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Ooops, that wasn't the story I thought it was. That was just a lazy, "best of" story I wrote one month when I didn't have anything new to say. The point I'm trying to make is in there, but not the math behind it.
I can't seem to find the original anywhere on the interwebs, so here's the draft I found on my HD (from Aug, 1999): Back in the June issue, I attempted to nail down an old rule of thumb about wheel weight. That rule, depending on whose thumb you listened to, said that because of the rotational inertia of a wheel, adding a pound to you wheels was like adding three, four, or even ten pounds to the car. Through the miracle of math, I managed to totally confuse the issue, adding obfuscation to speculation and ending up with what appeared to be empirical proof that all those thumbs were full of hot air. Instead of proving that the rule of thumb should be 2 pounds for every pound on the wheel, or 4 pounds if it’s at the rim, and 1.5 pounds near the center, or something of that nature, I proved that an 18 pound wheel was like 18.15-pounds on the car. In other words adding a pound on the wheel was like adding 1.008 pounds to the car—the rotational inertia of the wheel appeared to have a negligible effect on the car’s acceleration. When mathematical proof runs so contrary to common sense and personal experience, there’s a good chance something is wrong, so I invited our readers to shoot me down. Nothing gets people excited more than a good screw up, and there was a doosey hidden away behind all that math. Lance Schall, a weight-conscious Miata owner, was the first one to spot it. Before revealing the error, let’s take a quick look back at the argument. When an object is moving, it has a certain amount of kinetic energy. A rotating object that is spinning in place also has kinetic energy. Since a rolling object like a wheel is both spinning and moving in a straight line, the kinetic energy of a wheel is the sum of the kinetic energy from rotation and the kinetic energy from translation (moving in a straight line). That’s simple enough, but what I wanted to figure out was how significant the rotational portion of the wheel’s kinetic energy was, and how much non-rotating weight you would have to add to the car to have the same kinetic energy. My error was in the formula for kinetic energy from rotation. The formula is 
Where I is the moment of inertia (the rotational equivalent of mass) and w is the rotational speed. Nobody really likes equations in the first place, but there is nothing worse than equations with Greek letters in them. In an attempt to make all this geeky stuff seem interesting, I usually try to translate from Nerdish to English, but that translation is where I screwed up. Since nobody wants to try and remember what w is, I simply changed it to a more familiar expression of rotational speed: rpm. Unfortunately, the equation only applies if rotational speed is measured in radians/time. What the hell is a radian? Now you understand why I wanted to use rpm... Radians are just like degrees, only bigger. Where there are 360 degrees in a full circle, there are 2pi (about 6.283) radians in one revolution, so my solution last time ended up being off by a factor of 2pi^2 (about 39.5). In other words, for the 18-pound wheel that I tried to model last time, the rotational inertia of the wheel acts like 5.9 pounds sitting in the car. To figure out how much weight you would have to add to the car to have the same kinetic energy as the rotating wheel, you have to figure out the relationship between wheel rpm and speed. That relationship is simply rpm times the circumference of the tire (not the circumference of the wheel). Since rpm is w divided by 2pi, we can substitute w/2p*Circumference for velocity, and equate the rotational portion of the wheel’s kinetic energy with the linear kinetic energy of that rule-of-thumb mass we are looking for. In other words: 
If we solve for M, we get the formula for the rule of thumb: 
We calculated rotational inertia (I) for a theoretical wheel last time, but the reason this formula isn’t in general use is that it is extremely difficult to figure out the rotational inertia of any object more complex than a cylinder or a disc. Now that we have the formula, let’s look at some examples. Instead of modeling a whole wheel, lets just look at what an extra pound will do in various parts of a wheel. Take some relatively conservative wheels like a 15x7, a 16x7 and a 17x7, and wrap them in 205/50-15, 205/45-16, and 205/40-17 tires (remember, you have to know the tire size to know the rolling circumference). Adding a pound to the rim of the 15x7-inch wheel is like adding 1.37 pounds to the car (M=0.37). Adding a pound to the 16x7 wheel rim acts like 1.41 pounds, and a pound on the 17x7 acts like 1.47 pounds. (For all of these calculations I assumed that the rim is 1/2-inch in from the outer edge of the wheel, since the majority of the material is not actually at the edge.) If you add a pound to the tire on any of these, you will feel like you added about 2 pounds to the car. Our old rule of thumb, it appears, does exaggerate somewhat, but at least it means well. You should also consider that every pound added to a wheel or tire also has to be multiplied by four, unless you have mastered the art of driving on three wheels. The same logic used to look at spinning wheels can also be used to look at spinning flywheels, but the relationship between w and velocity is more complex. Because the gearbox is between the engine and the flywheel, you have to factor in the gear ratio and final drive ratio, and the relationship will be different depending on what gear you are in. For a flywheel, 
where “Gear†is the gear ratio, “Final†is the final drive ratio, and “Circ.†is still the circumference of the tire. This gives us a new rule for flywheels that goes like this: 
Play with this a little and you can see why a lighter flywheel can make a car faster. I consulted Dirk Starksen at Advanced Clutch Technology and learned that they take 7 pounds off a Civic EX flywheel. All of that weight comes out of a band that is toward the outer edge. Based on ACT’s specs, we assumed that all the weight came from a cylindrical band 9 1/8 inches in diameter. Using our new formula we learned that the effect is surprisingly large in the lower gears. Removing 7 pounds from the flywheel is like removing as much as 206 pounds from the car! Next month no math, I promise. |
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