[ohohyodafarted]

OK here is what to do.

First, measure the current on the B+ line you want to place the dropping resistor into.

Now using Ohm's law solve for R. R=E/I ..... Plug in the current (I) in AMPS and the voltage you want to drop (E) in volts (60) and the solve for R (the resistance needed).

So lets assume 500ma is the measured current, the equation would look like this:

R=60volts /.5 Amps so R=120 ohms

Then to figure out how many watts the resistor needs to be use P=IE Where P the power in watts = the current (I) in AMPS times the voltage drop across the resistor E (which in this case is 60 volts)

Lets assume the following for the Power equation

P (watts) = IE so P= .5 Amps * 60 volts Therefore P=30 watts

No guesswork needed if you know the current flowing and the voltage drop needed.

The value of the resistance will not likely be a standard value, so get one of those tubular ceramic types with the adjustable slider band so you can adjust to the needed value.

Good Luck!