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# How do I weigh snow across a circle?

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Maybe. If the snow were as dense as water ... ;-)

I get what you're saying, but that's not the point. If the question is, "I have two inches of snow on my roof, how much does it weigh?" Then the answer is "First, define the density of the snow relative to liquid water, and then you have an equivalent weight of water." If the density is X percent less than water, then the snow weighs (volume of snow * ((weight of water) - X%))."

So, asking what snow weighs is a little pointless without knowing the density, which is highly variable.

Which is what others said above, so ... screw it. ;-)

posted by Cool Papa Bell at 10:09 PM on January 13, 2010

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# How do I weigh snow across a circle?

January 13, 2010 6:18 PM Subscribe

I want to figure out how much a bunch of snow weighs. If I have a relatively even distribution of snow two-feet high and spread evenly across a twenty-four foot circle, where do I start?

This is hypothetical so I imagine I would need to assume the snow, per metric, weights more or less X pounds per unit at all times and without any constant melting and refreezing. Then I suppose I would seek the volume.

How much does a unit of snow weigh, assuming some regular density? What math should follow?

This is hypothetical so I imagine I would need to assume the snow, per metric, weights more or less X pounds per unit at all times and without any constant melting and refreezing. Then I suppose I would seek the volume.

How much does a unit of snow weigh, assuming some regular density? What math should follow?

Fresh snow is usually on the order of 5-15% of the density of water (1g/ml) and it'll get more dense over time.

posted by pappy at 6:31 PM on January 13, 2010

posted by pappy at 6:31 PM on January 13, 2010

You are describing a right cylindrical prism of snow. The volume of such a cylinder can be given by:

volume = height * (base area)

where

base area = pi * (diameter/2)^2

By "24 foot circle", I assume you mean a circle with a diameter of 24 feet. Therefore, the volume of snow is around 905 cubic feet.

The mass of the snow is given by:

mass of snow = volume of snow * mass density of snow

The density of snow varies quite significantly depending on how well it's packed. This article suggests that the density of new fallen snow is around 0.08 g*cm^-3, but it often compresses up to around 0.3 g*cm^-3. Using those limits, the mass of your snow is somewhere between:

4500 lbs and 17000 lbs

posted by Salvor Hardin at 6:32 PM on January 13, 2010 [3 favorites]

volume = height * (base area)

where

base area = pi * (diameter/2)^2

By "24 foot circle", I assume you mean a circle with a diameter of 24 feet. Therefore, the volume of snow is around 905 cubic feet.

The mass of the snow is given by:

mass of snow = volume of snow * mass density of snow

The density of snow varies quite significantly depending on how well it's packed. This article suggests that the density of new fallen snow is around 0.08 g*cm^-3, but it often compresses up to around 0.3 g*cm^-3. Using those limits, the mass of your snow is somewhere between:

4500 lbs and 17000 lbs

posted by Salvor Hardin at 6:32 PM on January 13, 2010 [3 favorites]

Wikipedia indicates snow density varies from 0.08g/cm

posted by ssg at 6:32 PM on January 13, 2010

^{3}to 0.50g/cm^{3}.posted by ssg at 6:32 PM on January 13, 2010

wikipedia gives new snow as roughly 1/13 the density of water, but as above 'it varies greatly'.

You may as well use 1/10 the density of water to make the arithmetic simpler.

So (2 feet)*pi*12

or to the degree of accuracy we can hope for, 'a couple of tons'.

posted by hexatron at 6:37 PM on January 13, 2010

You may as well use 1/10 the density of water to make the arithmetic simpler.

