...So I'm trying to rig up a pair of driving lights on my bike. just fried a switch (no fucking laughing). In the above diagram, i have everything connected as above except the "switch". So WTF does lead & load mean, and is this a 2 position (pole) switch or a 3 position switch. i'm really showing my 'dumbassness' on this one. I used a '2' wire on/off switch without the "LEAD" shown above. and of course when turned "on" nuthin lit up sept my blood pressure.
The way that diagram is written, it looks like you're taking the lead (+12v switched power) to both the relay coil AND ground. This is a recipe for, as you have discovered, a fried switch. Lose the ground at the switch, the source has a path to ground through the relay coil. Without +12v, your relay coil is grounded on both sides and there is no voltage to energize it.
By supplyling voltage to 85 and 86, you are closing the switch from 30 to 87. + to one side of switch - from switch to 85 Ground to 86 Positive to 30 positive lead from lights (be sure if LEDS) to 87 Negative lead from lights (again, be sure) to ground. No need for ground at switch unless you are lighting. You could also do ground through switch and + straight to the relay which is a little safer.
All this assumes a traditional mechanical relay. If you go solid state, it gets a tiny bit more complicated.
when the switch has a "ground" on it like that... they also have an LED to indicate the switch is on.... thats what the ground is for... it completes the path for the indicator LED, not the actual lights. so no LED in the switch means no need to ground (as you already know). I'm guessing wrong switch & diagram combo.
There is some good basic info on relays at http://www.bcae1.com/relays.htm that would probably be helpful
You are dealing with a 12V automotive single pole single throw (SPST) relay. Here is a web page with an explanation: http://www.eurekaboy.com/f250/relay101.htm To further describe the relay: applying 12V to pin 85 and with 86 grounded causes the main switch, or relay, to close between contacts 30 and 87. Stated another way: applying a (low current) 12V supply to pin 85 with pin 86 grounded will complete the 15 amp fused power to the headlamp between pins 30 and 87. The relay is nothing more than an electrically activated switch - so that a low current switch can activate a high current switch. The problem with the circuit diagram shown in this thread is the 'extra' ground on the switch supplying 12V to pin 85. Pin 85 just needs 12V. The diagram you have is extremely confusing regarding 'load' and 'lead' and a ground symbol - there should be none of these and the switch only directs 12V to pin 85. I recommend not using the diagram you have shown. Edit: the technique shown in the linked (eurekaboy) website puts the switch between the ground and pin 86 with pin 85 always energized. This is a good technique, although having a switch apply 12V to pin 85 also works. Some of the wiring depends on what you are wanting to do: if energizing external lights with 12V from the ignition switch is the plan then having 12V on one side of the switch with a connection to pin 85 on the other works. Good luck!
I second the switch ground not being need Switch should go from bat + or key + to relay socket no ground needed at switch unless it is lighted. normal wiring is next to the last dwg on relay page Stan_R80/7 quoted http://www.eurekaboy.com/f250/relay101.htm