how can i figure this out?

Question

if my location is 1150 feet above sea level

most of the surrounding towns and cities are at about 1000-1200 feet above sea level.

my geography around me is hilly.

my antenna is 50 feet above the ground and is 22 feet tall, it is an omnidirectional antenna.

i beleive the entire 22 feet radiates the R.F.

i am transmitting at 28.555mhz with a 150 watt deadkey and1000 peak @ 90% modulation

not counting for skipping off the ionsphere ( no skip)

how far away theroetically, could the signal be recived by a similar station.

how about the antenna at 70' which should be right at a 2 wavelengths which would be ideal.

i hope i have included enough info

it would be am mostly, but i have the oprtion of SSB and with FM it would only be 100watts

Answer 1

"The Earth's equatorial radius, or semi-major axis, is the distance from its centre to the equator and equals 6,378.135 km (≈3,963.189 mi)" To simplify your problem, lets consider this a constant radius, R:

Let h = height of your antenna72' and 92',
a = your altitude, and
d=a+h = distance above the earth, 1172' and 1192'.

We also have to assume that R defines mean sea level.

let s be the desired distance. then
s/2 is the distance to the horizon.
(s/2)^2 = (R+d)^2 - R^2
= (d/5280)^2 + 2Rd/5280 miles, d given in ft., R in miles
s=2*sqrt(d))*sqrt(2R/5280+d/5280/5280)
=2*sqrt(d)*sqrt(1.5012 + d/27878400)
1192/27878400=.000042757, which is insignificant compared to 1.5012, so I drop the 2nd term:
s=(6d)^0.5 miles makes a pretty good approximation
s1 = 83.9 or 84 miles
s2 = 84.6 or 85 miles

These are maximum distances with no hills or obstructions rising above your sight line. Otherwise, of course, your maximum theoretical distance is the distance to the obstruction.

BTW, according to Louis L'lamour, this relation was derived by sailors to determine how far you could seethings at sea. The simplification used allowed them to calculate distances in their head.

Answer 2

I presume you have an amateur radio license of at least General, Advanced, or Extra class, and therefore are doing this legally.

Since the antenna is omnidirectional and more than a quarter wave long at 10 meters, only about 8 feet are radiating, and the rest is isolated by a series trap. If this antenna is resonant and above a near-perfect ground plane, you have the equivalent of a dipole antenna, oriented vertically, at a height of 1200 feet ASL.

Even if you are running 150 watts DC input power with the key down in CW mode, I don't think there's any you can get a peak effective power (PEP) as high as 1000 watts. So figure the 150 watt CW signal.

The groundwave signal is limited approximately to line-of-sight. If Google Earth has reasonably accurate elevation information for your area (use the beta version), you can change your angle of view and zoom out to 50-100 miles and get a fair idea of where your signal can be detected. But you need an educated engineering estimate rather than a theoretical calculation. The effect of your antenna, immediate terrain, and general terrain is so great it's nearly impossible to calculate.

A nearly-ideal antenna 2 wavelengths above the nearby terrain is nearly optimal.

So now it's an estimate. A similar antenna at an arbitrary location 50 miles away has probably a 70% chance of detecting your signal. 75 miles away, probably 50%. 100 miles away probably 20%.

Answer 3

I can't say that I know anything about this but wouldn't the answer be within a range of having a 222 ft to 22 ft antenna on level ground?

By that I mean if the lowest surrounding town was at 1000 ft and the highest at 1200 ft, with your 22 ft antenna mounted 50 ft in the air would mean that the top of your antenna would at 1222 ft, 22 ft higher than the highest surrounding town and 222 ft higher than the lowest.

So wouldn't the furthest a signal would travel would be the same as having a 222 ft high antenna on level ground, or is that over simplifying?

Answer 4

You should be able to x-mit to every antenna with a clear line of site w/ your antenna. approx.(30 mi) Unfortunately, you would only be able to receive signals that had a minimum of 15W at your max range.(ideal conditions of course). Elementary calculation is point five watts * miles is max range, to horizon approx.(30 mi). [theoretically, if you transmitted from Mt. Everest, your signal could be received 75 mi away. Airplane 75 mi away with line of site could receive your signal also]

Answer 5

78 miles

Answer 6

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- Draw a picture of the information you know

- Mark the values you know (with arrows, lines etc)

- Write down the equations you know (decay, wavelength etc)

- Calculate unknowns using the knowns




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Answer 7

pretty far with an antina that big and that kid of elivation, i could give you an better gess if you told me what you are trying to pic up (am,fm,lsb,umd,tv......)

Answer 8

I have no idea what you are talking about but i like your monkey picture :)