need advice on wire size choice for power and ground

Mr Bright

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I am running two 450w GE landing lamps on my car, they have 6AWG wire which i THINK may be overkill, i want to redo my wiring to clean things up a bit. With that said, they are designed for 16.5v, and i realize a car is 12v.....so for one thing, how would you end up figuring out how many amps you now would accommodate for? I hope that makes sense, i know they will not be as bright if they arent powered by a true 16.5v, but the car when running is typically 14.3v.
Ok so after figuring out the new wattage, divided by voltage (no idea if you would use 12v or the 14.3), you get the amperage, i understand that....but so many wiring charts show different size requirements, some say 8AWG, 10AWG, etc....so i have no clue if i can get away with 10AWG wire.

Thanks for any advice!​
 

Lightdoctor

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Would not use anything smaller that 8AWG...to much risk of a melt down. When you build this harness, make sure the connections are tight.

(I sure hope that you're using these lights offroad only.)
 
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Mr Bright

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i hardly drive the car, i live in the middle of farm land and there are coyotes and other animals that you cannot see until its too late....long 5 mile stretches with vineyards hiding potential animals.
Would multi-strand wire be of any advantage heat-dissipation-wise aside from being more flexible to route? i have a soldering iron that i am learning how to use instead of crimping them on.
 

-Virgil-

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Multi-strand wire is the only acceptable kind in automotive service. Solid wire can't withstand the vibration.

Be advised that if you are running lamps with a rated voltage of 16.5v on 14v (I suspect they are probably actually seeing less than 13v with what is probably massive voltage drop in their circuit, but we'll go with your 14v figure for calculation) then they are producing only 57% of their rated light flux, and luminance will be very poor. You can do a lot better than your present setup with more thoughtful selection of lamps.
 

Norm

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Mr Bright please resize the pic in your sigline. Sigline images cannot exceed 500 pixels wide, 100 pixels high and 20k.

If you do happen to post in the wrong forum do not open another thread in the correct forum, click the triangle symbol directly under your post, this is the report button, explain your problem and send a report this will go to all Admins and Mods, someone will move your thread to the correct area. - Norm
 

ryguy24000

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I am not a auto wireman, but 900/16.5=54.6 amps In the construction world that would require at least a 6awg.
 

Mr Bright

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forgot to mention that each light has its own circuit off of the battery with its own fuse and 50A relay.
 

Alaric Darconville

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forgot to mention that each light has its own circuit off of the battery with its own fuse and 50A relay.

That's all fine and dandy, but mathematics are against you. The bulb is rated for intensity at 16.5V. You're feeding it, at the most, 14.0V.
(14.0/16.5)^3.4 = .57

Give it 0-gauge wire if you want, you're not going to conjure up more brightness until you conjure up more voltage.
 

HotWire

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I would consider the 6 AWG the minimum size. I use 2 AWG (not 2/0 AWG) for grounds and power. Be sure to use a proper hex crimp tool and use heat shrink tubing with adhesive at the lugs/connections. I would suggest you read some 4X4 forums and see what others are using for driving lights. Your fuses and relays are a must.
 

Alaric Darconville

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I would consider the 6 AWG the minimum size. I use 2 AWG (not 2/0 AWG) for grounds and power. Be sure to use a proper hex crimp tool and use heat shrink tubing with adhesive at the lugs/connections.

It's a non-starter. The filament luminance will be awful. The color will be nearly brown.
Current draw will be ginormous because the filament's resistance won't be maximal until the filament is hot enough, but it won't get hot enough at 14-15V.

I would suggest you read some 4X4 forums and see what others are using for driving lights.
Better yet, go to danielsternlighting.com and ask for recommendations there, or search this forum more carefully.

Your fuses and relays are a must.
As are bulbs designed for automotive voltages.
 

VegasF6

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That's all fine and dandy, but mathematics are against you. The bulb is rated for intensity at 16.5V. You're feeding it, at the most, 14.0V.
(14.0/16.5)^3.4 = .57

Give it 0-gauge wire if you want, you're not going to conjure up more brightness until you conjure up more voltage.

Alaric, can you please explain how you arrived at that formula? Is there a name for it?
 

Alaric Darconville

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-Virgil-

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It's just standard filament lamp math. The usual exponent is 3.4 for filament output with voltage change, 1.6 for filament power consumption with voltage change, and -13 (negative 13) for filament lifespan with voltage change. These aren't carved in stone; each individual lamp (bulb) design will have slightly different characteristics and so there's no one exponent that is exactly, precisely correct for all bulbs. But the exponents here are close to the middle of the range of variance, and will get you close to the middle of the ballpark for calculations like this on whatever kind of filament lamp you have in mind.

The generic form is m=(v2÷v1)^x where:

m = multiplier
v1 = original or rated operating voltage
v2 = changed or actual operating voltage
x = exponent as given above.

