View the thread, titled "Osg not to hand" which is posted in UK Electrical Forum on Electricians Forums.

If indeed I am talking with senior members who are the wisest of wise, is this your legacy? to cynically scroll this forum looking for confrontation?! I'm not looking for arguments over the internet, come on this is insane; I haven't gone out either to create any animosity. I genuinely needed to know two figures, nothing else. I would imagine either of you handing over your OSG and letting me get what figures I needed if it were face to face, why does the internet make demons of people.
 
If indeed I am talking with senior members who are the wisest of wise, is this your legacy? to cynically scroll this forum looking for confrontation?! I'm not looking for arguments over the internet, come on this is insane; I haven't gone out either to create any animosity. I genuinely needed to know two figures, nothing else. I would imagine either of you handing over your OSG and letting me get what figures I needed if it were face to face, why does the internet make demons of people.
You have the answer, post 2
 
In my reply (#20) I (a) missed the fact that you were asking for R1+R2 (my bad) and (b) (perhaps incorrectly) assumed that you didn't realise that mV/A/m == mOhm/m. If you already knew this, then apologies. If you didn't, you wouldn't be the first.

Now, as regards the question, "Is table I1 of the OSG [resistance/m] in the BGB?" then the answer is no, not directly, and not as simply presented as it is in the OSG. You could work it out from first principles using (e.g.) table 4D1B (p. 333) and the information given in the "NOTE" of BGB Appendix 4, section 6.1 (p. 313) to apply the correction factor between maximum conductor operating temperature (which 4D1B applies to) and ambient temperature at which you'd measure the mOhm/m (you could try using equation 6 on the same page if you felt some self flagellation was required, but 0.004/C is fine); not forgetting of course that the voltage drop/A is twice the resistance because you've got a resistance on both the L and N.

Or, to put it more simply: Take the numbers in mV/A/m in 4D1B, divide by 2, then divide by 1.2, to get resistance in mOhm/m. 1.2 comes from 1+[(70-20)*0.004] (and matches up with the number in the bottom left of table I3 in the OSG).

So I think your question "anyone got the OSG to hand?" is probably a fair one.
 
In my reply (#20) I (a) missed the fact that you were asking for R1+R2 (my bad) and (b) (perhaps incorrectly) assumed that you didn't realise that mV/A/m == mOhm/m. If you already knew this, then apologies. If you didn't, you wouldn't be the first.

Now, as regards the question, "Is table I1 of the OSG [resistance/m] in the BGB?" then the answer is no, not directly, and not as simply presented as it is in the OSG. You could work it out from first principles using (e.g.) table 4D1B (p. 333) and the information given in the "NOTE" of BGB Appendix 4, section 6.1 (p. 313) to apply the correction factor between maximum conductor operating temperature (which 4D1B applies to) and ambient temperature at which you'd measure the mOhm/m (you could try using equation 6 on the same page if you felt some self flagellation was required, but 0.004/C is fine); not forgetting of course that the voltage drop/A is twice the resistance because you've got a resistance on both the L and N.

Or, to put it more simply: Take the numbers in mV/A/m in 4D1B, divide by 2, then divide by 1.2, to get resistance in mOhm/m. 1.2 comes from 1+[(70-20)*0.004] (and matches up with the number in the bottom left of table I3 in the OSG).

So I think your question "anyone got the OSG to hand?" is probably a fair one.

Many thanks for the insight, I thought as much the answer would be there cryptically, but thank you for taking your time to divulge.
 
No worries. :) I always wondered where the numbers in that table came from. Since I'd piped up earlier (misunderstanding your question as I skim-read your first post) I thought I'd better work it out.

Every day's a school day, n all that.

(I could be wrong, of course...) :)
 

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