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Assume that you are on a cross-country flight at an altitude of 4,500 feet pressure altitude and your course is 275 degrees. The outside air temperature is 10 degrees Celsius, your indicated airspeed is 115 knots, the wind direction is 210 degrees, the wind velocity is 15 knots, and you passed your last checkpoint at 0942. If you have 218 nautical miles remaining, at what time will you arrive at your destination?

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First you need to find your True Air Speed (TAS). I used the formula:


TAS=IAS\sqrt{(\rho_0)/(\rho)}\\

where
\rho_0 is the density at mean sea level at 15°C which is 1.225 kg/m^3 and
\rho is the density of the air in which the airplane is flying. In this case with the given altitude and air temperature is about: 1.073 kg/m^3.

So TAS is equal to: 123 kt in this case.

To get the time of arrival you need to get your ground speed. For that you need to compensate for the wind speed and direction. With the drift angle you can get your ground speed.

In this case you get ground speed of 115 kt which means to travel 218 nautical miles you need 1 h and 53 min which means you will arrive at: 1135.

User Oliver Blue
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