Chapter 11

Determine the magnetic course from Airpark East Airport (area 1) to Winnsboro Airport (area 2). Magnetic variation is 6°30'E. A 075°. B 082°. C 091°.

A 075°. Your answer is CORRECT. (FAA-H-8083-25B Chap 16)To find the magnetic course from Airpark East Airport (lower left of chart) to Winnsboro Airport (right of 2 on Fig. 24), you must find true course and correct it for magnetic variation. Determine the true course by placing the straight edge of your plotter along the given route such that the grommet (center hole) is on a meridian (the north/south line with crosslines). True course of 082° is the number of degrees clockwise from true north. It is read on the protractor portion of your plotter at the intersection of the meridian. To convert this to a magnetic course, subtract the 6°30'E (or round up to 7°E) easterly variation and find that the magnetic course is 075°. Remember to subtract easterly variation and add westerly variation.

An aircraft departs an airport in the central standard time zone at 0930 CST for a 2-hour flight to an airport located in the mountain standard time zone. The landing should be at what time? A 1030 MST. B 0930 MST. C 1130 MST.

A 1030 MST. Your answer is CORRECT. (FAA-H-8083-25B Chap 16)Flying from the Central Standard Time Zone to the Mountain Standard Time Zone results in a 1-hour gain due to time zone changes. A 2-hour flight leaving at 0930 CST will arrive in the Mountain Standard Time Zone at 1130 CST, which is 1030 MST.

What is the estimated time en route for a flight from Denton (area 1) to Addison (area 2)? The wind is from 200° at 20 knots, the true airspeed is 110 knots, and the magnetic variation is 7° east. A 13 minutes. B 19 minutes. C 16 minutes.

A 13 minutes. Your answer is CORRECT. (FAA-H-8083-25B Chap 16)The requirement is time en route and not magnetic heading, so there is no need to convert TC to MC.1.To find the en route time from Denton (southwest of 1) to Addison (south of 2), use Fig. 25.2.Using the associated scale on the side of the chart, measure the distance to be 22 NM.3.TC = 125°.4.Mark up 20 knots with 200° under true index.5.Put TC of 125° under true index.6.Slide the grid so the pencil mark is on TAS of 110 knots.7.Read the groundspeed of 103 knots under the grommet.8.On the calculator side, place 103 knots on the outer scale over 60 minutes.9.Read 13 minutes on the inner scale below 22 NM on the outer scale.

An aircraft departs an airport in the central standard time zone at 0845 CST for a 2-hour flight to an airport located in the mountain standard time zone. The landing should be at what coordinated universal time? A 1645Z. B 1445Z. C 1345Z.

A 1645Z. Your answer is CORRECT. (FAA-H-8083-25B Chap 16)First convert the departure time to coordinated universal time (Z) by using the time conversion table in Fig. 27. To convert from CST to Z, you must add 6 hours. Thus, 0845 CST is 1445Z (0845 + 6 hours). A 2-hour flight would make the estimated landing time at 1645Z (1445 + 2 hours).

Estimate the time en route from Majors Airport (area 1) to Winnsboro Airport (area 2). The wind is from 340° at 12 knots and the true airspeed is 136 knots. Magnetic variation is 5° east. A 17 minutes 30 seconds. B 19 minutes. C 14 minutes 30 seconds.

A 17 minutes 30 seconds. Your answer is CORRECT. (FAA-H-8083-25B Chap 16)The requirement is time en route and not magnetic heading, so there is no need to convert TC to MC. Measure the distance between Majors Airport and Winnsboro Airport using the associated scale located on the side of the chart. You should find the distance to be about 41 NM. Use your plotter to find a true course of 100°. Using your flight computer, place 340° under the true index and mark a wind speed of 12 knots. Place 100° under the true index and slide the card so the true airspeed arc of 136 knots is under the wind dot. The flight computer should indicate a groundspeed of approximately 140 knots. Turn the flight computer over and place the pointer on 14 for 140 knots groundspeed. Follow the outer scale to 41 for 41 NM and read a time of approximately 17:30 below the 41 on the outer scale.

