Date   : Tue, 08 Sep 2009 09:05:36 +0100
From   : rs423@... (Mick Champion)
Subject: Leccy @ Acorn World '09

Sprow wrote:
> In article <4AA13E26.6010104@...>,
>    Mick Champion <rs423@...> wrote:
>   
>> Sprow wrote:
>>     
>>> In article <090903235504@...>,
>>>    Jonathan Graham Harston <jgh@...> wrote:
>>>   
>>>       
>>>>> because each socket  has two cables feeding it from either side of the
>>>>> ring effectively doubling up the rating.
>>>>>           
>>>>  
>>>> Nooooo. times SQR(2), not times 2. Check your parallel resistance
>>>> calculations.    
>>>>         
>>> Say one cable is 1 ohm, then the resistance of such 2 cables is
>>>
>>>  [ (1/1) + (1/1) ] ^-1
>>>
>>> which is 0.5 ohms. That's a straight factor of 2.
>>>   
>>>       
>> If the socket was exactly in the middle of the ring, this would be true 
>> I guess? Most sockets would be nearer on end of the ring than another.
>>     
>
> Resistance is a function of cross sectional area and length
>
>   bigger cross section -> lower resistance
>   longer length -> higher resistance
>
> It's linear with length, so half the length is half the resistance etc...
> Therefore a socket which was closer to the fuse box might be served by a
> short cable of 0.5 ohms, and a much longer one of 1.5 ohms for example (such
> that they add up to 2 ohms again).
>
> Then the resistance would be
>
>  [ (1/1.5) + (1/0.5) ] ^-1
>
> which is 0.375 ohms: better!
>
> I chose the centre point of the ring because that's the worst case,
> Sprow.
>
>   
Oh I see... I think? So as you move clockwise or anti-clockwise frim the 
centre of the loop, the resistance will increase on one cable, and 
decrease on the other cable giving overall the same resistance where 
ever you may be. Is that it? Please tell me I've got it - finally.





Mick