# Tag Info

21

The Sun rises and sets at a different point on the horizon every day. The change is small, so without careful observations, it may take several days or weeks to be fully aware of the change. Mathematically, the position of rising/setting can be found from the following formula: $$\cos(\theta) = -\frac{\sin(declination)}{\cos(latitude)}$$ where $\theta$ is ...

17

You're not silly1, it certainly swings back and forth (North and South) one full cycle every year. It's directly related to why days are longer in the summer and shorter in the winter. I am no expert, but some say that Stonehenge and other ancient "observatories" are supposedly set up to do exactly what it is you do, except much more carefully and ...

16

@Calc-You-Later's answer is correct in places where the path of the Sun is perpendicular to the horizon (at the equator during the equinoxes, somewhere between the tropics the rest of the time). There, the sunset lasts 2 minutes. However, at other latitudes, the sunset may last a bit longer. Let $\alpha$ be the angle the path of the Sun makes with the ...

11

This answer is a supplement to the existing answers. I looked around for nice graphs showing the sunrise azimuth over the year for various latitudes, but I couldn't find anything suitable. So I just wrote a couple of small Python scripts, using Sage / Matplotlib to do the plotting. Sunrise azimuths for various latitudes Sunrise times for various latitudes ...

10

We can start by converting between equatorial coordinates (right ascension $\alpha$ and declination $\delta$) to horizontal coordinates (azimuth $\text{Az}$ and elevation/altitude $a$). If you want to go past this, feel free to skip to the end. We draw a spherical triangle on the sky, with the three points being the object of interest at a given point $X$, ...

7

The Sun does indeed drift across the sky throughout the year, not only rising higher in the summer and lower in the winter, but also varying along an east-west axis. This can be shown by observing the Sun at the same time each day throughout the year, and seeing that it changes position. This shape is called an analemma, and is a result of the earth's axial ...

7

The Sun's apparent angular diameter is approximately 0.5 degrees - though atmospheric conditions can make this vary. We can calculate how fast the Earth rotates in degrees per minute like so: $24h\: /\: 360^{\circ} = 1h\: /\: 15^{\circ} \rightarrow 15^{\circ}\: /\: h$ $15^{\circ}\:/\:h = 15^{\circ}\:/\:60sec=0.25^{\circ}\:/\:min$ If the Earth is rotating ...

5

You have a good idea. You mentioned 4 effects already (clock accuracy, latitude, time of year, and clear horizon), but there is another effect that is larger than those: atmospheric refraction. Refraction causes a rising object to appear to be half a degree higher than reality. Refraction depends on atmospheric pressure and temperature, so it may be harder ...

5

For a quick and dirty approximation, let's assume a constant Sun angular diameter of 32' and no atmospheric refraction. Let $\varphi$ be the observer's geographic latitude, and $\delta$ be the Sun's declination (±23.4° at solstices, 0° at equinoxes), which you can get from the NOAA solar calculator. The time between the Sun's lower and upper ...

5

Short answers: It's not just angular velocity of Earth's rotation times elapsed time. You do need to factor in declination and latitude. Using your representative date and observer location: The apparent Sun, on the day of the summer solstice, crossed the Greenwich meridian at 12:02 GMT/UT in 2021, then proceeded westward directly above the Tropic of Cancer ...

5

The transformation from the equatorial coordinates to altitude (azimuth is not important) is given by $$\sin{a}=\cos{h}\cos{\delta}\cos{\phi}+\sin{\delta}\sin{\phi}$$ Let's say that $t=0\rm\, s$ at local solar noon. Then, the hour angle is given by $$h=\frac{360°\cdot t}{86400\rm\,s}$$ and we can reformulate the first equation using above expression to $$a=\... 4 Here's a partial answer: As mentioned in the comments, the transmission of the atmosphere depends quite a lot on local factors. But given a transmission T_0(\lambda) at zenith, the transmission at an angle \theta from zenith can be written$$ T(\lambda,\theta) = T_0(\lambda)\times X(\theta), $$where X is the air mass. The air mass can be approximated ... 3 Shalom! Strictly speaking, yes, the time of sunset does change, but the amount of change is imperceptible to humans over a lifetime. The change is due to the slowly varying obliquity of the Earth on its axis, the slowly varying eccentricity of Earth’s orbit around the Sun, and other such cyclic or secular changes. Other than that, you’ll notice a slight ... 3 I hope someone can come along and make this more precise, but I'm pretty sure the effect is greater the further you move away from the equator. So it would be more noticeable in Canberra (35° S) than in Darwin (12° S). Do you know the latitudes of the previous places you lived? 3 The mean angular radius of the Sun in radians is$$ \frac{R_\odot}{\mathrm{au}} = \frac{6.96 \times 10^5~\mathrm{km}}{1.496 \times 10^8~\mathrm{km}} = 4.65 \times 10^{-3}  and its mean angular diameter is twice that, 0.00930 radian or 0.533°. As the Earth's distance from the Sun annually varies by ±1.67%, the Sun's angular diameter varies ...

