Let there be light - Question

[shaley]

I think what "midlf" was saying was don't use NEC building construntion standards for wiring sizes on tractors and 12 volt vehicles. Under SAE specs 16 AWG is max at 25 amps. Ok for my tractor and boat but not my house.
http://www.westmarine.com/pdf/MarineWire.pdf

 

[gemini5362]

Mike is very correct the current rating of wire does not consider at all what the voltage going to the wire is. It is strictly a matter of the current flow. The difference that voltage makes is in the insulation. The higher the voltage the better the insulation needs to be to keep it from arcing through the insulation. But wire size is only about current and wire composition. Basically how many electons can you push through the wire before it gets too hot and melts

 

[norsker]

But wire size is only about current and wire composition. Basically how many electons can you push through the wire before it gets too hot and melts

Are you familiar with the water pipe analogy regarding current flow? The one that states that for a given pipe size, let`s say 1/2", and a given pressure which we can abritrarily state for the purpose of this discussion to be 10psig, we will get a flow of maybe .75gpm. If we increase pressure to 100psig we can get maybe 20gpm through the same 1/2" pipe. Can everyone see the relationship here?

So if we equate current flow to fluid flow and voltage to fluid pressure, the larger our flow requirements are, the higher the pressure we need through any particular conductor size, whether it be a pipe or wire. If pressure is not variable, the only thing we can increase to get more flow is increase the pipe size, same thing with wire, if the voltage is fixed at 12 we can`t get the same ampacity in a conductor that we would get if the voltage were 120.

All that I am trying to illustrate is that voltage is relevant to current flow through a particular wire size and I don`t mean to direct my statements to anyone in particular although I did quote gemini5362 merely as a point to jump into this discussion.

The notion that the primary difference between 12v conductors and 120v conductors for a given amperage is in the thickness of the insulation is misguided. With a dielectric strength of a few thousand volts per mil of insulation, insulation thickness is not relevant to this discussion. Composition and temperature rating of the insulation may be, but for other reasons related to the service environment of the installation.

Another thing, the chart from West Marine is misleading and easily misread if you aren`t careful to keep your colors in the proper order and the chart below it that suggests that 16 ga wire is sufficient for 25 amps does not delineate between momentary and continuous loads. A better chart can be found here: American Wire Gauge table and AWG Electrical Current Load Limits which shows that 16ga wire is only good for a continuous load of 3.7 amps.

Hope this helps.

 

[dtd24]

Going with 12ga in-line fuse from battery to 6 gang fuse block

16ga from fuse block to 20 amp switch
(2) seperate 16 ga from 20 amp switch to two front rops lights

16ga from fuse block to 20 amp switch
(2) seperate 16 ga from 20 amp switch to two rear rops lights

I am thinking about another light right on the arm of the backhoe, but I'm not sure about that one yet. I really don't think I will need it. 6 gang sounds like alot, but I want to use same fuse block for future electric diverter for grapple and maybe a horn or something.

 

[gemini5362]

Are you familiar with the water pipe analogy regarding current flow? The one that states that for a given pipe size, let`s say 1/2", and a given pressure which we can abritrarily state for the purpose of this discussion to be 10psig, we will get a flow of maybe .75gpm. If we increase pressure to 100psig we can get maybe 20gpm through the same 1/2" pipe. Can everyone see the relationship here?

So if we equate current flow to fluid flow and voltage to fluid pressure, the larger our flow requirements are, the higher the pressure we need through any particular conductor size, whether it be a pipe or wire. If pressure is not variable, the only thing we can increase to get more flow is increase the pipe size, same thing with wire, if the voltage is fixed at 12 we can`t get the same ampacity in a conductor that we would get if the voltage were 120.

All that I am trying to illustrate is that voltage is relevant to current flow through a particular wire size and I don`t mean to direct my statements to anyone in particular although I did quote gemini5362 merely as a point to jump into this discussion.

The notion that the primary difference between 12v conductors and 120v conductors for a given amperage is in the thickness of the insulation is misguided. With a dielectric strength of a few thousand volts per mil of insulation, insulation thickness is not relevant to this discussion. Composition and temperature rating of the insulation may be, but for other reasons related to the service environment of the installation.

Another thing, the chart from West Marine is misleading and easily misread if you aren`t careful to keep your colors in the proper order and the chart below it that suggests that 16 ga wire is sufficient for 25 amps does not delineate between momentary and continuous loads. A better chart can be found here: American Wire Gauge table and AWG Electrical Current Load Limits which shows that 16ga wire is only good for a continuous load of 3.7 amps.

Hope this helps.

