I'm not so familiar with the 3 phase voltage drop formula, but I would think the calculation should be pretty close to the correct answer.
However it seems to me that formula is for a single 3 phase load, rather than a series of single phase loads. So if you want to use it, I would suggest approximating the first 3 fixtures as a 3 phase load at 1200' out, and the second 3 fixtures as a 3 phase load at 3000' out. Maybe even wiring those first six as ABC CBA so that the total distances balance.
The seventh fixture is the odd one out (mod 3) and so I would think using the 2 wire circuit voltage drop formula would be appropriate for it.
So I guess I'm proposing:
Cmil = K / Vd * ( √3 * 1200 * 1.2 +√3 * 3000 * 1.2 + 2 * 4200 * 1.2)
Cheers, Wayne
ok. I think I am starting to get this. So would that final light still be added to Phase A on the Panel Schedule?
For the actual design, the distances and amperages are not nearly as uniform and there are separate branches off of the main run of conduit. I wouldn't be able to group them together so easily in terms of distance and amperage. Would I add each light at their actual distances and divide the distance and amperage each by 3?
What would I do for a branch where there are 2 extra lights (like if I had another B at a distance of 600' past the farthest A on this run)? Would I run 3 wires to A and then 2 wires to B and the calculation be: Cmil = K / Vd * ( √3 * 1200 * 1.2 +√3 * 3000 * 1.2 + 2 * 4200 * 0.6 + 2 * 4800 * 1.2)?
Thanks