Hello, I found this board while searching for e-fan ideas for my 1996 yukon 5.7L Vortec. Great board!

I saw a few threads about the Big 3 cable upgrade: where you replace factory cables for battery+ to alternator and starter, and battery- to engine, frame, and chassis.

The idea is that fatter cables are less resistive and therefore more efficient, so fatter cables are better.

Consider this:

Copper wire is efficient.

Fatter wire (solid, stranded, or fine stranded) is more efficient than thinner wire.

BUT - fatter wire only delivers real benefits when the conductor length is long and/or the load is large.

When the runs are short and the loads are small, the incremental benefit of fatter wire is small.

This concept is best demonstrated by a simple chart:

I made this chart using the free calculator at:

https://photovoltaic-software.com/solar-tools/dc-ac-drop-voltage-calculator

You can confirm the voltage drops using different calculators:

https://www.calculator.net/voltage-drop-calculator.html

or

https://grealpha.com/resources/dc-load-wiring-calculator/calc/voltage-drop/

etc.

I made the chart for 14vdc at 150°F, for 1 meter (39 inches) of wire.

The relationships are all predominantly linear.

When temperature decreases, conductive efficiency increases slightly, and vice versa.

When conductor length decreases or increases, conductive efficiency is linearly affected. Example: the power loss in a 13" wire will be 1/3 of the loss in the

Importantly, 1 meter of any size copper wire is not resistive. Because it is short.

A single meter of skinny 20g wire only has three hundredths of 1 ohm of resistance, and 1 ohm is nothing.

But, load matters. The higher the load, the more the resistance of the conductor matters in terms of power loss caused by resistance to current.

I made the chart more expansive than it needs to be, in order to show the big picture.

When your wire is too small for your load, a small increase in wire size makes a big improvement in power loss.

However, when your wire is suitable for your load, a small increase in wire size makes almost no improvement in power loss.

Using the '96 Yukon as an example:

Example 1:

The factory cable from battery+ to starter motor is 2awg copper cable.

Assuming this cable is 1 meter long, and assuming a 50 amp load, the power loss caused by the 2awg cable size is 2.9 watts, or less than 1% of the total wattage, which means that the cable is more than 99% efficient for that application.

If you change that 1 meter cable from 2awg to 2/0awg, and run the same 50amp load, the power loss associated with the new fatter 2/0 cable will be 1.4 watts, an even smaller fraction of 1% of the total load, still greater than 99% efficient.

If you pay $40 to change the factory 2awg cable to a fatter 2/0awg cable, then you have spent $40 to change a cable that is more than 99% efficient, to a cable that is more than 99% efficient.

Assuming a 100amp load and everything else the same, the power loss on the 2awg cable is 11.5 watts or 1% of the load, and the power loss on a 2/0 cable would be 5.7 watts and 0.5% of the load.

So the fatter 2/0 cable, at 100 amps steady draw, would change your system from 99% efficient to 99.5% efficient.

Example 2:

The factory cable from battery+ to the driver's side fuse block is 6awg copper cable.

Assuming this cable is 1 meter long, and assuming a 50 amp load, the power loss caused by the 6awg cable size is 6.3 watts, or 1% of the total wattage, which means that the cable is 99% efficient.

If you change that 1 meter cable to 2awg, and run the same 50amp load, the power loss associated with the new fatter 2awg cable will be 2.9 watts, or about a half of a percent.

So, by changing to the fatter 2awg cable, you have gone from 99% efficient to 99.5% efficient.

This cable never sees 100 amps, but you can do the same comparison at 100amps if you want to, and you will see that changing to a fatter wire makes very little real difference within a reasonable range of operating specifications.

Example 3:

The factory cable from battery+ to the alternator is 6awg copper cable.

Assuming this cable is 0.66 meter long, and assuming a 50 amp load, the power loss caused by the 6awg cable size is 4.2 watts (0.66 of 6.3 watts from the chart), which is less than 1% of the total wattage, which means that the cable is more than 99% efficient.

If you change that 0.66 meter 6awg cable to a

So, by changing the battery+ to alternator cable from 6awg to 2/0awg, you change the cable efficiency from greater than 99% to greater than 99%.

By now, you know how to use the table to compare 6awg to 2/0awg batt+toAlt cable at 100 amps.

6awg = 25.1watt loss x 0.66 = 16.6watt loss = 1.3% loss = 98.7% efficient at 100 amps.

2/0awg = 5.7watt loss x 0.66 = 3.8watt loss = 0.4% loss = 99.6% efficient at 100 amps,

which seems like Hey, maybe I want that 1% increase in efficiency, but you have to remember that you only get that 1% improvement when you make a massive jump in cable size

*******

The same basic concepts apply to ground cables (negative side cables).

