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Old Sep 22, 2006 | 11:56 AM
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Ok, not really FI related, but I doubt I can get an answer in another section.

I know there is someone here who can answer this question. I just can't seem to come to a clear conclusion on my own. I have some ideas, but need some expert input.

Let’s say you wanted to remote mount a battery. You were going to put a distribution block up front and then fuse the battery at the rear. The main concern here is getting the alternator (with the stock regulator) to properly charge the battery in the rear of the car.

You have the choice of:

1. Run a single 1/0 gauge wire from the distribution block back to the positive terminal on the battery (with single 200amp inline fuse).

2. Run two 2 gauge wires from the distribution block back to the positive terminal on the battery (with a 200amp inline fuse on each).

With all else being equal, which option will perform better overall and why?
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Old Sep 22, 2006 | 12:51 PM
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The audio section will have some input on this. I always ran single 1/0 to the back in my Vette because it was more "clean" from a wiring and routing perspective, but 2x 2 gauge should work just fine too.

With the lengths of wire we're talking about here and the current demand characteristics of your car (e.g., no EnglandGreen style subwoofers in the back), I doubt you'll notice a difference between either route. Save for the 2x 2 guage approach may be more easily concealed on the inside trim, and the 1x 1/0 gauge will most certainly require you to route the wire on the outside of the vehicle.
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Old Sep 22, 2006 | 12:52 PM
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Oh, and sorry, just to be clear I am clearly not an expert. I am just sharing my anectdotal experience having done it both ways in other vehicles.
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Old Sep 22, 2006 | 12:53 PM
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From NEC:

Max ampacities based on insulation, 1/0 = 230A and 2 = 170A.

Now direct current resistance, 1/0 = 0.127 ohm/kFT and 2 = 0.201 ohm/kFT

So by running two #2 gauge resistance will be lower then 1/0 do to parallel circuit. This will result in lower voltage drop.


Mike

Edit - Not an expert

Last edited by Skunkworks; Sep 22, 2006 at 01:29 PM.
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Old Sep 22, 2006 | 01:19 PM
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Originally Posted by QuickSilver2002
Ok, not really FI related, but I doubt I can get an answer in another section.

I know there is someone here who can answer this question. I just can't seem to come to a clear conclusion on my own. I have some ideas, but need some expert input.

Let’s say you wanted to remote mount a battery. You were going to put a distribution block up front and then fuse the battery at the rear. The main concern here is getting the alternator (with the stock regulator) to properly charge the battery in the rear of the car.

You have the choice of:

1. Run a single 1/0 gauge wire from the distribution block back to the positive terminal on the battery (with single 200amp inline fuse).

2. Run two 2 gauge wires from the distribution block back to the positive terminal on the battery (with a 200amp inline fuse on each).

With all else being equal, which option will perform better overall and why?
I know you said your main concern was charging the battery, but what about starting the car? Charging current provided to the battery is considerably less than starting current drawn from the battery. Option 1 has a single 200amp fuse, and option 2 has two 200 amp fuses. Charging or starting, current is going through those fuses. So option 1 is limited to 200 amps, while option 2 can deliver 400 amps (based on the fuses).

Admittedly, I have no idea how much current a C5 starter will draw, but I do know that some high torque GM starters draw around 250 amps (I think most fall somewhere between 60-150 amps). Anything near 200 amps may be problematic for option 1.

First thing I would do (and maybe you have already done this) is to determine the current draw for the starter and make sure you size the cabling and fuses accordingly.
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Old Sep 22, 2006 | 02:25 PM
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ok the state of texas calls me an expert on this stuff (texas Electrical engineer)

run the duel #2, lower V drop, increased ampacity. make sure that they are the same leingth and are terminated in exactly the same way at each end. avoid sharp bends in the wire (it seperates the strands and causes overheating of those strands)

the starting curent will be near to over 200 amps and with a single 200a fuse you may find your self stranded with a car that will not start. (carry spares)

More Than Zero
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Old Sep 22, 2006 | 02:33 PM
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Originally Posted by Skunkworks
From NEC:

Max ampacities based on insulation, 1/0 = 230A and 2 = 170A.

Now direct current resistance, 1/0 = 0.127 ohm/kFT and 2 = 0.201 ohm/kFT

So by running two #2 gauge resistance will be lower then 1/0 do to parallel circuit. This will result in lower voltage drop.


