Sunday, December 30, 2012

Reindeerpower

While watching football today I saw a commercial for the Chevy Silverado featuring the Santa Salesman. Here's a Youtube clip:



Santa is poised to calculate the reindeerpower of the Silverado, but sadly the commercial ends before he can give us an answer. Thus we are left to speculate the magnitude of 1 reindeerpower.

How would we go about doing that? We need to start off by setting a few ground rules. Santa and his reindeer are clearly capable of time travel and/or superluminal speeds, which makes estimating their power output somewhat troublesome from a theoretical perspective. Like good scientists, we have to limit ourselves to the reindeers' documented behavior.

The most authoritative eyewitness account makes the claim that:
More rapid than eagles his coursers they came...
Although the authorship of this account is apparently in dispute, the two possibilities are both American. We can assume that the author was most familiar with the American Bald Eagle, which Wikipedia says can fly at about 17 m/s. Let's estimate that Santa's reindeer can achieve something on the order of 20 m/s.

Now, the record here provides us with another useful tidbit as well:
As dry leaves that before the wild hurricane fly,
When they meet with an obstacle, mount to the sky.
So up to the house-top the coursers they flew,
With the sleigh full of Toys, and St Nicholas too.
We learn from this that the reindeer are ascending while traveling faster than eagles. This gives us a very easy way to calculate their power output.

But before we go any further, we have to figure out just what power is from a physics standpoint. Most of us encounter power in two forms: the horsepower of our cars, and the wattage of our electrical items. Both measure the same thing, just in different units. Power is the rate at which energy is transferred -- the standard unit being Joules/second, the watt. 1 hp is 746 W.

So how do we measure the rate of energy transfer? Well, we know that Santa's reindeer can ascend at 20 m/s, which means they can create 20 meters worth of gravitational potential energy in just 1 second. Gravitational potential energy, as we recall, is mass * gravity's acceleration * height. Gravity's acceleration is a constant and our height is 20 meters, which just leaves us with mass as an unknown.

There are about 2 billion Christians in the world. If every one of them receives an iPad or a Kindle for Christmas, we can estimate that (with packaging) this amounts to 2 billion kg worth of presents. We can safely assume that Santa, the sleigh, and even the reindeer add nothing significant to this estimate. All that's left to do is plug in the numbers, then.

The power output for the entire sleigh comes to about 400 gigawatts. (This is an encouraging result, because it's over 300 times the power needed by the DeLorean to travel through time. And as previously discussed, Santa must rely on time travel to deliver all his presents in a single night.)

If we assume that Santa derives all his motive power from the reindeer, and that there are only 8 reindeer (Rudolph, being a mutant, likely delivers all his power to his red nose and provides no motive contribution), then each reindeer is worth 50 gigawatts. In terms of horsepower, this is 67 megahorsepower. The Chevy Silverado talked about above delivers a pitiful 315 hp, which comes out to 4.8 &#181rp (microreindeerpower).

I must reiterate that this is only an estimate. Santa clearly knows the proper conversion himself, so any confirmation on his part would be swell.

Happy holidays everybody.

Thursday, December 27, 2012

Blog with the Wind

It's been pretty windy today, and I'm reminded of a problem I tried to work out when Hurricane Sandy came through in October. As the wind howled relentlessly, thumping against windows and bending trees, I wondered: how strong a gust of wind would it take to move my car? Unsurprisingly, this idle thought did nothing to ease my fears as Frankenstorm barreled down on my house. But occupying myself with the calculations did somewhat lessen my cabin fever.

So the first question is, what's keeping my car from moving? Gravity clearly plays a role, as does friction. Because it's a much simpler calculation, I first looked at this from the standpoint of friction. This is asking how strong a gust of wind it takes to slide my car across the ground.

To answer this, we need to know how strong the friction force is that prevents my car from sliding. Friction is pretty complicated, but it can be approximated simply. Friction = N&#956, where N is the normal force, and &#956 is the coefficient of friction, a measure of how hard it is to move an object composed of material A across a surface composed of material B. &#956 is determined experimentally, but it's easy to find estimates for common materials. Between the rubber of my tires and the pavement of my driveway, a decent estimate I found is 0.9. Now, the normal force pushing up against my car is just its weight. My Mazda3 comes in at about 1400 kg, which works out to 13,720 N. Multiply this by &#956 and the total force required to slide my car is something like 12,000 N, since it's probably a little wet out there anyway.

Okay, now how many newtons is a gust of wind? We want to know the pressure the wind exerts on my car. It turns out that we've seen wind pressure before in the form of the air resistance that kept our pets from becoming apocalyptically fast. We had a formula for terminal velocity, which was v = √(2mg)/(&#961AC). How does that help us? If we look at things from the perspective of the animal, then the animal is stationary and the wind is moving past it at a speed equal to the terminal velocity. We know terminal velocity means that the wind and gravity balance each other, so the force of that wind is equal to the weight of the animal. This means we have found an equation to determine the speed of a gust of wind based on the force of that wind (and vice versa).

