Continuity of ring final

[Hunter3006]

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

 

[Octopus]

Take a piece of paper, draw what you describe, sit back and think about it.

 

[Badged01]

Take a piece of paper, draw what you describe, sit back and think about it.

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

 

[SirKit Breaker]

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

 

[happysteve]

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:


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!)

 

[Silly Sausage]

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!

 

[telectrix]

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 .

 

[Geoffsd]

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.

 

[telectrix]

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).

 

[Bick]

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).

 

[Badged01]

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!

 

[ArseyDee]

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.

 

[happysteve]

PS "always show your working..."

View attachment r1_plus_r2_over_4 .txt

(also important to show you're working)

 

[ArseyDee]

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!!

 

[telectrix]

too much time on your hands. i'd start by typing the first line then give up. :behead: