Thanks cbxw34, that's just what I recall off the top of my head.
New things might have been discovered that affects voltage drop, but this is the formula we used, back in the day.
single phase
VD=(2 x D x I x K)/CM
3 phase
VD=(1.73 x D x I x K)/CM
VD/supplied voltage gives %VD.
Where
VD= volts dropped
2= number of current carrying wires in the run, typical for single phase circuits (hot & neutral).
1.73= conversion factor for number of wires in a 3 phase circuit because of shared load.
D= distance of run, measured in feet
I= total load of the circuit in amperes. For a motor it should be the name plate FLA (full load amps) at 100%.
K= resistance constant? = 12.9 for copper or 21.2 for aluminum.
CM= circular mills of the wire, found here.
http://www.engineeringtoolbox.com/awg-wire-gauge-circular-mils-d_819.html
I think the circular mills of a given size wire can be slightly different depending on how many strands it has. The above link might only be for solid wire. Best to look up the exact wire.
There is another factor involved to compensate for self inductance, but only applies to size 2/0 wire and above.
Note: Stranded wire has more current carrying capacity than the same size (awg) solid wire, but not sure what, if any affect it has on voltage drop.