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Calculating AM directional ERP based on field strength?

Well, I would expect the fence around the towers to be far enough out so that the field outside that fence would be less than the 614 V/m that the FCC specifies for MPE in most of the AM band.
Also, these radios were having difficulty even BEFORE taking them near the towers. My grandma is a whopping SIX THOUSAND millimeters, if not more, from 23kW 1300 KAZN and 50kW 1430 KMRB, and my radios, even before they were tested near towers, had trouble with locals on 1230, 1260, 1280, 1330, 1390 (a bit stronger than the others), 1460, 1480, 1580 (noisy even though it's 50kW) and 1650.

In this article, the predicted minimum distances (on pages 4-5) for a few situations are as follows:
50kW, ~640kHz - 5/8 wave = 4 meters, 1/4 wave = 4 meters, 1/10 wave = 13 meters.
5kW, ~1250kHz - 1/4 wave = 2 meters.
I was a good 5.2 meters away from one of the towers for 2.5kW 590 KTIE (DA-2 using 65.2° self-supporting radiators - I measured on Google Maps from under the center of the base), outside the fence and was only there for a few minutes. In that article, the minimum distance for 5kW on that frequency for a 1/4-wave radiator is 2 meters.
Another time, I was also 5.2 meters from 5kW 1290 KKDD (DA-2 with 90° self-supporters), again outside the fence, and only for a few minutes. Minimum distance in that situation, according to the linked article, is also 2 meters.

In no case do I believe I was exceeding MPE.
I would sometime like to experiment with how close I would have to get, using a small ferrite bar antenna, to get the same signal strength that I would get using a longwire beverage antenna and several foot tuned loop, assuming I'm using the same radio in both cases, just a different antenna configuration. Just as an example, using a Select-A-Tenna and untuned longwire with my PL-606, I got a comparable signal on a 50kW on 760 from 11.7 km away, that I had to be less than 6 meters from the 2.5kW on 590 to get the same signal just using its internal ferrite bar. I've used the longwire+loop within about 0.1 or 0.2 km of a 50kW station on 1170 (diplexing with a 5kW on 910), so I wonder how close I'd have to get to have the same signal level using only the built-in ferrite? I expect it would have to be over the MPE, so I'd either have to consult with a station engineer to assist in the experiment (and be expected to pay for his time at minimum), or just attempt to calculate it.

Do any consumer / enthusiast (not military grade or extremely high end / expensive) radios exist that can handle being within a dB of MPE without overloading?
 
Hah... yes, you're right. :) Thanks for the correction.
Although I WILL point out that your post is missing a 0... 600 meters is actually 600,000 mm.... or did you mean to say 60,000 cm?
 
boiseengineer said:
Sooo, are we figuring the "ERP" of the signal as referenced to an isotropic radiator?
Click to expand...

Nooo, we are figuring out the ERP of an AM (530-1700 kHz) directional station in a specific direction (looking at the field strength at 1 km listed in the FCC listing), as referenced to their transmitter power operating non-directionally (whose field is listed as the RMS field strength at 1 km). Basically, how much power would a particular station have to transmit, operating non-directionally, to have the same field at a particular location as it does when operating directionally/normally? That RMS field at 1 km will vary due to various antenna configurations. For example I know of a station (KTIE) that uses 65.2° towers, KFI uses an almost half-wave tower with a top hat (and probably a ground radial system, maybe 120 radials each 90° long), and KFBK uses towers that consist of two 180° elements stacked on top of each other, insulated from each other, with no ground system. Which of those is an isotropic radiator?
One original reason for asking the question was I wanted to find out the equivalent power / ERP / (suggest a better term?) that a few directional stations broadcast toward me. I was also wanting to figure out whose Franklin antenna was more efficient, KFBK which is 3,545.89 mV/m RMS @ 1 km for 50kW, or KSTP, which is 511.77 mV/m @ 1 km for 1 kW (but actually transmits 50 kW - when specifying the RMS for non-directional operation the FCC assumes a transmit power of 1 kW.)
 
tfcwings said:
boiseengineer said:
Sooo, are we figuring the "ERP" of the signal as referenced to an isotropic radiator?
Click to expand...

