I would request some comment on the following:
I come up with a more conservative estimate.
Take the board width W and length L for area and normalize
it for a length, (W*L)^0.5. The number of connections
being N means a minimum of N/2 traces will be used,
i.e. 2 connections produce 1 trace, 4 connections produce
2 traces, etc ... But, 3 connections producing 2 traces
divide the board into 3 equal parts. So, this normalized
length (W*L)^0.5 should be divided into (N/2 + 1).
So, for one layer we have the average pitch being
(W*L)^0.5
P(ave) = ---------
(N/2 + 1)
For M number of layers reserved for traces,
this effectively multiplies the spacings
for the pitch. I end up with
(W*L)^0.5
P(ave) = M* ---------
(N/2 + 1)
Which reduces to
2*M*(W*L)^0.5
P(ave) = -------------
(N + 2)
Comparing both equations for
Connections = 800
Board Length = 8 inches
Board Width = 12 inches
Pavg = (((XY)^0.5)*2.7M)/N gives 0.13 inches
and
P(ave) = (2*M*(W*L)^0.5)/(N + 2) gives 0.097 inches
I understand the limitations with each,
but anyone care to comment?
Side Note: I'm sure you all can appreciate that
either equation can be solved for
the number of routing layers needed
for a given set of criteria. Either
way, works pretty well on a spreadsheet.
Regards, Doug McKean
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