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Here goes a view-from-below picture of a 3 kilovolts DC railway catenary

http://fotki.yandex.ru/users/poduruev-petr/view/651405/?page=2

Here B is a suspension cable, C is a two-stranded contact wire - one which the train pantograph touches. Between B and C there's a thick jumper cable that is attached to both B and C. Such jumpers are installed every several dozen meters. Also not shown in the picture but there're hundreds of pieces of thin of wire located every several meters that are attached to the suspension cable, hang down to the contact wire and hold the contact wire at exactly the right height above the rails.

So a lot of effort is made to ensure that B and C have equal voltages at all times.

Now A is a cable that comes from a distribution cable running along the railway. The picture shows how that cable "feeds" energy into the contact wire. Such "feed" points occur about every kilometer.

So the "feed" cable A comes from the distribution cable and is connected to the jumper that is in turn connected to both the suspension cable and the contact wire. Clearly A, B and C will have the same voltage on them.

Finally there's that huge insulator. It separates A from B but both A and B have the same voltage so they could have been connected directly without any kind of insulator.

What's the use of that insulator?

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    \$\begingroup\$ This is the way to fix supply cables to the suspension cables. It needed to be fixed, and they just took the usual part. The insulater does not have a function, it just happens to be part of the hook. \$\endgroup\$ – posipiet Mar 6 '13 at 12:50
  • \$\begingroup\$ The very large "umbrella" indicates that that is an anti-lightning insulator. It's possible that there is some special behaviour of the other wiring in the case of a lightning strike. \$\endgroup\$ – pjc50 Mar 6 '13 at 15:51
  • \$\begingroup\$ Similar question: electronics.stackexchange.com/questions/33725/… \$\endgroup\$ – clabacchio Mar 6 '13 at 16:27
  • \$\begingroup\$ @clabacchio: Actually this is not a dupe of that question. That question is about splitting wires in AC systems with voltages of 220 kilovolts and higher, such splitting is not performed in systems with voltages around three kilovolts. \$\endgroup\$ – sharptooth Mar 7 '13 at 7:40
  • \$\begingroup\$ Indeed I didn't suggest for closing it, but I wanted to point out that there was a similar question, and to my knowledge they were similar :) Not that I now a lot about power distribution, though \$\endgroup\$ – clabacchio Mar 7 '13 at 7:59
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It is an insulator so that current does NOT flow through the mechanical link. The mechanical link is not designed to handle high current flows, and may fail if subjected to high current flows.

In a fault condition, you do NOT want to have your mechanical supports fail because they have high current flowing through them.

In non-fault conditions you still don't want to have current flowing through your mechanical supports: they would find an equilibrium where the small fraction of the current flowing through the steel would keep it hot enough to not conduct more current. The hot steel would be susceptible to corrosion, and the threaded connectors would be susceptible to creep.

Note that the voltage is irrelevent for this mechanical support. The insulator is there because, even at 3KV, trains take a lot of current out of the supply. You can get the same problems with Low Voltage (48V) supply systems.

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  • \$\begingroup\$ By "fault condition", you're saying that if the connection between A and the rest of the wiring fails, you want current to simply flow noplace rather than through the mechanical joint? That certainly makes sense. I'm curious, though: if there were a break in one of the pickup-level wires, would the system be designed to have current flow along the arched support wire or not? If the system were designed so only limited current could flow through the upper support wire, (just enough to keep it at the same potential in non-fault case), then even if a wire came down someplace... \$\endgroup\$ – supercat Mar 7 '13 at 16:06
  • \$\begingroup\$ ...and caused a short that would carry enough current to melt wires, such current would mechanically compromise only the lower wire, which would still be supported at intervals by the upper wire. Also, I wonder if that insulator has pivots on it? It's a lot easier to design a pivot that can handle zero current, than one that can handle lots. \$\endgroup\$ – supercat Mar 7 '13 at 16:09
  • \$\begingroup\$ --You can see a pivot in the picture -- it's the ring link connection. Dunno how the support wire works. \$\endgroup\$ – david Mar 8 '13 at 4:37

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