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February 2019 Discussion I


NorEastermass128
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39 minutes ago, Connecticut Appleman said:

FWIW - water reaches a maximum density at approx 4 C or 40 F.  The density of water decreases as you approach the freezing point.

Yup...below 4C the water molecules start to arrange themselves in hexagonal arrangements in preparation for freezing at 0C. I’ve found you need to get the wetbulbs above that 3-4C mark to really start melting the snow in a torch.

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1 hour ago, #NoPoles said:

@CoastalWx @OceanStWx @dendrite

Or any other math person. How do you calculate the weight of a cubic yard of snow? I googled a cubic foot of snow generically weighs 20lbs. How would you calculate for my area based on the water content of the storm that just fell?

I dunno ... 170 lbs...

figure 10:1 and go from there...  If 1700 lb is a cubic yard of liquid water, that implies 1700/10:1, which is the thus (1700/10) X 1 = 170

But ...real life seldom resembles real numbers...  If you "fill" snow into a yardXyardXyard bin... it's probably highly aerated and therefore, not really 10:1... more like 20 :1 or perhaps 15 ...or something less dense than 10:1 ... So you gotta kinda use your head.

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I think I see what she's asking for...

The weight of water in lbs as a function of snow depth and or volume (length,width,height)...

Density of water = 997 kg/m^3.

Assuming a volume of 1 cubic yard:

weight(depth) = ((997 kg/m^3)*[1 yard]^3*[(3 ft/yard)^3]*[(12 inch/ft)^3]* [( 2.54 cm/inch)^3]* [(1m/100cm)^3)])*2.20462 lb/kg == 1680 lb

You can also change it to (assuming length/width/height is measured in yards):

weight(length,width,height) = ((997 kg/m^3)*[length*width*height]*[(3 ft/yard)^3]*[(12 inch/ft)^3]* [( 2.54 cm/inch)^3]* [(1m/100cm)^3)])*2.20462 lb/kg

Just be careful with the units. I made these two conversions assuming you measured a depth/length/width/height in yards. You'll have to exclude some terms if you utilize a different unit of measurement.

edit: weight not wight...

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1 minute ago, MegaMike said:

I think I see what she's asking for...

The weight of water in lbs as a function of snow depth and or volume (length,width,height)...

Density of water = 997 kg/m^3.

Assuming a volume of 1 cubic yard:

weight(depth) = ((997 kg/m^3)*[1 yard]^3*[(3 ft/yard)^3]*[(12 inch/ft)^3]* [( 2.54 cm/inch)^3]* [(1m/100cm)^3)])*2.20462 lb/kg == 1680 lb

You can also change it to (assuming length/width/height is measured in yards):

weight(length,width,height) = ((997 kg/m^3)*[length*width*height]*[(3 ft/yard)^3]*[(12 inch/ft)^3]* [( 2.54 cm/inch)^3]* [(1m/100cm)^3)])*2.20462 lb/kg

Just be careful with the units. I made these two conversions assuming you measured a depth/length/width/height in yards. You'll have to exclude some terms if you utilize a different unit of measurement.

edit: weight not wight...

Adding a SWE term:

If snow water equivalent (SWE) is known (in inches)...

weight(depth) = ((997 kg/m^3)*[(SWE)^3]* [( 2.54 cm/inch)^3]* [(1m/100cm)^3)])*2.20462 lb/kg

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We ice! 

 

A wintry mess is forecast across much of Southern New England. A brief burst of snow and sleet followed by a longer period of freezing rain, especially across the interior and away from the coastline. Greater concern is ice accretion on all surfaces beginning Saturday night into the first-half of Sunday. Hazardous travel impacts. At times, reductions to visibility.

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11 minutes ago, ineedsnow said:

We ice! 

 

A wintry mess is forecast across much of Southern New England. A brief burst of snow and sleet followed by a longer period of freezing rain, especially across the interior and away from the coastline. Greater concern is ice accretion on all surfaces beginning Saturday night into the first-half of Sunday. Hazardous travel impacts. At times, reductions to visibility.

Where is this from 

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2 minutes ago, STILL N OF PIKE said:

We are gonna will Sunday into a festival of shotgun blasts over the interior that herald in the new favorable pattern for coastals...Rev is grabbing the keys to the Bus, if the STJ stays Active we may have some fun

Or we're going to see a crusty inch of snow/sleet followed by 5 hours of ZR and then 33F rain followed by a spike to 44F during FROPA and then everything refreezes into a thin glacier we can walk on top of. 

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40 minutes ago, Ginx snewx said:

Or use an already developed roof snow load calculator  

:lol: I thought she was looking for how the units canceled out and a simple function for the weight of snow/water on a flat plane. This is what I did. But, last point and I'll stop discussing this topic... If you're looking for the weight of snow/water on an incline/roof, you can take the weight of the snow *times* the cosine of theta (angle of the roof). So,

weight.perp.roof(depth[inches],theta[degrees]) = (((997 kg/m^3)*[(SWE)^3]* [( 2.54 cm/inch)^3]* [(1m/100cm)^3)])*2.20462 lb/kg)*cos(theta)

I'm sure the younger audience on this forum would find it helpful to see how the units cancel out.

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