At this point we’ve spent some considerable time doing the work to develop our rising time tables. Now Ptolemy answers the question: What can we do with them?
Ptolemy provides several algorithms:
Seasonal Hours (Alternative Method)
Seasonal Hours to Equinoctial Hours
Upper Culmination (Alternative Method)
The first answer he gives is that this is a handy way to calculate the length of a day. It requires a bit of prerequisite knowledge of where the Sun is on the day in question, but since $\frac{1}{2}$ of the day cycles through $\frac{1}{2}$ of the zodiac, we can take the total of the rising times for half of the appropriate table going forward through the signs or backwards for the night.
The translation I’m working with provides an example1 that I’ll follow along with. It considers the length of the night for Babylon when the sun is 28;18º into Sagittarius. The problem doesn’t give a specific latitude for Babylon, but it does state that they used the Rhodes table.
Unfortunately our table only gives the rising times in 10º intervals so we’ll have to do a bit of extra work. 28;18º is $\frac{8;18º}{10º}$ the way between the 20º and 30º we do have. Converting that fraction to a modern decimal is 0.83, or 83% of the way between the two. The rising time for the arc that corresponds to the 20º-30º points is 11;16º. Multiplying that by 0.83 we get 9;21º. Adding that to the accumulated rising time at the 20º mark (277;29º), we get 286;50º2.
Next, we need to repeat this procedure for the opposing sign directly across the zodiac, Gemini. Again, we will be 28;18º into that constellation, which is again, 83% the way between the 20º time and the 30º time. That difference is 8;46º. Adding that to the 60;41º that is the 20º accumulated rising time we get 69;27º.
It’s a bit tricky discerning the day from the night. Since we’re doing the night, we need to go backwards from 286;50º to 67;27º. In other words: $286;50º – 67;27º = 217;23º$.
If we were doing the day, we’d need to go forward around the zodiac. In which case we’d go from 286;50º to 360º, which is 73;10º, and then add the 67;27º to get 142;37º3.
But those are time-degrees. We want to convert them to hours. In modern math, I would divide them each by 360º to determine what percentage of a full day they are, and then multiply that by the 24 hours in a day, but Ptolemy recommends a shortcut: Just divide by 15. This works because $360º \div 15^{º/h} = 24h$.
Jumping back to the length of the night, we can state: $ 271;23 \div 15 = 14;30h$.
This brings us to another topic which I mentioned briefly in the introduction to this post: The unequal or seasonal hours. In this time keeping system, the length of hours is not consistent. Rather, there are 12 hours in the day, and 12 hours in the night4. As such, hours during the day are longer in the summer while the hours of night are shorter. The opposite is true in the winter. Since there are always 12 hours the length of them can be determined by taking the actual period and dividing by 12. To continue the example above, $\frac{14;30}{12} = 1 \; hour \; 12.5 \; minutes$. A cumbersome system to be sure.
However, dividing the equal hours5 by 12 to determine their relationship to unequal hours as I did above isn’t quite what Ptolemy meant. Rather, it was the degrees that we would divide by 12. In other words, $\frac{217;33}{12} = 18;7,45$º which is the angular distance the sun would travel in one of these unequal hours.
Seasonal Hours (Alternative Method)
Ptolemy then provides a formula for finding this more quickly:
[Take], from the above Table of Rising-times, the total rising-time corresponding to the sun’s degree for the day (or the degree opposite the sun for the night) both at the parallel beneath the equator [i.e. sphaera recta] and at the relevant latitude, and forming the difference. Take $\frac{1}{6}$th of the latter, and add it to the 15 time-degrees of one equinoctial hour for points on the northern semi-circle [of the ecliptic], or subtract it from 15º for the points on the southern semi-circle. The result will be the length of the relevant seasonal hour in time-degrees.
Another dense paragraph. So let’s continue with the example. The first thing we need to find is the “rising-time corresponding to the sun’s degree for the day” at sphaera recta. Since the example we were previously working was looking at the night, with the sun setting in Sagittarius. Thus, the rising would be the point in Gemini at 28;18º. Thus we go back to the rising time table, this time at sphaera recta. And again we have to do the math to interpolate the accumulated degrees since we’re between two values. The difference between the two is 10;55º and 83% of that is 88;9º.
From that, we subtract the rising time of the same point at the latitude in question which we established above as 69;27º to get 18;42º.
Divide that by 6 to get 3;7º.
