1 of 35

Here Comes the Sun

The Ptolemaic Solar Model

2 of 35

Solar Phenomena

The motion of the sun along the ecliptic varies throughout the year (slower in summer, faster in winter)

If the sun’s position in the sky at noon is plotted throughout the year, it makes a pattern known as the analemma
The length of the time between solstices and equinoxes in unequal

Start	End	Duration
Vernal Equinox	Summer Solstice	92 days, 18.5 hours
Summer Solstice	Autumnal Equinox	93 days, 15.5 hours
Autumnal Equinox	Winter Solstice	89 days, 20 hours
Winter Solstice	Vernal Equinox	89 days, 0 hours

There are a few notable observations related to the sun that any solar model will need to be able to explain.

The key observation here is that the sun moves faster, in relation to the background stars, at different times of the year. It moves fastest in winter and slowest in summer. We can’t see this directly, since the sun’s brightness blocks out the background stars against which we’d measure, but we can infer this from two other observations.

The first is that, if you were to plot the position of the sun in the sky throughout the year, it would make a funny figure eight shaped pattern known as the analemma. While the vertical component of this motion is easily explicable by noting that the sun’s path is tilted relative to the celestial equator (thus bringing the sun higher and lower throughout the year), the horizontal component is explained by the sun’s motion speeding up and slowing down. If the sun’s motion were perfectly uniform, then it would be directly south every day at noon. But it’s not. Sometimes it gets ahead of that average or mean motion, and sometimes it falls behind.

Another observation that we can make is that the length of time between solstices and equinoxes is not equal. In the year 2000, the time between the vernal equinox and the summer solstice was 92 days and 18.5 hours. From the summer solstice to the autumnal equinox was 93 days and 15.5 hours. From the Autumnal Equinox to the winter solstice was 89 days and 20 hours. And from the Winter Solstice to the 2021 vernal equinox was 89 days and 0 hours.

3 of 35

Platonic philosophy assumes heavenly objects move in perfect circles at perfectly uniform speeds about them.
Ptolemy argues that the Earth is the center of the cosmos.

Building a Model:

Philosophical Assumptions

But before starting on building a model, astronomers would need to include a few key assumptions. The first one came from Plato who argued that, while things here on Earth are corrupted and chaotic, things in the heavens must be perfect and ideal. Thus, he stated, objects in the cosmos must move in perfect circles which rotated at perfectly consistent speeds.

The second important assumption was that the Earth is the center of the cosmos and does not move about. Although there is evidence that not all astronomers agreed with this, Ptolemy argues for this using a few lines of reasoning.

The first is that, if the earth moved, then it would naturally get closer and further to some stars than others, making them appear brighter or fainter depending on the distance and meaning that we would see less than half of the celestial sphere at any given time. No such observation was made, so Ptolemy concluded the Earth was stationary.

Second, he argued that things fall “down”. Ptolemy defined “down” as “away from the celestial sphere” and thus, he argued, if something passed the center of the cosmos, then it would be falling “up” and should again try to fall “down”. Therefore, even if we did displace the Earth from the center of the cosmos, it would naturally return to its position at the center.

With these assumptions in hand, Ptolemy was ready to start considering various models that would explain the observations. There are two models he considered.

4 of 35

Epicyclic Model

“Circle on a circle”
The larger circle (centered on E) always rotates counter-clockwise at the mean speed of the object
The smaller circle (centered on M) could rotate either direction, as necessary, with the speed equal to the anomaly
On the portion of the small circle the object is moving the same direction as the large circle, it appears ahead of the mean. When it is moving the opposite direction, it appears behind.
Naturally creates points which the object is closest to the Earth (Perigee @ P) and furthest (Apogee @ A)

The first model, and most frequently known, is the Epicyclic model. This model uses two circles. A large circle (known as the deferent), here centered on the Earth at E, would rotate with the average period of the object. For the sun, this means once per year. Since all objects that move with respect to the background stars do so going right to left as viewed from our skies, it means that this circle must rotate counter-clockwise in this view.

At the edge of that circle, was another, smaller circle, known as the epicycle and carried the actual object. This one would be allowed to rotate either direction, depending on the object. Its speed was tied to what was known as the anomaly. “Anomaly” is just a short way to say the motion getting ahead of and behind the average motion we just discussed. When that period of getting ahead and falling behind starts over again, that is the period of anomaly at which the smaller circle rotates. For the sun, its period of getting ahead and falling behind exactly matches the period of the average motion: 1 year.

