I'm still hoping someone might explain how to know whether a potential replacement PT is suitable.
The PT delivers AC voltage, but we will convert to DC voltage. The filter caps tend to charge to the peak of the PT's AC voltage waveform.
The PT's AC voltage is described in "
Volts RMS" of a sine wave, and we assume the outlet voltage is a sine wave.
The linked article goes to some lengths to figure out what "RMS Volts" is of a wave whose instantaneous voltage we know at a number of points in time. It derives a formula for calculating the RMS voltage of the sine wave when we know the peak voltage of that wave, but we can go the other way: RMS Volts x √2 = Peak Volts
When we rectify the PT Voltage, we will get a DC Voltage equal to the PT Voltage's peak volts, minus some amout due to losses across the rectifier, or because the amp circuit is sucking current out of the filter caps. However, it's a start to know whether a transformer is suitable.
Tubenit's PT had a PT delivering 190-0-190v, so the peak voltage output is 190v x √2 = ~269v DC. This is exactly the value he indicates at Point A of his schematic. Tubenit probably got more like 268v, because there's about 1v of loss across the solid-state rectifier diodes he used, but this discrepancy is immaterial. A tube rectifier will have a much greater voltage-loss, but the calculation is very involved such that many folks just use rough-guess, or use various fudge-factors to estimate the reduction from peak voltage.
To estimate a suitable PT voltage when the desired power supply DC voltage is known, you can use the article's final formula to calculate RMS Volts from a known Peak Voltage: Peak Volts x 1/√2 = Peak Volts / √2 = Peak Volts x 0.7071.
For Tubenit's circuit, we calculate 269vdc x 0.7071 = 190.2v RMS ----> we use a bridge rectifier with a single winding of 190v AC, or we use a full-wave (not-bridge) rectifier of 190-0-190v AC.