> Am I right in assuming that on a 4-stroke engine like this, the TDC
> mark on the pulley turns at the speed of the crankshaft, so there are
> only two possibiilties (either correct, or 180 degrees off)?

To clarify:  By "180" I meant the distributor drive shaft would be
180 degrees off.  The crank would be 360 from TDC (i.e. TDC, but on
on the intake stroke rather than power stroke).  The distributor
drive runs at 1/2 crank speed, yes?