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

Mike wrote:
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


Sorry for the double post. Last send just "hung" so I resent thinking it
didn't send the first time.

--
Mike

Mike wrote:
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


The TCH is 46 feet, so the G/S is 842 feet about TDZ at 2.5 miles from
the threshold.

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.

3 degree G/s = 318.44 feet per mile. 2.5 (318.44) + 46' TCH = 842 feet.

"Sam Spade" wrote in message
news:A%Slg.179411$bm6.90388@fed1read04...

3 degree G/s = 318.44 feet per mile. 2.5 (318.44) + 46' TCH = 842 feet.


All these calculations assume the full ILS was used. The narrative refers
to a 376' minimum altitude, which was the localizer MDA at the time of the
accident.

Steven P. McNicoll wrote:
"Sam Spade" wrote in message
news:A%Slg.179411$bm6.90388@fed1read04...

3 degree G/s = 318.44 feet per mile. 2.5 (318.44) + 46' TCH = 842 feet.



All these calculations assume the full ILS was used. The narrative refers
to a 376' minimum altitude, which was the localizer MDA at the time of the
accident.


I doubt anyone knows whether he was using LOC or ILS minimuma.

The NTSB doesn't even understand the concepts:

"The ILS 36 has a minimum approach altitude of 376 feet above ground
level (AGL). The cloud ceiling was at 500 feet AGL. After the accident,
the ILS 36 was taken out of service to be tested. It was flight checked
on December 24, 1997, with no anomalies found."

What does "minimum approach alitude" refer to?

What does "376 feet above ground level" refer to?

If the field office investigator can't sort oout MDA, DA, and HAT, I
don't expect to figure out much about his or her's view of how the
approach was being flown.

"Sam Spade" wrote in message
...

I doubt anyone knows whether he was using LOC or ILS minimuma.

The NTSB doesn't even understand the concepts:

"The ILS 36 has a minimum approach altitude of 376 feet above ground level
(AGL). The cloud ceiling was at 500 feet AGL. After the accident, the ILS
36 was taken out of service to be tested. It was flight checked on
December 24, 1997, with no anomalies found."

What does "minimum approach alitude" refer to?


It could only be an MDA.



What does "376 feet above ground level" refer to?


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.

"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.