbjgC& pg = PA202&液化石油气= PA202& dq =估计% 20 w % 20 % 20ω% 20 qg&源= bl& ots = l3lT-YIsiq&团体= aOb6ZXDWv7f9Vp_IGUgxtqYDgWM& hl = en& sa = X& ved = 0 ahukewjk3rnptddrahxlriykhtzra1q4chdoaqgomai # v = onepage& q =估计% 20 w % 20从% 20ω% 20 qg& f = false”rel = " noreferrer " >第20页< / >:< / p > < blockquote > < p >这个流体静力平衡条件提供了一个很好的近似垂直的依赖的压力场在现实的气氛。只对飑线和龙卷风等强烈的小规模系统有必要考虑偏离流体静力平衡。< / p > < /引用> < p >这是一个松散的主意。已经有一段时间,但基本上我记住它,静水强烈时有效的垂直加速度(因此垂直运动时在地表附近)相比,规模要小得多的水平运动/加速度。所以实际上,这不是真正的龙卷风和对流。所以你非常强烈的雷暴的正确值。但它将帮助您诊断地区上升运动。

You can look a bit more towards breaking it down into more advanced detail and the mathematics of hydrostatic and how it fails in Doswell and Markowski's paper on buoyancy.

But, with such limited model output, not sure that you're looking to get into all of that detail, as I don't believe you'll find any extra help going those avenues with the variables you have.

With just slightly more information, most importantly temperature, you could use the ideal gas law:

 p = ρRT 

[where R is the ideal gas constant of the gas under consideration and T is temperature in Kelvin]

This would allow you replace the unknown density, and you'd then end up with:

w = −ωRT/pg

Which you could then either further approximate R to be that for dry air = Rd = 287 J/Kkg, or even work up the adjustment your moisture content variable causes (though don't think doing so would offer much useful improvement as you've already got greater errors due to the failings of the hydrostatic approximation).

Since you don't have temperature, it seems your only remaining option would be to plug in the density approximation for the level of interest from the approximated "Standard Atmosphere" here (use the table where ρ is in units of kg/m^3).

It has been a long while since I've done this stuff, but I don't see any better options. Unless somehow there's a way to work from the geopotential, maybe apply something like mass continuity in pressure coordinates, and get to something more complex to return something more correct for w. But if there is, it outpaces me!

So, long story short, it appears that your best hope is w = −ωRT/pg, but it won't be 100% reliable, particularly in strong convection.

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