Richard Fry wrote:
"The most common radiator height for Class A non-directional AM
broadcast stations operating at 50kW day and night is 195-degrees."

I won`t challenge that as I have conducted no survey. WJR Detroit is
shown on the broadcast allocations map book as an unlimited (day and
night) 50 kilowatt Class 1 station.

Class 1 stations operate in a clear channel (this does not mean alone on
the channel) with an assigned power between 10kW and 50kW

A Class 2 station operates in a clear channel with an assigned power
between 250W and 50kW. They must operate so as to not cause interference
to the Class 1 stations. There are 29 clear channels which permit class
2 station operation.

Class 3 stations share regional channels and operate with assigned
powers between 500W and 5kW. Thyere are 41 regional channels and more
than 2000 Class 3 stations. These numbers were taken before expansion of
the AM broadcast band which has grown the totals.
Class 4 stations operate in assigned local channels with no more than
1kW day and 250W night assignments. There are 6 local channels with 150
or more Class 4 stations on each channel.

Primary service area is the statiob`s groundwave coverage. Secondary
service is uninterfered skywave coverage. Intermittent service lies
between the primary and secondary service areas.

A clear channel has one or more high-powered stations which serve wide
areas. All primary service areas and a substantial portion of their
secondary service areas are cleared of objectional interference.

A regional channel has stations not exceeding 5kW which have coverage
contours which limit the primary service interference between these
stations.

A local channel has stations not exceeding 1 kW daytime and 250 watts at
night. Primary coverage is limited by interference. Assignments are made
to limit interference.

Radio waves are radiated into a hemisphere as space below the antenna is
hidden by the surface of the earth. This results in a formula for the
field power at one mile from a perfect radiator emitting 1 kilowatt:
P = 1000 / 16266419 = 0.00006 watts/sq mtr
E = sq rt (PR)
and R=377 ohms

Volts/mtr=152 at 1 mile from a perfect infinitely short uniform
hemispherical radiator.

From a 1/4-wave vertical, it`s about 195 millivolts per mtr at 1 mile.

From a 1/2-wave vertical, it`s about 236 millivolts per mtr at 1 mile.

From a 5/8-wave vertical, it`s about 267 millivolts per mtr at 1 mile.

Volts vary as the square root of the power. So, for 50 killowatts,
multiply the 1 kilowatt values by 7.07.

The field strengths are the inverse distance or lossless values. Real
earth has losses.

Best regards, Richard Harrison, KB5WZI

I found that increasing the number of segments had a significant change
in the input Z. The material I read on 5/8 antennas indicated the real
part of the Z was near 50 ohms. I could not get that result until I
increased the number of segments. Guess it is cause 146 MHZ antennas
are a good bit shorter than 3.5 MHZ antennas, and any small deviation
such as lenght, or # of segments will change the end results.
Gary N4AST

A source is spread out over an entire segment. So when you change the
number of segments, you change both the length and the effective
position of the source. When the source is at the bottom of a quarter
wavelength radiator, small changes in source position don't make much
difference in the impedance seen by the source. However, when the
antenna approaches a half wavelength, the source impedance changes quite
dramatically with source position. Consequently, you'll see substantial
changes in reported source impedance with segmentation in that case.
This might or might not be the cause of what you're seeing. As an
experiment, you might try moving the source up one segment and see how
big a difference it makes.

Whenever the result is very sensitive to small changes in the model, you
shouldn't expect a real antenna to come out exactly like the model
predicts, since small differences between the model and real antenna
will likewise cause significant differences.

The absolute length doesn't matter -- a 146 MHz antenna will be no more
or less sensitive to the same amount of change (in terms of percentage
of the antenna size or of the wavelength) than a 3.5 MHz antenna if both
are proportioned the same. In fact, 146 MHz antennas are typically
considerably fatter in terms of wavelength than 3.5 MHz antennas, and
this makes them less sensitive to small changes.

Roy Lewallen, W7EL

wrote:
I found that increasing the number of segments had a significant change
in the input Z. The material I read on 5/8 antennas indicated the real
part of the Z was near 50 ohms. I could not get that result until I
increased the number of segments. Guess it is cause 146 MHZ antennas
are a good bit shorter than 3.5 MHZ antennas, and any small deviation
such as lenght, or # of segments will change the end results.
Gary N4AST

Richard Fry wrote:
"The most common radiator height for Class A non-directional AM
broadcast stations operating at 50kW day and night is 195-degrees."

"Richard Harrison" responded (in part)
I won`t challenge that as I have conducted no survey. WJR Detroit is
shown on the broadcast allocations map book as an unlimited (day and
night) 50 kilowatt Class 1 station.
__________________

Your text is referenced to out-of-date versions of applicable FCC Rules.
The FCC adopted the metric standard over 15 years ago, and the
classification of AM broadcast stations no longer is defined as Class 1 to 4
but Classes A, B, C & D. The current versions of the applicable Rules are
contained in 47CFR Part 73, and for this topic are dated October 1, 2004.

