# Tag Info

15

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 ...

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 ...

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

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

I don't think this particularly useful way to think about it. The equator is the equator, Cancer is 23.4 North, and Capricorn is 23.4 South. The equator is a great circle on the sphere of the sky, but the Tropic of Cancer is not a great circle. These are fixed. Now (in Northern Hemisphere summer) the sun is directly above a point on (or near) the tropic ...

7

The area of a triangle enclosed by 3 stars on the celestial sphere, in square degrees, is given by: $$A = \frac{180}{\pi}\times E$$ Where E is the spherical excess and is equal to the sum of all angles of the triangle minus 180°. Problem is, we don't know the angles of the Summer Triangle and instead we have to use an alternate equation for the spherical ...

5

The answers to this question are very clear and much better than I could have written. I'm not sure what you mean by "the stellar time at Greenwich at 0h UTC is 22h20min", but I'm assuming that you mean that that is the amount of time since the culmination of Aries over the Prime Meridian (Greenwich). That being the case, the Right Ascension (RA) of your ...

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

Jacob's staff is probably all you need. It is a pole with a cross arm. By positioning the arm at a certain point on the pole and sighting down the pole one can get a position on the sky. A skilled user can achieve sub-degree accuracy with such a tool. The position of the cross beam is easy enough to calculate by trigonometry, or by careful construction of ...

4

DecRA is the decimal right ascension RAh, RAm, RAs are the hms form $${\rm DecRA} = {\rm RAh}\times 15.0 + {\rm RAm}/4.0 + {\rm RAs}/240.0$$ $${\rm RAh} = {\rm INT}({\rm DecRA}/15.0)$$ $${\rm RAm} = {\rm INT}(({\rm DecRA}-{\rm RAh}\times 15.0)\times 4.0)$$ $${\rm RAs} = ({\rm DecRA}-{\rm Rah}\times 15.0 - {\rm RAm}/4.0)\times 240.0$$ where INT is the ...

4

The position angle P of a body ($\alpha_1, \delta_1$) with respect to another body ($\alpha_2, \delta_2$) can be calculated from $$tan(P)={sin(\Delta\alpha)\over cos(\delta_2)tan(\delta_1)-sin(\delta_2)cos(\Delta\alpha)}$$ where $\Delta\alpha = \alpha_1-\alpha_2$. If the denominator is negative, the position angle lies in the range of 90 to 270 degrees. ...

4

You are correct that topocentric coordinates are for the position of "close" objects, corrected for observing from the Earth's surface instead of the theoretical center. The topocentric and geocentric coordinates would use the same system (J2000, true, etc.) There is no convention to use one in favor of another (but there may be preferences in some cases).

3

Constant for every star with different declination's. Though you would have to assume the the declination is such that it can have an azimuth of 90 each, not all stars will. Depending on your latitude some stars will never get to an azimuth of 90, they will either be always be further north or south.

3

0 RA is the direction from the center of the Earth to the vernal equinox. You've already imagined the plane in space created by projecting the equator into space. Now consider the plane formed by the mean orbit of the Earth; this is the ecliptic. The two planes intersect in a line. The ray from the center of the Earth toward where the Sun is in March is the ...

3

Short answer, no, they don't. Longer answer, it's complicated. There are, in essence, five different measures for the centre of a galaxy cluster, based on different physical properties of the cluster, and occasionally not in agreement with each other. BCG, the brightest cluster galaxy. It should sink to the bottom of the gravitational well, but that's only ...

3

Only a syzygial tide during/near an equinox is the strongest. This means, there must be either new or full moon. In general, syzygial tides are strong because three bodies (Earth, Moon, Sun) align near one line, and tidal effects of Moon and Sun on Earth become (nearly) collinear and sum to the maximal possible magnitude. The Moon’s orbit is inclined to the ...

3

The calculations are not trivial, but they are encapsulated in software libraries such as pyephem, which has examples of finding rise, set and transit If you want to understand how these are calculated, you can readnthe source, which is based on the xephem application. One detail which complicates the calculation is the refraction caused by the atmosphere, ...

3

The values are more steady for objects close to Ecliptic poles, but in genenral, they change by a few arcseconds every year. Yes, we lack an absolutely fixed frame of reference.

3

It's a bit late but hopefully it will help others in the future. The calculations can be found on this site, article from Keith Burnett (keith@xylem.demon.co.uk) The first problem is that you use J2000 as time reference but it's not what the site takes as time reference! "Get the days to Dec 31st 0h 2000 - note, this is NOT same as J2000" So you ...

3

The complexity of the formula depends on the precision you need. We can make a crude approximation with two sine waves: one for the Moon's travel eastward around the ecliptic over a tropical month $$\delta_1 = 23.4^\circ \sin {2 \pi (u - 10.75) \over 27.32158}$$ and one for the Moon's travel north or south of the ecliptic over a draconic month $$\delta_2 =... 3 Given a date and time, the position of the Moon can be calculated to provide the declination and right ascension. The sub-point of the Moon (the point on the Earth at which the Moon is at the zenith) is as follows: latitude = declination of the Moon longitude can be found by calculating the local mean sidereal time (LMST) that equals the Moon's right ... 3 The stars should fit r = \sqrt{x^2 + y^2 + z^2} but seem to be plotted in the x = r plane instead. Conventionally the x axis is at (α=0h, δ=0°), the y axis is at (α=6h, δ=0°), and the z axis is at δ=+90°. Using right triangles instead of isosceles triangles, Then h = r \cos \delta, and$$\begin{align} x &...

3

Wikipedia has formulas for the area of a celestial triangle in steradians (which they call the "excess" and write $E$) in terms of the angles, the side lengths, and side-angle-side. The angle formula is the simplest, but calculating the angles is tricky. It may be easier to calculate the side lengths and use the side-side-side formula. You can ...

3

Answer to your question RA changes just a little bit through time (because of minor effects like parallax), but we can just say it is constant for a specific non-Sun star. Same applies to declination. Why is that? The declination and right ascension of vernal equinox (where the Sun is in March) is defined to be 0°. This point is fixed with respect to other ...

2

The number 18.697374558 was the value of GMST (in hours) at noon UT on 1 January, 2000. The number 24.06570982441908 reflects the fact that it takes slightly less than 24 hours (23.934469591898 hours to be precise) for a star to the appear to rotate by 360 degrees. This is a consequence of the fact that the Earth orbits the Sun. Our 24 hour day is based on ...

2

The declination of the point overhead (zenith) is the same as the observer's latitude. The RA of the point transiting zenith at any given time is the equivalent of the local sidereal time. (Alternatively, the local sidereal time is RA of the observer's meridian.)

2

After quite an extensive search of dictionaries and books on astronomical nomenclature I came across this article that has a comprehensive review of celestial coordinate systems. And upon reading page 84 (8/14) I came across the answer to your question. It would appear that the reason declination is used, as opposed to inclination, is that declination was ...

2

You are forgetting parallax. Something which is very distant and lies on the celestial equator will have a declination of 0°, but if it is nearby then its declination will only be 0° to an observer located along Earth's equator. From your location in Zagreb, something like the ISS, even if it is crossing Earth's equator at a longitude of ~16° will appear ...

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