To do the diode trick, the heater string voltage needs to add up to about 85V. Not 60V. Because Power = (volts squared)/resistance. The diode cuts the power the resistance of the heater string gets by 1/2. It turns out that (85^2)/R=(1/2)(120^2)/R
If you're short on the 85V for the heater string, you can pad it some with a power resistor. Even so, you avoid a lot of waste heat.
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Never tried the single diode approach. If I were to try it I'd bring the set up on the variac while monitoring heater string voltage
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Problem is that your meter will likely be confused by the funny waveform, and give you a false voltage reading. Eyeballing the looks of the heater glow, coupled with the calculation above, should be sufficient confirmation that you got it right. If you do make an error and feed the tubes with too much heater power, a short period of time like that shouldn't damage them, as tubes were powered up like that when they were made at the factory.