Tag: mathematics

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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 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 – Example

In the last post, we derived equations to demonstrate that the right ascension (α) and declination (δ) of an object can be gotten by knowing four other variables: altitude (a), azimuth (A), sidereal time (ST), and latitude (φ).

In this post, I’ll do an example of using these equations to do just that. For my data, I’ve jumped into Stellarium and selected Altair1 as it’s a nice bright star that I’ll certainly be observing.

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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, astronomers2 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 works3. 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.

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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 slowly5 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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