cricri
Posts: 540
Joined: Sat Oct 02, 1999 12:10 am

Nice But Heavy Math Problem!

Tue Nov 18, 2003 11:48 pm

An imaginary aircraft is flying with constant speed at 360kts at an altitude of 30.000 feet (let's say 6nm). He spends a whole day airborne.

At sunrise, which occurs just in front of his nose on that June 21st day, the aircraft is at 45°N on greenwich.

During this day, the aircraft will keep the sun in front of him in his heading, turning right, so that at sunset, the sun keeps being on the longitudinal axe of the aircraft.

We will take the earth rotation into account

1 - tell what the flight of the aircraft will look like as if drawn on the ground

2 - give the flight duration, saying the flight starts at sunrise and stops at sunset.

Gentlemen, good luck, i failed and failed and failed....
 
cancidas
Posts: 3985
Joined: Thu Jul 03, 2003 7:34 am

RE: Nice But Heavy Math Problem!

Tue Nov 18, 2003 11:56 pm

i'm stumped. i'll be giving it to my physics professor. Big grin
"...cannot the kingdom of salvation take me home."
 
mjzair
Posts: 354
Joined: Tue Nov 30, 1999 12:10 pm

RE: Nice But Heavy Math Problem!

Wed Nov 19, 2003 12:32 am

This seems like it is a trick question..
If the plane is at 45N on June 21st, the sun cannot be aligned directly on the longditudinal axis of the aircraft...
On June 21st the sun reaches it point furthest North at 23.5N.
That is the only latitude that the sun can be directly in front of the aircraft's longditudnal axis.
Plus, at 45N, the sun will always be south of you. Assuming you are following the sun, you are travelling West, so making turns to the right as indicated will put you off to the North, and cannot keep the sun directly in front of you.

The only thing that I can add if this is inface a genuine question is:
sunrise at 45N on June 21st is at 04:13 Local Mean Time
sunset at 45N on June 21st is at 19:52 Local Mean Time
The sun travels at 15.04 degrees per hour.
The circumfurence of the earth at 45N is 11347NM

Anyone else?


[Edited 2003-11-18 16:39:06]
 
mjzair
Posts: 354
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RE: Nice But Heavy Math Problem!

Wed Nov 19, 2003 12:41 am

I take back my previous post, I think I understand a little better, I will work on it, and get back to you...
 
Guest

RE: Nice But Heavy Math Problem!

Wed Nov 19, 2003 12:44 am

Salut Cricri -
xxx
Good one, I had a similar problem to solve some 30+ years ago, when PanAm had the strange idea to give pilots navigator licence courses. Now you catch me on a layover - and I dont have my good old HO.216 reduction tables, and my slide rule... but basically you are going to end up with a beautiful clockwise spiral. This because initially you are going towards the sun for the "local" morning, with your 360 KTAS... then follow it during what will be a very long afternoon and sunset...
xxx
Without my tables, cannot give you the accurate flight time, but it will be around some 18+ hours for sunrise to sunset, and if wou started like you said at 45.00N and 0.00W, you will probably land in Labrador, Canada... If you had given us much higher TAS, could end at the North Pole, and the flight would never end...
xxx
Oh la la, vous me donnez des cheveux gris, les copains du A.net...  Big grin
Happy contrails -
(s) Skipper
 
Ikarus
Posts: 3391
Joined: Mon Jan 08, 2001 10:18 pm

RE: Nice But Heavy Math Problem!

Wed Nov 19, 2003 1:08 am

Of course the sun can rise directly ahead of the aircraft. It means the pilot isn't necessarily flying at 90 degrees or 270 degrees, but a bit less.

So the plane starts out flying East (well, slightly south of due East), always keeps the sun ahead and ends up flying West in the end.

At 360 knots, the distance travelled during a day (if the 15.5 hours or so quoted by Mjzair are correct) will be just above 10,000 kilometres. That may well put the plane in a place where the sun sets at a different time, especially as it will spend most of the day flying southwards (with some east- or west- element in it). If the plane travels a quarter of the earth's cirumference in distance, and a lot of that southwards, it could end up south of the Equator. On the other hand, because it starts out flying East, it is speeding against the sun, so the day will be shortened (and in the evening, when it chases the sun, the day will be lengthened. If there is no wind - and you have to assume there is none - then the two should be exactly equal in effect and the day ends up at the same length. That would put the plane back on the exact same longitude as it started out as, just much further South (unless it actually crossed the Equator, and flew north for a while)) So if this is correct (and it may not be) you know at least one thing about the flight path: Start and End will be along the same longitude.

