1. Determine whether or not your circuit is an AC or a DC circuit. Most likely, if the circuit you are working on is assisted by the battery, it is DC. If it is an AC circuit (moped headlight circuits are usually AC) you will need to add a diode on your positive line before beginning to take your measurements. A 1N4004 diode will work just fine for moped circuitry. The diode will convert the AC current to DC current, DC current is best for led purposes.
2. measure your moped's stator output voltage at higher rpms (don't forget your diode if not working on the battery circuit). 6-12 volts most likely at peak motor revs measuring from the light's power wire to ground. (Your actual measurement might not be exactly 6-12 volts DC, it may be quite different, but we will use 12 volts for this explanation because its an easy number).
3. take your led voltage drop (usually 1.7v for a single led, check your led packaging for this number)
4. Take the max amperage your led can take (usually anywhere from .015 - .030 amps or also read as 15-30 milliamps. check your packaging for this number.
so for this explanation, our stator outputs 12 volts our voltage drop is 1.7 and the max amperage our led can handle is .015 milliamps
take your max stator output and subtract your led's voltage drop 12 - 1.7 = 10.3 this will tell us how much voltage we need to get rid of with your resistor.
now divide the remaining voltage (10.3) by the amperage your led can handle (.015) to get the correct ohm value of the resistor you need.
10.3 ÷ .015 = 686.66
you need a resistor of 686.66 ohm, you wont find that exact, use next higher like 700 ohm.
now, take your voltage (12) and times it by your max amperage your led can handle (.015) and you get your wattage.
12 x .015 = 1.8 watts.
you need a 1.8 watt resistor, you probably wont find one, use next higher like a 2 watt, i would even go 3. choose a wirewound.
so, we need a 686.66 ohm (700 ohm), 1.8 watt (2 watt) resistor.
BONUS: If you have more than 1 led and connect them in series, you add their voltage drop (2 led's with voltage drops of 1.7 in series is 1.7 + 1.7 = 3.4) you keep your led's max amperage at .015 so: 12v - 3.4vd ÷ .015ma = 573.3 (575 ohm resistor)
If you wire in parallel, you keep your voltage drop at 1.7, but you add your max amperages (.015) together. (2 leds with max amperages of .015 is .015 + .015 = .030) so: 12v - 1.7vd ÷ .030ma = 343.33 (350 ohm)
you can get everything you need at DIGIKEY.COM or MOUSER.COM