Nothing more than a simple resistor:
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Technically any basic resistor will work as a voltage dropper as long as it is sized correctly.
First you need to know the listed wattage and rated voltage stamped on the timer motor itself. For this example I will use 4 watts at 120 volts. Next you will need to find the operating impedance of the motor circuit in ohms.
Resistance in ohms = (Volts x Volts / Watts)
3,600 ohms = (120 Volts x 120 Volts / 4 watts)
Knowing we need to drop 240 volts down to 120 gives us a ratio 1:1, so the resistor must be around the same impedance as the operating timer motor itself, in this case the resistor can not be less than 3,600 ohms.
(OEM manufacturers appear to add a safety factor of 105% to take into account a 240 volt supply operating near the upper bandwidth, so 3,600 ohms times 1.05 = 3780 ohms)
Second you need to find the amount of power (watts) the resistor will be dissipating by knowing the current draw of the timer motor.
Amps = watts / volts
0.03333 amp = 4 watts / 120 volts
Being a series circuit, this same current magnitude will be passing through the dropping resistor.
Power dissipated in watts = Current times Current times Resistance
P=I2R
4.1 watts = 0.03333 amps x 0.03333 amps x 3780 ohms
A 5 watt resistor rated, 3780 ohms resistor will work, however given given the environment and heat dissipation across such a small surface I typically like to over-size my resistors wattage wise. A 7 watt resistor would be my starting pick.
I like what you're doing, good luck!