What is the estimated time en route for a flight from Allendale County Airport (area 1) to Claxton-Evans County Airport (area 2)? The wind is from 100° at 18 knots and the true airspeed is 115 knots. Add 2 minutes for climb-out. A 30 minutes. B 27 minutes. C 33 minutes.

A 30 minutes. Your answer is CORRECT. (FAA-H-8083-25B Chap 16)The requirement is time en route and not magnetic heading, so there is no need to convert TC to MC.1.To find the en route time from Allendale County (north of 1) to Claxton-Evans (southeast of 2), use Fig. 23.2.Using the sectional scale located at the top of the chart, measure the distance to be approximately 56.5 NM.3.TC = 212°.4.Mark up 18 knots with 100° under true index.5.Put TC of 212° under true index.6.Slide the grid so the pencil mark is on TAS of 115 knots.7.Read the groundspeed of 120 knots under the grommet.8.On the calculator side, place 120 knots on the outer scale over 60 minutes.9.Read 28 minutes on the inner scale below 56.5 NM on the outer scale.10.Add 2 minutes for climb-out and the en route time is 30 minutes.

What is the estimated time en route from Sandpoint Airport (area 1) to St. Maries Airport (area 4)? The wind is from 215° at 25 knots, and the true airspeed is 125 knots. A 34 minutes. B 30 minutes. C 38 minutes.

A 34 minutes. Your answer is CORRECT. (FAA-H-8083-25B Chap 16)The requirement is time en route and not magnetic heading, so there is no need to convert TC to MC.1.You are to find the en route time from Sandpoint Airport (north of 1) to St. Maries Airport (southeast of 4) on Fig. 22.2.Using the scale at the top of the chart, measure the distance to be 59 NM.3.TC = 181°.4.Mark up 25 knots with 215° under true index.5.Put TC of 181° under true index.6.Slide the grid so the pencil mark is on TAS of 125 knots.7.Read the groundspeed of 104 knots under the grommet.8.On the calculator side, place 104 knots on the outer scale over 60 minutes.9.Find 59 NM on the outer scale and read 34 minutes on the inner scale.

How far will an aircraft travel in 2-1/2 minutes with a groundspeed of 98 knots? A 4.08 NM. B 3.35 NM. C 2.45 NM.

A 4.08 NM. Your answer is CORRECT. (FAA-H-8083-25B Chap 16)To determine the distance traveled in 2-1/2 minutes at 98 knots, note that 98 knots is 1.6 NM/minute (98 ÷ 60 = 1.633). Thus, in 2-1/2 minutes, you will have traveled a total of 4.08 NM (1.633 × 2.5 = 4.08). Alternatively, put 98 on the outer scale of your flight computer over the index on the inner scale. Find 2.5 minutes on the inner scale, above which is 4.1 NM.

Determine the magnetic course from Cooperstown Airport (area 2) to Jamestown Airport (area 4). A 030°. B 210°. C 218°.

B 210°. Your answer is CORRECT. (FAA-H-8083-25B Chap 16)Find the magnetic course from Cooperstown Airport (northeast of 2) to Jamestown Airport (south of 4). Because Jamestown has a VOR on the field, a compass rose exists around the Jamestown Airport symbol on the chart. Compass roses are based on magnetic courses. Thus, a straight line from Jamestown Airport to Cooperstown Airport coincides with the compass rose at 030°. Because the route is south to Jamestown, not north from Jamestown, compute the reciprocal direction as 210° (030° + 180°). The course, then, is approximately 210°.

While practicing S-turns, a consistently smaller half-circle is made on one side of the road than on the other, and this turn is not completed before crossing the road or reference line. This would most likely occur in turn A 1-2-3 because the bank is decreased too rapidly during the latter part of the turn. B 4-5-6 because the bank is increased too rapidly during the early part of the turn. C 4-5-6 because the bank is increased too slowly during the latter part of the turn.

