Would an old salt help me with an AM calculation?
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Thread: Would an old salt help me with an AM calculation?

  1. #1

    Would an old salt help me with an AM calculation?

    Would one of you old salts out there that did this over and over in the bad old days help me?

    If you had a 250 watt nondirectional daytime AM station using a 1/4 wave series fed antenna with a ground conductivity of 8 mhos on 1480 khz how far from the antenna would you expect the ground wave 4 mv/m contour to fall everything else being equal? Been too long, don't remember how to calculate it anymore Thanks
    Larry

  2. #2
    For the most accurate response, please post the physical details of the complete antenna system including elevation of the base of the monopole above physical earth, and the r-f ground reference against which that monopole is driven (buried radials, elevated counterpoise, or ?).

  3. #3
    We are talking a guyed vertical tower 12 inches on a face with the guys broken up by insulators. The tower itself is the radiator and it is series fed above a base insulator. Tower sections are welded on one leg at each section junction. Total height of the tower AGL is 1/4 wavelength for 1480 khz. The buried ground system is made up of 120 evenly spaced number 24 copper ground radials each 150 feet in length buried 12 inches deep. There is a wire mesh copper screen around the base of the tower and forms a square extending outward 10 feet from each corner of the square to the tower base. All are silver soldered together with the ground radials attached to a copper strap which extends around the outer edge of the wire screen.

  4. #4
    If that 1480 kHz antenna system is in good condition, with 250W of applied power it should generate an inverse distance groundwave field of about 153 mV/m at 1 km.

    The 4 mV/m groundwave contour for 8 mS/m earth conductivity would be located about 8.5 miles from the tower.
    Last edited by rfry; 11-04-2015 at 06:49 AM.

  5. #5
    Thanks so much for the help. I haven't been able to find my old copy of the curves that I used to calculate that on so your a big help thanks again

  6. #6

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