Cables don't drain batteries - it's the overall load that drains a battery.

Ideally cables would have zero resistance (aka super-conductors) so there is no power loss.

Using the water analogy, whether you water your garden with a 10' or 1/2" diameter hose doesn't make much difference to the flow - it's the nozzle at then end that determines that.

Fattening your pipe (cable) reduces your transmission resistance hence reducing voltage/power losses, hence your load (amplifiers) get more voltage/power.

As to "stable voltage" - this merely reduces voltage drops.

With a fixed resistance (ie, 1, 2 or 10 parallel 0G cables etc), the voltage drops vary with the load current.

So is that stable? The load voltage will still vary, but with lower transmission losses, its variation will less.

The battery will dip.

A large AGM battery might have 5mR internal resistance, hence it drops 200A x 0.005R = 1V.

A larger (Odyssey 75/86-PC1230) is 2.5mR hence a 0.5V drop at 200A.

Then there is the batteries capability...

For now, keep it simple and try to minimise your losses.

Later you can decide if are you concerned with voltage losses (resistance), or voltage dips (from varying current with power output)?

And if concerned about dips, why? (Keeping in mind you are already down at least 1.5V for high power outputs because your alternator isn't big enough.)

As to running a "direct ground to the battery for the sub amp" - only if you have a high resistance chassis or connections. But I would be surprised (read: worried) if a cable had a lower resistance than a car chassis. But all those voltages are easy to measure with a voltmeter anyway. And there is unlikely to be much difference in length between chassis and cable anyhow (who cares? and why?)

Normally I'd say the more parallel cabling the better, but not for a metal chassis (unless cable-connector-chassis joints have a higher resistance than end-to-end cable).

For more info though, I suggest experienced gurus. Whilst I keep hearing things that seem strange to me (like single cables not paralleled, or ground loops, or antennae planes etc), I don't have the practice.

And they can advise of the best equipment and fusing etc. I'd run the fattest cable as far as possible. And if lower gauge fanouts are short and secure, they need not be fused.

The rest of the crap below is optional detail. Feel free to ignore it!

EG: Lets assume you measure a 2V drop with the amps using 200A - ie, the battery as 12V; the amp as 10V; 12-10 = 2V.

Your Amp is consuming 10V x 200A = 2,000W = 2kW (P=VI).

Your cable is dropping 2V @ 200A = 2x200 = 400W (20% of 2kW the same as its 2V drop is 20% of 10V.)

What resistance is the cable (including fuses etc)?

V=IR hence R=V/I = 2V/200A = 0.01 Ohm = 10 milli-Ohm.

That's the equivalent of 100' of 0G cable (from 0G = 0.0001 Ohms per foot).

I'll use "R" to represent "Ohms".

You might then find that the distribution block is dropping 1.5V and the cable 0.5V.

So the d-block is 1.5V/200A = .0075R = 7.5mR.

The cable is 0.5V/200 = .0025R = 2.5mR.

The total "transmission" resistance is cable + d-block = 7.5mR + 2.5mR = 10mR (same 10mR as original (1.5V + 0.5V) 2V at 200A.

Say if you halve the transmission resistance.

Instead of 0.01R we have 0.005R.

The voltage drop is V=IR = 200A x 0.005R(ohms) = 1.0V.

That's half of the original 2.0V drop. (As we'd expect - we halved the resistance & since V=IR and we made 1/2 x R, that means 1/2 x V where V is the voltage (drop) across the Resistance R with current I running through it.

So now you have one extra volt for your amps. But that's looking at it in simple terms. The lowered cable & d-block resistance means more current will flow - eg V=IR hence I=V/R = 12V/(50+5)mR = 218Amps (where 50mR is the effective amplifier(s) resistance - ie, 10V/200A) so hence you now lose 1.1V across the cable & d-block, and the amps get 10.9V or 0.9V (~10%) more than it did. (1.1+10.9 = 12V = battery voltage.)

In practice, the battery voltage will drop a bit due to the increased current, and amplifiers may not behave like a resistor....

But the main point is that by reducing your transmission losses aka resistance, the amps get more power.