In the last chapter, Ptolemy walked us through a light proof on how to find the apparent position of a planet. But, immediately in this chapter, Ptolemy tells us that going through that process every time is troublesome:
[A]lthough that method is the only one which provides a fully accurate solution to the problem, it is too cumbersome to be convenient for [astronomical] investigations.
What is Ptolemy’s solution?
As you probably expect, it’s to make tables that allow us to calculate quickly:
We have constructed, for each of the five planets, a table which is easy to use as we could devise, while at the same time being very close to full accuracy.
In this chapter, we’ll follow along as Ptolemy explains how they are laid out.
However, Ptolemy’s description is not well explained, so I will interweave this with a more comprehensive explanation based on Neugebauer’s History of Ancient Mathematical Astronomy.
We have, again, arranged each of the tables in $45$ lines for the sake of symmetry, and we have arranged each in $8$ columns.
Common Numbers – Columns 1 & 2
The first $2$ columns will contain the numbers of the mean positions arranged as for the sun and moon [$III.6$ and $V.8$]: in the first column, the $180º$ beginning from the apogee, from the top down, and in the second, the remaining $180º$ of the [other] semi-circle, from the bottom up, in such a way that the number “$180$” is in the last line in both columns. And the increment in the numbers is $6º$ in the top $15$ [successive] lines, but $3º$ in the $30$ lines remaining below (for the differences between [successive] values for the anomalies remain almost constant for longer stretches near the apogee, whereas they change faster near the perigee).
As usual, these are the “input” columns. When we need to look up a value, we start by finding it in one of these columns and then reading across to the appropriate column.
Correction for Longitude – Columns 3 & 4
As for the next two columns, the third will contain the equations corresponding to the mean position in longitude (each to the arguments on the same line), computed for the greater eccentricity, but under the simplifying assumption that the centre of the epicycle is carried on the eccentre which produces the mean motion [i.e., that of the equant]. The fourth column will contain the corrections to the equations due to the fact that the epicycle centre is carried, not on the above circle, but on another.
The method by which each of these quantities [the equation and its correction], both in combination and separately, can be found geometrically has already been made plain by numerous preceding theorems.
What Ptolemy is saying here is that he won’t be walking us through the calculations for columns $3$ and $4$ because this is something we’ve covered enough.
In this place, since this is a [scientific] treatise, it was appropriate to display this way of separating the zodiacal anomaly, and hence to tabulate it in two columns. However, for actual use, a single column formed by combining these two will suffice.
Here, Ptolemy is telling us that columns $3$ and $4$ could be a single column. Indeed, when looking up values in both of these columns, we use the same value for input, and then we add the values we find together. Thus, we should just be able to save some work and combine these columns. And that’s exactly what Ptolemy does in his Handy Tables. But he doesn’t because, as a scientific treatise, we may want one of these values independently. Thus, preserving it as two columns is useful for such purposes.
But, what are columns $3$ and $4$ actually representing?
To explain, let’s look at a few diagrams from Neugebauer1.
In this figure, we have the center of mean motion (the equant) at $E$ and the observer at $O.$ What’s left off is the center of mean distance on which the circle (only partially displayed with $A$ and $C$ on its circumference) is centered.
The epicycle is carried on $C$ with radius $r$ and the planet being at point $P.$
Previewing where we’re going to be going a bit, we’re obviously going to want to know $\angle COP$ (also shown here as $\alpha$) as this is the anomaly.
To get this, Ptolemy adds two other angles. The first is $\angle FCP$ (also shown here as $\overline{\alpha}$) which is the angle about the epicycle from its true apogee (at $F$) when considered from the point of view of the equant.
This plus $\angle GCF$ which is a vertical angle with $\angle OCE$ (also shown as $\eta$) gives us $\angle GCP$.
We can find $\angle FCP$ from our mean motion tables, but the other angle is what we get from columns $3$ and $4$.
