Astronomy/angle of the sun

Advertisement

Expert: - 7/27/2010


Question
Can you tell me what the angle of the sun is at the equinoxes and solstices, for Palm Coast, FL 29.6N 82.1W ?  Is there a chart somewhere?  Thank you  K

Answer

Zenith diagram
Hello,

You haven't specified altitude or azimuth but I am going to assume you mean the maximum altitude of the Sun corresponding to the equinoctial and solstice positions of the Sun.  I'm unaware of any chart to work the angle out (apart from specific software for sky mapping etc.) but it can be easily done using basic geometry- and I append a simple diagram for reference.

The diagram positions the observer at the center, and fixes the NCP (North Celestial Pole) and CE (Celestial equator = 0 degrees declination) in relation to him/her.

Several key points from the geometry:

The altitude a = the latitude of any observer (in this case 29.6 deg)

The zenith distance z = a

The total angular distance from NCP to CE must be 90 deg.

Given all these, the angles you desire can be worked out for any location.

Thus, for the equinoxes, we know the Sun's declination is 0 deg since it is on the Celestial equator (CE). Thus, for the equinoctial conditions the Sun will be at 'CE' in the diagram. This implies z = 29.6 deg, and hence the "angle of the Sun" (actually its maximum altitude on those days) must be:

90 deg -  29.6 deg =  61.4 deg

(This applies to both equinox dates)

For the winter solstice, we know the Sun is at declination (-23.5 deg). This puts it 23.5 degrees south (i.e. BELOW) 'CE'. Therefore, its altitude A must be reduced by 23.5 degrees from what it was on either equinox date, or:

61.4 deg-  23.5 deg  =  37.9 deg

For the summer solstice, the situation is reversed as the Sun is now at declination +23.5 deg. So the maximum altitude on that date will be 23.5 deg higher than it was on either equinox date, or:

61.4 deg + 23.5 deg =  84.9 degrees

As I said all of these can be worked from the geometry of the diagram, and it is general enough to do for any latitude - OR any object (so long as one is referencing meridian transit, and no other time). Some obvious angular relations from the diagram:

90 - z = A

a + A = 90

In addition, any object passing through your zenith at meridian transit (whenever that may be- for a star, obviously at night) will have a declination = to your latitude.

Hope this sheds some light!

Philip Stahl

Expertise

I have forty years of experience in Astronomy, specifically solar and space physics. My specialties include the physics of solar flares, sunspots, including their effects on Earth and statistics as applied to astronomical investigations.

Experience

Astronomy: more than forty years experience starting with construction of my own simple telescopes. Worked at university observatory in college, doing astrographic measurements. M.Phil. degree in Physics/Solar Physics and more than ten years as researcher.

Organizations
American Astronomical Society (Solar Physics and Dynamical Astronomy divisions), American Mathematical Society, American Geophysical Union

Publications
Solar Physics (journal), The Journal of the Royal Astronomical Society of Canada, The Proceedings of the Meudon Solar Flare Workshop (1986), The Proceedings of the Caribbean Physics Conference (1985). Books: 'Selected Analyses in Solar Flare Plasma Dynamics', 'Physics Notes for Advanced Level'.

Education/Credentials
B.A. Astronomy, M. Phil. Physics

Awards and Honors
American Astronomical Society Studentship Award (1984), Barbados Government Award for Solar Research

©2012 About.com, a part of The New York Times Company. All rights reserved.