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

[Montblack]

("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

[M]

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

[Steven P. McNicoll]

"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.

[john smith]

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.

[Montblack]

("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

[Marty Shapiro]

"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)

[Sam Spade]

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.

[Mike]

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

[Steven P. McNicoll]

"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.

[Mike]

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

[Mike]

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

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

--
Mike

[Sam Spade]

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.

[Sam Spade]

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.

[Steven P. McNicoll]

"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.

[Sam Spade]

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.

[Steven P. McNicoll]

"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.

[Sam Spade]

Steven P. McNicoll wrote:
> "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.
>

How do you conclude that?

A decision altitude is a minimum approach altitude, too, in a broad use
of a term that lacks any official definition.

Besides, no one has any way of determining whether the pilot was flying
the LOC or ILS profile.
>
>
>>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.
>
>
Yes, but "above ground level" is a term of ignorance. Above ground
level at the crash site, at the runway, or at the DA point, or along the
entire length of the final approach segment where the LOC DMA could
resonably apply?

[Sam Spade]

Sam Spade wrote:

> Yes, but "above ground level" is a term of ignorance. Above ground
> level at the crash site, at the runway, or at the DA point, or along the
> entire length of the final approach segment where the LOC DMA could
> resonably apply?

MDA

[Steven P. McNicoll]

"Sam Spade" > wrote in message
news:A5Vlg.179419$bm6.137774@fed1read04...
>
> How do you conclude that?
>
> A decision altitude is a minimum approach altitude, too, in a broad use of
> a term that lacks any official definition.
>

The term would be "decision height" in the US, not "decision altitude".
It's an MDA because nothing else fits. The DH (Decision Height, the height
at which a decision must be made during an instrument approach where an
electronic glideslope is provided to either continue the approach or to
execute a missed approach) for the full ILS RWY 36 was 264 MSL, 200' above
the TDZE (Touchdown Zone Elevation, the highest elevation in the first 3000'
of runway) of 64'. The MDA (Minimum Descent Altitude, the lowest altitude
to which descent is authorized on final approach in execution of a standard
instrument approach procedure where no electronic glideslope is provided)
for the ILS RWY 36 to straight-in localizer minimums was 440 MSL, 376' above
the TDZE.

Think of an MDA as an altitude to be maintained while a DH is an altitude to
be flown through.


>
> Besides, no one has any way of determining whether the pilot was flying
> the LOC or ILS profile.
>

Yes, I already pointed that out.


>
> Yes, but "above ground level" is a term of ignorance.
>

Not at all, the term is used quite often in aviation.

[Sam Spade]

Steven P. McNicoll wrote:
> "Sam Spade" > wrote in message
> news:A5Vlg.179419$bm6.137774@fed1read04...
>
>>How do you conclude that?
>>
>>A decision altitude is a minimum approach altitude, too, in a broad use of
>>a term that lacks any official definition.
>>
>
>
> The term would be "decision height" in the US, not "decision altitude".
> It's an MDA because nothing else fits.

I can't make such an inference from an NTSB report that fails to use the
FAA definitiona, thus creating vagueness where precision is required.

As to the US, the FAA is in transition from DH to DA so they can
harmonize with the rest of the world. You need to catch up on your
reading; i.e., AIM 5-4-5-4:

4. Chart Terminology
(a) Decision Altitude (DA) replaces the familiar term Decision Height
(DH). DA conforms to the international convention where altitudes relate
to MSL and heights relate to AGL. DA will eventually be published for
other types of instrument approach procedures with vertical guidance, as
well. DA indicates to the pilot that the published descent profile is
flown to the DA (MSL), where a missed approach will be initiated if
visual references for landing are not established. Obstacle clearance is
provided to allow a momentary descent below DA while transitioning from
the final approach to the missed approach. The aircraft is expected to
follow the missed instructions while continuing along the published
final approach course to at least the published runway threshold
waypoint or MAP (if not at the threshold) before executing any turns.

The DH (Decision Height, the height
> at which a decision must be made during an instrument approach where an
> electronic glideslope is provided to either continue the approach or to
> execute a missed approach) for the full ILS RWY 36 was 264 MSL, 200' above
> the TDZE (Touchdown Zone Elevation, the highest elevation in the first 3000'
> of runway) of 64'. The MDA (Minimum Descent Altitude, the lowest altitude
> to which descent is authorized on final approach in execution of a standard
> instrument approach procedure where no electronic glideslope is provided)
> for the ILS RWY 36 to straight-in localizer minimums was 440 MSL, 376' above
> the TDZE.
>
> Think of an MDA as an altitude to be maintained while a DH is an altitude to
> be flown through.

Thanks for the flying lesson.

Having said that there is no logical reason for the NTSB investigator in
this report to have focused on the HAT of the sraight-in LOC MDA rather
than the HAT of the precision DA. Both should have been discussed since
the pilot could have been flying either the NPA or PA profile.
>
>
>
>
>
>>Yes, but "above ground level" is a term of ignorance.
>>
>
>
> Not at all, the term is used quite often in aviation.
>
>
But, not in reference to approach and landing minimums other than HAA
and HAT, which have precise definitions and are not valid as AGL values
anywhere along the final approach course (excepting CAT II RAs, which
are not applicable to this accident).

[Steven P. McNicoll]

"Sam Spade" > wrote in message
news:w5Ylg.179436$bm6.143818@fed1read04...
>
> I can't make such an inference from an NTSB report that fails to use the
> FAA definitiona, thus creating vagueness where precision is required.
>

It appears you confused yourself by snipping too much of the quote.


>
> As to the US, the FAA is in transition from DH to DA so they can harmonize
> with the rest of the world. You need to catch up on your reading; i.e.,
> AIM 5-4-5-4:
>

No. The IAP still uses DH. I'll continue to use current terminology.


>
> Thanks for the flying lesson.
>

You're welcome.