Running the equation gives you a multiplier which you then apply to whatever rating you're looking at. If you are using the lifespan exponent of -13 to figure the change in lifespan with changed voltage, you multiply the lamp's rated lifespan by the multiplier to get the expected lifespan at the new voltage. If you're using the 3.4 exponent to figure the change in light flux with changed voltage, you multiply the lamp's rated output by the multiplier. If you're using the 1.6 exponent to figure the change in power consumption with changed voltage, you multiply the lamp's rated actual wattage (watch out for inaccurate nominal wattages) by the multiplier.
 
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VegasF6

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That's all fine and dandy, but mathematics are against you. The bulb is rated for intensity at 16.5V. You're feeding it, at the most, 14.0V.
(14.0/16.5)^3.4 = .57

Give it 0-gauge wire if you want, you're not going to conjure up more brightness until you conjure up more voltage.

If you're using the 3.4 exponent to figure the change in light flux with changed voltage, you multiply the lamp's rated output by the multiplier. If you're using the 1.6 exponent to figure the change in power consumption with changed voltage, you multiply the lamp's rated actual wattage (watch out for inaccurate nominal wattages) by the multiplier.

Well that sure is handy to know. If I am understanding you then, is the formula being used correctly above? When you say the lamps rated output you are talking lumens, or luminous flux?
The only data given is watts at a specified voltage, so the only thing we could extrapolate would be the change in watts consumption with the 1.6 exponent?

*edit* after a re-read I guess Alaric was using the formula right, that resulted in his multiplier to figure approximate lumens. So, after you multiply the rated output by the multiplier, you then add the result to the original rating?
 
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-Virgil-

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Well that sure is handy to know. If I am understanding you then, is the formula being used correctly above?

Yes.

When you say the lamps rated output you are talking lumens, or luminous flux?

Er...say what? Luminous flux is stated in lumens. Sometimes, usually from American vendors, you will see it stated in MSCP or MSCD (mean spherical candlepower, sometimes labelled just "candlepower") instead; to get lumens multiply MSCP by 4π (12.57). Example: a 32cp 1157 bulb produces 32 ×4π = 402 lumens.

The only data given is watts at a specified voltage, so the only thing we could extrapolate would be the change in watts consumption with the 1.6 exponent?

The only specific value we can come up with given what we know is the power consumption. But we can come up with relative (percentage) values for luminous flux and for lifespan. Alaric's statement of ~57% of rated output is correct.

So, after you multiply the rated output by the multiplier, you then add the result to the original rating?

No. Why would you do that...? I don't understand your thinking here. Let's run through an example. Take a 9006 bulb rated 55 watts, 1000 lumens and 1000 hours at 12.8v. Let's run it at 11.9v and see what happens to…

…Luminous flux:

m = (11.9 ÷ 12.8)^3.4

m = (0.9296875)^3.4

m = 0.7805

0.7805 × 1000 lumens = 780.5 lumens at new operating voltage.

…Power consumption:

m = (11.9 ÷ 12.8)^1.6

m = (0.9296875)^1.6

m = 0.89

0.89 × 55w = 49w at new operating voltage

…Lifespan:

m = (11.9 ÷*12.8)^-13

m = (0.9296875)^-13

m = 2.58

2.58 × 1000 hours = 2580 hours at new operating voltage.

-----
It's worth keeping in mind that the decrease in luminous flux is not the only vector of reduced performance. Filament luminance nosedives, which sharply reduces the "punch" of the beam.
 

VegasF6

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I was trying to compute for over voltage, and I reversed the 2 voltages in the formula. So when I arrived at a # lower in flux I figured you had to add the two together. I see my mistake now. I will be sure to reference this page in the future.
 

-Virgil-

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Looks like the error is actually mine -- I've corrected my post above to
read (V2 ÷ V1) rather than the erroneous (V1 ÷ V2),
which would have had you dividing rated voltage by operating voltage. In fact, you need to divide operating voltage by rated voltage. I do these calculations so often it just goes through the spinal cord and I seldom ever have occasion to write them out in general form like that. Sorry!
 

deadrx7conv

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With vehicle running, measure voltage at light. Get an ammeter and measure current. Then, you should be able to figure out the gauge that you need, and the actual wattage that you're running those 450w bulbs at.

An option is to convert to a 16v battery/alternator for your lighting(either independent 2nd alternator/battery or vehicle conversion). Conversion will use several step down resistors to power the each of the car circuits at ~13v. Check with a competent car stereo shop or race shop. Higher voltage systems are common enough for amplifier setups or for more ignition power.
 

-Virgil-

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An option is to convert to a 16v battery/alternator for your lighting(either independent 2nd alternator/battery or vehicle conversion). Conversion will use several step down resistors to power the each of the car circuits at ~13v.

And while you're at it, you could travel from Los Angeles to San Diego by way of Calcutta, too -- which would be comparably cost-effective.

Just put in more thoughtfully-selected lights and be done with it!
 
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