Cross Country Flight Planning 8
PA.VI.A.K5 Plotting a course, to include: PA.VI.A.K5a a. Determining heading, speed, and course PA.VI.A.K5b b. Wind correction angle PA.VI.A.K5c c. Estimating time, speed, and distance PA.VI.A.K5d d. True airspeed and density altitude
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1. What is the effect of a temperature increase from 25 to 50° F on the density altitude if the pressure altitude remains at 5,000 feet? (Refer to Figure 8: Density Altitude Chart)
Answer (A) is correct. (FAA-H-8083-25B Chap 11) Increasing the temperature from 25°F to 50°F, given a pressure altitude of 5,000 ft., requires you to find the 5,000-ft. line on the density altitude chart at the 25°F level. At this point, the density altitude is approximately 3,800 ft. Then move up the 5,000-ft. line to 50°F, where the density altitude is approximately 5,400 ft. There is about a 1,600-ft. increase (5,400 ft. – 3,800 ft.). As temperature increases, so does density altitude; i.e., the atmosphere becomes thinner (less dense). Because a 1,600-foot increase is not an answer choice, 1,650-foot increase would be the best answer.
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Determine the density altitude for these conditions: (Refer to Figure 8: Density Altitude Chart) Altimeter setting = 29.25 Runway temperature = +81°F Airport elevation = 5,250 ft MSL
Answer (C) is correct. (FAA-H-8083-25B Chap 11) With an altimeter setting of 29.25" Hg, about 626 ft. (579 plus 1/2 the 94-ft. pressure altitude conversion factor difference between 29.2 and 29.3) must be added to the field elevation of 5,250 ft. to obtain the pressure altitude, or 5,876 feet. Note that barometric pressure is less than standard and pressure altitude is greater than true altitude. Next, convert pressure altitude to density altitude. On the chart, find the point at which the pressure altitude line for 5,876 ft. crosses the 81°F line. The density altitude at that spot shows somewhere in the mid-8,000s of feet. The closest answer choice is 8,500 feet. Note that, when temperature is higher than standard, density altitude exceeds pressure altitude.
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3. What is the effect of a temperature decrease and a pressure altitude increase on the density altitude from 90°F and 1,250 feet pressure altitude to 55°F and 1,750 feet pressure altitude? (Refer to Figure 8: Density Altitude Chart)
Answer (C) is correct. (FAA-H-8083-25B Chap 11) The requirement is the effect of a temperature decrease and a pressure altitude increase on density altitude. First, find the density altitude at 90°F and 1,250 ft. (approximately 3,600 feet). Then find the density altitude at 55°F and 1,750 ft. pressure altitude (approximately 1,900 feet). Next, subtract the two numbers. Subtracting 1,900 ft. from 3,600 ft. equals a 1,700-ft. decrease in density altitude.
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4. What is the effect of a temperature increase from 30 to 50 °F on the density altitude if the pressure altitude remains at 3,000 feet MSL? (Refer to Figure 8: Density Altitude Chart)
Answer (A) is correct. (FAA-H-8083-25B Chap 11) Increasing the temperature from 30°F to 50°F, given a constant pressure altitude of 3,000 ft., requires you to find the 3,000-ft. line on the density altitude chart at the 30°F level. At this point, the density altitude is approximately 1,650 feet. Then move up the 3,000-ft. line to 50°F, where the density altitude is approximately 2,950 feet. There is an approximate 1,300-ft. increase (2,950 – 1,650 feet). Note that 50°F is just about standard and pressure altitude is very close to density altitude.
