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Precision Approach Radar  
User currently offlineNovice From United Kingdom, joined Aug 2012, 90 posts, RR: 0
Posted (1 year 10 months 5 days ago) and read 4469 times:

Azimuth:
The azimuth element scans a very narrow beam (0.6 degrees) backwards and forwards over a sector covering the required minimum azimuth sector.

Elevation:
The elevation element has a narrow vertical beam width (0.6 degrees) but a broader azimuth beam width (up to 30 degrees). It scans vertically from an elevation of about 0.5 up to 8 degrees.

Why is it that it scans using a narrow beam how come it doesn't use a wider beam and less scaning? Is it that using a narrow beams makes it more sensitive/accurate? If so why is this?

Thanks

7 replies: All unread, jump to last
 
User currently offlineferpe From France, joined Nov 2010, 2800 posts, RR: 59
Reply 1, posted (1 year 10 months 4 days 21 hours ago) and read 4432 times:

The PAR operator gets a blip on his screen for each beam painting a hard subject reflecting an echo back. This painting on the screen of the 2 traces (elevation and side position) starts when the beams start painting the target and stops when it no longer does as the radard sweeps it's angular coverage in elevation and side angle. Now the radar can predict where the target is angle wise from the radar antenna by this return on echos, (lenght wise it is much easier, it is the round trip time for the pulsed radar wave). Imagine that the beam would be 10° wide, you would have a begin and end of the target return over a 10° sector. If it was not for a multipath phenomenon called target fading one could guess the target would lie in the middle of this arc but given that fading makes the return echo go up and down in strenght when the painting takes place you need to make this arch as short as possible to get precision.

The PAR is directing you when your are some 50-100 m from hard things, you'd like it to have a least a 10 m accuracy and that the guy who tells you if you are above or below the glidepath (or left or right) has the best blip possible. This is why they make the sharpest beam practicle in the direction where they want the precision.

So how god is 0.6°. Well sinus for 0.6° is 0.01 so you would have a 10 m blip at 1000 m range, just about the maximum uncertainty you want at that range because then you are at some 50m above threshold if your are on a 3° glideslope like for a standard ILS.

[Edited 2012-09-22 13:32:10]


Non French in France
User currently offlineNovice From United Kingdom, joined Aug 2012, 90 posts, RR: 0
Reply 2, posted (1 year 10 months 4 days 18 hours ago) and read 4363 times:

Thanks for that explanation ferpe still just a few queries if you could help
"Imagine that the beam would be 10° wide, you would have a begin and end of the target return over a 10° sector. If it was not for a multipath phenomenon called target fading one could guess the target would lie in the middle of this arc but given that fading makes the return echo go up and down in strenght when the painting takes place you need to make this arch as short as possible to get precision."
How come you would guess the target would lie in the middle of the arch if it was not for target fading? i am guessing that it is better then for the beam width to be narrow because that way it will not cover as much i.e. the unwanted area to the left/right of the aircraft?

"So how god is 0.6°. Well sinus for 0.6° is 0.01 so you would have a 10 m blip at 1000 m range, just about the maximum uncertainty you want at that range because then you are at some 50m above threshold if your are on a 3° glideslope like for a standard ILS."
Does 0.01 mean seconds yes? could you tell me how you calulated that and i am presuming that a blip is basically a scan cycle refreshed on the PAR controllers screen?

Thanks again for helping  


User currently offlineferpe From France, joined Nov 2010, 2800 posts, RR: 59
Reply 3, posted (1 year 10 months 4 days 16 hours ago) and read 4333 times:

Quoting Novice (Reply 2):
How come you would guess the target would lie in the middle of the arch if it was not for target fading? i am guessing that it is better then for the beam width to be narrow because that way it will not cover as much i.e. the unwanted area to the left/right of the aircraft?

