World Phones

"John Navas" <[email protected]> wrote in message
news:[email protected]...
> On Thu, 10 Aug 2006 03:04:45 GMT, DecaturTxCowboy <[email protected]> wrote in
> <[email protected]>:
>
>>John Navas wrote:
>>> Attenuation is a function of distance, not frequency.
>>> See <http://www.sss-mag.com/pdf/1propagation.pdf>.
>>> Methinks you have in mind reflection, diffraction, and scattering?
>>
>>Wrong...fundamentally WRONG...SOOO WROOOOONNNNGGGGG!
>>
>>Path Loss = 20* log(12.57*r/w) dB, where
>>r = distance between transmitter and receiver
>>w = wavelength (the inverse of the frequency)
>
> See <http://en.wikipedia.org/wiki/Friis_transmission_equation> for the
> actual Friis transmission equation. "The Friis transmission equation is
> used in telecommunications engineering, and gives the power transmitted
> from one antenna to another under idealized conditions." In other
> words, it's a model of available antenna power under ideal conditions,
> not just free space signal attentuation. Furthermore, "The ideal
> conditions are almost never achieved in ordinary terrestrial
> communications, due to obstructions, reflections from buildings, and
> most importantly reflections from the ground."
>
>

Let me get this right- the cellular tech guy is wrong and wikipedia is
right. Wikipedia is ripe with misinformation- a sad result of people
trhinking that they know whatthey are talking about.

And Novice- do you have any experience with the topic at hand?

---

› See More: World Phones

"Scott" <[email protected]> wrote in message
news:[email protected]...
>
> "John Navas" <[email protected]> wrote in message
> news:[email protected]...
>> On Thu, 10 Aug 2006 03:04:45 GMT, DecaturTxCowboy <[email protected]> wrote in
>> <[email protected]>:
>>
>>>John Navas wrote:
>>>> Attenuation is a function of distance, not frequency.
>>>> See <http://www.sss-mag.com/pdf/1propagation.pdf>.
>>>> Methinks you have in mind reflection, diffraction, and scattering?
>>>
>>>Wrong...fundamentally WRONG...SOOO WROOOOONNNNGGGGG!
>>>
>>>Path Loss = 20* log(12.57*r/w) dB, where
>>>r = distance between transmitter and receiver
>>>w = wavelength (the inverse of the frequency)
>>
>> See <http://en.wikipedia.org/wiki/Friis_transmission_equation> for the
>> actual Friis transmission equation. "The Friis transmission equation is
>> used in telecommunications engineering, and gives the power transmitted
>> from one antenna to another under idealized conditions." In other
>> words, it's a model of available antenna power under ideal conditions,
>> not just free space signal attentuation. Furthermore, "The ideal
>> conditions are almost never achieved in ordinary terrestrial
>> communications, due to obstructions, reflections from buildings, and
>> most importantly reflections from the ground."
>>
>>
>
> Let me get this right- the cellular tech guy is wrong and wikipedia is
> right. Wikipedia is ripe with misinformation- a sad result of people
> trhinking that they know whatthey are talking about.
>
> And Novice- do you have any experience with the topic at hand?

Scott,

The Wikipedia reference is correct.
If:
1) the antenna gains are set to unity to reflect only the path,
2) the power ratio is inverted to express the path loss as a positive
decibel quantity,
3} and the result is expressed in decibels,
the two equations are identical. Do not call the reference miss-information
unless you are sure of yourself.

David

"David" <[email protected]> wrote in message
news:[email protected]...

>
> Scott,
>
> The Wikipedia reference is correct.
> If:
> 1) the antenna gains are set to unity to reflect only the path,
> 2) the power ratio is inverted to express the path loss as a positive
> decibel quantity,
> 3} and the result is expressed in decibels,
> the two equations are identical. Do not call the reference
> miss-information unless you are sure of yourself.
>
> David
>

I'm sure enough of myself to state categorically that the equation cited by
Cowboy is indeed correct and actively used in the field of cellular
communication by all carriers. Is that being sure enough of myself?

