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remainder of what you bin drinking will do.
 
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.
 
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%.

(r1+r2)/4, the myth expelled? r1_plus_r2_over_4 - EletriciansForums.net

For all practical purposes, it's as near as Phuket is to swearing.
 
was already strarrrtiiing to see doublle .happpy stevesss graph finnishhed time for bed seddd zebeddeee. boiiing. .
 
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?
 
How have you measured R1 + R2, to compare to r1 + r2 / 4?

The main bit of the code is like this:

Code:
    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? :)
 
The main bit of the code is like this:

Code:
    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? :)
:oops:
 

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