I came across this question from the book "A Problem book in Astronomy and Astrophysics by Aniket Sule":-

At Brazil’s National Observatory, located at the city of Rio de Janeiro $(22◦54′ S, 43◦12′ W)$, there is a sundial above the door of the dome of the $32$ cm telescope, facing to the north. The dial lies on the plane East-Zenith-West and the rod is parallel to the Earth’s axis. For which declinations of the sun and during what period of the year (months and seasons) the clock (i) does not work during, at least, some fraction of the day? and (ii) does not work at all during the day? (I12 - T01 - B)

I thought that for (i) the upper culmination should be greater than $90◦$ (i.e., it must culminate north of zenith) since the Sun must pass through the meridian atleast once so that the shadow become zero on the sundial, plugging the values (according to the southern hemisphere convection):-
$H_u = 90◦ - \phi + \delta \geq 90◦$
$\delta \geq \phi$

comparing my answers, it seems I am going fundamentally wrong somehwere, and this does explain why I can't get how to solve (ii)

Please give hints to solve this question, I would be grateful, also please edit/suggest for edits if you find any

  • $\begingroup$ Consider what happens when the Sun's declination is further south than the observatory's latitude. $\endgroup$
    – PM 2Ring
    Commented May 2 at 16:14
  • $\begingroup$ @PM2Ring I added "lower culmination" just incase the second bit needed it to solve. I am removing it though $\endgroup$ Commented May 3 at 1:25
  • 1
    $\begingroup$ Yes, as John says, the azimuth of the Sun has to be within 90° of north for the Sun to shine on the vertical north-facing dial plate. So between the September and March equinoxes there are times during the day when the Sun is too far south. $\endgroup$
    – PM 2Ring
    Commented May 5 at 7:02
  • 1
    $\begingroup$ I have a program here that shows the azimuth & altitude of the Sun in 3D. The Rio observatory has site code 880, and it's in the UT-3 timezone. $\endgroup$
    – PM 2Ring
    Commented May 5 at 7:36
  • 1
    $\begingroup$ ssd.jpl.nasa.gov/api/… $\endgroup$
    – PM 2Ring
    Commented May 6 at 8:55

1 Answer 1


The question asks when is the azimuth of the Sun between 90 degrees (due east) and 270 degrees (due west). Since the face of the sundial faces due north, it will not show a time when the Sun is south of the East-West line.

Culmination is an altitude, the angle above or below the horizon. An object can never be at a higher altitude than 90 degrees. When an object passes through the zenith (altitude = 90), the azimuth changes by 180 degrees and the altitude starts to decrease.

  • $\begingroup$ I get it now, I am doing the necessary changes $\endgroup$ Commented May 5 at 1:40
  • $\begingroup$ can we solve it this way- if the sun is north of the equator then sun will always set north of west or north of prime vertical so sundial must work. but when sun crosses the equator it will set south of west or south of prime vertical so sundial wont work for parts of day so that is when declination of sun is south or 0 to -23.27 (because thats the lowest it can go in winter solstice) so answer is from sept 23 to march 21 $\endgroup$ Commented May 5 at 1:51
  • $\begingroup$ @DarthPseudonym No, it's not just because the sundial is attached to a building. If the azimuth of the Sun is more than 90° away from north it will shine on the south side of a vertical north-facing plane. If it's directly overhead, neither side gets direct sunlight. $\endgroup$
    – PM 2Ring
    Commented May 5 at 7:09
  • $\begingroup$ You should calculate the azimuth of the Sun throughout the day and year. Where is rises and sets may not be the only information required. Someone posted a question recently about websites to do that. (What that you?) Since I live in the Northern hemisphere, I have never investigated what the path of the Sun looks like from the Southern hemisphere. $\endgroup$
    – JohnHoltz
    Commented May 6 at 12:10
  • $\begingroup$ @stretch The face of the dial is on a vertical wall and faces north. Once the Sun is farther than 90 degrees in azimuth from north, the sun would be on the back of the dial. (Since there is a building in the way, there is no Sun on the back of the dial, even if it it had two faces (that would be neat!) $\endgroup$
    – JohnHoltz
    Commented May 7 at 2:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .