Oh, gosh. That schematic reminds me of the good old days at AT&T. I used to have mild arguments with drafting as to how a schematic should be shown on paper. Their approach was often that the schematic was just a graphical netlist and the goal was to use as few pages as possible. My position was that a schematic told a story, in graphic form, a story about how a circuit worked, so that the reader could easily understand its operation.
Anyway, here's a redrawn version of the schematic with some commentary below that. Please let me know if you spot any transcription bugs in the schematic.
A couple of things to note here. First, I don't see any reference to earth ground in the original schematic. To me, that means that this circuit "floats" with respect to earth ground, and the relationship with ground is determined by the applied 20mA (or 60mA) current loop (one side of which is likely grounded directly or through some finite resistance). When I call out a voltage on the schematic or text, that voltage is with respect to the Zero volt bus.
Second, the selector current is BIG, about 0.5 A. And when Q2 is conducting, Q2 is operating in its linear range, so it will get HOT.
Let's look at what happens in the MARKing condition. In the marking condition the 20 mA (or 60 mA) of positive current is shoved toward the base of Q1. In this condition the loop current flows through CR5 and the R3 (0.8 ohms) as well as through R1 into the diode network. Also in this condition the voltage at the base of Q1 is clamped to about +1.1V, and the BE junction of Q1 is cut off, so Q1 is out of the picture in the MARKing state. The base of Q2 is held at -5.4 volts by the diode network and the voltage the the emitter of Q2 is held at 4.7 volts (due the diode voltage at the base, less the diode drop across the Q2 BE junction). So with a constant voltage at the emitter of Q2, and a constant resistance between the emitter and the 0V bus, Q2 forms a constant current driver. If Q2 has a decent gain value, then Ic = Ie = 4.7V / (0.8 +1.5 + 8) = 456 mA. So that's the marking condition.
In the case of the spacing condition, the 20 mA loop current is disabled and Q1 is biased on via R1 and the diode network. This causes Q1 to turn on and current is sent from the collector of Q1 toward R8, which tends to rob current from the diode network, sending the voltage at the base of Q2 toward zero volts. It won't get to zero but it will be close (maybe -1 volts). With lower voltage at the base of Q2 (and 0.7 to 0.5V less than at Q2 emitter), that will reduce the selector magnet current significantly (but I don't think it will go to zero). In a working circuit the actual MARKing and SPACEing selector current can be determined by measuring the voltage across R7 and dividing by 14.
Paul, ad7i