I am sure you have answered this question before. I searched for sometime here on CB but couldn't find the full answer.
This is what I found here on CB posted by Derekleffew back in 2008
"At 208V, moving all motors inside a VL3000Spot (never tested a 3500Spot) will only increase the current draw approximately 1A, an almost insignificant amount.
Calculating the total current required when operating at 208V 3Ø ∆ is not as simple as the above posts would make it appear.
Assume a VL3000 draws 11A, across TWO of the three hot legs. For calculation purposes, consider twelve(12) units, distributed equally across all three legs: 4 on ØX, 4 on ØY, 4 on ØZ. The total on any one leg is (4*11=44A), right?
WRONG! The total(s) on each leg must be multiplied by the constant: 1.732 [=SQRT(3)], to determine the actual load, in order to account for the "across both legs" portion, (as It's actually: 4 on X-Y, 4 on Y-Z, 4 on Z-X).
Thus, assuming the legs are perfectly balanced, ~76.2 Amps/Leg, 120/208VAC 3Ø Wye-connected service, is required for 12x VL3000s.
Two quick ways to calculate, but ways that WILL get you into trouble--as MLs often don't come in multiples of three, nor are always circuited perfectly balanced; are:A) Multiply the total number of fixtures by the current draw of each, then divide the result by 1.732, [or SQRT(3) if you enjoy lots of insignificant digits].
OR,
B) Multiply the total number of fixtures by the current draw of each, divide by 3, and multiply that result by 1.732, [or SQRT(3) if you enjoy lots of insignificant digits]."
This all makes sense, but when I try to use the same formula using only 2 x VL3000 fixtures the total Amps per leg does not make any sense. I used Pual Pelletier's LD calculator to calculate the following. One VL3000 is pulling 11A on X and 11A on Y, The other VL3000 is pulling 11A on Y and 11A on Z. Why is the total load 11A on X, 19A on Y and 11A on Z? . I don't understand how it is calculating the 2 x 11A on the Y leg. I would assume you would add the two 11A from each fixture to get 22A total for the Y leg and then as mentioned above multiply this by 1.732. Also why do you not Multiply the other 2 legs by 1.732? Was the LD Calc wrong? Can you please explain how you would calculate total load per legs if you only had 2 x VL3000? Also when you look up the specs on a VL3000 Spot fixture it states the following:
Standard AC power distribution from 200– 264 VAC, 50/60 HZ. The unit requires 7 to 12 A depending on the AC supply voltage.
So I am trying to figure out how you get 11A total for the fixture. I came close to 11A using the following equation. 1200w lamp divided by 208v = 5.7A then multiply that by 1.732 = 9.8 then add 1A for the internal motor power and that gives you 10.8A round to 11A. Is this correct?
Thank you for your expertise.
Jon
This is what I found here on CB posted by Derekleffew back in 2008
"At 208V, moving all motors inside a VL3000Spot (never tested a 3500Spot) will only increase the current draw approximately 1A, an almost insignificant amount.
Calculating the total current required when operating at 208V 3Ø ∆ is not as simple as the above posts would make it appear.
Assume a VL3000 draws 11A, across TWO of the three hot legs. For calculation purposes, consider twelve(12) units, distributed equally across all three legs: 4 on ØX, 4 on ØY, 4 on ØZ. The total on any one leg is (4*11=44A), right?
WRONG! The total(s) on each leg must be multiplied by the constant: 1.732 [=SQRT(3)], to determine the actual load, in order to account for the "across both legs" portion, (as It's actually: 4 on X-Y, 4 on Y-Z, 4 on Z-X).
Thus, assuming the legs are perfectly balanced, ~76.2 Amps/Leg, 120/208VAC 3Ø Wye-connected service, is required for 12x VL3000s.
Two quick ways to calculate, but ways that WILL get you into trouble--as MLs often don't come in multiples of three, nor are always circuited perfectly balanced; are:A) Multiply the total number of fixtures by the current draw of each, then divide the result by 1.732, [or SQRT(3) if you enjoy lots of insignificant digits].
OR,
B) Multiply the total number of fixtures by the current draw of each, divide by 3, and multiply that result by 1.732, [or SQRT(3) if you enjoy lots of insignificant digits]."
This all makes sense, but when I try to use the same formula using only 2 x VL3000 fixtures the total Amps per leg does not make any sense. I used Pual Pelletier's LD calculator to calculate the following. One VL3000 is pulling 11A on X and 11A on Y, The other VL3000 is pulling 11A on Y and 11A on Z. Why is the total load 11A on X, 19A on Y and 11A on Z? . I don't understand how it is calculating the 2 x 11A on the Y leg. I would assume you would add the two 11A from each fixture to get 22A total for the Y leg and then as mentioned above multiply this by 1.732. Also why do you not Multiply the other 2 legs by 1.732? Was the LD Calc wrong? Can you please explain how you would calculate total load per legs if you only had 2 x VL3000? Also when you look up the specs on a VL3000 Spot fixture it states the following:
Standard AC power distribution from 200– 264 VAC, 50/60 HZ. The unit requires 7 to 12 A depending on the AC supply voltage.
So I am trying to figure out how you get 11A total for the fixture. I came close to 11A using the following equation. 1200w lamp divided by 208v = 5.7A then multiply that by 1.732 = 9.8 then add 1A for the internal motor power and that gives you 10.8A round to 11A. Is this correct?
Thank you for your expertise.
Jon