I have a Triumph Speed Triple. The headlights are constantly live once the ignition is switched on, even when the engine is cranking under starting. it has twin headlamps, I have found a plug in relay and harness for single lamps called Easy Start Battery Saver which says "Simply plug-in control module into existing headlight wiring, now when ignition is in “on” position, only position light will be turn on while headlight stays off; this will allow more battery power available to start the bike. Once engine’s been started, simply flip high-low beam switch to turn on bike’s headlight and headlight will stay on until ignition is turned off again. This product originally was design for customer to have the option of turning off headlight when starting the bike. It is particularly useful when trying to start the bike in cold climate, when battery does not have much cold cracking power left; it would also allow people to run ECU or ignition test on the bike without having to hook-up to a larger battery because headlight is turned off. Unfortunately the kit for just one head light is £30, and I would need two kits for twin lights, therefore I am hoping to construct my own for less cost.

I am presuming this is a latch relay type set up, but I can't figure out the wiring for it, or if I should be using another type of relay, can anyone advise me please?

Old Spark,

Ok so I groung the relay through the starter, as the ground or negative is common through the chassis of the bike how is that going to diconnect the lights whilst starting, as all I will be doing is moving the earth/ground for the headlight relay to a different location, the relay will still be energised and the lights on??

The bike has a starter solenoid, I'm still confused to how connecting the ground to the positive work, how will the lighting relay ground through the positive starter circuit, as when the the starter button is not pressed then the circuit will be open circuit.  Any chance of providing a diagram, as I'm still not clear on this?

well it'll be zero volts, but the ground from the lighting relay needs to be -12v to make the circuit for that relay, but the starter circuit when the starter switch is not operated will be zero volts with no continuity. Obviously I'm not understanding something here, do not all circuits need to have continuity to operate, and the component, in this case the lighting relay, needs to have a +12v to the internal electronmagnet and a -12v to the other side of the electromagnet in the relay to operate it, but going through an open starter circuit will not provide the -12v supply ??

I know OldSpark's idea seems to go against the flow of everything you know, but it is right and it will work.  A Tyco (formerly Bosch) relay draws only 160 milliamps.  The coil part of the solenoid on the starter draws at least 3 amps of current.  So if you wire both of these in series, the only one that will energize is going to be the relay. 

You can not judge a user by their screenname. The Old part is probably correct, but not so much on the Spark part.

Guys, I appreciate the help, but just to make sure I don't fry my loom or relays, I take the ground from the lighting relay and connect it to the +12v feed to the starter solenoid/relay, this feed comes from the starter button. Any chance of a diagram just to be sure?

Cheers

Karl