# Tag Info

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A system projected into the sky Pretend all the stars are painted on the inside of a large ball, and you are at the center of the ball. The imagined ball is called the Celestial Sphere. If the Earth wasn't blocking the lower half of your view, and the Sun wasn't making the blue sky so bright, you would be able to look at the stars in any direction. First, ...

14

The equatorial coordinate system is very similar to the system used on a globe or on maps. To specify a point on a sphere or a globe you need only two numbers. These are the longitudes and latitudes. On a globe of the earth you have longitudes from -180º to +180º and latitudes from -90º to +90º. The south pole lies at -90º latitude, the north pole at +90º ...

14

For locating objects in the sky, the horizontal and equatorial coordinate systems are commonly used. These systems describe the position of some object in the sky very well, but do not explain the position of the object in space (if you know the distance you know "where" the object is, but this is relative to the equatorial/horizontal plane and is tricky to ...

12

The easiest way to think of equatorial coordinates is by extending lines of latitude/longitude (the geographic coordinate system we usually learn in school) out into space. The earth's equator becomes the celestial equator, and the north and south poles become the celestial north and south poles, respectively. By doing this you have a fairly simple way to ...

12

The altazimuth system is directly linked to your position on the Earth. It is the “left-right/up-down” coordinates system; the azimuth being the “left-right” (with 0° for North, 90° East, etc.), and the altitude is the “up-down” (with 0° being the horizon, 90° being overhead, also called zenith). Of course, a star’s (or planet’s, comet’s, galaxy’s, etc.) ...

10

A map of all galaxies gets kind of unwieldy, like a map of all stars in the milky way or a map of every house in the country, or every grain of sand on a beach . . . you get the idea. Start here - Local Group Source Then Virgo Supercluster Source Then local superclusters Source And an article, even if it's a summary it's very much worth reading, with ...

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LST = 100.46 + 0.985647 * d + long + 15*UT They don't explain what the two constants are (100.46 and 0.985647), could anyone explain what those constants are and how they were calculated in the first place please? There are three constants there, 100.46, 0.985647, and 15. The value of 100.46 degrees is the value needed to make the expression yield ...

9

How big is one arcsecond at various distances? An arcsecond is a small angle, 1/3600 of a degree or about 5 millionths of a radian ($4.85\times10^{-6}$). To estimate the size of something that appears 1 arcsecond across you can use the small angle approximation to trigonometry: Multiply the distance to the object by $4.85\times10^{-6}$ Examples: One ...

9

Sort of... There is a system called the International Celestial Reference System (ICRS) which has center at the Solar System Barycenter (normally inside the Sun but not the same as the Sun's center) and which has the x and y axis in the plane of the Earth's equator and the z-axis ("North" if you will) pointing towards the Celestial Pole. The x-axis points ...

8

As Wikipedia explains in the article on the Zodiac, the Zodiac was originally developed with the assumption of fixed equinoxes. It is assumed this is because the Babylonians who developed the Zodiac were unaware of the precession of the equinoxes. At that time, about 2500 years ago, the vernal equinox did in fact lie in Aries. For this reason, in astrology, ...

8

The zero of Galactic latitude (i.e. the Galactic plane) was defined by a working group of the International Astronomical Union. The observations used to do this were 21 cm radio observations of atomic hydrogen, which are unaffected by dust absorption near the plane. The details of the observations are found in Gum, Kerr & Westerhout (1960). This paper ...

7

Read the fine print under the Horizons output when you select an OBSERVER table. It says (emphasis mine) R.A._(ICRF/J2000.0)_DEC = J2000.0 astrometric right ascension and declination of target center. Adjusted for light-time. Units: HMS (HH MM SS.ff) and DMS (DD MM SS.f) You could do the same, but it will take some extra work. The Mars you see at ...

7

There are multiple formulae for how to calculate this angle. The simplest is to construct the unit vectors: \begin{align} \hat{n}_i & = \left[\begin{array}{c} \cos \delta_i \cos \alpha_i \\ \cos \delta_i \sin \alpha_i \\ \sin \delta_i \end{array}\right] \end{align} where $\delta$ is the declination and $\alpha$ is the right ascension, then ...

7

To do so, I'm following what's described in this document and implementing C++ code. That is a very old document you are using. It's using the 1982 IAU precession model. There have been multiple updates since then. The latest set of changes are working toward eliminating the concept of Greenwich Sidereal Time. (In software terms, GMST and GAST are ...

