CDFA Calculation in Charts
 

Hello,

I'm trying to find the accurate formula which is used on a 2D and sometimes 3D approach which calculates the ALT you have to be at a specific distance for a CDFA. Using trigonometry yields a result which is off by 4' to 10' sometimes even more. Is the curvature of the Earth used for this calculations?

Ex.
LAX RWY 24R

My calculation to find 2,200' at 6.3nm from rwy threshold:
TDZE= 122'
Distance= 6.3nm
Alt= x

x= tan(3) * (6.3 * 6076.12)
x= 2006.1465' + 122'
ALT at 6.3nm = 2128.1465'... I'm off by roughly 70'+.

Any insight will be appreciated, thank you in advance.

Regards,
mabg

Won't it be on the approach plate every mile?

Thank you for the reply, yes in most of the charts you do have the altitudes for each mile. Never the less I would like to understand and know the actual calculation in order to implement it into a project of mine.

Thank you for your reply.
The number 57.2 that you are using after TAN represents what exactly?

Don't forget to add the TCH; 55ft for the ILS and RNP.

Oddly, no. We've had them here for decades.

Any idea if the Altitudes of the charts are rounded up or do they implement the curvature of the Earth in the calculations? I feel that the trigonometry applied for my calculation are off by a good margin compared to the actual altitudes of the published charts.

I suspect it’s relatively simple. The table in Jeppesen for gradient to distance suggests it’s 319’ per nautical mile. If there’s a more complicated formula behind the scenes, it’s not immediately obvious.

You're quoting a post that I deleted because I made a mistake with it. The 57.2 is rounded too much, which threw out my calculation.

The 57.2 is the rounded number of degrees in a radian. Excel uses radians for TAN calculations so you have to divide the degrees by 57.2 to get the radians, then work out the TAN of that. More accurately, radians = degrees/57.2958.

I don't know. It could be the curvature of the earth; it would be easy enough for you to work out. You can find out the diameter of the earth and work from there.

No, you're only off by 15ft. Add the TCH to your numbers, because that raises the whole approach.

Re rounding, definitely. Have a look at the rounding on this D/A chart from one of my hometown ILSs. 3>4 is 320ft, 6>7 is 310ft.

https://cimg2.ibsrv.net/gimg/pprune....c00c4f6fd5.jpg
The 2200 at KOBEE could be 2195, or there may even be a convention where the number is rounded up (from say 2192 to 2200) to be on the safe side.

you have to divide the degrees by 57.2 to get the radians

If you are going to the trouble of using the spreadsheet, better to use the normal conversion π/180.

I suggest, with your approach, that you are trying to measure a mile with a tape measure rather than an odometer ? End result is that you are playing with precision far finer than that inherent in the numbers to which you are making comparisons; with the result that you are getting needlessly confusing outcomes ?

J

JT, I have to get REALLY close to my screen to see that... pie? ;)

I hope my name isn't mentioned in the crash report.... :{

On a more serious note, I don't have a problem with people trying to understand what's going on on final and on the charts. The fact that the yanks don't have distance/altitude on some of their charts, I suspect, has contributed to prangs in the past and near misses in the recent past (SW).

I'm curious about the calculations because for the past 9 years I've operated in and out of VNKT shooting the VOR which works well. Now the airport has installed a LOC antenna and they have a LOC only approach. Charted Altitudes and distance to cross check with mcdu are totally off meaning waypoints based on distance are not matching the chart same goes with the altitudes. Was curious to see if the chart was the issue by trying to understand the calculations. Sorry I'm not allowed to post URLs, was gonna show the LOC only app into KTM.

Thank you all for the replies, much appreciated.

For the most common final approach angle (3 degrees): ideal height above touchdown is nm from touchdown x 320
So at 6.3nm, the ideal height would be:
(320 x 6.3) + 50ft TCH = 2066 ft

You mean VNKT, in Nepal, with lots of mountains? The LOC approach has a 5.4º gradient from the FAF to 5.5 DME and then a 3.0º gradient from 5.5 DME to the threshold.

The 3.0º gradient is approximately 320' x distance to the threshold plus 4320 runway elevation plus 50' TCH. That formula gets me 5586' at 5.5 DME vs the charted 5590'.

Hi Mabg

Generally earth curvature isn't considered within PANS-OPS unless you are creating an ILS approach with a final approach segment longer than about 5NM. There are tables included in PANS-OPS Volume II which give the height above threshold at various distances, whilst taking account of curvature of the Earth and an RDH of 49 ft:
https://cimg1.ibsrv.net/gimg/pprune....6f13d89c27.png
Additionally, PANS-OPS includes this formula:

https://cimg8.ibsrv.net/gimg/pprune....24e03dc01d.png
Note, the 'D2' in the SI units should be 'D squared' (the same as the Non-SI formula) and these formulae are for a quadratic equation i.e. for a parabola, so will be for a parabola that approximates the curvature of the Earth's surface over 'intermediate distances'.

Hope that helps

Cheers
TeeS

Hi again Mabg

Sorry, I missed your post saying which approach you were interested in, that is challenging flying!!

On the subject of rounding; up to next 10ft in final segment; up to next 100ft in the intermediate segment, which must also include a 1.5NM level section for Cat C and D (1.0NM for Cat A and B) if there is a descent; and up to next 100ft in initial segment.

If you are using Excel for your calculations, I would suggest using the Radians() function to convert degrees to radians within your formula.

Cheers
TeeS

Measuring with a micrometer, cut with an ax comes to mind. Good grief, 300’ per mile worked for decades and public math friendly

The KTM charts from the AIP are available here https://e-aip.caanepal.gov.np/_uploa...91d30d1ac7.pdf

Last page is the LOC chart.

