Given are:

• Wind (W/V) 030° / 15 kt
• MW (VAR) 0°
• Dev (DEV) -2°

Performance data from the flight manual for the planned pressure altitude are:

• % BHP = 70
• KTAS = 111
• (US)GPH = 8,1
• For warm-up, taxi, climb, descent, approach, and reserve an additional 6 USGal must be considered.

Response selection:

(A) KSK (CH) 323° / 10 USGal
(B) KSK (CH) 342° / 8 USGal
(C) KSK (CH) 337° / 11 USGal
(D) KSK (CH) 339° / 19 USGal

The correct answer is (C). 

Let's first determine the compass heading (KSK). To do this, we convert the course into a heading by taking the wind into account (wind correction angle WCA). The wind angle can be determined by drawing or by using a navigation computer.

At the same time, we can determine the speed over ground (GS), which we will need later for fuel calculation. We can then insert the determined values into the course scheme:

First, the true course (rwK) is converted into a true heading (rwSK), taking into account the wind correction angle (WCA). Subsequently, the magnetic variation (MW) is taken into account and the magnetic heading (mwSK) is determined.

Finally, the deviation (DEV) is taken into account and the result is the compass heading.

Now it is time to determine the minimum fuel requirement calculation: Previously we determined a speed over ground of 102 kt. This corresponds to a speed of 102 NM/hour.

Answer choice:

(A) 2 hr 00 min.
(B) 2 hr 08 min.
(C) 1 hr 52 min.
(D) 2 hr 20 min.

The correct answer is (B). 

In this task, there is a tailwind and then a headwind on the route from A to B. The wind is not blowing.

From the gut, most of the examinees would vote for the same time as when there is no wind; thus, also 2 hrs 00 min. Once you just have the wind from behind and on the way back you have the wind from the front. What I did well at the beginning will be taken away from me afterwards. That will then probably cancel each other out.

But this is not the case. Why, will amaze some.

Without wind influence the time for the outward and return flight amounts to 2 hours, or only 1 hour for the flight from A to B. The airplane's own speed (VE, TAS) is therefore 150 km/h or the equivalent of 81 knots.

Note: Since the wind is given in knots, we convert the airspeed and the distance also in knots.

If the wind on the route is now taken into account, we should start calculating: On the route from A to B we fly - due to the tail wind - with a speed over ground of 101 kt. For this we need for the given distance of 81 NM, a time of 48 minutes.

On the route from B to A we fly - due to the headwind - with a speed over ground of 61 kt. For the given distance of 81 NM, we need a time of 80 minutes.

Added up, this results in a pure flight time of 2 hours and 8 minutes for the outward and return flight.

Response selection:
(A) 20 NW

(B) 18 NM

(C) 12 NM 

(D) 16 NM

The correct answer is (C). 

Since the QNH value is equal to the standard pressure value at mean sea level, no correction needs to be made to the altimeter when passing through the transit altitude.
The difference between airfield altitude and FL 75 is equal to 6,000 ft.
At an average climb rate of 600 ft per minute, this requires 10 minutes.
Since ground speed is given here in MPH, it must be converted to knots (kt).
85 MPH is equivalent to 74 kt or 137 km/h.
If you travel 74 NM over ground in 1 hour, you cover a distance of 12 NM in 10 minutes.

Explanation of terms

Answer choice:

(A) No, the landing aircraft must only avoid gliders that are also on approach.

(B) Yes, gliders must always be avoided.

(C) No, only if it is higher than the glider.

(D) Yes, since it does not know how high the glider can climb.

Correct is answer (A). One can find the explanation in SERA.3210 Evasive action rules.

Despite the unambiguous description of the avoidance rules, there are always ambiguities. Frank Peter Dörner, a lawyer specializing in aviation law, has written a detailed article on the subject for the Aerokurier.  

>> Here << you can read the whole text. (With kind permission of the Aerokurier.)  

Maximum Elevation Figure can be translated as "highest elevation value". There are two calculation methods for determining this value: elevation point evaluation and obstacle evaluation.

Elevation point evaluation

The landscape elevation (elevation point) related to the mean sea level is used. In our example, the elevation of the landscape is 2331 ft AMSL.
A fictitious obstacle of 328 ft (100 m) height above ground is placed on this elevation point. The reason for this fictitious value is that obstacles up to 100 m are not entered in the ICAO chart.
Then a safety margin of 30 ft is added for possible changes in air pressure and setting errors on the altimeter.
The sum of these three values is rounded up to the nearest 100 ft value.