On the portion of the circle over which the object was moving the same direction as the larger circle, these motions would add together making the object appear to move faster. On the other half of the circle, the motions would be in opposite directions which would make the object appear to slow down. If the smaller circle rotated fast enough, the object could appear to be moving backwards – a phenomena which we observe for the planets, but not the sun.

This model also naturally creates two important points around the epicycle in which the object is closest to the Earth (known as Perigee at P) and furthest from (Apogee at A).

5 of 35

Eccentric Model

6 of 35

Eccentric Model

“Off center circle”

7 of 35

Eccentric Model

“Off center circle”
Naturally creates points which the object is closest to the Earth (Perigee @ P) and furthest (Apogee @ A)

8 of 35

Eccentric Model

“Off center circle”
Naturally creates points which the object is closest to the Earth (Perigee @ P) and furthest (Apogee @ A)

9 of 35

Eccentric Model

“Off center circle”
Naturally creates points which the object is closest to the Earth (Perigee @ P) and furthest (Apogee @ A)
From Apogee to Perigee (measured counter-clockwise) the sun appears to lag the position it would otherwise have if not offset
From Perigee to Apogee (measured counter-clockwise) the sun appears ahead of the position it would otherwise have if not offset
At Apogee and Perigee, the apparent position matches the position it would otherwise have if not offset

10 of 35

Equivalence of Models

In III.3, Ptolemy shows that both models can produce identical results depending on how they are set up.
Ptolemy adopts the eccentric model although there is no preference for either.

11 of 35

Setting Up the Models

Key components:

Mean position: Where the sun would be if it didn’t speed up and slow down
Anomaly: The offset caused by the sun speeding up and slowing down compared to the mean position

Key definition:

Epoch Date: The “start date” for the model

Key parameters:

Location of Apogee
Distance between Earth and center of sun’s Eccentre
Length of a Year
Position of sun on Epoch date

But before jumping in to setting up the model, I’ll preview how we’re going to go about things.

First, we’ll determine two key parameters: the location of

The position of apogee - This is important because, as we just discussed, it’s our reference point for the sun being ahead of and behind its mean position.
The position of the sun on a specific date - This is done from observation.

Then, using the value of the length of a year, we can “rewind” the model to calculate where the sun was on a certain date in the past which we’ll use as a consistent start date to calculate from once the model is configured. This sort of start date is known as the epoch date. While we could choose any date we want, it’s generally best to choose an important date that everyone will understand. For modern astronomical models, we typically use midnight GMT on January 1, 2000. This is known as the J2000 epoch.

Ptolemy chose for his epoch date, noon on the first day of the Egyptian month in which the reign of Nabonassar began (February 26, 747 BCE in our modern calendar).

However, because Ptolemy’s model isn’t perfectly accurate, it tends to drift. Rather notably after about a century. This was well known to astronomers throughout period who frequently had to do the mathematical equivalent of turning it off and turning it back on again by recalculating it just as we are about to do here.

In short, we’ll be following Ptolemy’s methods to rederive the same model but calibrated for use in the present. Specifically, following Ptolemy’s methods, I have chosen a “start date” from which we can determine the interval of time and, applying the sun’s speed as a constant, we can easily determine how far it has moved. For my epoch date, I chose noon on the day of the founding of the SCA, May 1, 1966. So, let’s start.

12 of 35

Determining the Key Parameters

Earth at E
Equinoxes & Solstices

Vernal Equinox: A
Summer Solstice: B
Autumnal Equinox: G
Winter Solstice: D

Which quadrant does the center of the sun’s circle belong in?

13 of 35

Determining the Key Parameters

Sun centered in quadrant between summer solstice and autumnal equinox at Z
Arc on this circle in each quadrant will directly correspond to proportion of time of a year between equinoxes and solstices.

We can determine this by looking back at length of time between successive equinoxes and solstices. Since the sun’s motion about this circle is constant, the amount of time will correspond directly to the length of the path in each quadrant.

The longest amount of time was between the summer solstice and the autumnal equinox, so we should put the center of the sun’s circle in this quadrant. Here, I’ve vastly overstated this to make things easier to see as we go.

As a quick note, this is only true for present day. In Ptolemy’s time, the center of the sun’s path was actually in the lower left quadrant between the autumnal equinox and winter solstice due to a phenomenon known as precession of the equinoxes.

