Service Volumes of VOR's make no sense

But Hilton, that doesn't explain the effect observed. What DOES explain it
is the antenna pattern that the FAA chose for the VORs. In their infinite
wisdom they never considered that aircraft would fly much above FL250.
Remember, this was in the late 40s and early 50s that the system was
designed.

Therefore, they "squished" the antenna pattern to squirt more signal at a
lower radiation angle than an isotropic ("all angles") radiator. Think of
it as a ball of dough (isotropic) that has been squished to become a pancake
(low angle radiation). If you are ABOVE the pancake, you receive less
signal strength than if you are in the dough, so to speak.

There are two effects here. One is "radio horizon" which limits low
altitude reception to what the antenna can "see". The equation for this is
that radio horizon (in miles) equals the square root of the aircraft
altitude above the VOR (in feet). Thus, an aircraft near San Diego
receiving SAN VORTAC (which is on an island near Pacific Beach, damn near as
close to sea level as you can get) at an altitude of FL180 will have a radio
horizon of 134 miles, almost exactly what the fellow said, and will be
almost in the dead center of the antenna "beam" pattern. However, take that
same aircraft in the same geographic spot and honk it up to FL500, the radio
horizon moves to 224 miles, but you have climbed yourself way above the beam
and the signal strength has dropped below usable..

Howzat?

(Signal strength, BTW, falls off as the SQUARE of the distance.)

Jim




"Hilton" wrote in message
.net...
However, the FAA has depicted cylinders of various diameters stacked
upon each other. Given that the VOR is line-of-sight, I did not
understand why, for example, a VOR would be received 130nm out at FL180
yet only be received 100nm at FL500. Doesn't it logically follow that at
the higher altitude the VOR would be able to be received further out?

No, then it wouldn't fit in the semi-sphere. Signal strength drops off
(non-linearly I believe) as you move away from its source, so the further
you go, the weaker it becomes, hence the semi-sphere. Since the sphere
tapers off at the top, so too do the cylinders.

"RST Engineering" wrote in message
...
But Hilton, that doesn't explain the effect observed. What DOES explain
it
is the antenna pattern that the FAA chose for the VORs. In their infinite
wisdom they never considered that aircraft would fly much above FL250.
Remember, this was in the late 40s and early 50s that the system was
designed.

The engineers may have been more farsighted than you give them credit.
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.

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.

"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.

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.

RST Engineering wrote:
But Hilton, that doesn't explain the effect observed. What DOES explain it
is the antenna pattern that the FAA chose for the VORs. In their infinite
wisdom they never considered that aircraft would fly much above FL250.
Remember, this was in the late 40s and early 50s that the system was
designed.

Therefore, they "squished" the antenna pattern to squirt more signal at a
lower radiation angle than an isotropic ("all angles") radiator. Think of
it as a ball of dough (isotropic) that has been squished to become a pancake
(low angle radiation). If you are ABOVE the pancake, you receive less
signal strength than if you are in the dough, so to speak.

There are two effects here. One is "radio horizon" which limits low
altitude reception to what the antenna can "see". The equation for this is
that radio horizon (in miles) equals the square root of the aircraft
altitude above the VOR (in feet). Thus, an aircraft near San Diego
receiving SAN VORTAC (which is on an island near Pacific Beach, damn near as
close to sea level as you can get) at an altitude of FL180 will have a radio
horizon of 134 miles, almost exactly what the fellow said, and will be
almost in the dead center of the antenna "beam" pattern. However, take that
same aircraft in the same geographic spot and honk it up to FL500, the radio
horizon moves to 224 miles, but you have climbed yourself way above the beam
and the signal strength has dropped below usable..

Howzat?

(Signal strength, BTW, falls off as the SQUARE of the distance.)

Jim

Very comprehensive and "engineer like" elaboration to Hilton's answer.
I think I finally get the picture. Thank you!

Antonio

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 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?