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

[Ian1981]

what's a wheel?

Before your time Tel.

Edit that makes zero sense

 

[Baddegg]

what's a wheel?

Cars don’t have em up there @telectrix

 

[Ian1981]

It’s what the scoucers replace with bricks

 

[KeenPensioner]

Having been an electrician for 19 years I can categorically say that the formula is correct.
If I have say 0.30 end to end for my line conductor and 0.50 end to end for my cpc’s (2.5mm T+E) then when I measure at each socket outlet with the cross connection method, then I will measure 0.20 ohms for my R1+R2 measurement.
The calculation/formula is indeed correct.

For same size conductors say 2.5mm then my R1+R2 will be half of my measured r1 ,rn and r2 end to ends, in this example 0.15ohms.
It’s a simple resisters in parallel calculation clearly explained in GN3.

We are splitting hairs here maybe I’ll measure 0.21 at some sockets maybe 0.20 at some and maybe 0.19.

Thank you replying. The way I see it is that there are two formula at play here. The recognised formula for resistors in parallel (1/R=1/R1 + 1/R2) and the derived formula for RFCs (R=(r1+r2)/4). Using your own example of 0.3 r1 and 0.5 r2, the recognised formula gives R as 0.225 ohms and the derived formula gives 0.2 ohms (as you say). This difference is negligible, would be within the 0.05 ohm tolerance and ignored in the real world. However the difference is there. If you take a more exaggerated example, say r1=4 and r2 =12 - the recognised formula gives 3 ohms but the derived formula gives 4 ohms - a significant difference. As "happysteve"says in #23 below " (Q) Does R1+R2 always equal (r1+r2)/4? (A) If half way round the circuit, OR CSAs are equal then yes, otherwise.....well not quite but close enough". That was the point if my original post.

 

[Deleted account]

Half way round the circuit is the furthest point. And that is the point we take our read as or the highest recorded.
So from this definition r1+r2/4 is equal to R1+R2

 

[Simon47]

Yes, half way round the ring is the worst case - and that's what matters for fault protection etc. Anywhere else on the ring will have a lower value - as demonstrated by the graph posted (whichever way up you look at it)
The different lines on the graph are irrelevant. The deviations from the simplistic formula only occur away from the centre point of the ring - where the resistance is lower than the worst case anyway.

But for a simple explanation ...
Consider a fault at the midpoint of the ring. From the MCB in the CU to the fault, you have two equal conductors in parallel - each of which is half the length of the ring and hence has half the resitance of the whole ring. So you have two resistors of value r1/2 in parallel, and two equal resistors in parallel give you half the value of each of them. So you have (r1 / 2) / 2, or r1 / 4.
The CPC has the same calculation, so from MET to fault you have r2 / 4.
So total is (r1 / 4) + (r2 / 4), which is better written as (r1 + r2) / 4
Again, this is the worst case, at any other pointbround thecring, the total resistance to the fault will be lower. So you don't need to considervthe effects of different line & CPC sizes etc. This even hold true if the cable CSA changes round the ring IF you consider "middle of the ring" to be in terms of cable resistance rather than distance.
It would be a diffetent case if the CSA variations were "odd" - eg if in one leg you had a larger line, and in the other leg you had a larger CPC. But that would be unusual !

 

[Risteard]

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

That is the correct explanation.

 

[Deleted member 26818]

Technically, by making a figure of 8, you are doubling the length of the conductors and halving the CSA.

 

[static zap]

You are stabilizing the length to be the same from any vantage point, so consider the middle , now thats where double the csa comes form (equal in both directions)