To break it down a little further, let’s look at another diagram from Neugebauer2:
Here, we can see that column $3$ is $\angle ODE$ and column $4$ is $\angle COD.$ Thus, $\angle DCO = 180º – \angle ODE – \angle COD.$
And since $\angle OCE = 180º – \angle DCO$ we can do a bit of substitution to find that $\angle OCE = \angle ODE + \angle COD.$
However, as stated above, Ptolemy could have just reported this angle directly but chose to break it apart. Perhaps it will be useful later. But it certainly isn’t for calculating planetary longitudes (which we’ll explore shortly).
Epicyclic Anomaly – Columns 4, 5, & 6
From the previous columns, we’ve worked out how far the planet would be from its apparent apogee on the epicycle. Now we need to turn that into an actual anomaly. And this is where the eccentricity comes to bite us because the actual anomaly will also be dependent on how far away the epicycle is.
To deal with this, Ptolemy calculates the anomaly for various angles (from the apparent apogee) at three distances of the epicycle: mean, greatest, and least distance.
Each of the next three columns will contain the equations due to the epicycle. These, again, are computed under a simplifying assumption, [namely] that the apogee or perigee of the epicycle is viewed along the line from the observer [to the epicycle centre]. The way in which this kind of demonstration is performed as been made plain by the previous theorems.
The midmost of these three columns (which is the sixth from the beginning) will contain the equations computed for the ratio [of epicycle radius to distance of epicycle centre] at mean distance.
The fifth will contain, [for each argument], the difference between the equation at greatest distance [of the epicycle] and the equation for the same argument at mean distance.
The seventh will contain the differences between the equations at least distance and the [corresponding] equations at mean distance.
The result for the mean he places in column $6$ unchanged.
But, for the greatest and least distances, he does not give the anomaly directly, but instead gives the difference between the anomaly at greatest/least distance and the anomaly at mean distance. The difference in anomaly between mean and greatest distance goes in column $5$ and the difference in anomaly between mean and least distance goes in column $7$.
A Few Reminders
Before explaining the next column, Ptolemy reminds us of a few parameters we’ve found:
For we have shown that, for the following epicycle sizes (from now on it would be best to list [the planets] in order from the outermost):
– Saturn: $6;30^p$
– Jupiter: $11;30^p$
– Mars: $39;30^p$
– Venus: $43;10^p$
– Mercury: $22;30^p.$
Next, he reminds us that the mean distance is, in all cases by definition, $60^p$. Then he examines the greatest and least distances (i.e., $60^p \pm$ the double eccentricity3):
the mean distance, i.e., the distance [equivalent] to the radius of the eccentre which carries the epicycle, is $60^p$ in all cases.
And the greatest distances (with respect to the centre of the ecliptic) are:
– Saturn: $63;25^p$
– Jupiter: $62;45^p$
– Mars: $66;00^p$
– Venus: $61;15^p$
– Mercury: $69;00^p.$The least distances (defined similarly) are:
– Saturn: $56;35^p$
– Jupiter: $57;15^p$
– Mars: $54;00^p$
– Venus: $58;45^p$
– Mercury: $55;34^p.$
Scale Factor – Column 8
Returning back to our train of thought, we still have to contend with the fact that the epicycle is not always at one of the three extremes. At mean distance, we can take column $6$ directly. At greatest, we can reduce that by the value in column $5$. And at least, we can increase it by the value in column $7$.
But at any position between these special cases, we’re going to take the maximum difference (columns $5$ and $7$) and scale them appropriately.
This is what column $8$ is going to be: a scale factor to tell us how much, proportionally of the correction in column $5$ or $7$ we will need to apply.
In Ptolemy’s own words
As for the remaining, eighth column, we provided it in order that one may find the applicable fraction of the above differences [in columns $5$ and $7$] when the planet’s epicycle is not exactly at mean, greatest or least distance, but in an intermediate position. The computation of this correction is based only on the maximum equation ([i.e.,] that formed by the tangent from the observer to the epicycle) at each intermediate distance. For the [fraction] of the difference to be applied for any particular position [of the planet] on the epicycle is not significantly different from that for the greatest equation.
Ptolemy realizes this comment is rather opaque, so provides clarification:
But, in order to make our meaning clearer, and to explain the actual method of computing the [fractions] to be applied, let us draw [the following diagram].
