906 headlight conversion

[duckboy]

Hi Pasofiles
i am looking at fitting a headlight from a fz 750 with the twin globes (86 to 88 model) to my 88 Paso
given the output (as stated in manual) of the alternator is is 12V-350W, will this have enough power to run both globes (on high beam) as well as the other tasks the electrical system is expected to do? (like run the clock :laugh)
there is already a highbeam relay in place so the extra demand would not be put on the (aging) original wiring harness.
what size (power) globes would i be able to use?

[Desmo_Demon]

i am looking at fitting a headlight from a fz 750 with the twin globes (86 to 88 model) to my 88 Paso

Good luck finding one of the lights, and if you find a second one, let me know. I've been looking off-and-on for about six months.

[gail]

Redpaso has posted on this before, has one fitted to his 906,(I believe).
Where are you RED?, off rallying somewhere no doubt! and going hard!
Hopefully he may be able to sort out your query.
..........Marty

[Mc tool]

I remember reading ( I can you know !) in a two wheels mag cicra 1990 that there is a hella light that can be fitted to the paso with little modification. The article even gave a part number ...... which I now cant find ( the wife is a very tidy person, in fact if you stand still to long in our house you are likely to have your top pollished and your bottom vacuumed ! ) I will try to find this # for you as I am sure now that the topic has come up I will need one myself soon

[duckboy]

HI
I have previously owned an '89 900 ss. when i broke the headlight on that I replaced it with the Hella Halogen insert (part No. 1042), which needed slight modification (eg. secured with duct tape). Apparently the FZ750 unit is an exact mach for the Paso light, and should screw in without modification.

[duckboy]

Im still looking for someone 'versed in the flow of the electron' to do the math regarding the ability of the wattage output to cope with the photon generation.

[Mc tool]

Hi Duckboy what you are looking for is a thing called "Ohm's law " This is a basic formula to calculate the watts , current flow ( measured in amps ) and the resistance ( to current flow, measured in ohms ) This formula can be transposed to calculate any of the components above
Your alternater has an out put of 350 watts , if you divide this by the voltage in the circut ( which is not nessessarily 12v , you will have to use a volt meter to measure this at the battery and should be somewhere between 12v and 14.5v depending on engine speed ) but for arguements sake lets call it 12v , so 350w divided by 12v is 29 amps which is the max load your alt can support . using the same formula we can take 100 watts ( headlight power ) and divide by 12v which give us 8.3 amps so we can see that of the available power from the alt ( 29 amps ) the headlight is giong to need 8.3 of these amps to work . you would need to apply this formula to all of the constsant loads on the elect system ie magic box (ignition module ) ,head and tail lights , fuel pump and coils . Blinkers , brake lights ,horn and charging system are not constant so we can lump them together and call it 5 amps or 60 watts ( 5 amps times 12volts is 60 watts )
To get the wattage of the coils and fuel pump you would need to measure the current ( amps ) to each and the mutiply by 12 volts to give you the watts , Add up all the constant loads , in either watts or amps, add in the 5 amps or 60 watts for the other stuff ( this is a bit of a guestimate ) and as long as this total doesnt exceed the total power output of the alt everything is cool
BUT the higher the load on the system the harder it is on the alt and the voltage regulator / rectifyer and like the motor itself , doesn't pay to run them at the red line all the time so I would limit this total load to 300 watts . In a nut shell the 100 watt headlight is a doddle , but I would abvise the use of a relay and a direct feed to the battery or you may fry the standard light switches
good luck

[Finnpaso]

BUT the higher the load on the system the harder it is on the alt and the voltage regulator / rectifyer

50% right. More You have load in circuits, less regulator makes heat(very good for longer living time to it)! Regulator put ALL that current direct to earth, what are not used in circuits!!! Rotor make more power output, when more RPMs(so its very stupid machine) and regulator feeds only that amount of current to battery, that is used from it, all other power it feed to ground and make more heat. If u want to "protect" little Your regulator(from too much heat), then use so much current in circuits,as it is possible(like use hole time low beam, not only parking lights, etc....). If drive only with parking lights, then current feed from regulator to ground is about 55W more(low beams)(what is making that amount of heat to regulator!)

Anyway power consumption and power outlet from system must be in balance. Injection takes very much current, what is not problem in carburated bikes.....

[persempre907]

If u want to "protect" little Your regulator(from too much heat), then use so much current in circuits,as it is possible(like use hole time low beam, not only parking lights, etc....). If drive only with parking lights, then current feed from regulator to ground is about 55W more(low beams)(what is making that amount of heat to regulator!)

So, have we to choice?
We can save the regulator, turning the lamps on, or save the battery, draining less current as it is possible...

Ciao

[Finnpaso]

there is enough power in battery to handle all normal stuff, so if some reason battery goes down, then MUST be some faults in system. But in 907 that 350W system is very critical and cant put so much other devices, what use current. 906 is carbied, so there is not such problem. I use allways low beams, when i drive and all function well.....

[duckboy]

Hi Duckboy what you are looking for is a thing called "Ohm's law " This is a basic formula to calculate the watts , current flow ( measured in amps ) and the resistance ( to current flow, measured in ohms ) This formula can be transposed to calculate any of the components above

Fantastic, that is the formula i am looking for!
now this one is more of a curiosity than anything, but does anyone know if 2 x 50W globes will give the same light as a single 100W globe?

[ducapaso]

I can tell this: a single 100W bulb gives you more light (expressed in "lumen" ) than a pair of 50w bulbs

[duckboy]

I can tell this: a single 100W bulb gives you more light (expressed in "lumen" ) than a pair of 50w bulbs

so how does that work then?

[ducapaso]

I can tell this: a single 100W bulb gives you more light (expressed in "lumen" ) than a pair of 50w bulbs

so how does that work then?

it works with the same wattage and a bit less light than a 100w single bulb

[paso750]

I guess it`s hard to compare 2 50W bulbs with one 100W if you`re considering the FZ headlight.
As you`re using 2 reflectors with 2 bulbs in the FZ headlight I would assume that the spread of light is larger than with one bulb only. Furthermore I made the experience with the TÜV (vehicle control institute) that the light can worsen if you use stronger bulbs simply because the reflectors and specially the lenses have not been developed to handle such power causing the light output to become diffuse. (but I guess this problem is more when you try to use i.e. HD/Xenon bulbs in a standard headlight)

G: