Tried many sets of bulbs on the 14r,.....even HID's which I thought were the hot set up...NOT! Took them off as the beams do not project correctly with stock projector beams...

PIAA 65 Watt Platinum White, are The ticket.....take your time centerstand the bike, put a white foam board up in front use black magic marker to cross hair the white sheet..then take your time and adjust the high beams first so that they are parallel........then check the lows....get the highs up a little higher and mine focus about 75 yards in front and are exactly level with each other...bright white light and will make you proud to have these babies in your Ninja.

look here on my zzr 1400 : http://www.youtube.com/watch?v=Ek6q1K0Ll2g

Full, yours look great, are these light bulbs the Piaa platinum white? Question, you also replaced the top ones what type of light bulbs are the original these (little ones on top), how much you paid for them?
Thanks, great video

real WHITE HID Look :)

full PIAA

H11 Precious White 4800k (croisement) : http://www.piaa.co.jp/piaa_news/information/car/bulb/201101311135.html
H9 Xtreme White 4150 K (plein phare) : http://www.piaa.co.uk/bulbs/bulbshow.asp?ID=xtwhite
LED Hyper Dimple Super LED 6000 K (feu de position) : http://www.piaa.co.jp/products/car/led/position_valve/img/img_position_07.jpg

for buy here : http://stores.ebay.com/tekimokas-garage

I just ordered on Amazon the Piaa xtreme white H11 55w bulbs, I will try them first and if I'm happy with the results I will order H9 xtreme white ones.
Thank you guys

alohaalex, your going to like the PIAA bulbs...I have the same setup that Full has on his ZX and I'm super happy with it.

for real WHITE color, you must go around 5000 k !!!

my H11 4800 k : OK

my H9 4150 k : not real good ... :( when brocken , i'll put the same of my wife on his Honda :

http://www.youtube.com/watch?v=Mjh5-fr9Ud8&list=UUK1-4H1my0vM9m5yEkGZJAg&index=7

PIAA Southern Star White : 5100 k , this is the best model for White HID effect

* Last updated by: Full on 12/17/2012 @ 12:55 AM *

Do the Piaa Southern Stars come in the H9 & H11?

only h11

Nice... No overheating problems with 65 watts

no problem :)

@Full:

The H4's you indicate with the link to the japanese site show this:

H11 Precious White 4800k (croisement) : http://www.piaa.co.jp/piaa_news/information/car/bulb/201101311135.html:

12V 55W -> 115W

...which I think means the original spec OEM bulbs are 60/55 watt, and the H11 Precious White draw 115 watt.

Is this correct?

The H11 "Southern Stars" show:

55W -> 110W

...which I think means the original spec OEM bulbs are 55 watt, and the H11 Southern Star draw 110 watt.

Right?

How can you not have an overheating problem?

They're both gonna draw at least twice as much current.

* Last updated by: mebgardner on 2/26/2013 @ 12:27 PM *

No...it means....55w is equal in lumens to the 110w.Not that the stock bulbs are 55....it's comparing the brightness difference.I have 55w Piaa's in all my bulbs...no overheating at all....I mean..that's how I interpreted it.My voltage was reading 14 something with those installed....running voltage.55W H11's.(high beam bulbs).I can't say that a running voltage means a certain amount of draw...it might.And I don't know if you can go by the voltage indicated on the display as going into the battery or not?It may have to increase the output with different bulbs...IDK.This is something I've never actually seen anyone answer here....displayed voltage while running and what it means.I suspect it is NOT actual battery voltage...but what's being put into the battery.I guess you could get actual voltage by turning the key on and reading that.

I've never had an actual 14 something starting volts.It's always been 12's,13's.

* Last updated by: Grn14 on 2/26/2013 @ 1:08 PM *

Your "14 something" volts is your alternator output

@Leo: +1, almost. The displayed voltage value is almost certainly the regulator / rectifier (RR) output to the battery. The displayed value varies with the engine speed. Turn more rev's, and watch the RR generate more DC voltage, depending on the load (in amperage) through the circuit. The displayed voltage will increase until the RR clamps it at some max value. It can also increase if the load goes down. (for example: you turn off all the lights).

The alternator generates AC, not suitable for a DC Voltage display...

@Grn:

....it's comparing the brightness difference

Thanks very much, that answer is clear to me.

I guess you could get actual voltage by turning the key on and reading that.

Yes, that's correct.

...a running voltage means a certain amount of draw...

Almost there. Voltage (in Volts) divided by Resistance (in Ohms) equals Current (in Amps).

