Last month, I had a post that briefly introduced the two primary coordinate systems for recording the position of objects on the celestial sphere: the Altitude-Azimuth (Alt-Az) and Right Ascension-Declination (RA (α)-Dec (δ)) systems. There, I noted that Alt-Az is quick and easy to use, but is at the same time nearly useless as objects fixed on the celestial sphere do not have fixed coordinates.
Instead, astronomers1 use the RA-Dec system because fixed objects have fixed positions. My modern telescope does allow for this system to be used rather directly because it has an equatorial mount which tilts the telescope to match the plane of the ecliptic instead of the plane of the horizon. Additionally, it is motorized to allow it to turn with the sky, thereby retaining its orientation in relation a coordinate system that rotates with the celestial sphere. Thus, once it’s set we’re good to go.
However, the quadrants Brahe used were neither inclined to the ecliptic nor motorized. Thus, measurements were necessarily taken in the Alt-Az system and would need to be converted to RA-Dec to be useful. Here, we’ll explore how that conversion works2.
So far in this post, we’ve only introduced four variables, Alt, Az, RA, Dec. However, for conversion, we’re going to need several more concepts so let’s go slowly and introduce them one at a time.
The first one we’ve already seen. It’s the Vernal Equinox. It is defined as the intersection of the ecliptic and celestial equator on the date of the Vernal Equinox and is represented by the symbol ♈︎.
Next up, the celestial sphere, much like the Earth has poles around which it appears to rotate. These are the North Celestial Pole (NCP) and South Celestial Pole (SCP). For our purposes here, I’ll just illustrate the NCP.
While we won’t be using these directly, we’ll be measuring things in relation to them. Specifically, the Hour Circle. This is defined as the great circle perpendicular to the celestial equator (and therefore going through the North Celestial Pole (NCP)) on which the object lies. The angular distance from the intersection of the Hour Circle and the Vernal Equinox is the definition of the Right Ascension. Adding that to our diagram:
Next up, let’s remember that the observer has a local meridian, the line from their north point on the horizon, through the zenith, to the south point on their horizon. Since the Earth is constantly spinning beneath the celestial sphere, it will be moving as well. But we can draw it in at an arbitrary point simply to illustrate the point.
You may notice I’ve added another new concept here, the Hour Angle (HA). This is defined as the angular distance, along the celestial equator between the hour circle and the observer’s meridian. Coincidentally, you’ll see I marked it in two places in two different ways. One (at the top) is a more direct angle. The other, along the celestial equator, is a side. But since we divide the full circle into 360º along the great circle as Ptolemy did, their measures are the same.
From there, we’ll add another new concept simply by adding two we already have together: the Sidereal Time (ST). Loosely speaking, sidereal time is time measured in relation to distant stars which means that it’s Earth rotating $360º$3. However, since our reference point here is the vernal equinox, that’s going to be our starting point, and our other point will be the intersection of the meridian and celestial equator. I’ve added that to our diagram:
So from this we can easily see that: $ST = \alpha + HA$.
We’ve got a good start, but now we’ll need to shift frames of reference a bit. Thus far, we’ve been looking at things from outside the celestial sphere, but since we’re concerned with converting from the Alt-Az system which in measured in relation to the observer’s horizon, we’ll need to switch to a view that includes that. So here we go:
At first glance it doesn’t look like we’ve changed much, so before moving on, let’s make sure we know what’s different. First, I’ve completely dropped the ecliptic. We only needed it to find the ecliptic to define a few things4.
The sphere as presented previously has been rotated a bit (counter clockwise as we’re looking at it) such that the NCP is now off at an angle suitable for an observer in the northern hemisphere5.
I’ve also marked the cardinal directions of the side of the sphere facing us (so left off east for clarity).
Next, let’s add a piece of a great circle that passes through the NCP through the object, and cut it off at the horizon.
This line is going to help us because our diagram, with that one line, now contains everything we need to visually draw in the definitions for Altitude ($a$) and Azimuth ($A$).
Recall that altitude is defined as the angular height above the closest horizon. In planar geometry, the shortest distance between two points is a straight line. But since we’re in spherical geometry, the shortest distance (the line closest to the horizon) is the great circle arc. Thus, what I’ve drawn in is the proper altitude.
Similarly, the definition of azimuth is the angular distance along the horizon, starting from north, going through east. Thus in this diagram, it starts at N, goes around the back, and ends back on the front. The dashed line would be $360º – A$.
