最近30从www.hoelymoley.com 2023 - 07 - 10 - t21:25:06z //www.hoelymoley.com/feeds/question/9428 https://creativecommons.org/licenses/by-sa/4.0/rdf //www.hoelymoley.com/q/9428 5 怪不得我 //www.hoelymoley.com/users/6643 2017 - 01 - 09 - t15:54:43z 2017 - 01 - 10 - t06:05:46z

< p >我用的输出< a href = " http://www.weather.bg/0index.php?koiFail=txt/textAladin&lng=1" rel="nofollow noreferrer">ALADIN meteorological model, in particular I need volumetric soil moisture. However, the output from ALADIN is in "funny" units - $\mathrm{kg/m}^2$ per top 1 cm. Is it acceptable to assume water density $\rho_w = 1000~\mathrm{kg/m}^3$ and calculate as follows: $\frac{\mathrm{kg}}{\mathrm{m}^2.\mathrm{cm}}*\frac{100}{1000} = \frac{\mathrm{m}^3}{\mathrm{m}^3}$? So basically just take the values and divide them by 10? I don't need to be extra scientific, but I don't want to miss anything crucial.

//www.hoelymoley.com/questions/9428/-/9432 # 9432 2 弗雷德 //www.hoelymoley.com/users/2470 2017 - 01 - 10 - t01:37:58z 2017 - 01 - 10 - t01:37:58z

< p >你的方法是好的,然而,< a href = " https://water.usgs.gov/edu/density.html " rel = " nofollow noreferrer " >水密度随温度< / >。< / p > < p >不知道你的目的到底是什么或如何准确/挑剔的你想要,你可能要考虑使用水的密度的温度,而不是接受平均1000公斤/ m ^ 3美元。< / p >

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