Welcome to the CPL Navigation question and answer forum. Please feel free to post your questions but more importantly also suggest answers for your forum colleagues. Bob himself or one of the other tutors will get to your question as soon as we can.
Could someone please assist with EOD and latest departure time. My understanding is to hit the Arc , and convert arrival location EOD back to UTC, then deduct flight time + 10mins which should be the answer in UTC.... for any location. If LST is required, then add either +10, 9.5 or 8 hours. Is this correct as I cannot find this example in the current Nav book. Thanks in advance
The working you describe would be correct. Find EOD at the destination and subtract the planned time interval + ten minutes for your latest departure time for a day VFR flight. Also, if the destination required holding, you would have to allow for that as well.
Thanks Bob. This is where i am unsure, from the calculated UTC, would you just add the hours to bring it back to standard time ie EST, CST or WST, or refer back to the ARC for an LST? What i'm trying to clarify is some of the states meridians may alter the time by maybe 60 minutes, and that the standard time has been set as an average time for that state. Thanks again.
The standard time for each time zone is not an 'average' for the zone, it is a particular meridian that that zone agrees to use as a standard. For example, in QLD everybody in QLD sets their clocks to 150°E for Eastern Standard Time. That's UTC + 10. . SA and NT uses 142.5° for Central Standard Time. That's UTC = 9.5. WA uses 120° for Western Standard Time. That's UTC plus 8 hours.
Once you've got the answer in UTC, you convert it back to the appropriate standard time by adding the standard time allowance of 8hours for WA, 9.5 hours for SA and NT, and 10 hours for VIC, NSW and QLD, (add an extra hour to these if you want summer time).
So to convert UTC back to a standard time in Australia, simply add either 8, 9.5 or 10 hours depending on the state you are interested in.