Immediately, by doing this we can tell that the apogee will be somewhere in this quadrant as well, but we’ll need to know precisely where. So let’s begin by adding a few points to help us with this.

14 of 35

Determining the Key Parameters

Here, we’ll add the points where the sun’s path intersect the lines for the division of the zodiac for the equinoxes and solstices.

This allows us to state that arc M-Theta is the portion of the path between the vernal equinox and summer solstice, Theta-K, between the summer solstice and autumnal equinox, K-L between the autumnal equinox and winter solstice, and LM between the winter solstice and vernal equinox.

The proportion of each of these arcs of 360º will be exactly the same as the proportion of each of these periods to a full year.

In other words, we said the time between the vernal equinox and summer solstice we said was 92 days, 18.5 hours (92.77 days). If we divide that by the length of a year, 365.246 days, we get that this period is 25.399% of a year. Taking that same proportion of 360º of a circle, we find that arc M-Theta = 91.44º.

Repeating the same calculation for the other arcs, we find that arc Theta-K is 92.3º, arc KL is 88.54º, and arc LM (which we won’t actually need), is 87.72º.

Next Ptolemy creates several triangles.

15 of 35

Determining the Key Parameters

Let’s first look at arc Theta-L around this circle, going counter-clockwise. This big arc is the sum of arcs Theta-K and KL. Since we know both of these, we can add them together to find that their sum is 180.84º. However, now look at arc NO, again counter-clockwise. Because this divides the circle in half, it must be 180º. This means that the sum of the two little arcs, Theta-N, and OL, must be the difference or 0.84º. And since these two arcs are equal, each one must be 0.42º.

Let’s concentrate on the upper one of these. Specifically Theta-N, and the angle it subtends, angle NZ-Theta. As a general geometry rule, the angle at the center of the circle, is equal to the arc it subtends. Thus, angle NZ-Theta is 0.42º as well.

Now, in modern math, we have the concept of a unit circle. These circles have a radius of 1 unit without the unit of length being particularly important. Ptolemy has the same concept, but instead of a radius of 1, he gives them a radius of 60 and calls them parts.

Thus, if we consider the sun’s eccentre one of these Ptolemaic unit circles, its radius is 60 parts. Thus, line Z-Theta = 60p.

We’ll now create right triangle TZ-theta and in this triangle, we know the hypotenuse, Z-Theta = 60, and the angle, TZ-Theta = 0.42º. Thus, we can solve this triangle for any other component we’d like.

To make things easier, I’ll solve this using a bit of modern trigonometry, but as a reminder, Ptolemy didn’t have this available, so he used a geometrical form of proto-trigonometry which I explored in the Mathematics of the Almagest class previously.

But using some SOH CAH TOA, we can determine that line T-Theta = 0.44 parts.

This is helpful because line XE is the same length. Why is that important? Well, looking ahead, this is part of triangle ZXE and the angle XEZ in it, will tell us where the apogee, H, is in angular distance from G, the autumnal equinox. Furthermore, that triangle also contains line ZE which is the distance between the center of the sun’s circle, and the earth. In short, that little triangle gives us both of the key parameters we need to set up the model.

And now that we’ve found one part of that triangle, if we find another, we can solve that whole triangle to get everything we need. Fortunately, that’s quite easy to do using a very similar trick to the one we just did.

16 of 35

Determining the Key Parameters

17 of 35

Determining the Key Parameters

We can now determine angle XEZ, which will tell us how far before the autumnal equinox apogee is. Recalling that the tangent is the opposite over the adjacent legs or ZX over XE, we can determine that the tan of ZEX is 4.48, and then taking the inverse tangent of that we find that <ZEX is 77.41º.

Thus, the sun’s apparent position the apogee, H, from the autumnal equinox, G, is 77.41º or 102.59º after the vernal equinox, at A.

Similarly, we can also use this triangle to determine the distance between the earth and the center of the sun’s circle, line EZ, simply using the Pythagoren theorem.

Doing so, we quickly find that ZE is 2.02 parts. Making some serious progress!

With this information in hand, we’re ready to start asking the question of how moving the sun’s eccentre off the Earth will impact the anomaly in a quantitative sense. To do so, we’ll start with a fresh diagram.

18 of 35

Effect of the Eccentricity on the Anomaly

In this new diagram, I’ve removed the equinoxes and solstices since we’re finished with them. Instead, since we now know where the apogee is, we’ll do everything in reference to this. This is because, as we showed qualitatively earlier, the apogee and perigee are where the anomaly flips from being ahead to behind and vice versa. Thus, it’s a convenient reference point.