In this, we have the line of apsides being $\overline{AD}$. Point $B$ is the center of mean motion (i.e., the equant) and $G$ the observer on Earth at the center of the ecliptic.
We have the epicycle centered on $E$ and have extended $\overline{BE}$ continuing it until it hits the opposite side of the epicycle at $Z$. We’ve also extended $\overline{GH}$ which is tangent to the epicycle at $H$ making $\overline{HE}$ perpendicular to this.
As I said above, this is the only part where Ptolemy shows any work4. But, as with previous tables, Ptolemy only shows one calculation as an example. So for this example, we’ll examine case in which the center of the planet’s epicycle ($E$) is $30º$ from apogee. This is $\angle ABE.$
NOTE: For the rest of the calculations, Ptolemy works through all five planets simultaneously.
Our goal here is going to be to determined $\angle EGH$ for various cases, starting with our example in which $\angle ABE = 30º.$
Example Angle
Our first step will be to determine $\overline{GE}$.
Ptolemy doesn’t explicitly tell us how to do this. Rather, he reminds us that this is essentially something we’ve done before, so interested readers can refer back to previous examples.
However, we should also note that because Mercury has the additional component to its model, the calculations we’re about to walk through do not work for Mercury. Thus, I’ll skip Mercury for now5, but walk through the general steps of the calculations for the other four planets. However, to avoid having to type it all out here, I’ve again done the calculations in a Google Sheet. I’ll just show the results of the steps that Ptolemy shows, and compare our results.
First, I’ll extend $\overline{ZB}$ until it meets a perpendicular dropped from $G$ at $\Theta.$ I’ll also drop a perpendicular onto this line from $K$ at $L$.
Since $\angle ABE = 30º$ the vertical angle $\angle KBL$ does as well.
We’ll now solve $\triangle KBL,$ but to save some time, I’ll just use some modern trig. However, we’ll need to know at least one side of this triangle.
Fortunately, for each of the planets, we calculated the eccentricity, i.e., $\overline{BK}$.
Once that small triangle is solved for $\overline{KL}$ and $\overline{BL}$ we can determine $\overline{EL}$. We’ll then add $\overline{L \Theta}$ which is equal to $\overline{BL}$ to that to find $\overline{E \Theta}$. Then, $\overline{G \Theta} = 2 \cdot \overline{KL}$ which gives us two of the side of $\triangle EG \Theta$, so we can solve for $\overline{GE}$.
Doing so, I find
– Saturn: $63;02^p$
– Jupiter: $62;26^p$
– Mars: $65;25^p$
– Venus: $61;06^p.$
Ptolemy writes each of these as a ratio to the radii of their epicycles.
From there, we can easily calculate $\angle EGH$ (the maximum amount of anomaly when the center of the epicycle is $30º$ away from its apogee) since we now know two sides of this triangle.
