I am kind of stuck for calculating a descent path on the ILS RWY 7 approach into KORL. Assuming the glide slope is not available and you have to use the localizer only, what is the best way to calculate the descent and Decision to Land Point?


As you can see in the approach chart, the MDA is 660 feet for the localizer only approach. Assuming a glide path of 3°, I calculated a descent of 318 feet per Nautical Mile. However, that would not correlate to the 1919 feet at the Outer Marker.

What would be the best way to calculate / plan the approach in this example?


3 Answers 3


The approach shows that the glideslope (if it were working) is a three degree descent angle. You can calculate your own visual descent point (VDP), since one isn't provided for you, by taking the height above touchdown (600 ft. in this case) and dividing it by 300 ft/NM. This gives you 2.0 miles from the runway. Since the chart shows the runway threshold at 0.2 DME, your VDP will be at 2.2 DME.

Assuming that your navigation equipment does not provide a pseudo-glideslope, your best bet is to descend at your normal rate (in a light airplane I would say 800-900 fpm) until you level at 660 ft. (the MDA), then fly to your calculated VDP (2.2 DME) and if you have the required visual cues then descend to the runway knowing that it will be a standard 3 degree descent (300 ft./NM). If you don't have it at this point, then you can start an early climb and fly the missed approach.

Put another way, first you calculate your VDP:

$$\frac{Height~Above~Touchdown}{Descent~Rate}=VDP~Distance~From~Runway$$ $$\frac{600ft}{300 ft/NM}=2.0NM$$ $$VDP=Runway~Threshold+VDP~Distance~From~Runway$$ $$0.2DME+2.0NM=2.2 DME$$

Then you fly the approach:

  • Cross HERNY at 2,000 ft.
  • Descend to minimums (660 ft.) at 800-900 fpm.
  • Fly to 2.2 DME at 660 ft.
    • At 2.2 DME, if you see the runway environment, start your descent and land
    • If you don't see the runway, start your climb and fly the published missed approach

I built a calculator to help with these types of calculations.


You can input the icao identifier for an airport. I pull up the current METAR for you. You input the altitude, altimeter setting, temperature, indicated/calibrated airspeed and wind velocity, degrees the wind is off from the runway heading, altitude to lose and the distance to lose it.

enter image description here

You have to first descent from 2000 to 1020 in 2.7 miles. With the current weather, you would have a 62 knot ground speed in a C172. Traveling 2.7 NM losing 980 feet is a 3.42 degree descent angle which requires a 375 fpm descent.

You could do a similar calculation from BUVAY to the VDP. 0.6 nm to lose 360 feet. That is a 5.64 degree descent angle and 620 fpm.


Ignore the 1919 glideslope intercept at the OM, as that is for an ILS approach only. For the LOC approach, in Cat A, you will cross the FAF at 2000 and descend to the 660ft MDA over the next 5.5nm.

There are (at least) two ways to plan and fly a non precision approach. You can just try to descend constant from the FAF to the MAP, hoping you reach the MDA and maybe being in a position to land. The alternative is to reach the MDA early, prior to the MAP, ensuring that you spend at least some time at the actual approach minimums.

constant rate to threshold

For a constant descent rate to the MAP, you should divide the altitude to descend (FAF minus MDA) by the distance to travel. In this case, that's (2000 - 660)/5.5, or 243 feet per nautical mile.

early descent to MDA

There is no published VDP, so I'll assume a self-determined point 1.0nm from the MAP; this means you must descend at 330 feet per nm: (2000-660)/4.5. At 90 knots that is around 450 fpm, which is about what I'd expect for a non precision approach.

As far as a decision-to-land point goes, that depends on the aircraft and your personal minimums more than something I can calculate.

  • $\begingroup$ Your constant rate # puts you over the threshold at the MDA, which is probably unlandable. Shouldn't the constant rate be based on FAF altitude minus TDZE? $\endgroup$
    – Ralph J
    Aug 20, 2016 at 19:25

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