Here, the sun’s apparent position at apogee and perigee are A and G respectively with their true positions on the eccentric circle as E and H respectively.

The earth is at the center of the celestial sphere at D, and the sun’s center at Theta.

We’ll take a scenario in which the sun is at Z, so some distance after the apogee. Then from Earth, its apparent position on the celestial sphere would be at B.

Lastly, we’ve extended line Z-Theta until it makes a right angle with a line extended from D which will come in handy shortly.

Here, let’s consider what happens when the sun’s true distance after the apogee is 30º. Putting that in math terms, arc EZ = 30º. In turn, that means that angle E-Theta-Z is also 30º.

What we want to calculate is the anomaly: how much ahead of or behind, the sun appears to be compared to that 30º, when viewed from Earth?

Without getting into the details, Ptolemy proves in III.3 of the Almagest that this can be determined by looking at angle DZ-Theta. So that’s what we want to solve here.

To figure that out, we’ll first look at the little right triangle, Theta-DK. In that triangle, recall that line Theta-D is the distance we just found between the Earth and the center of the sun’s eccentre or 2.02 parts. Furthermore, angle D-Theta-K is a vertical angle from angle E-Theta-Z which we said is 30º. Therefore, this angle is also 30º.

With these two parts in hand, we can solve for anything else in that right triangle. So, we’ll solve for line Theta-K using a bit of trig. Doing so we find it is 1.75 parts.

Then, using the Pythagorean theorem, we can quickly determine line DK = 1.01 parts.

19 of 35

Effect of the Eccentricity on the Anomaly

Now, let’s look at the bigger triangle, ZDK. In this, we can determine the length of line ZK as it’s the sum of Z-Theta and Theta-K. We just determined Theta-K to be 1.75 parts and Theta-Z is a radius of the circle, so 60 parts. Thus, the total length is 61.75 parts.

Also, line DK is part of this triangle, which means we now know two sides, so we can solve for any angle we want, including the one we were asking about, DZ-Theta. Again, applying a bit of trig, we find that angle DZ-Theta is 0.94º meaning the sun would appear 0.94º away from its true position about its eccentric circle when viewed from Earth.

But in which direction? Recalling from earlier, we remember that from apogee to perigee (going counter-clockwise), the sun appears to lag. Thus, the sun would appear to be behind the true position of 30º from the apogee by 0.94º. In other words, it would look like it’s 29.06º from apogee when viewed from Earth.

Ptolemy repeats this calculation for numerous angles from apogee (not just 30º) and collects a list of the values for the anomaly in a table.

As a note, recall that the anomaly depends on the distance between the Earth and the center of the sun’s eccentre, these values change over time. We found it to be 2.02 parts. However, in Ptolemy’s time it was 2.49 parts. Thus, the values Ptolemy calculated for his time would be different. As such, I’ve recalculated the table and a copy is available in the handout. If you’re not grabbing a copy of the handout but want one, you can find them in the Class Materials section of my website.

Taking stock of where we are, we now know by how much the center of the sun’s circle is removed from the Earth and how much that affects where the sun appears to be.

However, there’s one final step we’ll need to do before the model is complete. We’ll need to determine the epoch position. Since it’s in the past, this isn’t something we observe directly, but instead, will reverse-calculate using our model using an observation from the present.

Sadly, there’s a complication. As we’ve discussed, the sun’s apparent position is a combination of two factors: It’s true position (viewed from the center of its own circle) +/- the adjustment due to the anomaly. Thus, we’ll need to disentangle these two portions before we can do that so we know the sun’s true position about its own circle before we can rewind things.

This would be most easily done by simply observing the sun at apogee or perigee where the anomaly is zero. But determining the coordinates of the sun in such a manner is not something Ptolemy could do.

Instead, he leans on something we just looked at – a moment in time at which we already know the position of the sun: The autumnal equinox.

20 of 35

Effect of the Eccentricity on the Anomaly

Here, I’ve drawn the configuration for the sun at the autumnal equinox. Recall that, a few steps ago, we determined the angular distance of apogee from the autumnal equinox as viewed from the Earth. We determined this to be 77.41º so arc AB is 77.41º.