Plugging in the numbers (my car's profile is 6.6 m2, and its C is 1-ish), we come out with an answer of 54 m/s, or about a 120 mph gust of wind. Whew. But what about gravity? It was at this point that a friend of mine on Facebook pointed out that, according to a "reliable source," an 80 mph wind can roll over a 6-wheel army truck. Why such a disparity between the two answers? Because two different questions are being asked. When we try to overcome gravity rather than friction, we're attempting to roll the car, a task that requires torque.

This turns out to be a more difficult question to answer because at least one hidden assumption we made earlier becomes an important unknown now. The new variable is the duration of a gust of wind. We need the duration of the gust of wind because we're attempting to move the car a certain angular distance, which takes time. Specifically, we need to tip the car such that the center of mass is no longer over the wheel-base. This requires tipping the car to about 57 degrees. For now, we'll estimate that a gust of wind lasts a single second.

Now we need to figure out the torque of a gust of wind. The torque is equal to the force of the wind multiplied by the moment arm, which is the vertical distance from that force to the car's axis of rotation. The most logical axis of rotation is the tire on the side of the car opposite the wind. But where is the wind blowing? We can assume that the cumulative effect of an evenly spread out gust of wind acts at half the height of the car. If my car is 1.46 m tall, then this length is .73 m. In order to lift the car any distance, this torque must exceed the torque due to gravity. Gravity acts at the car's center of mass, which is a vertical distance of roughly .88 m from the axis of rotation, and has a force equal to the weight of the car. This means the force of the wind must be 1.2 (the ratio of the weight's moment arm to the wind's moment arm) times the weight of the car, or about 16,500 N.

But that amount of force will take approximately forever to tip my car. What matters is the force in excess of 16,500 N. To get that, we need to know the angular acceleration (&#945) required to move a particular angular distance (&#952). The equation is &#952 = &frac12&#945t2. t is 1 second, and it turns out that 57 degrees is about 1 radian, so the angular acceleration is 2 rad/s2. We multiply this by the moment of inertia of my car, which is a measure of how the mass is distributed about the axis of rotation, and we get a net torque of 5000 Nm. Adding this onto the minimum needed to move the car, and we're at 23,000 N, which is a wind speed of 73 m/s, or about 163 mph.

Wow, that's twice my friend's estimate of 80 mph, and also significantly different from an earlier calculation I did. I'm inclined to believe my previous calculation is incorrect (or at least more incorrect) because I did it earlier in the semester before actually getting to torque. But what explains the disparity between my result and my friend's reliable source? My friend's source could be wrong, but a quick perusal of Wikipedia shows that 163 mph is enough to demolish a house. My best guess is that a 1 second gust of wind is too short to be a good average.

How long would an 80 mph gust have to blow to roll my car? Well, an 80 mph gust of wind only produces 5,300 N of force, so it's never able to lift my car. The minimum 16,500 N is a 141 mph gust. A 145 mph gust would flip my car in 3 or 4 seconds. Again, Wikipedia indicates 111-135 mph is enough to lift cars off the ground. I don't know where its data come from, but that means my rough calculation is in the right neighborhood, even if a little high.

The conclusion? Unless you're dealing with tornadoes or very powerful hurricanes, your car should be safe from the wind. Unless, you know, I'm wrong.

Monday, December 24, 2012

Physics of the Apocalypse

Since we all just survived another apocalypse, let's take a look at what could have been. I've always associated impending apocalypses (apocalypsi?) with feline and canine precipitation. A quick googling shows that there isn't any particular reason for me to make this association. I think I'm probably conflating two things: raining animals generally is seen as an apocalyptic portent, and Bill Murray believed dogs and cats living together was a disaster of biblical proportions. I've learned over the years to trust Bill Murray on these sorts of things.

Anywho, just how bad would a rain of cats and dogs be? To answer that, we need to know how much energy a falling animal packs. The basic equation here is U = mgh, where U is the gravitational potential energy of a falling object, m is its mass, g is the acceleration due to gravity at the Earth's surface (~9.8 m/s2), and h is the height of the falling object above the surface. When the object falls, it loses potential energy as its height decreases. But conservation of energy tells us that this energy cannot simply be lost; it is transformed into other forms. Ideally, all of the potential energy has been turned into kinetic energy by the time the helpless animal hits the ground.

So, wikipedia says that cumulonimbus clouds, which are the kind of clouds we often see during thunderstorms, range from 2,000-16,000 meters in elevation. Now, I don't really know what sorts of clouds you'd have during the apocalypse, or what sorts of clouds precipitate animals, but let's just say your dogs and cats are falling from a nice round 10 km up.

If we have a fat cat, or a smaller breed of dog, the animal may have a mass in the neighborhood of 10 kg. This gives us a potential energy of 980,000 J. Let's just call it one megajoule. So, on a perfectly spherical, airless ball resembling the Earth, a falling animal hits you with the kinetic energy of one megajoule. What's that like? Well, that's about the same as getting hit by a car going 70 mph. That is to say, it kills you dead.