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Determine the density altitude for these conditions: (Refer to Figure 8: Density Altitude Chart) Altimeter setting = 30.35 Runway temperature = +25°F Airport elevation = 3,894 ft. MSL
Answer (B) is correct. (FAA-H-8083-25B Chap 11) With an altimeter setting of 30.35" Hg, 394 ft. must be subtracted from a field elevation of 3,894 to obtain a pressure altitude of 3,500 feet. Note that the higher-than-normal pressure of 30.35 means the pressure altitude will be less than true altitude. The 394 ft. was found by interpolation: 30.3 on the graph is –348, and 30.4 was –440 feet. Adding one-half the –92 ft. difference (–46 ft.) to –348 ft. results in –394 feet. Once you have found the pressure altitude, use the chart to plot 3,500 ft. pressure altitude at 25°F, to reach 2,000 ft. density altitude. Note that since the temperature is lower than standard, the density altitude is lower than the pressure altitude.
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6. 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? (Refer to Figure 27: Time Conversion Table)
Answer (B) 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.
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7. An aircraft departs an airport in the mountain standard time zone at 1615 MST for a 2-hour 15-minute flight to an airport located in the Pacific standard time zone. The estimated time of arrival at the destination airport should be (Refer to Figure 27: Time Conversion Table)
Answer (C) is correct. (FAA-H-8083-25B Chap 16) Departing the Mountain Standard Time Zone at 1615 MST for a 2-hour 15-minute flight would result in arrival in the Pacific Standard Time Zone at 1830 MST. Because there is a 1-hour difference between Mountain Standard Time and Pacific Standard Time, 1 hour must be subtracted from the 1830 MST arrival to determine the 1730 PST arrival.
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8. 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? (Refer to Figure 27: Time Conversion Table)
Answer (A) 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).
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9. 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? (Refer to Figure 27: Time Conversion Table)
Answer (A) 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.
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10. An aircraft departs an airport in the Pacific standard time zone at 1030 PST for a 4-hour flight to an airport located in the central standard time zone. The landing should be at what coordinated universal time? (Refer to Figure 27: Time Conversion Table)
Answer (C) 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 PST to Z, you must add 8 hours; thus, 1030 PST is 1830Z (1030 + 8 hours). A 4-hour flight would make the proposed landing time at 2230Z (1830 + 4 hours).
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11. An aircraft departs an airport in the eastern daylight time zone at 0945 EDT for a 2-hour flight to an airport located in the central daylight time zone. The landing should be at what coordinated universal time? (Refer to Figure 27: Time Conversion Table)
Answer (B) 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 eastern daylight time (EDT), add 4 hours to get 1345Z (0945 + 4 hours). A 2-hour flight would have you arriving at your destination airport at 1545Z.
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12. En route to First Flight Airport (area 5), your flight passes over Hampton Roads Airport (area 2) at 1456 and then over Chesapeake Regional at 1501. At what time should your flight arrive at First Flight? (Refer to Figure 20: Sectional Chart Excerpt)
Answer (C) is correct. (FAA-H-8083-25B Chap 16) The distance between Hampton Roads Airport (north of 2) and Chesapeake Regional (northeast of 2) is 10 NM. It took 5 min. (1501 – 1456) to go 10 NM. On your flight computer, place the 5 min. the first leg took on the inner scale under 10 NM on the outer scale. Then find 50 NM (60 NM total distance – 10 NM of the first leg) on the outer scale and read 25 min. on the inner scale for the time from Chesapeake Regional to First Flight. The distance from Chesapeake Regional to First Flight (right of 5) is 50 NM. Add 25 min. to the time you passed Chesapeake Regional (1501) to get 1526.
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13. 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. (Refer to Figure 25: Sectional Chart Excerpt)
Answer (A) 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.
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14. 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. (Refer to Figure 25: Sectional Chart Excerpt)
Answer (C) 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.
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15. 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. (Refer to Figure 24: Sectional Chart Excerpt)
Answer (B) 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.