Quickly going over the points, A is the position of apogee on the celestial sphere, E is its true position, Theta the center of the sun’s eccentre, and D the Earth. The sun is at Z and its apparent position on the celestial sphere is B. Lastly, we’ve created another handy little right triangle by extending a line from Theta onto line ZD so that it’s perpendicular where it meets, at point K.

What we’re really after here is either arc EZ or angle E-Theta-Z as this will tell us where the sun was in the context of its own circle at that moment, effectively removing any effect of anomaly.

To find that, we’ll first look at the little right triangle we made, triangle KD-Theta. In it, we know angle Theta-DK as it’s the same as the arc it subtends, arc AB, or 77.41º. Furthermore, line Theta-D is the distance between Earth and the center of the sun’s eccentre which we already determined to be 2.02 parts.

Thus, we can solve this for K-Theta. Doing so, we find it is equal to 1.97 parts.

Next, we can look at the larger right triangle, triangle ZK-Theta. We just determined one of the legs in this triangle and we know that line Theta-Z is a radius of the sun’s eccentre, and thus has a length of 60 parts. So again, we can use some trig to solve for any other part. So, we’ll solve for angle Theta-ZK which we find to be 1.88º.

As noted previously, this is the anomaly. Thus, the sun’s true position is not 77.41º away from apogee, but 77.41º +/- 1.88º. Which is it? Again, since the position of the sun is after apogee but before perigee, the anomaly’s effect was subtractive. Thus, to remove it, we add. Thus, the sun’s true position on its own circle was 79.29º which is angle E-Theta-Z.

Before proceeding, let’s take a brief moment of what this all means.

21 of 35

Solar Model: Sept 22, 2000�@11:30am

22 of 35

Mean Motion Table

Give a list of how far the sun moves in a given period of time (with full revolutions removed)
Periods

18 years
1 year
Months (30 days)
Days
Hours

23 of 35

Rewinding Time

The autumnal equinox happened at 11:30 am, Sept 22, 2000
This is a period of 34 years, 152 days (143 days in that calendar year + 9 leap days), and 23.5 hours since noon on May 1, 1966
Look up periods that add up to this in the table and add
Remove increments of 360º
Total increase was 142.53º

Interval	Degrees
18 years	355.6237794
16 years	356.1100261025
150 days	147.8452917663
2 days	1.9712705569
23 hours	0.9445671418
0.5 hours	0.0205340683
TOTAL	862.5154690358

To rewind time, we need to figure out how far the sun moved around its eccentre. This is made quite easy by use of the mean motion table if we know the interval of time.

When doing this, I generally calculate the number of years and days first, ignoring the leap years. Then, I add one day for each leap year that was included (I have to look those up online each time).

Doing so, between noon on May 1, 1966 and 11:30 am on September 22, 2000, there was a period of 34 years, 152 days, and 23.5 hours.

We’ll turn to the mean motion table and find intervals that add up to this going in order. We start with the table of 18 years and use the first row since the second (36 years) is too great. That leaves another 16 years we need to add. That’s on the next table for single years.

Next for the days. We’ll start with the period of 150 days from the months table (called months because the Egyptian calendar had each month as exactly 30 days plus 5 days that weren’t part of a month). That leaves 2 more days we’ll need from the single days calendar.

Lastly, the hours. We can look up the 23 from the table, but will have to get at the half hour by dividing a single hour in two.

Once we have all these, we add them all up to get that the sun moved 862.52º about its circle. In reality, it moved many more times than that, but Ptolemy handily already subtracted the full circles out for the multi-year increments. Thus, we need to do the same if the value was above 360 by subtracting out increments of 360º. We have to do that twice (i.e., subtracting 720º) to get a final value that the sun had moved 142.53º about its circle.

Thus, to determine where the sun was on that date, we subtract this from the position of the sun on the 2000 autumnal equinox. Recalling that the sun was at 181.88º in ecliptic longitude, this means that at noon on May 1, 1966, the sun was at 39.35º ecliptic longitude.

Or, stating this in a more helpful way (since we like to refer to apogee since it will tell us more about the anomaly), this is 63.23º before apogee.

Let’s draw that out to help us visualize.

24 of 35

Solar Model: May 1, 1966�@12:00pm

25 of 35

Finding the Anomaly

Determine the angular distance after apogee �(360º - 63.23º = 296.77º)
Look up in table, estimating between rows
Determine if additive (column 1) or subtractive (column 2)
Add or subtract from true position

First, determine the distance of the true sun before or after apogee. In this case, we just did and found it to be 63.23º.

