Not that I want to scare off or substitute for others...
(Wot? Me! That's as ridiculous as me rambling. And rambling!)
And (PS), I missed dragon's response. (And only skimmed the links - & I wondered if my aforementioned "copper (molecular) domains" was mentioned in those 1980's articles - ha ha.)
But IMO wiring is simple.
Find its resistance - eg,
powerstream.com-Wire_Size, though others hereon have links to others (better or more applicable?).
EG - Ohms per length times length (= Ohms resistance).
The voltage drop across it is simple Ohm's Law: V=IR (Voltage across something equals the current through it (I, in Amps) times the resistance of it.
FYI - it's Volts, Amps and Ohms. But that's equivalent to working in milli-Volts (mV), Amps, and milli-Ohms.
Decide if that voltage drop is acceptable - eg, 100mV (0.1V), 1V, 3V...?
The loss is also simple - P=VI, Power (Watts) = Voltages times Current (across & thru it). The Watts is heat for wires. (And heat & light for lights, etc.)
This next is a bit more complicating FYI stuff (though its use can make things much easier!)
The above V=IR and P=VI can be combined to yield P=VV/R or P=IIR. That's convenient if you know 2 of the 3 (V, I or R).
[ Ignore if too complex: Note that Power increases as a
square of voltage or current (VxV or IxI, aka VV or II) for a given resistance. Hence why (typically) 14.4V delivers 44% more power than 12V - ie, not simply 14.4/12 = 1.2 = 20% extra, but 1.2 x 1.2 or 14.4x14.4/(12x12) = 1.44 = 44% more! Sort of obvious if understood that V=IR means that V is proportional to I across a given resistance R - ie V=IR => V/I = R. But that assumes a constant resistance load. ]
This next may be complex, but illustrates how easy - or applicable - the above can be.
Will a 1.5V battery across a speaker blow it?
Let's see - worst case is highest current which means lowest impedance (~resistance) speaker - eg, 1 Ohm.
P = VxV/R = 1.5V x 1.5V /1 Ohm = 1.5 x 1.5 = 2.25W.
Round 2.25W up to 3W (to be conservative, aka add a safety margin - just in case).
Hence a 3W 1-Ohm speaker should handle a 1.5V battery.
So if your 1Ohm speaker is rated above 3W, it should be okay to use a 1.5V battery to test its connections (for continuity or (cone) polarity).
So please check that your speakers are above 3W. I'd hate you to blow your expensive subs etc. (That's a joke - ie,the "greater than 3W" - NOT the blowing bit!)
What voltage drop for 500W using 4G?
Let's assume 600W power input. (amp inefficiency)
4G is 0.2485 Ohms per 1000'; let's say 0.25. (From my link above.)
Hence from V=IR, V = I x .25R per 1000' (many use "R" instead of the Ohms Ω symbol) or mV = I x .25mR per foot - eg, 25mV per 100A per foot ( = 100A x .25mR = 100 x .25 = 25mV).
Though that may be too complex... (We need strike-thru formatting!)
Instead, 600W.
From P=VI, hence I=P/V = 600W/12V = 50A.
Assume 10' of 4G, from V=IR => V= 100A x 10' x 0.25R/1000' = 100 x 10 x .25/1000 = 1000/1000 x .25 = .25 => .25V
IE - 1/4 of a volt for 50A thru 10' of 4G.
Assume the same for the ground path => 2 x .25V = a 0.5V drop (between the alternator/battery and the amp).
If 0.5V is ok, then fine.
If close to ok, then consider that the might be less than he 4G resistance - ie, ground is a short length of amp-to-body 4G, 4G from battery to body (assuming you have ADDED that to existing battery-body grounds), and the body/chassis resistance
should be less resistance than the equivalent length of 4G.
Then again, the ground includes terminals/bolts to the body which have their (tiny?) resistance. But the 4G also has its "terminal" resistance.
That's where things get poopy - but we are ONLY considering the 4G resistance. It is up to the designer (you!) to factor in the (oft forgotten!) return/ground path, terminal resistances etc.
And experience soon learns connection resistances (ie, ignore. or are they significant?) and the importance of GOOD clean contacts!
But now I'm breaching into other contributions and the Rules Of Thmb or the knack of design that the experienced have.
Though in principle it is easy - THINK and break each "segment" down and find its contribution. (The same as with fault-finding.)
And I said "in principle" its easy - it may take a lot of time, but you soon(??!) learn what is important etc.
So, since I'm not one to ramble, I'll leave it here.
Besides, others and other links (on the12volt or other sources like
bcae1.com) do a far better job at taking people through the basics than I do. I prefer to "apply" those basics into reality - I often see (IMO)
the obvious that others seem to miss. (LOL - my today reply to a 2005 OP on mp3car.com (
here) being IMO an amusing case in point. Oh dear - the old complex yet flawed suggestions!)
And I have some common ROTs (Rules Of Thumb) and tricks - eg, divide amplifier output by 10 for input current. Much easier than /12V or /14.4V and it compensates for amp inefficiency. Though used for initial designs, its amazing how "real" it becomes (except in close-call and some cases) in part because you can't get the exact value you want (eg, wire resistance) hence you round up to what is available...
But I'm not one to ramble.
Hopefully you have censored the too complex bits. (My replies often take several re-reads! Same for their technical content. (LOL))