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Explanation - QB ID: 320386

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  • Explanation - QB ID: 320386

    The Explanation for QB ID: 320386 is confusing (at least for me). It starts out by converting the 100 KIAS into TAS and GS then calculates the "Still Air Distance" as 14948ft. (No issues up to that point!). However, it goes not continue to calculate the "Ground Distance" using the formula in CAP 698 para 3.2.3 (e) which outlines "Ground Distance = Still Air Dist x GS/TAS" (14948 x 113/108 = 15640ft - the correct answer shown in the QB).

    Instead the Explanation states that "The CQB answer is clearly wrong" and then proceeds to "prove this by taking the ROC from the graph 1080 fpm" to work out the "Ground Dist" - a rather complex method that includes conversions from ft>nm>ft. Was this calculation necessary to arrive at the "close enough" answer or is it just another (complex) method? This does not appear to be consistent with the Explanation given in QB ID: 320385 or the example in CAP 698 where the same formula ("Ground Distance = Still Air Dist x GS/TAS") is used . Am I missing something here????

    Appreciate any assistance in clarifying this.


  • #2
    No idea what's happened here but have corrected the explanation so that it uses the formula same as 320385.


    • #3
      I can't understand why it's necessary to go to the the lengths of working out the climb gradient in this question.
      Why can we not take the height to be gained, 1450 ft divide that by the ROC, which by my calculation from the graph comes to 1100 ft /min
      1450/1100 = 1.31818 mins and then multiply that by the ground speed in ft/min
      GS is TAS 108 min + TW 5kts = 113 kts
      (113/60) x 6080 x 1.31818 = 15094 ft
      My next question is why is this answer so different to that given as the correct answer.
      The less interrogation of the graphs the better in my mind, as there is more to go wrong, but why are the answers so different?


      • #4
        I too, like the above poster, would be very grateful (and interested!) to know why one cannot obtain the answer by use of ROC. There's no need to use the gradient. I get:

        Time = (height difference) / ROC = 1450 / 1080 = 1.34259 mins

        Therefore the ground distance covered is:

        Distance = GS * Time = (108 + 5) * 1.34259 / 60 = 2.528549 nm = 15373 ft

        This approach actually seems safer, because if you mess up the final line which gives you the gradient (which is fiddly to draw as it's between two curves), your answer will be in error. This is also correctly pointed out by the above poster. This answer is close enough to get it correct on the bank, but it is concerning that the two answers are off by a not-inconsiderable margin.
        Last edited by JWB; 17-09-2019, 12:59.


        • #5
          I'd be very grateful for an answer to this as I have performance next week!


          • #6
            I would stick with the method outlined in Section 2 (Page 8) of CAP 698. The gradient, and therefore the Still Air distance, is dependent on the TAS. I agree that just using the ROC will get you close enough in this case but I recommend you use the method outlined in the CAP for greatest accuracy.