I didn't want to jump in on this one, but now your post has been here for two weeks without a single response. I can just as well give and answer and hope for the more knowledgeable guys to correct me.
Without any warranty, no assumed liability and everything else you can think up for a disclaimer: Here is a formula I made up:
I =10^(3/50 x (35-AWG))
I could explain how I arrived at it, but I don't think anyone is interested nor do I have the time at the moment.
Anyway, it's not that simple. Much depends on the insulation and the cable construction just like the larger stuff. Unless you are running uninsulated wire in free air the correct source of information will be the manufacturer.
There are variations in the allowable ampacity depending upon specific conditions, just as the NEC restricts currents for bundling, ambient temperature, etc.
The low-voltage cabling used in marine/automotive applications, for example, often has somewhat higher ratings than for house wiring. Compare the NEC with this table of marine use, for example: http://www.cmsquick.com/Tech.html
NOTE: This information is provided as is without any express or implied warranties. While effort has been taken to ensure the accuracy of the information contained in this text, the author assumes no responsibility for errors or omissions, or for damages resulting from the use of the information contained herein. The ampacities listed might be totally inaccurate, inappropriate, or misguided. There is no guarantee as to the suitability of said circuits and information for any purpose.
Now we have at least four different answers, with a huge spread. There is a reason for this: It makes a difference if it is a network cable with 8 wires or a single wire inside a computer case. You need to derate for bundling, as you can see in Paul's first link. If you have one of those phone cables with dozens of pairs, it must be quite severe.
My little formula above simply approximates 310.16.
[This message has been edited by C-H (edited 11-11-2004).]