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16. What is the estimated time en route for a flight from Claxton-Evans County Airport (area 2) to Hampton Varnville Airport (area 1)? The wind is from 290° at 18 knots and the true airspeed is 85 knots. Add 2 minutes for climb-out. (Refer to Figure 23: Sectional Chart Excerpt)
Answer (C) is correct. (FAA-H-8083-25B Chap 16) Using the sectional scale located at the top of the chart, you will find the distance en route from Claxton-Evans (southwest of 2) to Hampton Varnville (east of 1 on Fig. 23) is approximately 57 NM. Use your plotter to determine that the TC is 045°. The requirement is time en route and not magnetic heading, so there is no need to convert TC to MC. Using the wind side of your computer, turn your true index to the wind direction of 290° and mark 18 knots above the grommet with your pencil. Then turn the inner scale so that the true index is above the TC of 045°. Place the pencil mark on the TAS of 85 knots and note the groundspeed of 91 knots. Turn your flight computer over and set the speed of 91 knots above the 60-minutes index on the inner scale. Then find the distance of 57 NM on the outer scale to determine a time en route of 37 minutes. Add 2 minutes for climb-out, and the en route time is 39 minutes.
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17. 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. (Refer to Figure 23: Sectional Chart Excerpt)
Answer (A) 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.
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18. While en route on Victor 185, a flight crosses the 248° radial of Allendale VOR at 0953 and then crosses the 216° radial of Allendale VOR at 1000. What is the estimated time of arrival at Savannah VORTAC? (Refer to Figure 23: Sectional Chart Excerpt)
Answer (C) is correct. (FAA-H-8083-25B Chap 16) The first step is to find the three points involved. V185 runs southeast from the top left of Fig. 23. The first intersection (V70 and V185) is about 1 in. from the top of the chart. The second intersection (V157 and V185) is about 1-1/2 in. farther along V185. The Savannah VORTAC is about 6 in. farther down V185. Use the sectional scale located at the top of the chart. From the first intersection (V70 and V185), it is about 10 NM to the intersection of V185 and V157. From there it is 40 NM to Savannah VORTAC. On your flight computer, place the 7 min. the first leg took (1000 – 0953) on the inner scale under 10 NM on the outer scale. Then find 40 NM on the outer scale. Read 28 min. on the inner scale, which is the time en route from the V185 and V157 intersection to the Savannah VORTAC. Arrival time over Savannah VORTAC is therefore 1028.
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19. What is the estimated time en route for a flight from St. Maries Airport (area 4) to Priest River Airport (area 1)? The wind is from 300° at 14 knots and the true airspeed is 90 knots. Add 3 minutes for climb-out. (Refer to Figure 22: Sectional Chart Excerpt)
Answer (C) 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. Time en route from St. Maries Airport (southeast of 4) to Priest River Airport (upper left corner) on Fig. 22. 2. Using the scale at the top of the chart, measure the distance to be 54 NM. 3. TC = 346°. 4. Mark up 14 knots with 300° under true index. 5. Put TC of 346° under true index. 6. Slide the grid so the pencil mark is on TAS of 90 knots. 7. Read the groundspeed of 80 knots under the grommet. 8. On the calculator side, place 80 knots on the outer scale over 60 minutes. 9. Find 54 NM on the outer scale and read 40 minutes on the inner scale. 10. Add 3 minutes for climb-out to get time en route of 43 minutes.
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20. 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. (Refer to Figure 22: Sectional Chart Excerpt)
Answer (B) 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.
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21. Determine the estimated time en route for a flight from Priest River Airport (area 1) to Shoshone County Airport (area 3). The wind is from 030 at 12 knots and the true airspeed is 95 knots. Add 2 minutes for climb-out. (Refer to Figure 22: Sectional Chart Excerpt)
Answer (A) 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 Priest River Airport (west of area 1) to Shoshone County Airport (area 3) use Fig. 22. 2. Using the scale at the top of the chart, measure the distance to be 48 NM. 3. TC = 143°. 4. Mark up 12 knots with 030° under true index. 5. Put TC of 143° under true index. 6. Slide the grid so the pencil mark is on TAS of 95 knots. 7. Read the groundspeed of 99 knots under the grommet. 8. On the calculator side, place 99 knots on the outer scale over 60 minutes. 9. Read 29 minutes on the inner scale below 48 NM on the outer scale. 10. Add 2 minutes for climb-out and the en route time is 31 minutes.
