<p>The other answers are correct that you need to know the soil depth to convert from mass units (kg m-2) to volumetric units (m3 m-3).  However, when comparing these variables you should also be aware that satellite soil moisture is typically measured only over the top few centimetres of soil. Total column soil moisture from a climate model will typically be the sum over several metres, so the dynamics of these two variables are likely to be quite different (contrast the red and light blue lines in the upper plot).</p>

<p>It's more common to compare satellite data with top-layer soil moisture and even then to normalize each dataset in some way, e.g, </p>

<p>$$NWI = \frac{SM - SM_{min}}{SM_{max} - SM_{min}}$$</p>

<p>or as normalised anomalies,</p>

<p>$$NWI = (SM - \overline{SM}) / \sigma_{SM}$$</p>

<p>I've used the latter to normalize the lower plot.  Normalization gets around the units problem, but doesn't get around the fact that you're comparing physically different quantities.</p>

<p><a href="https://i.sstatic.net/qp8vK.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/qp8vK.png" alt="Daily soil moisture"></a></p>

CC BY-SA 4.0
2018-09-03T12:59:04.253Z
2018-09-03T12:59:04.253Z
Deditos
<ol>
<li>Divide by the depth of the soil</li>
<li>Divide by the density of water, or something resembling density.</li>
</ol>

CC BY-SA 4.0
2018-08-01T21:24:32.673Z
2018-08-01T21:24:32.673Z
BarocliniCplusplus
<p>Not one that I can immediately see, in order to derive a per cubic metre figure (m<sup>3</sup>) you need to know how many cubes of soil you have in a given column unit, that requires the depth of the unit in question.</p>

CC BY-SA 4.0
2018-05-30T17:34:34.000Z
2018-07-30T23:33:15.497Z
2018-07-30T23:33:15.497Z
Ash
<p>I will cite <a href="https://medium.com/@wenzhao.li1989/soil-moisture-convert-mm-kg-m-2-and-m%C2%B3-m-3-e52be1fe98ef" rel="nofollow noreferrer">this medium post</a> referring to LDAS datasets</p>
<blockquote>
<p>Many of the output fields for water amounts (e.g., precipitation, evapotranspiration, soil moisture, etc.) are given in units of <span class="math-container">$[kg\ m^{-2}]$</span>. Many users prefer units of <span class="math-container">$[mm]$</span>. If the assumption is made that the density of the water in the soil is 1000 <span class="math-container">$[kg\ m^{-3}]$</span>, then the value in <span class="math-container">$[kg\ m^{-2}]$</span> is identical to the value in <span class="math-container">$[mm]$</span>:</p>
</blockquote>
<p>Also, the post explains some datasets present the data in</p>
<blockquote>
<p>The NCA-LDAS dataset currently provide soil moisture values in units of <span class="math-container">$[m^3\ m^{-3}]$</span> for volumetric soil moisture. To convert to units of <span class="math-container">$[kg\ m^{-2}]$</span> for the total soil moisture amount in each layer, all you do is multiply the <span class="math-container">$[m^3\ m^{-3}]$</span> value by the thickness of the layer in <span class="math-container">$[mm]$</span>.</p>
</blockquote>
<p>So. You can divide the data by the soil depth.</p>

CC BY-SA 4.0
2024-01-31T07:51:41.270Z
2024-01-31T07:51:41.270Z
Robert Chang
<p>I have soil moisture data from Regional Climate Models expressed in $kg/m^2$. Τhis quantity is the total soil moisture that is estimated for the land column. The depth of this column is dependent on the land surface model that was used.</p>

<p>How could I transform soil moisture units expressed in $kg/m^2$ to $m^3/m^3$ so that model outputs are comparable to remote sensing soil moisture data that are expressed in $m^3/m^3$? Is there a way of doing this transformation without taking into consideration the soil depth?</p>

