This is a timed quiz. You will be given 2700 seconds to answer all questions. Are you ready?
What is the name given to a line which crosses all meridians at the same angle?
A rhumb line is an arc crossing all meridians of longitude at the same path angle with a constant bearing, measured relative to true or magnetic North.
The angle between the Earthโs magnetic field and the horizontal is referred to as:
Magnetic dip is the angle between the horizontal axis and the Earthโs magnetic field lines. Dip will increase the closer your position to the magnetic poles.
What is the earliest time of day that a LAPL or PPL holder without a night rating can legally fly?
Official night time occurs between 30 min after sunset, when the centre of the disk is 6 degrees below the horizon, to 30 min before sunrise, when the centre of the disk is 6 degrees above the horizon.
If an aircraft flies due north along a meridian from latitude 05ยฐ30โS to 32ยฐ30โN. What is the distance flown?
1 degree of latitude is equal to 60NM. when crossing the equator from south to north, add up the difference in latitude. 5ยฐS + 32ยฐN = 37ยฐof latitude. 30โS + 30โN= 60โ = 1ยฐ. 38ยฐx 60NM = 2280NM. 1NM = 1.852 km, so 2280 x 1.852 = 4222.5km.
If the scale of a chart is given as 1:500,000 which of the following statements is true?
A scale or ratio is always represented in the same units, so 1cm on the chart equates to 500,000cm in real life.
Bangkok airport is located at 13ยฐ41โN100ยฐ45โE. If local mean time is 1500, what is the time in UTC?
One hour of time difference corresponds to 15ยฐof longitude, so 1ยฐof longitude equals 4 minutes. In this case the phrase longitude East = UTC is least and longitude West = UTC best works, because as you are travelling east, the local time in the east will be greater than the UTC time. 100ยฐof longitude x 4 minutes = 400 minutes = 6 hours and 40 minutes. 45โ is ยพ of a 1ยฐof longitude, so 4 x ยพ = 3 minutes. 15:00 – 06:43 = 08:17 UTC Be warned, although this question is seemingly irrelevant to PPL a very similar one does appear in the EASA PPL examinations.
A magnetic compass is designed to use the horizontal component of the earthโs magnetic field. The compass will be most accurate:
A magnetic compass is most accurate at low latitudes as itโs the furthest point away from the magnetic poles where the compass dip error has least effect.
An aircraft is maintaining a compass heading of 315ยฐC, the deviation is 1ยฐE, the true heading is 324ยฐT. What is the magnetic variation?
The โCadburyโs Dairy Milk Very Tastyโ mnemonic is very useful for this question. With a compass heading of 315ยฐ and a deviation of 1ยฐE our magnetic heading becomes 316ยฐ. If our true heading is 324ยฐ the difference is 8ยฐ. Using the mnemonic we can see we need to add and so are travelling East meaning A is the correct answer.
You are flying and maintaining a heading of 179ยฐC. from the compass card you know that the deviation is 4ยฐW, the magnetic variation in the area is 9ยฐW. What is the aircraftโs true heading?
When calculating in the direction from True to Compass, East is Least and West is Best. In this question we start with compass heading and want to know the true heading, so East and West are opposite. 179ยฐC – 4ยฐW = 175ยฐM, 175ยฐM – 9ยฐW = 166ยฐT
If you travel for 19 minutes with a ground speed of 66 knots ground speed, how far will you travel?
Speed = Distance / Time, so D = S x T, 19 min in hours is 19 : 60 = 0.3167 hour, 0.3167 hour x 66kts = 20.9NM
An aircraft is flying between two points which are 90nm apart, after 30nm the aircraft is found to be 5nm right of track. In order to route directly to the destination what heading direction is required?
Using the 1:60 rule, for every 60nm travelled if the aircraft is 1nm off track, the aircraft is 1ยฐ off course. This means at 30nm an aircraft will be 2ยฐ off course for every 1nm off track, so 2ยฐ x 5nm = 10ยฐ off course. A correction of 10ยฐ will get the aircraft onto a parallel track. To correct the heading back to point B apply the same rule from point B, so at 60nm the aircraft is 1nm off track, the aircraft is 1ยฐ off course, 1ยฐ x 5nm = 5ยฐ. The total correction back to point B will be 10ยฐ + 5ยฐ = 15ยฐ left, as the aircraft was right off track.
An aircraft has a maximum demonstrated crosswind component of 12kts. Given a 30kt wind speed by how many degrees can the wind direction differ from the runway direction before the aircraft crosswind limit is reached?
Use the CRP-1: use the wind side of the CRP at the bottom squared section for crosswind calculations. In this case there is no wind direction so place 0ยฐ under the INDEX and mark the wind speed (30kts) under the centre dot using the speed arcs. Twist the disk anti-clockwise till the dot for the crosswind component reaches 12kts. Under the INDEX mark read off the degrees (30ยฐ).
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