# How to convert soil moisture units from meteorological model?

I am using the outputs of the 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.

• If you are worried about solutes or temperature changing the water density, you may have to look into the underlying methods in the ALADIN model - it may assume pure water. I suspect the density correction wouldn't be significant relative to the model uncertainty. The other issue is how you are going to use the water content. It may not be appropriate to use the value for the top 1 cm - that gives you a boundary content but probably does not represent the mass present in a greater soil thickness. – haresfur Jan 10 '17 at 23:08
• @haresfur Thank you for your insight. As I commented on the answer below, I will be using the data as input for the MEGAN model (biogenic emissions). The documentation isn't too thorough, so I have to make assumptions - for example the variable description from the sample input data "volumetric soil moisture in top cm" suggests that the value for the top 1 cm should be just what I want. – Janina Jan 11 '17 at 8:28

Not knowing exactly what your aim is or how accurate/fussy you want to be, you may want to consider using the density of water for its temperature rather than accepted average of $1000 kg/m^3$.