<p>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.</p>
<p>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.</p>

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2024-05-05T01:40:38.510Z
I get it now, I am doing the necessary changes
Celestia Novalis
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2024-05-05T01:51:37.270Z
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
Celestia Novalis
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2024-05-05T07:09:41.047Z
@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.
PM 2Ring
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2024-05-06T12:10:38.723Z
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.
JohnHoltz
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2024-05-07T02:39:28.223Z
@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!)
JohnHoltz
CC BY-SA 4.0
2024-05-03T16:46:06.250Z
2024-05-03T16:46:06.250Z
JohnHoltz
<p>I came across this question from the book &quot;<em>A Problem book in Astronomy and Astrophysics</em> by Aniket Sule&quot;:</p>
<blockquote>
<p>At Brazil’s National Observatory, located at the city of Rio de Janeiro
<span class="math-container">$(22^◦54′ S, 43^◦12′ W)$</span>, there is a sundial above the door of the dome
of the <span class="math-container">$32$</span> 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)</p>
</blockquote>
<p>I thought that for (i) the upper culmination should be greater than <span class="math-container">$90^◦$</span> (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):-<br>
<span class="math-container">$H_u = 90^◦ - \phi + \delta \geq 90^◦$</span><br>
<span class="math-container">$\delta \geq \phi$</span> <br></p>
<p>comparing my answers, it seems I am going fundamentally wrong somewhere, and this does explain why I can't get how to solve (ii).</p>

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2024-05-02T16:14:47.497Z
Consider what happens when the Sun's declination is further south than the observatory's latitude.
PM 2Ring
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2024-05-03T01:25:49.180Z
@PM2Ring I added "lower culmination" just incase the second bit needed it to solve. I am removing it though
Celestia Novalis
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2024-05-03T01:34:09.303Z
@PM2Ring thank you for the hint, but I see no problems in that, the farthest Sun can go is -23.27 and it would still pass through the prime vertical, causing no shadow...?
Celestia Novalis
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2024-05-03T02:27:57.847Z
When the Sun is on the south side of the building the north side of the building is in shadow.
PM 2Ring
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2024-05-05T01:51:57.733Z
@PM2Ring 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
Celestia Novalis
CC BY-SA 4.0
2024-05-05T07:02:39.627Z
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.
PM 2Ring
CC BY-SA 4.0
2024-05-05T07:36:12.083Z
I have a program [here](https://astronomy.stackexchange.com/a/55736/16685) 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.
PM 2Ring
CC BY-SA 4.0
2024-05-06T08:55:54.330Z
https://ssd.jpl.nasa.gov/api/horizons.api?format=text&COMMAND=10&CENTER=880%40399&QUANTITIES=%274%2C34%27&OBJ_DATA=NO&START_TIME=%272023-Dec-22+5%3A00+UT-3%27&STOP_TIME=%272023-Dec-22+19%3A00%27&STEP_SIZE=15m
PM 2Ring
CC BY-SA 4.0
2024-05-08T11:59:09.530Z
@JohnHoltz Oops. I didn't read the question carefully enough. Deleted comment.
stretch
<positional-astronomy>
Problem on setting of sun
Celestia Novalis
# Problem on setting of sun

(Asked by: username_0 on 2024-05-02)

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 somewhere, and this does explain why I can't get how to solve (ii).

**Tags:** positional-astronomy

> **Comments:**
> **username_1**: Consider what happens when the Sun's declination is further south than the observatory's latitude.  
> **username_0**: @username_1 I added "lower culmination" just incase the second bit needed it to solve. I am removing it though  
> **username_0**: @username_1 thank you for the hint, but I see no problems in that, the farthest Sun can go is -23.27 and it would still pass through the prime vertical, causing no shadow...?  
> **username_1**: When the Sun is on the south side of the building the north side of the building is in shadow.  
> **username_0**: @username_1 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  
> **username_1**: 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.  
> **username_1**: I have a program [here](https://astronomy.stackexchange.com/a/55736/16685) 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.  
> **username_1**: https://ssd.jpl.nasa.gov/api/horizons.api?format=text&COMMAND=10&CENTER=880%40399&QUANTITIES=%274%2C34%27&OBJ_DATA=NO&START_TIME=%272023-Dec-22+5%3A00+UT-3%27&STOP_TIME=%272023-Dec-22+19%3A00%27&STEP_SIZE=15m  
> **username_2**: @username_3 Oops. I didn't read the question carefully enough. Deleted comment.  

---

## 1 answers

### Answer by username_3 on 2024-05-03. Score: 4

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.

> **Comments:**
> **username_0**: I get it now, I am doing the necessary changes  
> **username_0**: 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  
> **username_1**: @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.  
> **username_3**: 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.  
> **username_3**: @username_2 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!)  

---

