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# Automated Fixtures Amperes/Current Draw for a moving light.

Discussion in 'Lighting and Electrics' started by McCready00, Sep 10, 2008.

1. ### McCready00Active Member

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Hi everyone,

When you build a cable plan with moving lights in, how do you calculate the current needed for a moving lights. I'm actually working with some vl3500 spot on 208v. The manual says :

"Standard AC power distribution from 200– 264 VAC, 50/60 HZ.
The unit requires 7 to 12 A depending on the AC supply voltage."

We have HMI 1200w lamps into the units. I know the power needed changes constantly as the devices are constantly used.

Thank you

Last edited: Sep 10, 2008
2. ### FooterSenior TeamSenior TeamPremium Member

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Higher voltage, equals lower total amperage. If you are running at 208, I would calculate them each taking a max of 12 amps, because you are at the low end of the spectrum. If you could get 240 to them you could run 2. That being said, to be safe, strike one up and throw an meter on it and see what it draws depending on your specific power output, you might be able to get away with 2 per.

3. ### McCready00Active Member

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That part was good with me.

But is that good to say that my unit will be using more power at 208v than at 264?

7*264=1848w
12*200= 2400w

last question !

what is the peak of amp my unit can use if I am using all (the more are I can use at the same time) internal motors at the same time.

4. ### McCready00Active Member

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oh and by the way, I'll receive my new meter tomorrow. I'll check this out.

thanks

5. ### waynehoskinsActive Member

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Current. You mean "current needed for moving light". Here I came wandering into this thread trying to figure out how you were trying to connect power amplifiers to moving lights.

As to why the power consumption is as it is, I would suspect that it will operate on a slightly lower voltage (potential) than 200, but they spec the low end conservatively. Similarly, if the current numbers are worst-case numbers, that would skew the power consumption curve.

it were me, I'd plan on it realistically drawing 10A.

6. ### SonyActive Member

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A lot of equipment runs more efficiently at higher voltages, Especially Inductive Loads like things with motors and ballasts. While it's true for Purely Resistive loads that P = IV (P=Power, I=Current, V=Voltage), Partially Resistive and Inductive Loads follow a different law which is I = V/R (I=Current, V=Voltage, R=Resistance.)

7. ### McCready00Active Member

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perfect.. Thank you for the informations !

8. ### philhaneyCBModCB Mods

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P = IV and I = V/R are both parts of Ohm's law for purely resistive circuits.

When inductors and capacitors are intorduced, phase shifts between voltage and current flow are introduced and a new set of rules apply.

(Google ELI the ICE man )

9. ### derekleffewResident CurmudgeonSenior TeamPremium Member

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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].

Electricity Kills!
If in any doubt, consult a qualified electrician.

10. ### lenWell-Known Member

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I use this freeware Paul Pelletier - LD Calculator for that kind of stuff. Newer fixtures might not be in there, but a VL most likely will. there's a bunch of tools, including a dip switch calculator, a truck pack calculator, rigging weights, you name it.

11. ### derekleffewResident CurmudgeonSenior TeamPremium Member

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Thank you [user]len[/user]. The LD Calculator is indeed a fantastic resource. However, A) Paul has never ported it to Mac OS (but says he will, one day) and, B) it's important to be able to calculate this sort of thing by oneself.

As a matter of fact, I referred to the attached document in writing the above post.

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12. ### Jon8-)Member

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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 2 x 11A to get 22A total for the Y leg and then as you mention 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? My goal is to be able to calculate this or any number of fixtures without using the LD Calc. 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 got to 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 I am assuming you rounded it to 11A. Is this correct?

Thank you for your expertise.

Jon

13. ### Chris15CBModCB ModsPremium Member

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So let's look first at how the current draw gets calculated...
Using the example Derek attched earlier;
The MAC2000 has a measured wattage of 1450W and a power factor of 0.68 at 208V.
So first calculate the total power (VA) as real power (W)/ power factor. This gives 2132 VA. This includes both the real power, the watts, which are actually used up, and the apparent power, which basically just circulates in the distribution system and causes heat losses and means cables need to be thicker. This is part of why low power factors are bad.
So take 2132 and divide by 208V and you get 10.25A which has been rounded to 10.3.

Now onto why the 19A not the 11A...
In simple terms it's because the 2 lights aren't in phase, ie. they aren't drawing their power at the same time.

Have a look at the picture; the red and the blue lines would be indicative of each of the lights. The purple is the sum of the two.
If you calculate the RMS of the purple wave, you'd end up at 19A...

14. ### Jon8-)Member

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Chris,

Thank you for taking the time to respond to my post with the correct answer of spelling out the formula or equation.
This makes a lot more sense then just replying with what the dictionary says power is.

I do however have a question of how you would calculate the RMS. What is the formula for this. I want to be able to calculate any given number of fixtures not just two. There has to be a mathematical equation that I can use every time to figure out what my total amps per leg are. Can you please try to explain this in the simplest way possible?

Thanks again for taking the time to spell everything out for me.

Jon

15. ### Chris15CBModCB ModsPremium Member

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So RMS is an acronyn for Root Mean Square.
It's the square root of the mean (average) of the square of the waveform.
Mathematically it's sqrt( 1/n * integral from 0 to n[ X(t)]) where X could be voltage, current, or any other function.

