Almagest Book VI: Angles of Inclination at Eclipses

The final few chapters of Book VI are rather odd. Now that we’ve completed the discussion of eclipse prediction, Ptolemy wants to do an “examination of the inclination which are formed at eclipses.” However, he doesn’t appear to provide any motivation for doing so. Toomer and Neugebauer both indicate that the actual reason was likely weather prediction1, but the Almagest doesn’t contain any information on how this is to be used. Neugebauer indicates that,

[T]he technical term connected with this problem is “prosneusis”…developed from the original meaning of the verb νευειν (to nod, to incline the head, etc…). According to the terminology of hellenistic astrology, the planets or moon can, e.g., give their consent by “inclining” toward a certain position, i.e., by being found in a favorable configuration.

However, aside from these astrological purposes, these last few chapters are essentially left as a free-floating bit of material.

So what are “inclinations”? Ptolemy’s answer to this question is very poor. Ultimately what we’re wanting to determine is where a great circle drawn between the center of the sun and moon (for a solar eclipse) or center of earth’s shadow and moon (for a lunar eclipse) intersects the horizon.

To determine this, Ptolemy breaks the problem into two pieces:

the inclination of the eclipsed part [of the body] to the ecliptic and … the inclination of the ecliptic itself to the horizon.

The first of these is the angle I just mentioned between the two components of the eclipse and the ecliptic. The second of these, “the inclination of the ecliptic itself to the horizon” is highly misleading with respect to what Ptolemy actually ends up discussing. It implies that it’s the angle created between the ecliptic and the horizon. In actuality, what we’ll be looking at is the angle along the horizon between the ecliptic and the celestial equator. These two components of the final inclination are what we’ll explore in this post. We’ll be creating a new table and diagram which we’ll display in the next post. Then, we’ll qualitatively discuss how to use these two components to find the final inclination and, since the great circle will intersect the horizon at two points, which of those is “significant” for Ptolemy’s weather astrology.

Ptolemy goes on to say a bit about these angles:

Both of these angles, during the course of every eclipse phase, undergo great change as a result of the shift in position [of the bodies], in a way which could not be controlled if one wanted to undertake the task of computing the inclinations throughout the whole of the duration [of the eclipse], a superfluous task, since predictions on such a scale are not in the least necessary or useful. For, since the situation of the ecliptic relative to the horizon is determined from the position on the horizon occupied by its rising or setting points, the angled formed by the ecliptic at the horizon must necessarily change continuously during the course of the eclipse, as those points on the ecliptic which are rising or setting change continuously.

Ptolemy spends a lot of words here stating something very simple: The angle between the ecliptic and horizon is constantly changing2.

Similarly, since the inclination of the eclipsed part [of the body] to the ecliptic is determined from the great circle drawn through the two centers, [i.e.,] the centers of moon and shadow or the centers of the moon and sun, it is, again, a necessary consequence of the motion of the moon’s center during the course of an eclipse that the circle through the two centers occupy a continuously varying position relative to the ecliptic, and [hence] that the angle formed at their intersection vary continuously.

To get a better understanding, let’s draw a diagram:

Here, $arc \; A$ is the first inclination to which we referred: The angle along the horizon between the ecliptic and celestial equator.

And $\angle B$ is the angle between the great circle drawn through the ecliptic and the two components of the eclipse: The moon and the earths shadow, but could just as easily be the moon and sun in the event of a solar eclipse.

Therefore, [the need for] this kind of examination will be satisfied if it is carried out only for those points in [the progress of] the eclipse which have some significance, and only roughly for the inclinations with respect to the horizon.

Here, Ptolemy is stating that because these angles are constantly changing coming up with an exact solution will be difficult. How are they changing? Well, consider the above image. Note that the circle between the center of the moon and the earth’s shadow is not the lunar path. If I sketch that in and re-draw the image with the moon at a different time, we can see how $\angle B$ changes:

Here, I’ve drawn the moon earlier in the eclipse in red and we can see that great circle between the moon at that time and the center of the earth’s shadow is notably different. Thus, trying to come up with a complete solution that describes this angle at every point along the eclipse for every possible eclipse is going to be challenging. Fortunately, Ptolemy tells us we’re going to skip such a robust solution and and instead only worry about a few points in time he deems important. Which points are those?

Hold onto that thought. Instead of addressing that obvious question, it’s brief detour time.

[To achieve this kind of accuracy] people who actually observe the eclipse as it occurs could, merely by eye, estimate the important inclinations by looking at the relative positions in both cases [at eclipse and horizon], since, as we said, a rough notion [of the amount] is sufficient in such matters.

