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-   -   FISTS petition to the FCC (https://www.radiobanter.com/policy/26834-re-fists-petition-fcc.html)

Bob Brock September 13th 03 10:20 AM
On 13 Sep 2003 05:26:01 GMT, (Jeffrey Herman)
wrote:

In article ,
Mike Coslo wrote:

Perhaps if Mr. Brock does have me killfiled, someone could explain to
him that if he wants to find relevant hits on 1/4 wave dipole, he can
just put the words in quote marks on Yahoo or Google.

Then the search will return relevant hits, rather than simple
occurrences of the individual words, along with the hits wanted. There
is no doubt in my mind that he got those 39,500 hits. It is symptomatic
of poor database searching technique.

His 39,500 hit number, for all it's sound and fury, signifies nothing.

Then please tell him I will be happy to be filtered from his mail
forever if possible.

- Mike KB3EIA -

What's really scary, Mike, is that folks like Bob believe everything
they read on the web. A manufacturer might advertise a misnomer such
as "1/4-wave dipole" and someone like Bob believes such an item can
really exist, when in fact "1/4-wave" and "dipole" are contradictions.

Jeff KH6O

When about 10-15,000 of them get together and do it, it does give it
some creedence.

What I find really ironic is that, instead of simply providing
adefinition, someone would dance all around it and tell lies. Here,
let me do your homework for you...

http://rhyme.lycos.com/r/d?u=dipole_antenna

http://dict.die.net/dipole%20antenna/

http://www.hyperdictionary.com/dicti...dipole+antenna

There, do you feel better now??? The way the discussions go is that
if you make an assertion, it's up to you to support it. I'm getting
damn sick and tired of having to do you guys homework for you.

Brian September 13th 03 02:04 PM
(Jeffrey Herman) wrote in message ...
Mike Coslo wrote:
Jeffrey Herman wrote:
"Carl R. Stevenson" wrote:

How many *US* licensees are members of FISTS, Dick?

Geez, talk about a page from history! It was years ago that I
posed almost the same question to you, Carl, except I asked
"How many members of NCI are licensed" to which you replied,
"It doesn't matter..."

Jeffrey, the answer is "thousands, and growing every day."

It's too bad that a person that refuses to give out even the basic
numbers of devotees decides to ask for a numerical breakdown of a rival
groups numbers.

Assuming his accounting methods are acceptable, the answer is
thousands, and growing every day.


When I posed that question to Carl some years ago, he couldn't come
up with any membership roster; there were no records of who had joined
since there was no dues, nor did any application form exist.

Jeff KH6O

Jeff, people who get their mail in glass mailboxes shouldn't throw stones.

Brian September 13th 03 02:14 PM
(N2EY) wrote in message ...
In article , Bob Brock
writes:

I'm
here to see if the ham community will then move on to something really
important like revising the written tests and the test pools so that
when someone passes the test, they actually know how to use a radio,
set up or build an antenna, the protocols on the frequencies they are
authouized to use, how to minimize RF exposure and stay within safe
levels.

The test pools are under constant revision. Anyone can submit proposed Q&A to
the QPC for consideration.

Smith Charts, again?

If someone doesn't know what all those buttons do on their
radio, they shouldn't be licensed to use it.

Then we'd need rig-specific licenses.

Bad idea.

And what would you do about homebrew rigs like mine?

Lets not get all superior. Elecraft manuals are on-line.

If they don't know
enough not to use repeater imputs for simplex operation, they
shouldn't be licensed to use the frequency.

Now that makes sense! But such things are already addressed in the question
pools.

The problem is that what needs to change is the test methodology. By lumping
all of the subjects into a one-size-fits-all written test, prospective hams can
have huge holes in their knowledge and still pass because of strenghts in other
areas. One answer to that is subelements.

The big problem is convincing FCC that testing at such a level is needed. For
more than 25 years, FCC's view towards amateur radio testing is to reduce the
license requirements and make the licenses easier to get, not harder. That's
one of the reasons some folks defend the code test so diligently - they know
that if it goes, it will not be replaced by any other test, nor will the test
methods used for the written be improved.

But Jim, the Morse Exam was supposed to be a Morse Exam, not a "Lets
make it harder for people to get into amateur radio" Exam.

It's my hope that when it is all over with Morse, that the ham
community will address the real important issues and Morse ain't it.

Morse will continue to be a major part of amateur radio with or without a test.
The test, however, is symbolic of the changes that have been ongoing for a long
time.

The Morse Exam is symbolic of only the Morse Code. The VECs threw
that out the window when they subbed a Farnsworth Exam at 13-15WPM for
Morse at 5WPM as specified by the FCC in PArt 97.

It is at best a scapegoat that hams can argue about while the more
important issues of licensing inept operators is ignored. IMO, the
ham community has some really screwed up priorities and hopefully,
with the endgame for code in sight, they may....just may come to grips
with some real issues.

I'm hopeful, but not expectant.

I think you will find that the exact same problems will arise in connection
with any move to increase license requirements. The "Smith chart solution" post
in another thread is only half in jest.

There are already folks like W5YI campaigning for less WRITTEN testing, saying
the Tech test is too hard.

The Tech test And privs are insane for an entry level license.

Kim W5TIT September 13th 03 02:35 PM
"Bob Brock" wrote in message
...
On 13 Sep 2003 04:50:38 GMT, (Jeffrey Herman)
wrote:

Bob Brock wrote:
If you can't accept the fact that people build, sell, and buy 1/4 wave
dipoles for use in UHF, get over it. Just don't try to make stuff up
and post it.

Again, by definition "dipole" means two current or voltage poles; it
does not refer to the number of elements of the antenna.

You can't have two (di-) current or voltage poles in just a 1/4-wave
segment. Calling it such doesn't make it so. It takes a 1/2-wavelength
for two (di-) such poles to appear.

