There is nothing crappy about the wiring diagram at all, I would say on the contrary, it is very clear and depicts how the thing is wired, which is the point of it in the first place...
How to test the ballast for an internal short came to my mind:
It is based on quality factor (aka losses) measurement, but allowing you to use just basic instruments you likely have.
First measure the DC resistance of the coil. That would determine what losses are to be expected as "normal".
Then connect the ballast in series with an incandescent lamp (disconnect the ignitor from the ballast and socket, connect the socket to the ballast directly or via an Ameter when measuring the current, put the incandescent into the socket).
Measure:
- The circuit current
- The mains voltage
- The voltage across the incandescent
- The voltage across the ballast coil.
All three voltages must be measured as close in time as possible and with the same meter on the same range, so they depict as accurately as possible the ratios at the same time.
Try to pose the results here, then we (myself or RRK) may do the math, I guess that would be too much for you if you still did not go through any electrical course yet...
Now the theory behind:
Now the 2'nd Kirchoff law says the sum of the lamp and ballast voltages must equal the mains voltage. Because we are on an AC circuit, that sum must be a vector one, taking into account the phase relations. But the meter only shows the absolute values, the sum of just those won't match.
But still we may calculate the phase relations (vs the circuit current) out of them. Plus we know the incandescen is a resistor, so we know its phase shift is zero.
Then calculate the phase shift in the ballast itself, with the current reading its exact impedance, subtract the anticipated lossesfrom the wire resistance and then look at what losses remain.
If the remaining losses is something smaller than the wire resistance, the ballast is very likely good.
But if the calculated remaining losses are way higher, there is very likely an internal short circuit within the ballast, as a shorted turn forms in fact a secondary winding (that turn) with the turn's wire resistance as its "load".