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NTSB report - ILS and ATC. How does it all come together?
("Matt Barrow" posted this link in a different thread)
http://www.ntsb.gov/ntsb/brief.asp?ev_id=20001208X09256&key=1 (WARNING: Long confused post ...????) "The minimum altitude for the approach was 376 feet above the ground." (Is that for a point 2.5 miles out? ...where the power line is 150 feet AGL?) "The approach controller stated he did not notice anything unusual with the airplane after handing it off to the tower. He stated the airplane's altitude appeared normal, and he did not see it deviate from the localizer. According to the supervisor, he saw a low altitude alert for the airplane, which was followed shortly by the interruption of power to the building." "The local air traffic controller stated that shortly after clearing the accident airplane to land, the tower had a power interruption which caused the radar to blink and get skewed. She then noticed the airplane's data block disappeared. Prior to the power outage, she had been looking out the window to check the weather conditions, and did not notice any problems with the airplane." "The reported weather consisted of a 500 feet overcast and 3 miles visibility." (The plane is 3 miles out, at 200 feet AGL (guessing) ...and it was missed by 3(?) people with radar/transponder info - and missed by the pilot? I don't understand the interaction between a plane, on an ILS approach, and ATC? Is an ILS approach doomed from the onset if the plane's altimeter is set wrong?) (From the "Full narrative available" link in the NTSB report) "The ATC controller cleared the airplane for the ILS runway 36 approach at 1813:23. The last radio contact with the airplane occurred at 1814:53, when the airplane was cleared to land. Minutes later, the ATC Tower experienced a power outage. When power was restored about 9 seconds later, the airplane had disappeared from the radar. ATC attempted to contact the airplane but was unsuccessful. The airplane was located about 2.5 miles south of runway 36." http://www.digitaldutch.com/unitconverter/ 75 knots (guessing) = 125 ft /second. 9 seconds (power outage in tower) = almost 1/4 mile of travel 3°(?) glide slope = 3.75 ft/second alt loss x 9 seconds = 34 ft of altitude loss, while the power was out in the tower - using the entire 9 seconds. Heck, they might have hit the 150 foot high power lines 2 seconds into the power outage? 2.5 miles out (power lines) = 13,200 ft from touchdown At 125 ft/second ...13,200 ft = (105 seconds out @ 75 knots?) 105 seconds out @ 3° glide slope = approx 400 ft (394-ft) of altitude to lose. or.. (100%)13,200 ft out from the threshold (10%)1,320 ft (1%) 132 ft (3%) 396 ft altitude to lose, from 2.5 miles out @ 3° glide slope. (See where I'm going with this? I don't get it. Duh! 376-ft was the "minimum altitude for approach." So, what does that mean - where does 376-ft start?) (I checked http://204.108.4.16/d-tpp/0606/05048ILD36.PDF and see 2° on the 2006 airport chart - which is even lower over the power lines - I think?) (Tripped/Googled over this bit - a possibility in this crash?) http://www.allstar.fiu.edu/aero/ILS.htm False signals may be generated along the glide slope in multiples of the glide path angle, the first being approximately 6º degrees above horizontal. This false signal will be a reciprocal signal (i.e. the fly up and fly down commands will be reversed). The false signal at 9º will be oriented in the same manner as the true glide slope. There are no false signals below the actual slope. An aircraft flying according to the published approach procedure on a front course ILS should not encounter these false signals. (Overall, are my numbers right? How does someone on an ILS "end up" at 150 ft (AGL) 2.5 miles from the threshold? When does the "Low Altitude Alert" buzzer alert ATC? In 20 seconds the plane is going to lose 75 ft of altitude with a 3° glide slope @ 75 knots. That's 9 seconds for the power outage and 11 seconds to reorientate themselves in the tower. Remember, the Low Altitude Warning came BEFORE the power outage.) I'm not considering the altimetor setting in the plane - I'm mostly looking at it from the ATC side of things. How's this all work?) Montblack |
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NTSB report - ILS and ATC. How does it all come together?
