Re: How fast do snowflakes settle? How much snow is in the air?

I don't know, but I can give an estimate. My analysis of the problem will be rough: don't expect an answer more accurate than a factor of 3.

First, what is the rate of accumulation for a typical snow fall? In my experience, a typical snowfall accumulates at about a rate of 3 cm per hour or so (give or take a factor of 2). From Table 1.3 of this site, 3 cm of snow per hour corresponds very roughly to .15 cm /hour of liquid water (the rest is air). Water weighs 1 gram / cm^3, which means that the mass of water falling is about 1.5 kilograms per square meter per hour, or .4 grams per square meter per second

Now, at what speed does a snowflake fall? I'll make a rough "eyeball" estimate of .3 m/sec. Now, the mass of snow suspended in air per cubic meter (s=unknown), the rate at which it falls (v=.3 m/s), and the rate of accumulation per unit area (a=.4 g/(m^2 s)) are given by the equation

a = s * v

s

This number will vary widely depending on how hard it's snowing.

Let's try a different way to approach the same problem. We can estimate how much snow is in the air by using the fact that it's hard to see in heavy snow: the visibility is reduced. You may skip the next paragraph, explaining my methods of calculation, if you like.

During a snowfall, visibility is limited, often to around 200 meters. Imagine light rays coming out from a light source. They will travel outward for a significant distance before striking a snowflake. At a certain distance, almost none of them will have avoided snowflakes. This distance is the visibility. During a snowstorm, the visibility is often only 300 meters or so. It turns out (see me for calculations) that the average distance a light ray travels before striking a snowflake is about 1/3 the visibility distance, or 100 meters. This distance is called the "mean free path". Imagine a box of air one square meter across and 100 meters long. If we make the very rough approximation that snowflakes are opaque spheres 2 mm in radius, I find that if there are 80,000 snowflakes in the box, then each light ray passing through the box will strike one on average. So the number of snowflakes in a 1 cubic meter box is about 800. Now, assume that the density of these "spherical" snowflakes is about .1 g/cm^3. (This is less than the true density of ice (.9 g/cm^3), to roughly correct for the fact that snowflakes aren't spheres). Each snowflake thus weighs about .008 grams, so the total weight of snow is about 6.4 grams per cubic meter.

From the rate-of-snowfall calculation, we get about 1.3 grams per cubic meter, while from the visibility calculation, we get about 6.4 grams per cubic meter. The visibility calculation is much rougher (mostly because it depends on the shape and structure of snowflakes), so I'd guess the first number is more accurate... but doing the calculation two ways lets you see how large the uncertainty is. I also think my estimate of 300-meter visibility implies a heavier snowfall than the 3 cm/hour accumulation I assumed in the first part.

References:

• definition of visibility
• basic meteorological parameters
• Measuring snowfall, depth, and liquid equivalent