Take This Air Navigation Quiz - 1

page 88

ICAO's definition of a nautical mile is that it is a measure of distance of 1852 meters. A nautical mile is a unit used in aviation and maritime navigation to calculate distances over water. It is based on the circumference of the Earth and is equal to one minute of latitude. This standardized measurement allows for accurate and consistent calculations in aviation and maritime operations.

The GC bearing of Q from P is given as 060deg, and the CA (Course Angle) is given as 4deg. In the northern hemisphere, the RL (Relative Bearing) is obtained by adding the CA to the GC bearing. Therefore, the RL bearing of Q from P would be 060deg + 4deg = 064deg. Similarly, the RL bearing of P from Q would be the opposite direction, which is 180deg + 064deg = 244deg. Finally, the GC bearing of P from Q is obtained by subtracting 180deg from the RL bearing of Q from P, resulting in 244deg - 180deg = 064deg.

The ground speed of an aircraft can be calculated by dividing the distance traveled by the time taken. In this case, the distance between points A and B on the chart is given as 6.6cm. Since the scale of the chart is 1:2,000,000, we need to convert this distance to the actual distance traveled by multiplying it by the scale factor. The actual distance traveled is therefore 6.6cm * 2,000,000 = 13,200,000cm.

The time taken to cover this distance is given as 15 minutes and 12 seconds, which can be converted to seconds by multiplying the number of minutes by 60 and adding it to the number of seconds. In this case, the time taken is 15 * 60 + 12 = 912 seconds.

To calculate the ground speed in knots, we need to convert the distance traveled from centimeters to nautical miles. Since 1 nautical mile is equal to 185,200 centimeters, we can divide the actual distance traveled by 185,200 to get the distance in nautical miles. In this case, the distance in nautical miles is 13,200,000cm / 185,200 = 71.28 nautical miles.

Finally, we can calculate the ground speed by dividing the distance in nautical miles by the time taken in hours. Since 1 hour is equal to 3600 seconds, we can divide 71.28 by 912 / 3600 = 0.25 hours.

Therefore, the ground speed of the aircraft is 71.28 / 0.25 = 285.12 knots, which is closest to the given answer of 281.3.

page 89

The heading at True (Hdg(T)) is given as 258 degrees and the convergence is 20 degrees East. To find the heading at Grid (Hdg(G)), we need to subtract the convergence from the True heading. Therefore, 258 - 20 = 238 degrees. However, since the question asks for the heading in the opposite direction, we need to add 180 degrees. Therefore, 238 + 180 = 418 degrees. However, since headings are always expressed as 3-digit numbers, we subtract 360 degrees. Therefore, 418 - 360 = 58 degrees. Hence, the correct answer is 278 degrees.

To find the change in longitude between 160°W and 158°E:

Convert the longitudes to a common reference:

160°W is 160 degrees west of the Prime Meridian.

158°E is 158 degrees east of the Prime Meridian.

Calculate the total change in longitude:

Since the longitudes are on opposite sides of the Prime Meridian, add their absolute values to find the total change.



Therefore, the change in longitude between 160°W and 158°E is 318 degrees.

page 64

textbook page 26-27

The given question asks for the distance in kilometers for a flight that is planned from 40degS 110degE to 40degS 102degE along the rhumb line track. The rhumb line track is a line of constant azimuth or bearing, which means that the flight will follow a path with a constant compass direction. To calculate the distance, we can use the formula: distance = (longitude difference) x (cos(latitude)) x (radius of the Earth). In this case, the longitude difference is 110degE - 102degE = 8 degrees, the latitude is 40degS, and the radius of the Earth is not given. However, since the answer choices are all in kilometers, we can assume that the radius of the Earth is in kilometers. Using the given value of cos(40deg) = 0.766, we can calculate the distance as: distance = 8 x 0.766 x (radius of the Earth). Since the answer is 681.5 km, we can conclude that the radius of the Earth is approximately 89 km.

The scale of the chart can be determined by calculating the ratio of the distance shown on the chart (1.4 inches) to the actual distance flown by the aircraft in 40 seconds (unknown value). Since the ground speed of the aircraft is given in knots, which is a unit of speed, we need to convert it to a unit of distance. 1 knot is equal to 1.852 kilometers per hour. Therefore, the actual distance flown by the aircraft in 40 seconds can be calculated by multiplying the ground speed (480 knots) by the time (40 seconds) and then converting it to kilometers. The scale of the chart can then be determined by dividing the distance shown on the chart (1.4 inches) by the actual distance flown by the aircraft. After performing the calculations, the scale is determined to be 1 cm to 2.778 km.

textbook page 22

The scale of a chart represents the ratio between a distance on the chart and the corresponding ground distance. In this case, at 60degN, a 30 cm distance on the chart measures a ground distance of 390nm. To find the scale at the equator, we can use the fact that the scale of a chart remains constant along a line of latitude. Therefore, the scale at the equator would be the same as the scale at 60degN, which is 1: 4810000.

The scale on the chart at 48°N is 1 : 977664. This means that one unit on the chart represents 977,664 units in reality. The scale is determined by the distance between the meridians on the Mercator projection. Since two meridians one degree apart are spaced at a distance of 7.62 cm, the scale can be calculated using the formula: scale = (distance on chart) / (distance in reality). In this case, the distance on the chart is 7.62 cm and the distance in reality is 1 degree. Therefore, the scale is 1 : 977664.

textbook page 10

page 39

Based on the given information, the aircraft's grid heading is 310 degrees and the variation is 10 degrees E. The grid heading is calculated by adding or subtracting the variation from the magnetic heading, depending on whether the variation is east or west. In this case, the magnetic heading (Hdg(M)) is 340 degrees, and since the variation is east, we subtract 10 degrees from the magnetic heading to get the grid heading. Therefore, the grid heading is 330 degrees. Since the grid datum is 40 degrees W, we subtract 40 degrees from the grid heading to get the aircraft's longitude. Therefore, the aircraft's longitude is 10 degrees E.

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