Your ground cables need to be matched to your load for each and all circuits, but after that, small changes in ground cable size produce the same small changes in power transmission efficiency.

*******

Sooooooo,

If you buy a $200 Big-3-4-5-6 cable kit that has fatter wires for everything, you are changing the factory wiring from 99+% efficient to 99+% efficient.

That's it.

Additional comments:

1 - The chart above displays linear relationships that are inescapable. If you think you don't understand electronics, study the relationships for a while and you will see that physics correlates with common sense: nothing is free. There is no magic.

2 - The 1990's engineering capability at general motors company was not "poor". They did not design and build and sell trucks that had puny inadequate wiring.

3 - Temperature, and fine vs. coarse strands, and alloys, and lug connections, all matter.

A little.

Only a little.

The linear relationships of ohms law applied to conductive materials do not change just because you want them to.

If you have a really crappy dirty loose greasy crimp on a lug, it will cause problems. Clean up the connection, and the problems will go away. You don't need a 2/0awg wire for a 2awg job. Period.

4 - The OBS family is susceptible to problems caused by deteriorating ground connections. Find every ground connection and take it apart and clean all the surfaces. If the cable strap is cracked or corroded, replace it with a similar gauge of wire. A good ground is a good ground. You don't have to triple the wire size to maintain a good ground. A huge wire connected to a rusty greasy ground will be a crappy ground. Oil and dirt are not conductive, at all.

5 - There is such a thing as a ground loop. If you mod your harness like a madman, and then discover that you have voodoo electrical issues, do some research on a ground loop. Maybe you installed too many ground connections and made a weird loop. It happens.

6 - If your cables are old and the lug crimps are dirty and bent over and stressed out, then they will be more resistive than when new. Fix and clean them, or replace them, but don't think that tripling the cable size will create magic. Your problem was dirty bent-over stressed connections, not too-skinny wire.

7 - If you install a 500 amp alternator and a 2000 watt music amplifier in a car, you will need to use fatter wires everywhere that matters. Common sense.

But if you change a total nominal 50 amp load to a total nominal 60 amp load and you have relays and fuses that can handle startup spikes, you don't have to change any wires.

2 cents.

Thx.

DS

I saw a few threads about the Big 3 cable upgrade: where you replace factory cables for battery+ to alternator and starter, and battery- to engine, frame, and chassis.

The idea is that fatter cables are less resistive and therefore more efficient, so fatter cables are better.

Consider this:

Copper wire is efficient.

Fatter wire (solid, stranded, or fine stranded) is more efficient than thinner wire.

BUT - fatter wire only delivers real benefits when the conductor length is long and/or the load is large.

When the runs are short and the loads are small, the incremental benefit of fatter wire is small.

This concept is best demonstrated by a simple chart:

You must be registered for see images attach

I made this chart using the free calculator at:

https://photovoltaic-software.com/solar-tools/dc-ac-drop-voltage-calculator

You can confirm the voltage drops using different calculators:

https://www.calculator.net/voltage-drop-calculator.html

or

https://grealpha.com/resources/dc-load-wiring-calculator/calc/voltage-drop/

etc.

I made the chart for 14vdc at 150°F, for 1 meter (39 inches) of wire.

The relationships are all predominantly linear.

When temperature decreases, conductive efficiency increases slightly, and vice versa.

When conductor length decreases or increases, conductive efficiency is linearly affected. Example: the power loss in a 13" wire will be 1/3 of the loss in the

__same__wire 39" long, and the power loss in a 117" wire will be triple the loss in the__same__wire 39" long, all else the same.Importantly, 1 meter of any size copper wire is not resistive. Because it is short.

A single meter of skinny 20g wire only has three hundredths of 1 ohm of resistance, and 1 ohm is nothing.

But, load matters. The higher the load, the more the resistance of the conductor matters in terms of power loss caused by resistance to current.

I made the chart more expansive than it needs to be, in order to show the big picture.

When your wire is too small for your load, a small increase in wire size makes a big improvement in power loss.

However, when your wire is suitable for your load, a small increase in wire size makes almost no improvement in power loss.

Using the '96 Yukon as an example:

Example 1:

The factory cable from battery+ to starter motor is 2awg copper cable.

Assuming this cable is 1 meter long, and assuming a 50 amp load, the power loss caused by the 2awg cable size is 2.9 watts, or less than 1% of the total wattage, which means that the cable is more than 99% efficient for that application.

If you change that 1 meter cable from 2awg to 2/0awg, and run the same 50amp load, the power loss associated with the new fatter 2/0 cable will be 1.4 watts, an even smaller fraction of 1% of the total load, still greater than 99% efficient.

If you pay $40 to change the factory 2awg cable to a fatter 2/0awg cable, then you have spent $40 to change a cable that is more than 99% efficient, to a cable that is more than 99% efficient.