Doing so, I find
– Saturn: $5;55,09º$
– Jupiter: $10;36,55º$
– Mars: $37;08,48º$
– Venus: $44;57,27º.$
To which Ptolemy does some creative rounding (especially for Venus):
– Saturn: $5;55,30^p$
– Jupiter: $10;36,30^p$
– Mars: $37;09^p$
– Venus: $44;56,30^p.$
Mean Distance
Next, Ptolemy calculates the greatest anomaly at mean distance, which is simply when $\overline{GE} = 60^p.$ For this, I get:
– Saturn: $6;13,09º$
– Jupiter: $11;03,00º$
– Mars: $41;10,22º$
– Venus: $46;00,31º.$
Ptolemy’s rounding is a bit better this time, although he’s rounded Venus down oddly:
– Saturn: $6;13º$
– Jupiter: $11;03º$
– Mars: $41;10º$
– Venus: $46;00º.$
Greatest Distance
Ptolemy’s next demonstrations are for when the epicycle is at its greatest distance. This is when it is at $60^p$ plus the eccentricity for each of the planets. Calculating this, I find:
– Saturn: $5;52,58º$
– Jupiter: $10;33,36º$
– Mars: $36;45,02º$
– Venus: $44;48,37º.$
Ptolemy, after his rounding, for which he’s strangely rounded Venus down again:
– Saturn: $5;53º$
– Jupiter: $10;34º$
– Mars: $36;45º$
– Venus: $44;48º.$
Least Distance
Then, at least distance which is $60^p$ minus the eccentricity:
– Saturn: $6;35,47º$
– Jupiter: $11;35,16º$
– Mars: $47;01,46º$
– Venus: $47;17,10º.$
And after Ptolemy’s rounding in which Mars has been rounded down instead of up:
– Saturn: $6;36º$
– Jupiter: $11;35º$
– Mars: $47;01º$
– Venus: $47;17º.$
Difference between Mean and Greatest Distance
Ptolemy then calculates the difference between the planet at mean and greatest distance. Again, I find:
– Saturn: $0;20,10º$
– Jupiter: $0;29,23º$
– Mars: $4;25,20º$
– Venus: $1;11,53º.$
And Ptolemy, after rounding (accurately this time):
– Saturn: $0;20º$
– Jupiter: $0;29º$
– Mars: $4;25º$
– Venus: $1;12º.$
Difference Between Mean and Least Distance
Ptolemy then does the same for the difference between mean and least distance. My results:
– Saturn: $0;22,37º$
– Jupiter: $0;32,16º$
– Mars: $5;51,24º$
– Venus: $1;16,39º.$
And after Ptolemy’s rounding (again, accurate):
– Saturn: $0;23º$
– Jupiter: $0;32º$
– Mars: $5;51º$
– Venus: $1;17º.$
Difference Between Epicycle @ $30º$ from Greatest and Mean Distance
We then take the difference between the anomaly when the planets are $30º$ from apogee and at their mean distance. My values:
– Saturn: $0;17;59º$
– Jupiter: $0;26,04º$
– Mars: $4;01,33º$
– Venus: $1;03,03º.$
Ptolemy, after some additional creative rounding:
– Saturn: $0;17,30º$
– Jupiter: $0;26,30º$
– Mars: $4;01º$
– Venus: $1;03,30º.$
Ratio Between Mean and Greatest Distances
This time, Ptolemy’s math is a little different. Let’s first look at how he describes it. He tells us that we will find the values
expressed as sixtieths of the above total difference between [the equations for] mean and greatest distance.
This is… not particularly clear, but what Ptolemy is doing here is the following (using Saturn as the example):
$$\frac{0;17,30^p}{0;20^p} = \frac{x}{60}$$
Here, the $0;17,30º$ is the difference between the maximum anomaly at greatest (apogee) and mean distance; the $0;20º$ is the difference between the maximum anomaly mean and least distance (perigee).
Then, Ptolemy expresses the result in sixtieths. Doing this (using my unrounded values), I find:
– Saturn: $53;29,31$
– Jupiter: $53;13,09$
– Mars: $54;37,24$
– Venus: $52;37,23.$
Ptolemy’s values are notably different after all the consecutive rounding:
– Saturn: $52;30$
– Jupiter: $54;50$
– Mars: $54;34$
– Venus: $52;55.$
So, those are the values, in sixtieths, which we put in the $8^{th}$ column of the appropriate table, on the line containing the number $30$ for the mean motion in longitude.
These are the scale factors (expressed in sixtieths) we’ll apply to the adjustments from columns $5$ and $7$ to determine the final anomaly.
Ptolemy then tells us he repeated this calculation for each interval for his tables of anomaly which I’ll lay out in the next post.
- This is $Fig. \; 184$ from his History of Ancient Mathematical Astronomy.
- This is $Fig. 185$ from HAMA.
- I’ve been somewhat sloppy about how I refer to the eccentricity. Specifically, the eccentricity is the distance between each of the centers. In other words, the distance between the observer and the center of mean distance, which is equal to the distance between center of mean distance and the equant. Thus, the double eccentricity is the distance between the observer and the equant.
- Even though he doesn’t show all steps.
- Hopefully I’ll come back to it later.