Online Ground School
Airline Transport and IATRA

Part 5: Navigation General

5-028 (Answer). TAS is approximately 419 kts depending upon the flight computer/method used to solve the problem.

To solve this problem using a CR type circular flight computer, the first step is to determine the CAS, so 240 kts IAS + 2 kts = 242 kts CAS. Next, since the CR circular computer already compensates for compressibility error and temperature rise, we ignore the compressibility and temperature rise data given in the problem. Thus, we simply input the given CAS of 242 kts and place this opposite FL 350 in the CAS/PA window (don't use the smaller TAS/PA/TEMP window closer the centre of the computer!). Next we take the cursor and move it over the indicated temperature of -28º C at the point the arcing Ct 1.0 lines up with the inner spiral line. Reading the cursor at the point below where it crosses the TAS scale yields a TAS of 419 kts.

To solve this problem using an electronic computer, select the function "ACT TAS", then enter pressure altitude 35,000 feet and OAT -28º C. The computer yields an actual TAS of 418 kts. Using this method, there is no need to apply the compressibility data or temperature rise data provided in the problem.

To solve this problem using a standard E6B type flight computer, since the standard E6B's do not compensate for compressibility or for temperature rise, we need to use all of the data provided in the problem. Use the sequence "ICE-T" = Indicated, Calibrated, Equivalent, True i.e. first calculate CAS = 242 kts, then subtract the compressibility error of -10 kts for an equivalent EAS of 232 kts. Then determine the true outside air temperature = -28º C minus a further 20º C for compressibility heating of the temperature probe = -48º C. Finally, as per the standard TAS calculation method using an E6B, line up the pressure altitude 35,000 feet with OAT -48º C in the TAS window, find EAS 232 on inner scale of outer edge and read actual TAS of 420 kts on outer scale. [Ref: ATP GSC s. 5.02]

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