I don't understand your calculation. At 2.5 miles from the touch-down zone (assuming that's what it is), the GS should be about 750 feet above the touch-down zone elevation. The pilot was way below the glideslope. (Simple and quick approximate calcuation method: 2.5mi = 15000 feet. The 3 degree ILS is approximately 1:20 approach ratio. So 15000 / 20 = 750 ft). Montblack wrote: (Overall, are my numbers right? How does someone on an ILS "end up" at 150 ft (AGL) 2.5 miles from the threshold? When does the "Low Altitude Alert" buzzer alert ATC? In 20 seconds the plane is going to lose 75 ft of altitude with a 3° glide slope @ 75 knots. That's 9 seconds for the power outageand 11 seconds to reorientate themselves in the tower. Remember, the Low Altitude Warning came BEFORE the power outage.) I'm not considering the altimetor setting in the plane - I'm mostly looking at it from the ATC side of things. How's this all work?) Montblack |
#3
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NTSB report - ILS and ATC. How does it all come together?
"Montblack" wrote in message ... ("Matt Barrow" posted this link in a different thread) http://www.ntsb.gov/ntsb/brief.asp?ev_id=20001208X09256&key=1 (WARNING: Long confused post ...????) "The minimum altitude for the approach was 376 feet above the ground." (Is that for a point 2.5 miles out? ...where the power line is 150 feet AGL?) At the time of this accident 440 MSL was the MDA for the S-LOC 36, that's 376 feet above the TDZE of 64 feet. That MDA applied from the LOM to the runway threshold. "The approach controller stated he did not notice anything unusual with the airplane after handing it off to the tower. He stated the airplane's altitude appeared normal, and he did not see it deviate from the localizer. According to the supervisor, he saw a low altitude alert for the airplane, which was followed shortly by the interruption of power to the building." "The local air traffic controller stated that shortly after clearing the accident airplane to land, the tower had a power interruption which caused the radar to blink and get skewed. She then noticed the airplane's data block disappeared. Prior to the power outage, she had been looking out the window to check the weather conditions, and did not notice any problems with the airplane." "The reported weather consisted of a 500 feet overcast and 3 miles visibility." (The plane is 3 miles out, at 200 feet AGL (guessing) ...and it was missed by 3(?) people with radar/transponder info - and missed by the pilot? I don't understand the interaction between a plane, on an ILS approach, and ATC? Is an ILS approach doomed from the onset if the plane's altimeter is set wrong?) Well, on a full ILS you'd have the glideslope, but it appears the approach was made to localizer minimums only suggesting the airplane did not have a working GS receiver or the GS was out of service. (From the "Full narrative available" link in the NTSB report) "The ATC controller cleared the airplane for the ILS runway 36 approach at 1813:23. The last radio contact with the airplane occurred at 1814:53, when the airplane was cleared to land. Minutes later, the ATC Tower experienced a power outage. When power was restored about 9 seconds later, the airplane had disappeared from the radar. ATC attempted to contact the airplane but was unsuccessful. The airplane was located about 2.5 miles south of runway 36." http://www.digitaldutch.com/unitconverter/ 75 knots (guessing) = 125 ft /second. 9 seconds (power outage in tower) = almost 1/4 mile of travel 3°(?) glide slope = 3.75 ft/second alt loss x 9 seconds = 34 ft of altitude loss, while the power was out in the tower - using the entire 9 seconds. It's a 3 degree glideslope, but the reference to the 376' MDA suggests the glideslope was not being used. Heck, they might have hit the 150 foot high power lines 2 seconds into the power outage? The impact with the powerline might have caused the power outage. The powerline doesn't appear on the chart as an obstacle. Perhaps the 150' height is MSL, making the powerline a more reasonable 90' or so. 2.5 miles out (power lines) = 13,200 ft from touchdown At 125 ft/second ...13,200 ft = (105 seconds out @ 75 knots?) 105 seconds out @ 3° glide slope = approx 400 ft (394-ft) of altitude to lose. or.. (100%)13,200 ft out from the threshold (10%)1,320 ft (1%) 132 ft (3%) 396 ft altitude to lose, from 2.5 miles out @ 3° glide slope. (See where I'm going with this? I don't get it. Duh! 376-ft was the "minimum altitude for approach." So, what does that mean - where does 376-ft start?) If it's the localizer MDA it stars at WAKUL, 4.1 miles from the threshold, and localizer MDA is the only way it makes sense. (I checked http://204.108.4.16/d-tpp/0606/05048ILD36.PDF and see 2° on the 2006 airport chart - which is even lower over the power lines - I think?) The current NACO chart shows 3 degrees and a DH of 264 MSL, just as it did then. |
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NTSB report - ILS and ATC. How does it all come together?
