[aprssig] The Current Meaning of WIDEn-N (incorrect -again)
bruninga at usna.edu
Mon May 19 14:55:06 EDT 2014
Neither is true. It is not the level, nor is it a sum of anything. The
"n" is the original value of "N". Except in the "special case" of
"WIDE2-1" where it is simply a "special case" to achieve 1 hop without
triggering local "WIDE1-1" digis.
From: aprssig-bounces at tapr.org [mailto:aprssig-bounces at tapr.org] On Behalf
Of Kenneth Finnegan
Sent: Monday, May 19, 2014 2:16 PM
To: TAPR APRS Mailing List
Subject: Re: [aprssig] The Current Meaning of WIDEn-N (incorrect)
On Mon, May 19, 2014 at 10:47 AM, Robert Bruninga <bruninga at usna.edu>
>> If the first "n" is a "1", then *EITHER* a home fill-in digi --OR-- a
>> high-level true wide can handle the hop.
>> If the first "n" is other than "1", then ONLY a true wide will
>> respond to and handle the hop.
> Simply not true. The first "n" is the original N hop when the packet
> was initiated. The exception (which causes all this confusion) is the
> special "WIDE2-1" which is simply a cheating way to get ONE HOP
> without keying up all surrounding fill-in WIDE1-1 digis. For example,
> if I want to transmit a one-hop packet, but I don't want to bring up
> my neighbors WIDE1-1 digi, and the one that might be in a neighboring
> car (heaven forbid), I would use the single hop path of WIDE2-1.
> Which has always been a special case of a WIDEn-N packet only intended
> to go one hop and without bringing up all the old-fill-in-digis that can
only operate on "WIDE1-1".
>> Today the *total* number of digipeater relay hops requested by the
>> user is the sum of the first N of the first clause (normally "1") and
>> the first N of the second clause (typically "2"), minus any
>> "pre-decrementing" indicated by the second "N" of a clause initially
being smaller than the first. I.e.
> Simply not true.
So which one is it? You just said that the first number doesn't
differentiate between wide and fill-in digis, and also doesn't indicate
the total number of hops requested from the original user. I thought you'd
at least completely agree with Stephen's second statement.
Kenneth Finnegan, W6KWF
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