How far apart are the sunset points at solstices by angle around the horizon at my latitude?
01/07/2014
This autumn as I noticed the sun starting to set earlier, and not so far round the sky when it did, I wondered just how far the setting direction moves over the year. I know the height of the sun at mid-day varies according to the angle of Earth's axis to the plane of Earth's orbit around the sun. That is currently 23.4 degrees, and is technically named the obliquity of the ecliptic to the celestial equator. The Earth's annual orbiting of the sun means the noon height of the sun varies by twice that, 46.8 degrees.
Then things get tougher to figure out. It is obvious the sunset direction variation must depend on latitude, because at the poles it varies from never setting to never rising over the course of the year. At the equator the noon height must move 23.4 degrees either side of the zenith, thus the sunset point would be moving back & forth by the same angle, that is 46.8 degrees between summer and winter setting points. This would put the Arctic and Antarctic circles at 23.4 degrees?
Well, I must be doing something right because looking that up I find the Arctic Circle lies at 66.5622 deg. North, i.e. 23.4378 deg. South of the North Pole!
Then things get tougher to figure out. It is obvious the sunset direction variation must depend on latitude, because at the poles it varies from never setting to never rising over the course of the year. At the equator the noon height must move 23.4 degrees either side of the zenith, thus the sunset point would be moving back & forth by the same angle, that is 46.8 degrees between summer and winter setting points. This would put the Arctic and Antarctic circles at 23.4 degrees?
Well, I must be doing something right because looking that up I find the Arctic Circle lies at 66.5622 deg. North, i.e. 23.4378 deg. South of the North Pole!
Apparently at spring and autumn equinoxes the sun should set exactly due West, so at the solstices it will set on paths parallel to the path at equinox, but 23.4 degrees either side, measuring this angle at right-angle to the path at equinox, not to either side along the horizon.
Now, the angle of the path of the sun to the plane of the horizon is dependent on the latitude. My latitude is 36.87 S approx.
You can see from the diagram that the ">23.4 deg" distance will get greater as you go South as the path of the sun flattens with respect to the horizon, until at the pole it is parallel.
This is where it has to go to Mathematics for a value for my latitude, and I see if I can make this highly capable computer actually do a few computations for me!
A check on the web gives me first, an approximation: -
North and South of the Equator the deviation is larger by a factor 1/cos(Latitude) The approximation breaks down at large latitudes - it does pretty much OK up to about 50 degrees North or South. (We should be OK with this at 36.87 deg. Sth.)
Then the full formula: -
The full calculation using spherical geometry (derivation) gives the angle from due west on the solstice as arcsin(sin(23.4)/cos(Lattitude)). ( I don't think we need to go there do we?)
Thanks to Windows' bundled Calculator, which I have just discovered has a Scientific mode (Apparently it has been there since Windows 3.0, 1990, 24 years and I have not noticed!!), Cos(36.87) = 0.7999989281485085293299423897698 and the reciprocal is 1.2500016747701993040919584188775, lets call it 1.25 near enough :-)
Now, the angle of the path of the sun to the plane of the horizon is dependent on the latitude. My latitude is 36.87 S approx.
You can see from the diagram that the ">23.4 deg" distance will get greater as you go South as the path of the sun flattens with respect to the horizon, until at the pole it is parallel.
This is where it has to go to Mathematics for a value for my latitude, and I see if I can make this highly capable computer actually do a few computations for me!
A check on the web gives me first, an approximation: -
North and South of the Equator the deviation is larger by a factor 1/cos(Latitude) The approximation breaks down at large latitudes - it does pretty much OK up to about 50 degrees North or South. (We should be OK with this at 36.87 deg. Sth.)
Then the full formula: -
The full calculation using spherical geometry (derivation) gives the angle from due west on the solstice as arcsin(sin(23.4)/cos(Lattitude)). ( I don't think we need to go there do we?)
Thanks to Windows' bundled Calculator, which I have just discovered has a Scientific mode (Apparently it has been there since Windows 3.0, 1990, 24 years and I have not noticed!!), Cos(36.87) = 0.7999989281485085293299423897698 and the reciprocal is 1.2500016747701993040919584188775, lets call it 1.25 near enough :-)
So the ANSWER is:
(23.4 + 1.25) x 2 = 49.3 degrees
Postscript:
I've been thinking about the ridiculous 31 decimal places in the calculator results.
To put it in perspective, 10^31 in metres is the lower bound of the (possibly infinite) radius of the universe.
I've been thinking about the ridiculous 31 decimal places in the calculator results.
To put it in perspective, 10^31 in metres is the lower bound of the (possibly infinite) radius of the universe.