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Range of VOR?

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Old 4th Aug 2008, 20:09
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Range of VOR?

Does anyone know the actual range of a VOR. I have had so many conflicting reports
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Old 4th Aug 2008, 20:18
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Max range at which you will receive a signal/indication will depend on tx power, altitude, rx sensitivity, atmospheric conditions and various other intangible factors.

The range and altitude within which a reliable signal will be received is defined by the Deisignated Operational Coverage (DOC) as published for the specific VOR - they vary depending on what the aid is there for.
 
Old 4th Aug 2008, 20:32
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I can't give you the range because a VOR is line-of-sight and is dependent upon ones altitude and terrain and other factors.

What you're probably looking for is standard service volume (SSV). This defines the unrestricted reception limits usable for random or unpublished route navigation.

The standard service volume (SSV) class designators of a VOR are:

Terminal (T)
1,000 up to and including 12,000 AGL out to 25NM.

Low Altitude (L)
1,000 up to and including 18,000 AGL out to 40NM.

High Altitude (H)
1) 1,000 up to and including 14,500 AGL out to 40NM.
2) 14,500 up to and including 60,000 AGL out to 100 nm.
3) 18,000 up to and including 45,000 AGL out to 130 nm.

Reception below 1,000 AGL is based on line-of-sight. Reception within the SSV is restricted by vertical angle coverage limitations. Distance information from DME and TACAN, and azimuth information from VOR, is normally usable from the radio horizon to elevation angles of at least 60 degrees. Azimuth information from TACAN is normally usable from the radio horizon to elevation angles of at least 40 degrees. At higher elevation angles — within the so-called cone of ambiguity — the NAVAID information may not be usable.
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Old 5th Aug 2008, 06:40
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The max theoretical range in nautical miles was always worked out thus:
1.25 √Receiver Ht (in feet) + √Transmitter Ht (in feet)

As has been said before, this is obviously subject to the fact that VHF is a line of sight transmission etc
kishna is offline  
Old 5th Aug 2008, 08:17
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kishna, for whatever reason, the accepted formula is now 1.23√Receiver Ht (in feet) + 1.23√Transmitter Ht (in feet) Don't ask me how/why, I just do what the book says.
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