No one actually knows how to lculate what speaker cable they need

Ohms law is saying that too, as any relation between U (voltage), I (current) and R (resistance). U= I x R
This can be also R = U / I or I = U / R. The power P = I x U in AC also x2.
What about the resistance (R) of a cable, if you need to keep it the same, but also to extend that cable from 8' to 20'?
To keep the voltage drop on a resistor (the speaker cable), if you make it longer, you need to increase its cross section to keep it the same R.
this is exactly what 3 AWG at 8’ would become 3x0 AWG at 20'.
Go to the AWG table and do your calc.

Your "Smart" quote from Google is showing how little you understand this subject. Way less than you need for an argue with it.

Mr. kosst_amojan
Speaker cables are not supose to have any inductive or capacitance values. It is a cooper wire. You wrote: " 16g conductors in a round braid" By putting them into paired parallel lines, twisted 6 fabric insulated 16g conductors, get them some small values of impedance. My cables are two separate cables, so no inductance or capacitance are involved.
You are telling a tell of a wire you made, but how you ended up with that particular value of cable resistance, to fit your system!
Was it a divin revelation in your dream, that instructed you to build those cables, as the rest of the arc?
You call me a liar, at a time you can not tell the diference between an interconnect and a speaker cable.
A power amp. input resistance is usually 10kOhms, and it is pasive resustance. The speaker is 4-8ohms, a complex coil loaded impedance.
Interconnects have a shield to ground and a capacitance developed between the two. Good speaker cables, unless you twist them, have none. An interconnect need to pass milliamps, a speaker cables pass 1000 to 10,000 more current.
Giving a jumpstart with good 4-0 AWG speaker cables would do. Doing that with an interconnect cable...?
Well they are not the same, and bever were.
I’m not a liar, but you do not understand a thing in electronics or audio. 
Calling me that again, will end up with a comlpain and no more answers!

I experience the same results with the silver wire ladder design driving a second system consisting of a Hegel H200 integrate amp driving Aria A speakers (Joe D’Appolito three-way design employing Cabasse cone woofer and Accutron ceramic drivers).
I buy 16AWG naked .999 pure solid silver wire from a jewelry industry supplier in New Mexico for $2.87 per foot and poly sleeve on Amazon for $.14 per foot.  It only takes 30 minutes to assemble these cables (cut silver and sleeve to length, insert wire in sleeve, attach spacers to cables and attach wire to terminals).
Measuring the differences between conventional materials and designs when compared to the ladder described, with instruments other than my ears, is beyond my ken.
Invest $70 and thirty minutes and let your ears be your guide.
For over thirty years, I to, was a cable denier.  I have $9,500 of various unsatisfying cables sitting in drawer or were resold.  I wonder if the measurement sciences have caught up to the reality of how electrons behave when transmitted via different materials, and in proximity to the signal and ground wires.
Happy listening.

Mr. keppertup
Siver has a 9% better conductivity than cooper. It should cost 94 times more. A 16 AWG silver wire would have a very similar resistance to a 16 AWG cooper wire. For the price of the silver ( $2.87 per foot ) you should buy it by the mile and sel by the ounce. 
Your Hegel H200 integrate amp has a Df of 1,000. Very good.
http://www.hegel.com/images/discontinued/H200manualenglish.pdf
That would call (by calculation) to a 2x 0 AWG cable @ 8' long.
How long is your cable?
You might have $9,500.- value of speaker cables that were all purchased without any serious guidance, to become obsolete inventory.
What you need is a set of 2 x 0 AWG, if 8' long, that should cost you less than $2,500.-