If you leave Atlanta at...

that's actually fairly complicated if you want precision. What time of year, what course (great circle, I assume), what altitude, definition of "dark", etc.

Practical answer is probably "until they raise the shades".
 
a rough estimate would suffice

and I hope the pilots wouldn't have shades up front.
 
Well:

Earth (mean radius, nm): 3442nm

Sunrise, Paris France: 5:57am local. That's 0357Z.

Civil twilight is 6 degrees west of that, and at latitude 49deg, a degree of longitude is roughly 42nm. So, twilight begins 6 * 42 = 252nm west of paris at 0357Z.

At 49N lat, the twilight boundary is moving west at 514kts.

Where's the jet? flightaware is currently showing 509kts. Great circle calculator for the atlanta - paris trip gives a distance of 3819nm.

position of jet (wow, this is rough): x = vt + x0, (509)t + 0 = 509t
position of twilight: x = vt + x0 = -514t + 3819 - 252 = -514t + 3567

509t = -514t + 3567
t= 3.48 = 3:38

By this hot mess of assumptions.... 3:38 hours into the trip, the sky gets brighter.

Meh.. too many beers.

Edit: You'd better believe I'm editing... I keep finding errors.
 
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WRONG. crap. I assumed the plane started at sunrise.

2100L = 0100Z. The plane gets to travel for 2:57 in darkness before the sunrise hits Paris. so..

x = vt + x0 = 509t + (3 * 509) = 509t + 1527.

509t + 1527 = -514t + 3567
t = 1.99 (already I'm suspicious)
2:57 + 2 = 4:57 call it right at 5 hours of the 7:41 minute flight in pre-civil twilight (dark).
 
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Aaaand that's the exact reason I posted the question on here. I was at work and watched Air France depart. A coworker wondered how long he would be in the dark seeing as he was flying towards the sunrise.

I quickly saw it wasn't going to be a quick calculation.

Thanks for the effort.
 
Can't have help ya, but I once departed Heathrow at sunset, and we were the over Canada before the sun went out of view.
 
Mistake Not... said:
2100L = 0100Z. The plane gets to travel for 2:57 in darkness before the sunrise hits Paris.
Click to expand...

Sunrise in Paris has nothing to do with it.

It's actually a pretty simple math problem if you have all the info. Kind of like the math problem you get in middle school where a train leaves one station doing x speed while another train leaves another station doing Y speed. How long will it take them to meet?

The airplane is traveling East at speed X and the sunrise is moving west at Y speed. How long will it take them to meet?

After having said that, I can't give an answer because I don't know the values for X and Y.
 
Sunrise in Paris is crucial, because that's the moment we know where the sunrise is, and how fast it's moving west.

But no, it's not at all simple. My very, very crude approximation is probably in the ballpark, but you have to realize that the sunrise travels different speeds at different latitudes, the terminator is not perfectly north/south (in fact, this time of year as we head into the summer solstice, it's nearing 23ish degrees to the pole).

Worse, the plane isn't flying along a line of constant latitude. It's a great circle route. What you want to find is the point on the route where the elevation of the sun matches whatever your definition of "dark" is. In this case, I chose start of civil twilight, 6 degrees below the horizon.

This probably has no closed form solution, and requires an iterative solution. I admit the calculations above are a mess, but they essentially attempt to do your "plane travels east, sun travels west" idea, with some reasonable assumptions about the time the plane starts, the time the sun starts, how fast each is travelling, and when we consider them to have met.

Actually, you could set this up fairly easily with an astronomy library.... Generate points along the great circle route and get the elevation of the sun at each step. It's been a few decades since numerical analysis, but a simple binary search looking for the desired solar elevation ought to converge really fast.


I was tempted to stay up and watch flight aware and the global day/night map and see how close I got, but... meh.
 
I was figuring about 4-5 hours. Thanks again for the effort.

The flight plan, time of year, speed of the sunrise, and rotation of the earth were too many variables for me at work.
 
If it was just recent that your coworker and you were discussing a particular flight at this time of year, then I think 4-5 hours is too soon to see the sunrise.

From ATL, starting at 2100 local, the flight would be about halfway over the Ocean (ave time is 8 hours from ATL-CDG).
http://flightaware.com/live/flight/AFR689/history/20140527/0045Z/KATL/LFPG

In 4 hours since departure, it will be about 1am EDT and 7am CET. The sun just rose an hour prior (in Paris), still a good 2200 miles ahead.

I suspect the sunrise will be somewhere around the 6 hour mark from departure, when sun has risen over Ireland and is working it's way West where the flight will be roughly positioned.
 
Mistake Not... said:
But no, it's not at all simple. My very, very crude approximation is probably in the ballpark,
Click to expand...

This whole thread reminds me of the old saw, "Measure with a micrometer, cut with an ax."

Given all the variables, just how close is the rough guess? Hours? Minutes? Seconds? I would think with this discussion, anything within about 15 minutes is close enough.
 
5 hours in . . .+/-

Will be in Astro twilight by the time you hit Labrador . . .
 
Did anyone take into account the altitude?
 
It's silly to even attempt this by hand. So many variables. The sunrise/sunset routine in a decent astronomical calc library would take altitude and other parameters (like atmospheric refraction due to temperature) into account.
 
I seem to recall from my differential equation class that we have problems launching a rocket and having it hit a precise spot on the moon. The math is known but the computations are too complex. IIRC we made in flight adjustments as the flight got closer to the moon and the variables became less relevant. This seems like a similar exercise.
 
Thursday.
 
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