In other words, if you’re watching the eclipse live, just eyeball it!

Nevertheless, not to pass over this topic all together, we shall try to set out some ways of achieving the kind of result desired as conveniently as possible.

So… about those points…

The points in [the progress of] the eclipse which we too take into consideration as deserving to be though significant are:

[1] the point of the start of obscuration, which coincides with the very beginning of the whole eclipse;

[2] the point of the completion of obscuration, which coincides with the beginning of the phase of totality;

[3] the point of greatest obscuration, which coincides with the middle of totality;

[4] the point of the start of emersion, which coincides with the end of the whole total phase;

[5] the point of the completion of emersion, which coincides with the end of the whole eclipse.

While Ptolemy is, in theory, discussing this for both lunar and solar eclipses, for a solar eclipse, really only $1$, $3$, and $5$ apply as the moments of $2$, $3$, and $4$ are practically the same for a solar eclipse.

The inclinations [with respect to the horizon] which we take into consideration as being more reasonable and more significant are those bounded by the meridian and also those bounded by the rising and setting points of the ecliptic at the equinoxes and at the summer and winter solstices.

The meaning of this phrase isn’t entirely clear to me, likely because it seems to be hinting at the weather prognostication by talking about which angles are “reasonable and significant.”

As for the points bounding the various ‘wind directions’, they may be understood in many different ways by many people; nevertheless, if desired, they can be indicated by means of the angles we have set out along the horizon.

Toomer notes that the term “wind directions” is another way of indicating compass directions that was common amongst the Greeks. By and large, this isn’t used in the Almagest.

Considering the intersections of meridian and horizon, let us make the following definitions:

the northern intersection is the ‘northpoint’;

the southern intersection is the ‘southpoint’.

This definition strikes me as purely tautological and hardly worth mentioning at this point. So why include it? I would guess that Ptolemy meant for these chapters to stand alone and may have readers jumping in at this point instead of working through the previous material.

Considering the rising and setting [points of the ecliptic, let us make the following definitions]:

the intersections of the beginning of Aries or Libra with the horizon are known as ‘equinoctial rising’ and ‘equinoctial setting’; these are always the same distance, [i.e.,] a quadrant, from the point where the meridian intersects [the horizon];

Let’s dissect this before continuing on. The first thing to note is that the first points of Aries and Libra are the two equinoxes – the points where the ecliptic intersects the celestial equator. Since the celestial equator always intersects the horizon due east and west, as these two points cross the horizon, they will necessarily be due east and west as well. Since due east and west are both always $90º$ away from the meridian, these points will also always be $90º$ away from the meridians when they cross the horizon. Ptolemy gives them the special names of “equinoctial rising” and “equinoctial setting” depending on which horizon they’re on, but presumably both terms are used for either point.

the intersections of the beginning of Cancer [are known] as ‘summer rising’ and ‘summer setting’, and the intersections of the beginning of Capricorn as ‘winter rising’ and ‘winter setting’.

The distances [from the meridian intersection] of these last [four] points vary according to the latitude in question. The inclinations are sufficiently characterized by saying that they are at one of the above situations or between some pair of them.

Here, things get a bit more complicated. This is because at the summer solstice, the beginning of Cancer, is when the sun is $23;30º$ north of the celestial equator. However, that is measured perpendicularly to the celestial equator and since that tilt is dependent on your latitude, that means where the sun actually intersects the horizon at that moment will change as well.

Ptolemy doesn’t go into more detail here, but simply names the points again with the sun on the eastern horizon during the summer solstice “summer rising”, the western horizon “summer setting”, and similarly for the winter solstice rising and setting.

Regardless, the solstical points do set bounds, and thus, the actual rising/setting of the sun on any given date will fall between the extremes at the solstices.

To enable one to determine the position of the ecliptic relative to the horizon for any given situation, we computed, by the method indicated in the first books of our treatise, the distance along the horizon, at rising and setting, of the beginning of each zodiacal sign from the points where the equator intersects [the horizon, computing them] on either side of it [ie., north or south].

Finally, we’re ready to get to the first of the two angles (which is really more of an arc) that Ptolemy discussed. The good news is that we’re not going to actually go through any calculations because we’ve already shown how this is calculated way back in II.2. But if you paid close attention, Ptolemy showed us how to do the calculation in that chapter, but then never created one of the reference tables like we often do. Rather, the display of this is saved for the next chapter in this book. I’ll follow suit and save it for a separate post and save Ptolemy’s discussion of it for that one.

Continuing on, we’ll now turn our attention to the second of the angles in question: The angle between the great circle defined by the center of the sun or shadow and center of the moon, and the ecliptic.