Now what happens in between? My initial guess was a sort of semi-ellipse, but I suspect I'm wrong, because the thing will be distorted by the location of the sun.

Hmmm. Let me think. Let's start with three lines: a vertical one (for the start & end location) and the 23.5th parallel (the path of the sun) and the 45th parallel (for the plane). Add two lines: One, 20,000 km to the East, and one 20,000 km to the West. Where these intersect the 45th parallel, the sun will rise and set. The sun travels at a constant speed. Maybe the easiest way to solve this would be with a computer program? Numerically? I know I'm ignoring the altitude of the plane and the Earth's curvature, but I think for a first estimate, this should be a feasible approach. Program it so that there's two points, at two coordinates. Let one travel on a constant path at constant speed, and let the other always head towards the first, at constant speed but naturally changing direction at every time step. Set the time step to ten minutes, and the total time to 12 hours. (err, why is the damn day longer than 12 hours? I hate 3D!!!!).

Ah, let's just say I failed miserably...

(Now wait for people more clever than myself to come in and say how wrong I am!  Wink/being sarcastic )

Good Luck

Regards

Ikarus
 
mjzair
Posts: 354
Joined: Tue Nov 30, 1999 12:10 pm

RE: Nice But Heavy Math Problem!

Wed Nov 19, 2003 1:10 am

I respectfully disagree with the skipper. I dont think it is possible to end up in Labrador Canada..
Since the sun stays directly overhead 23.5 north and is aligned with the longditudinal axis of the aircraft.. Here is what I think will happen..
The plane is in the Southwest of France at sunrise.. and the plane is heading Southeast.. since sunrise is directly in front of him...
The aircraft will then follow a path southeast for the early morning hours, and will be in a constant but slow right turn as the sun is moving faster than the aircraft... the sun is moving 15.04 degrees per hour, and at 360 kts the aircraft is only moving 6 degrees per hour (of course, this number will fluctuate based on the latitude the aircraft is at.)
The aircraft will then make its final turn to a westerly heading on 23.5N somewhere over Africa at noon or just there after. Then, it will travel at 23.5N at 360kts for quite sometime as the sun races the plane accross the sky..
Well, that is all I can add to this.. the trig involved is too much for me..
Skipper, like I said I respectfully disagree with you, and I hate doing so.... I hope what I posted makes sense.
 
Guest

RE: Nice But Heavy Math Problem!

Wed Nov 19, 2003 1:48 am

Dear Mjzair -
A lot of your numbers are correct the 23.5, 15.04... 360 KTAS makes you move at 6 degrees per hour, but this in latitude, not longitude, except at the equator. Initially you would have a "sunrise heading" of about 120 degrees, this increasing to 180 degrees at local noon time, ending up heading 245 degrees as you chase the sun to its sunset... Would love to be at home and plot this on an old map... try do it per segments of 1 hour flight... looks like the end result is a perfect spiral to me... Fast enough, chasing the sun, we would end at the North Pole, and fly forever...
xxx
And... your are NOT disrespectful, my friend...  Smile
Happy contrails -
(s) Skipper
 
Guest

RE: Nice But Heavy Math Problem!

Wed Nov 19, 2003 1:50 am

Boy am I glad...
that there are plenty of folks here, much smarter than myself, who have problems with questions like this. Aren't you glad problems like this aren't on the FAA's ATP written test or used as oral questions? We'd all be looking for work.  Big thumbs up

Jetguy
 
CO2BGR
Posts: 506
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RE: Nice But Heavy Math Problem!

Wed Nov 19, 2003 4:31 am

Heres my input:

The flight path woul take it to northwest Egypt where it woud turn to slightly south of west and chase the sun. The plane would never make it to 23.5degN but quite close, and sunset would be over the middle of the atlantic at about 30degW. The path would look like a backward Cwith an extended flat bottom.

I think....
There are too many self indulgent weiners in this town with too much bloody money" Randal Raines- Gone in 60 Seconds
 
timz
Posts: 6100
Joined: Fri Sep 17, 1999 7:43 am

RE: Nice But Heavy Math Problem!

