The Calculation Example 1 and Calculation Example 2 on NECHB 2014 page 1078 use a the term circuit length. What exactly is "circuit length" - one way or two way?
I have seen in other circuit analysis questions that length is to be the sum of the line and the neutral conductor lengths.
In Example 1, "circuit length" is the 1-way length of the circuit.
When you calculate percentage voltage drop based on line-to-neutral voltage, your formula is as follows, when you only consider pure resistive effects:
Vdrop = I*r*L/Vpn
Where:
I = per phase current
r = resistance per unit length from the table, divided by 1000 ft to make the units consistent.
L = 1-way length
Vpn = phase-to-neutral voltage
The volts measured phase-to-phase is equal to sqrt(3) * the volts measured phase-to-neutral. Percentage voltage drops are equal, no matter which voltage you base it off of, albeit calculated with a slightly different formula. Therefore, your percentage voltage drop formula is as follows, when you input the phase-to-phase voltage instead. You can derive this by multiplying the previous formula by 1 in a fancy way:
1 = Vpp/Vpp
Vpp = sqrt(3)*Vpn
1 = sqrt(3)*Vpn/Vpp
Vdrop = I*r*L/Vpn
I*r*L/Vpn * 1 = I*r*L/Vpn * (sqrt(3)*Vpn/Vpp)
Cancel Vpn, which gives us the resulting formula:
Vdrop = I*r*L*sqrt(3)/Vpp
The way to think about this is that when the phases are balanced, the remaining two phases carry the return current of the first phase, without any additional current. So each phase wire only "sees" its outgoing current, and the voltage driving this current to flow, is the phase-to-neutral voltage.
This is the principle advantage of three phase power. That the remaining phases carry the return current, without additional current imposed on them. If all phase currents are balanced, the neutral is not needed (in concept). So instead of each phase having a line and a neutral wire, you simply have one line wire for each phase, and the remaining phases' line wires serve the purpose of its neutral. In practice, we often still have the neutral for numerous reasons. So what would take the equivalent total KCMIL to do in single phase, you can do with only 2/3rds of the total KCMIL in 3-phase for the same load and the same voltage drop.