  Which of those is an isotropic radiator?
Click to expand...

Neither... In reality, there is no such thing! Only theory...

[/quote]
I was also wanting to figure out whose Franklin antenna was more efficient, KFBK which is 3,545.89 mV/m RMS @ 1 km for 50kW, or KSTP, which is 511.77 mV/m @ 1 km for 1 kW (but actually transmits 50 kW - when specifying the RMS for non-directional operation the FCC assumes a transmit power of 1 kW.)
[/quote]

If you use RSS calculation, KFBK wins; if RMS KSTP wins. KFBK uses two directional patterns and mutual impedances (slewing) are at play where KSTP is non directional. From a peak horizontal gain standpoint at a given azimuth, KFBK wins.

w/
 
Watt Hairston said:
tfcwings said:
boiseengineer said:
Sooo, are we figuring the "ERP" of the signal as referenced to an isotropic radiator?
Click to expand...

Which of those is an isotropic radiator?
Click to expand...

Neither... In reality, there is no such thing! Only theory...

I was also wanting to figure out whose Franklin antenna was more efficient, KFBK which is 3,545.89 mV/m RMS @ 1 km for 50kW, or KSTP, which is 511.77 mV/m @ 1 km for 1 kW (but actually transmits 50 kW - when specifying the RMS for non-directional operation the FCC assumes a transmit power of 1 kW.)
Click to expand...

If you use RSS calculation, KFBK wins; if RMS KSTP wins. KFBK uses two directional patterns and mutual impedances (slewing) are at play where KSTP is non directional. From a peak horizontal gain standpoint at a given azimuth, KFBK wins.

w/
Click to expand...

Ahh... well I was noticing that KFBK uses two 180° stacked elements per tower, whereas KSTP uses two 179.4° stacked elements. I was wanting to figure out whose was more efficient on groundwave assuming non-directional operation. I was thinking KFBK was more efficient, but when I went to figure out what KFBK's field (based on that field) would be at 1 kW, using the inverse distance formula, it looked like KSTP was slightly better. Maybe I was doing it wrong? So what should be the efficiency of the best possible AM antenna? 512 mV/m @ 1 km @ 1 kW? 515 mV/m @ 1 km @ 1 kW? Some other figure? Is 511.77 mV/m @ 1 km @ 1 kW the best that you can get?

Also what about the efficiency of other shorter Franklin-type antennas? For example how would two stacked 90° elements or two stacked 45° elements rank relative to a single 180° over a groundplane or a single 90° over a groundplane?
Or if we want to go REALLY short, what about two stacked 1.5 meter elements (for part 15 operation) compared to a single 3-meter element with 8 1.5-meter ground radials (or with NO ground radial system at all), for example?
 
tfcwings said:
So what should be the efficiency of the best possible AM antenna?
Click to expand...

Here is a link to an FCC chart showing this for typical broadcast-type monopoles for heights of 0.05 to 0.675 wavelength.

Fields for applied powers other than 1 kW may be calculated by multiplying the fields shown in the chart by the square root of the dividend of the other power (in kW) and 1 kW.

For example, a 1/4-wave monopole produces an inverse distance field (IDF) of about 306 mV/m at 1 km with 1 kW of applied power. If the applied power is 50 kW then the 1 km IDF will be 306 mV/m x SQRT(50) = 2,164 mV/m.

Note that the groundwave field at 1 km and greater distances over a real earth path will be less than the IDF value, depending on ground conductivity and frequency.

http://i62.photobucket.com/albums/h85/rfry-100/Monopole_Fields.gif
 
Interesting chart. So IDF of a 1/2-wave antenna is about 381 mV/m @ 1 km, and for a 5/8-wave antenna it's about 442 mV/m @ 1 km, both assuming 1 kW of applied power. It's also interesting to note that once you go below approximately 1/8 to 1/10 wavelength, the efficiency drops more rapidly.

So where does KSTP factor into that chart, though? They have an IDF of 511.77 mV/m @ 1 km for 1 kW according to the FCC. (As I'm sure you know, FCC assumes an applied power of 1 kW for non-directional operation when specifying the IDF.) I'm not finding where on the curve that IDF is.

