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Apparent Time While Flying  
User currently offlineCoolGuy From United States of America, joined Jan 2005, 414 posts, RR: 0
Posted (6 years 9 months 2 weeks 1 day 10 hours ago) and read 3139 times:

Suppose an aircraft is flying JFK-LHR, takeoff at midnight, and heading roughly northeast of course. How does one determine the local time during the entire flight (assuming it is traveling over the shortest path, for example).

For example, at what flight time elapsed would sunrise occur, etc.

9 replies: All unread, jump to last
 
User currently offlineZeke From Hong Kong, joined Dec 2006, 9105 posts, RR: 75
Reply 1, posted (6 years 9 months 2 weeks 1 day 10 hours ago) and read 3130 times:

Not hard at all, the local time is determined by the present position, the time zone at the resent position, plus the UTC time gives local time.

Sunrise can be calculated fairly easily with software, or the old fashion way with sunrise/sunset tables.



We are addicted to our thoughts. We cannot change anything if we cannot change our thinking – Santosh Kalwar
User currently offlineCoolGuy From United States of America, joined Jan 2005, 414 posts, RR: 0
Reply 2, posted (6 years 9 months 2 weeks 1 day 10 hours ago) and read 3120 times:



Quoting Zeke (Reply 1):
Not hard at all, the local time is determined by the present position, the time zone at the resent position, plus the UTC time gives local time.

Sunrise can be calculated fairly easily with software, or the old fashion way with sunrise/sunset tables.

That would be fine; however is there a way to find out my position throughout the flight? For example, JFK-LHR, after 1000 miles elapsed (if it were following the great circle path).


User currently offlineFuturecaptain From , joined Dec 1969, posts, RR:
Reply 3, posted (6 years 9 months 2 weeks 1 day 10 hours ago) and read 3116 times:

Let me see if I can't tackle this.

New York is at roughly 40 degrees N latitude. London is at roughly 51 degrees N latitude. Lets use 45 degrees as a rough average for the flight....flown as the crow flies of course.

At 45 degrees N latitude each degree of latitude equals approximately 47.3 miles. Every 15 degrees of latitude is one time zone. So, every 710 miles you fly your current time jumps ahead one hour.
As the crow flies NYC-LON is about 3500 miles. Say you have a ground speed of 600 mph average for the sector.
You stated you took off at midnight,
After one hour of flight you are still in the same time zone and have covered 600 mi, so it's 1 am.
After 2 hours you have crossed one time zone and have covered 1200 mi so it is now 3 am.
After 3 hours in flight you crossed one more time zone, covered 1800 miles and it's now 5 am.
After 4 hours in flight you crossed another time zone, covered 2400 miles and it's now 7 am.
After 5 hours in flight you crossed another time zone, covered 3000 miles, and it's now 9 am.
After 6 hours in flight you are in the same time zone as before, covered 3600 miles, have landed, and it's 10 am.

Hope I got this down right.


User currently offlineKELPkid From United States of America, joined Nov 2005, 6388 posts, RR: 3
Reply 4, posted (6 years 9 months 2 weeks 1 day 10 hours ago) and read 3112 times:



Quoting CoolGuy (Reply 2):
That would be fine; however is there a way to find out my position throughout the flight? For example, JFK-LHR, after 1000 miles elapsed (if it were following the great circle path).

If you're a pilot, yes, you almost universally have a moving map in front of you these days...as a passenger, well, that depends on how nice the airline is to you  Smile Many of them have moving maps in front of you on the PTV system now, especially on Trans-Atlantic flights.

Personally, when I travel as a passenger, I just leave my watch to local time until I arrive at the destination and I'm given the local time (usually in the first announcement after landing and before the plane comes into the gate).

When flying, I have a clock (digital) on my kneeboard that I set to Zulu time (which you have to know for flight planning purposes, anyways...).



Celebrating the birth of KELPkidJR on August 5, 2009 :-)
User currently offlineCoolGuy From United States of America, joined Jan 2005, 414 posts, RR: 0
Reply 5, posted (6 years 9 months 2 weeks 1 day 9 hours ago) and read 3094 times:



Quoting Futurecaptain (Reply 3):
New York is at roughly 40 degrees N latitude. London is at roughly 51 degrees N latitude. Lets use 45 degrees as a rough average for the flight....flown as the crow flies of course.

That calculation is probably a good approximation (I am actually looking for JFK-LHR and JFK-TLV). I wonder if the flight path makes any difference since JFK-LHR goes farther north than 51 degrees. It's especially relevant on polar routes. I wonder what it's like to go EWR-SIN near the summer or winter solstices.


User currently offlineZeke From Hong Kong, joined Dec 2006, 9105 posts, RR: 75
Reply 6, posted (6 years 9 months 2 weeks 1 day 9 hours ago) and read 3079 times:



Quoting CoolGuy (Reply 2):
That would be fine; however is there a way to find out my position throughout the flight? For example, JFK-LHR, after 1000 miles elapsed (if it were following the great circle path).