CC BY-SA 4.0
2018-05-30T22:06:46.650Z
Can you look at how the model came up with the soil moisture numbers in those weird units? The only thing I can think of is that it is taking precipitation input over a period of time and subtracting losses so the soil moisture is essentially computed from mass balance. Then you need to consider what depth the remote sensing assumes. See how well they compare if you use the same depth.
haresfur
CC BY-SA 4.0
2018-07-29T20:00:53.417Z
Possible duplicate of https://earthscience.stackexchange.com/questions/9428/how-to-convert-soil-moisture-units-from-meteorological-model
BarocliniCplusplus
<soil-moisture>
Transformation of soil moisture units
Maria Karypidou
# Transformation of soil moisture units

(Asked by: username_0 on 2018-05-30)

I have soil moisture data from Regional Climate Models expressed in $kg/m^2$. Τhis quantity is the total soil moisture that is estimated for the land column. The depth of this column is dependent on the land surface model that was used.

How could I transform soil moisture units expressed in $kg/m^2$ to $m^3/m^3$ so that model outputs are comparable to remote sensing soil moisture data that are expressed in $m^3/m^3$? Is there a way of doing this transformation without taking into consideration the soil depth?

**Tags:** soil-moisture

> **Comments:**
> **username_1**: Can you look at how the model came up with the soil moisture numbers in those weird units? The only thing I can think of is that it is taking precipitation input over a period of time and subtracting losses so the soil moisture is essentially computed from mass balance. Then you need to consider what depth the remote sensing assumes. See how well they compare if you use the same depth.  
> **username_2**: Possible duplicate of https://earthscience.stackexchange.com/questions/9428/how-to-convert-soil-moisture-units-from-meteorological-model  

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## 4 answers

### Answer by username_3 on 2018-09-03. Score: 4

The other answers are correct that you need to know the soil depth to convert from mass units (kg m-2) to volumetric units (m3 m-3). However, when comparing these variables you should also be aware that satellite soil moisture is typically measured only over the top few centimetres of soil. Total column soil moisture from a climate model will typically be the sum over several metres, so the dynamics of these two variables are likely to be quite different (contrast the red and light blue lines in the upper plot).

It's more common to compare satellite data with top-layer soil moisture and even then to normalize each dataset in some way, e.g,

$$NWI = \frac{SM - SM\_{min}}{SM\_{max} - SM\_{min}}$$

or as normalised anomalies,

$$NWI = (SM - \overline{SM}) / \sigma\_{SM}$$

I've used the latter to normalize the lower plot. Normalization gets around the units problem, but doesn't get around the fact that you're comparing physically different quantities.

[![Daily soil moisture](https://i.sstatic.net/qp8vK.png)](https://i.sstatic.net/qp8vK.png)


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### Answer by username_5 on 2024-01-31. Score: 1

I will cite [this medium post](https://medium.com/@wenzhao.li1989/soil-moisture-convert-mm-kg-m-2-and-m%C2%B3-m-3-e52be1fe98ef) referring to LDAS datasets

> Many of the output fields for water amounts (e.g., precipitation, evapotranspiration, soil moisture, etc.) are given in units of $[kg\ m^{-2}]$. Many users prefer units of $[mm]$. If the assumption is made that the density of the water in the soil is 1000 $[kg\ m^{-3}]$, then the value in $[kg\ m^{-2}]$ is identical to the value in $[mm]$:

Also, the post explains some datasets present the data in

> The NCA-LDAS dataset currently provide soil moisture values in units of $[m^3\ m^{-3}]$ for volumetric soil moisture. To convert to units of $[kg\ m^{-2}]$ for the total soil moisture amount in each layer, all you do is multiply the $[m^3\ m^{-3}]$ value by the thickness of the layer in $[mm]$.

So. You can divide the data by the soil depth.


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### Answer by username_2 on 2018-08-01. Score: 0

1. Divide by the depth of the soil
2. Divide by the density of water, or something resembling density.


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### Answer by username_4 on 2018-05-30. Score: 0

Not one that I can immediately see, in order to derive a per cubic metre figure (m3) you need to know how many cubes of soil you have in a given column unit, that requires the depth of the unit in question.


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