Pinewood derby: What a drag! January 16, 2012Posted by mareserinitatis in engineering, physics, science, younger son.
Tags: air resistance, cub scouts, drag, energy, energy conversion, friction, gravitational energy, physics, pinewood derby
husband son competed yesterday in my son’s his scouting group’s pinewood derby race. For those of you who have never had a kid in cub/boy scouts, they hand out these blocks of wood that you get to make into a car. The idea isn’t to win the race: it’s supposed to be that dads and their boys spend some time bonding over manly things like woodworking.
(One year, Mike was out of town during this whole thing, so I got to be manly and help the older boy build his car. All I can say is that Dremel tools are awesome.)
If you look on the web, you’ll find a lot of advice on how to prep pinewood derby cars and make them faster. One thing that consistently bugs me is that one should pay attention to aerodynamics of the car and give it a low profile.
This makes me nuts.
To understand the following, you might want to know what a track looks like. So here you go:
Let’s start out with the specifications. Most pinewood derby tracks have a height of about 4 feet and a length of 32 feet. I also will note that the ones we’ve raced on were wood, not aluminum.
Most of the pinewood derby cars I watched made it the whole 32 feet, though not all did. So let’s say that, on average, they travel 32 feet. The *fastest* ones traveled at an average speed of 10 ft/s (or 3 m/s). The maximum they can weight 5 oz. or 140 g.
What we’re going to do here is look at how much energy the system puts into overcoming air resistance versus friction. It’s very hard to figure out exactly how much goes into friction simply using equations, so we’re going to figure out the total energy and the energy lost to drag forces. Once we have those two quantities, we can subtract the drag forces from the total energy and assume that the difference is equal to the frictional losses. Finally, we can compare the drag and frictional losses.
First things first: what is our total energy? It starts out entirely as potential energy as the cars are placed at rest on the top of the ramp. Potential energy is, fortunately, very easy to calculate. It is simply the product of the height of the object, its mass, and the gravitational acceleration. In other words,
We know the mass of the car (0.14 kg), the height is approximately 1.22 m, and the graviational acceleration is 9.8 m/s2. This results in a total potential energy of approximately 1.67 J.
The potential energy is equal to the total energy in the system since the cars start with no other kind of energy. In a frictionless and drag-free system, all of this energy would be converted to kinetic energy and the cars would drive forever at the same speed once they reached the bottom of the ramp. Obviously, however, that’s not what happens. Eventually, all of the energy is converted to friction and drag, and the cars stop.
Now we need to determine the drag on the cars. The drag equation is:
The drag force is proportional, therefore, to the density of the fluid (ρ), velocity (v), drag coefficient (CD), and cross-sectional area (A).
The density of the fluid (air) is approximately 1.2754 kg/m3, and the velocity is 3 m/s. The cross-sectional area of the car, at maximum, is 2.75 x 3 in. In real *ahem* units, this is 0.00532257 m2.
The drag coefficient for a long cylinder, according to Wikipedia, is 0.82. Given the cars are all sorts of different shapes, I think this would probably be the closest approximation, although for some cars, this will be high.
All of this put together gives us a force of 0.025 N. Over a distance of 32 feet (or 9.75 m), this gives an energy of 0.24 J.
If we assume that all of the potential energy is converted to either drag or friction, that means 0.24 J is converted to drag while 1.43 J goes into friction. This means that roughly 14% of your energy losses are due to drag while the other 86% are due to friction.
My advice, therefore, is to not worry so much about drag and the aerodynamics of the car and to worry far more about the ways to reduce friction.
What should you do to help your pinewood derby car to be fast? There are three things:
1) You want to maximize your potential energy, so being as close to the 140 gram maximum weight is good. You can get all sorts of weights commercially to assist with this. Some people have argued that putting the weights near the back of the vehicle helps even more. I wouldn’t doubt this as you’re putting the weight at greater height, giving the car more initial energy to start with.
2) The biggest issue is reducing the friction between the wheel and the axle, and there are a few easy ways to do that: sanding and polishing the axles as well as using graphite or teflon as a lubricant. (Our troupe doesn’t allow graphite because it marks the floors, but teflon is allowed.)
3) Reduce the friction between the car and track. The best way to do this is to make sure your wheel alignment is as straight as possible as this will keep it from rubbing against the center of the track.
The winner of our pinewood derby actually had a very blocky car designed to look like a platypus. I’m guessing the winner didn’t spend much time focusing on aerodynamics and instead chose to minimize their frictional losses.
The force is weak with this one… June 16, 2011Posted by mareserinitatis in electromagnetics, geology, geophysics, physics, science.
Tags: electromagnetic energy, gravitational energy
I had an interesting question from someone today: why do we use electromagnetics to study so many things? Why can’t we use gravity or something similar? Specifically, they were wondering about non-invasive methods for studying the human body.
It’s easiest to start with Newton’s Law of Gravitation, which tells us how much gravitational force one object (M) exerts on another (m):
and Coulomb’s Law (which explains the force of attraction between electrical charges, Q and q):
If we want to find the ratio of gravitational force to electrical force, we end up with something like this:
Now, let’s imagine we’re just looking at the gravitational and electrical forces between two electrons from 1 m away. We use G=6.673•10-11 m3 kg-1 s-2 (the gravitational constant), M=m=9.11•10-31 kg (the masses of the two electrons), εo=8.854•10-12 C2 kg-1 m-3 s2 (the permittivity of free space, although may be just as easy to think of it as an electrical constant), and Q=q=1.602•10-19 C (this being the charge of an electron). Using these values, all our units will disappear, which is good because we’re looking at a ratio of two forces and shouldn’t have any units, and we end up with a value of about 2•10-43.
What this means is that the gravitational force is 43 orders of magnitude smaller than the electrical force…or you could put a decimal point with 42 zeros and then a 1 behind it, and that’s how much smaller the force is. When it’s already difficult to measure current values that contain many, many electrons (as compared to the two electrons we examined), it’s going to be impossible to find something that exerts a force that is 43 orders of magnitude smaller than what we can already pick up.
You can pick up small changes in gravitational forces when talking about large geophysical features – like ore deposits and mountain ranges. In fact, they use this principle a lot in exploration geophysics, where they use gravimeters to look for mineral resources. Our bodies are less sensitive than that, though, and can only pick up gravity when we are talking about changes in the size of planets or moons. However, we are sensitive to changes in acceleration, so you can feel changes in gravitational pull when riding on an elevator, but that is because the change is both fast and of a reasonable size.
Anyway, the huge difference is why we are permeated (ba dum ching!) by devices that detect and use changes in electromagnetic radiation but not in gravitational energy.