Voltage drop on 120v landscape lights over 1000' run.

[wwhitney]

Now that makes more sense.

I said the same thing both times (after fixing the cubic -> square error), so if it makes more sense the second time, I must not have explained it well the first time. Or perhaps you misread it the first time.

Another way to do it that works for equally and unequally spaced is to multiply the load at each point in the run by the distance from the source. Sum the results and divide by the total load. Use the resulting distance and total load to calculate voltage drop

Yes, that will work. You can also extend the method to adjust for different conductor sizes in each segment, by picking one size as the base size and multiplying the distances on the segments of other sizes by the ratio of the unit resistance on that segment to the base size unit resistance.

Cheers, Wayne

 

[Smart $]

I said the same thing both times (after fixing the cubic -> square error), so if it makes more sense the second time, I must not have explained it well the first time. Or perhaps you misread it the first time.

I believe I misunderstood what you were comparing in your first explanation... but I'm not going to reread it just to try and confirm one way or the other.

 

[FionaZuppa]

just use worst case scenario, if you put 28w x 17 at the end of 1k ft, that's 2k ft of wire amps path (approx 3.2 ohms), you'll get about 12v drop using 12ga wire.

108v on the fixtures will likely not be an issue.

if using 14ga wire you would be at ~100v at the end, which should still be fine.

14ga wire (5ohm @ 2k ft) seems ok to use for your 17 fixtures spread across a 1k ft distance.

my guess though, the fixtures are not using 28w each.

 

[GoldDigger]

A small but noticeable factor, if the LED drivers try for constant power, is that like motors they will draw more current as the voltage decreaes.
That will add about 1% to a calculated VD of 10%.

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