Wed Nov 19, 2003 8:29 am

Nobody's going to come up with an actual answer, but still...

The guys that are saying the plane will start out headed southeast need to think a little more. If you're on the ground at 45 North at sunrise on 21 June, where does the sun rise? Well north of east, right? Actual azimuth would be somewhere around 55-56 degrees. The plane starts out headed that way; I suppose it ends up headed WNW, but could be wrong about that.
 
mjzair
Posts: 354
Joined: Tue Nov 30, 1999 12:10 pm

RE: Nice But Heavy Math Problem!

Wed Nov 19, 2003 10:02 am

Timz...
I think maybe we are reading the problem a little differently.
It says that the plane is already in the air, and at sunrise the plane is directly facing the rising sun.
We established that the sun will rise at 23.5 N on June 21st, so from a position of 45N the sun must rise in the south east.
 
timz
Posts: 6100
Joined: Fri Sep 17, 1999 7:43 am

RE: Nice But Heavy Math Problem!

Wed Nov 19, 2003 10:23 am

Like I said, it rises 55-56 degrees east of north at latitude 45 North on 21 June.

It rises north of east everywhere in the world on that day-- just as it rises south of east everywhere in the world on 21 December. On 21 March (or thereabouts) it rises in the east, everywhere.
 
CO2BGR
Posts: 506
Joined: Thu Oct 02, 2003 11:30 am

RE: Nice But Heavy Math Problem!

Wed Nov 19, 2003 12:14 pm

Timz:
If you are at a higher latitude than the sun it will rise south of east, lower latitude it will rise north of east. Bangor, ME is at about 45degN and it never rises north of east here even in 21Jun.
There are too many self indulgent weiners in this town with too much bloody money" Randal Raines- Gone in 60 Seconds
 
vikkyvik
Posts: 11801
Joined: Thu Jul 31, 2003 1:58 pm

RE: Nice But Heavy Math Problem!

Wed Nov 19, 2003 12:18 pm

Timz,
How do you get the 55-56 deg E of N at lat 45N on June 21? I agree that it would rise north of east on that date...would the sun rise at 23.5 deg N of E at the equator on that date?
~Vik
I'm watching Jeopardy. The category is worst Madonna songs. "This one from 1987 is terrible".
 
CO2BGR
Posts: 506
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RE: Nice But Heavy Math Problem!

Thu Nov 20, 2003 1:28 am

The sun would rise due east if one was at 23.5degN. If one is north of that it would rise South of East, If one was south of 23.5degN the sun would rise North of east. Therefore from that plane the sun would be rising south of east at about 105deg.
There are too many self indulgent weiners in this town with too much bloody money" Randal Raines- Gone in 60 Seconds
 
timz
Posts: 6100
Joined: Fri Sep 17, 1999 7:43 am

RE: Nice But Heavy Math Problem!

Thu Nov 20, 2003 3:26 am

Okay, CO2BGR, try this. You agree that if you're at 70 degrees north latitude on 21 June the sun never sets, right? At "midnight" the sun will be pretty much north, right? And the same if you're at 67 degrees north: at "midnight" the sun will be just above the northern horizon.

So now move down to 66 degrees north latitude. If we ignore atmospheric refraction the sun will be just below the northern horizon at midnight-- then it will rise a few minutes later, just east of north.

Vikkyvik: the complication in calculating sun azimuth at sunrise/sunset is atmospheric refraction. If we were at sea level and the bottom of the sun were just touching the horizon, and for some reason the earth's atmosphere suddenly disappeared, the sun would disappear too. It would actually be just below the horizon, though previously the atmospheric refraction had made it appear to be above it.

But if we ignore that it's pretty simple. (If we also ignore the fact that the apparent horizon is slightly below a horizontal line from our eye.) Yes, at the equator on 21 June the center of the sun will rise 23.5 degrees north of east. If you're not at the equator-- well, I think there's a simpler formula than the following, but don't recall it offhand.

To specify the position in the heavens of the sun or any other celestial object we (in effect) give the position of the point on earth that's directly beneath it at that moment. If, at a given moment, the sun is directly above 23.5 degrees north, 60 degrees west, then the terminology is: the sun's declination is 23.5 degrees north and its Greenwich hour angle is 60 degrees. Declination is just like latitude, measured from 0 to 90 degrees north or south, but GHA goes from 0 to 360 degrees in a westerly direction.

