Auxiliary Light Hook Up?

[sledgehammer1990]

So I just bought a pair of off road lights for my front bumper. I know where to install the switch and where to run the wires through the fire wall, but I don't know how exactly to wire them up. Do I need relays or anything? I just want a separate switch for the lights so I can turn them on as I wish.

 

[IZwack]

Its a sticky:
http://www.explorerforum.com/forums/showthread.php?t=203037

 

[sledgehammer1990]

Cool, thanks... I figures everything out, but I don't know what fuse to get. I am running 2x 130W lights.

 

[fixxxer]

easy formula, power (in watts) equals voltage (in volts) times current (in amps)
260 watts = 12volts x 21.66 amps
I'd recommend using a relay as most 12 volt auto switches are not rated for 30 amps (need some headroom)
30 AMP relay with 30 AMP fuse.
Actually, I'd use a 40 amp relay and a 30 amp fuse. Place the fuse in the power lead circuit as close to the battery as possible. The fuse is there to protect the wiring in case of a short circuit.

 

[sledgehammer1990]

Oh, thanks. I guess I will have to stop back tomorrow and exchange mine. I grabbed some 20 AMP fuses. I stopped at radio shack and grabbed a few things...

4 Tap-In connectors (Splices wires together)
3 20-AMP mini blade fuses
1 30A In-Line Fuse Holder (Mini Blade)
1 30A Illuminated Rocker Switch
20' 12-Gauge Wire

Here is the diagram I am going off of.

 

[PedroDaGr8]

It is NEVER a good idea to run that much amperage through a switch. You are just ASKING for trouble. There is a reason that relays were invented and it is so you have minimal current near anything someone touches, sensitive components etc. Don't slack off on it, especially when using a relay is super simple.




The above keeps the current in the switch area to under an amp (more than likely around half an amp).

This is SUPER easy to make, keeps the high current wires outside the ****pit.


One more IMPORTANT reason to use a relay. That much amperage should be using around a 10Gauge power wire. Give or take. Otherwise you are sacrificing output as the wire isn't thick enough and the voltage is dropping. I will post more about this in detail tomorrow.

 

[IZwack]

It is NEVER a good idea to run that much amperage through a switch. You are just ASKING for trouble. There is a reason that relays were invented and it is so you have minimal current near anything someone touches, sensitive components etc. Don't slack off on it, especially when using a relay is super simple.

For what its worth - in off road situations, relays are half the time not used because relays are mechanical devices and are subject to vibration damage. In desert race trucks for example, you'll see switches driving the lights or whatever directly w/o a relay. Of course these switches have very high current ratings and are sealed off from dust and so on.

 

[sledgehammer1990]

I will swap out the 12 gauge for some 10 gauge tomorrow too then. The only thing I don't understand about that image is I see Positive running from the battery to the relay, then to the switch. Yet, there is another 12V running to no where. That bottom 12V is the supply though.

Also, is this the type of relay you are talking about?
http://www.radioshack.com/product/index.jsp?productId=2062477&tab=summary

 

[Bill Kemp]

And I'de swap out the 12 for 14 guage wire. See: http://www.explorerforum.com/forums/showthread.php?t=28683 (post #2) and make your own decision. Me, for most all lights in the front grill area (130 watt included) use 14ga wire. Only reason to use heavier guage is if the wire length is longer.

The 12 volt source wire going to nowhere is if your switch needs another source to light the switch. Hook that to your interior lights if so.

 

[PedroDaGr8]

And I'de swap out the 12 for 14 guage wire. See: http://www.explorerforum.com/forums/showthread.php?t=28683 (post #2) and make your own decision. Me, for most all lights in the front grill area (130 watt included) use 14ga wire. Only reason to use heavier guage is if the wire length is longer.

The 12 volt source wire going to nowhere is if your switch needs another source to light the switch. Hook that to your interior lights if so.

Hmmm, interesting link. While it is a GOOD chart, he then negates EVERYTHING he has said at the end with this statement:

Wire sizes for lighting is more critical than for other applications — The rated output of any lamp is figured at 13.5 volts, not 12 volts as you might expect. So with a ½ volt drop you are at 13.0 volts. And at 95% of the rated voltage, you are only putting out 80% of the rated luminous intensity - for a 100 watt lamp that’s only 80 watts!! Get what you pay for and figure to the high side when you are sizing wire for lighting.​

I am going to expand on this a little bit, also sorry Bill if you feel I am stepping on your toes or anything. I am just letting everyone know what I have learned and honestly mean no offense by it.

First off, I use this site here OFTEN for calculating voltage drop for wires:Voltage Drop Calculators

The site has four different calculators and I usually use the second one as it is the most useful for vehicle applications. It is called Any Voltage Voltage Drop Calculator, and I use it with THESE settings:
Material = Copper (change to alluminum if you are using alluminum wiring)
SelectPhase and No. of wires = 1-phase - 2 Wires
Enter Voltage = 14.0 (This just fills in the output voltage, it doesn't change the voltage drop value).
The rest I fill in with the relevant needed for calculation:
In this case I chose a Circuit Length of 6-Feet, the Amperage was 21.66 (as calculated before) and then varied the Gauge:

The results are:

18 Gauge: Voltage Drop = 2.07 Output Voltage = 11.93
16 Gauge: Voltage Drop = 1.31 Output Voltage = 12.69
14 Gauge: Voltage Drop = 0.82 Output Voltage = 13.18
12 Gauge: Voltage Drop = 0.51 Output Voltage = 13.49
10 Gauge: Voltage Drop = 0.32 Output Voltage = 13.68

Its Effect on Light output:

In this section, I am using a bulb that has an output of 1200 Lumens at 12.8V. All calculated Light outputs are based on that. The last entry is the theoretical maximum (assumes 0 voltage drop).

18 Gauge: Output Voltage = 11.93 Output Light = 944.60 Lumens
16 Gauge: Output Voltage = 12.69 Output Light = 1165.30 Lumens
14 Gauge: Output Voltage = 13.18 Output Light = 1325.50 Lumens
12 Gauge: Output Voltage = 13.49 Output Light = 1434.52 Lumens
10 Gauge: Output Voltage = 13.68 Output Light = 1504.39 Lumens
Theoretic: Output Voltage = 14.00 Output Light = 1627.43 Lumens

Lastly, time to compare the percent differences from 14.0V for voltage and light output:

18 Gauge: %Diff Voltage = -14.79 %Diff Light = -41.96
16 Gauge: %Diff Voltage = -9.36 %Diff Light = -28.40
14 Gauge: %Diff Voltage = -5.86 %Diff Light = -18.55
12 Gauge: %Diff Voltage = -3.64 %Diff Light = -11.85
10 Gauge: %Diff Voltage = -2.29 %Diff Light = -7.56

Personally, I would like to keep the Light output drop under 10%.

Note: This does work does NOT apply to devices like HID ballasts as quality ones maintain a precise output voltage no matter the input voltage. Instead, for a relay harness for HIDs, I usually just use 14Ga or so and call it a day as that has been shown to be a high enough Gauge to handle the initial start-up current (which can over 100-200A for a few milliseconds).


This makes me want to put together a rough website based off of the aforementioned calculator that will allow a user to see a calculated light output based on wire gauge. Time to see if I can make it work. If I can Bill, I will post it and maybe you can use that link to show people. That is I haven't pissed you off with this long lecture of a post.

 

[Bill Kemp]

Toes, lecture, pissed..??? This site is about information & education not personalities.
Whatever you can add is appreciated.
That was the only site I could find 5 yrs. ago that had any kind of chart that a simple guy like me could understand.
Makes me wonder though, due to the seeming differences in voltage drop info, what and where would you find "facts" as opposed to info from someone who claims to know, on the web.
It would be nice to know the definitive answer.

Edit: Unless I'm missing something a 130 watt bulb draws about 10.83 amps not 21.66. When you figure that most applications are 55 or 100 watt bulbs 14 guage is a pretty good size.

 

[sledgehammer1990]

Will 12 be fine? Radio Shack doesn't sell 14 gauge in store.

 

[rileyrs]

when i did my light i actually bought a 20ft line of 14 guage at home depot in the electrical section for about $6.

 

[Bill Kemp]

Will 12 be fine? Radio Shack doesn't sell 14 gauge in store.

Sure will. It'll work fine.

