Tag: chords

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Almagest Book II: Calculation of Rising Times at Sphaera Obliqua for 10º Arcs

With the previous theorem about the ascensional differences complete, it’s time to move on to determine how to figure out the rising time of arcs of the ecliptic for 10º segments at various latitudes using what Ptolemy promises to be a shortcut in the math. In the modern sense it really doesn’t seem to be much of a shortcut, but that’s because with the assistance of calculator’s, the equations we were using previously seem much less daunting. If it had to be done by hand, I’m sure it would be far more tedious.

Instead, Ptolemy reduces the number of calculations by going through the proof regarding ascensional differences as well as making use of some previously calculated values to avoid having to do other calculations.

To get started, Ptolemy revises the previous drawing, making it a bit simpler by removing the ecliptic and renaming a few of the points, as well as changing a few definitions.

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Almagest Book II: Ascensional Difference

Not content to simply figure out how long it would take a zodiacal constellation to rise at latitudes other than the equator, Ptolemy sets out to further divide the ecliptic into 10º arcs and he’s promised an easier method than what we’ve done previously. But before we can get there, Ptolemy gives a brief proof which he’ll make use of later.

To start, we begin with the vernal equinox on the horizon:

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Almagest Book II: Ratio of Gnomon Equinoctial and Solsticial Shadows

First off, what’s a gnomon?

Apparently it’s the part of a sundial that casts shadows. Now you know.

To start this next chapter, Ptolemy dives straight into a new figure, but I want to take a moment to justify it first. To begin, let’s start with a simple diagram. Just a side view of the meridian, the horizon, and the north celestial pole. The zenith is also marked (A).

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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: 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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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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Almagest Book I: Menelaus’ Theorem

So far in these preliminary theorems, we’ve looked at some that were based on triangles and some that were based on circles. We’ll be going one step further with this next one and work with spheres. Thus far, we’ve briefly touched on spheres in this post discussing the celestial sphere. If great circles and spherical triangles aren’t familiar to you, I suggest reading over that post.

But since this is the first time we’ve encountered math in 3D if you’ve been following along, I want to build this up more slowly2 and will be trying to add some 3D elements to make the visualization a bit easier.

So let’s get started.

[L]et us draw the following arcs of great circles on a sphere: BE and GD are drawn to meet AB and AG, and cut each other at Z. Let each of them be less than a semi-circle.

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