Calculations on electrical wiring. One of the problems with boats as opposed to cars is that you have no common earth in the form of the bodywork. As a result the ‘weight’ of the wire is important because you have twice as much. If you use lighter wire to compensate you get another problem. I’m talking of voltage drop over distance. A formula I recall runs: Watts = Volts x Amps. Bearing in mind that the wattage of a particular load/appliance remains, for the most part, constant, if the voltage drops you draw more current; to the detriment of your power supply. It’s fine if you have loads of battery power being regularly recharged but on a Senior this may not be the case. So you have to try to keep the voltage at the appliance as close to that of the source as possible. Weight of the wiring, its physical size, cost etc. have to be taken into consideration. To give an example: on early Minis the battery was in the boot. The starter motor was obviously at the front and the nonsolenoid starter switch was a floor push button. The wiring between the back and the front was huge to keep the voltage drop at high starter motor currents as low as possible. Plenty of tables are available to show what cable should be used for what application but to give you an example: Take a 10 watt mast steaming light. Even on a Senior it would not be impossible to use 12 metres of cable to get to it. In normal circumstances: Current = watts/volts or 10/12 = 0.833 amps. From tables I am able to say that 1.0mm^{2} cable loses 40 millivolts per metre per amp. 1.0 mm^{2} cable will happily carry a maximum current of 11 amps but what is the voltage drop? The drop for the steaming lamp = (0.833x12x40)/1000 = 0.4 volts. The voltage at the lamp is 11.6 volts. Therefore by recalculating actually the current at the lamp is 0.862 amps and so the voltage drop is .413 volts and so the current used is 0.863 amps. The calculation stabilises but that’s just for the one nav. light. Adding all of the other lights in you can get a substantial voltage drop and therefore a rise in current demand if you use too light a cable over the distances involved on a boat. If you up the cable size to 1.5mm^{2} the voltage drop is nearly halved… This was what amongst other things determined that I rewired using slightly heavier cable. Apart from that I wanted to reroute and tidy up the cable runs. I used plenty of fine cable ties. (Buy a B+Q Value Pack of ties. They are very good value!!) Much of the inside of "Mikros" is varnished. I used an ordinary staple gun to attach the cable ties to the boat before tying them around the cable bundle. If you are worried about the staples rusting, plonk a smear of varnish on it before inserting the cable bundle. To assist: A table giving what cable will take what current at what voltage drop: Cable Size (mm^{2}) Max. Current (Amps.) Voltage drop (dV) (millivolts/amp/metre) 1.0 11 40 1.5 13 25 2.5 18 15 4.0 24 10 6.0 30 7 10.0 40 4 (Source: "The small boat skippers handbook" Geoff Lewis.) A formula to calculate current: Current = Watts / Volts A formula to calculate total voltage drop: Voltage drop = (I x L x dV) / 1000 Where: I is the current (Watts / Volts) L is the length of cable dV is the voltage drop from the above table.
Geoffrey Hyde Fynn. YMS 779 "Mikros" Senior Advisor, see Advisors Page. March 2003, updated Nov 2004