B 4-5-6 because the bank is increased too rapidly during the early part of the turn. Your answer is CORRECT. (FAA-H-8083-3C Chap 7)Note that the wind in Fig. 66 is coming up from the bottom rather than from the top. The consistently smaller half-circle is made when on the upwind side of the road, i.e., 4-5-6. The initial bank is increased too rapidly, resulting in a smaller half-circle. Then an attempt is made to widen the turn out in the latter stages. Thus, the recrossing of the road is done at less than a 90° angle.

What is the estimated time en route from Mercer County Regional Airport (area 3) to Minot International (area 1)? The wind is from 330° at 25 knots and the true airspeed is 100 knots. Add 3-1/2 minutes for departure and climb-out. A 52 minutes. B 48 1/2 minutes. C 45 minutes.

B 48 1/2 minutes. Your answer is CORRECT. (FAA-H-8083-25B Chap 16)The requirement is time en route and not magnetic heading, so there is no need to convert TC to MC. Using Fig. 21, the time en route from Mercer Co. Reg. Airport (lower left corner) to Minot (right of 1) is determined by measuring the distance (60 NM measured with the associated scale at the bottom of the chart), determining the time based on groundspeed, and adding 3.5 minutes for takeoff and climb. The TC is 012° as measured with a plotter. The wind is from 330° at 25 kt. On the wind side of your flight computer, place the wind direction 330° under the true index and mark 25 kt. up. Rotate TC of 012° under the true index. Slide the grid so the pencil mark is on the arc for TAS of 100 kt. Read 80 kt. groundspeed under the grommet. Turn to the calculator side and place the groundspeed of 80 kt. on the outer scale over 60 minutes. Find 60 NM on outer scale and note 45 minutes on the inner scale. Add 3.5 minutes to 45 minutes for climb for en route time of 48.5 minutes.

In flying the rectangular course, when would the aircraft be turned less than 90°? A Corners 1 and 2. B Corners 1 and 4. C Corners 2 and 4.

B Corners 1 and 4. Your answer is CORRECT. (FAA-H-8083-3C Chap 7)When doing a rectangular course, think in terms of traffic pattern descriptions of the various legs. In Fig. 62, note that the airplane is going counterclockwise about the rectangular pattern. While on the base leg (between corners 3 and 4), the airplane is crabbed to the inside of the course. Thus, on corner 4, less than a 90° turn is required. Similarly, when the airplane proceeds through corner 1, it should roll out such that it is crabbed into the wind, and, again, a less-than-90° angle is required.

(Refer to area C.) How should the flight controls be held while taxiing a tailwheel airplane with a left quartering tailwind? A Left aileron up, elevator neutral. B Left aileron down, elevator down. C Left aileron down, elevator neutral.

B Left aileron down, elevator down. Your answer is CORRECT. (FAA-H-8083-3C Chap 2)When there is a left quartering tailwind, the left aileron should be held down so the wind does not get under the left wing and flip the airplane over. Also, the elevator should be down, i.e., controls forward, so the wind does not get under the tail and blow the airplane tail over front.

What information should be entered into item 16, "Destination Aerodrome," for a VFR day flight? A The destination airport identifier code and name of the FBO where the airplane will be parked. B The destination airport identifier code. C The destination city and state.

B The destination airport identifier code. Your answer is CORRECT. (AIM Para 5-1-5)In item 16, "Destination Aerodrome," of the flight plan form in Fig. 51, enter the ICAO four-letter location identifier.

Determine the magnetic heading for a flight from Fort Worth Meacham (area 4) to Denton Muni (area 1). The wind is from 330° at 25 knots, the true airspeed is 110 knots, and the magnetic variation is 7°E. A 017°. B 023°. C 003°.

C 003°. Your answer is CORRECT. (FAA-H-8083-25B Chap 16)1.This flight is from Fort Worth Meacham (southeast of 4) to Denton Muni (southwest of 1) on Fig. 25.2.TC = 019°.3.MC = 019° - 7°E variation = 012°.4.Wind magnetic = 330° - 7°E variation = 323°.5.Mark up 25 knots with 323° under true index.6.Put MC 012° under true index.7.Slide grid so pencil mark is on 110 knots TAS.8.Note that the pencil mark is 10° left.9.Subtract 10° from 012° MC for 002° MH. The closest answer choice is 003°.