In article .com,
"M" wrote: I don't understand your calculation. At 2.5 miles from the touch-down zone (assuming that's what it is), the GS should be about 750 feet above the touch-down zone elevation. The pilot was way below the glideslope. (Simple and quick approximate calcuation method: 2.5mi = 15000 feet. The 3 degree ILS is approximately 1:20 approach ratio. So 15000 / 20 = 750 ft). 20:1 is the floor of the protected airspace along the approach. The G/S centerline is going to be above that by 100 ft or more. |
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NTSB report - ILS and ATC. How does it all come together?
("M" wrote)
(Simple and quick approximate calcuation method: 2.5mi = 15000 feet. The 3 degree ILS is approximately 1:20 approach ratio. So 15000 / 20 = 750 ft). My quick method: 5200 + 5200 = 10,400 + 2,600 = 13,000 13,000 ft = 2.5 miles 1,300 ft = 10% 130 = 1% 390-400 ft = 3% %%%. Oops... NOT 3 degrees!! My mistake. Montblack |
#6
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NTSB report - ILS and ATC. How does it all come together?
"Montblack" wrote in
: ("M" wrote) (Simple and quick approximate calcuation method: 2.5mi = 15000 feet. The 3 degree ILS is approximately 1:20 approach ratio. So 15000 / 20 = 750 ft). My quick method: 5200 + 5200 = 10,400 + 2,600 = 13,000 13,000 ft = 2.5 miles 1,300 ft = 10% 130 = 1% 390-400 ft = 3% %%%. Oops... NOT 3 degrees!! My mistake. Montblack Are you using statute miles? A NM is slightly over 6,000'. Why not use the simple 3:1 rule-of-thumb? For every 3 NM traversed, you will descend 1,000' on a 3 degree glide slope. -- Marty Shapiro Silicon Rallye Inc. (remove SPAMNOT to email me) |
#7
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NTSB report - ILS and ATC. How does it all come together?
john smith wrote:
In article .com, "M" wrote: I don't understand your calculation. At 2.5 miles from the touch-down zone (assuming that's what it is), the GS should be about 750 feet above the touch-down zone elevation. The pilot was way below the glideslope. (Simple and quick approximate calcuation method: 2.5mi = 15000 feet. The 3 degree ILS is approximately 1:20 approach ratio. So 15000 / 20 = 750 ft). 20:1 is the floor of the protected airspace along the approach. The G/S centerline is going to be above that by 100 ft or more. That is not correct. The protected surfaces for an ILS are much more shallow than that. A 3 degree G/S itself is 19:01 to 1. |
#8
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NTSB report - ILS and ATC. How does it all come together?
M wrote:
I don't understand your calculation. At 2.5 miles from the touch-down zone (assuming that's what it is), the GS should be about 750 feet above the touch-down zone elevation. The pilot was way below the glideslope. (Simple and quick approximate calcuation method: 2.5mi = 15000 feet. The 3 degree ILS is approximately 1:20 approach ratio. So 15000 / 20 = 750 ft). Using trigonometry, I get ~ 785ft 2.5 miles out from the touch down zone, so your method is pretty accurate. Here's my calculation: Assuming: Distance = 15,000 ft Slope: 3 degrees Height = Distance * sin(Slope) = 785.04 ft. Mike -- Mike |
#9
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NTSB report - ILS and ATC. How does it all come together?
"Mike" wrote in message . .. Using trigonometry, I get ~ 785ft 2.5 miles out from the touch down zone, so your method is pretty accurate. Here's my calculation: Assuming: Distance = 15,000 ft Slope: 3 degrees Height = Distance * sin(Slope) = 785.04 ft. A 3 degree glidepath descends 318 feet per nautical mile. 318 x 2.5 = 795. |
#10
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NTSB report - ILS and ATC. How does it all come together?
Steven P. McNicoll wrote:
"Mike" wrote in message . .. Using trigonometry, I get ~ 785ft 2.5 miles out from the touch down zone, so your method is pretty accurate. Here's my calculation: Assuming: Distance = 15,000 ft Slope: 3 degrees Height = Distance * sin(Slope) = 785.04 ft. A 3 degree glidepath descends 318 feet per nautical mile. 318 x 2.5 = 795. Sorry, the original calculation was based on bad data. 15,000 feet is not 2.5nm as stated in the original post. 1nm = 6,076ft 2.5nm = 15,190ft Elevation = 15,190 * sin(3-degrees) = 795 ft -- Mike |
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