|Grand Prix Racing -||The Science of Fast Pinewood Cars|
It was quite surprising to find that a car with lower speed can actually win a race. But can a car with lower weight and other things being equal pull it off? We'll start with some basic principles of physics.
Newton's second law says that a force acts on it by changing its momentum. Simply stated, momentum is an object's mass times its speed. There are two kinds of momentum, linear momentum and angular momentum or rotational momentum.
It is the combined linear and angular momentum of your car that will carry it to the end of the track when the force of your car's weight is being supported on the flat by the track. Making sure your car has as much weight as possible helps insure that the highest momentum will be attained.
This potential energy or energy made available to your car can be written
E = mgH
|E||is the available energy to convert to speed in ozin|
|m||the mass of your car in ozs2/in (its weight in ounces divided by g)|
|g||the acceleration of gravity in inches per second squared (g = 386.088 in/s2)|
|H||the vertical measure of the height your car's center of mass falls on the track in inches|
This expression does not depend on the shape of the track or even if there is a track. In fact, the value of E does not change during the entire race. It is the bud of the Law of Conservation of Energy that opens by the Work-Energy Theorem.
Look at the expression for E on the right hand side of the equal sign. It has one term. Once the race begins and energy is spent, the right hand side of the equal sign will become cluttered with more terms expressing where the energy is spent. It is like a flower unfolding, and each petal holds the measure of something that will help us make a faster car! But if the value of E via mgh is less than it should be, the flower will be stunted and smaller than it could have been.
If you allow m to be less than the maximum allowed, you limit mgh and therefore E. The amount of energy available to be converted into linear momentum will be less.
We can also look at this through Newton's Second law. It tells us that more mass will take more force to get it moving as fast as a smaller mass. It sounds as if light cars will be propelled down the track faster.
This is not so. Remember the famous cannon ball toss off of the leaning Tower of Pisa? Galileo showed dramatically that the lighter ball landed at the same instant as the heavier. It's because the acceleration of gravity acts on the mass to produce a larger force on the heavier ball and a smaller force on the lighter one and because the difference in air resistance between the cannon balls was small compared to the force of gravity.
The same happens with Grand Prix cars. In fact, if only tread and axle friction operated, the weight of the car wouldn't matter a bit, they would always tie. This is because the model without wheels and air resistance does not depend on the mass of the car. But because the cars have spinning wheels that store energy and air resistance that don't depend on the mass, everything else being equal, the heavier one must always finish first!
Let's look at the wheels first. Suppose two cloned cars race on a frictionless track, except one is light (m) and other heavier (M). We assume that the wheels will spin, even though that wouldn't happen on a "real" frictionless track. If both cars attained the same speed at the end of the race, the energy stored in the wheels would be the same for both cars (2Iv2/r2) - because it depends on the speed, not the weight of the car. But the potential energy of the lighter car (mgH) at the starting line is less than the heavier car (MgH). So the difference of their linear kinetic energies ((M-m)v2/2) must equal the difference in potential energy at the start ((M-m)gH).
Their speed at the end of the race would be v = \[2gH]. But that is the speed of a car on a frictionless track with no wheels! So the two cars can't have the same speed since they do in fact have wheels. A correct expression for the speed in this race with no friction but spinning wheels is
v = \[2gHr2/(r2+4I/m)]
Note as m decreases, the denominator (r2+4I/m) grows making v smaller and the race time longer, not shorter.
Since the same two cars also have the same size, they have the same air resistance at the same speed. Now race them on a frictionless track without spinning wheels but with air resistance. A similar argument shows again that the only way they can tie is for them to have the frictionless, no air resistance race speed, v = \[2gH]. The conclusion is the same, they have to have different speeds at the finish line. A mass dependant speed equation can be found. Again it shows the lighter car losing.
Note, Galileo's cannon balls were different sizes, so the smaller one had proportionally less air resistance, not the same as in our case. His experiment would have failed if he had dropped Grand Prix cars - and the dark ages would have continued in confusion!
On the other hand, an increase in weight causes an:
If we clone the standard test car a few times and take off a bit of weight from each, we can get an idea of how much difference (split difference) it makes in a virtual race in inches at the finish line.
Effect of Weight Race Specification File: weight.xml ======================= Race Standings ====================== Car Name Time Split Distance Speed Accel Lane Place -------------------- Finish Line Summary -------------------- 16 oz 2.853 0.000 155.80 -10.14 5 1 5.2 oz 2.891 5.839 153.65 -9.86 4 2 5.0 oz 2.893 6.177 153.52 -9.85 3 3 4.8 oz 2.895 6.541 153.39 -9.83 2 4 3.0 oz 2.931 11.879 151.44 -9.56 1 5
To verify these results from the model, a derivative was taken of race time with small changes in weight to quantify more generally how much race time changed. The derivative was set to zero to find where the "best" weight is. As you might have guessed,
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|Grand Prix Racing -||The Science of Fast Pinewood Cars|
|Copyright © 1997, 2004 by Michael Lastufka, All rights reserved worldwide.|