JFK (40 degrees 38'23"N 73 degrees 46'44"W)
LHR (51 degrees 28'39"N 0 degrees 27'41"W)


In radians JFK is

lat1=(40.639751)*pi/180=0.709297462, lon1=(73.778926)*pi/180=1.287685177 and LHR is (lat2=0.898451866,lon2=1.287762)

The distance from JFK to LHR is

d = 2*asin(sqrt((sin((lat1-lat2)/2))^2+cos(lat1)*cos(lat2)*(sin((lon1-lon2)/2)^2))
= 2*asin(sqrt((sin(0.709297462-0.898451866)/2))^2+cos(0.709297462)*cos(0.898451866)*(sin((1.287685177-0.008052757)/2)^2))
= 0.869478789 radians
= 0.869478789*180*60/pi=2989nm
or

d = acos(sin(lat1)*sin(lat2)+cos(lat1)*cos(lat2)*cos(lon1-lon2))
= acos(sin(0.709297462)*sin(0.898451866)+cos(0.709297462)*cos(0.898451866)*cos(1.287685177-0.008052757))
= 0.869478789 radians
= 0.869478789*180*60/pi=2989nm

The initial true course out of JFK is:

sin(0.008052757-1.287685177) < 0 so

tc = acos((sin(lat2)-sin(lat1)*cos(d))/(sin(d)*cos(lat1)))
= acos((sin(0.898451866)-sin(0.709297462)*cos(0.869478789))/(sin(0.869478789)*cos(0.709297462))
= 0.896117182 radians
= 51.3437 degrees

An enroute waypoint 100nm from JFK on the 51.3437 degree radial (100nm along the GC to LHR) has lat and long given by:


100nm = 100*pi/(180*60)=0.029088821 radians

lat = asin(sin(lat1)*cos(d)+cos(lat1)*sin(d)*cos(tc))
= asin(sin(0.709297462)*cos(0.0290888)+cos(0.709297462)*sin(0.0290888)*cos(1.150035))
= 0.727241168 radians
= 41.667850 N

lon = mod(lon1-asin(sin(tc)*sin(d)/cos(lat))+pi,2*pi)-pi
= mod(1.287685177- asin(sin(0.896117182)*sin(0.869478789)/cos(0.727241168))+pi,2*pi)-pi
= 1.257276036 radians
= 72.036611 W



We are addicted to our thoughts. We cannot change anything if we cannot change our thinking – Santosh Kalwar
User currently offlineBond007 From United States of America, joined Mar 2005, 5417 posts, RR: 8
Reply 7, posted (6 years 9 months 2 weeks 1 day 8 hours ago) and read 3072 times:



Quoting Zeke (Reply 6):
100nm = 100*pi/(180*60)=0.029088821 radians

lat = asin(sin(lat1)*cos(d)+cos(lat1)*sin(d)*cos(tc))
= asin(sin(0.709297462)*cos(0.0290888)+cos(0.709297462)*sin(0.0290888)*cos(1.150035))
= 0.727241168 radians
= 41.667850 N

lon = mod(lon1-asin(sin(tc)*sin(d)/cos(lat))+pi,2*pi)-pi
= mod(1.287685177- asin(sin(0.896117182)*sin(0.869478789)/cos(0.727241168))+pi,2*pi)-pi
= 1.257276036 radians
= 72.036611 W

I'm assuming you've allowed for any time dilation differences  Wink


Jimbo



I'd rather be on the ground wishing I was in the air, than in the air wishing I was on the ground!
User currently offlineCoolGuy From United States of America, joined Jan 2005, 414 posts, RR: 0
Reply 8, posted (6 years 9 months 2 weeks 1 day 8 hours ago) and read 3067 times:



Quoting Zeke (Reply 6):
100nm = 100*pi/(180*60)=0.029088821 radians

lat = asin(sin(lat1)*cos(d)+cos(lat1)*sin(d)*cos(tc))
= asin(sin(0.709297462)*cos(0.0290888)+cos(0.709297462)*sin(0.0290888)*cos(1.150035))
= 0.727241168 radians
= 41.667850 N

lon = mod(lon1-asin(sin(tc)*sin(d)/cos(lat))+pi,2*pi)-pi
= mod(1.287685177- asin(sin(0.896117182)*sin(0.869478789)/cos(0.727241168))+pi,2*pi)-pi
= 1.257276036 radians
= 72.036611 W

That makes sense! Now all I have to do is come up with the apparent local time at that point and I'm all set!

I miss math class so much.  Sad


User currently offlineMark5388916 From United States of America, joined Aug 2007, 377 posts, RR: 0
Reply 9, posted (6 years 9 months 2 weeks 1 day ago) and read 2954 times:

Personally on the one international trip I took LAX-LHR-CDG I had my watch set to Zulu the whole way. I'm heading LAX-ZRH-FCO on the 26th and plan on using the same plan.

Mark



I Love ONT and SNA, the good So Cal Airports! URL Removed as required by mod
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