Cable sizes

[Theo]

In order to move the new fuse box to the correct place I need to extend the cables. Obviously I need to match the cable sizes when I add the new lengths. Is there a quick way of calculating what is there without counting strands, measuring strand csa and doing the sums.

 

I have a vernier calliper so measuring the total diameter should not be a problem.

 

In hopes

 

Nick

Edited July 30, 2007 by Theo

[chris w]

If you need to extend the cables, then you will need to use larger cross-sectional area cables.

 

If you can let me know the diameter of the cables you have (ie: the diameter of the copper if possible) and their present length and the current they are carrying and the length by which you need to extend them, I will calculate the new cable requirements for you.

 

Chris

Edited July 27, 2007 by chris w

[Theo]

If you need to extend the cables, then you will need to use larger cross-sectional area cables.

 

If you can let me know the diameter of the cables you have (ie: the diameter of the copper if possible) and their present length and the current they are carrying and the length by which you need to extend them, I will calculate the new cable requirements for you.

 

Chris

 

 

Thanks for the offer Chris. I will do some measurements tomorrow. The extra length is going to be in the region of a metre there and back so 2m total. Will that be enough to require an increase in size?

 

Nick

[chris w]

Thanks for the offer Chris. I will do some measurements tomorrow. The extra length is going to be in the region of a metre there and back so 2m total. Will that be enough to require an increase in size?

 

Nick

 

 

It depends on the answer to my questions. It may be that the original cables are marginal in size in which case an increase may be necessary or they may be well within voltage-drop bounds in which case the same size may be OK. Again it will be down to the maths.

 

Chris

[AllanW]

Thanks for the offer Chris. I will do some measurements tomorrow. The extra length is going to be in the region of a metre there and back so 2m total. Will that be enough to require an increase in size?

 

Nick

Hiya Nick...

 

I can see you're being given excellent advice by Chris already - however you may also find these couple of pages interesting (if only to better understand why Chris is saying what he is...)

 

http://www.tb-training.co.uk/MarineE03.html

 

and

 

http://www.reuk.co.uk/AWG-to-Square-mm-Wir...e-Converter.htm

 

Allan

[chris w]

Two points to note on Tony Brooks' site re volts drop.

 

1. The length in his formula is in metres which he states. The Area has to be entered in mm² which is not made clear.

 

2. Although he states that a 1.2v drop doesn't sound like much out of 12v, it is not the voltage which is important in terms of running a 12v device but the POWER it receives. Since power is proportional to voltage² then if the voltage drops by 10% (ie: to 90% as in his example) then the power that the device receives drops to 0.9² or about 80%.

 

 

That's the reason for limiting the volts drop to about 3% (~0.4v). There will also be around a further 0.2v drop due to battery internal resistance and connection resistance so the total volts drop might be 0.6v for the usual lowish amperage devices on board. 0.6v drop from a 12.6v fully charged battery gives about 95% voltage, ie: 90% power at the device.

 

Chris

 

PS: Just to satisfy anyone's curiousity, the 0.0164 figure in the equation comes from the fact that the resistivity of copper is 16.4 x 10^-9 ohm metres (note: ohm metres NOT ohms per metre). Since the length is in metres then, to be able to enter the area in mm², we must insert a correction factor of 10⁶ since 1 metre² = 10⁶ mm. Hence the 16.4 x 10^-9 figure becomes 16.4 x 10^-3 or 0.0164. The resistivity is usually given the greek symbol ρ ("rho"). Most sources quote the actual number as closer to 17 x 10^-9 ohm metres but the difference is academic here.

Edited July 28, 2007 by chris w

[Theo]

Hi there

 

After a good day on Theodora, scraping rust and scale off bits of the engine bay, I did the measuring. Unfortunately I could not measure the diameter of the conductors as that would mean removing them from the present very dodgy fuse holders. Anyway I need to get cable labels from VWP before I start disconnecting things.

 

Anyway the results from the vernier calliper are that the insulation diameter are 3.5mm, 5mm and 5.5mm.

 

I hope that these are recognisable sizes

 

Nick

[chris w]

My "guesstimate" (as obviously I don't know the true insulation thickness) is that the actual cable cross-sectional areas would therefore be:

 

4mm²

10mm²

12.5mm²

 

Clearly, all of these are too large to connect to the VWC 16 way fuse block you mentioned.