Looking up at the triangle near the top, there’s a few other things we should add in:
Here, we’ve noted where the hour angle ($HA$) is. And for completeness, note that angle $Z$ is the angle that subtends that $360º-A$6
Next, we need some sort of understanding of how tilted our NCP is from the zenith.
Here we’ve drawn in $\phi$ which is the angular distance of the NCP above the horizon. This can be measured fairly accurately by measuring the altitude of the North Star (Polaris) and is also your latitude. In addition we can mark the other segment from $arc \; NZ$ as $90 – \phi$ since the zenith is at a right angle from the horizon.
There’s still something conspicuously missing. We haven’t added anything directly from the RA-Dec system. As we noted earlier, we will be determining RA from the HA and ST, so the only one we need now is the Dec ($\delta$) which is the great circle arc from the object to the celestial equator.
I’ve also marked in the portion from the NCP to the object as $90º – \delta$ as we’ll be using that.
Now we’re done with diagrams and it’s time to get down to some math. First we’ll use the spherical law of cosines and apply it to that little triangle up at the top7.
$$cos(90 – \delta) = cos(90 – \phi)cos(90-a) + sin(90 – \phi)sin(90-a)cos(360 – A)$$
This simplifies a lot because when $cos(90 – x) = sin(x)$ and vice versa. Also, $cos(360 – x) = cos(x)$:
$$sin(\delta) = sin(\phi)sin(a) + cos(\phi)cos(a)cos(A)$$
Taking a good look at our variables here, we happily have everything on the right side. Altitude and azimuth are things we will measure directly, and φ is your latitude. Thus, we can solve for the declination with just those three pieces of information.
Recall we said we’re going to get the RA by way of the ST and HA. Since as we noted before, ST is something we get off a clock, what we really need next is the HA, which we can get by another application of the spherical law of cosines. Before doing that, let’s draw in one final piece to the diagram for clarity:
Here, I’ve added a definition for one more side to that little triangle at the top. Following the same arc that the altitude is on, the remaining part extending it to the zenith must be $90º – a$ since the zenith is $90º$ from the horizon. So applying another spherical law of cosines, this time with the side being the one between Z and the object we get:
$$cos(90-a) = cos(90 – \phi)cos(90 – \delta) + sin(90 – \phi)sin(90 – \delta)cos(HA)$$
Again this simplifies handily:
$$sin(a) = sin(\phi)sin(\delta) + cos(\phi)cos(\delta)cos(HA)$$
We can then solve this for the variable we’re after:
$$cos(HA) = \frac{sin(a) – sin(\phi)sin(\delta)}{cos(\phi)cos(\delta)}$$
From there, all we need to do is remove the $cos$ from the hour angle by taking the inverse and we’ve solved for the hour angle:
$$HA = arccos (\frac{sin(a) – sin(\phi)sin(\delta)}{cos(\phi)cos(\delta)})$$
We can solve our earlier equation for RA as we now have everything we need to plug into it:
$$\alpha = ST – HA$$
However, there’s one more thing to note which is that, as I’ve drawn it, the object is to the west of the observer’s meridian. If you imagine a situation in which the meridian is west of the object, then,
$$\alpha = ST + HA$$
In the next post, I’ll work an example that will hopefully help clarify things.
- By which I mean modern astronomers. While there’s no reason that RA-Dec couldn’t have been used in period, it doesn’t appear in star catalogs until the $1700$s. Although I haven’t gotten far enough in my study to confirm it, this implies that astronomers in period probably used the Ecliptic coordinate system which is very similar to RA-Dec except that the primary “left-right” axis is the ecliptic instead of the celestial equator. While I may use that system later, for the time being I’m going to stick with RA-Dec because it allows for more easy comparison with the modern values so I can better understand the capabilities of my quadrant.
- This method is the one I covered in my ASTR 591 class at KU. I’m not certain if it’s one used in period or not.
- There’s another kind of time measurement called Solar Time in which time is measured in relation to the Sun. However, since Earth orbits around the Sun, it must rotate a bit beyond 360º for the Sun to be in the same relative position, such as on the meridian. Thus, a solar day is slightly longer than a sidereal day.
- Specifically, we needed it to define RA and ST.
- If the NCP was at the Zenith, you would be at the North Pole. If it were on the North horizon, you would be on the equator.
- Because this is on the surface of a sphere, the three angles in that triangle won’t add up to $180º$, so we can’t find the third by subtraction, but it’s not particularly important anyway.
- The one with corners of the NCP, Zenith, and object.