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

"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.

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

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 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.