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My headlamps were not aimed at all properly from the factory. Problem 1. Problem 2 is the highbeam just seams weak to me. The low beam is adequate. I commute half city half highway, and the low beam is ok in either scenario. The high beam just doesn't seem to light the distant stuff up well. Maybe it is a bit too diffuse, or maybe I need to work on aiming it lower still.

Problem 3 is energy draw. Now that I have heated gear (jacket and gloves), at engine rpm below about 2500 there isn't enough alternator output with high beams on, so the battery is getting drained when sitting at a stop light. At highway speeds there is enough alternator power, but just. Meaning there is no spare power with high beams on if I were to add heated pants or aux lighting or anything else that needs much power at all. Reducing the power draw on the headlights would free up some power for elsewhere. This is not a big issue at all to me, but I could see for some folks who hang all kinds of aftermarket stuff on their bikes it could become an issue.

wow is there a beefier stator available??
 

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wow is there a beefier stator available??
Yes and No.

Some swear by modding the stator to a different configuration to keep it cooler. The OEM stator has some failures though I am not convinced the rate is terrible (but if it happens to you it is 100%). The mod does not increase the power output, so it doesn't solve the power limitations. It just reduces the power being converted to heat internally, which should improve long term reliability.

However, I find that without the high beam on there is enough of power for my heated gear even at idle. The high beam draws another 55W and that is what the alternator cannot handle at idle rpm. At higher rpms, something like 3000+, which is the normal range while riding, there is just enough power for heated gear plus high beam.

So I wouldn't say the alternator output is problematic unless you're looking to have 2 riders with full heated gear.

However, a lower power headlamp and high beam would be nice just to increase the margin of electrical power available.
 

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Dumb newbie questions.

What's the difference between the F2 and the H7? My 2015 V650 uses H7 bulbs, 55W. Does the Evitek F2 fit this or is it something totally different?

Do LED bulbs put out similar lumens? Apparently the law restricts halogen H7 bulbs in the USA to about 1200-1500 lumens. So would an LED replacement really be any brighter? If they are brighter, are they legal?

Power. Why are the LED bulbs not a very big difference in power? LED bulbs for home use only draw about 20% of the power for the same lumens. Or are they lying to us?
The difference is most require a driver, which creates losses, many of the cheap 120 VAC LED are just that, not dim-able and many have a overrated lumen output, also from a efficiency standpoint they aren't very efficient, the saying you get what you pay for, there is some truth to that.As to the F2, I checked current at 14.2 VDC and it was around there stated 36 watts.
I have that one in my 2015 low beam. If you read post #124, #132 #135 of this thread you would have your answer.

Here are the specs, and from memory the 55Watt OEM is 1500 lumens,or 27 lumens per watt consider the F2 is 6000 lumen at 36 watt= 166 lumens per watt, if we do the math and the lumens in the spec for LED are correct, it is 6 times more efficient than the OEM.

 

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The difference is most require a driver, which creates losses, many of the cheap 120 VAC LED are just that, not dim-able and many have a overrated lumen output, also from a efficiency standpoint they aren't very efficient, the saying you get what you pay for, there is some truth to that.As to the F2, I checked current at 14.2 VDC and it was around there stated 36 watts.
I have that one in my 2015 low beam. If you read post #124, #132 #135 of this thread you would have your answer.
Well I'm still not sure on whether the F2 is an H7 bulb or something different, and vice versa. The F2 description sounds as if the geometry is very close to the OEM bulb and should have the same beam pattern, but how does that relate to the H7?

Also, 6000 lumens is about 4x the legal limit in the USA according to what little I can find online. 1500 +/- 10% is commonly mentioned on the internet. Though I'm not sure it is an actual max limit rather than the minimum, and then I can't find any numbers in the CFR about headlights (but what a maze the regulations are...). I'm hoping somebody might know if there is an actual max brightness allowed.
 

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Well I'm still not sure on whether the F2 is an H7 bulb or something different, and vice versa. The F2 description sounds as if the geometry is very close to the OEM bulb and should have the same beam pattern, but how does that relate to the H7?

