Figure 39 
(Refer to figure 39.) Determine the total distance required to land 
over a 50-ft. obstacle.  

Pressure altitude 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3,750 ft
Headwind 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 kts
Temperature 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . Std





   ANSWER: 816 feet. 

   The total distance to clear a 50-ft. 
   obstacle for a 3,750-ft. pressure altitude is required. Note 
   that this altitude lies halfway between 2,500 ft. and 5,000 ft. 
   Halfway between the total distance at 2,500 ft. of 1,135 ft. 
   and the total distance at 5,000 ft. of 1,195 ft. is 1,165 ft. Since 
   the headwind is 12 kt., the total distance must be reduced by 
   30% (10% for each 4 kt.).

            70% x 1,165 = 816 ft. 

Figure 39 
(Refer to figure 39.) Determine the approximate landing ground roll 
distance.  

Pressure altitude 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5,000 ft
Headwind 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . Calm
Temperature 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101°F





   ANSWER: 545 feet. 

   The ground roll distance at 5,000 ft. is 
   495 ft. According to Note 2 in Fig. 39, since the temperature 
   is 60°F above standard, the distance should be increased by 
   10%.

         495 ft. x 110% = 545 ft. 

Figure 39 
(Refer to figure 39.) Determine the approximate landing ground roll 
distance.  

Pressure altitude 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1,250 ft
Headwind 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 kts
Temperature 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . Std





   ANSWER: 366 feet. 

   The landing ground roll at a pressure 
   altitude of 1,250 ft. is required. The difference between 
   landing distance at sea level and 2,500 ft. is 25 ft. (470 - 445). 
   One-half of this distance (12) plus the 445 ft. at sea level is 
   457 ft. The temperature is standard, requiring no adjustment. 
   The headwind of 8 kt. requires the distance to be decreased 
   by 20%. Thus, the distance required will be 366 ft. (457 x 
   80%). 

Figure 39 
(Refer to figure 39.) Determine the total distance required to land 
over a 50-foot obstacle.  

Pressure altitude 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7,500 ft
Headwind 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 kts
Temperature 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32°F
Runway 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . Hard surface





   ANSWER: 1,004 feet. 

   Under normal conditions, the total 
   landing distance required to clear a 50-ft. obstacle is 1,255 ft. 
   The temperature is standard (32°F), requiring no adjustment. 
   The headwind of 8 kt. reduces the 1,255 by 20% (10% for 
   each 4 kt.). Thus, the total distance required will be 1,004 ft. 
   (1,255 x 80%). 

Figure 39 
(Refer to figure 39.) Determine the total distance required to land 
over a 50-foot obstacle.  

Pressure altitude 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5,000 ft
Headwind 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 kts
Temperature 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41°F
Runway 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . Hard surface





   ANSWER: 956 feet. 

   Under standard conditions, the 
   distance to land over a 50-ft. obstacle at 5,000 ft. is 1,195 ft. 
   The temperature is standard, requiring no adjustment. The 
   headwind of 8 kt., however, requires that the distance be 
   decreased by 20% (10% for each 4 kt. headwind). Thus, the 
   landing ground roll will be 956 ft. (80% of 1,195). 

What should pilots state initially when telephoning a weather 
briefing facility for preflight weather information? 





   ANSWER: Identify yourself as a pilot. 

   When telephoning for a weather 
   briefing, you should identify yourself as a pilot so the 
   person can give you an aviation-oriented briefing. Many 
   nonpilots call weather briefing facilities to get the weather 
   for other activities. 

What should pilots state initially when telephoning a weather 
briefing facility for preflight weather information? 





   ANSWER: The intended route of flight and destination. 

   By telling the briefer your intended 
   route and destination, the briefer will be able to provide you 
   a more relevant briefing. 

When telephoning a weather briefing facility for preflight weather 
information, pilots should state 





   ANSWER: whether they intend to fly VFR only. 

   When telephoning for a weather 
   briefing, one should identify oneself as a pilot, the route, 
   destination, type of airplane, and whether one intends to fly 
   VFR or IFR to permit the weather briefer to give you the 
   most complete briefing. 

Figure 21 
(Refer to figure 21, area 3.) Determine the approximate latitude and 
longitude of Currituck County Airport. 





   ANSWER: 36°24'N - 76°01'W. 

   On Fig. 21, find the Currituck County 
   Airport, which is northeast of area 3. Note that the airport 
   symbol is just to the west of 76° longitude (find 76° just 
   north of Virginia Beach and at the bottom of the chart). 
   There are 60 min. between the 76° and 77° lines of longitude, 
   with each tick mark depicting 1 min. The airport is one tick to 
   the west of the 76° line, or 76°01'W.

   The latitude is below the 30-min. latitude line across the 
   center of the chart. See the numbered latitude lines at the top 
   (37°) and bottom (36°) of the chart. Since each tick mark 
   represents 1 min. of latitude, and the airport is approximately 
   six ticks south of the 36°30'N latitude, the airport is at 
   36°24'N latitude. Thus, Currituck County Airport is located 
   at approximately 36°24'N - 76°01'W. 

