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I am studying the US Terminal Instrument Procedures manual (TERPS) and as I was skimming through the guidelines about Sloping Obstacle Clearance Surface, I came across the image below from section 2-1-4 (the red circle remark is mine):

Figure 2-1-3. PA/APV Glidepath Descent

The problem is, I can not understand how the lowest point of the OCS, which I have marked with the red circle, is determined. It seems to be a point close to Departure End of Runway (DER), but where exactly?

The only things TERPS states is:

Descending on a PA/APV glidepath. 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 precise final approach fix (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.

Could someone point me in the right direction?

EDIT: What I mean is, why (and how do I calculate the extra distance) does the first image apply, rather than the second?

enter image description here

enter image description here

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3 Answers 3

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I can't yet find where this is indicated in TERPS, but it appears that the OCS surface starts at the point before the threshold where the GP is 250 feet above the threshold height.

I'm inferring that from this document about calculating VNAV criteria.

enter image description here

It has the above diagram and says:

The distance from the runway threshold where the OCS surface starts (Section 1) is defined by the formula D = (250-HATh)/.052407779

where HATh is the same as TCH, which is a standard 50', and the number .05240779 is the tangent of 3°. So far I'm unable to reference that back to any TERPS definition, but maybe you can find it.

There's an interesting article discussing how TCH is defined over at Code7700.com

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  • $\begingroup$ Thank you for your answer. I will edit my question in order to make it clearer. As far as I have understood, actually the GPA is based on the OCS and not the other way around, but in any case I am having trouble figuring out why the OCS does not start at the Departure End of Runway (DER), rather at an arbitrary distance from it. $\endgroup$
    – Vector Zita
    Commented Nov 1, 2018 at 8:36
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See Formula 10-2-2, "Slope Origin Distance," and Figure 10-2-3. The distance is equal to 200 feet or

$$ 1154 - \frac{\rm TCH}{\tan({\rm GPA})} $$

feet, whichever is greater.

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The probable reason why the obstacle clearance surface line is shown terminating prior to the threshold is because the glide slope transmitter antennas are located about 50 feet in front of the threshold.

When a plane lands using instruments, it uses what is called the "instrument landing system" (ILS), and nearly all commercial airports have them. The ILS transmits guidance signals to the aircraft through an array of antennas. Some of those antennas, including the glide slope antenna, are generally located about 50 feet in front of the threshold.

As a side note, this is one reason why it is bad to have to land short, because if you do, then your plane will plow through all those antennas.

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    $\begingroup$ You've got the wrong antennas in the wrong place. The glide slope antenna is a tower about 500-700 feet past the threshold (just prior to the touchdown zone) and 100-150 feet off to the side. The localizer antenna array is off the end of the runway at the far end. So if you're landing over a localizer antenna, it's for the other end of the runway or you're flying a LOC back course approach. And realistically, the bigger obstacles in that zone are the approach lighting towers. $\endgroup$
    – Gerry
    Commented Nov 2, 2018 at 12:29

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