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22. 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. (Refer to Figure 21: Sectional Chart Excerpt)
Answer (C) 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 knots.
On the wind side of your flight computer, place the wind direction 330° under the true index and mark 25 knots up. Rotate TC of 012° under the true index. Slide the grid so the pencil mark is on the arc for TAS of 100 knots. Read 80 knots groundspeed under the grommet.
Turn to the calculator side and place the groundspeed of 80 knots 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.
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On a cross-country flight, point A is crossed at 1500 hours and the plan is to reach point B at 1530 hours. Use the following information to determine the indicated airspeed required to reach point B on schedule. Distance between A and B 70 NM Forecast wind 310° at 15 kt. Pressure altitude 8,000 ft. Ambient temperature –10°C True course 270° The required indicated airspeed would be approximately
Answer (A) is correct. (FAA-H-8083-25B Chap 16) First determine the required groundspeed to reach point B at 1530 by placing 70 NM on the outer scale over 30 minutes on the inner scale to determine a groundspeed of 140 kt. On the wind side of the computer, put the wind direction of 310° under the true index and put a pencil mark 15 kt. up from the grommet. Next, turn the inner scale so the 270° true course is under the true index and put the grommet over the groundspeed. Note that to obtain the 140-kt. groundspeed, you need a 152-kt. true airspeed. Next, on the computer side, put the air temperature of –10°C over 8,000 ft. altitude. Then find the true airspeed of 152 kt. on the outer scale, which lies over approximately 137 kt. indicated airspeed on the inner scale.
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24. How far will an aircraft travel in 2-1/2 minutes with a groundspeed of 98 knots?
Answer (C) 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.
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25. The line from point C to point A of the wind triangle represents (Refer to Figure 68: Wind Triangle)
Answer (A) is correct. (FAA-H-8083-25B Chap 16) The line from point C to point A on the wind triangle represents the wind direction and velocity line.
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26. The line from point C to point B of the wind triangle represents (Refer to Figure 68: Wind Triangle)
Answer (B) is correct. (FAA-H-8083-25B Chap 16) The line from point C to point B, on the wind triangle, represents the true course and groundspeed line.
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27. The line from point A to point B of the wind triangle represents (Refer to Figure 68: Wind Triangle)
Answer (A) is correct. (FAA-H-8083-25B Chap 16) The line connecting point A to point B on the wind triangle represents the true heading and airspeed line.
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28. Determine the compass heading for a flight from Claxton-Evans County Airport (area 2) to Hampton Varnville Airport (area 1). The wind is from 280° at 8 knots, and the true airspeed is 85 knots. Magnetic variation is 7°W. (Refer to Figure 58: Compass Card, and Figure 23: Sectional Chart Excerpt)
Answer (A) is correct. (FAA-H-8083-25B Chap 16) 1. This flight is from Claxton-Evans (left of 2) to Hampton Varnville (right of 1) on Fig. 23. 2. TC = 045°. 3. MC = 045° TC + 7°W variation = 052°. 4. Wind magnetic = 280° + 7°W variation = 287°. 5. Mark up 8 knots with 287° under true index. 6. Place MC 052° under true index. 7. Move wind mark to 85 knots TAS arc. 8. Note that the pencil mark is 4° left. 9. Subtract 4° from 052° MC for 048° MH. 10. Subtract 4° compass deviation (obtained from Fig. 58) from 048° to find the compass heading of 044°.
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29. Determine the magnetic heading for a flight from Allendale County Airport (area 1) to Claxton-Evans County Airport (area 2). The wind is from 090° at 16 knots and the true airspeed is 90 knots. Magnetic variation is 7°W. (Refer to Figure 23: Sectional Chart Excerpt)
Answer (C) is correct. (FAA-H-8083-25B Chap 16) 1. This flight is from Allendale County (above 1) to Claxton-Evans County Airport (left of 2) on Fig. 23. Variation is shown on Fig. 23 as 7°W. 2. TC = 212°. 3. MC = 212° TC + 7°W variation = 219°. 4. Wind magnetic = 090° + 7°W variation = 097°. 5. Mark up 16 knots with 097° under true index. 6. Place MC 219° under true index. 7. Move wind mark to 90 knots TAS arc. 8. Note that the pencil mark is 9° left. 9. Subtract 9° from 219° MC for 210° MH.