Assume for the moment that you're on the Greenwich meridian at a latitude L, where north is positive. Then the sine of the sun's altitude (the angle between it and the horizon) is

(cosine GHA times cosine L times cosine declination)
plus (sine L times sine declination).

So if latitude is 45 degrees, GHA is 60 degrees and declination is 23.5 degrees, altitude comes out 37.3 degrees.

At sunset altitude is zero (we're assuming), so the above simplifies to

cosine GHA = -(tangent L times tangent declination)

so if we're still at latitude 45, the sun will set (not rise) when its GHA is 115.8 degrees.

Once we know its GHA, then the tangent of its azimuth is

sin GHA, divided by

[(cosine GHA times sine L) minus (cosine L times tangent declination)]

If you run through that you'll get azimuth -55.7 degrees-- i.e. 55.7 degrees west of north, at sunset.
 
timz
Posts: 6100
Joined: Fri Sep 17, 1999 7:43 am

RE: One More Thing

Thu Nov 20, 2003 3:43 am

"Bangor, ME is at about 45degN and it never rises north of east here even in 21Jun."

Take a look next year.
 
timz
Posts: 6100
Joined: Fri Sep 17, 1999 7:43 am

RE: Nice But Heavy Math Problem!

Thu Nov 20, 2003 7:26 am

I knew I forgot something simple: for any object, seen from any latitude on the Greenwich meridian,

cosine altitude times sine azimuth = -cosine declination times sine GHA

So at sunrise/sunset, once you've calculated GHA you can use that to get the sun's azimuth.

And what if you're not on the Greenwich meridian? Well, longitude can start anywhere-- the formulas will remain the same. If the sun's GHA is 60 degrees at a given moment, and you're at 100 degrees west longitude, then the sun's "GHA with respect to you" (otherwise known as LHA, local hour angle) is 320 degrees.
 
cancidas
Posts: 3985
Joined: Thu Jul 03, 2003 7:34 am

RE: Nice But Heavy Math Problem!

Thu Nov 20, 2003 12:07 pm

this one stumped my physics prof.
"...cannot the kingdom of salvation take me home."
 
rendezvous
Posts: 531
Joined: Sun May 20, 2001 9:14 pm

RE: Nice But Heavy Math Problem!

Thu Nov 20, 2003 2:25 pm

The plane lands about 100 miles after the engines run out of kero to drink.
 
SkyGuy11
Posts: 532
Joined: Fri Oct 26, 2001 7:09 am

RE: Nice But Heavy Math Problem!

Thu Nov 20, 2003 4:08 pm

I think you'd end up in a graveyard spiral after burning your eyes out trying to stare at the sun all day long.
.
 
Guest

RE: Nice But Heavy Math Problem!

Fri Nov 21, 2003 12:29 am

OK, Friends -
xxx
Means we end up with a spiral...
And somewhere like Labrador, or Newfoundland... ?
That is what I volunteered to say in reply 4 above...
xxx
By the way I live at latitude 30 South.
Went out on the roof at sunrise last morning with a compass...
The azimuth of the sun was approximately 105/110 degrees.
We are one month away from our summer equinox here.
xxx
I showed this one to a retired navigator...
He describes this as a "constructive brainstorm"...
But a great one, Cricri...
xxx
Happy contrails  Smile
(s) Skipper

 
timz
Posts: 6100
Joined: Fri Sep 17, 1999 7:43 am

RE: Math Problem

Sun Nov 23, 2003 8:07 am

Couple of boneheaded mistakes in the calculation posted above (or maybe deleted by now). Try again:

Ignoring refraction, and assuming "sunrise/sunset" is when the sun is 90 degrees from the zenith, and assuming a spherical earth on which the aircraft covers 6 degrees per hour, sun's declination constant 23.44 degrees and its hour angle increasing at exactly 15 degrees per hour, then:

2.06 hours from start, position 48.67 N 17.20 E (farthest north)
5.22 hours 38.56 N 37.45 E (farthest east)
5.76 hours 35.42 N 36.42 E (start chasing sun instead of approaching it)
9.47 hours 29.16 N 12.66 E (farthest south)
11.31 hours 29.76 N 0.00 W
18.75 hours 43.26 N 51.49 W (sunset)

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