"Scott" <[email protected]> wrote in message
news:[email protected]...
>
> "David" <[email protected]> wrote in message
> news:[email protected]...
>
>>
>> Scott,
>>
>> The Wikipedia reference is correct.
>> If:
>> 1) the antenna gains are set to unity to reflect only the path,
>> 2) the power ratio is inverted to express the path loss as a positive
>> decibel quantity,
>> 3} and the result is expressed in decibels,
>> the two equations are identical. Do not call the reference
>> miss-information unless you are sure of yourself.
>>
>> David
>>
>
> I'm sure enough of myself to state categorically that the equation cited
> by Cowboy is indeed correct and actively used in the field of cellular
> communication by all carriers. Is that being sure enough of myself?
Scott,
Here is a quote from you about the reference being discussed:
---
"Let me get this right- the cellular tech guy is wrong and wikipedia is
right. Wikipedia is ripe with misinformation- a sad result of people
trhinking that they know whatthey are talking about.

And Novice- do you have any experience with the topic at hand?"
---
That was the origin of my comment.

David

John Navas wrote:
> On Thu, 10 Aug 2006 03:04:45 GMT, DecaturTxCowboy <[email protected]> wrote in
> <[email protected]>:
>
>> John Navas wrote:
>>> Attenuation is a function of distance, not frequency.
>>> See <http://www.sss-mag.com/pdf/1propagation.pdf>.
>>> Methinks you have in mind reflection, diffraction, and scattering?
>> Wrong...fundamentally WRONG...SOOO WROOOOONNNNGGGGG!
>>
>> Path Loss = 20* log(12.57*r/w) dB, where
>> r = distance between transmitter and receiver
>> w = wavelength (the inverse of the frequency)
>
> See <http://en.wikipedia.org/wiki/Friis_transmission_equation> for the
> actual Friis transmission equation. "The Friis transmission equation is
> used in telecommunications engineering, and gives the power transmitted
> from one antenna to another under idealized conditions." In other
> words, it's a model of available antenna power under ideal conditions,
> not just free space signal attentuation. Furthermore, "The ideal
> conditions are almost never achieved in ordinary terrestrial
> communications, due to obstructions, reflections from buildings, and
> most importantly reflections from the ground."


[YAWN] Your dodging the issue by playing scrabble and back pedaling to
cover your ass (as usual).

So I'm going to take it one step at a time and dissect what you said to
illustrate how you are wrong and how you are playing scrabble to cover
your ass.

1) Lets stick to what you originally said - "Attenuation is a function
of distance, not frequency."

2) The well known Friss equation *does indeed take into account the
frequency* (which you conveniently omitted in your above Wiki quote).

3) The Wiki quote is not a relevant excuse to cover your ass. All the
Wiki quote is saying is *there are more additional factors that affect
the path loss in addition to the increased attenuation due to the higher
frequency*...most noticeable the first Fresnel zone clearance that
applies to terrestrial line of sight and near line of sight
communications (obviously not applicable to satellite line of sight
communications).

4) You are back pedaling by diluting the importance of frequency in the
path loss by quoting other factors. Frequency does come into play, but
you are now saying it is less important. That's your dodge.

5) Those other factor that Wiki is referring to would include:
Tropospheric ducting under 200 MHz, but above 20 Mz (generally
speaking). Inadequate Fresnel zone clearance (more critical above 900
MHz due to the increased diffraction of higher frequencies). Land cover
(trees in the summer and prairie grasses in the spring) at VHF
frequencies and above. Man made noise below VHF frequencies. Rain
attenuation at 10 GHz and above. Multipathing (reflection) at VHF and above.


Too many unexperienced people (radio expert wannabes) pull out a
topographic map and start making their cover predictions based on free
space loss, but find their systems don't work because of other factors
(again, most notably to to inadequate Fresnel zone clearance and land
cover issues).

