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I'm trying to figure out what's wrong with plugging the solonoid back in and zip tieing it up under the tank and out of the way?
 

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Putting the solenoid under the seat is fine and all, nothing wrong with it, I just want to go one step further and totally remove it.
 

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That does not sound right to me. The higher the resistance, the lower the current draw (Current =Voltage / Resistance, I=E/R). This implies that the higher the resistance, the less heat is going to be generated.

The highest resistance you can get is infinite resistance, which means that nothing is connected to the voltage source, and that means that the least power is drawn (none) and the least heating takes place (none).

If you did not burn out the 10 ohm 1/2 watt resistor, you should be twice as safe with a 20 ohm 1/2 watt resistor. You may even be able to go to 1/4 watt, but since I have waffled about everything else in this post, I certainly am not going to recommend the lower power.

I am not an electrical engineer, I only play one on TV.

Jan
Yeah, someone explained that to me before. But you are changing 2 variables, resistance and current (which of course is very applicable here). If all things are kept constant ('cept the resistance) then resistance is proportional to heat, look at a light bulb for example...maybe a bad example...but if you restrict the flow of energy it has to go somewhere, i.e. heat.

Here's a reply from a EE I know...

P = V^2 / R and P = I^2 * R

If the numbers were right for your solenoid, you would have 12v^2 / 20ohm = 7.2W, obviously much higher than the 50mW ( 0.05 watt from Redfire) you mentioned. This would also work out to 600mA current draw.

Or if you assume the 50mW wattage you mented is right then the numbers would be as follows: 0.05W = 144 / R, therefore the real resistance would be 2.88kOhms. So my point is, something doesn't line up with those numbers.

If I had to guess, your 20ohm number sounds about right for a solenoid coil. This means that the 7.2W number would apply and the 600mA would also apply.

Given that assumption, your 50ohm resistor(s) would be as follows: 144 / 50 = 2.88W. The reason your guy had a fire is this: 144 / 4.7 = 30.63W, or about 60x the rated wattage.

Another fairly intangible thing to consider is that 4 resistors in parallel have a MUCH higher surface area than one single resistor and can therefore dissipate much more heat, but they obviously take up much more space.


So I went with a 50 ohm, 5 watt, heat-sinked mountable resistor. I just haven't installed it yet (Saturday's project).
 

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I am still not happy with your case, but I will probably take quite a long time figuring out why.

Until that time, here is a circuit diagram to decipher while you wait:

xkcd: Circuit Diagram

Jan
 

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heat and electrical resistance

The higher the resistance, the more heat is going to be generated.
DrBLVD,

In order to illustrate that the lower the resistance, the more heat is generated, here is a thought experiment:

There is a 1200 watt heater plugged into a 120 VAC outlet.
Power (1200 watts) = voltage (120VAC) x current (amps), therefore
current = power/voltage = 1200/120 = 10 amps.

Ohms law says that resistance of the heater = voltage / current = 120 VAC / 10 amps = 12 ohms.

Now plug in another, identical heater into the outlet right below the one above (the thought-house has 20 amp circuits). That one will also draw 10 amps because it has a resistance of 12 ohms. These two heaters are in parallel, as are all the devices plugged into a given circuit of wall outlets, so their resistances are in parallel. The formula for the equivalent resistance of devices in parallel is:

Equivalent resistance of parallel resistances (ARRL Handbook) = 1/ (1/first resistance) + (1/second resistance)) = 1/ (1/12) +(1/12)) = 1/(2/12) = 1/(1/6) = 6 ohms.

Since the heat increases as you plug in more heaters, and the equivalent resistance of all the heaters together decreases (from 12 to 6 ohms) as you plug in more heaters, higher heat is associated with lower resistance.

The math is the same for a power source of 120 VDC. In fact, if Edison had won the power war instead of Westinghouse and Tesla, it would have been DC.

Jan
 

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DrBLVD,

In order to illustrate that the lower the resistance, the more heat is generated, here is a thought experiment:

There is a 1200 watt heater plugged into a 120 VAC outlet.
Power (1200 watts) = voltage (120VAC) x current (amps), therefore
current = power/voltage = 1200/120 = 10 amps.

Ohms law says that resistance of the heater = voltage / current = 120 VAC / 10 amps = 12 ohms.