Assuming a 100amp load and everything else the same, the power loss on the 2awg cable is 11.5 watts or 1% of the load, and the power loss on a 2/0 cable would be 5.7 watts and 0.5% of the load.

So the fatter 2/0 cable, at 100 amps steady draw, would change your system from 99% efficient to 99.5% efficient.

Example 2:

The factory cable from battery+ to the driver's side fuse block is 6awg copper cable.

Assuming this cable is 1 meter long, and assuming a 50 amp load, the power loss caused by the 6awg cable size is 6.3 watts, or 1% of the total wattage, which means that the cable is 99% efficient.

If you change that 1 meter cable to 2awg, and run the same 50amp load, the power loss associated with the new fatter 2awg cable will be 2.9 watts, or about a half of a percent.

So, by changing to the fatter 2awg cable, you have gone from 99% efficient to 99.5% efficient.

This cable never sees 100 amps, but you can do the same comparison at 100amps if you want to, and you will see that changing to a fatter wire makes very little real difference within a reasonable range of operating specifications.

Example 3:

The factory cable from battery+ to the alternator is 6awg copper cable.

Assuming this cable is 0.66 meter long, and assuming a 50 amp load, the power loss caused by the 6awg cable size is 4.2 watts (0.66 of 6.3 watts from the chart), which is less than 1% of the total wattage, which means that the cable is more than 99% efficient.

If you change that 0.66 meter 6awg cable to a

**2/0awg huge cable**, and run the same 50amp load, the power loss associated with the new fatter 2/0awg cable will be 0.9 watts (0.66 of 1.4 watts), or way less than 1% of the total wattage, meaning that the new fatter cable is more than 99% efficient.So, by changing the battery+ to alternator cable from 6awg to 2/0awg, you change the cable efficiency from greater than 99% to greater than 99%.

By now, you know how to use the table to compare 6awg to 2/0awg batt+toAlt cable at 100 amps.

6awg = 25.1watt loss x 0.66 = 16.6watt loss = 1.3% loss = 98.7% efficient at 100 amps.

2/0awg = 5.7watt loss x 0.66 = 3.8watt loss = 0.4% loss = 99.6% efficient at 100 amps,

which seems like Hey, maybe I want that 1% increase in efficiency, but you have to remember that you only get that 1% improvement when you make a massive jump in cable size

**your automobile is so screwed up that your alternator is constantly pushing 100 amps at your battery, in which case you will soon have a fire and you don't have to sweat that 1% anymore because your car burned down.**__AND__*******

The same basic concepts apply to ground cables (negative side cables).

Your ground cables need to be matched to your load for each and all circuits, but after that, small changes in ground cable size produce the same small changes in power transmission efficiency.

*******

Sooooooo,

If you buy a $200 Big-3-4-5-6 cable kit that has fatter wires for everything, you are changing the factory wiring from 99+% efficient to 99+% efficient.

That's it.

Additional comments:

1 - The chart above displays linear relationships that are inescapable. If you think you don't understand electronics, study the relationships for a while and you will see that physics correlates with common sense: nothing is free. There is no magic.

2 - The 1990's engineering capability at general motors company was not "poor". They did not design and build and sell trucks that had puny inadequate wiring.

3 - Temperature, and fine vs. coarse strands, and alloys, and lug connections, all matter.

A little.

Only a little.

The linear relationships of ohms law applied to conductive materials do not change just because you want them to.

If you have a really crappy dirty loose greasy crimp on a lug, it will cause problems. Clean up the connection, and the problems will go away. You don't need a 2/0awg wire for a 2awg job. Period.

4 - The OBS family is susceptible to problems caused by deteriorating ground connections. Find every ground connection and take it apart and clean all the surfaces. If the cable strap is cracked or corroded, replace it with a similar gauge of wire. A good ground is a good ground. You don't have to triple the wire size to maintain a good ground. A huge wire connected to a rusty greasy ground will be a crappy ground. Oil and dirt are not conductive, at all.

5 - There is such a thing as a ground loop. If you mod your harness like a madman, and then discover that you have voodoo electrical issues, do some research on a ground loop. Maybe you installed too many ground connections and made a weird loop. It happens.

6 - If your cables are old and the lug crimps are dirty and bent over and stressed out, then they will be more resistive than when new. Fix and clean them, or replace them, but don't think that tripling the cable size will create magic. Your problem was dirty bent-over stressed connections, not too-skinny wire.

7 - If you install a 500 amp alternator and a 2000 watt music amplifier in a car, you will need to use fatter wires everywhere that matters. Common sense.

But if you change a total nominal 50 amp load to a total nominal 60 amp load and you have relays and fuses that can handle startup spikes, you don't have to change any wires.

2 cents.

Thx.

DS

Last edited: