Reverse engineering a power supply characteristics
You apparently have 18 V on the output of a full wave bridge.  Such a bridge has two diode drops.  Figure 700 mV per diode, so 1.4 V drop thru the bridge.  That means the peaks of the AC wave were 19.4 V.  Assuming a sine, the RMS voltage is a factor of sqrt(2) lower than the peaks, so 13.7 V.

Of course the AC line voltage can vary a bit.  The 18 V figure may also not be the nominal.  Sometimes you put the worst case on a schematic to remind yourself that the rest of the circuit must be able to tolerate that.

Any circuit using a transformer directly from the line, then followed by a full wave bridge, is going to have some tolerance for the resulting voltage.  My guess is that the transformer is probably rated for "12 V" output, or maybe "12.6 V".  The latter was a common fillament voltage for tubes.