Also, 6000 lumens is about 4x the legal limit in the USA according to what little I can find online. 1500 +/- 10% is commonly mentioned on the internet. Though I'm not sure it is an actual max limit rather than the minimum, and then I can't find any numbers in the CFR about headlights (but what a maze the regulations are...). I'm hoping somebody might know if there is an actual max brightness allowed.
Post 124 of this thread ; Safego F2 led lamps
So Here are some photos of the Safego F2 that is in my 2015 650 ABS Versys low beam, I need to point out if used on a 2007 to 2014 no problem, on a 2015 or newer, unless you deliver pizza or like to direct air traffic , don't bother, only advantage( besides reducing wattage) is they can be partially focused and light output on the road is equal to my Osram 65 Watt , but at 14.2 VDC my Osram is closer to 100 watts, so I gain between 30 ( OEM ) and 60 watts using this on low beam.



Remember a filament is 360' of light, even though the placement and length of the led is similar, it exists on two opposite sides only, the reflector on the 2015 has too many angles, no two horizontal reflectors exist on the same plane as the led, so you may light up to the right and front, but no way to project light farther forward without blinding a driver coming towards you as well and lighting up the trees on your left. The problem is the reflector, no LED produced will make a difference, unless they can duplicate a filament at that wattage.






SafeGo mounting;note the map gas torch above photo


This is the after map gas heating and mod, OEM light will fit in no problem, the taper of the SafeGo wasn't clearing the spring before the mod


Another view and the boots already had holes cut out and were given to me for free


 
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LED Headlight High Beam

Just a update. I intend to try adding my SafeGo ( when purchased it was sold in pairs) to the high beam location, presently I have the Osram 65 watt which is roughly 75 to 80 watts . My only concern is beam pattern, since the high beam is aimed right, not concerned about oncoming traffic since it gets dimmed on traffic approach, my main concern is will I get more useful light?? I will report back, early 2020, once I can get my bike out.
 

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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:
 
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Lucky / Originally a PM

Thanks @onewizard. I'll have to read this a few times to understand it completely...:)
Yes I had to read it several times, then cut it out of my PM , then edited it. Then I looked up the wiring diagram, added more stuff, then finally stopped as many won't have a clue what I am talking about. Any questions feel free to ask. I have many ways of explaining things.
Formulas;
E = Volts
I=current expressed in amps
R = resistance in ohms
W or Watts is E times I
We will use a example here. 100 volts , 2 amp , 200 watts , 50 ohms ( one 20 ohm and one 30 ohm resistor connected in series)

Volts is equal to I X R or; the current times ohms so 100 volts or 2 amp times 50 ohms = 100 volts

Current or amps ( current is work done) is I equals E over R voltage divided by ohms 2 amp found by dividing volts of 100 by 50 ohms =2 amp

Resistance is equal to E over I or Voltage divided by current 50 ohms found by dividing 100 volts by 2 amp = 50 ohms

Watts is E X I or voltage times current so 200 watts or 100 volts times 2 amp = 200 Watts

Voltage drop, found by measuring the current through times the ohms, so the 20 ohm has 2 amp time 20 ohms = 40 volts

The 30 ohm resistor is 2 amp times 30 ohms or 60 volts.
Not going to get into three phase power, power factor correction, harmonics, series parallel circuits or electronics, converting AC to DC, the difference between RMS AC power and DC power.

As a side note I have a Ariens 8/24 snow blower that I bought used, paid way too much, but got it working pretty reliably for now. One thing is it had a light on it, basically useless. I converted it to DC with a filter capacitor, installed one of my original led headlight bulbs into a emergency light head, ( this was designed to operate on 9 to 32 VDC ) I can now read a newspaper in front of my blower:grin2: yes under load it barely makes it past 9 VDC but man it works real well>:)
 

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Formulas;
E = Volts
I=current expressed in amps
R = resistance in ohms
W or Watts is E times I
[/QUOTE

Thank you Onewizard for using this formula to explain to us DIY electronic enthusiasts the beauty and simplicity of electricity. Ohms law,isnt it? For anyone reading this who has ever been mystified by electricity, like me, this is the magic formula. Armed with a Multi-meter and these four expressions you can figure out pretty well any electronics circuit,big or small! :thumb:
 

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One clarification; in your original response, you mention (mainly) "in regards to switching to a Series regulator...".