Lets say the azimuth (side angular position) antenna is on a left sweep from it's rightmost position. The painting of the echo on the screen starts at the beams angular screen position (to get the side position the screen ray mimics the beam painting of the nature and starts painting an echo on the screen at it's copy of the antennas angular position) when the leftmost part of the lobe hit the planes closes = left wing and stops painting when the rightmost part of the lobe leaves the other side=right wing. As it is a symmetrical event and the lobe is symmetrical one could say to the controller "assume that the plane has the center position of the achlike radar echo" and it would be right. The fading make the center of the echo wandering around however, thus such a conclusion is dangerous, therefore in practice you need to have a narrow beam so that the wandering around of the radar echo does not generate interpretation faults, the radar echo is so small that you take it as the aircraft position and that is good enough.

(The first ship radars in WW2 had very broad beams but short enough pulses that they helped with the range for the ship to ship gunning but not with bearing as the echo was wandering around in bearing due to fading, you still used you optical method for that)

Quoting Novice (Reply 2):
Does 0.01 mean seconds yes?

No, it is practical to have math. To know the distance that an angle paint at eg 1 or 2 km you take Sinus of the angle (0.6°) and the result is a factor (0.1047...) that you then multiply with the distance and voila now you know what the angular coverage is at that distance.

[Edited 2012-09-22 18:56:05]

[Edited 2012-09-22 18:58:34]


Non French in France
User currently offlineDavid L From United Kingdom, joined May 1999, 9523 posts, RR: 42
Reply 4, posted (1 year 10 months 4 days 7 hours ago) and read 4255 times:

Quoting Novice (Reply 2):
Does 0.01 mean seconds yes?

To expand on what Ferpe said: because the angle is small and the distance is large, you can treat the radar beam as a right-angled triangle where the hypotenuse is the line between the radar antenna and the target and the opposite is the actual width, in metres, of the beam at the target. Thus, the sine of the beam angle (opposite divided by hypotenuse) gives the ratio between the beam width and the target distance "as near as dammit".

The sine of 0.6o is 0.01 so, at any given point along its length, the beam width is 1/100th of the distance from the radar antenna. That's what the "0.01" is.


User currently offlineNovice From United Kingdom, joined Aug 2012, 90 posts, RR: 0
Reply 5, posted (1 year 10 months 4 days 6 hours ago) and read 4233 times:

Thanks guys for them explnations using trig it has helped me understand where the 0.01 comes from:
Example assuming the range of the radar is 1000m and the angle is 0.6 degrees, the sine of 0.6 is 0.01 and you then times that by 1000 which give a width of 10m.

Quoting ferpe (Reply 3):
. As it is a symmetrical event and the lobe is symmetrical one could say to the controller "assume that the plane has the center position of the achlike radar echo" and it would be right. The fading make the center of the echo wandering around however, thus such a conclusion is dangerous, therefore in practice you need to have a narrow beam so that the wandering around of the radar echo does not generate interpretation faults, the radar echo is so small that you take it as the aircraft position and that is good enough.

I'm still a bit sketchy on this though, but i guess i just need to top up on my knowledge on this fading concept thanks for the help  


User currently offlineferpe From France, joined Nov 2010, 2800 posts, RR: 59
Reply 6, posted (1 year 10 months 3 days 21 hours ago) and read 4166 times:

Quoting Novice (Reply 5):
I'm still a bit sketchy on this though, but i guess i just need to top up on my knowledge on this fading concept thanks for the help

Poeple who use radar sometimes also don't understand the concept of fading, they do know however that the echo of an aricraft on the radar scope changes from scan to scan. Some scans the echo is strong, next time it is almost lost, this is fading in action. The reason for it is quite simple, different parts of the aircraft reflect the radar energy back. They all have different distances to the radar antenna/reciever. As the wavelength of a typcical PAR is some 3 cm a path difference of the same lenght between two reflecting surfaces would mean they are cancelling each other. Others 6 cm apart or any other multiple of 3 cm would reinforce each other, you have a stronger echo from these reflectors.

That is why the art of radar is a long term phenomenon with a need for several paintings (scans) to denote a target reliably, it is also why a radar screen has a very long afterglow phosphor or for a modern screen a paint memory, you like to add (integrate) several scans of the target to get a good blip.



Non French in France
User currently offlineNovice From United Kingdom, joined Aug 2012, 90 posts, RR: 0
Reply 7, posted (1 year 10 months 3 days 21 hours ago) and read 4163 times:

Thanks for that explanation Ferpe, i understand the concept alot better now cheers   

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