Sunrise/Sunset
The visual distance (in nm) to the horizon is given = 1.2 * SQRT(altitude)
So work out the distance to your horizon, and then the sunset time for that number of miles to the west for the sunset time at your current position and altitude.
So work out the distance to your horizon, and then the sunset time for that number of miles to the west for the sunset time at your current position and altitude.
Last edited by Checkboard; 19th Jul 2010 at 16:09.
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Sunrise/Sunset
Please go to this website, this should be a bit easier than the formula.
http://www.usno.navy.mil/USNO/astronomical-application
Regards,
Boeing Aloft
http://www.usno.navy.mil/USNO/astronomical-application
Regards,
Boeing Aloft
Correct link for the above is:
Astronomical Applications — Naval Oceanography Portal
... and no, it doesn't come close to answering the question.
Astronomical Applications — Naval Oceanography Portal
... and no, it doesn't come close to answering the question.
For a spherical earth, try:
acos[R/(R+z)] * 60/15 = minutes offset.
R is the spherical planet's radius, z is the aircraft altitude above the surface. Remember to use the same units. The sunrise time at altitude leads the time at the surface by the value of the offset and lags behind the surface time value during sunset.
There is obviously no account of refraction: normally at the surface 0.833° is used (3"20') but I have no idea how this should be varied with altitude. Secondly, the offset should be multiplied by the cosine of the subsolar point latitude. The worst case scenario only requires the offset to be reduced by about 8% - corresponding to the solstices.
Here's a plot of the above. In the lower plot the abscissa/ordinate axes are the same as the upper one - forgot to add the labels.
acos[R/(R+z)] * 60/15 = minutes offset.
R is the spherical planet's radius, z is the aircraft altitude above the surface. Remember to use the same units. The sunrise time at altitude leads the time at the surface by the value of the offset and lags behind the surface time value during sunset.
There is obviously no account of refraction: normally at the surface 0.833° is used (3"20') but I have no idea how this should be varied with altitude. Secondly, the offset should be multiplied by the cosine of the subsolar point latitude. The worst case scenario only requires the offset to be reduced by about 8% - corresponding to the solstices.
Here's a plot of the above. In the lower plot the abscissa/ordinate axes are the same as the upper one - forgot to add the labels.
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So work out the distance to your horizon, and then the sunset time for that number of miles to the west for the sunset time at your current position and altitude.
In mid-summer in the southern hemisphere there will be a latitude with almost 24 hours of daylight, which provides an extreme example. The sun will set just to the west of South, rising a few minutes later just to the east of South. On a high enough mountain there, the sun would not set at all - you would be looking 'over the top of' the pole.
Another famous example is the final flight of Amelia Earhart, toward Howland Island. Her navigator Fred Noonan timed sunrise and so established a 'line of position' running 157/337, rather than North and South.
Unfortunately, I don't know the answer to the OP's question.
