OT(ish) - cross sectional area of round wire.

I am sizing up some wire to connect two 12V leisure batteries together, also teminals, fuse holder etc.

I noted that the wire thicknesses were quoted in mm^2 so I went looking for an explanation on how to relate wire thickness (diameter) to mm^2.

Found this:

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said, amongst other things "Here is another example. Assume that a conductor is 3/8 inch thick and 4 inches wide. The 3/8 inch can be expressed in decimal form as 0.375 inch. Since 1 mil equals 0.001 inch, the thickness of the conductor will be 0.001 ´ 0.375, or 375 mils. Since the width is 4 inches and there are 1,000 mils per inch, the width will be 4 ´ 1,000, or 4,000 mils. "

Now is the mil completely different from the mm? Or is something horribly wrong here? Looks like the terminology here is mil = a thousandth of an inch.

Given that the UK spoken shorthand for a millimetre is a mil (as in 10 mil spanner) there is enormous scope for confusion.

You say tomayto and I say tomarto?

Now trying to estimate the thickness of the conductor if the conductor plus sheathing is about 6mm.

Cheers

Dave R

Reply to
David WE Roberts
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Yeah well, going back to basics ..... !

area of a circle is pi * r^2 (where r = half of the diameter)

pi = 3.14159

r^2 = r x r

so the cross sectional area of your wire is 3.14159 x d/2 x d/2

where d is the diameter in mm of the wire

the result is in mm^2

Reply to
Phil Jessop

To be fair the article does define a mil where there are 1,000 to the inch. It's an American and not a European measurement. Here, as you rightly say, mils generally refer to mm, but would still be written as mm.

Reply to
Fredxx

Yeah, can do that and have also found a web site with a calculator.

The important part of the phrase was "plus sheathing". I can calculate the cross sectional area of the (wire+sheathing) but this doesn't give me the area of the conductor, but I don't want to disassemble the current installation and strip back the insulation to get the true cross section.

I guess if I assume that the sheathing is 5% of the diameter of the wire that should be close enough.

Cheers

Dave R

Reply to
David WE Roberts

Or pi * d^2 / 4

which is what you've described here.

Assuming you start in mm in the first place.

That website above is, I guess, from the other side of the pond... They seem to use the term 'mil' or 'mills' to represent thousandth of an inch.

It's not one I'm familair with, but maybe that's how it's done over there.

Gordon

Reply to
Gordon Henderson

Correct, in USA mechanical slang the mil is one thousandth of an inch. Other abbreviations include point and thou (and the thou is increasingly being used in the USA to reduce confusion).

Oh yes, as the Hubble telescope found out.

Reply to
Peter Parry

Never mind. Just have a pint of beer instead...

Reply to
Ron Lowe

There really is no safe way of knowing without - the sheath thickness can vary quite significantly depending on the wire in question.

Reply to
John Rumm

Depends what the cable was orgininally intended for. A wire designed for use on low voltages the insulation may only be 0.5mm or less thick. For mains maybe 2mm, high voltages 5mm or more. Thickness of the insulator will also depend a bit on what that material is, though these days it's nearly always PVC or similar.

The only real way of knowing the conductor size is to measure it.

Reply to
Dave Liquorice

Assuming 5% would leave you miles out. Some wire types have common diameters, though they can vary.

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?title=3DflexNT

Reply to
Tabby

Not just in the USA. My father went through an apprenticeship and on upwards here in the UK and mil was (and in some areas still is) in perfectly common UK useage.

SteveW

Reply to
Steve Walker

It was used for many years by the GPO for adjustment and specification of relay and electromagnet springs and contact spacings in telephone and telex exchange (as well as some subscribers' apparatus)..

Spring pressures were measured in grams and spacial dimensions in mils (thousandths of an inch) for literally billions of relay-type stuiff.

Mind you - size of wire used in cable and jumper wire was expressed in pounds (weight) per mile (of copper).

I should imagine that the reason for tis was that it was easier for a local GPO stores to determine how much wire was on a particular hank by weighing it than by flaking it all out to measure its length.

6¼ lb equated more or less to 25 swg
Reply to
Frank Erskine

You would have a hard time guessing conductor size from od of insulation. for instance auto jumper leads often cheat with very thick insulation to fool you into thinking that they are heavy duty.

Reply to
F Murtz

Just to restate - this is for linking 12V leisure batteries together so it is similar to the cable used in jump leads. In fact, I am tempted to use some old jump leads where the crocodile clips are broken. However the cable quality in the cheaper jump leads is sometimes suspect so this could be false economy. Whatever, the cautious method of sizing the conductor is to pretend the sheathing is conductor also and thus get a slightly oversize cable. This would give a 28mm^2 conductor for a 6mm diameter cable. [ According to

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] From the Durite catalogue the nearest flexible starter cable is 25mm^2 to carry 170 amp.

However given that the earth lead from the battery to the chassis is protected by a 50 amp fuse (the 50 amp Maxi blade fuse in the Durite catalogue is rated 50 amp continuous 100 amp blow) 170 amp cable seems a bit over the top.

20mm^2/135 amp seems reasonable for 50 amp continuous (not that I'm likely to be pulling anything like 600 watts) and a slightly slimmer cable will be easier to route.

The existing cable does look remarkably slim compared to the usual automotive starter cable - will check the external diameter in the car - so perhaps I should be working back from the load the fuse will support instead of trying to replicate the existing cable.

My original problem remains - I can't measure the conductor diameter on a cable which is fully sheathed and with an integral ring connector at the end also covered with insulation without removing some insulation and thus damaging the existing cable. Which I obviously don't want to do.

Anyway, the discussion is helping me to get my head around the issues, so thanks to all.

Cheers

Dave R

Reply to
David WE Roberts

I've just confused myself again. I assume that the rating shown on a fuse is the continuous load and not the blow rating.

Reply to
David WE Roberts

At low voltages loses in the cable become significant. V = I * R.

Say you are drawing 50A through the 170A rated cable which, for the sake of argument, has a resistance of 0.01 ohm. That gives a volt drop of:

50 * 0.01 = 0.5V (4% drop) device gets 11.5V not 12.

Lets assume that the 50A rated cable has three times the resistance:

50 x 0.03 = 1.5V (12.5% drop) device gets 10.5V not 12.

The more current you draw the worse the volt drop in the cables becomes.

This is LV stick a pin in it and measure how far the pin goes in. The hole won't be a problem unless the cable is in a damp location. If you are worried wrap a bit of self amalgamting tape around where the pin went in.

Reply to
Dave Liquorice

Welding cable is often a good choice for battery linking.

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you need a more flexible cable the stuff used for arc welding torches is many strands of fine wire and has both low resistance and flexibility, eg

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Reply to
Peter Parry

Actually, no. Hubble's problem wasn't units.

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're thinking of Mars Climate Orbiter.

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Reply to
Andy Champ

Cables often have thinner insulation if they use something better than PVC. Raychem's PVDF cables are a case in point. Tiny overall diameter, cost a packet and will run at 140 deg C for years.

I'd never guess what was inside a cable. A vernier caliper is about 10 quid from one of the cheapy supermarkets/ebay

Reply to
The Other Mike

Use a stanley knife to cut two rings round the cable an inch or so apart. Make one cut longways and remove the insulation. Untwist the lay of the cable to enable a check of individual stand size with vernier caliper. Count number of strands.

Replace removed insulation. Cover with an overlap of half an inch back to good insulation with adhesive lined heatshrink or self amagamating tape.

Then look up the cable size online or report back here with the dimensions

Reply to
The Other Mike

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