|Grand Prix Racing -||The Science of Fast Pinewood Cars|
Rotations manifest themselves in many unexpected ways. Often vibrations are rotations. Zig-zag movements may include rotations that switch direction long before completing a full cycle. So it is with your car's body, not only do wheels roll, but to some extent, so does the car body. Unlike a wheel, the "axle" it rotates around is not visible. It doesn't even need to be "in" the car. It's the car's center of mass!
This rotation of the car body as it transitions from the ramp of the race track to the flat is a rather subtle event. The axis of this rotation lies along the width of the car, its pitch axis. There is only a partial rotation. Perhaps it has escaped your notice until now. Never-the-less, it is as interesting a feature of your car's behavior as any other and it uses up a bit of energy.
Another reason to know what the moment of inertia of your car is is to be able to determine how big a bump in the track could cause your car to jump up.
Below, a model for determining the moment of inertia of your car is presented. You can use the worksheet to evaluate the model to get the moment of inertia for your car. Using the evaluation for one car from the example below, we will see how much energy is being used and what it means in terms of forces.
In order to estimate your car's moment of inertia about the pitch axis, the car is broken up into smaller parts. These parts include the wheels, various parts of the body and any decorations attached to it. The body is the most difficult part to divide up. There are many shapes for which a moment of inertia formula has been determined. Some listed below are simple shapes that our Grand Prix cars can be divided into. Success of this model depends on how well your car can be broken up into these shapes and how good your measurements are from the center of mass of your car to the center of mass of each part as described below.
Here is a list of shapes and formulas for determining the moment of inertia and center of mass for some three dimensional shapes. Each shape's volume is assumed to have uniform density so that its center of mass is in its geometric center. Note that the width dimension is not drawn. Only the side having length and height is drawn. The center of mass location is along the length and height axies of the shape. For each shape, the axis of rotation passes through it's center of mass and is at right angles to the side drawn. The center of mass locations (cmh,cmx) are measured from the lower left of the shape drawing.
|SHAPE||NAME||cmh||cmx||MOMENT OF INERTIA|
The shapes in the table are matched to your car to determine the parts to divide it into. This is not easy, nor exact. The better it is done, the more acurate the result will be. But don't worry if it is not perfect.
Pinewood has a weight density of about 0.28 ounces per cubic inch. So you can measure the volume of each of your shapes and estimate its weight by multiplying its volume by 0.28 oz/in3 or just weigh the thing. To get mass, we divide by 386.088 in/s2.
Here's an example for a car that looks like this:
This car is broken up into: a wedge, the front; a block, the back; a cylinder, the cockpit on top; two axle rod pairs; and two pairs of wheels. Here's what it looks like on paper. Note the wheels are dealt with separately in the next chart. CMh = 0.7 inches and CMx = 4.4 inches
We'll use AWANA kit axles so we can get the weight (0.0416 oz each) and moment of inertia. Now, we can make a chart like:
|Part||SHAPE||m ozs2/in||h (or r) in||l in||cmh in||cmx in||MOMENT OF INERTIA ozins2|
We'll use AWANA kit wheels so we can get the weight and moment of inertia. Now, we can make a chart like:
|Wheel Pair||m ozs2/in||MOMENT OF INERTIA ozins2|
A measurement in height and length will be needed from the center of mass of each part (cmh,cmx) to the center of mass of the car (CMh,CMx). These measurements are labeled Dh and Dx. They are used to get the moment of inertia of each part about the center of mass of the car. The physical theorem that allows us to do this is called the parallel axis theorem. The final chart looks like this:
|Part||m ozs2/in||MOMENT OF INERTIA ozins2||Dh in||Dx in||I+m(Dh2+Dx2)|
Now, we total up the moment of inertia column to get the moment of inertia for the whole car about its pitch axis through its center of mass.
The amount of energy in this rotation can be studied by noting the usual expression for the energy of a rotating object,
T = Iw2/2 where T represents rotational kinetic energy, I is the moment of inertia and w is the rotational rate of the car.
From the above example, we set I to 0.02320680 ozins2. For w, consider the abrupt transition of the simple closed race model and note that a car will pass over it from front wheel to rear in about 4 hundredths of a second at typical speeds. Thus the car body rotates through 21.2 degrees in 0.04 seconds. Converting degrees to radians, we get w = 0.369/0.04 = 9.225 radians per second.
T = 0.02320680(9.225)2/2 = 0.99 ozin.
The total available potential energy at the start of a race is about 200 ozin. So the rotational energy of the body is about 0.99/200 = 0.5 percent of that available! That's pretty small. And that's as big as it gets! On a smooth transition, the same rotation angle is moved through, but it happens over a much longer time period, perhaps a second even. Then the rotation rate, w, would only be 0.369 radians per second and T becomes, 0.0016 ozin!
Starting with the energy relation, we can take the time derivative, then divide out velocity to get the force of rotation.
dE/dt = I(dw2/dt)/2 = Iwdw/dt = Iw(dw/dy)(dy/dt) = Iwvdw/dy
When the v is divided out, this becomes the "force" of body rotation in the equations of motion: Iwdw/dy. dw/dy is the change in the rate of body rotation with respect to the change in distance along the track. This is 0 on the ramp and flat. On the transition, it begins positive, then dips past zero to negative values before returning to zero on the flat. The energy stored during the positive thrust is expressed as force on the axles during the negative or deccelerating part of the change in rotation rate.
For the abrupt transition, dw/dy is about (9.225 rad/s)/(6 in) = 1.537 rad/ins. Because the change in rotation rate is not constant, this is a "ball park" estimate probably close in terms of the order of magnitude of the answer.
Iwdw/dy = 0.02320680(9.225)1.537 = 0.3290 oz.
On the smooth transition, dw/dy could be about 0.369/120 = 0.003 rad/ins
So, Iwdw/dy = 0.02320680(0.369)0.003 = 0.000026 oz.
Of course, these are just rough estimates, but in general they show that your typical smooth track race will not be greatly affected by the car body rotation energy loss.
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|Grand Prix Racing -||The Science of Fast Pinewood Cars|
|Copyright © 1997, 2004 by Michael Lastufka, All rights reserved worldwide.|