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NTSB report - ILS and ATC. How does it all come together?



 
 
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  #1  
Old June 19th 06, 10:32 PM posted to rec.aviation.ifr,rec.aviation.piloting,rec.aviation.student
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Default 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

  #2  
Old June 19th 06, 11:19 PM posted to rec.aviation.ifr,rec.aviation.piloting,rec.aviation.student
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Default 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  
Old June 19th 06, 11:34 PM posted to rec.aviation.ifr,rec.aviation.piloting,rec.aviation.student
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Default 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.


  #4  
Old June 20th 06, 12:07 AM posted to rec.aviation.ifr,rec.aviation.piloting,rec.aviation.student
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Default 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.
  #5  
Old June 20th 06, 02:40 AM posted to rec.aviation.ifr,rec.aviation.piloting,rec.aviation.student
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Default 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  
Old June 20th 06, 10:51 AM posted to rec.aviation.ifr,rec.aviation.piloting,rec.aviation.student
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Default 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  
Old June 20th 06, 11:13 AM posted to rec.aviation.ifr,rec.aviation.piloting,rec.aviation.student
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Default 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  
Old June 20th 06, 01:03 PM posted to rec.aviation.ifr,rec.aviation.piloting,rec.aviation.student
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Default 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  
Old June 20th 06, 01:14 PM posted to rec.aviation.ifr,rec.aviation.piloting,rec.aviation.student
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Default 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  
Old June 20th 06, 01:45 PM posted to rec.aviation.ifr,rec.aviation.piloting,rec.aviation.student
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Default 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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