(r1+r2)/4, the myth expelled?

[KeenPensioner]

Hi, I'm not an electrician but a keen amateur. I understand the testing on RFC and how it's done but couldn't get my head round the (r1+r2)/4 calculation. I've trawled this site (and others) and it seems to me that a lot of people struggle to explain this fully. The truth is that this equation only holds up when the CSA of the conductors are equal - which is not the case with the line and cpc. I attach a document that I think explains it reasonably but I'm sure more experienced professionals can comment (which would be appreciated)

 

[Deleted member 26818]

Post #7 you disagree with

Yes.
We are not eliminating R1.

 

[haptism]

Yes.
We are not eliminating R1.

We would be if you need R2

 

[Deleted member 26818]

We would be if you need R2

Not if the CSA or resistance of R1 was different to R2.
If we have a resistance of 1 Ohm for R1 and a resistance of 0.5 Ohms for R2, adding them together and dividing them by 2 would not give us the resistance of R2.

 

[telectrix]

but it's only 3 wires.

 

[haptism]

I do enjoy my Saturdays lol.
r1+r2 is halved cause its half the length (agreement on that), then the formula needs another division of 2 to satisfy (division of 4). So in that case its half the remainder

 

[telectrix]

remainder of what you bin drinking will do.

 

[Deleted member 26818]

I do enjoy my Saturdays lol.
r1+r2 is halved cause its half the length (agreement on that), then the formula needs another division of 2 to satisfy (division of 4). So in that case its half the remainder

Yes, which will give us the R1 + R2.
The second halving is required because we have two conductors in parallel which reduces the resistance by half.

 

[happysteve]

I wrote a Matlab script a while ago to determine the deviation from equality between R1+R2, and r1+r2/4, when the cpc and line conductors are different sizes.

For the different sizes of T&E available, the maximum discrepancy at the very edges of the ring was 20% (with 4+1.5mm cable); for 2.5+1.5mm, the maximum discrepancy was about 6%. If you've got old 2.5+1mm, you might get as much as 18%.

For all practical purposes, it's as near as Phuket is to swearing.

 

[Midwest]

Some very strange things happen on this forum of late. I'm perturbed.

 

[telectrix]

Some very strange things happen on this forum of late. I'm perturbed.

you can get medication for that.

 

[Midwest]

you can get medication for that.

I'm drinking it now.

 

[mart spark]

From my understanding it's r1 +r2/4 because by making the figure of 8 you are effectively halving the length of the conductors and doubling their CSA and when you measure at each socket you are effectively measuring a quarter of the circuit at any point.

 

[Baddegg]

My understanding is....it says it in the book so f&£kit that’s why I do it

 

[KeenPensioner]

Thank you all for taking the time to respond. "happysteve" in post 23 is agreeing with my original post.....ie the formula does through up discrepancies/variations if the live conductors and cpc are of different csa (as his graph clearly shows)

 

[telectrix]

was already strarrrtiiing to see doublle .happpy stevesss graph finnishhed time for bed seddd zebeddeee. boiiing. .