Refer to figure or CAP 697 Figure 4.5.3.1.
Given the following information, for a Commercial Air Transport operation, what is the Trip Fuel?
Pressure altitude: 32 000 ft
Cruise procedure: Long range
Ground distance: 300 NM
HWC: 25 kt
OAT: ISA
Mass at start: 40 000 kg
Refer to figure.
Plot starting mass 40 000 kg
Read TAS 415.
The question states ISA conditions, thus no correction required.
NGM to NAM = (NGM*TAS) / Ground Speed
Ground Distance = 300 nm
Headwind = 25 ks thus Groun dspeed = 415-25 = 390 kts
(300*415) / 390 = 319 nm NAM
Initial NAM = 1100 nm
1100 - 319 = 781 nm
Find 781 nm in table. Interpolation is required between 777 NAM and 798 NAM.
781 is within 777 (38 500 kg) and 798 (38 600 kg).
First find the difference between NAM at 38 500 kg (777) NAM at 38 600 kg (798) = 21NM => for an "interval" of 100 kg
If you divide 100 by 21 you will get the corresponding mass for each 1 NM.
100 / 21 = 4.8 kg
Which means that each NM corresponds to 4.8 kg.
Now we need to find the corresponding mass to 781:
781 - 777 = 4 NM
4 (NM) x 4.8 kg = 19.2 kg
38 500 kg + 19.2 kg = 38 519 kg
Start mass - new mass = fuel burn
40 000 kg - 38 519 kg = 1481 kg.
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