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30. 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. (Refer to Figure 25: Sectional Chart Excerpt)
Answer (B) 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°.
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31. When converting from true course to magnetic heading, a pilot should
Answer (B) is correct. (FAA-H-8083-25B Chap 16) When converting true course to magnetic heading, you should remember two rules. With magnetic variation, east variation is subtracted and west variation is added. With wind corrections, left correction is subtracted and right correction is added.
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32. 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
Answer (C) 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).
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33. Determine the magnetic heading for a flight 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. (Refer to Figure 22: Sectional Chart Excerpt)
Answer (A) is correct. (FAA-H-8083-25B Chap 16) 1. This flight is from Sandpoint Airport (above 1), to St. Maries (below 4) on Fig. 22. 2. TC = 181°. 3. MC = 181° – 15°E variation (14°30E rounded up) = 166°. 4. Wind magnetic = 215° – 15° (14°30E rounded up) = 200°. 5. Mark up 25 knots with 200° under true index. 6. Put MC 166° under true index. 7. Slide grid so pencil mark is on 125 knots TAS. 8. Note that the pencil mark is 6° right. 9. Add 6° to 166° MC for 172° MH.
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34. Determine the magnetic heading for a flight from St. Maries Airport (area 4) to Priest River Airport (area 1). The wind is from 340° at 10 knots and the true airspeed is 90 knots. (Refer to Figure 22: Sectional Chart Excerpt)
Answer (B) is correct. (FAA-H-8083-25B Chap 16) 1. This flight is from St. Maries (just below 4) to Priest River (upper left corner) on Fig. 22. 2. TC is 345°. 3. MC = 345° – 15°E variation (14°30E rounded up) = 330°. 4. Wind magnetic = 340° – 15° (14°30E rounded up) = 325°. 5. Mark 10 knots up when 325° under true index. 6. Put MC 330° under true index. 7. Slide grid so pencil mark is on 90 kt. TAS. 8. Note that the pencil mark is 1° left. 9. Subtract 1° from 330° MC for 329° MH. A magnetic heading of 330° is the best answer of the choices given.
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35. What is the magnetic heading for a flight from Priest River Airport (area 1) to Shoshone County Airport (area 3)? The wind is from 030° at 12 knots and the true airspeed is 95 knots. (Refer to Figure 22: Sectional Chart Excerpt)
Answer (B) is correct. (FAA-H-8083-25B Chap 16) On Fig. 22, begin by computing the true course from Priest River Airport (upper left corner) to Shoshone County Airport (just below 3) by laying a flight plotter between the two airports. The grommet should coincide with the meridian (vertical line with cross-hatchings). Note the 143° true course on the edge of the protractor. Next, find the magnetic variation that is given by the dashed line marked 14°30E (rounded to 15°E), slanting in a northeasterly fashion just to the east of Shoshone County Airport. Subtract the 15°E variation from TC to obtain a magnetic course of 128°. Because the wind is given true, reduce the true wind direction of 030° by the magnetic variation of 15°E to a magnetic wind direction of 15°.
Now use the wind side of your computer. Turning the inner circle to 15° under the true index, mark 12 knots above the grommet. Set the magnetic course of 128° under the true index. Slide the grid so the pencil mark is on 95 knots TAS. Note that the pencil mark is 7° left of the center line, requiring you to adjust the magnetic course to a 121° magnetic heading (128° – 7°). Subtract left, add right. That is, if you are on an easterly flight and the wind is from the north, you will want to correct to the left.
