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#31
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On Sat, 14 May 2005 18:53:06 -0400, "CryptWolf"
wrote in 1116111148.11781a37a6d5a6e2b697012478f45470@teran ews:: Since the radiation pattern is reduced at higher altitudes, there is less chance of frequency congestion and receiving a signal you don't want. With limited frequencies available, you have to depend on other limits to prevent unwanted reception of other signals. The radiation pattern also puts more energy where it was needed, as that energy that would be radiated upward is directed laterally instead. |
#32
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Um, that's just not true. SInce the radiation pattern is reduced at higher
altitudes, the chance for frequency congestion is every bit as probable. If all signals are reduced proportionally, then the RELATIVE signal strengths remain constant. Jim Since the radiation pattern is reduced at higher altitudes, there is less chance of frequency congestion and receiving a signal you don't want. With limited frequencies available, you have to depend on other limits to prevent unwanted reception of other signals. |
#33
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And a screw-up on the equation:
Radio horizon (in miles) equals the square root of [TWO TIMES the aircraft altitude above the VOR (in feet)]. Jim "Antoņio" wrote in message ... The equation for this is that radio horizon (in miles) equals the square root of the aircraft altitude above the VOR (in feet). |
#34
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On Sat, 14 May 2005 at 17:50:55 in message
, Peter Duniho wrote: "David CL Francis" wrote in message ... The real earth is not of course that flat except over the oceans! Actually, as long as we're being pedantic, it's not even flat over the oceans. It's much flatter, but the Earth simply is not an ideal "smooth sphere" anywhere on its surface. I did not realise that I was being pedantic! Also the further away you go the closer the horizon distance gets to being the same as the height. It is obvious that from the moon you can almost see the entire hemisphere. Almost. But the difference is significant enough to matter when you really care whether you can see the entire hemisphere (astronomy, for example). Of course. But the point is not important when considering VORs! -- David CL Francis |
#35
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On Sun, 15 May 2005 at 09:05:01 in message
, RST Engineering wrote: And a screw-up on the equation: Radio horizon (in miles) equals the square root of [TWO TIMES the aircraft altitude above the VOR (in feet)]. [1] Jim, Doesn't that assume that the number of feet in a mile is the same as the radius of the earth in miles? The formula for the tangential horizon distance is [2] d=(2*r*h + h^2)^0.5 where d is the tangential horizon distance, r is the radius of the earth and h is the height of the object above the surface. DKr and h all in the same units. Because the heights we are talking about are small compared to the radius of the earth the h^2 term can be ignored leaving what you said. E&OE :-) -- David CL Francis |
#36
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And you asked in another post if you were being pedantic? PEDANTIC? My
guess is that you stay up late at night worrying about whether anal(-)retentive is hyphenated or not. Jim "David CL Francis" wrote in message ... On Sun, 15 May 2005 at 09:05:01 in message , RST Engineering wrote: And a screw-up on the equation: Radio horizon (in miles) equals the square root of [TWO TIMES the aircraft altitude above the VOR (in feet)]. [1] Jim, Doesn't that assume that the number of feet in a mile is the same as the radius of the earth in miles? |
#37
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"RST Engineering" wrote in message ... Um, that's just not true. SInce the radiation pattern is reduced at higher altitudes, the chance for frequency congestion is every bit as probable. If all signals are reduced proportionally, then the RELATIVE signal strengths remain constant. Jim Under normal atmospheric conditions, excluding anomolies, as the frequency or distance increases, the required transmitter power increases while the recieved signal strength remains the same. At some point, even if you use the same frequencies, a VOR or any radio signal will vanish into the background noise. The reciever sensitivity is limited by background noise. If you like, I'll look up the specific math and we can really get technical. All but a few would understand it or care. This would be the other limit that I didn't explain previously. All you have to do is space VOR's of the same frequency far enough apart and they won't interfere. The fact that the radiation pattern is reduced for higher altitudes seems to imply that the radiation pattern was designed to reduce transmitted power and limit reception distances at those altitudes where line of sight would not be a factor. For reference you might want to find: Schoenbeck, Electronic Communications Modulation And Transmission It's one of the books I used when I was working on my electronics degree. Since the radiation pattern is reduced at higher altitudes, there is less chance of frequency congestion and receiving a signal you don't want. With limited frequencies available, you have to depend on other limits to prevent unwanted reception of other signals. |
#38
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"Antoņio" wrote in message ... I am not sure. Is it because the radiation pattern is spherical and not line-of-sight? You're looking at just one VOR at a time. There are about 1000 VORs in the US and just 100 VOR frequencies. The service volume has to ensure not only usable reception of the desired VOR, but non-reception of undesired VORs on the same frequency. As an example, let's say you're flying from EAU VOR to LAN VOR at 15,500'. They're about 320 miles apart but at that altitude line-of-sight distance is about 180 miles so you should receive LAN before losing EAU, even though you're well outside of the standard service volume of forty miles. When you're about halfway you switch from EAU to LAN, but you're unable to get a reliable signal. The problem may be that you're closer to ESC and RFD VORs than you are to LAN, and they all operate on 110.8. |
#39
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For reference you might want to find:
Hamsher, "Communications Systems Engineering Handbook" ; Jasik, "Antenna Engineeiring Handbook"; Kraus "Antenna Design"; MIT Radiation Lab Series; any of the ARRL publications on antennas which is what I recommend that you use when I teach you when you are working on your electronics degree... Jim For reference you might want to find: Schoenbeck, Electronic Communications Modulation And Transmission It's one of the books I used when I was working on my electronics degree. |
#40
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Steven P. McNicoll wrote:
You're looking at just one VOR at a time. There are about 1000 VORs in the US and just 100 VOR frequencies. The service volume has to ensure not only usable reception of the desired VOR, but non-reception of undesired VORs on the same frequency. Ok...I think I'm with you (though feeling like I might be exhibiting that I am a little obtuse at this point in the discussion). I assume that your post means you think I haven't quite got it yet? Are you saying that the service volume diagrams are artist renderings of the reception distances that the FAA have tested and will *guarantee* to be usable?? So the "double/inverted wedding cake" structure (which, as you recall, was the basis of my original question) really has little to do with the *actual* signal propagation distances of a particular VOR but, rather, provide approximations by taking into consideration the real world interference of other VORs, spherical wave radiations, curvature of earth, etc. ? So I would not necessarily loose VOR service if, while using a Standard High Altitude Service Volume station, I were to climb above FL45, with a depicted service distance of 130nm, to FL46 with it's depicted service distance of 100nm? Antonio (thinking he had it, then....) |
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