So (2 feet)*pi*12

^{2}*62.4/10 (water is 62.4 lbs/ft^{3}) is about 5600 lbs,or to the degree of accuracy we can hope for, 'a couple of tons'.

posted by hexatron at 6:37 PM on January 13, 2010

Even newly fallen snow varies greatly in density. When it's near freezing, an inch of snow cover is extremely heavy to shovel, but when it's closer to zero (F), an inch hardly weighs anything.

posted by advicepig at 6:46 PM on January 13, 2010

posted by advicepig at 6:46 PM on January 13, 2010

By the way, if this is real snow that you have access to, you can figure out the density for yourself - just scoop out a chunk of snow that you know the volume of (say a 2'x2'x2' piece) and weigh it (melt it in a pail first if that makes it easier to weigh)

posted by Salvor Hardin at 6:51 PM on January 13, 2010

posted by Salvor Hardin at 6:51 PM on January 13, 2010

In this table, from this paper referenced in the Wikipedia article for snow, the authors compute that the snow-to-liquid ratio for the Boston, MA area (assuming that's where you are based on your profile location) was an average of 10.7, with a 25th and 75th percentiles of 6.8 and 13.3. This gives an average estimate of 5,276 lbs, with the 25th to 75th percentiles corresponding to 4,244 lbs and 8,302 lbs respectively. So, basically it's probably a little over two and a half tons, but could reasonably be expected to be anywhere from two tons to four tons, depending on conditions.

posted by mhum at 6:56 PM on January 13, 2010

posted by mhum at 6:56 PM on January 13, 2010

Isn't snow just frozen water? Wouldn't it then weigh about 8.5. per gallon? Why does density play a factor at all?

posted by Cool Papa Bell at 7:11 PM on January 13, 2010

posted by Cool Papa Bell at 7:11 PM on January 13, 2010

Cool Papa Bell: "

Because it doesn't pack into a solid sheet of ice when it falls. It always has some air trapped in it (making it a pretty decent insulator, btw). Depending on the conditions both of when the snow is formed and those on the ground it will be more or less packed, and therefore more or less dense.

posted by ArgentCorvid at 7:16 PM on January 13, 2010 [1 favorite]

*Isn't snow just frozen water? Wouldn't it then weigh about 8.5. per gallon? Why does density play a factor at all?*"Because it doesn't pack into a solid sheet of ice when it falls. It always has some air trapped in it (making it a pretty decent insulator, btw). Depending on the conditions both of when the snow is formed and those on the ground it will be more or less packed, and therefore more or less dense.

posted by ArgentCorvid at 7:16 PM on January 13, 2010 [1 favorite]

Cool Papa Bell:

Imagine two buckets, one filled with water and one with snow. Would they weigh the same?

ArgentCorvid:

And, in fact, ice is approximately 10% less dense than water. So, there's that too.

posted by mhum at 7:26 PM on January 13, 2010

*Isn't snow just frozen water? Wouldn't it then weigh about 8.5. per gallon?*Imagine two buckets, one filled with water and one with snow. Would they weigh the same?

ArgentCorvid:

*Because it doesn't pack into a solid sheet of ice when it falls.*And, in fact, ice is approximately 10% less dense than water. So, there's that too.

posted by mhum at 7:26 PM on January 13, 2010

Measure the deflection of your trampoline with a known weight. When it snows, this will give you a hint as to how close the snow is to that weight.

posted by odinsdream at 7:31 PM on January 13, 2010

posted by odinsdream at 7:31 PM on January 13, 2010

There are people who spend their lives cutting and measuring and blasting snow for avalanche control. They would know a lot about this. Parks Canada spends $1.7 million annually doing this. Maybe contact them.

posted by weapons-grade pandemonium at 7:47 PM on January 13, 2010

posted by weapons-grade pandemonium at 7:47 PM on January 13, 2010

My father and I once calculated (during the 2003 Colorado blizzard) that the snow on our (round) trampoline weighed somewhere north of a few tons, just using basic volume/density/mass equations.

posted by dantekgeek at 8:11 PM on January 13, 2010

posted by dantekgeek at 8:11 PM on January 13, 2010

*Imagine two buckets, one filled with water and one with snow. Would they weigh the same?*

Maybe. If the snow were as dense as water ... ;-)

I get what you're saying, but that's not the point. If the question is, "I have two inches of snow on my roof, how much does it weigh?" Then the answer is "First, define the density of the snow relative to liquid water, and then you have an equivalent weight of water." If the density is X percent less than water, then the snow weighs (volume of snow * ((weight of water) - X%))."

