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take 50% headwind into account on graph questions

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  • take 50% headwind into account on graph questions

    Good evening

    I have a question regarding those 50% headwind and 150% tailwind rules.

    Do we need to take this rules into account when we solve graph questions on the cap 758 ?
    The thing is for example Figure 3.44 Page 56 on the cap 758 shows the sample arrow goes direct to 30kts headwind as statet in the given part. All question are so far right if you dont take the 50% headwind rule into account.

    So but now I`ve been told from a student whos studying with Phil Croucher they were teached that the cap 758 is wrong and at the exam we have to take the rule into account. (This apply only for helicopters not for fixed wing)

    Would be great if I get an answer for clarification. Actually I just want to know for what I should go at the exmas, its quite confusing me now.

    Thanks in advance


  • #2
    This is an old question, but is still valid.

    Regrettably the answer is not as clear as I would like, because we cannot be certain how EASA question writers have interpreted this. However, I can tell you the facts, and then if you lose larked over this you have the evidence for any challenge...

    1. The general principle is that Perf A performance data is factored - this applies to fixed-wing and rotary.
    2. CAP 758 para 5.2.5 Note states "The performance data given in the figures are determined from the actual wind speed without the application of any correction factor". This could mean the graphs reproduced assuming you will factor the wind, so they are based on actual data, or the data used in the examples is unfactored - i.e. you don't need to factor!
    3. Para Example 2a gives a headwind component of 20 kt. It then says in sub-para (f) 'drop vertically to intersect the headwind component gridline for 20 kt. This supports the second interpretation in (2) above - i.e. you don't need to factor.
    4. Using figure 3.44 it is possible to compare the data for different wind components to prove whether they are factored:
    Using the given example the wind components v gradient figures are 0 kt, 12%, 10 kt, 14.5%, 20 kt 17.2%, 30 kt, 21.7%, 40 kt, 29%.
    If not factored then the gradient at various groundspeed should be proportional because the rate of climb will be the same. Taking the 10 kt and 40 kt headwind unfactored gives GS of 60 kt and 30 kt respectively, based on Vy 70 kt. Starting with the 10 kt gradient we can show that 60 x 14.5 29 = 30 (more simply, if the GS is halved, then gradient should be doubled). This supports the view that the graphs themselves are not factored, and you should halve the headwind. If you want to test this further, the RoC which equates to all the headwind components should be close if the winds are unfactored. However, if they are factored (halved) then the higher headwinds should result in a lower RoC because the gradient decrement is increasing with wind. The figures all arrive at a RoC between 860 and 870 ft/min, which confirms that the graphs themselves are not factored.

    This leads to a real problem, because feedback over many years suggests that the questions are written applying the conclusions in 2 and 3 above, whereas I would argue that the data agrees with Phil (what a shock, Phil turns out to be spot on).

    What should you do? Comment on the exam if you get one of these questions, to the effect that there is no certainty whether the wind should be factored because the examples on the graphs themselves, and the example in para clearly states you don't factor, while the data says you do. If I were sitting it I guess I would factor, because this is factually correct, and not factoring would be the dangerous choice - but I would comment and appeal if they marked it incorrectly.