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Discussion Starter #1
Does anyone know offhand the wattage of Monster turn signals?

Also: what Ohm resistor would I use going to a 6 watt bulb from a Monster bulb on the turn signal to retain the stock blink rate?

Thanks in advance!
Rideon
 
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I don't know if this would suit your case, but some people hide a stock bulb somewhere to provide the required resistance
 

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Discussion Starter #6
Let me provide some information regarding my problem:

The replacement turn signals are Micro 1000's. The directions are designed for Jap bikes, and state the following:

When replacing a 20 watt bulb with the Micro 1000 6 watt, install a 10 ohm resistor in parralel for each bulb (signal). My limited math abilities put that at a 14 watt differential.

The Ducati turn signal is 10 watt. I am making up a 4 watt differential by going to a 6 watt bulb. If the relationship is linear, then I think I would be using a 2 or 3 ohm resistor.

As I am totally ignorant on this, can someone plug the info into the appropriate equation and let me know what to use?

Thanks in advance!!
Rideon
 

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the resistance of the bulbs would change eith heat which makes calculations difficult. radio shack sells variable resistors. that will allow you to experiment.
 
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Hrm you guys are going way into the wrong direction with guessing here but needless to say Rideon is right but he doesnt know why.

If you want to calculate resistance you need ohms law:

V=I*R (or I=V/R or R=V/I)

where:
V = voltage (battery)
I = current (amps)
R = resistance (the bulb)

(dont get confused over watts. it is voltage divided by current P=V/I which if you notice is the same as R=V/I so wattage IS resistance, common confusion)

In this case we know R (the bulb) was 10 watts and V is 12 volts which means in a "stock" system I = 1.2 amps which seems about right. If someone is feeling froggy go take a volt meter out to their Duc and measure it. Filaments take some amps to light up nice and bright.

So now we need to figure out what we get with the new bulbs values.

I = V/R
I = 12 / 6
I = 2 amps

since we do not want to almost double our amps over the wire and I dont think a 6 watt bulb will like 2 amps, we need to drop about .8 amp by raising resistance. In this case we would need about a 4 ohm resistor rated at no less than 10 watts.

Why a 10 watt rating? Longevity and heat dissipation. You dont want a mini heater, well unless you mount the resistor inside the handgrips. You can get some wirewound ceramics at RatShack which are rated for 25w if I recall correctly. They wont even be warm to the touch.

As for resistance changing isn't really true. A filament acts as a valve. It's the essential component of a vaccuum tube (one of my other expensive hobbies). When cold they have very low resistance allowing current to "rush in". As current passes through they heat up extremely quickly causing resistance to increase (It's how electric stove elements do their thing.)
At any particular point based on voltage/current/environment the filament will glow and resistance is stable, here it is 12v/1.2a in a vaccuum. The time from cold to hot is extremely quick given the very small size of the bulbs.

I won't go into thermonic emission or incandescence but look it up, it's very very cool. It is amazing to see those little buggers burning at 2000-3000 degrees inside a vaccuum.
 

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my post should have read 3 ohms, not .3 (typo), and Rideon, you are correct, parrallel. in series, you would have a dimly lit bulb due to voltage drop. mspgs2's calculations are correct.
 

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Discussion Starter #10
Thanks Guys! Especially MSPGS2!

The response make sense.

I am paranoid of Italian Electrics- going for coffee this morning I noticed a Alfa Veloce with a boom box on the passenger seat. You just know that if he attempted to plug in a normal radio, his whole system would go deader than Mussolini.

I will now attempt to marry the Italian electrics to the Kraut turn signals!

Regards,
Rideon
 
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