Tag: algebra

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Almagest Book II: Angle Between Ecliptic And Horizon – Calculations

We’ll continue on with our goal of finding the angle the ecliptic makes with the horizon. Fortunately, this task is simplified by the symmetries we worked out in the last post meaning we’ll only need to work out the values from Aries to Libra. Unfortunately, this value will change based on latitude as well as the position on the ecliptic, but we’ll still only do this for one location. And for that location, Ptolemy again uses Rhodes.

First we’ll start with angles at the equinoxes:

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Almagest Book II: Angle Between Ecliptic and Meridian – Symmetries

In my last post, I mentioned that entered a paper based on the rising sign calculations presented in this post into an A&S competition. This was a very interesting piece to do because it showed how well woven the roots are, as doing so made use of almost every section we’ve gone through previously. As such, it felt like a good capstone for book II. But it doesn’t end there.

Rather, Ptolemy decides to go on for several more chapters as this book is focused on the great circles on the celestial sphere. While we’ve covered the ecliptic and celestial equator pretty extensively, we have done less with the horizon and meridian which is where Ptolemy seeks to go for the last few chapters in this book. Specifically, we’ll be covering:

  • The angles between the ecliptic and meridian
  • The ecliptic and horizon
  • The ecliptic and an arc from horizon to the zenith (an altitude circle)

All followed by another summary chapter at various latitudes. As the title of this post may have indicated, we’ll be covering the first of these in this post1. Continue reading “Almagest Book II: Angle Between Ecliptic and Meridian – Symmetries”

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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.

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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.

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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:

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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.

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Exploring Menelaus’ Theorem

In the last post, we used several theorems we’d developed to arrive at Menelaus’ theorem. However, at the very end Ptolemy simply mentions another version of the theorem, but does not derive it. I simply took his word that it worked, but as that alternative form is used first thing in the next chapter, I want to make sure at the very use, we know how to use it, even if we don’t go through how it’s derived.

First, let’s set up a generic Menelaus configuration on a sphere which is the intersection of the arcs of four great circles:

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