 

Assuming for the moment that these areas are correct, then the volts drop per metre per amp calculate out to:

 

4mm²......................... 4.25 millivolts per metre per amp

10mm²....................... 1.7 millivolts per metre per amp

12.5mm²..................... 1.4 millivolts per metre per amp

 

Thus the maximum length for each of the cables (total out & return length) for a volts drop of 0.4v (ie: 3%) is as follows:

 

4mm²...................94 metre amps

 

10mm².................235 metre amps

 

12.5mm²..............294 metre amps

 

eg: if you used the 10mm² cable to carry a current of say 40 amps, the maximum TOTAL length would be 235/40 = 6 metres ie: 3 metres each way for a volts drop of 0.4v.

 

Note that for battery charging, even a drop of 0.4v is undesirable and you would want to use even larger cable.

 

Chris

Edited July 29, 2007 by chris w

[Theo]

My "guesstimate" (as obviously I don't know the true insulation thickness) is that the actual cable cross-sectional areas would therefore be:

 

4mm²

10mm²

12.5mm²

 

Clearly, all of these are too large to connect to the VWC 16 way fuse block you mentioned.

 

Assuming for the moment that these areas are correct, then the volts drop per metre per amp calculate out to:

 

4mm²......................... 4.25 millivolts per metre per amp

10mm²....................... 1.7 millivolts per metre per amp

12.5mm²..................... 1.4 millivolts per metre per amp

 

Thus the maximum length for each of the cables (total out & return length) for a volts drop of 0.4v (ie: 3%) is as follows:

 

4mm²...................94 metre amps

 

10mm².................235 metre amps

 

12.5mm²..............294 metre amps

 

eg: if you used the 10mm² cable to carry a current of say 40 amps, the maximum TOTAL length would be 235/40 = 6 metres ie: 3 metres each way for a volts drop of 0.4v.

 

Note that for battery charging, even a drop of 0.4v is undesirable and you would want to use even larger cable.

 

Chris

 

That will be very useful, Chris. Thanks a lot.

 

 

Would it be reasonable to fit short (100mm or so) tails of smaller cable just to fit in the fuse box? I am happy to do the calculations myself but I rather think that you will know off the top of your head.

 

If, as I suspect, I am having to go for a different fuse box do you have any in mind?

 

TIA

 

Nick

[chris w]

A very short tail won't add anything significant to the volts drop BUT you then come across another issue which is the maximum current carrying capacity of the cable. Normally this factor is of no consequence in 12v systems because the volts drop issue is far more predominant.

 

The fusebox which you are considering will not take larger than 2.5mm² cables. The maximum current carrying capacity of this cable will be in the region of 15A which will be OK for some circuits but not others.

 

Why don't you look at one of the commercially available switch panels which contain most of what you need in terms of all the general circuits. They come with resettable thermal fuses and a whole bank of labels etc.

 

Try here

 

or here

 

Chris

Edited July 29, 2007 by chris w

[Theo]

A very short tail won't add anything significant to the volts drop BUT you then come across another issue which is the maximum current carrying capacity of the cable. Normally this factor is of no consequence in 12v systems because the volts drop issue is far more predominant.

 

The fusebox which you are considering will not take larger than 2.5mm² cables. The maximum current carrying capacity of this cable will be in the region of 15A which will be OK for some circuits but not others.

 

Why don't you look at one of the commercially available switch panels which contain most of what you need in terms of all the general circuits. They come with resettable thermal fuses and a whole bank of labels etc.

 

Try here

 

or here

 

Chris

 

More useful info, thanks.

 

Nick

[ChrisPy]

A very short tail won't add anything significant to the volts drop BUT you then come across another issue which is the maximum current carrying capacity of the cable.

 

agreed, but introducing a tail involves a joint and there will be a slight addition to the voltage drop at every joint, the value of which will depend on how good the joint is.

 

as you say, the fuse box entries should match the cable sizes.

[chris w]

I agree with you Chris. It's not good enginnering practice to use tails.

 

Chris

[T.A]

That will be very useful, Chris. Thanks a lot.

Would it be reasonable to fit short (100mm or so) tails of smaller cable just to fit in the fuse box? I am happy to do the calculations myself but I rather think that you will know off the top of your head.

 

If, as I suspect, I am having to go for a different fuse box do you have any in mind?

 

TIA

 

Nick

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