 

[PedroDaGr8]

Toes, lecture, pissed..??? This site is about information & education not personalities.
Whatever you can add is appreciated.
That was the only site I could find 5 yrs. ago that had any kind of chart that a simple guy like me could understand.
Makes me wonder though, due to the seeming differences in voltage drop info, what and where would you find "facts" as opposed to info from someone who claims to know, on the web.
It would be nice to know the definitive answer.

Edit: Unless I'm missing something a 130 watt bulb draws about 10.83 amps not 21.66. When you figure that most applications are 55 or 100 watt bulbs 14 guage is a pretty good size.

He said TWO, that's where the 21.66 comes from. Depending on where he forks the line, it could end up being less or more of an issue. If he forks at the beginning versus having a long run of single wire then a split.

Yeah, this particular site is mainly geared towards electricians. They have all sorts of calculators based on number of cores, is it in raceway, temperature, insulation etc etc etc. Seems pretty in depth and reliable.

From what I am reading, the data comes from the NEC values. It uses information based on Ohms per 1000Ft of cabling etc. Aparently, you can also use K values, which are based on the radius of the wire and what the wire is made of. I am just NOW starting to find sites to piece things together to make sense.


The light output formula from what I have seen (at several places) is a pretty good approximation.
The formula is Light Output = Coeff *(Voltage^3.4), where the coeff is unique to the light bulb in question. The good thing is, all you need is one data point to find the coeff and you can find ALL of the light outputs across the range of voltages.

OK, some more information:
This is taken from wikipedia, which itself is sourced from a Textbook for engineers. These let you know how different aspects of the bulb change with voltage.

# Light output is approximately proportional to V^3.4
# Power consumption is approximately proportional to V^1.6
# Lifetime is approximately proportional to V^−16
# Color temperature is approximately proportional to V^0.42

 

[PedroDaGr8]

12-gauge would be reasonable as you are sub5% voltage drop.

 

[Bill Kemp]

He said TWO, that's where the 21.66 comes from. Depending on where he forks the line, it could end up being less or more of an issue. If he forks at the beginning versus having a long run of single wire then a split.

I guess I thought that a person would run a single wire to each lamp as shown in most lighting diagrams.

From what I am reading, the data comes from the NEC values. It uses information based on Ohms per 1000Ft of cabling etc. Aparently, you can also use K values, which are based on the radius of the wire and what the wire is made of. I am just NOW starting to find sites to piece things together to make sense.
The light output formula from what I have seen (at several places) is a pretty good approximation.
The formula is Light Output = Coeff *(Voltage^3.4), where the coeff is unique to the light bulb in question. The good thing is, all you need is one data point to find the coeff and you can find ALL of the light outputs across the range of voltages.
OK, some more information:
This is taken from wikipedia, which itself is sourced from a Textbook for engineers. These let you know how different aspects of the bulb change with voltage.
# Light output is approximately proportional to V^3.4
# Power consumption is approximately proportional to V^1.6
# Lifetime is approximately proportional to V^−16
# Color temperature is approximately proportional to V^0.42

As I mentioned, the site I have used is easy for a simple guy, like me, to understand. (As to the above info, like Sgt. Schultz, "I know nothing".)

 

[sledgehammer1990]

The lights have been working fine, but I couldn't get the relay to work. I have them directly wires at the moment. Your diagram with the relay doesn't work because I need that 12v source to go to the battery, but one already goes to the battery.

 

[Rofocowboy84]

The lights have been working fine, but I couldn't get the relay to work. I have them directly wires at the moment. Your diagram with the relay doesn't work because I need that 12v source to go to the battery, but one already goes to the battery.

That "positive wire to nowhere" would more accurately be "from", not "to". That's the wire coming from a 12v source to light up the switch and send a little bit of electricity to the relay to trip it. That's also why your relay isn't working, you don't have any power for the switch to trip it. Since it's such a small flow, you can probably get away with running a small wire to the cabin fuse box.

That's my understanding, but I'm not an automotive electricity expert, so if someone wants to correct me, feel free. Like Bill said, this forum is about knowledge and not feelings.

 

[sledgehammer1990]

Would it be ok to tap into the 12v ignition wire? Would I need a fuse to between the two? How would I hook it into the fuse box?