using aircraft landing lights

mr_sound

Member
i've found a great deal on aircraft landing lights, too good to pass up, but the whole power issues behind these confuses me. the lamps are 28 volts 450 watts. from what i understand, if you wire up 4 in a series it makes a standard 120 volt circuit. but do you add up the wattage with it, or does that stay the same? would 4 of these lights be 450 watts, or 1800 watts? power is the biggest issue where i'm using these at....of course adding another dimmer or two is another problem.

Mayhem

Senior Team Emeritus
Wiring them up in series will give you a total of:

112 volts and 1800 watts load

When ever you add lights to a circuit the wattage is always added.

By the way - welcome to the site.

mr_sound

Member
thanks for the welcome.

well i figured you added wattage as well as voltage, i was just hoping i was wrong. looks like i won't be using these lights anytime soon...i cant afford another dimmer (or 2)....and i'm about out of power as it is anyways.

Mayhem

Senior Team Emeritus
How good is the price? Is it worth you getting them now for future use, or in shows where you may not be using as many other lights?

mr_sound

Member
hmmm, looks like i'll have to hold off on those. i'm still keeping my eye on the other auction that popped up..it is a great price...but first i need to find out just how much i am actually getting paid for this one show. we didn't really talk money...though i'm willing to take a pay cut in order to show this promoter that i know what i'm doing and convince him to hire me for a lot more of his shows. i'm sure he'll love the idea of not having to rent a truck to bring in his sound and lighting rig...surprisingly he can fit his whole sound rig in one van. i don't know how..his is 5 times the size of mine, and mine takes up an entire cargo van. but now i'm just rambling.

i didn't realize that was how dimmers worked. that does make me a little paranoid about my setup for this show. though i'm not going over the rated load for the dimmer packs, nor am i going over the amps i have available, i am right at the limit. i have 8 300w lamps per each 2.4kw dimmer pack, and once i get everything fired up and running i doubt i'll have exactly 120 volts.

that brings me to another question. i already know that volts times watts gives you amperage, and watts divided by volts gives you amperage...and all the other calculations. but, if i have a 20 amp circuit, running 2400 watts on it, and my voltage drops to 110 because of the load...do i lose amperage or wattage or both? basically i'm trying to figure out if having less volatage is going to make all the circuit breakers blow. this theater is infamous for it's horrible electrical system. not only is the power noisy, but we've got circuit breaks, and sometimes actual fuses still, in 4 different locations in the building. 2 circuits on the stage go to the basement, 1 more goes to the boiler room under the lobby (which is a real pain to get into), and 5 more are up in a loft that you need to climb a ladder to get up to. it's a nightmare. so if i don't pop any circuits..or at least not often...then everything will be good.

ship

Senior Team Emeritus
Volts times amps still equals watts. If you have 110 volts coming into the building, short of a step up transformer or something similar to it in dimmer able to provide a larger voltage than given to it, volts times amps still equals watts. In other words, at 110v, your lamps at FF on the dimmers will be about 110v less any loss due to the dimmer and voltage drop due to the cable. 110v is still in most conditions considered within the design specification of most gear in being safe so that it's not over working in the case of a computer to bring the screen up to full intensity. Sort of like a light activated calculator. With less power it's either going to work, work slow or not at at all.

On the other hand, 8x300 is 2,400 and not leaving you much for headroom as we call it for problems, start up currents necessary to heat filaments, voltage drop etc. On large rock shows, 80% of maximum is the norm. Granted your circuit breakers and most fuses won't blow until either the sustained draw for a few minutes is 127% of the rated draw, or it's spiked to a current well over the rating, you are still playing it close on the wire feeding the dimmers etc. Best to knock off a can given it's all the same 20 amp circuit feeding the dimmer.

Hmm, 300w, using PAR 56 at 300w each?

In any case while it might be safe to put all lights up to full and leave them on for ten minutes during the test to see if anything goes pop, or given the system is rated for the amperage already, just limit the maximum output of all or some in getting back to that 80%, it is far more simple to exclude a can. Besides what is most commonly melting down is not the breaker though it can melt down without popping, it's the wire feeding it, receptacles etc. Smell burning plastic and it's something that really does need to be investigated.