Then we’ll look that up in the table of anomaly. In this case, the first two column are the Angle from Apogee measured counter-clockwise, i.e., “after”. Thus, being 63.23º before the apogee is the same as being 296.77º after it. Thus, we would attempt to look this up in the table, but the table is not to such specificity.

Thus, Ptolemy expects that we will estimate between rows. He never specifies how exactly to do so, but here’s how I approach it.

The number of degrees between 294 and 300 is 6. We are 2.77º past the lower number which is 2.77/6 ~ 46%. I apply that same percentage to the difference between the two values.

1.738 – 1.642 = 0.096 and 46% of that is 0.04.

Noting that as the angle increases the anomaly decreases, we should subtract this. Therefore I come up with an estimated anomaly of 1.74-0.04 = 1.7º for the anomaly.

Again, we must ask if this is additive or subtractive.

Referring to the chart makes it quite easy. If we found the angles we looked up in the first column, these are from apogee to perigee where the anomaly is subtractive. If they were in the second column (as was the case here) this is from perigee to apogee where it’s additive. Thus, we add this to the true position and the sun’s apparent position (as viewed from Earth) is 41.06º ecliptic longitude.

26 of 35

Solar Model: May 1, 1966�@12:00pm

27 of 35

Calculate Mean Position

Determine interval of Time (37 years, 0 months, 23 days, 9.75 hours)
Break up the interval into values found in the solar mean motion table and compile them into a table and sum them
Subtract out any full rotations of 360º. This is how much the sun advanced since the epoch date.
Add this to the position of the sun at epoch to find the mean position of the sun on the date in question.

Interval	Degrees
36 years	351.2475587
1 year	359.7568766314
23 days	22.6696114042
9 hours	0.3696132165
0.75 hours	0.0308011121
TOTAL	734.0744610642

734.07º – (360º x 2) = 14.07º

39.35º + 14.07º = 53.42º

First, we’ll need to determine the interval of time between noon on May 1, 1966, and 9:45pm on May 15, 2003. This is actually the hardest part of this whole calculation.

Short on Time

[I’ll leave it to you to figure out how to do it, but with the reminder that leap years matter]

Good on Time

[I typically do this by first determining the interval of years. If the day in the year at the end is after the start date, I can simply subtract the years. Since that’s the case here (i.e., May 15 > May 1) I can just subtract: 2003 – 1966 = 37. If the date at the end of the interval was earlier in the year, we’d subtract one from this.

Next, days is rather annoying. To begin, you could start by thinking about how many days each month has and working it out that way. But my trick is to have an Excel sheet with each day of the year in a column for two years' worth of days in case the end date is earlier in the year than the start date. I’ll then highlight the cell the day after the start date and go down the column until I get to the end date, including it if the time is later in the day than the time on the start date, but staying one shy if it’s not. Excel will then display how many cells I’ve highlighted which makes it quite easy. In this example, we don’t even need to get that complicated as it’s 14 days (note: we don’t count the start day but do count the end day).

However, there’s a wrinkle: This doesn’t include leap days. As such, I usually find a website that lists which years were leap years. In this case, there were 9 leap years so we’ll need to add another 9 days for a total of 23 days.

Lastly, we need to figure out the interval in hours. From noon to 9:45pm is 9.75 hours (expressed in decimal).]

The total time between these two is 37 years, 0 months, 23 days, and 9.75 hours.

Next, we’ll refer to the table of mean motion located at the end of the handout. Again, this handy pre-calculated table that tells how much the sun moved in ecliptic longitude over various intervals of time in the case that the sun’s circle were centered on Earth (i.e., no worrying about the anomaly). We can use the interval we just calculated and break it up into values in this table and get how much the sun moved over all these sub-intervals, and then add them back up to determine the total motion of the sun over the full interval (because yay! Addition is easy math!) As a reminder, if the sun made more than a full circle, the extra intervals of 360º are already subtracted out.

Once we have the total motion of the sun, we’ll go ahead and subtract out and full 360º circles so we have an easier number to work with. In this case, we have two full circles of 360º to subtract out for a final increase in ecliptic longitude of 14.07º. That the, gets added to the ecliptic longitude the sun had on our start date of 39.36º.

Let’s pause here and take a look at what this looks like as a diagram.

