Previously, when discussing eclipses, we’ve discussed the amount that is obscured in terms of “digits” where each digit is $\frac{1}{12}$ of the diameter of the object. However, Ptolmey indicates that not everyone necessarily estimates the magnitude of eclipses in this way, stating,
most of those who observe [eclipses]… measure the size of the obscuration, not by the diameters of the disks, but, on the whole, by the total surface area of the disks, since, when one approaches the problem naively, the eye compares the whole part of the surface which is visible with the whole of that which is invisible1.
To deal with this Ptolemy provides “another little table” which will allow us to convert between the linear diameter obscured and the area of either the sun, or moon.
The first column in this table will have the number of digits, from $1-12$ representing the linear obscuration. The next column will represent the area of the sun obscured, again in $12^{th}$s so that $\frac{1}{12} \approx 8.33…\%$ but this time it will be area digits.
We computed these amounts only for the sizes [of the apparent diameters] for the moon at mean distance, since very nearly the same ratio will result [at other distances], given so small a variation in the diameters. Furthermore, we assumed that the ratio of the circumference to the diameter is $3;08,30 : 1$ since this ratio is about half way between $3 \frac{1}{7} : 1$ and $3\frac{10}{71} : 1$ which Archimedes used as rough [bounds]2.
Having now clued us in on some of the numbers he intends to use, Ptolemy creates the following diagram to represent solar eclipses:
In it, $circle \; ABGD$ is the sun’s disc on center $E$. The other circle, $AZGH$ on center $\Theta$, is the disc of the moon, intersecting at $A$ and $G$. We’ll also draw a line through both of their centers, $\overline{BE \Theta H}$. Points $Z$ and $D$ are added where this center line intersects the limbs.
With this much drawn in, we can now get an understanding of what Ptolemy is wanting to achieve with the calculations we’re about to jump into. For this worked example, we’ll assume that $\frac{1}{4}$ of the sun is eclipsed, indicating $\overline{ZD} = 3$ digits when it’s the sun’s diameter being measured as having $12$ digits. What we’re wanting to determine is the area of shape $AZGD$ when the total area of the sun, circle $ABGD$ is $12$ area digits3.
In the case that the sun’s diameter is $12$ linear digits, the moon’s diameter, $\overline{ZH} \approx 12;20$ linear digits. How does Ptolemy come up with this?
Well, the sun’s radius is always taken to be $0;15,40º$ whereas the moon’s (at syzygy) varies between $0;15,40º$ and $0;17,40º$4. Thus, at the mean distance Ptolemy told us he was going to use above, the moon’s radius is is $0;16,40º$. Taking the ratio of the moon to the sun and using that to determine the diameter of the moon in the same units as the sun (digits where one digit is $\frac{1}{12}$ of the solar diameter) we come up with a lunar diameter of $12;46$ digits. Not $12;20$ as Ptolemy describes. According to Toomer, this is a well known error on Ptolemy’s part that even Pappus commented on. While this incorrect value will impact every calculation in the table, ultimately, Toomer notes, “the results are so crudely rounded that it is of little importance.”
Adopting Ptolemy’s values for now, we can then state that $\overline{E \Theta} = 9;10$. This isn’t immediately obvious, but to understand, first imagine the two circles as just touching and not overlapping as drawn. In that case, the sun’s radius is $6$ digits while the moon’s is $6;10$ digits for a total distance of $12;10$ digits. Then we allow it to overlap by $3$ which gets us to $9;10$.
Ptolemy then calculates the circumference of each body5. For the sun he finds it to be $37;42$ digits and for the moon, $38;46$ digits. Then, he multiplies this value by $r$ again, thus having twice the actual area at this point since this effectively works out to $2 \pi r^2$. Thus, dividing by $2$ gets it back to the area. For the sun he then comes up with $113;06$ square linear digits and for the moon, $119;32$ square linear digits. This gives us our basic parameters we need to proceed.
We’ll now add a few lines to our initial drawing by connecting points $A$ and $G$ to $E$ and $\Theta$ as well as connecting $A$ to $G$ which forms a perpendicular to the line drawn through the centers of the two bodies at $K$.
Now, $\overline{AE} = \overline{EG} = 6$ linear digits each since they are both diameters of the sun. Similarly, $\overline{A \Theta} = \overline{\Theta G} = 6;10$ linear digits since these are radiuses of the moon.
Now let’s consider right triangle $\triangle{AK \Theta}$. Using the Pythagorean theorem, we can state,
$$\overline{AK}^2 + \overline{K \Theta}^2 = \overline{A \Theta}^2$$
Similarly for $\triangle{EAK}$ we can state
$$\overline{EK}^2 + \overline{AK}^2 = \overline{AE}^2$$
We can then subtract the second of these from the first, getting $\overline{AK}^2$ to cancel out and leaving us with
$$\overline{K \Theta}^2 – \overline{EK}^2 = \overline{A \Theta}^2 – \overline{AE}^2$$
Factoring the left side of this equation gives us
$$(\overline{K \Theta} + \overline{EK})(\overline{K \Theta} – \overline{EK}) = \overline{A \Theta}^2 – \overline{AE}^2$$
We can then note that this first term, $\overline{K \Theta} + \overline{EK} = \overline{E \Theta}$. Thus, the equation can be rewritten as
$$\overline{E \Theta} \cdot (\overline{K \Theta} – \overline{EK}) = \overline{A \Theta}^2 – \overline{AE}^2$$
If we then divide both sides by $\overline{E \Theta}$, we end up with
$$(\overline{K \Theta} – \overline{EK}) = \frac{\overline{A \Theta}^2 – \overline{AE}^2}{\overline{E \Theta}}$$
We have all the values for the right side of this equation, so we can substitute in to determine $\overline{K \Theta} – \overline{EK} = 0;13,16$ linear digits implying that $\overline{K \Theta}$ is $0;13,16$ linear digits greater than $\overline{EK}$. Since the sum of these two segments is $9;10$ linear digits, we can determine that $\overline{EK} = 4;28,22$ linear digits and $\overline{K \Theta} = 4;41,38$ linear digits.