Hope it helps with my opinion. Have done the max load and max dimmer rate game a lot in the past even to the point of taking my largest draw lighting cue as the maximum amperage used for this figure. I can be done, even exceeded, but it's a question of safety and the necessity for it also. Playing such games is also very dangerous and can spell the end of a show before it's over. Much less half of my lights are working but the other half randomly seem not to be to which you find that half a double pole breaker melted down etc. 20amp breakers most likely won't get hot enough to melt down before tripping given they are functioning properly but it's not professional to take the chance.

HOpe it helps.

DMXtools

Active Member
mr_sound said:
i already know that volts times watts gives you amperage, and watts divided by volts gives you amperage...and all the other calculations. but, if i have a 20 amp circuit, running 2400 watts on it, and my voltage drops to 110 because of the load...do i lose amperage or wattage or both?
One must consider the difference between rated power (watts) and actual power. A given lamp may be rated at 300 watts/120 volts. If the actual voltage being supplied to it differs, the power the lamp will actually draw will also differ in the same direction, but by the square of the difference.

The one thing nobody seems to have mentiopned is the resistance of the lamp. That's a constant, for all practical purposes. For our 300 watt/120 volt lamp, it'll be 48 Ohms.

How do we get 48 Ohms? The basic power formula says P=E * I: Watts equals volts times amps. Ohm's Law tells us that I= E/R: Amps equals volts divided by Ohms. We can substitute E/R for I in the original equation, so P = E * E/R. Multiply both sides of the equation by R and you get P*R = E*E. Then divide both sides of the equation by P and you get R=E*E/P. Now we can plug our rated numbers into this equation: R=120*120/300 = 14400/300 = 48: The resistance of the lamp is 48 Ohms.

As I said, resistance is a constant. For our purposes it's the only constant. Voltage is the main variable - the electrical "pressure" causing a certain number of Amps to flow through this fixed resistance. If the voltage is actually 120, it will cause 2.5 amps to flow (I=E/R). If bad wiring causes the voltage to drop to 100 volts, the R remains the same at 48, so I=100/48= 2.083 Amps. The power drops to P=E*I = 100*2.083 = 208.3 watts. We onlly "lost" 20 volts, but we lost 91.7 watts!

John

Mayhem

Senior Team Emeritus
Thanks John - that was something that I must admit I had forgotten all about when considering lamp loads. It also reminded me to ask for a bit more info on voltage drop as it is something that I do not have a full grasp on.

Most of us have probably metered the wall outlet and found the voltage to be either higher or lower than expected, but what factors influence this? Especially with regards to cable lengths, coils etc?

DMXtools

Active Member
Mayhem said:
Most of us have probably metered the wall outlet and found the voltage to be either higher or lower than expected, but what factors influence this? Especially with regards to cable lengths, coils etc?
Copper is a good conductor, but it's not perfect: any wire has some finite amount of resistance. Typically, #12 AWG (standard for 20A circuits) is around 0.00162 ohms per foot. It might not seem like much, but it's pretty common to have 100 feet or more between a distribution panel and the last outlet on the circuit. And it's got to make a round-trip - 200 feet total, 100 there and 100 back, yields a resistance of 0.324 ohms in series with the load.

Lets assume that the load is eight of those 300W/120V PAR-56 cans I was using in my last example. They're 48 ohms each, but wired in parallel the total is 6 ohms... plus the 0.324 ohms of the cable. Our 120 volt source sees a resistance of 6.324 ohms. I=E/R, so 120 volts will push 18.975 amps through the circuit instead of the 20 we expected. The voltage measured at the lamps is only 113.85 volts instead of the 120 they're rated for. The other 6.15 volts is being used up in the resistance of the wire. The wire itself is dissipating P = E * I = 6.15*18.975 = 116.7 watts, just turning it into heat.

The same basic idea applies to all the wires between the power plant and the lamps. The power companies try to allow for those voltage drops, but that's the reason that sometimes, when loading is light, the voltage at a given outlet may actually be higher than the nominal 120 volts.

It gets to be even more confusing when you try to account for power losses in transformers and the effects of power factor in reactive loads (electric motors, etc.) Throw in a few solid-state light dimmers and the switchmode power supplies of a few computers and you have a real mess. I'm amazed the utility companies handle it as well as they do.

John