Mike,

The distance between centers of the leads (a.k.a. the pitch) of a standard DIP is 0.100" (2.54 mm). Clearly if you put a trace between adjacent pins that is this wide, you will contact both pins. Moreover, there is the dimension of the pads to which the pins solder. Half of one pad impinges upon that 2.54 mm gap from each side, so the gap is reduced by 2 x 1/2_Pad_Width, or by one full pad width.

Thus the first question you must ask is "How wide are the solder pads?" Once you know that, you have an exclusive upper limit to the maximum trace width you can put between adjacent pins. Note I said "exclusive," not "inclusive." I mean that this distance is an upper limit which you can not include in your allowable range. In other words, your trace width must be less than that value.

To find your maximum achivable trace width, find the smallest isolation you can produce (ETCH_min). In other words, what is the thinnest width of copper you can remove without leaving any stray bits in the channel to short the two sides together. You must remove at least this much from each side of the trace, or twice this ammount (2*Etch_Min).

So, the absolute maximum trace width (Trace_Width_Max) you can place between adjacent pins is

Trace_Width_Max = Pitch - Pad_Width - 2*Etch_Min

Note that this is true for any pitch, not just the standard 0.100" (2.54 mm) DIP. This is assuming, of course, that the Trace_Width_Max is both meaningful, which is to say greater than zero, and achievable, which is to say greater than or equal to Trace_Width_Min (the thinnest trace that you can produce).

To further generalize this formula, consider that you must add one factor of Etch_Min for each additional trace you wish to place in the same space. Thus, in the general case of n traces to be placed between adjacent pads,

Trace_Width_Max = Pitch - Pad_Width - (n + 1)*Etch_Min

subject to the same aforementioned constraints of being both meaningful and achievable.