# Power for rig3



## derekleffew (Mar 18, 2011)

How many amps per leg, 120/208VAC 3ØY, do I need for this?

> 8 High End Cyberlight Turbo Luminaire
> 16 High End Studio Beam Automated Luminaire
> 6 High End x.SPOT Automated Luminaire
> 2 Martin Atomic 3000DMX Strobe Fixture
> ...


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## xander (Mar 23, 2011)

No one is even going to try? Well, derekleffew, looks like you picked a good one!


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## avare (Mar 23, 2011)

A problem is that to answer the question effectively for the worst case scenario requires research on power consumption. One honest response would be: 

Insufficient data. Which units will be used simultaneously? Enough to cover those units. Is this for a generator supplied source? How long will the units be on simultaneously?

Andre


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## erosing (Mar 23, 2011)

Hey derekleffew, no fair changing things  (though I was wondering why the color blasts and DFD supply were under 208V), now I have to redo my math for the Parlites...


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## derekleffew (Mar 23, 2011)

avare said:


> ...Insufficient data. Which units will be used simultaneously? Enough to cover those units. Is this for a generator supplied source? How long will the units be on simultaneously?


With questions such as this, the only sufficient course must be to assume that all units will be run at Full, for over three hours.


Arez said:


> Hey derekleffew, no fair changing things ...


Sorry, but not my fault. I was trying to avoid renting a 120V distro as well as an ML PD. Then the shop tells me the ParLites CAN run at 208V, but they don't have any L620-Ed adapters. Now they're saying they don't know if the DFD/CB supply is auto-ranging or not, so I'd best keep it 120V (I think it can run at 208V, but one must open it and change a jumper inside). [EDIT: Yes, from the DFD LED300 data sheet: "It is internally switchable to nominal 115 volt or 230 volt operation."]

Oh, and the designer just told me he MUST have control of the two DF-50 s and fans from the console. Luckily the shop has an R20 module, and dimmers 23&24 are open.


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## Pie4Weebl (Mar 24, 2011)

derekleffew said:


> Oh, and the designer just told me he MUST have control of the two DF-50 s and fans from the console. Luckily the shop has an R20 module, and dimmers 23&24 are open.


Would he find it acceptable to have the fan always on and just use a DMX version of the DF-50?


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## derekleffew (Apr 4, 2011)

It's been over two weeks and no one has even attempted an answer. 

There will be no new QotDs (and we've got some good ones in the queue) until this one is answered satisfactorily.


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## erosing (Apr 6, 2011)

Attention: LightTech, Bull, Chausmen, Denny586, Nuggety, CSCTech, Dvyjones, LXPlot, Jmorrison120, Kylevdk, RFazz15, tayklor

You have all been specifically called upon now because you decided to join this forum from last week to last year, you are the latest crop of high school and college students that showed interest in lighting during your new member board posts. Guess what guys, it's game time.

ReelFF-DF50


Parlite

DFD LED300

Cyberlight

Atomic 3000

Source 4

Studio Beam

X.Spot


Molefay - ControlBooth

ACL - ControlBooth


The formula's needed are all floating around this forum, books you might have, and in the following links, but be warned, the game will be spoiled for you if you use these links, I suggest you try on your own first. 

http://www.controlbooth.com/forums/question-day/10774-power-rig.html
http://www.controlbooth.com/forums/question-day/14560-power-rig2.html


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## 65535 (Apr 6, 2011)

I worked out about 150A per leg. Probably spec no less than a 200A 120/208VAC 3ØY circuit. 

I may be doing something wrong with my loading, adding between the phase and to neutral loading. Don't do a whole lot of 3Ø calculations.


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## erosing (Apr 7, 2011)

65535 said:


> I worked out about 150A per leg. Probably spec no less than a 200A 120/208VAC 3ØY circuit.
> 
> I may be doing something wrong with my loading, adding between the phase and to neutral loading. Don't do a whole lot of 3Ø calculations.



Nope.

65535, and to everyone else who will be attempting to answer, if you show your work, we can help you more.


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## Techfiend (Apr 7, 2011)

Cyberlight 8(6.3A)=50.4A
Studio Beam 16(4.35A)=69.6A
X.Spot 6(5.8A)=34.8A
Atomics 2(20A)=40 (Martin says 16A to 20A circuit for each M15 Lamp on high power, but the spec sheet says 8A?)

575W Lamps (6S4+12Par)(4.8A)=86.4
1kW Cyc 5(8.3A)=41.5A
ACLs (In 120V series) 3(10A)=30A 
Moles 3(10.8A)=32.4A
LED Power Supply 2(5.5A)=11A
LED 8(.5A)=4A
DF-50 2(4.6A)=9.2

Total of 409.3A over 3 legs=136.4A?

Input is welcome, willing to take a second stab at it if not correct. Surprisingly these are the problems I like to solve.


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## erosing (Apr 8, 2011)

Techfiend said:


> Cyberlight 8(6.3A)=50.4A
> Studio Beam 16(4.35A)=69.6A
> X.Spot 6(5.8A)=34.8A
> Atomics 2(20A)=40 (Martin says 16A to 20A circuit for each M15 Lamp on high power, but the spec sheet says 8A?)
> ...