I haven’t had a looked at the MCDU for the approach, by any chance are these numbers what you are seeing ?

3.1nm 4860 ft
4.5nm 5550 ft
5.5 nm 6050 ft
7.5 nm 7030 ft
10.5 nm 8510 ft
11.5 nm 9000 ft

I worked those out based on a constant 4.637 degree slope between the FAF and the VDP, getting rid of the change from 5.31 to 3 degrees at 5.5 nm.

What does the coding look like in the MCDU, where does it say the descent starts ?
What decent angle is in the database ?
What are the database altitudes vs waypoints ?
Are you flying xLS or LOC-FPA ?
Is the software version on your MCDU temperature compensated for the approach ?

How well do you think that would work in Nepal? Did you look at the required glidepath angles for VNKT?

I extracted the LOC RWY 02 plate from the huge IAP document for easier reference. Note that the glidepath is expressed as percent not degrees.

https://cimg3.ibsrv.net/gimg/pprune....ab7aa327b7.png

The degrees are given in the notes for the recommended profile.

It clearly wouldn’t, but the initial post was KLAX 24, not a special aircraft with special crew qualification where one would expect this information would be trained and any required tables issued. Few are going to run Excel spreadsheets on final approach.

In any case, the 1:60 rule says the 5.4* path is 540 ft/nm times ground speed in nm/min. 150 knots equals about 1350 fpm.

Why wouldn't one fly an approach like this in FPA mode? Is it that few aircraft have an autopilot FPA mode or that the mode is there but crews are not trained to use it. (MD-11, B-717, and MD-10 all had an FD/AP FPA mode which is ground referenced, not airmass referenced. No need to observe GS and compute required VS. Just dial in FPA -5.4 at the FAF and change it to -3.0 at 5.5 DME. Yes, of course altitude monitoring is still required.)

We flew the VOR 02 in NAV-FPA, as an early stabilized approach, landing config and Vapp at the FAF, make Vapp a speed constraint at the FAF. The approach path was too steep to fly the Airbus CDA where you configure during descent to reach Vapp at 1000 ft. Even being fully configured at the FAF, often with 20 kts of tailwind, you are using a fair bit of speed brake to prevent a flap over speed.

the 1:60 rule

(With the risk of my being labelled a pita pedant), one needs to keep in mind that the 1:60 rule is a pilot-friendly approximation of the 1:57. (whatever decimals float your boat) rule.

Thank you TeeS, I will keep your reference in my notes!!!

Thank you all for the replies.

I will take a pic of the actual data base for LOC 02 at VNKT since I'm going there in a few days. We stopped shooting the LOC there cuz the coding of the mcdu leaves you high on the approach flying a A333 with 1T bellow max landing weight (usually). Unlike the VOR which has the correct data in the mcdu to xcheck the chart alt and distance with the actual flight path of the aircraft. As mentioned in another post, we fly it NAV-FPA (No FLS installed) and xchecking the recommended altitudes of the chart while the mcdu has a different coding reduces your situation awareness. Any how, I just wanted to make sure that I was not missing something regarding the math. I'll post the pic of the coding once I have it for those who are interested.

I made the post to confirm I was not missing any other data for the calculations of the altitudes in the charts to identify and understand if the chart of the data of the mcdu is wrong. Plenty of training and years of going there 2 or 3 times per month for the past 9 years. This is not about flying the approach with less than 10ft of precision while on NAV-FPA. As mentioned before this is about understanding the math and making sure I was not missing any other criteria for the calculations. Thank you for your time and replies.

The coding I have
GURAS /+11500
C022º 5NM
CL02/+9300
C022º TRK022º 3-5.3º
FF02/+9000
C022º 1-5.3º
LOC11/+8420
C022º 3-5.3º
75LOC/+6720
C022º 2-5.3º
55LOC/+5590
C022º 1-5.3º
45LOC/+5280
C022º 2-5.3º
KTM/4750

This corresponds to the charted altitudes. Linear regression through the altitude/distance table on the Jepp chart works out that they used 5.21 degrees.


That to me looks like the coding inside 55LOC is wrong, should be -3.0º, looked at a machine with xLS, the approach is not in the data base.

This time of year its ISA+25 or more, so your true altitude at the FAF is more like 9500ft despite it indicating 9000ft on the altimeter, the non xLS machines are not temperature compensated. Something like FPA -5.9 will need to be flown to meet the profile. This should be flown in LOC-FPA.

Hi SWH
I'm no expert on the coding but that almost looks like someone used the gradient of 5.3% in place of the 3° as that kept the numbers constant?
Cheers
TeeS

The chart has 5.31 degrees between FF02 and 55LOC, then 3 degrees from 55LOC to 3.1 DME for the VDP/MDA. If you were to draw a straight line from the FAF to the KTM VOR it would be a FPA of 4.25 degrees. The chart is designed so that the last 1200 feet is flown at 3 degrees from 55LOC.

Cargo rates into KTM are pretty good, and fuel is expensive (tanker), so it’s common to land near MLW. With the higher density height, TAS is up around 155-160 kts, plus a tailwind, on a 3 degree path you are pushing 900 fpm V/S on short final.

Hi swh
Sorry, I was a bit rushed with my last response. I was commenting on your statement of: "That to me looks like the coding inside 55LOC is wrong, should be -3.0º" (which I agree with from what you have shown). I was wondering whether some manual data input was involved (unlikely, I know) and a gradient of 5.3%, which equates (roughly) to 3° was copied and pasted in error, just a thought.
Cheers
TeeS