David wrote:
> The Wikipedia reference is correct.
> If:
> 1) the antenna gains are set to unity to reflect only the path,
> 2) the power ratio is inverted to express the path loss as a positive
> decibel quantity,
> 3} and the result is expressed in decibels,
> the two equations are identical. Do not call the reference
> miss-information unless you are sure of yourself.

Wiki is correct, but the point is Navis used it to dilute the importance
of frequency in the attenuation.

On Fri, 11 Aug 2006 12:51:36 GMT, DecaturTxCowboy <[email protected]> wrote in
<s3%[email protected]>:

>[SNIP]

What I posted is correct, but you seem to be unable to be civil, and I'm
not going to waste more time in a nasty argument, so I'm giving you the
last word.

--
Best regards, FAQ FOR CINGULAR WIRELESS:
John Navas <http://en.wikibooks.org/wiki/Cingular_Wireless_FAQ>

John Navas wrote:
> On Fri, 11 Aug 2006 12:51:36 GMT, DecaturTxCowboy <[email protected]> wrote in
> <s3%[email protected]>:
>
>> [SNIP]
>
> What I posted is correct, but you seem to be unable to be civil, and I'm
> not going to waste more time in a nasty argument, so I'm giving you the
> last word.

And that dear readers was classic Navas dodge.

I was certainly being civil. Pointing out a person's error is not being
uncivil. Calling you a pompous cellular expert wannabe that wants to run
with the big dogs would be uncivil.

"David" <[email protected]> wrote in message
news:[email protected]...
>
> "Scott" <[email protected]> wrote in message
> news:[email protected]...
>>
>> "David" <[email protected]> wrote in message
>> news:[email protected]...
>>
>>>
>>> Scott,
>>>
>>> The Wikipedia reference is correct.
>>> If:
>>> 1) the antenna gains are set to unity to reflect only the path,
>>> 2) the power ratio is inverted to express the path loss as a positive
>>> decibel quantity,
>>> 3} and the result is expressed in decibels,
>>> the two equations are identical. Do not call the reference
>>> miss-information unless you are sure of yourself.
>>>
>>> David
>>>
>>
>> I'm sure enough of myself to state categorically that the equation cited
>> by Cowboy is indeed correct and actively used in the field of cellular
>> communication by all carriers. Is that being sure enough of myself?
> Scott,
> Here is a quote from you about the reference being discussed:
> ---
> "Let me get this right- the cellular tech guy is wrong and wikipedia is
> right. Wikipedia is ripe with misinformation- a sad result of people
> trhinking that they know whatthey are talking about.
>
> And Novice- do you have any experience with the topic at hand?"
> ---
> That was the origin of my comment.
>

And while this is an instance where the information contained is correct,
all you need to do is look at the wiki refernce that Navas uses in his sig
to see an entry that contains questionable material, to put it politely.
John Navas has become a pro at googling and posting links that contain
information that is beyond his knowledge. As a result, most of it, while
correct information, does not apply to the conversation at hand. This is
one of those cases- he posted factual information that he has no experience
with- it simply is not the correct information for the discussion being
conducted.

"DecaturTxCowboy" <[email protected]> wrote in message
news:[email protected]...
> John Navas wrote:
>> On Fri, 11 Aug 2006 12:51:36 GMT, DecaturTxCowboy <[email protected]> wrote in
>> <s3%[email protected]>:
>>
>>> [SNIP]
>>
>> What I posted is correct, but you seem to be unable to be civil, and I'm
>> not going to waste more time in a nasty argument, so I'm giving you the
>> last word.
>
> And that dear readers was classic Navas dodge.

Oh, what a surprise- he got in over his head again.

>
> I was certainly being civil. Pointing out a person's error is not being
> uncivil. Calling you a pompous cellular expert wannabe that wants to run
> with the big dogs would be uncivil.
>
>

Careful, Cowboy- words like that could cause the AARP to charge you with
dinosaur hunting.