3

The easiest way to measure the sun would be to use a pin-hole camera. Use a piece of card with a pin hole. Hold it up so the sun shines through the pin hole and onto a piece of paper. (Don't look at the sun through the hole - eye damage) You will see a circle of light on the paper, this is the image of the sun. If the paper is 100cm from the card and the ...

3

The further north you go, the time between sunset and darkness becomes longer, no matter the season. The reason is due to the velocity at that latitude. If there is 10 min of twilight on the equator, then there is $10\sqrt{2}$ min at 45° latitude, 20 min at 60° latitude, ... Added: I used the website that @barrycarter listed above and discovered that there ...

2

Is that possible? Yes! You can see bright stars and planets through a telescope in the evening and in some cases during the day. As magnification increases the brightness of the sky or any extended object decreases with the square of the magnification. For example, a patch of sky will have a certain brightness per square degree, but at 100x magnification ...

2

Assuming you mean this in practical sense: Just use one of the numerous free websites that give this data: Eg. https://keisan.casio.com/exec/system/1224686065 Assuming you want to know about the math: Check out the sunrise equation

2

It is the first one, i.e. as soon as the heavenly body can be first seen. Here a definition from http://www.ga.gov.au/scientific-topics/astronomical/astronomical-definitions Sunrise is defined as the instant in the morning under ideal meteorological conditions, with standard refraction of the Sun's rays, when the upper edge of the sun's disk is coincident ...

2

The further the moon is from the sun the easier it is to see. There are two reasons. Firstly, when it is close to the sun it gets lost in the glare from the sun. The sky close to the sun is very bright, and there isn't enough contrast to see the moon against the bright sky. Secondly, as the moon gets further from the sun, more of the illuminated side of ...

1

Conventionally, 0 is North, 90 is East, 180 is South and 270 is West; in both the Northern and Southern Hemisphere. These are directions. This is nothing to do with GPS, which is about location, not direction.

1

I'm no SPICE expert but here are some potential solutions (unless of course you want to try Skyfield's Almanac methods!) Firsrt possibility, but all these may not work this answer links to this comment links to SPICE Tutorials, Updated December 11, 2019 links to the 69 slide presentation SPICE Geometry Finder (GF) Subsystem; Searching for times when ...

1

Will you know the date and location from where the image is taken? If you are sucessfully using the big dipper already you probably using this info. There is a "degeneracy" between date and time. The sky rotates once a siderial day (23hrs 56 min) and once a siderial year (which is 20min longer than tropical year: sun to same position). So the sky tonight at ...

1

When were these definitions formulated? Multiple religions have rather strict requirements based on sunrise / sunset, for example, when one must start or can stop fasting, or when exactly to sacrifice a lamb (or a human in some religions). The definition has very deep roots. Is this done purely because sunrise and sunset are the main examples of such ...

1

The thing that keeps the Earth from falling into the Sun is that it orbits the Sun, once a year. We usually don't say that the Sun orbits the Earth. Orbits have a direction or sense in space. What that direction is called is an issue of history and linguistics, The Earth rotates around its own axis, this rotation also has a sense, and the same caveats about ...

1

Answering my own question here. I spent quite a bit of time investigating this, this is what I came up with. As regards interpreting the rise and set values: If m1 or m2 go negative, I have found only that this means that the rise or set occurs on the previous Zulu day. And if the values go above 1, it/they occurred on the following Zulu day. I was surprised ...

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