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N993928 Question 11 General Nav PT06

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  • N993928 Question 11 General Nav PT06

    Hi there

    Please could you assist me with the answer to the definition of density altitude. The explanation appears similar to one of the answer options but gives the actual answer as pressure altitude corrected for the prevailing temperature.

    Page 7.6 of the general nav hard copy notes also says it is the altitude in the ISA to which the actual density corresponds. There appears to be no direct explanation towards temperature. Also, doesn't prevailing temperature mean the average temperature? When you do the hand calc, one would use the deviation value from ISA.


  • #2
    I quote this issue as well. According to theory the answer you should go for is "The altitude in the standard atmosphere at which density is equal to density in the standard atmosphere" (see attached Photo 1). This is confirmed by the explanation of the answer but not by the answer quoted as correct by the corrector of the quiz (see attached Photo 2).
    I think this is just an error of the corrector.

    Thanks a lot,


    • #3
      Density altitude is the altitude in the standard atmosphere (ISA) at which the air density at the place of observation would be found. It can also be considered as the pressure altitude adjusted for non-standard temperature. An increase in temperature at any given location will cause an increase in density altitude.

      As an example, if the temperature at 3000 ft were Plus 9 the pressure altitude and density altitude would be the same (3000ft) as Plus 9 is the ISA temperature at 3000 ft.

      However if the temperature were Plus 25 the pressure altitude would still be 3000ft but the density altitude would be just under 5000 ft . This means that the air is behaving as if you were at 5000 ft because the temperature is much higher than that to be expected in an ISA situation and its density is much less.

      As a rule of thumb use 120 ft per degree difference from the ISA temperature. In my example the ISA temperature is 9 and the actual temperature is 25 - a difference of +16. So the correction is 16 x 120 = +1920 ft. The density altitude is therefore 3000 + 1920 = 4920ft.

      In this case the word "prevailing " with regard to temperature means the actual temperature. (There are several definitions of the word but in this case it is the definition "predominant" that is being used).The density altitude is therefore the pressure altitude corrected for the prevailing temperature.

      Hope this helps


      • #4
        Thanks for your explanation JJ. Unfortunately with high probability the wording of this question is not such to let you choose a unique answer: I say that because I agree with what you said, even if I agree with the explanation of the quiz as well (as reported in the attached .jpg). I hope not to find this question in the exam.