Jeff KH6O

You must have replied to the wrong post. But out of curiousity, how
many current or voltage poles is a "J" as in J-pole.


*GASP*!!!! Bob! Don't get them going with the word i-m-p-e-d-a-n-c-e!!!!
; )

Kim W5TIT

WA8ULX September 13th 03 03:37 PM
You must have replied to the wrong post. But out of curiousity, how
many current or voltage poles is a "J" as in J-pole.

You Morron a J-Pole is a half wave ant , not a quarter wave. It appears that
Brock knows nothing about ants, my guess is hes nothing more then a Dumb Down
CBplusser Knuckle Dragger, at Im sure a member of NCI

Dee D. Flint September 13th 03 05:34 PM

"Brian" wrote in message
m...
(N2EY) wrote in message
...
There are already folks like W5YI campaigning for less WRITTEN testing,
saying
the Tech test is too hard.

The Tech test And privs are insane for an entry level license.

I do hope you mean the test is insanely simple and the privileges insanely
high for the level of testing done.

Dee D. Flint, N8UZE

Brian September 13th 03 08:25 PM
(Jeffrey Herman) wrote in message ...
Bob Brock wrote:
If you can't accept the fact that people build, sell, and buy 1/4 wave
dipoles for use in UHF, get over it. Just don't try to make stuff up
and post it.

Again, by definition "dipole" means two current or voltage poles; it
does not refer to the number of elements of the antenna.

You can't have two (di-) current or voltage poles in just a 1/4-wave
segment. Calling it such doesn't make it so. It takes a 1/2-wavelength
for two (di-) such poles to appear.

Jeff KH6O

Hmmm. Wonder why earth or radials are necessary for 1/4 wave
antennas? And if the earth is already a dipole...

Bob Brock September 14th 03 03:18 AM
On Sat, 13 Sep 2003 08:35:53 -0500, "Kim W5TIT"
wrote:

"Bob Brock" wrote in message
.. .
On 13 Sep 2003 04:50:38 GMT, (Jeffrey Herman)
wrote:

Bob Brock wrote:
If you can't accept the fact that people build, sell, and buy 1/4 wave
dipoles for use in UHF, get over it. Just don't try to make stuff up
and post it.

Again, by definition "dipole" means two current or voltage poles; it
does not refer to the number of elements of the antenna.

You can't have two (di-) current or voltage poles in just a 1/4-wave
segment. Calling it such doesn't make it so. It takes a 1/2-wavelength
for two (di-) such poles to appear.

Jeff KH6O

You must have replied to the wrong post. But out of curiousity, how
many current or voltage poles is a "J" as in J-pole.


*GASP*!!!! Bob! Don't get them going with the word i-m-p-e-d-a-n-c-e!!!!
; )

Kim W5TIT

ROTFLMAO...

Ryan, KC8PMX September 15th 03 08:20 AM
Yep, and I didn't see it specifying whether it was a half-wave or not.
Regardless of the fact I DO know how to calculate it or any fraction of a
full wavelength.

If the question was "What is the length of a quarter-wave dipole for
14.240Mhz?" or "What is the length of a half-wave dipole for 14.240Mhz?" or
whatever then I would have answered right of the bat. Yeah, it is simple to
do that formula, but to give the answer would be hard unless you gave ALL of
the possibilities such as below:


one-sixteenth wave @ 14.240Mhz = 4.108146
one-eighth wave @ 14.240Mhz = 8.216292
three-sixteenth wave @ 14.240Mhz = 12.324438
quarter-wave @ 14.240Mhz = 16.432584
five-sixteenth wave @ 14.240Mhz = 20.54073
three-eighths wave @ 14.240Mhz = 24.648876
seven-sixteenths wave @ 14.240Mhz = 28.757022
half-wave @ 14.240Mhz = 32.8651685
nine-sixteenths wave @ 14.240Mhz = 36.973314
five-eighths wave @ 14.240Mhz = 41.08146
eleven-sixteenths wave @ 14.240Mhz = 45.189606
three-quarter wave @ 14.240Mhz = 49.297752
thirteen-sixteenths wave @ 14.240Mhz = 53.405898
seven-eigths wave @ 14.240Mhz = 57.514044
fifteen-sixteenths wave @ 14.240Mhz = 61.62219
full wave dipole @ 14.240Mhz = 65.730337


All of the above answers would be correct (barring accidental typos), but
the original poster of the question in the first place DID NOT SPECIFY.
See, it is important, unless you consider ALL antennas to be exactly
one-half wavelength in length.



--
Ryan, KC8PMX
FF1-FF2-MFR-(pending NREMT-B!)
--. --- -.. ... .- -. --. . .-.. ... .- .-. . ..-. .. .-. . ..-.
... --. .... - . .-. ...


As you know from studying for your Technician license (and the material
also
appears in the General study guide), you can calculate the length of your
half-wave dipole directly knowing only the frequency. The equation is:

Length in feet = 468 divided by the frequency in megahertz.

Also from these same study guides, you can calculate the wavelength by the
following equation.

Wavelength in meters = 300 divided by the frequency in megahertz
(of course if you wish you can then convert the answer to feet).


Dee D. Flint, N8UZE

Brian September 16th 03 02:11 AM
Bob Brock wrote in message . ..

Kim, within a year the only debate in here about CW will be about the
consequences of it no longer being required. The whining will be
terrible for awhile, but like the no-code tech discussions, they will
eventually disappear.

Ooooh, huge, huge miscalcualtion, Brock. You give these guys way too
much credit. These guys have memories like an elephant, and they will
carrry the debate to their graves. Which is why I say that the only
cure are the actuarial tables. They are as unyielding as the PCTA,
till the bitter end.


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