## Thursday, December 27, 2012

### A Ton Of Snow

The other day my students were taking their pre-calc exam and due to my writing it too long, many needed to stay after class was over to finish. So many in fact, that I emailed "Wow, I have a ton of students needing to stay after!" to my principal.

After hitting send, I was instantly annoyed that I had used "ton" as an exaggeration -- something I have been known to chide my students about. But I was comforted when I did a quick estimation and realized that I was actually fairly accurate:
$\frac{1 ton}{ } \frac{2000 lb}{1 ton} \frac{1 student}{120 lb} \approx 17 students$

As it turned out, I had 14 students in my room, and many of them were the bigger athletes (certainly bigger than 120 lb) so I'm sticking to my original statement as literally true. And is 120 lb an appropriate average weight of a student?  Who knows.

Well, today I was out shoveling snow for the first time this season, and I caught myself again saying "This is a ton of snow!"  Well, was it? Was it really?

To calculate the weight of snow, I need to find the volume of snow on my driveway, and multiply it by the density of snow.  The volume of snow is easy enough to approximate -- I'll assume I have an even rectangular driveway with the same height of snow everywhere, and so volume of this rectangular prism is just length * width * height:
The density of snow is a little more difficult, as there is heavy snow, light fluffy snow, solid ice, etc.  Wikipedia says snow has a density of anywhere from 8% - 50% of water, depending on many things, but mostly on how compacted it is as it melts, freezes, melts, refreezes and more snow falls on top of it. This was relatively new snow, and so I'll assume it has a density of about 25% of water.

The density of water is 1 kg/L.  This "coincidence" of having such a clean number is actually not a coincidence at all, but was by design -- as 1 kg was originally defined to the weight of 1 liter of water as the metric system was being invented. Since then we have defined the kg more precisely than that using more complicated methods. Though it's no longer exactly 1, this value remains accurate enough for our purposes (I think it's 1.003 or something close?)  Converting kg to pounds and L to cubic inches is a tough exercise:
$\rho_{water} = \frac {1 kg}{L} \frac {2.21 lb}{1 kg} \frac {1 L}{1000 cm^{3}} \left (\frac {2.54 cm} {1 in} \right )^{3} = 0.036 \tfrac{lb}{in^3}$

If my snow was 25% the density of water, than:
$\rho_{snow} = .25*\rho_{water}= .25*0.036 \tfrac{lb}{in^3} = .009 \tfrac{lb}{in^3}$

The weight of the snow on my driveway is then:
$W = V_{snow}* \rho_{snow} = \left (276,480 in^3 \right ) * \left (.009 \tfrac{lb}{in^3} \right ) = 2503 lb$

Looks like I wasn't exaggerating after all! I moved over a ton of snow today!