寻找数学所以我可以运行模拟气候变化,地球科学堆栈交换江南电子竞技平台江南体育网页版 最近30从www.hoelymoley.com 2023 - 07 - 07 - t01:41:06z //www.hoelymoley.com/feeds/question/23089 https://creativecommons.org/licenses/by-sa/4.0/rdf //www.hoelymoley.com/q/23089 2 寻找气候变化模拟数学所以我可以运行 perep1972 //www.hoelymoley.com/users/14509 2021 - 11 - 05 - t10:12:42z 2022 - 11 - 24 - t16:48:52z < p >对不起,愚蠢的问题,但是我无法回答它甚至浪费时间谷歌一段时间后…我知道有几个气候变化的场景,我知道几个气候数据库提供表面(位图)与场景。然而,我想尝试一些计算我自己,因为我在寻找这些模型背后的数学(不同的场景)或一些代码(最好是在R)。任何建议吗? < / p > //www.hoelymoley.com/questions/23089/-/23090 # 23090 13 由丹尼尔回答。heydebreck寻找数学所以我可以气候变化模拟运行 daniel.heydebreck //www.hoelymoley.com/users/5594 2021 - 11 - 05 - t10:37:00z 2021 - 11 - 05 - t11:04:28z <标题>全面气候模型< / h1 > < p >大气气候模拟,提出了政府间气候变化专门委员会报告是由非常大而复杂的模型。Running these models needs a "supercomputer" (high performance cluster). Additionally, certain "forcing data" and "boundary conditions" are needed.

Boundary conditions are the top of the atmosphere and the Earth's surface/oceans. Partly, atmospheric climate model simulations are directly coupled to ocean model simulations. This makes the models even larger and harder to setup/run. "Forcing data" means thinks like CO2 emissions, emissions of particulate matter, solar radiation ... .

Having said that: it is quite unrealistic to setup und run a "real" climate model just for fun.

Further reading on recent climate modeling activities

Simulation results of the couple model intercomparison project (CMIP) form the basis for the IPCC reports. The CMIP Phase 6 (CMIP6) simulations form the basis of the most recent IPCC reports. There is an overview paper of CMIP6 which might indicate how large the overhead for running "real" climate models in a comparable way is (Eyring et al., 2016).

One of the models that were used for the CMIP5 and CMIP6 simuations was the MPI-ESM model. An overview of this model is given in Giorgette et al. (2013). Detailed descriptions are available in this special issue.

There are several more models used in CMIP5 and CMIP6. The MPI-ESM is just an example.

Climate models made for training

There are some simple climate models that are made for students and which are based on a few governing equations. Seems models show the general features of climate models but can be run on a normal end-user computer. One of these models is the "Monash simple climate model". An instance of that model which can be run via a web-GUI is available here. There is a documentation available in which you will find links to a publication describing the model and to a repository to download the source code.

There is another climate model made for training which is called Planet Simulator. Download details and documentation are offered on that web page.

//www.hoelymoley.com/questions/23089/-/24549 # 24549 0 答案由bpl1960寻找气候变化模拟数学所以我可以运行 bpl1960 //www.hoelymoley.com/users/12453 2022 - 11 - 24 - t16:48:52z 2022 - 11 - 24 - t16:48:52z < p >你可能尝试latitudinally-resolved气候模型。这些都是1969年发明的,看看全球变暖或冷极地冰冠的边界。结果是略有减少太阳常数使冰冻大地;这是因为模型不包括carbonate-silicate周期稳定地球的温度从长远来看。但对于研究微小的变化可能是好的。< / p > < p >的基本方程是一个权力制衡(通常错误地称为“能源balance"): < / p > < p >气(1 - Ai) = A + B Ti + C (Ti - T) < / p > < p >气是日晒,平均一年,在一个给定的纬度带,带我——你需要大量的球面天文学和三角计算出来的,但有些文章(例如沃伦和施耐德1979或Chylek和麻省1975)都可以使用预先计算的表。Ai的反照率在乐队——你可以设置这个翻转从高价值(明亮,冰冷的)低价值(黑暗,揭露了)当温度超过临界值时,如273 K。整个等式的左边,然后,是气候的太阳能吸收带。< / p > < p > A + B Ti是乐队的红外电源发出的空间。A和B是常数(不是Ai反照率),和Ti是乐队的温度,通常与一个等温地球开始,然后进行迭代,直到模拟稳定。A和B的值不同文献中,但在最近的一篇论文我使用一个= 214.5 W m ^ 2, B = 1.57 W m ^ 2 K ^ 1。

The last term is the energy moved from one band to another, in a very simple (read "crude") "diffusion" model. C is a constant, Ti is and temperature, and T is the average temperature of the whole globe (I'll tell you how to calculate that in a minute). Again, values for C vary. Recently I used C = 3.74 W m^-2 K^-1.

You can find the mean global albedo or temperature or what-have-you by multiplying the zonal values by the zonal area fraction and adding up all the figures. You find the fraction of a hemisphere's area by subtracting the sine of the more equatorial boundary from the sine of the more polar boundary. For instance, the area of a hemisphere between 30 and 40 degrees is sin(40) - sin(30) = 0.1428.

P. Chýlek, J.A. Coakley, Jr. J. Atmos. Sci. 32, 675-679 (1975).

S.G. Warren, S.H. Schneider. J. Atmos. Sci. 36, 1377-1391 (1979).

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