Departures

25.10.2004

Submission:

I am currently studying PANOPS departure procedures (VOL 2 Part 2) so as I can present a brief to my squadron on the subject. However I am a little confused regarding the Minimum Obstacle clearances and Procedure Climb Gradients used in a turning departure for Areas 1 and 2. It is relatively easy to establish the above areas for the straight departure, OIS + MOC (0.8%) = PDG, standard of 3.3% provides 2.5% OIS and 0.8% MOC. However a turning departure is not so easy. >From the PANSOPS notes I have deduced several answers.

- Area 1 (which is effectively the Turn Initiation Area) utilises a standard MOC regardless of distance from DER of 295ft. This would give a PDG of Max obstacle height + 295ft (expressed in a percentage), which would be applicable all the way to the turning point.

OR

- Area 1 (which is effectively the Turn Initiation Area) utilises a PDG the higher of 0.8% MOC + OIS (the same as for a straight departure) or Max obstacle height + 295ft at the turning point.

Now, for area 2. I understand that max obstacle height must be less than TNA/H + do PDG - MOC, and that MOC is the greater of 0.008(dr + do) or 90m(295ft), differing dr depending on whether obstacle is before or after TP. Is the PDG used in the equation, the same as the PDG used in Area 1? (which would also depend on your above answer). Also, how is the PDG for area 2 derived? Is it the same as for a straight departure i.e 0.8%MOC + OIS (as required)? Or is it similar to my suggestions above about area 1, i.e. Max obstacle clearance + 295 ft until above MSA?

Any help would be greatly appreciated as I am fairly stuck on the subject.

Lee Taylor, Royal New Zealand Airforce

Answer or Commentary:

(I.W

You are right the departure criteria can be rather confusing at times. As far as I can see Area 1 and Area 2 in a straight departure are the same thing, the splay is the same (15°) and the MOC (0.8%) is the same. So for simplicity you may think of the departure area as a single entity where certain things may or may not occur at certain positions. Lets take the important points step by step.

1.) Straight portion

The OIS (2.5%) is merely a PDG of 3.3% minus the MOC of 0.8%. So normally in a straight departure I determine a surface that gets over all the obstacles, then find the angle of the surface, Tan it, times it by 100 to get the MOC line and finally add 0.8 to get the PDG. If the PDG is below 3.3% do nothing if it is above promulgate it.





2.) Track divergence up to 15°

Widen the straight 15° splay area on the inside of the track adjustment by the amount of track divergence. Keep the outside splay area at 15° from the runway track until 1.9 nm and then bring the outside area in by the amount of divergence. MOC remains 0.8% throughout.

3.) Turns of > 15°

These turns are not permitted until an aircraft reaches a height of 120m (395 ft). The MOC is still 0.8% up until the earliest point that the turn can be commenced.





4.) Turns of > 15° at a fix

Determine the fix tolerance (i.e. DME tolerance...) add 6 seconds flying time (c) at the maximum IAS for the largest aircraft category. Now you have the turn initiation area, the ETP and the LTP. All obstacles within this area and the rest of the departure must be crossed with either a MOC of 0.8% or 90m whichever is greater.

Note that the 0.8% MOC does not reach 90m until 6.4 nm from the DER.

Note2 the longer the departure procedure the greater the 0.8% MOC becomes, a departure of 21.3 nm has a MOC of 1000 ft.

5.) Turns of > 15° at an altitude

These turns are a bit more complicated for a procedure designer (avoid them if you can). Firstly the previous point 2. Using the PDG you developed in point 1 determine the position of the turn altitude. Draw a straight line across the straight-ahead obstacle protection area. From this line onwards you must have a MOC of either 0.8% or 90m, which ever is greater, over all subsequent obstacles. Now the distance to climb varies depending on where the obstacle is. For simplification the areas can be divided in to sectors.

Sector 1 (the easy one)

Basically the distance to climb, for the 0.8% MOC calculations, is the straight-ahead distance.

So the MOC = (straight-ahead distance * 0.008) or 90m

Sector 2 (slightly more complicated)

The distance to climb to the obstacle is the distance from the DER to the line where an aircraft would reach the turn altitude using the PDG you established in point 1, plus the distance from point "ETP" to the obstacle.

The MOC = [ (straight-ahead distance to TP + distance "ETP" to the obstacle) * 0.008] or 90m.

Sector 3 (the tricky one)

This sector is only valid if you are using turn at an altitude, if not it becomes part of sector 2. Draw a line from the obstacle to perpendicular to the outside line of the straight-ahead obstacle protection area (X). Now a line from this position, on the area splay line, to a position perpendicular to the nominal straight-ahead track (Y) and finally back to the DER.

The MOC = [ (distance DER to "Y") + (distance "Y" to "X") + (distance "X" to the obstacle) ] * 0.008

or 90m

If you are lucky all obstacles will be crossed with the required obstacle clearance. If not you will have to work backwards from the obstacle to determine the new PDG. Another alternative is to change the turn altitude, which will vary the respective distances to climb to the obstacle. Also in some instances secondary areas will come into play. In this case you will have to calculate the primary MOC at that position, using the distance to climb, and then apply the secondary area criteria to determine the actual MOC required.

As you can see turning at an altitude is quite a complicated business for a procedure designer. At least with turning at a fix the distance to climb to an obstacle remains static. I hope this helps.

Has anybody else got any ideas on the subject?)