Let's stop the techno-babble gobbledygook and get back to answering the OPs real question.
For a set of circumstances, size/insulation of refer box, compressor characteristics, how much beer
is inside, desired temperature, environment
, yada, yada, yada, it's going to take X amount of power to keep the beer/eggs cold. It doesn't matter if the compressor slows down/runs longer or draws more current to maintain speed as the batteries are depleted, that's included in the compressor characteristics.
A portion of X is consumed to heat the wire from the power source to the unit in question. Since heating
of wire isn't normally considered a desired feature, it behoves one to eliminate as much of the wire heating
Let's pick a number of 50ah/day using 18awg wire to keep the beer
to the desired temperature. Let's use the voltage drop calculator and ampacity tables at:
Genuinedealz - Technical - Calculators
It really doesn't matter if the calculator/tables are totally accurate, so don't reply stating you disagree with them, they do demonstrate the main point, the effects of voltage drop in wire.
Back to the 50ah number. Per the tables, ABYC allows 18awg wire to handle 20amps when placed outside the engine
space. We also see that 18awg copper wire on a 20 foot one-way run @ 12v & 5amps has 10.97% voltage drop. In our theoretical scenario, that means that 5.5ah/day is consumed heating wire, leaving 44.5ah for keeping the beer cold. If you substitute 8awg wire for the smaller 18awg we see only 1.08% voltage drop, equating to .54ah/day heating wire, added to the 44.5ah for cooling
the box and you are now consuming ~45ah/day for your refrigerator
. Yes, heating wire costs you $$.
So, pick your poison, leave the small wire (and drink the beer) or spend the $$ and do the work to install bigger wire (and save the beer for tomorrow).