Tracing the history of medieval astronomy
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:
In the last post, we explored the rising time for one zodiacal sign which would comprise 30º of the ecliptic. Since the preliminary math is now out of the way, we can quickly do 60º of the ecliptic which constitutes Aries and Taurus. From there, we’ll be able to more quickly compute the remaining constellations as well. Continue reading “Almagest Book II: Calculation of Rising Times at Sphaera Obliqua for Remaining Arcs”
Now that we’ve done a bit of preliminary theory, we’re ready to jump into an example problem. As with before, Ptolemy selects Rhodes at a latitude of 36º. Since we’ve been seeing several of the same diagrams here, I’m going to forego the buildup and give a mostly completed diagram:
Continue reading “Almagest Book II: Calculation of Rising Times at Sphaera Obliqua”
In the last post, we proved that two arcs of the ecliptic that are equidistant from the same equinox rise in the same amount of time. In this post, we’ll prove something similar for what happens with arcs of the ecliptic equidistant from the same solstice.
It’s been awhile since I’ve been able to update the blog with anything from the Almagest. As noted in the last post from the book, this section is not one of the better written ones. Indeed, it’s taken me the better part of a month to really work out how the diagram is put together.
Ultimately the trouble stemmed from the fact that it’s not a single diagram; it’s actually two pasted together1, so instead of throwing it all at you at once like Ptolemy did, let’s work through each piece in turn before pasting it together.
To begin, let’s start with a simple celestial sphere:
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.
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).
Continue reading “Almagest Book II: Ratio of Gnomon Equinoctial and Solsticial Shadows”
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”
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.
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”
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”