Also how would I figure out the IDF of a segmented antenna (like WHO, KSTP or KFBK) or a top-loaded antenna (like KFI)?
 
tfcwings said:
So where does KSTP factor into that chart, though?. They have an IDF of 511.77 mV/m @ 1 km for 1 kW according to the FCC.
Click to expand...

They don't. As shown on the chart, it applies only to standard (non-sectionalized) vertical radiators using at least 120 x 1/4-wave (buried) radials.

But if their daytime, non-D IDF for 1kW at 1 mile is 511.77 mV/m, then their 50 kW IDF at 1 km may be found using the simple math relation stated and illustrated in my last post above.

Also how would I figure out the IDF of a segmented antenna (like WHO, KSTP or KFBK) or a top-loaded antenna (like KFI)?
Click to expand...

The values for those stations are available through AM Query, just as you found for KSTP.
 
R. Fry said:
They don't. As shown on the chart, it applies only to standard (non-sectionalized) vertical radiators using at least 120 x 1/4-wave (buried) radials.

But if their daytime, non-D IDF for 1kW at 1 mile is 511.77 mV/m, then their 50 kW IDF at 1 km may be found using the simple math relation stated and illustrated in my last post above.
Click to expand...
Actually it's at 1 km, but otherwise that equation would work. For KSTP, if I calculated correctly, their IDF at 1 km for 50kW is 3,618.76 mV/m. KFBK's is 3,545.89 mV/m (2 towers, stated in FCC data), iirc, which works out to an IDF for 1kW of 501.46mV/m @ 1 km. KFBK's antenna has two 179.4° segments, while each of KFBK's two towers has two 180° segments. IIRC, WHO has two 150° segments, with an IDF for 1kW of 471.54 mV/m @ 1 km.

The values for those stations are available through AM Query, just as you found for KSTP.
Click to expand...
I realize that. I just named them as examples, as i'd like to figure out how to calculate IDF for other electrical lengths of segmented antennae.
 
tfcwings said:
Actually it's at 1 km, but otherwise that equation would work.
Click to expand...

Yes I was wrong when writing 1 mile when quoting you (the result of an old habit from the days before the FCC converted to the metric system).

But in any case, the equation calculates the new IDF for a change in applied power with respect to the reference value for 1 kW at that distance, whether the distance is given in miles, kilometers, furlongs or any other linear unit of measure.

i'd like to figure out how to calculate IDF for other electrical lengths of segmented antennae.
Click to expand...

You could model those in NEC, and if properly done it could give you some idea. Normally such numbers are produced by the broadcast engineering/consulting firm who designed the sectionalized antenna.
 
Afterthoughts:

Just to note that it is not very likely now that an AM broadcast station in the U.S. not already using a sectionalized ("Franklin") antenna will elect to install one, or a directional array using one or more of them.

Given the competitive environment that AM broadcast stations now face from many other claimants for listeners, the cost of such in initial engineering, zoning/FCC/FAA clearances, hardware, installation, and maintenance probably has surpassed the commercial value of such an investment.

Years ago at least one, prominent, 24/7, 50 kW, non-D AM broadcast station in the U.S. formerly using a sectionalized, Franklin-type radiator has replaced it with a simple, non-sectional monopole.

So, tfcwings, maybe this will or should affect your wish to "figure out how to calculate IDF" for other stations using "segmented antennae" ?
 
Tfcwings:

About 50 years ago, perhaps longer, Charles Jeffers of WOAI San Antonio Texas had an extensive write-up in the IEEE journals that was an excellent treatment of the design of sectionalized antennas. These antennas behave like vertically stacked phased arrays and their synthesis revolve around that concept of a two-tower phased array. I saw another similar write up by A.D. Ring consulting engineers and it could have been in collaboration with Jeffers. It is a through treatise of the subject and a must read if you want a good basis for what is involved, might want to make a visit to your local library. Following that, you then can use NEC MOM modeling to develop the input geometry and run the model. Outside that, the complex equations, their execution and results is far beyond the scope of this discussion board. Reading this as well as other threads on this topic, all assumes your have the adequate mathematical, engineering and experience background necessary to fully comprehend the science. The ultimate answer will not be gained here in a few replies to Radio-Info.

Best of luck.
w/
 
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