(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)

 

[Baddegg]

Steve’s graph scared me....

 

[happysteve]

Steve’s graph scared me....

Is this one any better? A bit more like a smiley face, now...

 

[Baddegg]

i struggled with it sober @happysteve no chance now

 

[Deleted member 26818]

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%.

View attachment 50886

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

How have you measured R1 + R2, to compare to r1 + r2 / 4?

 

[happysteve]

How have you measured R1 + R2, to compare to r1 + r2 / 4?

The main bit of the code is like this:

    for c=1:length(percentage_round_loop); % c keeps track of which array index we're at,
                                           % as we go round the percentage
                                           % loop
        R_ratio = 1 / csa_ratio; % resistance ratio is the inverse of CSA ratio
        
        % (LS = Line at socket, ES = Earth at socket)
        %     LS        ES
        %     |         |
        %   __|__     __|__
        %   |   |     |   |
        %  _|_ _|_   _|_ _|_
        %  | | | |   | | | |
        %  | | | |   | | | |
        %  |A| |B|   |C| |D|
        %  | | | |   | | | |
        %  |_| |_|   |_| |_|
        %   |   |     |   |
        %   |   |     |   |
        %   L1  L2    E1  E2
        %   |   |     |   |
        %   |   |_____|   |
        %   |_____________|
        %        (CU)
        %
        % ... which is exactly the same as:
        %            _____      _____
        %        ---|__A__|----|__D__|---
        % LS ----|   _____      _____   |---- ES
        %        ---|__B__|----|__C__|---
        
        
        A = percentage_round_loop(c);
        B = max_percentage_around_loop - percentage_round_loop(c); % has to add up to 100%
        C = percentage_round_loop(c) * R_ratio; % CPC is smaller, so higher resistance
        D = (max_percentage_around_loop - percentage_round_loop(c)) * R_ratio;

        scaling_factor = ( max_percentage_around_loop * (1 + R_ratio) ) / 4;
        % scaling_factor normalises the graph, so that if R1+R2 ==
        % (r1+r2)/4, the answer would be 1

        A = A / scaling_factor;
        B = B / scaling_factor;
        C = C / scaling_factor;
        D = D / scaling_factor;
        r1 = A + B; % r1 is your P loop resistance
        r2 = C + D; % r2 is your CPC loop resistance

        R1_plus_R2(c) = ((A+D) * (B+C)) / (A+B+C+D);
        % resistors in parallel: product over sum (resistors in series
        % added, ie (A+D) in parallel with (B+C)
    end

Are we all clear?

 

[Des 56]

Sorry to interrupt
Came here reading this thread looking for simplicity,unfortunately couldn't find it

Best If I be on my way

 

[telectrix]

scaling factors...????? we getting onto kettles now???

 

[Baddegg]

The main bit of the code is like this:

    for c=1:length(percentage_round_loop); % c keeps track of which array index we're at,
                                           % as we go round the percentage
                                           % loop
        R_ratio = 1 / csa_ratio; % resistance ratio is the inverse of CSA ratio
       
        % (LS = Line at socket, ES = Earth at socket)
        %     LS        ES
        %     |         |
        %   __|__     __|__
        %   |   |     |   |
        %  _|_ _|_   _|_ _|_
        %  | | | |   | | | |
        %  | | | |   | | | |
        %  |A| |B|   |C| |D|
        %  | | | |   | | | |
        %  |_| |_|   |_| |_|
        %   |   |     |   |
        %   |   |     |   |
        %   L1  L2    E1  E2
        %   |   |     |   |
        %   |   |_____|   |
        %   |_____________|
        %        (CU)
        %
        % ... which is exactly the same as:
        %            _____      _____
        %        ---|__A__|----|__D__|---
        % LS ----|   _____      _____   |---- ES
        %        ---|__B__|----|__C__|---
       
       
        A = percentage_round_loop(c);
        B = max_percentage_around_loop - percentage_round_loop(c); % has to add up to 100%
        C = percentage_round_loop(c) * R_ratio; % CPC is smaller, so higher resistance
        D = (max_percentage_around_loop - percentage_round_loop(c)) * R_ratio;

        scaling_factor = ( max_percentage_around_loop * (1 + R_ratio) ) / 4;
        % scaling_factor normalises the graph, so that if R1+R2 ==
        % (r1+r2)/4, the answer would be 1

        A = A / scaling_factor;
        B = B / scaling_factor;
        C = C / scaling_factor;
        D = D / scaling_factor;
        r1 = A + B; % r1 is your P loop resistance
        r2 = C + D; % r2 is your CPC loop resistance

        R1_plus_R2(c) = ((A+D) * (B+C)) / (A+B+C+D);
        % resistors in parallel: product over sum (resistors in series
        % added, ie (A+D) in parallel with (B+C)
    end

Are we all clear?

 

[Ian1981]

It’s like people are trying to reinvent the wheel

 

[telectrix]

what's a wheel?

 

[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