On Wed, 09 Aug 2006 23:56:37 GMT, John Navas
<[email protected]> wrote in
<[email protected]>:

>On Wed, 09 Aug 2006 15:25:31 -0700, "Paul Hovnanian P.E."
><[email protected]> wrote in <[email protected]>:
>
>>John Navas wrote:
>
>>> In metro areas there is no significant technology difference between 850
>>> and 1900 bands (due to less than maximum range spacing), and even in
>>> non-metro areas the difference, due to shorter range from lower
>>> permitted maximum power for 1900 MHz, tends to be relatively small:
>>>
>>> * Maximum power in the 800 band is 3 watts.
>>> * Maximum power in the 1900 band is 2 watts.
>>>
>>> It's not intuitively obvious, but that's only about 18% less range for
>>> 1900, and then only when range is limited only by power (not by
>>> terrain).
>>
>>What about the increased attenuation of the higher frequencies (shorter
>>wavelengths)?
>
>Attenuation is a function of distance, not frequency.
>See <http://www.sss-mag.com/pdf/1propagation.pdf>.
>Methinks you have in mind reflection, diffraction, and scattering?

Oops -- also absorbtion.

--
Best regards, FAQ FOR CINGULAR WIRELESS:
John Navas <http://en.wikibooks.org/wiki/Cingular_Wireless_FAQ>

DecaturTxCowboy wrote:
> David wrote:
> Wiki is correct, but the point is Navis used it to dilute the importance
> of frequency in the attenuation.

Exactly!

3) The Wiki quote is not a relevant excuse to cover your ass. All the
Wiki quote is saying is *there are more additional factors that affect
the path loss in addition to the increased attenuation due to the higher
frequency*...most noticeable the first Fresnel zone clearance that
applies to terrestrial line of sight and near line of sight
communications (obviously not applicable to satellite line of sight
communications).

Scott wrote:
> "DecaturTxCowboy" <[email protected]> wrote in message
> news:[email protected]...
>> John Navas wrote:
>>> What I posted is correct, but you seem to be unable to be civil, and I'm
>>> not going to waste more time in a nasty argument, so I'm giving you the
>>> last word.
>> And that dear readers was classic Navas dodge.
>
> Oh, what a surprise- he got in over his head again.

Is it my imagination or does Navas,

a) Beat an off-NG topic (the Walmart thread comes to mind) to death with
opinions backed by Googled citations (relevant or not) until everyone
gives up responding to his nonsense.

b) Runs off into hiding when the facts don't support his delusions with
parting comments like...

Attack the man when you can't attack his logic (or to that effect).

One liners with no explanation what he is disagreeing with (called
leaving the door open for back pedaling).

Have a nice day.

Scott wrote:
> John Navas has become a pro at googling and posting links that contain
> information that is beyond his knowledge. As a result, most of it, while
> correct information, does not apply to the conversation at hand. This is
> one of those cases- he posted factual information that he has no experience
> with- it simply is not the correct information for the discussion being
> conducted.

Typical of a wannabe....

"DecaturTxCowboy" <[email protected]> wrote in message
news:[email protected]...
> Scott wrote:
>> "DecaturTxCowboy" <[email protected]> wrote in message
>> news:[email protected]...
>>> John Navas wrote:
>>>> What I posted is correct, but you seem to be unable to be civil, and
>>>> I'm
>>>> not going to waste more time in a nasty argument, so I'm giving you the
>>>> last word.
>>> And that dear readers was classic Navas dodge.
>>
>> Oh, what a surprise- he got in over his head again.
>
> Is it my imagination or does Navas,
>
> a) Beat an off-NG topic (the Walmart thread comes to mind) to death with
> opinions backed by Googled citations (relevant or not) until everyone
> gives up responding to his nonsense.
>
> b) Runs off into hiding when the facts don't support his delusions with
> parting comments like...
>
> Attack the man when you can't attack his logic (or to that effect).
>
> One liners with no explanation what he is disagreeing with (called
> leaving the door open for back pedaling).
>
> Have a nice day.
>

It looks like you have a grip on things- no imagination needed.