So, asking what snow weighs is a little pointless without knowing the density, which is highly variable.

Which is what others said above, so ... screw it. ;-)

posted by Cool Papa Bell at 10:09 PM on January 13, 2010

To figures out your specific snow, you need one (additional) peice of information.. the density.

So, grab a bucket. Fill it with snow, without compacting it (this method of doing this is left up to the reader)

Bring it inside. it will melt. Take the water and measure the amount. (Measuring cups, etc.). You now have your density! You know the volume of the bucket, and you know the weight of the snow (the weight of the melted snow, which is the volume of the water x it's density).

posted by defcom1 at 11:59 PM on January 13, 2010

So, grab a bucket. Fill it with snow, without compacting it (this method of doing this is left up to the reader)

Bring it inside. it will melt. Take the water and measure the amount. (Measuring cups, etc.). You now have your density! You know the volume of the bucket, and you know the weight of the snow (the weight of the melted snow, which is the volume of the water x it's density).

posted by defcom1 at 11:59 PM on January 13, 2010

keep in mind that as snow melts or rain falls on it, it becomes saturated with water/ice and will get MUCH heavier.

posted by subarctic_guy at 7:53 AM on January 14, 2010

posted by subarctic_guy at 7:53 AM on January 14, 2010

The whole 'filling a bucket with snow' thing wouldn't work very well either (unless the snow falls directly in the bucket), because any scooping you do packs it. Snow that has been shoveled (or plowed) previously is heaver than fresh stuff. Just ask anyone that has had to re-shovel the end of their driveway after the plow goes by.

posted by ArgentCorvid at 9:55 AM on January 14, 2010

posted by ArgentCorvid at 9:55 AM on January 14, 2010

The responses above based on the range of snow densities are about as close as you'll get to a broadly applicable estimate. Once you have a actual instance of an "even distribution of snow two-feet high and spread evenly across a twenty-four foot circle" you can get a result for that particular instance, and with more confidence, by sampling it:

- Weigh an empty bucket

- Carefully cut a core sample of snow - vertical sided, and a shape you can easily measure; say a 1' x 1' square.

- Weigh the bucket to get the increase due to the snow. We no longer care about how the snow packs into the bucket, or even whether it melts. It won't affect the weight, and we know what the original volume was (2 cubic feet, in this case.)

- What fraction is the 2 cubic feet of the original area? Divide the weight of snow in the bucket by that fraction - that's your estimate for the 24' circle.

You needn't even account for the snow depth, based on your assertion that the that it's an even distribution. You can do the fraction calculation above with the ratio of just the areas: (1 [sq ft] / ((24^2 * Pi) / 4) [sq ft]) * the weight of the snow in the bucket.

posted by TruncatedTiller at 2:38 PM on January 14, 2010

- Weigh an empty bucket

- Carefully cut a core sample of snow - vertical sided, and a shape you can easily measure; say a 1' x 1' square.

- Weigh the bucket to get the increase due to the snow. We no longer care about how the snow packs into the bucket, or even whether it melts. It won't affect the weight, and we know what the original volume was (2 cubic feet, in this case.)

- What fraction is the 2 cubic feet of the original area? Divide the weight of snow in the bucket by that fraction - that's your estimate for the 24' circle.

You needn't even account for the snow depth, based on your assertion that the that it's an even distribution. You can do the fraction calculation above with the ratio of just the areas: (1 [sq ft] / ((24^2 * Pi) / 4) [sq ft]) * the weight of the snow in the bucket.

posted by TruncatedTiller at 2:38 PM on January 14, 2010

This thread is closed to new comments.

posted by walleeguy at 6:23 PM on January 13, 2010