On Fri, 11 Apr 2003, Richard B. Johnson wrote:
> Every three-connection connector supplies power to two drives.
>
> |--------D1
> -----|--------D2 ________D3
> |______________|_______D4
> |_______Continue
Here's the way I thought of it.
|--x -- 3
|--------X----|--x -- 3
| |--x -- 3
|
| |--x -- 3
-------|--------X----|--x -- 3
| |--x -- 3
|
| |--x -- 3
|--------X----|--x -- 3
|--x -- 3
I now have 1 + 3 + 9 = 13 splitters, giving me 27 connections, out of 1.
etc, etc. Same numbers I'd have doing it your way, yours would be 13
levels deep instead.
I think I just went for the massively parallel method of hooking
these up and from there got massively lost.
Mike
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This archive was generated by hypermail 2b29 : Tue Apr 15 2003 - 22:00:24 EST