Discuss (r1+r2)/4, the myth expelled? in the UK Electrical Forum area at ElectriciansForums.net
Yes, which will give us the R1 + R2.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
you can get medication for that.Some very strange things happen on this forum of late. I'm perturbed.
you can get medication for that.
How have you measured R1 + R2, to compare to r1 + r2 / 4?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?
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
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?
Reply to (r1+r2)/4, the myth expelled? in the UK Electrical Forum area at ElectriciansForums.net
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