Tag: ecliptic

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Almagest Book II: Symmetry of Rising Times – Arcs of the Ecliptic Equidistant from the Same Equinox

In this next chapter, Ptolemy’s goal is to

show how to calculate, for each latitude, the arcs of the equator… which rise together with [given] arcs of the ecliptic.

To do this, we’ll do a bit of convenient math, breaking the full ecliptic into its traditional 12 parts. However, since these signs are not of equal size, Ptolemy takes an even 30º for each sign, beginning with Aries, then Taurus, etc…

The first goal will be to prove that

arcs of the ecliptic which are equidistant from the same equinox always rise with equal arcs of the equator.

Continue reading “Almagest Book II: Symmetry of Rising Times – Arcs of the Ecliptic Equidistant from the Same Equinox”

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Almagest Book II: For What Regions, When, and How Often the Sun Reaches the Zenith

The fourth chapter in book two is a very short one. In fact, it’s a single paragraph so I almost didn’t dedicate an entire post to it but ultimately decided to as it didn’t really fit with either the previous or next chapter.

In this, Ptolemy says,

it is a straighforward computation to determine for what regions, when, and how often the sun reaches the zenith. For it is immediately obvious that for those beneath a parallel which is farther away from the equator than the 23;51,20º (approximately), which represents the distance of the summer solstice, the sun never reaches the zenith at all, while for those beneath the parallel which is exactly that distance, it reaches the zenith once a year, precisely at the summer solstice.

Continue reading “Almagest Book II: For What Regions, When, and How Often the Sun Reaches the Zenith”

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Almagest Book II: Symmetries of Arcs and Day/Night Lengths

For the past few Almagest posts, we’ve been working on the following diagram:

As a refresher, AEG is the celestial equator and BED is the horizon of an observer. Z is south. H is the point of the winter solstice as it crosses the horizon (or rises).

In the first post we determined $arc \; EH$. In the second post, we used that to determine $arc \; BZ$. In the third, we’ve determined $arc \; E \Theta$ and then came full circle and showed another way to get $arc \; EH$.

Now, Ptolemy wants to generalize. Continue reading “Almagest Book II: Symmetries of Arcs and Day/Night Lengths”

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Almagest Book II: Difference Between Length of Solstice Day vs Equinox From Latitude

The next demonstration Ptolemy does is actually a reverse of what we did in the past 2 posts. Now, given latitude Ptolemy asks what the difference in length between the longest (or shortest) day and the daylight on the equinox would be. Again, Ptolemy has actually already given us the answer for the case we’re considering of the Greek city of Rhodes which is at 36º N latitude for which the longest day (the summer solstice) is 14.5 hours. Since the length of the day on the equinox is 12 hours, the answer will be 2.5, but Ptolemy simply wants to demonstrate that this can be achieved mathematically.

Again, we’ll use Menelaus’ Theorem II.

Continue reading “Almagest Book II: Difference Between Length of Solstice Day vs Equinox From Latitude”

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Almagest Book II: From Length of Longest Day Finding Elevation of the Pole

In our last post, we explored how to find the angular distance around the horizon from the ecliptic and celestial equator. In this chapter, we explore another value that can be derived from knowing the length of the longest day (i.e. on the summer solstice): the elevation (or altitude) of the celestial pole (which is also the latitude).

Once again, we’ll start with the same diagram we’ve been using for awhile now:

Continue reading “Almagest Book II: From Length of Longest Day Finding Elevation of the Pole”

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Almagest Book II: Arcs of the Horizon between Equator and Ecliptic

Let us take as a general basis for our examples the parallel circle to the equator through Rhodes, where the elevation of the pole is 36º, and the longest day $14 \frac{1}{2}$ equinoctial hours.

Immediately starting the second chapter, we’re given a lot to unpack. First off, Ptolemy chooses to work through this problem by means of an example, selecting Rhodes, a city in Greece as the exemplar. I’m assuming that the “elevation of the pole” is the latitude, as Rhodes’ latitude is 36.2º. But what of these equinoctial hours? Continue reading “Almagest Book II: Arcs of the Horizon between Equator and Ecliptic”

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Data: Converting Alt-Az to RA-Dec – Derivation

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. Continue reading “Data: Converting Alt-Az to RA-Dec – Derivation”

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Almagest Book I: Rising Times at Sphaera Recta

We’ve finally hit the last chapter in Book I. In this chapter our objective is to “compute the size of an arc of the equator”. At first pass, that doesn’t seem to have much to do with the title. Arcs of equator vs rising times?

However, Earth is a clock, rotating once every 24 hours. Thus, if we know the length of an arc, we know something about when an object following that arc through the sky will rise and set because it’s a certain proportion of 360º per 24h. Notice that if you actually complete that division, it comes out to an even 15º/hr. That’s not a coincidence.

Fortunately, to work on this problem, we won’t even need a new diagram. We can recycle the one from last chapter. Again this time we’ll be wanting to determine all sorts of arc lengths, but we’ll start with the one where $arc \; EH = 30$º.

Continue reading “Almagest Book I: Rising Times at Sphaera Recta”