Discuss Continuity of ring final in the Periodic Inspection Reporting & Certification area at ElectriciansForums.net

Welcome to ElectriciansForums.net - The American Electrical Advice Forum
Head straight to the main forums to chat by click here:   American Electrical Advice Forum

Hi guys I am new and struggeling. Can any one pleas tel me why r1+r2 have to be divided by 4 pleas

It is the product of a sum. You have your end to end resistance of r1 + r2 connected together, and then have to consider parallel resistances. It is well explained in GN3, a lot better than i can explain in words.

And here is another tip for ring continuity. If you can't identify visually the correct ends to join up to measure R1+R2, then connect any 2 opposing ends, and if you have got it wrong it will read half of your expected R1+R2. There reason behind this, is that instead of connecting 4 conductors, you are only connecting 2, draw a picture and you will see what i mean.

Cheers…………..Howard
 
Ruthless Murdoch -- You can draw that, sit back and think about it for hours and still not understanding it!

I agree, and this is precisely what I did last night. I spent a good few hours, and many sheets of paper trying to mathematically prove it was always true, and I couldn't. As I finally nodded off, I realised why:

It's not always true.

The IET are pulling a fast one.

Folks, R1+R2 does NOT always equal (r1+r2)/4.

It does in two situations:

(1) The CSA of your CPC and P are the same (eg 1/1, 2.5/2.5). If this is the case, doesn't matter where around the ring you measure it, R1+R2 always equals (r1+r2)/4.

(2) You measure R1+R2 exactly half way around the ring. If this is the case, doesn't matter what the CSA ratio is (1.5/1, 6/2.5 etc), R1+R2 always equals (r1+r2)/4.

However, if you're using summat like 4/1.5 and you measure at a socket close to the CU, then R1+R2 will be less than (r1+r2)/4.

To illustrate it, I wrote a program in Matlab, and got it to print a graph. Here it is:

Continuity of ring final r1_plus_r2_over_4 - EletriciansForums.net
This shows the ratio of (R1+R1) : (r1+r2)/4, for different testing points around the ring (expressed as "percentage around ring" ie 50% is exactly half way round). The flat black line at the top is the case where the CSA of CPC and P is the same - this is a flat line at height 1, showing that R1+R2 = (r1+r2)/4. All the other coloured lines are for different CSA combinations. Think of your height as "how close to correct" you are: 1 = "spot on", 0.8 = "20% error".

Note that the very lowest number on the height scale, for the cable with the most extreme ratio of CSA, is about 0.8, when your socket is VERY close to the CU.

So although R1+R2 does not stictly always equal (r1+r2)/4:
- for some cases it does (equal CSAs, or half way round the ring), and
- for other cases, it's pretty close.

Every day's a school day :)

(I appreciate this doesn't answer the OP's question - I can write an explanation for why (roughly) it works if need be... I'm sure others could too if they want to help him out, to be honest I'm pretty kn***ered!)
 
Liking your work there Steve!
I once worked it out at the connected ends and mid point to see the difference.
No way was I going to type it all out on here though!
 
I agree, and this is precisely what I did last night. I spent a good few hours, and many sheets of paper trying to mathematically prove it was always true, and I couldn't. As I finally nodded off, I realised why:

It's not always true.

The IET are pulling a fast one.

Folks, R1+R2 does NOT always equal (r1+r2)/4.

It does in two situations:

(1) The CSA of your CPC and P are the same (eg 1/1, 2.5/2.5). If this is the case, doesn't matter where around the ring you measure it, R1+R2 always equals (r1+r2)/4.

(2) You measure R1+R2 exactly half way around the ring. If this is the case, doesn't matter what the CSA ratio is (1.5/1, 6/2.5 etc), R1+R2 always equals (r1+r2)/4.

However, if you're using summat like 4/1.5 and you measure at a socket close to the CU, then R1+R2 will be less than (r1+r2)/4.

To illustrate it, I wrote a program in Matlab, and got it to print a graph. Here it is:

View attachment 21401
This shows the ratio of (R1+R1) : (r1+r2)/4, for different testing points around the ring (expressed as "percentage around ring" ie 50% is exactly half way round). The flat black line at the top is the case where the CSA of CPC and P is the same - this is a flat line at height 1, showing that R1+R2 = (r1+r2)/4. All the other coloured lines are for different CSA combinations. Think of your height as "how close to correct" you are: 1 = "spot on", 0.8 = "20% error".

Note that the very lowest number on the height scale, for the cable with the most extreme ratio of CSA, is about 0.8, when your socket is VERY close to the CU.

So although R1+R2 does not stictly always equal (r1+r2)/4:
- for some cases it does (equal CSAs, or half way round the ring), and
- for other cases, it's pretty close.