My bike has the original Kawasaki factory R/R... I don't know all of the detailed differences between the Polaris and the original Kawi R/R, but as I understand your response, it seems wiring the headlights directly to the battery (controlled by a relay of course) would still yield a lumen increase in the incandescents... correct?
 

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Separately Wired Headlight Wiring

One clarification; in your original response, you mention (mainly) "in regards to switching to a Series regulator...".

My bike has the original Kawasaki factory R/R... I don't know all of the detailed differences between the Polaris and the original Kawi R/R, but as I understand your response, it seems wiring the headlights directly to the battery (controlled by a relay of course) would still yield a lumen increase in the incandescents... correct?
Some have done it in the UK Versys forum. To be honest, why not just change the bulb if you want more lumens. I have about 5 of the 65 watt , about 600 lumens more than stock with the same 500 hour life span, a 40% increase in lumens, with voltage drop it may be a 35% increase, no screwing around.Think about the difference , the Osram has very close specs for the OEM 55 Watt of 1500 lumens https://www.osram.com/ecat/OSRAM OR...Cars-Automotive Lighting/com/en/GPS01_1057023
/PP_EUROPE_Europe_eCat/ZMP_61232/

The Osram 65 watt 64217 PX26D is rated at 2100 lumens and 500 hours, hard to find data anymore. I have several of these and if my mod works I will sell off what I have.

Want to see what wiring changes would make?? Take 16 or 14 gauge wire, first find your ground wire at the lights, connect a wire from there and start your bike, the touch the other end to your negative battery or engine ground, see if there is a difference. Next take a second wire and connect this to you positive battery, if you have a wire connector marette , can twist https://www.homedepot.ca/product/ideal-can-twist-wire-connector-300-pack-/1000665625 strip the insulation and place connector, take a second wire and connect to the positive headlight wire, with the bike running watch the light output and join the two positive wires, you are in parallel with the original wiring, bypassing the fuse and relay, see if you notice a increase in lumens. Be aware using a 16 gauge wire in parallel is equal to a 13 gauge wire, if using a 14 gauge you are closer to a 12 gauge wire in parallel. After screwing around, consider the fact that if you are lucky, you might get a 80 lumen increase . I will send a link for the UK site.
 
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Ok here's a new thought @onewizard or anyone else with more knowledge than myself.



If a switch were installed on the Yellow/Red wire (connecting the main fuse to pin#11 on the relay box), is the following correct?

1)When the switch is in the "off" position, the bike will not start, as the starter circuit is broken.

2)Once running, the switch will disable headlights but allow the bike to continue running, as the starter circuit relay is only required for starting.

3)The switch, when "off" would also eliminate the parasitic drain through the headlight relay.


Wiring intrigues me, but my electronics theory is much weaker than my ability to read a wiring diagram.
 

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Option #3

Ok here's a new thought @onewizard or anyone else with more knowledge than myself.



If a switch were installed on the Yellow/Red wire (connecting the main fuse to pin#11 on the relay box), is the following correct?

1)When the switch is in the "off" position, the bike will not start, as the starter circuit is broken.

2)Once running, the switch will disable headlights but allow the bike to continue running, as the starter circuit relay is only required for starting.

3)The switch, when "off" would also eliminate the parasitic drain through the headlight relay.


Wiring intrigues me, but my electronics theory is much weaker than my ability to read a wiring diagram.
In option #3 https://www.kawasakiversys.com/forums/1633961-post8.html adding a relay to the ground wire of the starter solenoid and connecting the coil circuit of that relay to ground and the tail light would accomplish what you suggest. FYI pin #11 goes to the starter solenoid positive side of the solenoid

I copied the relay diagram to this post, putting a switch in on pin 11 would solve the parasitic drain, however , it needs to be in the on position for the headlight to work , so if what you are asking, a 5th way of eliminating the drain, yes that would work.On a side note, my relays cost about $4, C/W harness, are reasonably waterproof, and unlike a switch, you forget to turn it off or the contact fails, both cases your bike ain't starting.

