If you are starting a network from scratch, you will have to decide whether to use thin ethernet (RG58 co-ax cable with BNC connectors) or 10baseT (twisted pair telco-style cables with RJ-45 eight wire `phone' connectors). The old-fashioned thick ethernet, RG-5 cable with N connectors is obsolete and rarely seen anymore.
See Type of cable... for an introductory look at cables. Also note that the FAQ from comp.dcom.lans.ethernet has a lot of useful information on cables and such. FTP to rtfm.mit.edu and look in /pub/usenet-by-hierarchy/ for the FAQ for that newsgroup.
Thin ethernet cable is pretty inexpensive. If you are making your own cables solid-core RG58A is $0.27/m. and stranded RG58AU is $0.45/m. Twist-on BNC connectors are < $2 ea., and other misc. pieces are similarly inexpensive. It is essential that you properly terminate each end of the cable with 50 ohm terminators, so budget $2 ea. for a pair. It's also vital that your cable have no `stubs' -- the `T' connectors must be attached directly to the ethercards.
There are two main drawbacks to using thinnet. The first is that it is limited to 10Mb/sec - 100Mb/sec requires twisted pair. The second drawback is that if you have a big loop of machines connected together, and some bonehead breaks the loop by taking one cable off the side of his tee, the whole network goes down because it sees an infinite impedance (open circuit) instead of the required 50 ohm termination. Note that you can remove the tee piece from the card itself without killing the whole subnet, as long as you don't remove the cables from the tee itself. Of course this will disturb the machine that you pull the actual tee off of. 8-) And if you are doing a small network of two machines, you still need the tees and the 50 ohm terminators -- you can't just cable them together!
There are also some fancy cable systems which look like a single lead going to the card, but the lead is actually two runs of cable laying side-by-side covered by an outer sheath, giving the lead an oval shaped cross-section. At the turnaround point of the loop, a BNC connector is spliced in which connects to your card. So you have the equivalent of two runs of cable and a BNC T, but in this case, it is impossible for the user to remove a cable from one side of the T and disturb the network.
Twisted pair networks require active hubs, which start around $50, and the raw cable cost can actually be higher than thinnet. You can pretty much ignore claims that you can use your existing telephone wiring as it is a rare installation where that turns out to be the case.
On the other hand, all 100Mb/sec ethernet proposals use twisted pair, and most new business installations use twisted pair. Also, Russ Nelson adds that `New installations should use Category 5 wiring. Anything else is a waste of your installer's time, as 100Base-whatever is going to require Cat 5.'
If you are only connecting two machines, it is possible to avoid using a hub, by swapping the Rx and Tx pairs (1-2 and 3-6).
If you hold the RJ-45 connector facing you (as if you were going to plug it into your mouth) with the lock tab on the top, then the pins are numbered 1 to 8 from left to right. The pin usage is as follows:
Pin Number Assignment
---------- ----------
1 Output Data (+)
2 Output Data (-)
3 Input Data (+)
4 Reserved for Telephone use
5 Reserved for Telephone use
6 Input Data (-)
7 Reserved for Telephone use
8 Reserved for Telephone use
If you want to make a cable, the following should spell it out for you. Differential signal pairs must be on the same twisted pair to get the required minimal impedance/loss of a UTP cable. If you look at the above table, you will see that 1+2 and 3+6 are the two sets of differential signal pairs. Not 1+3 and 2+6 !!!!!! At 10MHz, with short lengths, you *may* get away with such errors, if it is only over a short length. Don't even think about it at 100MHz.
For a normal patch cord, with ends `A' and `B', you want straight through pin-to-pin mapping, with the input and output each using a pair of twisted wires (for impedance issues). That means 1A goes to 1B, 2A goes to 2B, 3A goes to 3B and 6A goes to 6B. The wires joining 1A-1B and 2A-2B must be a twisted pair. Also the wires joining 3A-3B and 6A-6B must be another twisted pair.
Now if you don't have a hub, and want to make a `null cable', what you want to do is make the input of `A' be the output of `B' and the output of `A' be the input of `B', without changing the polarity. Tha means connecting 1A to 3B (out+ A to in+ B) and 2A to 6B (out- A to in- B). These two wires must be a twisted pair. They carry what card/plug `A' considers output, and what is seen as input for card/plug `B'. Then connect 3A to 1B (in+ A to out+ B) and also connect 6A to 2B (in- A to out- B). These second two must also be a twisted pair. They carry what card/plug `A' considers input, and what card/plug `B' considers output.
So, if you consider a normal patch cord, chop one end off of it, swap the places of the Rx and Tx twisted pairs into the new plug, and crimp it down, you then have a `null' cable. Nothing complicated. You just want to feed the Tx signal of one card into the Rx of the second and vice versa.
Note that before 10BaseT was ratified as a standard, there existed other network formats using RJ-45 connectors, and the same wiring scheme as above. Examples are SynOptics's LattisNet, and AT&T's StarLAN. In some cases, (as with early 3C503 cards) you could set jumpers to get the card to talk to hubs of different types, but in most cases cards designed for these older types of networks will not work with standard 10BaseT networks/hubs. (Note that if the cards also have an AUI port, then there is no reason as to why you can't use that, combined with an AUI to 10BaseT transceiver.)
Thick ethernet is mostly obsolete, and is usually used only to remain compatible with an existing implementation. You can stretch the rules and connect short spans of thick and thin ethernet together with a passive $3 N-to-BNC connector, and that's often the best solution to expanding an existing thicknet. A correct (but expensive) solution is to use a repeater in this case.