One day while riding two abreast, a small rock flew out from beneath my rear tire and hit my companion's chainstay. I apologised and explained that it was a rock bullet. Later I realized that I didn't really know anything about them from a physical standpoint. I did not find anything on this forum or the web, so here's my first pass at a 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 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. What are your observations?

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. Anyone seen one higher? Any ever hit your leg? Can you estimate the angle?

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. Forgive me, but I'm using the = sign to mean "proportional to" in the following equations to avoid using constants.
1. FORCE = PRESSURE X DIAMETER^2.
The work done on the rock as it is launched is equal to the FORCE on the rock times the distance moved while under that force. The maximum distance the force can act on the rock is the DIAMETER of the rock.
2. WORK = FORCE X DISTANCE
3. WORK = PRESSURE X DIAMETER^3
3. WORK = Kinetic Energy
4. PRESSURE X DIAMETER^3 = 1/2 MV^2
The mass of a rock is proportional to the cube of the DIAMETER, so the kinetic energy is proportional to DIAMETER^3 X VELOCITY^2
5. PRESSURE is proportional to VELOCITY^2
6. VELOCITY is proportional to the square root of the tire PRESSURE
This is not what I expected at all! The velocity of a rock bullet does not depend on the size or mass of the rock - only the tire pressure! They all shoot out at the same speed.

Size Limitations:
Obviously we 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? Can you estimate the largest/smallest one you have seen?

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 rock bullet observations!