CoolGuy
Topic Author
Posts: 366
Joined: Tue Jan 18, 2005 8:13 am

### Apparent Time While Flying

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.

zeke
Posts: 10931
Joined: Thu Dec 14, 2006 1:42 pm

### RE: Apparent Time While Flying

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.
Human rights lawyers are "ambulance chasers of the very worst kind.'" - Sky News

CoolGuy
Topic Author
Posts: 366
Joined: Tue Jan 18, 2005 8:13 am

### RE: Apparent Time While Flying

 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).

futurecaptain
Posts: 1918
Joined: Sat Sep 09, 2006 1:54 am

### RE: Apparent Time While Flying

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.
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KELPkid
Posts: 5247
Joined: Wed Nov 02, 2005 5:33 am

### RE: Apparent Time While Flying

 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   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 :-)

CoolGuy
Topic Author
Posts: 366
Joined: Tue Jan 18, 2005 8:13 am

### RE: Apparent Time While Flying

 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.

zeke
Posts: 10931
Joined: Thu Dec 14, 2006 1:42 pm

### RE: Apparent Time While Flying

 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)

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*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*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))
= 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:

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))
= 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
= 72.036611 W
Human rights lawyers are "ambulance chasers of the very worst kind.'" - Sky News

bond007
Posts: 4425
Joined: Mon Mar 14, 2005 2:07 am

### RE: Apparent Time While Flying

 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

Jimbo
I'd rather be on the ground wishing I was in the air, than in the air wishing I was on the ground!

CoolGuy
Topic Author
Posts: 366
Joined: Tue Jan 18, 2005 8:13 am

### RE: Apparent Time While Flying

 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.

mark5388916
Posts: 290
Joined: Thu Aug 09, 2007 7:35 am

### RE: Apparent Time While Flying

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
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