Service Volumes of VOR's make no sense

On Sat, 14 May 2005 18:53:06 -0400, "CryptWolf"
wrote in
1116111148.11781a37a6d5a6e2b697012478f45470@teran ews::

Since the radiation pattern is reduced at higher altitudes, there is less
chance
of frequency congestion and receiving a signal you don't want.
With limited frequencies available, you have to depend on other
limits to prevent unwanted reception of other signals.

The radiation pattern also puts more energy where it was needed, as
that energy that would be radiated upward is directed laterally
instead.

Um, that's just not true. SInce the radiation pattern is reduced at higher
altitudes, the chance for frequency congestion is every bit as probable. If
all signals are reduced proportionally, then the RELATIVE signal strengths
remain constant.

Jim



Since the radiation pattern is reduced at higher altitudes, there is less
chance
of frequency congestion and receiving a signal you don't want.
With limited frequencies available, you have to depend on other
limits to prevent unwanted reception of other signals.

And a screw-up on the equation:

Radio horizon (in miles) equals the square root of [TWO TIMES the aircraft
altitude above the VOR (in feet)].

Jim



"Antoñio" wrote in message
...

The equation for this is
that radio horizon (in miles) equals the square root of the aircraft
altitude above the VOR (in feet).

On Sat, 14 May 2005 at 17:50:55 in message
, Peter Duniho
wrote:
"David CL Francis" wrote in message
...
The real earth is not of course that flat except over the oceans!

Actually, as long as we're being pedantic, it's not even flat over the
oceans. It's much flatter, but the Earth simply is not an ideal "smooth
sphere" anywhere on its surface.

I did not realise that I was being pedantic!

Also the further away you go the closer the horizon distance gets to being
the same as the height. It is obvious that from the moon you can almost
see the entire hemisphere.

Almost. But the difference is significant enough to matter when you
really care whether you can see the entire hemisphere (astronomy, for
example).

Of course. But the point is not important when considering VORs!


--
David CL Francis

On Sun, 15 May 2005 at 09:05:01 in message
, RST Engineering
wrote:
And a screw-up on the equation:

Radio horizon (in miles) equals the square root of [TWO TIMES the aircraft
altitude above the VOR (in feet)].

[1] Jim, Doesn't that assume that the number of feet in a mile is
the same as the radius of the earth in miles?

The formula for the tangential horizon distance is

[2] d=(2*r*h + h^2)^0.5

where d is the tangential horizon distance, r is the radius of the earth
and h is the height of the object above the surface. DKr and h all in
the same units. Because the heights we are talking about are small
compared to the radius of the earth the h^2 term can be ignored leaving
what you said.

E&OE :-)
--
David CL Francis

And you asked in another post if you were being pedantic? PEDANTIC? My
guess is that you stay up late at night worrying about whether
anal(-)retentive is hyphenated or not.

Jim



"David CL Francis" wrote in message
...
On Sun, 15 May 2005 at 09:05:01 in message
, RST Engineering
wrote:
And a screw-up on the equation:

Radio horizon (in miles) equals the square root of [TWO TIMES the aircraft
altitude above the VOR (in feet)].

[1] Jim, Doesn't that assume that the number of feet in a mile is
the same as the radius of the earth in miles?

"RST Engineering" wrote in message
...
Um, that's just not true. SInce the radiation pattern is reduced at
higher
altitudes, the chance for frequency congestion is every bit as probable.
If
all signals are reduced proportionally, then the RELATIVE signal strengths
remain constant.

Jim

Under normal atmospheric conditions, excluding anomolies, as the frequency
or distance increases, the required transmitter power increases while the
recieved signal strength remains the same. At some point, even if you use
the same frequencies, a VOR or any radio signal will vanish into the
background noise. The reciever sensitivity is limited by background noise.
If you like, I'll look up the specific math and we can really get technical.
All but a few would understand it or care. This would be the other limit
that
I didn't explain previously.

All you have to do is space VOR's of the same frequency far enough apart
and they won't interfere. The fact that the radiation pattern is reduced for
higher altitudes seems to imply that the radiation pattern was designed
to reduce transmitted power and limit reception distances at those altitudes
where line of sight would not be a factor.

For reference you might want to find:
Schoenbeck, Electronic Communications Modulation And Transmission
It's one of the books I used when I was working on my electronics degree.

Since the radiation pattern is reduced at higher altitudes, there is
less
chance
of frequency congestion and receiving a signal you don't want.
With limited frequencies available, you have to depend on other
limits to prevent unwanted reception of other signals.

"Antoñio" wrote in message
...

I am not sure. Is it because the radiation pattern is spherical and not
line-of-sight?


You're looking at just one VOR at a time. There are about 1000 VORs in the
US and just 100 VOR frequencies. The service volume has to ensure not only
usable reception of the desired VOR, but non-reception of undesired VORs on
the same frequency.

As an example, let's say you're flying from EAU VOR to LAN VOR at 15,500'.
They're about 320 miles apart but at that altitude line-of-sight distance is
about 180 miles so you should receive LAN before losing EAU, even though
you're well outside of the standard service volume of forty miles. When
you're about halfway you switch from EAU to LAN, but you're unable to get a
reliable signal.

The problem may be that you're closer to ESC and RFD VORs than you are to
LAN, and they all operate on 110.8.

For reference you might want to find:

Hamsher, "Communications Systems Engineering Handbook" ; Jasik, "Antenna
Engineeiring Handbook"; Kraus "Antenna Design"; MIT Radiation Lab Series;
any of the ARRL publications on antennas

which is what I recommend that you use when I teach you when you are working
on your electronics degree...

Jim





For reference you might want to find:
Schoenbeck, Electronic Communications Modulation And Transmission
It's one of the books I used when I was working on my electronics degree.

Steven P. McNicoll wrote:

You're looking at just one VOR at a time. There are about 1000 VORs in the
US and just 100 VOR frequencies. The service volume has to ensure not only
usable reception of the desired VOR, but non-reception of undesired VORs on
the same frequency.

Ok...I think I'm with you (though feeling like I might be exhibiting
that I am a little obtuse at this point in the discussion). I assume
that your post means you think I haven't quite got it yet?

Are you saying that the service volume diagrams are artist renderings of
the reception distances that the FAA have tested and will *guarantee* to
be usable?? So the "double/inverted wedding cake" structure (which, as
you recall, was the basis of my original question) really has little to
do with the *actual* signal propagation distances of a particular VOR
but, rather, provide approximations by taking into consideration the
real world interference of other VORs, spherical wave radiations,
curvature of earth, etc. ?

So I would not necessarily loose VOR service if, while using a Standard
High Altitude Service Volume station, I were to climb above FL45, with a
depicted service distance of 130nm, to FL46 with it's depicted service
distance of 100nm?

Antonio (thinking he had it, then....)