| Symbol | Meaning | Typical units |
|---|
| SWE | Snow‑water equivalent (depth of liquid water if the snowpack melted) | m w.e. (metres of water) or mm w.e. |
| ρsnow | Bulk density of the snowpack | kg m⁻³ |
| ρwater | Density of liquid water (≈ 1000 kg m⁻³ at 0 °C) | kg m⁻³ |
| D | Snow depth you want to find | m |
The mass of snow per unit ground area can be written in two equivalent ways:
$
\text{mass per area} = \rho_{\text{water}} ; \text{SWE}
= \rho_{\text{snow}} ; D
$
Solve for D:
$
\boxed{D = \frac{\rho_{\text{water}}}{\rho_{\text{snow}}}; \text{SWE}}
$
-
Express SWE in metres of water equivalent (m w.e.)
Example: 250 mm w.e. = 0.25 m w.e.
-
Convert the snow density to kg m⁻³
Fresh powder: 50‑100 kg m⁻³
Settled seasonal snow: 200‑400 kg m⁻³
Old firn / wind‑packed: 400‑600 kg m⁻³
-
Insert the numbers:
D = (1000 kg m⁻³) / (ρ_snow) × SWE
| Example | SWE | ρsnow | Depth D |
|---|
| Light powder | 0.25 m | 80 kg m⁻³ | 3.1 m |
| Settled snow | 0.25 m | 300 kg m⁻³ | 0.83 m |
| Dense firn | 0.25 m | 500 kg m⁻³ | 0.50 m |
| Issue | How to avoid |
|---|
| Mixed units (e.g., SWE in mm but density in g cm⁻³) | Convert everything to SI: metres and kg m⁻³. |
| Using “snow‐density ratio” values (e.g., 0.25 means 25 % of water density) | Then use $D = \text{SWE} / \text{ratio}$ directly, because the ratio already accounts for ρwater. |
| Assuming ρwater other than 1000 kg m⁻³ | Only matters for very precise work; at 0–5 °C the variation is <0.2 %. |
- Depth (m) ≈ 1 × 10³ / ρsnow (kg m⁻³) × SWE (m w.e.)
If you supply numeric values for SWE and density, you can plug them straight into the boxed equation to get the snow depth.