 
Nope, would you like to try again?

Perhaps someone who hasn't responded yet would like to help Techfiend?


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## seansbar (Apr 8, 2011)

The 208v lights are fed with 2 legs, thus they must be doubled for this calculation

I figure: total of 551.2 Amps, 183.7 Amps per leg.

I'm going with 8A for the strobes

Cyberlight 8(6.3A)=100.8A
Studio Beam 16(4.35A)=139.2A
X.Spot 6(5.8A)=69.6A
Atomics 2(8A)=32 (Assuming 8A)

575W Lamps (6S4+12Par)(4.8A)=86.3
1kW Cyc 5(8.3A)=41.7A
ACLs (In 120V series) 3(8.3A)=25A 
Moles 3(10.8A)=32.5A
LED Power Supply 2(5.5A)=11A
LED 8(.5A)=4A
DF-50 2(4.6A)=9.2


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## derekleffew (Apr 8, 2011)

Newguy seansbar is close enough to be declared the winner, but not entrely accurate. I got ~182 per leg.

To calculate draw for 208V units, what I do is take the listed draw, multiply by √3 (1.732), then divide by 3 for the "per leg."


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## mstaylor (Apr 8, 2011)

Many forget the 1.732 in a 3phase calculation.


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## jonslilbro (Apr 8, 2011)

derekleffew said:


> Newguy seansbar is close enough to be declared the winner, but not entrely accurate. I got ~182 per leg.
> 
> To calculate draw for 208V units, what I do is take the listed draw, multiply by √3 (1.732), then divide by 3 for the "per leg."View attachment 4800


 
I've been shown that way, but I've also been shown that you can convert everything to watts and then divide by your line to neutral volts and divide by the 3 legs. I.e. Add all of your line to line current and multiply by 208v, then add all of your line to neutral current and multiply by 120v, then add these two numbers, divide by 120v and then divide by 3 legs. So with adding the amperage's you posted you get 193.4a for line to line and 211.8a for line to neutral. So ( (193.4a*208v) + (211.8a*120v) )= 65643.2 Watts. Then divide by 65643.2W/120v= 547.03a to get your total amps then divide by 3 legs to get 547.03a/3legs= 182.343a per leg. I notice this is slightly different from your answer, but is this a valid way to answer the question, or should I stray away from this method and stick to way you showed.


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## derekleffew (Apr 9, 2011)

jonslilbro, between this and the IES Dimmer Curve thread, my brain is hurting from all the math!

jonslilbro said:


> I've been shown that way, but I've also been shown that you can convert everything to watts and then divide by your line to neutral volts and divide by the 3 legs. I.e. Add all of your line to line current and multiply by 208v, then add all of your line to neutral current and multiply by 120v, then add these two numbers, divide by 120v and then divide by 3 legs. ...



Attached is a document I use every time I have to think too hard about this topic. It's 90 Mac2000s at 208V.

Using your "method": 
90 x 10.25 x 208 = 191880 "watts"
191880 /120 /3 = 553 amps per leg.

My (and Paul Pelletier's LD Calculator Lite's) "method":
((90 x 10.25) x 1.732) /3 = 532.59 amps per leg.

So there's a 21A difference in our methods. When dealing with 922A, a 2% difference. 

jonslilbro said:


> ...but is this a valid way to answer the question, or should I stray away from this method and stick to way you showed.


By you first using 208 and then later dividing by 120 (208/120=1.73...), we're both doing the same thing. 
Instead of using 208, if you use 207.84 (120 x 1.732), you'll come up with 532.59, same as me.

Yours is probably close enough for government work, as they say.

(Anyone want to input all of the original problem into LD Calculator and let us know the "real" answer? Of course that means you'd actually have to assign things to specific phases.)
View attachment M2K_Pelletier_load_calcs.pdf


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## erosing (Apr 9, 2011)

derekleffew said:


> (Anyone want to input all of the original problem into LD Calculator and let us know the "real" answer? Of course that means you'd actually have to assign things to specific phases.)



Already did out of curiousity, biggest difference is he uses a draw of 22A for each Atomic3000, X 186.3A Y 188.5A Z 188.7A.


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## jonslilbro (Apr 9, 2011)

derekleffew said:


> Using your "method":
> 90 x 10.25 x 208 = 191880 "watts"
> 191880 /120 /3 = 553 amps per leg.
> 
> ...


 
Just checking, but every time I do the math using "my method" I get the same watts as you, but you lose me here:
"191880 /120 /3 = 553a"

When I do that math I get 533 amps per leg everytime, leaving just a difference of .2amps per leg, or .6 amps on the total service from your 1599.7 amps, making that a .05% difference.

But I now see how essentially our "methods" are the exact same and that's why we're so close, thanks for the insight.

Sent from my DROIDX using Tapatalk


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## derekleffew (Apr 9, 2011)

You are, of course, correct. I misread the calculator. I should only use Excel.


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## xander (Apr 9, 2011)

Yes, both ways are valid, for the reason [user]derekleffew[/user] points out: 120√3 = ~208. You are each accounting for the 120deg phase difference at a different point in your calculations. Personally, I like [user]jonslilbro[/user]'s method because then you get your kVA right there if you need a genny.


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