Now plug in another, identical heater into the outlet right below the one above (the thought-house has 20 amp circuits). That one will also draw 10 amps because it has a resistance of 12 ohms. These two heaters are in parallel, as are all the devices plugged into a given circuit of wall outlets, so their resistances are in parallel. The formula for the equivalent resistance of devices in parallel is:

Equivalent resistance of parallel resistances (ARRL Handbook) = 1/ (1/first resistance) + (1/second resistance)) = 1/ (1/12) +(1/12)) = 1/(2/12) = 1/(1/6) = 6 ohms.

Since the heat increases as you plug in more heaters, and the equivalent resistance of all the heaters together decreases (from 12 to 6 ohms) as you plug in more heaters, higher heat is associated with lower resistance.

The math is the same for a power source of 120 VDC. In fact, if Edison had won the power war instead of Westinghouse and Tesla, it would have been DC.

Jan
Sir, I stand corrected, Thank you. Good analogy!.....ummmm...nice schematic too, but where is the Flux Capacitor?

As for my little project, I installed the 50 ohm, 5 watt and idled the bike for some time (raining outside, I hate to ride this purdy thing in the rain). I did not notice any heat generation by touch nor infrared thermometer and the FI light did not illuminate.

It's suppose to get a touch warmer this week and will ride it into work a couple days. If anything happens I will post.
 

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Ok can any one tell how I can get my FI light to go out after 4 1/2 min.?
 

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Ok can any one tell how I can get my FI light to go out after 4 1/2 min.?
I'm not sure if this will work, but it does on OBDII systems in cars and trucks.

Disconnect the battery, turn the key to the one position and hold the starter button for 4-5 seconds. This should drain the computer of any power being held in capacitance and/or storage clearing the memory.

Tell us if it works.

There is also methods of shorting certain terminals, like in GSXR's, but one would need very exact directions so no damage is done.
 

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I will this in the AM and will let you know.

Thank
 

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Ok can any one tell how I can get my FI light to go out after 4 1/2 min.?
The display in your speedo (odometer display) is storing a fault code. The FI light will not go out until the code is downloaded and the fault is corrected. To download the code, there is a white connector with a rubber boot over it down near your ECU behind the right side cover. You dealer has a simple switch that plugs into this connector and that will display the code in your odometer display. Disconnecting the battery will not do it. This link shows how you can make your own switch with a paper clip. Doing this will display the code in your odometer window. If the anomoly that caused the fault is repaired, the code will dissapere and the FI light will not illuminate. If the problem is not fixed, the FI light will again remain on until the code is downloaded and the fault fixed. Mine was code C29, secondary throttle position indicator. Mine had a broken terminal inside. Here is the link, oops, hang on I gotta go find it again.
The link http://www.motorcycle-journal.com/forum/gixxer-mixer/25721-secondary-throttle-valve-actuator-defect.html
 

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Well I try the disconnecting the battery and turning on the bike and hold the start button. Did not clear it. I know the code C49 the pair valve. After the short out and I repaired the wires and put the solenoid back on the FI light stay on. Then after driving for three weeks the FI light starts going out after 4-1/2 min. of riding. So I know that I do not have a short in the wires because the FI light is off for 4-1/2 min. If any one can help please do.
 

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Too bad the old battery-disconnect trick didn't work.
Did you try the method Viper posted for clearing the code?
Maybe you damaged the solenoid in the initial short?

All I can suggest is the resistor mod (50 ohm, 5 watt, heat sink type) once you clear the codes from memory. I've put on about 3 hours of riding and about 30 minutes total of idling and so far no light, no codes.
 

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It's just anyone's quess as to what effect the short he had has done. If the FI light goes out after a few minutes, the system is at least trying to work normally. Could it be somthing that takes 4 1/2 minutes to heat (I'm thinking build sufficient resistance?) and when it reaches that point it turns the FI light out?
 

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OK, finished reading the thread. And I decided to leave mine zip tied under the tank. I wrapped it with the soft side of velcro so as not to scratch against anything. Has been doing fine since Aug 2008.
 

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All I can suggest is the resistor mod (50 ohm, 5 watt, heat sink type) once you clear the codes from memory. I've put on about 3 hours of riding and about 30 minutes total of idling and so far no light, no codes.
To answer a PM from Daycruiser.
So far so good. 500 miles before the snow ended my riding season. No problems whatsoever.
 

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Just an update:
Got 1000 miles (or so) on the Resistor Mod, no FI lights...
I also just purchased a '07 C90T (pics to come shortly), the Resistor Mod works on it too. Got about 200 miles on it so far (just got it Sunday, PAIR off within 1 hour), no FI Light.
 

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Sorry to resurrect a dead thread, but it sounds like a resistor like the ones I used for LED conversion are the ones needed? 6ohm/50 watt?
 
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