1/10/2024 0 Comments Horizon line calculator![]() ![]() Realistically, the theoretical horizon for your aircraft at 40,000' is around 280 miles. This situation is less than 1 degree above horizontal, not 2 to 5 as you state. Improved my TV signals quite a bit.Ī ridge a mile away and 70' higher than your aerial you will see next to no difference in reception from an aircraft at 40,000' if you removed the ridge (ie. My UHF TV antenna even has a beam tilt setting, which I use to raise the forward end of the antenna so it points at the top of the ridge, rather than directly into it. The vertical shoots the signal right over the close in stations, but excels out at the long ranges. Same reason a 1/2 wave dipole is much more useful for short range comms on 40 and 80 Meters than a vertical. If I had to go up an additional 100' to do it, it just would not be cost effective.Īs for angle of radiation, if the antenna has all its gain at 0.5 degrees, even if its receiving only, that means it would have very little gain, or a loss, at 5 degrees. Since the ridge is 1 mile east of me, I was just trying to figure out what range to expect, so if I had a mathematical formula, I could plug in numbers of higher antenna height, to see if maybe an additional 20' would get me out 150 miles to the east. It worked, actually all the antenna's I have tried worked, they all just cut the planes off at about 70 miles going East, where they work out to 150-200 miles north, west and south. The one I built was not the "coax" version, it was the 1/4 wave under two 5/8 wave version. Receiver RF gain set at max (49.5db) to obtain the highest data rate.Īnything is possible. Station A is using a 1/2 wave dipole (coaxial), with 50' of LMR400 coax. ![]() What I am trying to determine, is if that range (120-150) is the actual radio horizon, or if a higher gain antenna would improve this range. In reality, I am getting about 120-150 miles. Using real world testing, the calculator shows the distance to be 253 miles. What I need to figure out, is if I am not pointing at the horizon, but instead pointing 2 degrees above the horizon (to clear the hill), at what distance would a receiver at 40,000 feet no longer be in the site of the antenna, but would be in the shadow area of the hill? This would cause a "shadow" area behind it. Now I know because of the ridge, the "line" to the horizon is tilted up about 2 degrees or more if I point at the top of the hill. Station B, 259' ASL, 40,000' antenna height (airplane), 1090 Mhz operating frequency Ridge at 550' ASL at 1 mile from station A, blocking signals approx 2 to 5 degrees from horizontal. 450' ASL, 30' antenna height, 1090 Mhz operating frequency. I failed geometry, and never took calculus, so I dont even know how to go about doing this.but if someone is good at math, and can help out, I am all ears. I have used the various calculators online to generate a radio horizon chart, but they are all based on both stations being on level ground, with no hills, and apparently, all at the same altitude. ![]()
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