Thanks for your comments, Steve.

I, like you, consider myself fairly good at understanding 3-dimensional relationshiphs, and yet, like you also, had to work at understanding the beach balls. All that worked for me was simply working at it. I doubt that I could explain it better than the USGS site does. Perhaps John V. would care to take a stab at it. The best way to begin your comprehension of it is to take the simplest example, a pure strike-slip quake on a vertical fault. Let's have the fault run straight N-S. To simplify further, assume the quake occurs at the surface, and draw a line perpendicular to the fault (E-W) so that it crosses through the location of the quake. If you then place seismometers a short distance north of the quake on the fault's east side, then one south of the quake on the east side, then duplicate that on the west side of the fault, you will have a seismometer in each of the quadrants formed by your two lines.

The beach balls graphically show the very first motion (dark for compression, and light for tension). In our example, you can dispense with the third dimension, and you will just have a circle with each adjacent quadrant a contrasting color. In the case of a right-lateral slip, I think you can see that the lower right quadrant would, at the very first motion, experience compression, as would the upper left quadrant. The other two quadrants would have a tensional force.

After you think about that and figure you understand it, then it is easier to understand the third dimension necessary for the beach balls. The actual hypocenter of a real earthquake is below ground, and the fault can also dip at angles other than vertical. It is also necessary to realize you are looking at the upper hemisphere of a sphere (the beachball), since, of necessity, the seismometers are at or near the surface of the Earth. Finally, in nearly all cases, the beachball presents two possible solutions. The actual fault would correspond to one of the two axis shown on the beachball. Which solution is the correct one is usually determined by an expert looking closely at the data (that process is probably computerized for smaller, less significant quakes). But even an amateur can usually figure it out correctly if the quake occurs on a known fault with fairly simple geometry and which historically has quakes that are almost entirely in one sense of motion (i.e. right-lateral strike slip, or dip-slip with north side down, etc.). For instance, the beachball for a standard, off-the-shelf right-lateral strike slip on the San Andreas Fault could also be interpreted as a pure dip-slip quake. But you can with near certainty rule the latter solution out.

I haven't looked at the quake under discussion here (the M4.5), but, if it did indeed occur on the same fault as Northridge, or very near Northridge quake's location, then I would certainly expect its mechanism to be similar (i.e. - as your colleague sensed, a good deal of vertical motion).

Just so you're clear, I'm an amateur, and wouldn't be at all surprised if my explanation above was in error at least in part.

Mike W.
93420