Mike

Edit - Not an expert
I must not be calculating right or not understand parallel circuts.

taking the above data and 10ft I get the following resistance in ohms

0ga = 0.00127
2ga = 0.00201

Given a 100amp load on 1 0ga = 100 * .00127 = .127 voltage drop

Given a 100amp load across 2x 2ga = ((100/2) * .00201) * 2 = .201 voltage drop

Is that right, i'm just assuming current would be spit across the two wires.

The other thing I'm not sure about is the resistance exposed by each fuse connection and tap. It seems like those might add up to be an issue in the multi wire approach.

Agree on the comment about the amp load on startup. The 2x wires would support more current and give more fuse options.

Last edited by QuickSilver2002; Sep 22, 2006 at 02:36 PM.
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Old Sep 22, 2006 | 02:37 PM
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Originally Posted by M_T_0
ok the state of texas calls me an expert on this stuff (texas Electrical engineer)

run the duel #2, lower V drop, increased ampacity. make sure that they are the same leingth and are terminated in exactly the same way at each end. avoid sharp bends in the wire (it seperates the strands and causes overheating of those strands)

the starting curent will be near to over 200 amps and with a single 200a fuse you may find your self stranded with a car that will not start. (carry spares)

More Than Zero
Why is the exact same length and termination important. It seems like the load would naturally balance. Sorry, I'm a bit confused here.

Last edited by QuickSilver2002; Sep 22, 2006 at 02:40 PM.
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Old Sep 22, 2006 | 03:06 PM
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Originally Posted by QuickSilver2002
I must not be calculating right or not understand parallel circuts.

taking the above data and 10ft I get the following resistance in ohms

0ga = 0.00127
2ga = 0.00201

Given a 100amp load on 1 0ga = 100 * .00127 = .127 voltage drop

Given a 100amp load across 2x 2ga = ((100/2) * .00201) * 2 = .201 voltage drop

Is that right, i'm just assuming current would be spit across the two wires.

The other thing I'm not sure about is the resistance exposed by each fuse connection and tap. It seems like those might add up to be an issue in the multi wire approach.

Agree on the comment about the amp load on startup. The 2x wires would support more current and give more fuse options.
RT = (R1*R2)/(R1+R2)

RT = Resistance total
R1 = first #2 (calc based on ft)
R2 = second #2 (calc based on ft)

Use RT in dual #2 clac...for v drop.

I'll let MTO handle length and terminations... If not I'll chime in later.


Mike
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Old Sep 22, 2006 | 03:09 PM
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Originally Posted by QuickSilver2002
Ok, not really FI related, but I doubt I can get an answer in another section.

I know there is someone here who can answer this question. I just can't seem to come to a clear conclusion on my own. I have some ideas, but need some expert input.

Let’s say you wanted to remote mount a battery. You were going to put a distribution block up front and then fuse the battery at the rear. The main concern here is getting the alternator (with the stock regulator) to properly charge the battery in the rear of the car.

You have the choice of:

1. Run a single 1/0 gauge wire from the distribution block back to the positive terminal on the battery (with single 200amp inline fuse).

2. Run two 2 gauge wires from the distribution block back to the positive terminal on the battery (with a 200amp inline fuse on each).

With all else being equal, which option will perform better overall and why?

I have had problems with bringing the alt feed to the back of the car. The distance couses charging issues after the car heats up. Leave the alt wire at the distibution block and run a single wire 1/0 to the battery, grounding it to the frame in the rear. I have ran my vert (and several others) like that for years trouble free.
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Old Sep 22, 2006 | 03:11 PM
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Not an EE by any means but those two fuses, if that's the route you choose, will still give you more of a voltage drop than the single 1/0 cable. If it ever gets shorted out, check out the spots where it'll be warmest. It'll be where the biggest resistance is.
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Old Sep 22, 2006 | 03:17 PM
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skunkworks=
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Old Sep 22, 2006 | 03:25 PM
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Originally Posted by QuickSilver2002
I must not be calculating right or not understand parallel circuts.

taking the above data and 10ft I get the following resistance in ohms

0ga = 0.00127
2ga = 0.00201

Given a 100amp load on 1 0ga = 100 * .00127 = .127 voltage drop

Given a 100amp load across 2x 2ga = ((100/2) * .00201) * 2 = .201 voltage drop

Is that right, i'm just assuming current would be spit across the two wires.
Placing the two in parallel effectively cuts the resistance in half (assuming both have the same resistance to begin with). Thus, the effective resistance of two 2ga wires (10 ft) in parallel is 0.001005 ohms. The error in your calculation above is that you are multiplying by 2 for the 2 ga example. Get rid of this and the voltage drop will be .1005 volts.