7

Assuming you mean the angle between the meridian line through A and the great circle that goes through points A and B, then it goes something like this. Define vectors from the origin to A and B assuming they lie on a unit sphere, such that $x_A= \cos \delta_A \cos \phi_A$, $y_A = \cos \delta_A \sin \phi_A$ and $z_A= \sin \delta_A$, and similar for B. Here ...

7

Near the pinned location, at 24.0893°S 69.9306°W, Bing Maps shows a building like the one in the University of Antofagasta photos, with signs of recent construction. It's probably a matter of time until Google updates their satellite imagery in that region. The Minor Planet Center maintains a list of observatory codes with positional information. ...

6

Well, your problem is that you have a sphere. To compensate for the polar declination skew, you just calculate $$\sin^{-1}(\mathtt{rand})$$ Where $\mathtt{rand}$ is a number in an evenly divided -1 to 1 range. Alternatively, you can use another trigonometric function based on what you have available, but you get the idea. RA can obviously be divided evenly.

6

This code reads coordinates as equatorial (ra, dec) and transforms them to galactic (l, b): eq = SkyCoord(xarr[:], yarr[:], unit=u.deg) gal = eq.galactic The contents of 'galacticwperiod.csv' are already in galactic coordinates and should not be transformed. Something like this may give better results: gal = SkyCoord(xarr[:], yarr[:], frame='galactic', ...

6

Apparently, Stellarium's "J2000.0" reports the coordinates of stars in J2000 frame, but at the epoch of the date you specify, instead of reporting the coordinates of the star at epoch 2000.0, which can be misleading to say the least. So the difference in J2000 coordinates that you noticed corresponds to the proper motion of the star. And the star you picked, ...

6

Yes, the orientation of the Sun will be different from the Earth's northern and southern hemisphere, just like your example of the Moon. I would not say that a sunspot in the "northern" hemisphere would appear to be in the "southern" hemisphere just because of the change in orientation. North is fixed on the Sun and Moon, just like they are on the Earth. (...

5

For computer software, the easiest way to take a sphere (and/or hemisphere) and flatten it into a flat shape (usually a rectangle) is the equi-rectangular projection (also known as the plate carrée), because it has the simplest formula relating pixels and coordinates: $$x = w \times \frac\lambda{360} + \frac w2$$ $$y = -h \times \frac\phi{180} + \frac h2$$ ...

5

In order to know when a star will be above horizon, you'll need an equation with times, not "celestial latitudes". So you end up adding RA and sidereal time. EDIT In order to find out the position of a star you need to find out how its Hour Angle varies with time. The Hour Angle is how far the star is from the observer's meridian: In this image, it is ...

5

What you are looking for is the navigation method used by ships and aircraft before the advent of GPS. It requires not only an instrument for measuring the angle between the sun or a star (such as Polaris) and the horizon, but also an accurate time measurement, and charts that can be used to interpret the numbers. And of course an accurate chronometer -- ...

5

Right ascension is a historical oddity. To specify a point in the sky you need a coordinate system, the one which we have come to use has it's origin at the Point of Aries on the Equator and the Ecliptic (a reasonable choice), It uses the Equator for one axis, and the meridian through the point of Aries for the other, again these are convenient and ...

5

Sure you can! Galactic coordinates have the same origin as other J2000.0 systems; the solar system barycenter (center of mass). This is very close to the Sun, usually but not always inside the Sun, because the larger planets, especially Jupiter, pull it around a little bit. You can read a little more here for example, and also read @zephyr's excellent answer....

5

You're asking two different, but somewhat related questions: One has to do with a practical way of describing the expansion of the Universe; the other has to do with a way of dealing with our ignorance of the exact expansion rate of the Universe. I will answer your specific question in the last paragraph. Comoving and physical coordinates A comoving ...

5

TL;DR: Calculating the right ascension (i.e. converting from ecliptic to equatorial coordinates) requires some spherical trigonometry, but fortunately there are internet tools that will do this, such as here. The resulting RA coordinates are: Aries 0h 0m 0s; Taurus 1h 51m 39s; Gemini 3h 51m 16s; Cancer 6h 0m 0s; Leo 8h 8m 44s; Virgo 10h 8m 21s; Libra 12h 0m ...

5

You can think of the position given in your title as the coordinates of the star that is directly over the Sun's north pole. (Similar to how Polaris is almost directly over the Earth's north pole.) Likewise, the ecliptic plane has a north pole which can be located by right ascension and declination. 7.25° to 7.26° appears to be the correct calculation, ...

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