2331 ft + 328 ft + 30 ft = 2689 ft
Rounded up: 2700 ft AMSL

Obstacle evaluation

The height of a known obstacle that protrudes prominently from the landscape is used. In our example, the obstacle height is 2463 ft AMSL.
A safety margin of 60 ft is added to this for any changes in air pressure and setting errors on the altimeter.
The sum of these two values is rounded up to the nearest 100 ft value.

2463 ft + 60 ft = 2523 ft
Rounded up: 2600 ft AMSL 

Note: Note: By the way, for the indication on the ICAO chart Germany and the V500 charts the higher of the calculated values is used. An explanation including the calculation methods can be found on the back of the charts.

There is the following question in the PPL questionnaire:

For an obstacle with the altitude of 2850 ft AMSL, which Maximum Elevation Figure would result?  Using the obstacle evaluation calculation method, the result can be calculated as follows:

Obstacle height 2850 ft AMSL + surcharge 60 ft = 2910 ft AMSL

Rounded up: 3000 ft AMSL

What is the practical significance of the Maximum Elevation Figure for the pilot?

First things first: the Maximum Elevation Figure is not the same as the minimum safety altitude. But it helps pilots in planning their flight routes and especially in situations where it is necessary to deviate from the planned flight corridor.

When planning a flight, the minimum safety altitude must be taken into account. The MEF assists in determining this.

When a MEF of 13 is given, there is no natural or man-made obstacle higher than 1,300 ft in this 30 x 30 minute rectangle, calculated on the meridian and parallel. This value has been rounded up to the nearest hundred.

Laut SERA.5005 Buchstabe f (im deutschen Luftrecht umgesetzt in §37 der LuftVO) muss eine Mindesthöhe von 500 ft (150 m) über dem Boden oder Wasser oder 500 ft über dem höchsten Hindernis in einem Umkreis von 150 m um das Luftfahrzeug eingehalten werden. Über According to SERA.5005 letter f (implemented in German air law in §37 of the LuftVO), a minimum altitude of 500 ft (150 m) above the ground or water or 500 ft above the highest obstacle within 150 m of the aircraft must be maintained. Over cities, other densely populated areas and outdoor crowds, it is even 1000 ft above the highest obstacle within 600 m of the aircraft. These altitudes may only be undershot during takeoff and landing. Gliders, hang gliders and paragliders, for which it is necessary due to their design, are allowed to fly below the minimum altitude. This can also be read in the Aeronautical Information Publication AIP VFR in chapter ENR 1-16.

If you now add the MEF with the safety minimum altitude, you get a safe flight altitude. Of course, the airspace structure must still be taken into account. But especially in unexpected deviations of a planned flight route, the value is a very helpful indication for a continued safe flight.

The display of the MEF can only be found on the ICAO chart and the European charts of DFS. Charts of other manufacturers show directly the safety altitude. There, the highest point plus a safety distance of 1,000 or 2,000 ft is directly indicated.

The altimeter in the aircraft displays altitude above the air pressure value set on the altimeter in the reference pressure window.

In the area around the aircraft, the air pressure is 10 hPa lower. The air pressure decreases upwards and increases downwards. Thus, the reference area 1020 hPa is below the -area 1010 hPa. That is, we measure the altitude of 2500 ft above a surface that is currently below mean sea level.

The vertical distance between two pressure surfaces is about 27 ft (8 m) per hPa in the lower layers. The difference in our case is therefore 270 ft (80 m).

The altimeter would indicate a flight altitude of 2230 ft AMSL with the QNH value set correctly.

The 1:60 rule is based on the fact that if an aircraft has a heading error of 1° after 60 NM of flight, it will have a separation of 1 NM.

Formular 1:60 rule: 

This means:

• if an airplane is 1 NM off the course line after 60 NM flight distance, the course error is 1°
  
• if an airplane is after 60 NM flight distance 5 NM beside the course line, the course error is 5°
  
• etc.

After the aircraft has been offset, the 1:60 rule can be used to determine the angle by which the heading must be improved in order to fly directly to the destination (B). To do this, the 1:60 rule is applied twice:

Note: The 1:60 rule applies to heading offsets up to 20°.

Fig. 1: Course improvement (α) on parallel course.

Fig. 2: Additional course improvement (β) to the destination.