Figure 22 
(Refer to figure 22, area 2.) Which airport is located at 
approximately 47°39'30"N latitude and 100°53'00"W longitude? 





   ANSWER: Crooked Lake. 

   On Fig. 22, you are asked to locate an 
   airport at 47°39'30"N latitude and 100°53' longitude. Note 
   that the 101° longitude line runs down the middle of the 
   page. Accordingly, the airport you are seeking is 7 min. to 
   the east of that line.

   Each crossline is 1 min. on the latitude and longitude lines. 
   The 48° latitude line is approximately two-thirds of the way 
   up the chart. The 47°30N latitude line is about one-fourth of 
   the way up. One-third up from 47°30' to 48° latitude would be 
   47°39'. At this spot is Crooked Lake Airport. 

Figure 22 
(Refer to figure 22, area 3.) Which airport is located at 
approximately 47°21'N latitude and 101°01'W longitude? 





   ANSWER: Washburn. 

   See Fig. 22. Find the 48° line of 
   latitude (2/3 up the page). Start at the 47°30' line of latitude 
   (the line below the 48° line) and count down nine ticks to the 
   47°21' mark and draw a horizontal line on the chart. Next find 
   the 101° line of longitude and go left one tick and draw a 
   vertical line. The closest airport is Washburn. 

Figure 23 
(Refer to figure 23, area 3.) Determine the approximate latitude and 
longitude of Shoshone County Airport. 





   ANSWER: 47°33'N - 116°11'W. 

   See Fig. 23, just below 3. Shoshone 
   County Airport is just west of the 116° line of longitude (find 
   116° in the 8,000 MSL northwest of Shoshone). There are 60 
   min. between the 116° line and the 117° line. These are 
   depicted in 1-min. ticks. Shoshone is 11 ticks or 11 min. past 
   the 116° line.

   Note that the 48 line of latitude is labeled. Find 48 just 
   northeast of the 116°. The latitude and longitude lines are 
   presented each 30 min. Since lines of latitude are also 
   divided into 1-min. ticks the airport is three ticks above the 
   47°30' line or 47°33'. The correct latitude and longitude is 
   thus 47°33'N - 116°11'W. 

Figure 27 
(Refer to figure 27, area 2.) What is the approximate latitude and 
longitude of Cooperstown Airport? 





   ANSWER: 47°25'N - 98°06'W. 

   First locate the Cooperstown Airport 
   on Fig. 27. It is just above 2, middle right of chart. Note that 
   it is to the left (west) of the 98° line of longitude. The line of 
   longitude on the left side of the chart is 99°. Thus, the 
   longitude is a little bit more than 98°, but not near 99°.

   With respect to latitude, note that Cooperstown Airport is 
   just below a line of latitude that is not marked in terms of 
   degrees. However, the next line of latitude below is 47° (see 
   the left side of the chart, northwest of the Jamestown 
   Airport). As with longitude, there are two lines of latitude for 
   every degree of latitude; i.e., each line is 30 min. Thus, 
   latitude of the Cooperstown Airport is almost 47°30', but not 
   quite. Accordingly, Cooperstown 
   Airport's latitude is 47°25'N and longitude is 98°06'W. 

Figure 28 
(Refer to figure 28.) 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? 





   ANSWER: 1545Z. 

   First convert the departure time to 
   coordinated universal time (Z) by using the time conversion 
   table in Fig. 28. To convert from eastern daylight time (EDT), 
   add 4 hr. to get 1345Z (0945 + 4 hr). A 2-hr. flight would have 
   you arriving at your destination airport at 1545Z. 

Figure 28 
(Refer to figure 28.) 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? 





   ANSWER: 1030 MST. 

   Flying from the Central Standard Time 
   Zone to the Mountain Standard Time Zone results in a 1-hr. 
   gain due to time zone changes. A 2-hr. flight leaving at 0930 
   CST will arrive in the Mountain Standard Time Zone at 1130 
   CST, which is 1030 MST. 

Figure 28 
(Refer to figure 28.) 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? 





   ANSWER: 1645Z. 

   First convert the departure time to 
   coordinated universal time (Z) by using the time conversion 
   table in Fig. 28. To convert from CST to Z, you must add 6 
   hr., thus 0845 CST is 1445Z (0845 + 6 hr.). A 2-hr. flight 
   would make the estimated landing time at 1645Z (1445 + 2 
   hr.). 

Figure 28 
(Refer to figure 28.) 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 





   ANSWER: 1730 PST. 

   Departing the Mountain Standard 
   Time Zone at 1615 MST for a 2-hr. 15-min. flight would result 
   in arrival in the Pacific Standard Time Zone at 1830 MST. 
   Because there is a 1-hr. difference between Mountain 
   Standard Time and Pacific Standard Time, 1 hr. must be 
   subtracted from the 1830 MST arrival to determine the 1730 
   PST arrival. 


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