# How much obstacle clearance is provided within full scale deflection below glide slope on ILS?

Basically the question in the title. The military has regulations that say to obey the localizer minimums if more than have scale deflection below glide slope, but for civilians how much obstacle clearance is provided if they are below glide slope and not quite full scale deflection?

In the USA, FAA Order 8260.3 specifies the Standard for Terminal Instrument Procedures. The defining concept of terminal procedures is the Required Obstacle Clearance (ROC) between the nominal flight path and the Obstacle Clearance Surface (OCS). From Paragraph 2–1–4a:

The obstacle evaluation method for descent on a glidepath is the application of a descending OCS below the glidepath. The vertical distance between the glidepath and the OCS is the ROC; thus ROC = (glidepath height) - (OCS height). The ROC decreases with distance from the PFAF as the OCS and glidepath are converging towards the landing surface (see Figure 2–1–3). The OCS slope and glidepath angle (GPA) values are interdependent: OCS Slope = 102/GPA; or GPA = 102/OCS slope.

Note that the TERPS convention is to define "slope" as "run over rise." Thus a standard glide path angle of 3.00º corresponds with an OCS angle of 1.685º.

This is almost enough to answer your question, which boils down to: What is the height of full-scale-deflection-low above the OCS? In order to answer that, we also have to know at what point the OCS begins.

Chapter 10 of the 8260.3 deals with precision approaches (and LDAs with glideslope). From 10–2–1a, the OCS associated with the Final Approach Segment is divided into the "Y," "X," and "W" Surfaces, mirrored about the Final Approach Course. The "W" and "X" surfaces are considered the primary area and the "Y" surface is considered the secondary area.

Side view of the runway threshold. Click to enlarge. The rising W surface begins at a point given by Formula 10–2–2:

$$d_\mathrm{origin}~(\mathrm{ft}) = \max \left[ 200, 1154 - \frac{TCH}{\tan \left( GPA \right)} \right] \tag{1}$$

There are many pages of instructions on how to determine the actual slopes of the surfaces according to the known obstacles; the TERPS specialists will adjust the TCH and OCS slope to find an acceptable compromise which keeps the GPA within the limits imposed by Table 2–6–1. But we don't need to concern ourselves with that; we know the resulting numbers already because they are published on the instrument approach plate.

So we can calculate the Required Obstacle Clearance at a point $$x$$ feet from the threshold (where $$x$$ is equal to or greater than $$d_\mathrm{origin}$$):

\begin{align} H_\mathrm{GP} (x) &= x \cdot \tan (GPA) + TCH \tag{2} \\ H_\mathrm{OCS} (x) &= (x - d_\mathrm{origin}) \cdot \tan \left( \cot^{-1} \left[ \frac{102}{GPA} \right] \right) \\ &= (x - d_\mathrm{origin}) \cdot \frac{1}{102 / GPA} \\ &= (x - d_\mathrm{origin}) \cdot \frac{GPA}{102} \tag{3} \\ ROC (x) &= H_\mathrm{GP} - H_\mathrm{OCS} \\ &= x \cdot \tan (GPA) - (x - d_\mathrm{origin}) \cdot \left( \frac{GPA}{102} \right) + TCH \tag{4} \end{align}

For example, consider the "typical" ILS approach with a GPA of 3.00º and a TCH of 50', leading to a $$d_\mathrm{origin}$$ of 200'. At a point 1NM from the threshold,

\begin{align} ROC (6076) &= 6076 \cdot \tan (3.00º) - 5876 \left( \frac{3.00}{102} \right) + 50 \\ &= 318 - 173 + 50 \\ &= 195~\mathrm{ft} \end{align}

But we aren't looking for nominal obstacle clearance; we want to know the obstacle clearance at full-scale deflection. The FAA's Instrument Flying Handbook implies (but does not say outright) that full-scale deflection is 0.7º off of GPA. Thus:

\begin{align} H_\mathrm{FSD-} (x) &= x \cdot \tan (GPA-0.70º) + TCH \tag{5} \\ OC_\mathrm{FSD-} (x) &= x \cdot \tan (GPA-0.70º) - (x - d_\mathrm{origin}) \cdot \left( \frac{GPA}{102} \right) + TCH \tag{6} \end{align}

Again considering the typical ILS approach from before, the function is bounded by

\begin{align} OC_\mathrm{FSD-} (50200) &= 50200 \cdot \tan (2.30º) - 50000 \cdot \left( \frac{3.00}{102} \right) + 50 = 596~\mathrm{ft} \\ OC_\mathrm{FSD-} (200) &= 200 \cdot \tan (2.30º) - 0 \cdot \left( \frac{3.00}{102} \right) + 50 = 58.0~\mathrm{ft} \end{align}

Note that the above numbers only hold true when the aircraft is within the W surface. The X and Y surfaces rise higher than the W surface, therefore obstacles may be closer to the glide slope height and the aircraft will be closer to those obstacles at full-scale-deflection-low.

See also Paragraph 2–6–6, setting the standard for the Vertical Guidance Surface, which "must be evaluated for all PA and APV approach procedures (except helicopter APV PinS procedures)." This Vertical Guidance Surface extends from the Landing Threshold Point (LTP) to the Decision Altitude (DA) point, i.e. it is inside of the expected "pilot flies visually" point.

• +1 very nice. Note that I didn’t check it. I just trust you. ;-)
– Jim
Nov 18, 2022 at 4:25