apt100 wrote:

^^ the cross sectional area (eg 40mm2 or 35mm2 or whatever) is the area of the copper as this is what carries the current. It doesn't include the insulation

and as you say it is just the area of a circle, so area= pi x radius x radius (and radius is half the diameter)

so, as above... 40mm diameter is 20mm radius, so area = pi x 20 x 20 = 3andabit x 400 = 1200mm2 andabit

edit: my reply was to steve

blackwolf wrote:

apt100 wrote:

^^ the cross sectional area (eg 40mm2 or 35mm2 or whatever) is the area of the copper as this is what carries the current. It doesn't include the insulation

and as you say it is just the area of a circle, so area= pi x radius x radius (and radius is half the diameter)

so, as above... 40mm diameter is 20mm radius, so area = pi x 20 x 20 = 3andabit x 400 = 1200mm2 andabit

edit: my reply was to steve

Whilst in practical terms this it true there is another factor which in critical applications may be significant (not in ours, though) which is that the number of strands has a bearing on the CSA as well.

If you have a single strand of copper which is 10mm diameter, the CSA will be pi times the square of the radius using the traditional simple formula. If you have a cable made up of, say, six smaller strands of copper laid up in a way where the overall outside diameter of the bunched conductors is still 10mm, the CSA will actually be less since you can never lay up circular conductors without gaps between them. So the downside is a reduction in CSA for the same overall size, but the upside is flexibility. As the individual conductors get smaller, so does the difference in CSA since the gaps get smaller.

On very heavy duty power cables etc you will find that the individual conducturs are not circular in cross section but shaped to minimise the inter-conductor gap.

Incidentally it is the gaps between the individual strands of copper in the injector harness of the TD5 that provides the conduit for the passage of oil from the engine to the ECU. If the conductors were solid, this phenomenon would not happen.

Indeed. Quoted CSA are usually nominal, and depending upon the stranding the actual area will vary.
This table shows an example (it is American based, because I have easy access to it at home, but the principle is the same)



These wires are all nominal 8 AWG.
The first row is a solid conductor. Diameter is approx. 3.3mm and the area of the copper is approx. 8.4mm2. (This roughly matches the simple A=pi r2).
Row 3 shows a course stranded wire made up of 19 strands of wire, each of which are 21 AWG.
The diameter of the wire (excluding insulation) now goes up to 3.6mm, but the copper area comes down, to 7.8mm2.

In this table the true areas vary from 7.82mm2 to 8.63mm2, but they are all the same nominal size.

This was carefully calibrated for in my original formula by varying the "andabit"

A practical consideration is that when crimping lugs to wire, the datasheet will often state that for finely stranded wires, such as welding cables, the next size up terminal might be needed to get the wire to fit.