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36. Determine the magnetic heading for a flight from Mercer County Regional Airport (area 3) to Minot International (area 1). The wind is from 330° at 25 knots, the true airspeed is 100 knots, and the magnetic variation is 10°E. (Refer to Figure 21: Sectional Chart Excerpt)
Answer (B) is correct. (FAA-H-8083-25B Chap 16) On Fig. 21, begin by computing the true course (TC) from Mercer Co. Reg. (lower left corner) to Minot Int’l. (upper left center) by drawing a line between the two airports. Next, determine the TC by placing the grommet on the plotter at the intersection of the course line and a meridian (vertical line with cross-hatchings) and the top of the plotter aligned with the course line. Note the 012° TC on the edge of the protractor. Next, subtract the 10° east magnetic variation from the TC to obtain a magnetic course (MC) of 002°. Because the wind is given true, subtract the 10° east magnetic variation to obtain a magnetic wind direction of 320° (330° – 10°).
Now use the wind side of your computer to plot the wind direction and velocity. Place the magnetic wind direction of 320° on the inner scale on the true index. Mark 25 knots up from the grommet with a pencil. Turn the inner scale to the magnetic course of 002°. Slide the grid up until the pencil mark lies over the line for true airspeed (TAS) of 100 knots. Correct for the 10° left wind angle by subtracting from the magnetic course of 002° to obtain a magnetic heading of 352°. This is intuitively correct because, given the magnetic course of 002° and a northwesterly wind, you must turn to the left (crab into the wind) to correct for it.
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37. Determine the magnetic heading for a flight from Majors Airport (area 1) to Winnsboro Airport (area 2). The wind is from 340° at 12 knots, the true airspeed is 136 knots, and the magnetic variation is 6° 30'E. (Refer to Figure 24: Sectional Chart Excerpt)
Answer (B) is correct. (FAA-H-8083-25B Chap 16) On Fig. 24, begin by computing the true course (TC) from Majors Airport (northeast of area 1) to Winnsboro Airport (east of area 2) by drawing a line between the two airports. Next, determine the TC by placing the grommet on the plotter at the intersection on the course line and a meridian (vertical line with cross-hatchings) and the top of the plotter aligned with the course line. Note the TC of 101° TC on the edge of the protractor. Next, subtract the 6° east magnetic variation from the TC to obtain a magnetic course (MC) of 095°. Because the wind is given true, subtract the 6° magnetic variation to obtain a magnetic wind direction of 334° (340° – 6°). Now use the wind side of your computer to plot the wind direction and velocity. Place the magnetic wind direction of 334° on the inner scale on the true index. Mark 12 kt. up from the grommet with a pencil. Turn the inner scale to the magnetic course of 095°. Slide the grid up until the pencil mark lies over the line for true airspeed (TAS) of 136 kt. Correct for the 4° left wind angle by subtracting from the magnetic course of 095° to obtain a magnetic heading of 091°. This is intuitively correct because, given the magnetic course of 095° and a northwesterly wind, you must turn to the left (crab into the wind) to correct for it.
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38. Determine the magnetic course from Airpark East Airport (area 1) to Winnsboro Airport (area 2). Magnetic variation is 6°30'E. (Refer to Figure 24: Sectional Chart Excerpt)
Answer (C) 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.
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39. Determine the magnetic course from Cooperstown Airport (area 2) to Jamestown Airport (area 4). (Refer to Figure 26: Sectional Chart Excerpt)
Answer (B) 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°.
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40. Determine the magnetic course from First Flight Airport (area 5) to Hampton Roads Airport (area 2). (Refer to Figure 20: Sectional Chart Excerpt)
Answer (C) is correct. (FAA-H-8083-25B Chap 16) You are to find the magnetic course from First Flight Airport (lower right corner) to Hampton Roads Airport (above 2 on Fig. 20). True course is the degrees clockwise from true north. Determine the true course by placing the straight edge of your plotter along the given route with the grommet at the intersection of your route and a meridian (the north/south line with crosslines). Here, TC is 320°. To convert this to a magnetic course, add the 11° westerly variation (indicated by the dashed magenta line that parallels the coastline north/south), and find the magnetic course of 331°. Remember to subtract easterly variation and add westerly variation.
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