BAS,
I make that 26mm squared for a single conductor?
Pi x r² = 2.5
r² = 2.5/Pi = 0.796
r = 0.892mm
d = 1.784mm
c = 1.784 x Pi = 5.6mm
5.6mm x 10mm = 56mm²
So 2 x 2cm bared & bent over, including the ends of the conductors is a total surface area of 229mm²...
(Actually a bit more because of the deformation of the cylindrical conductors where they bend, but I think we can ignore that)
Thanks BAS - knew I was going wrong somewhere.
Had substituted CSA for circumference in C=Pi X D and arrived at the wrong radius.
Therefore, even better contact area.
It might be amusing to you, but I'm disappointed that nobody seems to have understood.THRIPSTER said:Or do you believe that that the only electrical conduction takes place through the copper where the conductors touch and that there is none through the solder?
Not to start on Ban again, but, what's wrong with through crimps and a bit of heatshrink?
thanks for the info plugwash, im going to be going to maplin for some heatshrink tubing so i think i might just do it ur way
You triggered a long lost memory there Saxondale - took me right back to Day Release for a minute.thanks for the info plugwash, im going to be going to maplin for some heatshrink tubing so i think i might just do it ur way
can you wrap it to make a T shape ?
thats how we were taught to do it in college
You triggered a long lost memory there Saxondale - took me right back to Day Release for a minute.thanks for the info plugwash, im going to be going to maplin for some heatshrink tubing so i think i might just do it ur way
can you wrap it to make a T shape ?
thats how we were taught to do it in college
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