# GN8 Chapter 6 Extraneous-Conductive-Parts and their connections Equation 6.1

In Guidence Note 8 where the equation 6.1 is used to illustrate the range of values of resistance RCP for the currents IB of 0.5mA, 10mA & 30mA in equations 6.2, 6.3 & 6.4, what is the second value of 1000 thats deducted from the value of RCP for? Looking at equation 6.1 the first value of 1000 thats deducted is for the impedance of the human body ZT taken to be 1000Ω from BS IEC 60479-1:2018 for the purpose of this calculation, but I'm struggling for this moment to understand what the subtraction of the second value of 1000 is for?

It looks to me like the equations have been incorrectly printed and the sum of each should be 459,000Ω, 22,000Ω & 6670Ω respectively and the second value of 1000 should be dividing the sums by 1000 to convert the values to KΩ?

If this is the case then these equations have been miss printed and miss represented in GN8 for the last 20+ years! Not to mention the publications and documents that they've been copied across to over these years...

Any thoughts on this would be much appreciated...

Parents
• should be dividing the sums by 1000 to convert the values to KΩ?

Not really. 459,000Ω is exactly the same as 459kΩ - mathematically there's no need to divide by 1000. The k prefix itself means x1000.

Strictly speaking, In the metric/SI system the there's only one "unit" for a dimension (in this case Ω for resistance)  Unlike the imperial system where you might divide by 12 to convert between feet and inches, in metric there is only the metre - so really the prefixes are just to 'scale' the number to something more convenient as your write them down (or display them).

- Andy.

• There are always exceptions to the rule.

Whilst it is true that, according to SI units, 1 km2 = (1,000m)2 = 1,000,000 m2, this would lead us to believe that 100,000 A2s ≠ 100 kA2s ... but when reading standards and data sheets of let-through energy of protective devices, you would be a factor of 1000 out if you assumed SI rules are applied ... it seems the "great and the good" of our industry are happy to consider [A2s] as a 'derived unit' and hence use  1 kA2s = 1 k[A2s]

Confused yet?

• There are always exceptions to the rule.

Indeed - probably the biggest blunder is that the SI system uses kg rather than g as a base unit for mass (or perhaps that the metric system adopted a unit that was 1000x too small for consistency). If we knew then what we know now things would certainly have been done differently. But for resistance at least it should be fairly straight forward.

- Andy.

• hence use  1 kA2s = 1 k[A2s]

The unit could be Amp-squared-kiloseconds [A² ks].

Note that there should be a space between the different fundamental (or derived) units so the let-through energy is measured in [A² s] or indeed, [s A²] but not [A²s].

• really its just as valid expressed in joules per ohm of course, then the problem of the meaning of the k being squared or not disappears.

And that has the advantage of really makes it look like a let through  energy as well.. But text books seem frightened by joules for some reason.

I do sometimes wonder if things are written in obscure units so the authors can make something simple look more complex that it is.

Mike.

• really its just as valid expressed in joules per ohm of course, then the problem of the meaning of the k being squared or not disappears.

And that has the advantage of really makes it look like a let through  energy as well.

Agreed - it does look more like energy, but of course it is not energy at all. Joules and Ohms are derived units, so that is probably why A² s is preferred.

1 J = 1 kg m² s‾² and 1 Ω = 1 kg m² s‾³ A‾². Therefore to get to 1 J Ω‾¹ the kg and m cancel out and we are left with A² s.

• Whilst it is true that, according to SI units, 1 km2 = (1,000m)2 = 1,000,000 m2,

On reflection, I think that these distances must be the anomaly. Along similar lines, strictly, one square millimetre (1 mm²) should be 1μm².

Once again, the problem can go away with derived units. So I can fairly easily exert a force of 25 N with my finger tip, which probably has an area of 50 mm². This gives a pressure of 0.5 MPa, which sounds a lot, but is approximately 70 psi.

• I'm not sure I agree - 1 mm² is really short for 1 (mm)² if you like -a square of 1mm by 1mm . As one who occasionally also dabbles in square microns when working with the chip people, 1μm².  is a square of 1μm.  by 1μm.

A Square amp is an oddity as it has no physical manifestation - while an current density of one amp per square metre certainly does as does an H field of one amp per metre.

I agree Pascals are an unnaturally sized unit, but in many ways so is the newton, A force that made g one unit per unit mass would have been helpful.  I will admit to using PSI for real engineering pressures as it is easier to handle the numbers.

Mike