Minimum radiator heights in meters for Class A, B and C AM broadcast
stations are shown in Figure 7 of 47CFR73.190. Radiator efficiency for
Class A stations (other than in Alaska) must be such as to produce a ground
wave of least 362 mV/m at 1km for 1kW of antenna input power. A 90° omni
radiator cannot do that.

RF

Richard Fry wrote:
"Radiator efficiency for Class A stations (other than in Alaska) must be
such as to produce a groundwave of at least 352 mV/m at 1 km for 1kW of
antenna input power. A 90-degree omni radiator cannot do that."

Most such antennas have long been in place. Surely they are
grandfathered until the station is modified.

Best regards, Richard Harrison, KB5WZI

Richard Harrison"
Richard Fry wrote:
"Radiator efficiency for Class A stations (other than in Alaska) must be
such as to produce a groundwave of at least 352 mV/m at 1 km for 1kW of
antenna input power. A 90-degree omni radiator cannot do that."

Most such antennas have long been in place. Surely they are
grandfathered until the station is modified.
________________

The minimum efficiency for Class A is 362mV/m as I posted, not 352mV/m as
you "quoted me" above.

As far as the 50kW / non-D / 24-hr operations are concerned, their
facilities have avoided the use of 90° radiators nearly from their start,
due to the commercial/competitive need to generate a very strong groundwave
while avoiding self interference to that groundwave at night from their own
high-angle radiation. The advantages of using greater fractional wavelength
radiators to accomplish this has long been known and confirmed by field
measurements.

RF

I'm trying to understand the shunt tapped inductor as a circuit.

How does it work??? As long as the inductor cancels out the reactance of
the radiator you just need to find the 50 ohm point on the coil ... or for
that matter any feedline impedance???

Pardon my lack of knowledge.

de ka2pbt

"Richard Fry" wrote
As far as the 50kW / non-D / 24-hr operations are concerned, their
facilities have avoided the use of 90° radiators...
___________________

The list below illustrates my statement above. This data for most of the
50kW, 24-hr stations with the same radiation pattern day and night was taken
from a secondary source linked to the FCC database, current as of October
2004 (typos, if any, excepted).

The average height of these radiators is 188.8°, not including
WFAN and KDKA.

Data Format =
Freq, Call Letters, Tower Height in Degrees, Notation
640 KFI 175.7
650 WSM 190.3
660 WFAN 155.3 (top loaded)
670 WSCR 181.5
700 WLW 189.3
710 WOR 177
720 WGN 195
750 WSB 179.3
760 WJR 194.7
770 WABC 180.3
780 WBBM 194.1
810 WGY 182.9
820 WBAP 192.1
830 WCCO 194.4
840 WHAS 201.1
850 KOA 207.8
870 WWL 182.1
880 WCBS 207.1
890 WLS 189.3
1020 KDKA 280.28 (sectionalized)
1030 WBZ 188.5
1050 WEPN 186
1060 KYW 180
1070 KNX 193.5
1100 WTAM 185.2
1120 KMOX 192.7
1140 WRVA 185
1160 KSL 193.2
1180 WHAM 177.1
1200 WOAI 193.2
1210 WPHT 186

RF

Richard Fry wrote:
"The list below illustrates my statement above."

Yes, and it proves your point that non-directional 50 KW broadcast
stations in the U.S. use tall towers in terms of wavelength.

The r-f is precious and it makes sense to make the most of it. Thank you
for the list of high-powered stations which can be heard far and wide. I
still regret the end of WLW`s 500 KW days.

Best regards, Richard Harrison, KB5WZI

john doe wrote:
I'm trying to understand the shunt tapped inductor as a circuit.

How does it work??? As long as the inductor cancels out the reactance
of
the radiator you just need to find the 50 ohm point on the coil ...
or for
that matter any feedline impedance???

Pardon my lack of knowledge.

de ka2pbt

Hi John, Is the shunt tapped inductor you are refering to is a coil
with one end connected to the 5/8 radiator, the other end to the
radials, the coax shield to the radials, and the center conductor
tapped up the coil from the radials? If this is the case, according to
the Smith Chart, you can only get a perfect 50 ohm match if the real
part of the impedance you are trying to match less than 50 ohms. Our
models indicate this is not the case with a 5/8 radiator.
The shunt tapped inductor is actually a 2 element matching system,
with both elements being inductors. As you stated, the inductor in
series with the radiator cancels the reactance, and the tapped shunt
inductor provides a 50+j0 point. Works only if the real part of the
impedance is 50 ohms.
Gary N4AST