Hi all
This question is the updated question I dowloaded the pdf from website. It seems doesn't make sense about the wind direction in the explanation ( please find the attached files for question and result explanation). [b]Track is 220 wind 220/15 so it is HW so GSo should be 95 GSh is 125 however uncle Bob work out is different[/b]
I have the attached my work as well. The reason I have different answer because my wind was different compare to uncle Bob. Can anyone let me know if I did it correctly for the wind and whole thing?
Question 8, the wind seems isn't correct as track is 060 and wind is 240M/10 it means we will have TW so out GS should be TAS 160+10 =170. In Bob Tait explanation it was HW.
Thank you for reading
According to the latest issue of the Performance book, the wind in Question 7 is a direct headwind (track 220 and wind 220). So it seems that your book has them as a tailwind (wind from 020). If that is the case, I apologize. Also, I noticed that in your working you gave the time to continue to the destination on one engine, whereas the question asked for the time to return to the departure point.
Also, in Question 8, in the current issue is 060, that makes it a direct headwind, which is what the working shows.
Thanks for your feedback.
Uncle Bob
Dear uncle Bob
Thank you for your reply
However I wonder at Q8, if the track is 060 and wind 240/10 it is supposed to be a TW instead. How comes it shows as a HW in the explanation
Also you wrote " Also, I noticed that in your working you gave the time to continue to the destination on one engine, whereas the question asked for the time to return to the departure point. "
I can't get it quite clear what you meant. As the question ask for dis and time from the asymmetric ETP back to Alpha. So I did use
the provided TAS for asy is 110 knot. As I work I can have a same result as you did. Is it something that I missed out?
Thank you uncle Bob