For a sine wave, it's sqrt(2) * the peak voltage of the sinusoid.
Where it gets complicated is when you start dealing with either dimmed waveforms or loads with non trivial harmonic components.
Anything with an electronic power supply is a likely contender for having harmonics, moving lights included.

The best result for calculating the RMS draw is to physically test it. You need an RMS reading meter. You can make your life easy with something like this: Wattage Meter | Electric Power Meters | Fluke 345 Power Quality Clamp Meter but it will need someone with deepish pockets...
That will give you RMS voltage, current, real power, total power, power factor, harmonics and all sorts of things. But it's expensive...

There are some times when just using the tools other people have already put the hard work into creating is the more sensible move

16. ### Jon8-)Member

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Chris,

Thanks again for taking the time to respond. I do own a Fluke 334. It works great but it doesn't help me when I am trying to balance my load when I have over 100 fixtures. This needs to be done in advance. I attached a PDF file as an example. in this example there are 5 x vl1000 fixtures. The total amperage of each leg is 11.1, 11.1, 14.6. If I want to calculate this without using the LD Calculator App, how would I do it? I understand that there is going to be cancelation between legs but I guess I don't know which ones are being canceled. Its easy to figure out if there are an even amount of fixtures. Lets say there were 6 of the vl1000 fixates. I would take 4.21A multiply by 2 (the number of fixtures per leg) then multiply this by 1.732 which = 14.58 or 14.6. I just don't understand how to calculate it when there are an uneven amount of fixtures or different fixture types.

Thanks again for your time and help on this.

Jon

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17. ### Chris15CBModCB ModsPremium Member

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So the trick is vectors...
What's your exposure to them so far?
Start to get your head around them, and adding / subtracting them, such as: Euclidean vector - Wikipedia, the free encyclopedia

Then we'll talk about how they fit into this picture...

18. ### derekleffewResident CurmudgeonSenior TeamPremium Member

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@Jon, (annoying that our smilie code obfuscates your screen name; we'll look into that.) I think you're over-complicating matters. If you're powering 100 moving lights, you'll automatically end up with around 33 on each phase, provided you don't make circuit#6 the spare on every break-out. And even that is not the worst thing in the world.

While calculating the exact current draw of each phase is beneficial, it's not even truly necessary. Even the laudable goal of balanced legs, unless dealing with a generator. And in that case, the generator should be over-sized enough to not be bothered by an inadvertent imbalance of a few amperes.
In all instances in the above quotes, substitute "moving light" for "dimmer" in the above.

I used to be overly anal about calculating my loads to the exact amp, but eventually gave up when I realized I was never going to get it right. Too many variables working against me: line voltage that day, voltage drop due to length of feeder, new lamps/old lamps, phase of the moon, solar flares, and so on.

Calculate just to make sure no one leg is over the capacity, then measure once everything is up and running. The more you do it, the more you learn.

19. ### Jon8-)Member

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@Chris15 - I know very little about Vectors. Thank you again for taking the time to reply to my post. With that said I agree with Derek.

@derekleffew - Thats exactly what I was trying to figure out, the exact load per phase to make sure that it was A.) Balanced and B.) more importantly not over drawing the 80% of each phase.

I realized that instead of trying to calculate the rig per Breakout I need to look at it as a whole. I was getting stumped when trying to calculate 5 lights on a 6 circuit breakout. Using a power calculated app I couldn't figure out how it was getting the final amps per leg. What I should be looking at is how many fixtures of the same type are on each phase for the entire lighting rig. As long as I evenly distribute all of the fixture types across the three different phases it will be easy to calculate the total amps per phase.

Just to make sure I understand, if I wanted to quickly calculate a rig that has 24 x VL3000 Spot and 35 x Mac 2k Spot I would do the following to calculate the total amps per leg. Please correct me if I am wrong in my calculations.

1. Take 24 & 35 and Divide them both by 3 which will give me 8 and 11.6. Would it be better to just round this number up? Doing so would give me an even load across all 3 phases.
2. Take 8 x VL3000 and 12 x Mac 2K per phase and multiply those two numbers by the Amps per fixture. VL3000 = 11A , Mac 2k = 10.3A. This gives me VL3000 at 88A per phase and Mac 2k 123.6 per phase.
3. Add the two fixture totals together to get 211.6A per phase.
4. Multiply the result by 1.732 (Square Root of 3) to give me the total amps per phase.
5. So the total amps would be 367A 120/208v 3 Phase. I would need to order a 1 x 400A & 1 x 60A 120/208v 3 Phase system keeping in mind that each service should only be used to its 80% capacity. The result would give me 320A on the 400A service and 46 amps on the 60A service.

If one did not multiply the total amps per leg by 1.732 then they would think they could fit the entire rig on single 400A system, or if they didn't take into account for the 20% buffer they would think it would fit on a single 400A service.

Once again please correct me if I am wrong in any of the above calculations. Is there any circumstances that I would not want to calculate a rig in the fashion I used above?

Thanks for your time and expertise,

Jon