28 of 35

Solar Model: May 15, 2003�@9:45pm

29 of 35

Finding the Anomaly

Determine the angular distance after apogee (310.83º)
Look up in table, estimating between rows�(1.43º)
Determine if additive (column 1) or subtractive (column 2)�(additive)
Add or subtract from true position for apparent position�(54.85º)

Again, this clearly isn’t in the table. It falls between the rows for 306º and 312º. As such, we’ll need to estimate between the respective values of 1.530 and 1.402.

Good on time

I’ll go ahead and do another example of estimating between rows:

First off, the difference between 306º and 312º is 6º. Our value is 4.84º past 306º or 4.84º into that 6º interval. If we express that as a fraction and convert to a decimal that’s 0.8066 or 80.66% of the way through that interval.

We can apply the same proportion to the interval in degrees. The difference in degrees is 1.530º – 1.402º = 0.128º and if we apply that proportion, we get 0.10º. Note that, going from 306º to 312º is decreasing, so we’ll need to subtract this from the value at 306º.

Thus, we determine the anomaly to be 1.43º.

This, then, gets either added to or subtracted from the mean position we calculated. Recall that if from apogee to perigee (i.e., 0º - 180º which was the lookup values in the first column) the displacement makes the sun appear behind where it otherwise would, so we subtract. However if from perigee to apogee (i.e., 180º - 360º which was the lookup values in the second column) the displacement makes the sun appear ahead of where it otherwise would, so we add.

Since the values we looked up were in the second column, we need to add.

30 of 35

Solar Model: May 15, 2003�@9:45pm

31 of 35

Solar Model: Ptolemaic Era

32 of 35

Solar Model: Present

33 of 35

Dates	Ecliptic Longitude	Astrological Sign	Actual Sign
March 21 – April 19	0° − 30°	Aries	Pisces
April 20 – May 20	30° − 60°	Taurus	Aries
May 21 – June 21	60° − 90°	Gemini	Taurus
June 22 – July 22	90° − 120°	Cancer	Gemini
July 23 – August 22	120° − 150°	Leo	Cancer
August 23 – September 22	150° − 180°	Virgo	Leo
September 23 – October 23	180° − 210°	Libra	Virgo
October 24 – November 21	210° − 240°	Scorpio	Libra
November 22 – December 21	240° − 270°	Sagittarius	Scorpio
December 22 – January 19	270° − 300°	Capricorn	Sagittarius
January 20 – February 18	300° − 330°	Aquarius	Capricorn
February 19 – March 20	330° − 360°	Pisces	Aquarius

34 of 35

Accuracy of the Model

Lastly, we should ask how good this model is?

To answer that question, I used Excel to predict the position of the sun each day at noon and then used a modern solar model calculator to do the same. I compared the difference between the two as a function of time. Effectively the y-axis is the amount of error.

When it is lower, the Ptolemaic model lags the position of the true sun and when it is higher, it is ahead of the true sun.

What we can easily notice about this is that the model is roughly best right around the year 2000. This is unsurprising since this is the year we chose the equinoxes and solstices from to calibrate the whole model. Thus, while we chose an epoch date of 1966, the model is clearly calibrated for the year 2000. In truth, I did this intentionally, as this epoch is commonly used by modern astronomers and so I wanted to be able to easily compare to it.

Still the amount of error here is relatively small. I’ve graphed this from 1966 to 2029 and in that time period, the error goes from about -0.15º to +0.1º. This is a reasonably small amount and well within the margin of error for most observations before the invention of telescopes.

Overall, the model gains about 0.12º every thirty years. This is primarily due to the length of the year being ever so slightly incorrect (off by about 6.5 minutes out of the length of a year or 0.012% off – an astoundingly accurate value for 2000 years ago).

There’s also some additional periodicity that we can easily notice. There’s two primary reasons for this:

The first is that the earth’s orbit is not circular, but elliptical. This variation is the yearly one we can see.

Second, the moon tugs the earth around as it orbits, which is the roughly 1-month variation we see.

Overall, while modern data clearly shows the deficiencies of this model, it would have taken a few centuries for the error to become large enough that it would be obvious to astronomers. Specifically, after about a century, it would be off by about ½ of a degree. This is the diameter of the moon, and thus, the model would no longer correctly predict eclipses. However, astronomers would be more likely to attribute that to errors with the lunar model (which we’ll explore in future classes I’m still working on) as it was the more complicated.

35 of 35

Thank you for attending

Learn more at: JonVoisey.net