Full scale deflection on CDI
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Full scale deflection on CDI
Hello, I'm getting slightly confused about the theoretical 'width' of an ILS localiser beam but hoping an expert here can possibly explain what I'm missing.
It's often quoted that a full scale deflection of the Course Deviation Indicator (e.g. 2 dot deviation on a typical instrument) is equivalent to 2.5 degrees either side of the ILS beam centre.
But then you also hear about the "Course Width" of the ILS, the angle of which varies as a function of runway length, but is set such that it provides a FSD of the indicator at 350ft either side of beam centre when you pass over the landing threshold. So (presumably) for a very long runway FSD would be equivalent to a smaller angle from centre, compared to FSD when approaching a shorter runway.
So my question - is the 'often quoted' figure of 2.5 degrees just an approximate/average value of what a CDI's FSD represents? I hope my question makes sense.
many thanks
It's often quoted that a full scale deflection of the Course Deviation Indicator (e.g. 2 dot deviation on a typical instrument) is equivalent to 2.5 degrees either side of the ILS beam centre.
But then you also hear about the "Course Width" of the ILS, the angle of which varies as a function of runway length, but is set such that it provides a FSD of the indicator at 350ft either side of beam centre when you pass over the landing threshold. So (presumably) for a very long runway FSD would be equivalent to a smaller angle from centre, compared to FSD when approaching a shorter runway.
So my question - is the 'often quoted' figure of 2.5 degrees just an approximate/average value of what a CDI's FSD represents? I hope my question makes sense.
many thanks
As I recall, the bean originally was 2.5º either side of the centreline, however when autolands became common, it was necessary to have a set beam width to runway ratio to ensure correct autoland performance, hence the change. As every runway is of different length, and the localizer antennae is at the far end of the runway, the LLZ beams are now of different angular spread for each runway.
jr,
Please disregard the above post.
On the standard CDI, full scale deflection is indeed 2.5 degrees in ILS mode.
This has absolutely nothing to do with ILS localiser azimuth coverage whatsoever.
Angular deviation left or right is based purely on the difference of depth of modulation of the 90Hz AM frequency and the 150Hz AM frequency.
Please disregard the above post.
On the standard CDI, full scale deflection is indeed 2.5 degrees in ILS mode.
This has absolutely nothing to do with ILS localiser azimuth coverage whatsoever.
Angular deviation left or right is based purely on the difference of depth of modulation of the 90Hz AM frequency and the 150Hz AM frequency.
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The LOC course sector width does depend on runway length, and you are correct that the nominal width is to have full deflection 105 m (350 ft) either side of the threshold*.
The width is, however, limited to six degrees. On shorter runways, this creates a problem which is solved by instead having full deflection equal to the same displacement at point B, which is 1050 m (3500 ft) from the threshold.
This is specified in ICAO Annex 10 Vol I §3.1.3.7.
The 2.5 degree figure often given is just another "close enough and won't confuse the public" figure.
Cheers,
Fred
*) Or the ILS reference datum, if we are to be strict about the definitions. That's where you'll be when crossing the threshold.
The width is, however, limited to six degrees. On shorter runways, this creates a problem which is solved by instead having full deflection equal to the same displacement at point B, which is 1050 m (3500 ft) from the threshold.
This is specified in ICAO Annex 10 Vol I §3.1.3.7.
The 2.5 degree figure often given is just another "close enough and won't confuse the public" figure.
Cheers,
Fred
*) Or the ILS reference datum, if we are to be strict about the definitions. That's where you'll be when crossing the threshold.