While I admit I was trying to simplify things too much I do believe that for the things that most people reading this will be doing , my statement was very valid. Wire size is installed generally
to carry a certain amount of current. a wire that is rated at 20 amps will melt when 100 amps is ran through it. It does not matter if the 100 amps is generated out of a 6 volt circuit. a 12 volt circuit, a 120 volt circuit, 440 volt circuit or 13.2kv circuit if you put 100 amps through it then it is going to melt. The composition of the wire makes a difference aluminum wire is generally rated at half of the current capcity as copper for a given wire size.

 

[SPIKER]

While I admit I was trying to simplify things too much I do believe that for the things that most people reading this will be doing , my statement was very valid. Wire size is installed generally
to carry a certain amount of current. a wire that is rated at 20 amps will melt when 100 amps is ran through it. It does not matter if the 100 amps is generated out of a 6 volt circuit. a 12 volt circuit, a 120 volt circuit, 440 volt circuit or 13.2kv circuit if you put 100 amps through it then it is going to melt. The composition of the wire makes a difference aluminum wire is generally rated at half of the current capcity as copper for a given wire size.

Hi Guys:

while it is true, wire size determines the current carrying ability of the wire, there are also wire composition that also determines it as well as insulation types.

there is no 100% way to determine what is and is not correct other than following the wire manufactures recommendations. vehicles wiring generally is made up of FINE stranded wires, this is due to lower voltages, vibration resistances as well as bending ability. these are also generally run in free air and not inside a conduit or panel box to get ratings. Battery cables like WELDING cables are made up of many very small dia strands of wire, building wires are made up of solid wire. 12 ga wire for building is not as good of current carrying as 12 ga. wire made of small stranded wire. for the reason that current actually flows over the out side surface of the wire not through the center like many people think. all the little strands have a greater surface area so can carry more current before heating gets too bad, BUT heating resistance is not as good for all the little strands of copper because melting a very small strand is easier than melting one larger strand.

to figure the amp required for you're lights be sure to use Ohms Law

P=i*e (power in watts) = i (voltage) times E (Current) it is easy to find for AMPS (Current) by algebra using the formula as follows
E = P/I Amps = 55 watts/12 volts
SO Amps = 4.6 amps per light

the pic/diagram provided up above (PINERIDGE was given credit for this but was posted by someone else) is a good drawing. several notes that I would like to add to this to help make it clear for someone with less electrical experience.

1. the BLACK lead from the SWITCHES will go to the relay coils.
2. the BLUE leads go to the relay contact common
3. the Yellow wires are the relay Switched Normally Open leads.
4. to make these work only when the head light switch is ON use the Switched side of the headlight switch and run power from that to the hot side of the Aux Switch Bank. (hot side of the BLACK wires) leave the BLUE wires going through the added Fuse Blocks from the AUX system or even from the battery if a higher output is needed to run these.

5. note above when someone said use the hot key switched on bank to run the light in either case so that when the key is off the lights are off and that way no run down battery... This is drawn this way on the diagram that pineridge drew.

Mark M

 

[gemini5362]

Mark your post is good I do have some questions for you though. 1. what type of wire insulation affects the current carrying capacity of wire. I am aware of different types of insulation that allow you to run temps in different outside environments. T.hese are to keep the wire from being melted for instance, if it is in a high temp environment. I am not aware of any wire insulation that affects the current carrying capacity of the wire.

2. When current is carried on the outside of the wire this is called skin effect. To the best of my knowledge this is a high frequency occurence. Sometimes when you run high freguency RF through a wire they silver plate the outside of the wire because the high frequency only runs on the outside of the wire and the silver has less resistance to electron flow. To the best of my knowledge wire at 60 cycle or DC runs through the whole wire. Do you have any websites that I could go to show me what you are talking about ?

 

[gemini5362]

Hi Guys:

P=i*e (power in watts) = i (voltage) times E (Current) it is easy to find for AMPS (Current) by algebra using the formula as follows
E = P/I Amps = 55 watts/12 volts
SO Amps = 4.6 amps per light

Mark M

Mark I believe you have a typing error in your formula

P = I * E is correct

E= P/I Amps = 55watts/12volts you went on to do the math right I am assuming you meant to say that

I= P/E I (amps) = P (power) / E (volts) = Amps = 55 watts/12 volts

 

[shaley]

American Wire Gauge table and AWG Electrical Current Load Limits
Maybe I'm reading this chart wrong but doesn't the maximum current for chasis wiring (next to last column) for 16 AWG listed as 22A.

 

[638]

I don't believe that the reference to 'chassis' is referring to an automobile, or tractor chassis. More like a battery charger chassis, or TV chassis. The fine stranded wire is more flexible, and less likely to break due to vibration. Not too much vibration in a building. That's why 'romex' is solid wire. Wire that is ran in conduit is almost always stranded, a lot easier to pull into place. Amps is amps. If you use the ampacities that I listed, you will not have any problem with overloaded wires.

ron