One day while riding two abreast, a small rock flew out from beneath my rear tire and hit my partner's chainstay. I apologised and explained that it was a tire bullet. Later I realized that I didn't really know anything about them from a physical standpoint. I did not find anything on any cycling forums or the web. I even looked into the theory of tiddly winks! Having found nothing at all, here is my attempt at a physical model:
Mechanism:
Since the rocks shoot out at high velocity, I assume the tire does not simply hit the rock and push it aside, but stores up energy in tire deformation which is suddenly released. The tire must run over the top of the rock for the tire to deform. The rock must also have access to the outside. If the tire covers it
completely, it can't fly out.
Launch Direction:
If the rock is launched at or near the point of maximum force, that would be when it is under the contact patch of the tire. The front/rear forces would then be the same and cancel, leaving a net force out to the side. This matches most of my observations, straight out to the side. If the rock launched before maximum
force was reached, it would shoot out a little toward the forward direction, and if it launched as the contact patch was lifting, the rock would shoot out a little toward the rear. And indeed, I have seen some fly out at an angle, but not more than about 20 degrees either way.
Launch Angle:
The force on the rock certainly has a downward component, but can't travel that direction because of the road. It can only travel out to the side and upward. If it starts out traveling parallel to the road while still touching the road, it will hit something and bounce upward at some angle. Is there a characteristic launch angle for rock bullets? Do the ones with low trajectories go unnoticed? My friend and I were about 3 feet apart, and her chain stay was about 1 foot high, so the angle = ATAN(1/3) = 18.4 degrees.
Launch Speed:
At the point of maximum tire deformation, the force on a spherical rock is proportional to both the tire PRESSURE and the cross-sectional area (square of the DIAMETER) of the rock. The velocity of a tire bullet is proportional to the square root of the tire pressure, and does not depend on the size of the rock. They all shoot out at the same speed!
Size Limitations:
Obviously, cyclists don't run over big rocks for fear of pinch flats, but what about the smaller rocks? Do they get launched, too, or do they just go unnoticed? Is there a characteristic size for a rock bullet? The most common size (of those I notice) is around .25 to .4 inch.
Well, I think that's enough for a first pass. I hope you have fun thinking about rock bullets! I sure did! Please share your tire bullet observations!
Update:
During the 2005 TDF commercials, I experimented by sticking small objects between my rear tire and the roller on my trainer while slowly turning the tire. (What some people will do for fun!) The best object I found was a white plastic airsoft BB with a couple of flats filed on opposite sides. If I got it just right, it would shoot out the side at high velocity, even though I was barely turning the wheel. That observation made me absolutely certain that it's the tire pressure that governs the velocity. I wish I could measure the angle and distance, but it always hits the side of the flywheel. Since it hits, The angle must be less than about 45 degrees. It would have been nice to plot the distance vs tire pressure.
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