Ok so by the sounds of it I don't have much to worry about, this is more for interest/educational sake now.
The wire resistance can be calculated from this information
22 gauge wire can carry 5 amps and it has 16 ohms per 1000 feet
Calculate wire length for 1.2 ohms
1000 * 1.2/16 = 75 feet of wire
If your relay needs 100mA to switch then the voltage lost in 75 feet is
v = ir = 0.1Amp * 1.2 ohms - 0.12 volts lost from 5 volts
5.0 - 0.12 = 4.88 volts delivered to the relay 75 feet away.
I know theres an equation to calculate resistance based on wire length and diameter. But I was wondering more how to overcome this extra resistance, thanks for the information though I'll no doubt use it.
V=IR so an increase in resistance means increase in voltage, but the arduino pins only supply 5v? Still trying to wrap my head around it.
driving DC relays using a transistor connected to an Arduino pin.
http://playground.arduino.cc/uploads/Main/relays.pdf, the transistor being used to increase the voltage to more then 5V
Thanks this is what I was looking for I think.. But this is using the transistor as a switch right? So where is the relay +V coming from? Wikipedia says this to use a transistor as an amplifier http://upload.wikimedia.org/wikipedia/commons/thumb/8/8c/NPN_common_emitter_AC.svg/200px-NPN_common_emitter_AC.svg.png
What is the downside to using thicker wire though, what happens if you use too large a guage? I cant be bothered remembering physics right now but wouldent it affect the charge carrier density, or just draw more current?