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

[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. .

 

[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?