The "draw" is the current. With a steady state resistance and voltage, then the current will be steady. Increase the voltage, keep the resistance the same (dont turn on any more lights...), and the current (the "draw") will increase.

So, make more revs, watch the displayed voltage go up, and know that the lights get a bit brighter because of that.

At the same time that the RR is satisfying the electrical "draw" on the system (keeping the lights lit), the RR is also replacing energy that may have been drained from the battery (if there is excess capacity in the generating system, and a shortage in the battery). In that scenario, the battery is indicating it has a "shortage" by it's output voltage being slightly (or greatly) lower than "normal" (normal by specific gravity of chemistry). If the RR has excess capacity by indicating it's generated voltage is greater than "normal" (with the RR electrically "tuned" to the circuit by the designers), then the RR output voltage will fill energy into the battery until the battery output voltage increases over time, and thereby indicates it's "full".

* Last updated by: mebgardner on 2/26/2013 @ 5:05 PM *

Okay...so....if the load increases..say...more wattage needed for a set of gloves or vest...then the display is going to show more 'volts' while those are plugged in?Basically...it's showing that the stator is operating okay....can I assume this?I've had mine fluctuate between 13 and 14 on various rides.But normally it will display 14+ on there...consistently.I've also noticed that any of my bulb systems...on this bike...or my 07...they never dimmed or got brighter in any voltage move....And I used to ride a lot at night...I didn't see any effect upping the revs or slowing the revs.They stayed bright all the time.

* Last updated by: Grn14 on 2/26/2013 @ 9:16 PM *

I am waiting on these, ordered last week.

http://www.ebay.com.au/itm/PIAA-5100K-Southern-Star-White-H11-JDM-Headlight-Fog-Light-Bulbs-MADE-IN-JAPAN-/310420583932?pt=Motors_Car_Truck_Parts_Accessories&hash=item48468235fc

http://www.ebay.com.au/itm/PIAA-4800K-Precious-White-H9-JDM-Cornering-Lamp-Light-Bulbs-MADE-JAPAN-H-786-/310568857178?pt=Motors_Car_Truck_Parts_Accessories&hash=item484f58ae5a

http://www.ebay.com.au/itm/PIAA-5100K-Southern-Star-White-T10-194-168-FRONT-SIDE-MARKER-Light-Bulbs-H-531-/310568870418?pt=Motors_Car_Truck_Parts_Accessories&hash=item484f58e212

Hopefully be ok.

@Grn:

...more wattage needed for a set of gloves or vest...then the display is going to show more 'volts' while those are plugged in?

No. The RR will attempt to "regulate" the system as best it can. If the load on the system increases from a set of gloves or vest, then *if* there is surplus generating capacity, the RR will tap some of that surplus to supply the new additional load.

Think about these two simple math formula: 1) Volts / Resistance = Amps, 2) and Volts * Amps = Watts.

If there was no "regulator" in the system, and there is 12 V, supplying 6 A into the circuit. (1) indicates the resistance is 2 ohms (12 / 2 = 6), and (2) indicates 72 watts (12 * 6 = 72). Thats the "initial conditions".

Now, lets add a set of gloves that requires 40 watts. Remember, no regulator in the system. Here's what happens:

72 + 40 = 112 watts is now whats needed to supply the "new" system. From (2), we get 12 / ? = 112, ? = 9 amps in the new system. It increased from 6 A to 9 A.

The new circuit will need to supply 9 amps now, *if it can*, at 12 volts.

If it can not, then the supply voltage (the battery / generator combination: the "system supply") will "droop", go down, until the system stabilizes at the voltage and current that it *is* capable of supplying.

It "droops" because of this: The "new" system requires 112 watts (12 V @ 9 A) at steady state. From (1) we get the resistance of the "new" system: 12 V / ? = 9 A, ? = 1.33 ohms. We see that adding the gloves has decreased the system resistance (from 2 ohms to 1.33 ohms), and increased the required current (from 6 A to 9 A).

This "new" resistance after adding the gloves does not change (well, not in this discussion). Its steady at it's new value of 1.33 ohms.

This is why the system supply voltage will "droop" if it can not supply the required current. Something *has* to change. Looking at (1) math formula shows that *if* the resistance is not changing (from its "new", after gloves added value of 1.3 ohms), *and* the system is not able to supply the required "new" current value of 9 Amps (there's no surplus capacity, but can supply some smaller number of amps), *then* the only other system circuit value that *can* change is the system supply voltage. It *has* to decrease.

Thats why the system voltage droops when it's not able to supply enough current to support all the load.

