**PM Question regarding Voltage Drop / Relays**

I was asked about adding a relay and switches for the headlights in regards to switching to a Series regulator.

**This post is primarily for the guys converting to Polaris or CompuFire Series regulators, and those modifying lighting.**

Two important things, the headlight is relay driven already and it has it's own fusing. What many do not understand and make the mistake of moving existing wiring to a direct battery connection or moving the positive output when converting to a series regulator to the battery instead of the original location of the main fuse. Each time I explain this I come up with a different way, today I am feeling pretty good.

So , Versys 650, V1000;X300; Ninja 650;KLR650 and more are all wired the same way. The battery, consider it as a load in relation to the charging system. The battery serves two purposes, to start the bike and to supply a auxiliary source of power when at idle if the charging system isn't able to keep up.

**So technically if you add the Polaris positive output directly to your battery, you have increased the continuous design current of the wiring system connected to the battery, now every load will be forced to travel to the battery over it's wiring. ** I point this out because original wiring with the regulator, the positive output is directly connected to the main fuse, the key switch and a few other locations all branch circuits stem from this point.

Adding any wiring directly to the battery rather than say auxiliary output wiring relay will again increase the current over the battery wiring, this increase will result in a slight voltage drop. Yes I have , a 20 amp circuit for the Denali horn, a 5 amp for the Oxford grips, a 15 amp for auxiliary power socket, 10 amp for my Denali lights , all on relays , all with 10 gauge or 12 to 14 gauge wire.

Two things, if you have converted to LED headlights, I was running Osram 65 watt ( actual 100 watt at 14.2 VDC, to be honest, it is 16 gauge wire, going to 14 gauge I probably would see a increase of lumens by about 10%) .** However** with my LED conversion it is something like 6000 lumens at 36 watt, be aware incandescent and LED with drivers act in reverse to a voltage increase, so on the first with 36 watt it would be a total waste of time, I doubt you would see even a 1% difference using LED, simply because original 55 watt is at 12 VDC , a increase in voltage ( by reducing voltage drop with a wire increase ) increases the actual wattage on incandescent. LED light with drivers, will output a fixed lumen capacity, many are designed around 9 to 32 volts , so the current through the wire at 12 volts is greater than the current at 14.2 VDC, also the current at 9 VDC is greater than that at 12 VDC. The losses in the wiring on the LED headlight conversion I am running . I will calculate.

Some info for you, 16 gauge wire which is what the headlights are fed by is 4.016 ohms per 1000 feet. Assume we have a total of 10 feet between ground and the fuse block / low and high beam switches, without considering losses on contacts, my 36 watts divided by 14.2 VDC =2.53 amp. The voltage drop on 10 feet = 2.53 X .04016 or 0.10 VDC voltage drop, unless you have a really good meter, you won't be able to measure this.

Let's go one step further, I am putting my LED Safego into the High Beam as well ( presently it is equal to 100 watt at 14.2 VDC ). So with both lights on, and we will throw in a additional curve, running heated gear and heated grips, put my voltage at say 13.5 VDC ( it isn't , but lets say it is), then current will be 36 divided by 13.5 =2.666 amp times two = 5.333 amp X .04016 =0.214186 VDC and that is a total wattage of 5.33 amps X 0.214186 VDC = 1.1416watts total loss in the wiring with both lights on.

A rough estimate of the OEM 55 watt I would say you have the same loss of 1.14 watts, with both on that might translate to 3 watts .

However if I went back to my original set up, using the Osram 65 watt ( actual with both on would be 160 watts at 14.2 VDC ) so current through would be 11.2 amp. Voltage drop on the same wire would be 11.2 X .04016 = 0.44979 Volts and 11.2 amp X .04016 VDC = a loss of 5.03 watts in the wiring.

I think you can see an advantage or what appears as a advantage for my original 160 watt lighting, thinking that you could reduce the loss of 5 watts in the wiring. You would be disappointed if it was the loss you were after by say going to 14 or 12 gauge wire and a relay. What would happen is a increase in lumen output of the incandescent lights.:grin2: