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I will explain some simple tips here. The cable must be of the correct type for the voltage. This is related to the insulation breakdown voltage. Cable is specified by the manufacturer as being suitable for use up to a certain voltage. This specification will be written on the cable drum. As long as this specification is higher than the system voltage everything is fine. That is to say it is perfectly acceptable to use1000 volt cable on a24 volt system. Obviously a cable rated for24 volts would be totally unsuitable for use on a1000 volt system.
Secondly, The physical strength and durability of the cable. This one really is down to nothing more than common sense. For instance, cables inside mobile telephones and calculators are tiny. Suppose there is a piece of equipment at the bow of a boat that draws extremely low current but needs powering from the stern. Say a0.001 amp load. A tiny cable of around0.1mm2 as found in a mobile 'phone will handle the current. But physically this cable is not going to survive very long on a boat due to vibration and chaffing etc. The size of the cable and the physical strength of the insulation must be up to the job.
Further, the cable insulation must be tolerant of any other chemicals it may come into contact with. For instance cables in engine bays should be oil and fuel resistant. Those of us on the UK Inland Waterways are more than aware of the problems with PVC insulation when in contact with expanded polystyrene.
Thirdly, The cable must be able to safely handle the current without overheating the cable and/or it's insulation. This specification can be calculated from the current through the cable and the resistance of the cable (which will show how much heat will be generated). This can then be used with further figures relating to the type and make up of the cable, the ambient air temperature etc to calculate the cable temperature rise. Fortunately this has been made much simpler for us as international standards bodies have drawn up tables which show this in a simple tabulated format. One simply looks up the cable size, the table then shows the maximum safe current for a cable in free air or a cable in a conduit etc. Even more fortunate for us is that the cable suppliers take the worst examples from these tables and specify that as being the current capability for each particular wire.
For instance2.5mm2 cable is usually specified by the standards bodies as being suitable for30 amps in free air or20 amps when in a conduit. The manufacturers therefore specify this size cable as being safe for use up to20 amps.
Any cable you buy should have the current carrying capacity specified on it's packaging. This is sometimes referred to as the cable's "ampacity". Staying within this specification ensures that the cable will not be overheated.
Fourthly, For volt drop purposes the required cable size in mm2=
18/((volt drop [volts]*1000/current [amps])/length [metres])
= 18/((volts*1000/amps)/metres)
The final rule is one that usually means a much larger cable than that specified by rule3 must be used. This is almost always the case for low voltage (i.e.12 or24 volt) systems.
The reason for this is that rule3 takes in to account the possibility of overheating the cable only. Rule4 is regarding the acceptable volt drop down the cable at a certain current. This is usually more of a problem in low voltage systems than it is in higher voltage systems.
An explanation is in order here.
As stated above,2.5mm2 cable is specified as being safe up to20 amps.
Now suppose we have a230 volt load drawing20 amps on the end of20 metres of this2.5mm2 cable. The resistance of this40 metres (20 metres each way) of cable is approximately0.288 ohms. This sounds like nothing. Using Ohm's law (V=I*R : V=volts, I=amps, R=resistance) we can calculate the total volt drop to be20amps*0.288Ohms =5.8 volts. So our230 volt load at the end of the cable will see224.8 volts instead of230 volts. This is well within the specification for a230 volt supply (the acceptable voltage for a230 volt supply ranges from216 volts to253 volts).
So this cable is perfectly acceptable for a20 amp230 volt load.
However, irrespective of the voltage the system is running at, a20 amp load will drop5.8 volts down a total of40 metres of this cable. So if our load at the end is a12 volt load drawing20 amps then by the time the power gets there, it will now be at12 -5.8 =6.2 volts. Obviously this is no use to us whatsoever! The cable, although being run within it's rating is dropping too much voltge. It was OK at230 volts, but no use at all at12 volts.
There are large, complicated tables available that one can carry round showing the resistance of various sized cables, the volt drop per km (or per metre or furlong or whatever) at various current draws etc etc etc. One never seems to have one at hand when it is needed.
Fortunately, since metrication, things have become very simple. This is because the resistance of cable (and therefore the volt drop) is directly, inversely, proportional to the cross sectional area of the cable. And the cross sectional area of cable is now how they are specified and sold.
So. Decide on the acceptable volt drop for the job in hand. For instance a12 volt light really needs to run on a minimum of11.0 volts to operate correctly. So in this case the maximum allowable volt drop is1.0 volt. In a split charge system the cables between various batteries ideally need to drop no more than0.05 volts in order to allow the system to operate at it's best.
So anyway, decide on the acceptable volt drop in volts.
Multiply this by1000.
Divide this by the current in amps.
Now divide the result by the actual total cable run length (both positive and negative) in metres.
Now divide18 by the result of the above. Hey presto, that is the required cable size in mm2. Obviously most of the time this will come out to a silly required wire size so you just choose the next up, standard, available, wire size.
Finally choose whichever wire is the largest from rule3 and rule4. On low voltage systems rule4 will almostalways dictate the wire size. On high voltage systems rule3 will usually dictate the wire size.
I told you it was simple. The trick is that the resistance of copper wire is roughly18/size=Ohms/Km
cable size [mm2]=18/((volt drop*1000/amps)/metres)
This rather messy looking formula was kept in this format because it is easy to work with as the above example shows. However it is much nicer and easier to remember when rearranged thus:-
cable size[mm2]=18*metres*amps/(V*1000)