Originally Posted by QuickSilver2002
Why is the exact same length and termination important. It seems like the load would naturally balance. Sorry, I'm a bit confused here.
I wouldn’t worry about it being “identical”. Just get it as close as you can. Ideally, if the two runs are the same length and have the same termination, they will have the same resistance and therefore share the same current. If one has a slightly higher resistance than the other (either because the lengths and/or terminations are different), more current will flow through the one with the lower resistance (and thus not be balanced).

Last edited by NigelTufnel; Sep 22, 2006 at 03:31 PM.
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Old Sep 22, 2006 | 03:28 PM
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Originally Posted by NICK YOSKIN
skunkworks=

Hey not niece… I drink beer, go to strip clubs and on occasions scratch my a$$.

Put two equal resistances in parallel and it gets cut in half… there are you happy now.


Mike

Edit answered above.

Last edited by Skunkworks; Sep 22, 2006 at 03:31 PM.
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Old Sep 22, 2006 | 04:37 PM
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Ok, got it, I knew I was screwing something up in my calculation. So the voltage drops are not additive across the two wires (I should have known that).

Now the question is how much resistance is introduced by the fuses and terminals.

I think I'm going to go with the two wire approach and see what happens. It is a heck of a lot eaiser to work with and conceal.
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Old Sep 22, 2006 | 05:05 PM
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Just curious... What would the voltage drop be for a single 1/0 run?
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Old Sep 22, 2006 | 05:17 PM
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Originally Posted by QuickSilver2002
Ok, got it, I knew I was screwing something up in my calculation. So the voltage drops are not additive across the two wires (I should have known that).

Now the question is how much resistance is introduced by the fuses and terminals.

I think I'm going to go with the two wire approach and see what happens. It is a heck of a lot eaiser to work with and conceal.
Same rule applies for parallel fuse blocks, so I wouldn’t worry about.

Also you need a very good meter to read very low resistance, but as a guess 0.01ish. Now if you do have problems and you can see fuse element, the one that really vaporized is not the culprit… or it had to absorb full load after first one let go.

I would personally opt for two #2 myself do to ease in working with it.

The clamping or terminations is what I would concentrate on more so then fuse block resistance. You do want to have balanced loads as mentioned above.

Maybe you can do 1 fuse block as apposed to 2, although I haven’t looked into fuses in some time so don’t know what’s currently available in small and inexpensive form factor.


Mike

Last edited by Skunkworks; Sep 22, 2006 at 05:31 PM.
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Old Sep 22, 2006 | 05:26 PM
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Originally Posted by QuickSilver2002
I must not be calculating right or not understand parallel circuts.

taking the above data and 10ft I get the following resistance in ohms

0ga = 0.00127
2ga = 0.00201

Given a 100amp load on 1 0ga = 100 * .00127 = .127 voltage drop

Given a 100amp load across 2x 2ga = ((100/2) * .00201) * 2 = .201 voltage drop

Is that right, i'm just assuming current would be spit across the two wires.

The other thing I'm not sure about is the resistance exposed by each fuse connection and tap. It seems like those might add up to be an issue in the multi wire approach.

Agree on the comment about the amp load on startup. The 2x wires would support more current and give more fuse options.
the par. resistance = r1*r2/r1+r2 .00201x.00201/.00402=.001005

for 2 equal resistances it is equal to 1/2 the resistance

vd = 100a*.001005= .1005v

More Than Zero
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Old Sep 22, 2006 | 05:33 PM
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Originally Posted by diynoob
Just curious... What would the voltage drop be for a single 1/0 run?
Quick had 1/0 calc right.
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Old Sep 22, 2006 | 05:35 PM
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Originally Posted by QuickSilver2002
Why is the exact same length and termination important. It seems like the load would naturally balance. Sorry, I'm a bit confused here.
the problem is that the resistance of the terminations can be much greater than the resistance of the wire and unbalance the current flow. this may cause 1 fuse to blow, and this will lead to the second fuse overloading, this all depends on a lot of things so the simple way to avoid problems is to make an effort to keep them the same (as much as possable).

I have seen large ac systems show as much as 60% diff. in badly done par. circuits.

More Than Zero
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