In order to have tabulated the apparent inclinations of the actual phases to the ecliptic, i.e., the angles formed between the ecliptic and the great circle joining the centers in question at each of the significant points mentioned above, we computed these too, for [successive]j positions of the moon corresponding to a difference of $1$ digit in obscuration. However, we did this only for lunar positions at mean distance (since that is sufficient), and under the assumption that those arcs of the ecliptic and the moon’s inclined circle which we consider for the obscurations are sensibly parallel to each other.

Here, Ptolemy is stating we’re going to make a simplification such that the moon’s path is essentially parallel to the ecliptic over the short distances while the moon is within the earth’s shadow.

Here, $\overline{AB}$ is the ecliptic with $A$ being the center of earth’s shadow or the sun. Similarly, $\overline{GDE}$ will be the moon’s path with $G$ as the center of the moon at mid-eclipse (i.e., point $3$ above), $D$ as the center at the first or last moment of totality (i.e., points $2$ or $4$), and $E$ the moon’s center at the moment of first or last contact (i.e., points $1$ or $5$).

We’ll join $A$ to each of these other points. So long as we’re considering these two paths parallel, $\angle{BAG}$ and $\angle{AGE}$ will both be right angles. This means $\angle{BAD}$ will be the angle we’re looking for for cases $2$ and $4$, or as Ptolemy states, “the end of [the partial phase of] the eclipse and at the beginning of the emersion”. Similarly, $\angle{BAE}$ will be the angle “at the beginning and end of the eclipse” which was situations $1$ and $5$.

Because of this:

it is immediately clear that $\overline{AE}$ represents the sum of the radii of both circles3 and $\overline{AD}$ is their difference.

Now that we’re all set up, we’ll work through an example

in which half of the sun’s diameter is obscured at mid-eclipse.

This means that $A$ will be the center of the sun’s disc. In that case,

$$\overline{AE} = 0;15,40º + 0;16,40º = 0;32,20º$$

Next, we’ll consider $\overline{AG}$. It is less than $\overline{AE}$ by half the sun’s diameter or just $0;16,40º$. Since $\triangle{AEG}$ is a right triangle, we can now solve for $\angle{AEG}$ which is equal to $\angle{EAB}$, which is the angle we’re looking for, since they’re alternate interior angles.  Doing so, I find $\angle{AEG} = \angle{EAB} = 31;02º$.

As a reminder, this is a solar eclipse so we need only consider this one situation since points $1$ & $2$ and $4$ & $5$ are virtually indistinguishable as stated above.

Next, Ptolemy considers the case of a lunar eclipse. Here, we’ll need to consider cases $1$ & $2$ and $4$ & $5$ separately.

We’ll let $A$ represent the center of the earth’s shadow. If we again assume that the moon is at mean distance,

$$\overline{AE} = 0;43,20º + 0;16,40º = 1;00,00º$$

As with before $\overline{AD}$ will always be the radius of the earth’s shadow minus half the moon’s diameter:

$$\overline{AD} = 0;43,20 – 0;16,40º = 0;26,40º$$

In this example, we’ll consider the moon eclipsed by $18$ digits4. This indicates $\overline{AG}$ will be less than $\overline{AD}$ by half the diameter of the moon5

$$\overline{AG} = 0;26,40º – 0;16,40º = 0;10,00º$$

We now have everything we need to calculate the angles we’re searching for in both circumstances. For the moment when the shadow first/last touches the lunar disc, this would be $\angle{BAE}$ which I then find to be $9;36º$.

Similarly, we can find the angle at the first/last moment of totality, $\angle{BAD} = 22;01º$.

As usual, with the examples completed, Ptolemy walks through the layout of the table for which he lays out the completed calculations for other circumstances, but I’ll save that to put alongside the table in the next post!



 

  1. Neugebauer states that, “the moon’s sickle or the moon’s latitude are considered significant factors for storms and weather according to the part of the horizon toward which they are ‘inclined’ since the segments of the horizon are naturally associated with the winds and the weather they bring.” Here, he cites Ptolemy’s Tetrabiblos, which is his his work on astrology.
  2. In contrast, the angle between the celestial equator and horizon is constant for a given observer.
  3. The circles here being either the sun/moon or the earth’s shadow/moon.
  4. If you’ve forgotten how the moon can be eclipsed by more than $12$ digits (i.e., its own diameter), refer to this post.
  5. If this were not a total eclipse, it would be the radius of the shadow plus the uneclipsed portion of the moon, but since it’s more than $12$ digits, it is minus the extra number of digits (which in this case is $6$) so half the diameter.