With the solar eclipse tables complete, we’ll now turn our attention to the lunar eclipse tables. These largely follow the same format and calculations as the solar eclipse tables, but will include an additional column for the “half totality”.
As with before, the table is incremented in terms of digits. However, since the diameter of earth’s shadow is significantly larger than the moon, Ptolemy allows for the number of digits to go past $12$.
The argument in these tables is again, the argument of latitude (i.e., the distance from the node along the lunar circle) and is calculated in the same manner. For example, the first entry in the table is for when the edge of the moon just touches the earth’s shadow. In that case, the distance between their centers is $0;15,40º + 0;40,40º = 0;56,20º$. We can then multiply this by $11;30$ to determine the distance from the nodes. For the first and last entry, it would be $10;48º$.
The minutes of immersion is similarly calculated as we did in the last post. As an example, Ptolemy again uses when the moon is eclipsed by $3$ digits.
This time, Ptolemy begins with a value of $\overline{AB}$ of $0;56,24º$1. There, $\overline{AG}$ is less that by $0;07,50º$ ($\frac{1}{3}$ of the diameter of the moon at apogee) or $0;48,34º$. We then solve for $\overline{GB}$ using the Pythagorean theorem to find it to be $0;28,40º$ which Ptolemy evidently rounds up to $0;28,41º$ and is what gets entered into the table, again expressed in arcminutes instead of degrees.
He then gives another example for the moon at least distance in which he uses a value of $1;03,36º$2 for $\overline{AB}$. Then,
$$\overline{AG} = 1;03,36º – 0;08,50º = 0;54,46º$$
We then use the Pythagorean theorem to find $\overline{GB} = 0;32,24º$. Ptolemy comes up with $0;32,20º$ due to rounding on the intermediate steps and enters that into the table.
Now, we’ll explore the new column and to do so, we’ll construct a new diagram, although quite similar to the one we used for the last post.
Here, we again have the moon moving along its path from right to left. At $B$ it just begins touching earths’ shadow. At $G$ is the first moment of totality and $E$ is the last. Point $Z$ is when the eclipse ends. Our goal will be to find $\overline{DG}$ which Ptolemy refers to as “half of totality” and is equal to $\overline{DE}$. Meanwhile, $\overline{BG}$ is “the immersion.” Finally, $\overline{EZ}$ he calls the “emersion” and will be equal to $\overline{BG}$.
Quite obviously, this will make no sense in any context in which there is not a total eclipse as points $G$ and $E$ cannot exist if it is not. Thus, we will only be concerning ourselves with the value for this column when the number of digits is $12$ or greater.
But what exactly does that mean? I find it easier to understand if we ignore all points except $D$ and move vertically instead of horizontally.
Here, I’ve drawn out what things look like for several types of obscurations and listed the number of digits. For the last, corresponding to $15$ digits (which is what Ptolemy gives as an example here in a moment), is when the moon’s center is $3$ digits inside the shadow towards the shadow’s center. Or as Ptolemy states it, $15$ digits is when
the moon’s centre [at mid-eclipse], lies $1 \frac{1}{4}$ lunar diameters inside the boundary set by the limits of the eclipse.
Similarly, this will mean that our previous definition of the Minute of Immersion doesn’t make sense in this context. Previously, it was from the moment of contact to mid-eclipse, but since Ptolemy takes the diameter of the sun and moon to be equal when the moon is at apogee, this means that this moment is also when the moon completely covers the sun. For this situation, when the moon is within earth’s shadow for an extended period, we’ll need to adjust our definition to be $\overline{GB}$ (the distance from first contact to full immersion).
Now let’s consider a few of the lines in our above diagram starting with $\overline{AB}$. This is the shadow’s radius plus the lunar diameter or $0;56,24º$.
Similarly, $\overline{AG}$ is the shadows diameter minus the lunar diameter or $0;25,04º$3.
Lastly, we can look at $\overline{AD}$ which is the shadow’s radius minus the lunar radius and the number of digits. So for $15$ digits, this would be:
$$0;40,44 – (0;15,40 + 0;07,50º) = 0;17,14º$$
Using these, we can solve $\triangle{ADB}$ using the Pythagorean theorem in which we determine $\overline{DB} = 0;53,42º$.
We can do the same for $\triangle{ADG}$ to find $\overline{DG} = 0;18,12º$. Which is what gets entered in the table for the column in question.
Ptolemy goes on to demonstrate that we can get the immersion, $\overline{GB}$ by subtracting $\overline{DB} – \overline{DG}$ which is $0;35,30º$ and we will put in the column for the Minutes of Immersion when there is a total eclipse.
So that completes the example for apogee. Now we’ll repeat it for perigee again for the case of $15$ digits.
In that case, $\overline{AB} = 1;03,36º$, $\overline{AG} = 0;28,16º$, and $\overline{AD} = 0;19,26º$4.
Solving, we find $\overline{DB} = 1;00,34º$ and $\overline{DG} = 0;20,32º$ (which gets entered into the column for Half Totality) and, by subtraction, $\overline{GB} = 0;40,02º$ (which gets entered in the column for Minutes of Immersion).
In the next post we’ll work on a
convenient way of obtaining the fraction of the difference [between values derived from the first and second tables] for positions of the moon on the epicycle in between greatest and least distances.
- You’re probably wondering why there’s a $4$ arcsecond difference between this value and what I used previously. This is because Ptolemy is a bit loose with the values he uses. The value I cited above of $0;56,20º$ comes from the values at apogee from this post which are based on our calculations. However, in this post, I included in a footnote that Ptolemy sometimes uses a value of $0;56,24º$ which is based on the ratio of earth’s shadow to that of the moon’s. I used the value I did above because it rounds appropriately to $10;48º$ which is the value in the table. But for this calculation, Ptolemy explicitly uses the value cited here and it works out closer to the value he comes up with for the table.
- This disagrees with the values we’d calculated in previous chapters by a $4$ arcminutes for the same reason as my previous footnote.
- Again, we have a $4$ arcsecond discrepancy with how Ptolemy defines the radius of the shadow at apogee in various places.
- Note that in this case, the moon’s diameter is larger and thus, $\frac{3}{12}$ of it will be as well. Thus the amount we subtract out for the digits will be $0;08,50º$.