Readers may find this article of interest:


VHF Marine Radio Communication

http://continuouswave.com/whaler/reference/VHF.html

In article ,
James Hebert wrote:

http://continuouswave.com/whaler/reference/VHF.html

I find your math to be very good, but your conclusions don't track
very well with REAL World experience.

Bruce in alaska
--
add a 2 before @

Would you care to be more specific?


"Bruce in Alaska" wrote in message ...
In article ,
James Hebert wrote:

http://continuouswave.com/whaler/reference/VHF.html

I find your math to be very good, but your conclusions don't track
very well with REAL World experience.

Bruce in alaska
--
add a 2 before @

On Sun, 9 Jan 2005 07:47:25 -0800, "Chuck Tribolet"
wrote:

Would you care to be more specific?


"Bruce in Alaska" wrote in message ...
In article ,
James Hebert wrote:

http://continuouswave.com/whaler/reference/VHF.html

I find your math to be very good, but your conclusions don't track
very well with REAL World experience.


off the top of my head, looking at the article, i think there's an
error. he says EACH antenna needs to be 12.4 feet above the surface,
but i think the SUM of the antenna heights needs to be this...IOW each
antenna needs to be 6.2 feet high.

---------------------------
to see who "wf3h" is, go to "qrz.com"
and enter 'wf3h' in the field

In article , (Bob) wrote:

On Sun, 9 Jan 2005 07:47:25 -0800, "Chuck Tribolet"
wrote:

Would you care to be more specific?


"Bruce in Alaska" wrote in message
...
In article ,
James Hebert wrote:

http://continuouswave.com/whaler/reference/VHF.html

I find your math to be very good, but your conclusions don't track
very well with REAL World experience.


off the top of my head, looking at the article, i think there's an
error. he says EACH antenna needs to be 12.4 feet above the surface,
but i think the SUM of the antenna heights needs to be this...IOW each
antenna needs to be 6.2 feet high.

---------------------------
to see who "wf3h" is, go to "qrz.com"
and enter 'wf3h' in the field

If two vessels are ten miles apart, they will each
need antennas 12.4 feet high in order for their
radio horizons to be in view (line-of-sight) of
each other.

The case presented shows how much margin there
is in a typical circuit. There are many poor
radio installations aboard recreational vessels
which can barely talk to the marina office
from its dock own gas dock, but this does
not constitute a negation of laws of physics.

On Sun, 09 Jan 2005 23:52:46 -0500, James Hebert
wrote:

In article , (Bob) wrote:


off the top of my head, looking at the article, i think there's an
error. he says EACH antenna needs to be 12.4 feet above the surface,
but i think the SUM of the antenna heights needs to be this...IOW each
antenna needs to be 6.2 feet high.

---------------------------
to see who "wf3h" is, go to "qrz.com"
and enter 'wf3h' in the field

If two vessels are ten miles apart, they will each
need antennas 12.4 feet high in order for their
radio horizons to be in view (line-of-sight) of
each other.

why is that? the equation sez 12.4 feet for antenna height. does that
mean 1 antenna could be on the ground and the other at 12.4 feet?
---------------------------
to see who "wf3h" is, go to "qrz.com"
and enter 'wf3h' in the field

In article ,
James Hebert wrote:

If two vessels are ten miles apart, they will each
need antennas 12.4 feet high in order for their
radio horizons to be in view (line-of-sight) of
each other.

The case presented shows how much margin there
is in a typical circuit. There are many poor
radio installations aboard recreational vessels
which can barely talk to the marina office
from its dock own gas dock, but this does
not constitute a negation of laws of physics.

this is where the Practical and empirical evidence shows that
the math isn't showing what really is hapopeneing.

I have two 1 watt Vhf handhelds, with rubber antennas.
I can talk 16 miles over water with these two radios.
Both myself and my wife are less than 6.5 ft tall.
We are both standing at sealevel. (water lapping at our feet)
Now how does you MATH explain this empirical DATA?

Do my handhelds have receive sensitivity lower than atmospheric
noise? Maybe the Laws of Physics cease to apply north of 58 degrees?
Since this is a perfectly viable Path 24/7 and we have used it
daily for the last 15 years, what is your explanation?

Me who actually does know the answer.........

"Me" wrote in message
...
this is where the Practical and empirical evidence shows that
the math isn't showing what really is hapopeneing.

I have two 1 watt Vhf handhelds, with rubber antennas.
I can talk 16 miles over water with these two radios.
Both myself and my wife are less than 6.5 ft tall.
We are both standing at sealevel. (water lapping at our feet)
Now how does you MATH explain this empirical DATA?

VHF waves tend to follow the curvature of the earth slightly (just like
light diffracts around sharp corners), which indeed gives you more range
than the maths tells you.

Meindert

Sealevel is not a word. If you mean sea level, that does not equate to
"water lapping at our feet". If you are claiming a 16 mile range, direct,
when you are both at 6.5 feet above sea level, I say your full of crap. If
this is a riddle, then I say your both at a seperate body of water with a
valley between, or your communicating via a repeater.

Eric

"Me" wrote in message
...
In article ,
James Hebert wrote:

If two vessels are ten miles apart, they will each
need antennas 12.4 feet high in order for their
radio horizons to be in view (line-of-sight) of
each other.

The case presented shows how much margin there
is in a typical circuit. There are many poor
radio installations aboard recreational vessels
which can barely talk to the marina office
from its dock own gas dock, but this does
not constitute a negation of laws of physics.

this is where the Practical and empirical evidence shows that
the math isn't showing what really is hapopeneing.

I have two 1 watt Vhf handhelds, with rubber antennas.
I can talk 16 miles over water with these two radios.
Both myself and my wife are less than 6.5 ft tall.
We are both standing at sealevel. (water lapping at our feet)
Now how does you MATH explain this empirical DATA?

Do my handhelds have receive sensitivity lower than atmospheric
noise? Maybe the Laws of Physics cease to apply north of 58 degrees?
Since this is a perfectly viable Path 24/7 and we have used it
daily for the last 15 years, what is your explanation?

Me who actually does know the answer.........

On Mon, 10 Jan 2005 17:38:06 GMT, (Bob) wrote:

On Sun, 09 Jan 2005 23:52:46 -0500, James Hebert
wrote:

In article , (Bob) wrote:


off the top of my head, looking at the article, i think there's an
error. he says EACH antenna needs to be 12.4 feet above the surface,
but i think the SUM of the antenna heights needs to be this...IOW each
antenna needs to be 6.2 feet high.

---------------------------
to see who "wf3h" is, go to "qrz.com"
and enter 'wf3h' in the field

If two vessels are ten miles apart, they will each
need antennas 12.4 feet high in order for their
radio horizons to be in view (line-of-sight) of
each other.

why is that? the equation sez 12.4 feet for antenna height. does that
mean 1 antenna could be on the ground and the other at 12.4 feet?
---------------------------
to see who "wf3h" is, go to "qrz.com"
and enter 'wf3h' in the field

It means that the horizon is halfway between. If one antenna was at
sea level and the other at 12.4 feet you would only talk half the
distance.

Regards
Gary