What effect does high density altitude have on aircraft 
performance? 





   ANSWER: It reduces climb performance. 

   High density altitude reduces all 
   aspects of an airplane's performance, including takeoff and 
   climb performance. 

Which combination of atmospheric conditions will reduce aircraft 
takeoff and climb performance? 





   ANSWER: High temperature, high relative humidity, and high 
   density altitude. 

   Takeoff and climb performance are 
   reduced by high density altitude. High density altitude is a 
   result of high temperatures and high relative humidity. 

Figure 8 
(Refer to figure 8.) Determine the density altitude for these 
conditions:  

Altimeter setting 
. . . . . . . . . . . . . . . . . . . . . . .   30.35
Runway temperature 
. . . . . . . . . . . . . . . . . . . . . . .   +25°F
Airport elevation 
. . . . . . . . . . . . . . . . . . . . . . . 3,894 ft MSL





   ANSWER: 2,000 feet MSL. 

   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 ft. 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 
   ft. Adding one-half the -92 ft. difference (-46 ft.) to -348 ft. 
   results in -394 ft. 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. 

Figure 8 
(Refer to figure 8.) 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? 





   ANSWER: 1,300-foot increase. 

   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 ft. Then move up the 3,000-ft. line to 
   50°F, where the density altitude is approximately 2,950 ft. 
   There is an approximate 1,300-ft. increase (2,950 - 1,650 ft.). 
   Note that 50°F is just about standard and pressure altitude is 
   very close to density altitude. 

Figure 8 
(Refer to figure 8.) Determine the pressure altitude at an airport that 
is 3,563 feet MSL with an altimeter setting of 29.96. 





   ANSWER: 3,527 feet MSL. 

   Note that the question asks only for 
   pressure altitude, not density altitude. Pressure altitude is 
   determined by adjusting the altimeter setting to 29.92" Hg, 
   i.e., adjusting for nonstandard pressure. This is the true 
   altitude plus or minus the pressure altitude conversion 
   factor (based on current altimeter setting). On the chart, an 
   altimeter setting of 30.0 requires you to subtract 73 ft. to 
   determine pressure altitude (note that at 29.92, nothing is 
   subtracted because that is pressure altitude). Since 29.96 is 
   half way between 29.92 and 30.0, you need only subtract 36 
   (-73/2) from 3,563 ft. to obtain a pressure altitude of 3,527 ft. 
   (3,563 - 36). Note that a higher-than-standard barometric 
   pressure means pressure altitude is lower than true altitude. 

Figure 8 
(Refer to figure 8.) 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? 





   ANSWER: 1,700-foot decrease. 

   The requirement is the effect of a 
   temperature decrease and a pressure altitude increase on 
   density alti tude. First, find the density altitude at 90°F and 
   1,250 ft. (approximately 3,600 ft.). Then find the density 
   altitude at 55°F and 1,750 ft. pressure altitude (approximately 
   1,900 ft.). Next, subtract the two numbers. 3,600 ft. minus 
   1,900 ft. equals a 1,700-ft. decrease in density altitude. 

Figure 8 
(Refer to figure 8.) Determine the pressure altitude at an airport that 
is 1,386 feet MSL with an altimeter setting of 29.97. 





   ANSWER: 1,341 feet MSL. 

   Pressure altitude is determined by 
   adjusting the altimeter setting to 29.92" Hg. This is the true 
   altitude plus or minus the pressure altitude conversion 
   factor (based on current altimeter setting). Since 29.97 is not 
   a number given on the conversion chart, you must 
   interpolate. Compute 5/8 of -73 (since 29.97 is 5/8 of the way 
   between 29.92 and 30.0), which is 45. Subtract 45 ft. from 
   1,386 ft. to obtain a pressure altitude of 1,341 ft. Note if the 
   altimeter setting is greater than standard (e.g., 29.97), the 
   pressure altitude (i.e., altimeter set to 29.92) will be less than 
   true altitude. 

Figure 8 
(Refer to figure 8.) 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? 





   ANSWER: 1,650-foot increase. 

   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 
   3,850 ft. Then move up the 5,000-ft. line to 50°F, where the 
   density altitude is approximately 5,500 ft. There is about a 
   1,650-ft. increase (5,500 - 3,850 ft.). As temperature increases, 
   so does density altitude; i.e., the atmosphere becomes 
   thinner (less dense). 

Figure 8 
(Refer to figure 8.) Determine the pressure altitude with an 
indicated altitude of 1,380 feet MSL with an altimeter setting of 
28.22 at standard temperature. 





   ANSWER: 2,991 feet MSL. 

   Pressure altitude is determined by 
   adjusting the altimeter setting to 29.92" Hg, i.e., adjusting for 
   nonstandard pressure. This is the indicated altitude of 1,380 
   ft. plus or minus the pressure altitude conversion factor 
   (based on the current altimeter setting).

   On the right side of Fig. 8 is a pressure altitude conversion 
   factor schedule. Add 1,533 ft. for an altimeter setting of 28.30 
   and 1,630 ft. for an altimeter setting of 28.20. Using 
   interpolation, you must subtract 20% of the difference 
   between 28.3 and 28.2 from 1,630 ft. (1,630 - 1,533 = 97 x .2 = 
   19). Then, 1,630 - 19 = 1,611 and add 1,611 ft. to 1,380 ft. to 
   get the pressure altitude of 2,991 ft. 

Figure 8 
(Refer to figure 8.) Determine the density altitude for these 
conditions:  

Altimeter setting 
. . . . . . . . . . . . . . . . . . . . . . . . 29.25
Runway temperature 
. . . . . . . . . . . . . . . . . . . . . . . . +81°F
Airport elevation 
. . . . . . . . . . . . . . . . . . . . . . . . 5,250 ft MSL





   ANSWER: 8,500 feet MSL. 

   With an altimeter setting of 29.25" Hg, 
   about 626 ft. (579 plus ½ 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 ft. Note 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 ft. The closest 
   answer choice is 8,500 ft. Note that, when temperature is 
   higher than standard, density altitude exceeds pressure 
   altitude. 

Figure 41 
(Refer to figure 41.) Determine the total distance required for 
takeoff to clear a 50-foot obstacle.  

OAT 
. . . . . . . . . . . . . . . . . . . . . . . . . Std 
Pressure altitude 
. . . . . . . . . . . . . . . . . . . . . . . . . 4,000 ft
Takeoff weight 
. . . . . . . . . . . . . . . . . . . . . . . . . 2,800 lb
Headwind component 
. . . . . . . . . . . . . . . . . . . . . . . . . Calm





   ANSWER: 1,750 feet. 

   The takeoff distance to clear a 50-ft. 
   obstacle is required. Begin on the left side of the graph at 
   standard temperature (as represented by the curved line 
   labeled "ISA"). From the intersection of the standard 
   temperature line and the 4,000-ft. pressure altitude, proceed 
   horizon tally to the right to the first reference line, and then 
   move parallel to the closest guideline to 2,800 lb. From there, 
   proceed horizontally to the right to the third reference line 
   (skip the second reference line because there is no wind), 
   and move parallel to the closest guideline all the way to the 
   far right. You are at 1,750 ft., which is the takeoff distance to 
   clear a 50-ft. obstacle. 

Figure 41 
(Refer to figure 41.) Determine the approximate ground roll distance 
required for takeoff.  

OAT 
. . . . . . . . . . . . . . . . . . . . . . . . . 100°F
Pressure altitude 
. . . . . . . . . . . . . . . . . . . . . . . . . 2,000 ft
Takeoff weight 
. . . . . . . . . . . . . . . . . . . . . . . . . 2,750 lb
Headwind component 
. . . . . . . . . . . . . . . . . . . . . . . . . Calm





   ANSWER: 1,150 feet. 

   Begin on the left section of Fig. 41 at 
   100°F (see outside air temperature at the bottom). Move up 
   vertically to the pressure altitude of 2,000 ft. Then proceed 
   horizontally to the first reference line. Since takeoff weight is 
   2,750, move parallel to the closest guideline to 2,750 lb. Then 
   proceed horizontally to the second reference line. Since the 
   wind is calm, proceed again horizontally to the right-hand 
   margin of the diagram (ignore the third reference line 
   because there is no obstacle, i.e., ground roll is desired), 
   which will be at 1,150 ft. 

Figure 41 
(Refer to figure 41.) Determine the total distance required for 
takeoff to clear a 50-foot obstacle.  

OAT 
. . . . . . . . . . . . . . . . . . . . . . . . . . Std
Pressure altitude 
. . . . . . . . . . . . . . . . . . . . . . . . . . Sea level
Takeoff weight 
. . . . . . . . . . . . . . . . . . . . . . . . . . 2,700 lb
Headwind component 
. . . . . . . . . . . . . . . . . . . . . . . . . . Calm





   ANSWER: 1,400 feet. 

   Begin in the left section of Fig. 41 by 
   finding the intersection of the sea level pressure altitude and 
   standard temperature (59°F) and proceed horizontally to the 
   right to the first reference line. Then proceed parallel to the 
   closest guideline, to 2,700 lb. From there, proceed 
   horizontally to the right to the third reference line. You skip 
   the second reference line because the wind is calm. Then 
   proceed upward parallel to the closest guideline to the far 
   right side. To clear the 50-ft. obstacle, you need a takeoff 
   distance of about 1,400 ft. 

Figure 41 
(Refer to figure 41.) Determine the approximate ground roll distance 
required for takeoff.  

OAT 
. . . . . . . . . . . . . . . . . . . . . . . . . . 90°F
Pressure altitude 
. . . . . . . . . . . . . . . . . . . . . . . . . . 2,000 ft
Takeoff weight 
. . . . . . . . . . . . . . . . . . . . . . . . . . 2,500 lb
Headwind component 
. . . . . . . . . . . . . . . . . . . . . . . . . . 20 kts





   ANSWER: 650 feet. 

   Begin with the intersection of the 
   2,000-ft. pressure altitude curve and 90°F in the left section 
   of Fig. 41. Move horizontally to the right to the first 
   reference line, and then parallel to the closest guideline to 
   2,500 lb. Then move horizontally to the right to the second 
   reference line, and then parallel to the closest guideline to 
   the right to 20 kt. Then move horizontally to the right, 
   directly to the right margin because there is no obstacle 
   clearance. You should end up at about 650 ft., which is the 
   required ground roll when there is no obstacle to clear. 

Figure 36 
(Refer to figure 36.) What fuel flow should a pilot expect at 11,000 
feet on a standard day with 65 percent maximum continuous 
power? 





   ANSWER: 11.2 gallons per hour. 

   Note that the entire chart applies to 
   65% maximum continuous power (regardless of the throttle), 
   so use the middle section of the chart which is labeled a 
   standard day.

   The fuel flow at 11,000 ft. on a standard day would be 1/2 of 
   the way between the fuel flow at 10,000 ft. (11.5 GPH) and 
   the fuel flow at 12,000 ft. (10.9 GPH). Thus, the fuel flow at 
   11,000 ft. would be 11.5 - 0.3, or 11.2 GPH. 


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