But a car is significantly more massive than a dog or cat, which suggests that the falling animal is going much faster than 70 mph when it hits. Indeed, we can calculate the speed of the animal's descent from the energy equation. The kinetic energy of a moving object is the familiar K = &frac12mv2. A megajoule of energy and a mass of 10 kg works out to a speed of 450 m/s, or about a thousand miles per hour. That is very, very fast. It's also entirely unrealistic.

(Readers may notice that the initial energy, mgh, is equal to the final energy, &frac12mv2. Since both terms contain mass, the mass cancels out. Consequently, the final speed of a falling object turns out to be √2gh, which is independent of mass, just as Galileo and Newton told us.)

We're all familiar with the concept of terminal velocity, and it would seem to play a role here. Objects in an atmosphere can only fall so fast, because there comes a point at which the upward force from air resistance is equal to downward force of the object's weight. At that point, there are no net forces, which means no acceleration according to our good friend Newton. Modeling air resistance turns out to be somewhat hard (and involves differential equations, which your humble physics student hasn't gotten to yet), but one of the interesting things about air resistance is that it increases as the speed of the object increases. In fact, it generally increases with the square of the speed. The upshot here is that falling objects reach their terminal velocity very quickly, so the height from which our animals are falling doesn't really matter all that much.

While figuring out the specifics of air resistance can be difficult, the formula for terminal velocity pops right out if you have a model for air resistance. The basic formula is v = √(2mg)/(&#961AC), where &#961 is the density of air, A is the cross-sectional area of the animal, and C is the drag coefficient, a constant that has to do with the shape of the animal. Working this all out, the terminal velocity for our 10 kg animal is around 25 m/s (56 mph). This is lower than the terminal velocity of a human-like object, and this fact has been put forward as one of the reasons why cats seem particularly capable of surviving falls from great distances.

So, with a maximum speed of 25 m/s, the kinetic energy of our raining animal is a measly 3 kJ, or roughly equivalent to the kinetic energy of a bullet. Getting shot is a bad thing, I'm told, but bullets are much smaller than cats and dogs, which means that the force exerted by the falling animal at any one point is significantly lower than that of the bullet. Getting hit by a falling cat or dog would certainly hurt, but it would be unlikely to kill you. This turns out to be a fairly mild apocalyptic event. The real danger probably arises from the fact that, once the rain has ended, there will be a surprisingly large number of living, and very pissed off, cats and dogs roaming the streets.

The bigger the animal, of course, the more dangerous it gets. Because terminal velocity is dependent on mass, the kinetic energy of a falling animal is proportional to mass3/2. Cloud-bound elephants would be quite deadly, for example.

Now, some people might be wondering what happened to the other 9,997 kJ our falling furry friend was supposed to have. After all, energy is conserved, right? The answer can be framed a couple different ways. One is to say that the air does negative work on the falling animal. If this negative work is leaving the air, then the air gains energy. While this is a valid interpretation, it leaves one wondering just what negative energy is supposed to mean. The other point of view is that the falling animals push the air out of the way as they fall -- that is, they do work on the air. From this vantage point, the energy isn't lost but transferred to the air. Where it goes from there is beyond the scope of this example, but thanks to Emmy Noether, we know the energy will do just fine on its own.

Hello, world.

So, one of my absolute favorite quotes comes from Descartes. It goes:
If you would be a real seeker after truth, it is necessary that at least once in your life you doubt, as far as possible, all things.
This directive has informed much of the philosophy that underpins my life. I research everything, I (try to) correct my own mistakes, and I sincerely believe that just about the only thing worthwhile in the universe is the accumulation of knowledge.

Until recently, however, what I've actually done with my life has had very little to do with this idea. I'm a relatively smart and well-read guy, but my knowledge of any particular subject is limited. This past spring, I decided it was time for that to change. I retaught myself all the high school math I'd forgotten and returned to my local community college with the intention of studying physics.

After successful summer and fall semesters, I'm now on break until the spring semester begins at the end of January. Calc 2 and Newtonian mechanics are behind me; Multivariable Calculus and E&M await me.

And what is this blog doing here? It's a chronicle of my descent into the world of science, mathematics, and education. I've always loved science, and I want to share my exploration of it with others. That may come in the form of applying physics to everyday situations, talking about neat things I've learned, or ranting about my school's stupid policies. We'll see where the blog goes.

There is a little more to me, however. I work full-time during the day (banging on a keyboard at the request of the FDA), which means I'm juggling work and school for the time being. I'm an avid science fiction fan, and I've been writing science fiction since I was a small child.

And there's also the matter of that life philosophy thing I talked about at the beginning of the post. I have very strange ideas about how things are and how things should be. Over the years, I've discovered that many people are quite offended by my beliefs, whereas others just find them quirky. If you fall into the former group, you probably won't stick around; if the latter, we may have some fun together.

My next post will, with any luck, have a significantly higher science to long-winded musing ratio. So until next time... uh... live long and prosper?