Next, consider $\overline{AK}$ and $\overline{KG}$ which are part of triangles $\triangle{EAK}$ and $\triangle{EKG}$ respectively. Both of these triangles share $\overline{EK}$ in common and their hypotenuses are both $6$ digits which means that the third side of each, $\overline{AK}$ and $\overline{KG}$, must also be equal. We can quickly use the Pythagorean theorem to determine $\overline{AK} = \overline{KG} = 3;59,57 \approx 4;00$ linear digits.
Ptolemy then considers the areas of triangles $\triangle{AEG}$ and $\triangle{A \Theta G}$. Each one of these is twice the area of $\triangle{EAK}$ and $\triangle{EK \Theta}$ respectively. Calculating, I get the area of $\triangle{AEG} = 17;53,28$ square linear digits6 and $\triangle{A \Theta G} = 18;46,32$ square linear digits7.
We’ll set the question of areas aside for a moment and again consider the circles and segments. In our setup, we had the diameter of circle $ABGD = 12$ linear digits. For circle $ZH$ it was $12;20$ linear digits. When that’s the case, $\overline{AG} \approx 8$ linear digits. If we now switch from digits to parts, the diameter of circle $BD = 120^p$ in which case $\overline{AG} = 80^p$.
We can also ask what happens if we let $\overline{ZH} = 120^p$. In that case, $\overline{AG} = 77;50^p$.
In both cases, $\overline{AG}$ is a chord in each of the circles and as such, we can look up the corresponding chords. This gives us $arc \; ADG = 83;37º$ when the diameter of its circle is $120^p$ and $arc \; AZG = 80;52º$ when the diameter of its circle is $120^p$.
Next Ptolemy lays a theorem on us:
the ratio of a circle to one of its arcs equals the ratio of the area of the whole circle to the area of the sector beneath that arc.
In other words, the proportion of a wedge taken out of a whole circle, is the same as the arc for that wedge to the circumference. Our ratio for the sun, circle $ABG$, is $83;37º : 360;00º$. If we multiply that ratio by the area of the sun we derived above ($113;06$ square linear digits) we get the area of sector $AEGD = 26;16$ square linear digits.
Similarly, for the moon, our ratio would be $80;52º : 360;00º$. Multiplying by the area of the moon from above, $119;32$ square digits, we get the area of sector $A \Theta GZ = 26;51$ square linear digits.
Within each of these wedges, we have above calculated the triangle formed within them and their chords. Thus, we can subtract $\triangle{AEG}$ from wedge $AEG$ and determine the area of the little bit out near the edge, $ADGK = 26;16 – 17;52 = 8;24$ square linear digits. Similarly, $AZGH = 26;51 – 18;48 = 8;03$ square linear digits.
These two little slices can then be added together to determine the area of shape $AZGD$ which will be the area eclipsed in terms of linear digits. In this case, that’s $16;27$ square linear digits squared being eclipsed of a possible $113;06º$ (for the sun). If we instead consider the area of the sun to be $12$ area digits, the amount eclipsed would then be $1;44$ area digits which Ptolemy rounds to $1 \frac{3}{4}$.
This is what we set out to demonstrate: When the sun is eclipsed by $3$ linear digits, the area eclipsed is $\approx 1 \frac{3}{4}$ area digits which is what will get entered into the table we’re working on.
I’ll quit here because this post is already getting quite long, but in the next post, I’ll walk through Ptolemy’s example on performing this same calculation for a lunar eclipse.
- Toomer notes he has not reason to doubt Ptolemy’s statement that other astronomers measure eclipses in this way, but there is no extant evidence of this being the case.
- A rare citation of anyone other than Hipparchus! Here, Ptolemy is referring to Archimedes “Measurement of the Circle”.
- This term of “area digits” is one I’m inventing here because there is some serious sloppiness in this discussion. In particular, Toomer’s translation consistently refers to everything as “parts” with his normal abbreviation of a superscript p. I’m not certain if this is his doing or Ptolemy’s but for certain, it’s not really correct as our previous definition for parts were when the radius of a circle is $60$ parts. Since that’s not the case here, I’m being more careful. However, even in more appropriately referring to the divisions as digits, I still want to be careful because, as we’ll see, even the definition of “digits” will get fuzzy towards the end of the calculation. Specifically, the area digits I’m referring to here do not correspond to $1$ square linear digits as there are $\pi \cdot 6^2 \approx 113$ square linear digits in the area of the body, but still only $12$ area digits. Without being consistent about which kind of digits we’re talking about, it’s easy to get confused.
- I always refer to this post which has a handy table detailing this for the moon.
- As a reminder, the formula for the circumference of a circle is $2 \pi r$.
- Ptolemy has $17;52$ here.
- Ptolemy gets $18;48$.