An aircraft departs an airport in the mountain standard time zone at 1515 MST for a 2-hour 30-minute flight to an airport located in the Pacific standard time zone. What is the estimated time of arrival at the destination airport? A 1845 PST. B 1745 PST. C 1645 PST.

C 1645 PST.Your answer is CORRECT. (FAA-H-8083-25B Chap 16)Departing the Mountain Standard Time (MST) Zone at 1515 MST for a 2-hour 30-minute flight would result in arrival in the Pacific Standard Time (PST) Zone at 1745 MST. Because there is a 1-hour difference between MST and PST, 1 hour must be subtracted from the 1745 MST arrival to determine the 1645 PST estimated time of arrival at the destination airport.

If a true heading of 135° results in a ground track of 130° and a true airspeed of 135 knots results in a groundspeed of 140 knots, the wind would be from A 200° and 13 knots. B 019° and 12 knots. C 246° and 13 knots.

C 246° and 13 knots. Your answer is CORRECT. (FAA-H-8083-25B Chap 16)To estimate your wind given true heading and a ground track, place the groundspeed under the grommet (140 knots) with the ground track of 130° under the true index. Then find the true airspeed on the true airspeed arc of 135 knots, and put a pencil mark for a 5° right deviation (135° - 130° = 5°). Place the pencil mark on the centerline under the true index and note a wind from 246° under the true index. The pencil mark is now on 153 knots, which is about 13 knots up from the grommet (153 - 140).

Estimate the time en route from Addison (area 2) to Dallas Executive (area 3). The wind is from 300° at 15 knots, the true airspeed is 120 knots, and the magnetic variation is 7° east. A 14 minutes. B 11 minutes. C 8 minutes.Your answer is CORRECT. (FAA-H-8083-25B Chap 16)The requirement is time en route and not magnetic heading, so there is no need to convert TC to MC.1.To find the en route time from Addison (south of 2) to Dallas Executive (area 3), use Fig. 25.2.Using the associated scale on the side of the chart, measure the distance to be 18 NM.3.TC = 186°.4.Mark up 15 kt. with 300° under true index.5.Put TC of 186° under true index.6.Slide the grid so the pencil mark is on TAS of 120 kt.7.Read the groundspeed of 125 kt. under the grommet.8.On the calculator side, place 125 kt. on the outer scale over 60 min.9.Read 8.5 min. on the inner scale below 18 NM on the outer scale.

C 8 minutes. Your answer is CORRECT. (FAA-H-8083-25B Chap 16)The requirement is time en route and not magnetic heading, so there is no need to convert TC to MC.1.To find the en route time from Addison (south of 2) to Dallas Executive (area 3), use Fig. 25.2.Using the associated scale on the side of the chart, measure the distance to be 18 NM.3.TC = 186°.4.Mark up 15 kt. with 300° under true index.5.Put TC of 186° under true index.6.Slide the grid so the pencil mark is on TAS of 120 kt.7.Read the groundspeed of 125 kt. under the grommet.8.On the calculator side, place 125 kt. on the outer scale over 60 min.9.Read 8.5 min. on the inner scale below 18 NM on the outer scale.

If more than one cruising altitude is intended, which should be entered in item 15, "Level," of the flight plan? A Lowest cruising altitude. B Highest cruising altitude. C Initial cruising altitude.

C Initial cruising altitude. Your answer is CORRECT. (AIM Para 5-1-5)If more than one cruising altitude is intended, enter the planned cruising level for the first (initial) portion of the route to be flown.

(Refer to area B.) How should the flight controls be held while taxiing a tailwheel airplane into a right quartering headwind? A Right aileron down, elevator neutral. B Right aileron up, elevator down. C Right aileron up, elevator up.

C Right aileron up, elevator up. Your answer is CORRECT. (FAA-H-8083-3C Chap 2)When there is a right quartering headwind, the right aileron should be up to spoil the excess lift on the right wing that the crosswind is creating. The elevator should be up to keep weight on the tailwheel to help maintain maneuverability.