Do batteries need to be next to each other?

ghostlymoron

ghostlymoron

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With 6v traction batteries wired in series, for convenience one is located by the companionway and one in the forepeak. Connected by 35mm2 cables. Will this cause any problems with charging?
 
Graham_Wright said:
35/3.142 =11.14
Square root of 11.14 = 3.33 (radius)
Diameter = 6.6

Sounds a bit thin to me! (Or are my sums wrong?)
Click to expand...

Sums look correct to me, but the result feels a little thin. Possibly the difference between solid conductor and stranded?

Just looked up the spec of the stuff I buy and the overall diameter (including the insulation) is 11.8mm

Pete
 
Graham_Wright said:
35/3.142 =11.14
Square root of 11.14 = 3.33 (radius)
Diameter = 6.6

Sounds a bit thin to me! (Or are my sums wrong?)
Click to expand...

Your calculation is for a solid conductor. Multistrand flexible cable is less densely packed and has air voids inside it, so the apparent conductor diameter would be greater than 6.6mm.

Stranded_and_solid_core_cable.png
 
VicS said:
For a small boat which is not likely to have high current accessories such as a bow thruster or electric anchor winch etc. it is certainly generous, for what is presumably just the domestic battery, but far better than being under sized.
Click to expand...
The boat in question does have a electric windlass but a friend was concerned about voltage drop between the 6v batteries when charging.
 
ghostlymoron said:
The boat in question does have a electric windlass but a friend was concerned about voltage drop between the 6v batteries when charging.
Click to expand...

Relevant but perhaps not as important as the volts drop when the windlass is in operation.

Determine the correct cable size from the highest current that the 2x6volt battery will supply, or be charged at, and the cable run lengths.
 
ghostlymoron said:
The boat in question does have a electric windlass but a friend was concerned about voltage drop between the 6v batteries when charging.
Click to expand...

Charging current is likely to be pretty low (say 10 or 20 amps) so voltage drop will not be too serious, but even a small 600 watt windlass draws 50 amps on full load so the voltage drop does become a real issue. It would be far better to get both batterieds into the forepeak closest to the heaviest load and just use the 35mm cables to take the charging current to the batteries and to supply midships ordinary domestic loads from them.

EDIT Conventional wisdom says put batteries close to the alternator, but the alternator is only going to be supplying full current if the batteries are pretty discharged, whereas the windlass draws a lot every time you press the switch. If using a mains battery charger the current will be low and the charger can be put near the batteries anyway. There are formulae for voltage drop per metre of cable, according to load, so its important to know what the maximum current is and how long the cables are. I once found a good calculator tool on the web, but have now lost the link.
 
Last edited:
xeitosaphil said:
Link to voltage loss Calculator as long as the amps and cable length is not too extreme

http://www.solar-wind.co.uk/cable-sizing-DC-cables.html
Click to expand...
Iv'e just posted this on another thread:

The problem with these online calculators is they don't make it clear if you take the single length or the length there and back to take into account the losses in both the positive and negatives cables. All table calculators tell you to double the lengths, these online calculators take that into account, but don't tell you.

Try this one from BlueSeas that does tell you to double the length and offers more options. They also do a much better version as an App.

http://circuitwizard.bluesea.com/
 
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