 

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The 5th Element

>:)Ok; so to confirm... once the bike is running... the switch would become a "headlight off switch"...

When I saw that Pin#11 serves the headlight relay coil, it made me curious.

This option, call it #5, would give you a "hidden" kill switch as some like to install... also eliminating the drain and giving you an option for "stealth mode" >:)

Of course, as stated, the switch need be robust and properly installed.
 

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>:)Ok; so to confirm... once the bike is running... the switch would become a "headlight off switch"...

When I saw that Pin#11 serves the headlight relay coil, it made me curious.

This option, call it #5, would give you a "hidden" kill switch as some like to install... also eliminating the drain and giving you an option for "stealth mode" >:)

Of course, as stated, the switch need be robust and properly installed.
It is a unusual circuit. A simpler solution without removing the gas tank would be to cut the ground wire coming from the start solenoid and connecting a switch to it. The end result would be the same, that circuit is described in option #3.
 

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Option #3 / Solenoid Ground / Relay Added

Copied from FastEddie , for his complete install on a 2015 Versys 650 go here ;https://www.kawasakiversys.com/foru...mk-1-2-3-650s-revised-2019-a.html#post1366202

Once I had crimped the connectors (3) onto the stator wires, I looked at WHERE (and HOW) I could route them, then after a 'trial-fitting', attached everything.





Glen also included some of that 'wire-loom' plastic so I enclosed the wires in it (THIS after trial-fitting the side fairing to ensure that everything would fit OK)...!





I also made this HELPFUL note:



...so I took a pic looking DOWN at the gas tank, right side....



On the Gen 1 Vs I'd had to cut a wire (going to the main fuse area) to stop an electrical drain, then add a diode to stop the drain, but on the Gen 3 there's been changes to some wiring.


Option #3

Actually I ( Onewizard)came up with a easier method of isolation after a member got a little snarly and thought there should be a plug and play method., I intend to do the alternative method post in this thread, the advantage of this method is no tank removal.

This requires that you pull the MAIN (30 amp) FUSE, disconnect the plug that goes to the small covered area where the main fuse is, then as the "parasitic drain" happens thru the GROUND (black/yellow again) wire, you need to make it ONLY pass current when you want it to (while running the engine). The method is to cut that black/yellow wire (AGAIN leaving enough wire BOTH sides of the cut) to solder about 24" of wire to each end. (THIS 'breaks' that 38 milliamp drain.)

Option #3 Cut Solenoid ground, connect to relay, coil cct use tail light

What you're going to do w/ the two 24" wires, is to run them to an "ORDINARILY OPEN" relay, and 'triggered' by either your running lights or stop lights, which I decided to put to the rear of the battery, towards the right side. Here are the two wires soldered, then 'shrink-wrapped' w/ them leading to the underseat area.



I used the 'rear running lights' as the trigger



then attached them thus: the "black/yellow" 2 wires go to #87 and #30, then are triggered by the running lights attached to #86, w/ the ground supplied by another wire going from a frame ground to #85.

HOW it works is, that by cutting that "black/yellow" wire you disable the drain, then by 'firing' (triggering) the relay by something ONLY powered ignition ON, everything operates normally on start-up.



And here's the left side "buttoned-up".



HOPE this is helpful to some of you....
 
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It is a unusual circuit. A simpler solution without removing the gas tank would be to cut the ground wire coming from the start solenoid and connecting a switch to it. The end result would be the same, that circuit is described in option #3.

Ok I'm seeing it now... cutting the ground wire has the same effect as cutting the positive side (Pin#11)
wire... because the ground wire feeds the Pin#11 wire... right?

Thanks for indulging my questions:)
 

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Cover seal source

182826
 
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