Let's not drag this out.

So, I've taken another shot at the snow-sticking-to-my-windshield problem. After scouring the internet for answers, I decided to ask the good people at physicsforums.com if they had any idea what might be the cause of my observed phenomenon. Their answer? Friction and drag.

In other words, I was thinking too hard about the problem. Drag from the air pushes a snowflake up the windshield. Gravity pulls it down. And friction resists its movement. It's just Newton's laws, of course.

But it's not quite as simple as that, because the math for determining whether or not the snowflake sticks involves approximating some constants based on the properties of the snowflake, my windshield, etc.

The biggest one is this: what is the coefficient of friction between a snowflake and glass? That is, how easy is it to move a snowflake across a windshield? It turns out there's no simple or single answer to this question, because it depends very much on the properties of pretty much everything: the temperature of the windshield, the moisture of the snow, the shape of the snowflake, etc.

There's a fair amount of research out there on the friction between skis and snow, but not much about snowflakes on glass. The best I could find was this paper, but their one example was a little unclear on the total mass of snow present, so it was difficult to extract a figure for μ.

Anywho, eventually I decided I would simply test the hypothesis that friction and drag are the culprits. Once again, the equation for terminal velocity appears, but this time in the form of the drag equation. It is: F = ½ρv²C_DA. We need to know the area of a snowflake and its drag coefficient. Like a good physicist, I'm going to imagine a spherical snowflake, giving it a drag coefficient of .1 and an area of (assuming a 1 cm diameter) 8x10^-5 m. All in all, assuming a speed of 10 mph, this works out to a force of 10^-4 N pushing on the snowflake.

This force is pushing directly on the snowflake, however, and the snowflake is sitting on an inclined surface. So we have to find the component of that force pointing in the direction of the surface, which means we have to multiply that result by the cosine of the angle of my windshield (about 40°), giving us a force of 7.8x10^-5 N.

Because this drag is pushing the snowflake up the windshield, gravity opposes the motion. The force here is much smaller, amounting to 1.9x10^-6 N. Because the drag force is so much greater than the weight, we know that the snowflake will move upwards and be resisted by friction. The question is, how much friction does it take to resist motion? Here we must use Newton's first law, which says that objects experiencing no net force experience no acceleration.

Therefore, 7.8x10^-5 - 1.9x10^-6 - μN = 0, which we can rearrange as μ = (7.8x10^-5 - 1.9x10^-6)/N to find μ. N is the normal force, which is the windshield's equal and opposite reaction to the snowflake's weight. Solving this equation, I get a μ of 40. What does that mean? Well, it means that my answer is probably wrong. But if it were correct, it would mean the friction holding the snowflake down would have to be extremely strong to resist any motion at all.

There are a couple possible solutions here. One is that there is a lot of friction between a snowflake and a plate of glass because an individual snowflake is a jagged crystal. Another possibility is that the critical speed is much lower than 10 mph. This is possible, except that bringing μ down to what I would expect means the critical speed would need to be something like 1 mph, which is not really what I observed. The only other variable with a lot of uncertainty is the drag coefficient, which I'm quite sure is wrong. A snowflake isn't really a smooth sphere. But a more realistic drag coefficient would probably increase the drag force, leading to an even higher value for μ.

All in all, I'm not satisfied with this answer. The result is less unreasonable than my last one, but still not exactly a perfect match for the data. My conclusion is that there's more going on here than just drag and friction. But I think that's about all the energy I have for this problem.