Every day's a school day :)

(I appreciate this doesn't answer the OP's question - I can write an explanation for why (roughly) it works if need be... I'm sure others could too if they want to help him out, to be honest I'm pretty kn***ered!)

nice diagram there. you almost got a full rainbow. all we need now is zippy ( geordie spark ) and bungle ( glennspark ) to complete .
 
The 'easy' answer to the OP's question is because (at the midpoint of) the conductor -

R1 (or Rn or R2) path is half the length - therefore half the resistance
on double the csa - therefore half the resistance
of r1 (or rn or r2).

Half of a half is one quarter.
 
or double the csa to divide by 2 as you basically have 2 conductors in parallel, and divide by 2 again as R1+R2 is mid-point. ( same result at every point if cross connected legs).
 
Assume conductors of the same CSA for a second. Measure r1 end to end (leg1 to leg 2) then add this to r2 end to end measurement. Now if you cross connect you are in effect halving the loop size (half the resistance) and also doubling the CSA of the conductors as they have twice as much copper for a current to flow down ( remember twice the CSA of a conductor will equal HALF the resistance) .....so you have half the loop resistance x half the resistance due to DOUBLING the combined CSA.....so a 1/2 x 1/2= 1/4 of the r1(end to end measurement) + r2 (end to end measurement).
 
Last edited:
I reckon you just have to accept somethings as just being correct and not question the logic/calcs behind it -- like Einstein's theory of relativity. I tried once many, many years ago but its long well forgotten, so I just settle for you divide by 4!
 
Nice work happysteve! Hope the OP appreciates your efforts.

Have just spent ages typing a basic explanation and worked example for the OP, but lost it all due to internet connection crash.

May re-do later.
 
Following on from the work of happysteve and others:-

OP,

Treat the ‘divide by 4’ as a quick piece of on-site mental arithmetic to prove correlation between your measured values of r1 & r2 and your measured value of R1+R2. The alternative, of course, is to check accuracy with the formula for resistances in parallel.

As for the question “Why divide by 4?”, the following seems a good way to explain:-

r1 is Line end to end, so each Line leg to the mid point would be half r1.
r2 is CPC end to end, so each CPC leg to the mid point would be half r2.
Current would flow in both Line legs, so you have two Line resistances in parallel.
Earth fault current would flow in both CPC legs, so you have two CPC resistances in parallel.

As a worked example:-

Say 2.5mm T&E with r1 of 0.40 ohms & r2 of 0.67 ohms.

Note 1

Line legs to mid point would be 0.20 & 0.20.
CPC legs to mid point would be 0.34 & 0.34 (with rounding up).

Note 2

Parallel resistance of the two Line legs gives us 0.10.
Parallel resistance of the two CPC legs gives us 0.17.


During Note 1 we halved the r1 & r2 values.
Then we halved again in Note 2 for parallel resistances.
½ x ½ equals ¼ (or 0.25 if you prefer).
Four quarters equals a whole.
Hence the (r1+r2) ‘divided by 4’ to check against our measured (R1+R2) in the cross connected test.

So, (R1+R2) = (0.10+0.17) = 0.27 ohms.
Then, as a quick check, original (r1+r2) divided by 4 = 0.27 ohms.
And to work the result backwards, 0.27 multiplied by 4 = 1.08 which is the r1+r2 original total value.

This seems to be a logical way to interpret the ‘divide by 4’ rule and achieve some clarity. Works for me anyway!!
 
too much time on your hands. i'd start by typing the first line then give up. :behead:
 
too much time on your hands. i'd start by typing the first line then give up. :behead:


I did. Gave up twice last night. Changed my mind and re-started today. Promptly lost it all somehow. Then did it again from scratch. Hope it's accurate cos I was losing the will to live.

That said, I'd do anything for a 'like'! Well, almost anything.
 
OK . given you a "like". now can we get back to silly inane posts?
 
steady on archie. he's from kent. they all swell heads down there. ever since the battle of britain was fought over their turf.
 

Reply to Continuity of ring final in the Periodic Inspection Reporting & Certification area at ElectriciansForums.net

OFFICIAL SPONSORS

Electrical Goods - Electrical Tools - Brand Names Electrician Courses Green Electrical Goods PCB Way Electrical Goods - Electrical Tools - Brand Names Pushfit Wire Connectors Electric Underfloor Heating Electrician Courses
These Official Forum Sponsors May Provide Discounts to Regular Forum Members - If you would like to sponsor us then CLICK HERE and post a thread with who you are, and we'll send you some stats etc
This website was designed, optimised and is hosted by Untold Media. Operating under the name Untold Media since 2001.
Back
Top
AdBlock Detected

We get it, advertisements are annoying!

Sure, ad-blocking software does a great job at blocking ads, but it also blocks useful features of our website. For the best site experience please disable your AdBlocker.

I've Disabled AdBlock