What the RR brings to the table is the ability to stabilize the system voltage over a range of load conditions (a range of varying resistance), given some amount of surplus system capacity. That is, if the system has no excess generating capacity, then the RR can not do its job. Run the electrical system at max amps, nothing to spare, and drain the battery because the RR can not "route" the excess to the battery. The load wants it all...

Hope this helps.

Ya...I understand it...thanks Meb!

The RR, or regulator/rectifier unit does just 2 things. It converts the AC output of the alternator to DC and pegs the max voltage at 14.2 - 15.2v. It has nothing to do with load sensing. The output from the alternator (with no regulator) will be well over 12v, typically anything between 40-50v AC, obviously way too high to run the vehicles electrics directly. The alternator has a fixed power output, on the 14 it's rated at 490watts. If the bikes electrics require more than this then power will start to be drawn from the battery rather than being supplied solely by the alternator. You don't really need to get involved with resistances or equations etc. All you need to do is compare the rating (in watts) of your accessories, plus the OE electrics, lights etc (excluding the starter), with the output rating of the alternator. Realistically you want a max continual draw of no more than 90% of the alternators capacity on a regular basis. More will be OK for short periods with time for the battery to recover.

@pegscraper:

Are you *sure* there's no load sensing feedback? Car alternators do that by adjusting the field strength of the magnetic field being used to generate the electricity. That is, the "spinning magnet" gets it's field strength adjusted if max power is not being needed by the system. The designers do this for fuel economy, and to keep the alternator from burning up. (dont burn fuel to develop max electric power if'n ya dont gotta).

I realise this is not a car alternator, but it *is* an alternator. If the spinning magnet really is just a magnet, then yes, there's likely no feedback loop for alternator output voltage adjustments (but does not completely rule it out, just makes sensing it more difficult).

The alternator has a fixed power output, on the 14 it's rated at 490watts.

That value represents the *max* available electric power, at some (high) engine rpm that will support the max electric load of 490W.

The RR's job is to "regulate" the system output voltage when it (the system, including the engine) does not require all the power (amps * volts) being developed (this is the surplus power idea I mentioned).

One can either: Not generate a surplus, or burn it off as waste heat, or shunt it to a storage device (a battery).
If the battery is full or nearly full, this last idea does not work...

If there's truely no alternator output voltage adjustment feedback loop, and the alternator develops full power the entire time it can (at some enngine speed), then where does the surplus power go?

If it's not used by an electric device, then it's either stored, or dissipated as heat in some other load as waste heat (its gotta go somewhere) (see kirchoff's law :).

So, again if there's truely no alternator output voltage adjustment feedback loop, and the alternator develops full power the entire time it can (at some enngine speed), then its' going into a "dummy load" resistor, or it's retained as waste heat in the alternator itself.

Still wanna assert there's no load sensing feedback? I'll listen to any plausible ideas you've got.

Now, will you agree with me that a bit of knowledge about ohm's law will not hurt anyone's understanding about simple electric circuits? If you can perform multiply and divide, you can understand it.

I wanted to make it (the math and the circuit ideas) as easy and as simple as I could for folks to understand. That's all...

I think the excess is converted to heat at the regulator... no provision for modulating field strength at the point of generation.

@Hagrig:

no provision for modulating field strength at the point of generation.

Thanks, Hagrig. Thats a key peice of info.

@Pegscraper:

The output from the alternator (with no regulator) will be well over 12v, typically anything between 40-50v AC...

Thanks, my friend, that's also key info.

I think the excess is converted to heat at the regulator

So, Hagrig may be right.

Since the output of the Alternator/Rectifier can vary from, say, 20 VDC to prolly about 60 VDC (from 50 VAC max, from pegscraper). This Alt/Rect output is input to the Regulator.

I think the Regulator wants to act like a big 'ole variable resistor to maintain 12 VDC output. That means the Reg. has to "drop" the voltage difference across it. (The difference from the varying rectifier output / regulator input to the regulator output).

Lets say the engine is throttled to make steady revs for alternator/rectified output of 50 VDC, and that the load is 12 VDC @ 9 A, also steady (after adding the heated gloves :)

So, the regulator must drop a voltage difference of (50 - 12 = 38 VDC) across the regulator. 38V * 9 A (through the regulator) = 340 Watts.

That's pretty toasty, and should fry it in short order.

So, I'm still missing some key info, but I dont know what that might be.

Maybe I'm wrong... I'm not seeing the circuit that primes the field with low tension awaiting multiplication from mechanical energy source.

You know...you two are treading dangerously close to the "no non-motorcycle" discussions here.....LOL!!!! Where's scottfarm when ya need him;)...carry on .

* Last updated by: Grn14 on 2/28/2013 @ 12:21 PM *