Can't figure ring final r1 + r2 divided by 4

[KeenPensioner]

Hi...keen amateur here......don't crucify me for asking!! I understand Ohm's law and I understand the mechanics of measuring end to end on ring final and then joining incoming live to outgoing neutral etc, and I know that this configuration gives a figure of 8 (in fact can be straightened to give a complete circle) but I can't figure out the "calculated method" of dividing by (r1+r2)/4....why?
If you meter across the mid point of the "circle" then yes...
if K= r1+r2 (the complete circle) then...since in parallel...
1/R = 1/(K/2) + 1/(K/2) which simplifies to 1/R = 4/k, then R=K/4....(r1+r2)/4........but what happens if you measure across the circle at any other point other than the mid point?

 

[Pete999]

Hi...keen amateur here......don't crucify me for asking!! I understand Ohm's law and I understand the mechanics of measuring end to end on ring final and then joining incoming live to outgoing neutral etc, and I know that this configuration gives a figure of 8 (in fact can be straightened to give a complete circle) but I can't figure out the "calculated method" of dividing by (r1+r2)/4....why?
If you meter across the mid point of the "circle" then yes...
if K= r1+r2 (the complete circle) then...since in parallel...
1/R = 1/(K/2) + 1/(K/2) which simplifies to 1/R = 4/k, then R=K/4....(r1+r2)/4........but what happens if you measure across the circle at any other point other than the mid point?

Far to difficult for a Sunday Afternoon

 

[cliffed]

Ring main cabled in parallel,so the csa of the cable is half & the resistance of the cable is half.
So 1/2 x 1/2 = 1/4.
So we divide by 4 to get r1 & r2.

 

[Pete999]

Ring main cabled in parallel,so the csa of the cable is half & the resistance of the cable is half.
So 1/2 x 1/2 = 1/4.
So we divide by 4 to get r1 & r2.

R1+R2 to be pedantic again. But then again if we don't get the basics right we get all confused.

 

[JAWS]

Try this..
photobucket.com/gallery/user/kingofrocknroll/media/cGF0aDovUmluZzFhcGRmLmpwZw==/?ref=

 

[cliffed]

R1+R2 to be pedantic again. But then again if we don't get the basics right we get all confused.

How true

 

[cliffed]

Try this..
photobucket.com/gallery/user/kingofrocknroll/media/cGF0aDovUmluZzFhcGRmLmpwZw==/?ref=

Error loading page

 

[Madocks]

John warde ring final testing explained

 

[cliffed]

Too long

 

[happysteve]

I did the maths in the thread link below (post 13 ish), there's also a graph in post 5 that shows the percentage error depending on difference in line/cpc csa. Continuity of ring final - https://www.electriciansforums.co.uk/threads/continuity-of-ring-final.84314/#post-859026

R1+R2 only = (r1+r2)/4 if either (a) line and cpc have the same csa, or (b) you're exactly half way round the ring.

(but in other cases, it's pretty close).

 

[Risteard]

Ring main cabled in parallel,so the csa of the cable is half & the resistance of the cable is half.
So 1/2 x 1/2 = 1/4.
So we divide by 4 to get r1 & r2.

Actually cross-sectional area is doubled and distance is halved hence a factor of four.

 

[Ian1981]

To the furthest point or mid point 2 cables are run and the length of cable to the furthest point is half the length of cable in the ring giving a multiplier of 4 for the installed length of cable.
I just think of it as to mid points and 2 cables in parallel = multiplier of 4

 

[Deleted member 26818]

Ring Fimal Circuits are basically two sets of conductors wired in parallel.
If you measure r1 + r2, you are in fact measuring two sets of parallel conductors, but in series.
If you had a Radial circuit 50m long and then decided to turn it into an RFC, all you would need to do is run another 50m cable from the Consumer Unit to the last socket.
So although you might use 100m end to end of cable to wire an RFC, the circuit is in fact only 50m long or only 1/2 as long.
Because an RFC is in fact two conductors in parallel, the resistance is halved another 1/2.
Multiply these two 1/2s together and you get 1/4.
Therefore when we measure r1 + r2 in an RFC, we divide by 4 to obtain R1 + R2.

 

[KeenPensioner]

Thank you for your replies......food for thought.......even on a Sunday! I'll study these answers later and get back to you if I still don't get it.

 

[telectrix]

an easy way to simplify it is to consider a length of single cable around 30m of 1.5mm
it's resistance measures at o.36 ohms. now add an identical length in parallel.resistance is halved, so 0.18 ohms

then cut both in half and parallel 2 of the 15m lengths. halved again = 0.9 ohms.

 

[westward10]

0.09 Tel

 

[telectrix]

correct. it was a sticking "0" key on the laptop, honest.

 

[KeenPensioner]

Thanks every body.....I understand it now

 

[johnduffell]

the resistance at the CU will be approximately 0.94x that at the midpoint, depending on the accuracy of the cable CSAs.

... Edited in case I'm being silly!

 

[Wilko]

Thanks every body.....I understand it now

Now you'll have to explain it to me