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Converted with kit
NS402S with ROTAX 912 S3 and
HOFFMANN HO-V352F/170FQ+10
FOR ISA+16 THE RESULT IS THEREFORE
(
T
=
x
roll
−
T
2
T
1
16
C
=
x
roll
−
22
C
0
C
T
=
x
D
15
−
T
2
T
1
16
=
x
D
15
−
22
C
0
C
These are for calm wind conditions. So, we need also wind correction as before:
12
kts
=
10
%
13
%
9
kts
Ground roll:
Takeoff distance:
CRUISE FLIGHT
• Total distance ................................................. 260NM
• Pressure altitude ............................................ 5500ft
• Temperature ................................................... 17°C (13°C above STD)
• Expected wind enroute ................................... 10kts Headwind
The cruising altitude should be selected based on a consideration of trip length, winds aloft,
and the airplane's performance. A typical cruising altitude and the expected wind enroute have
been given for this sample problem. However, the power setting selection for cruise must be
determined based on several considerations. These include the cruise performance
characteristics presented in figure 5-9 with corresponding consumptions and ranges, as well
as the charts for climb, fuel and distance in climb to cruise altitude.
Pressure altitude of 5500 ft at 20 °C (16°C over ISA) gives density of 7000 ft, for this
determination, see Fig. 5 - 1 . From cruise performance charts the following values can be
determined for density altitude if 7000ft.
Power setting ......................... 55% (2050 Rpm / 21.5 inHg)
Range available: .................... 412NM
Endurance: ............................ 4.9 hours. (without reserves)
TAS: ...................................... 96mph (84kts)
Consumption ......................... 17.4 ltr/h
Considering a headwind of 10kts at 5500ft the ground speed affects the range as follows:
Ground speed ........................ 84kts-10kts = 74kts
Range available ..................... 74kts*4.9 hours. = 363NM
Manual Cessna F150L
)
−
+
x
x
x
−
−
roll
2
roll
1
roll
(
)
−
+
310
m
251
m
251
(
)
−
+
x
x
x
−
−
D
15
2
D
15
1
D
(
)
−
+
611
m
495
m
495
294m –0.13*294 = 256m
579m –0.13*579 = 504m
Airplane Flight
−
1
=
m
294
m
−
15
1
=
m
579
m
Nr. 160
5-3
Page:
Revision 1
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