Next comes the part about adding or or subtracting it from 15º depending on where it is on the semi-circle of the ecliptic6. Since we’re referring to Gemini, which is north of the celestial equator, that means we add 15º to get 18;7º, which agrees (to the degree of precision presented) with the answer we derived above.
Seasonal Hours to Equinoctial Hours
Next, one can convert seasonal hours for a given date into equinoctial hours by multiplying them by the length in time-degrees of the hour of the day in question at the relevant latitude (if they are hours of the day), or by the length in time-degrees of the hour of the night in question (if they are hours of the night). Then division of that product by 15 will give the total of equinoctial hours. Vice versa, one can convert the equinoctial hours to seasonal by multiplying by 15 and dividing by the length of the hour of the relevant interval in time degrees.
The next thing Ptolemy tells us we can do, is to convert the above found seasonal hours to equinoctial hours. Continuing from above, we found that the unequal hour was 18;7º. As an example, we can ask from that, what is $5 \frac{1}{2}$ seasonal hours after midnight in equal hours?
As the above algorithm describes we begin by taking the number seasonal hours and multiplying it by the length of a seasonal hour in time degrees and then divide by 15. In other words:
$$5\frac{1}{2} \cdot \frac {18;7}{15} = 6;38$$
We then add that to the time in question (midnight) which means that waiting $5 \frac{1}{2}$ seasonal hours after midnight would land you at 6;38 equinoctial hours or 6:38am.
Ptolemy’s next application is to use a time, place and date (more specifically, the location of the sun on that date) to determine what point on the ecliptic was on the horizon at that place and time. Although Ptolemy doesn’t specifically call it out as such, this is a method that would be used for casting horoscopes7. Ptolemy states the method as follows:
[G]iven a date and any time whatever, expressed in seasonal hours, on that date, we can find, first, the degree of the ecliptic rising at that moment. We do this by multiplying the number of hours, counted from sunrise by day, and from sunset by night, by the relevant length of the [seasonal] hour in time-degrees. We add this product to the rising-time at the latitude in question of the sun’s degree by day (or the degree opposite the sun by night): the degree [of the ecliptic] with rising-time corresponding to the total will be rising at that moment.
For this example, we’re provided with a location of Alexandria, a time of $2 \frac{1}{4}$ seasonal hours after midnight, and a day on which the sun is 13;17º into Scorpio. Alexandria is at a latitude of about 31ºN which means we’ll need to be switching to to Lower Egypt table. The example as provided treats the length of a seasonal hour as a given, but for some extra practice, I’ll do the derivation here. I’ll be using the first method which involves finding the length of the day/night.
To do so, I first calculate what percentage 13;17º is between 10º and 20º:
$$\frac{3;17º}{10º} = 0.3283$$
Multiply that by the difference between the two:
$$11;54º \cdot 0.3283 = 3;54º$$
Which then gets added to the total rising time for the 10º mark:
$$226;34º + 3;54º = 230;28º$$
Taurus is the sign opposite Scorpio on the zodiac, so again determine the proportion between 10º and 20º:
$$8;2º \cdot 0.3283 = 2;38º$$
Add that onto the 10º mark:
$$28;26º + 2;38º = 31;04$$
Take the difference of the two to determine the length of the night (since the problem is asking in relation to midnight):
$$230;28º – 31;04º = 199;24º$$
Divide this by 12 to find the length of the night time seasonal hour:
$$199;24º \div 12 = 16;38º$$
Now we can begin applying the new method. First we multiply the number of hours past sunset by the length of each seasonal hour. But notice there’s a disconnect with the information we have. We were given a number of hours after midnight, but we need to do things in terms of number of hours after sunset.
Fortunately, since there are 12 hours in seasonal time for night, midnight is always 6 seasonal hours after sunset. So $2 \frac{1}{4}$ seasonal hours after midnight is $8 \frac{1}{4}$ seasonal hours after sunset. So multiply that by the length in time-degrees:
$$8 \frac{1}{4} \cdot 16;38º = 137;14º$$
Next, “we add this product to the rising-time at the latitude in question of the sun’s degree by day (or the degree opposite the sun by night)”. Since this is night time, we need to add in Taurus’ rising time to that point which we calculated above8:
$$137;14º + 31;04º = 168;18º$$
This is the total rising time which we can then look up on the rising time table for this same latitude (for Lower Egypt). We see that this falls in Virgo just before the 20º mark. Thus, Virgo was the constellation rising at the given time and location.
For the sake of completeness, we can calculate just how far into Virgo it is. Ptolemy doesn’t give a recommendation on how to do this and I can visualize a few different ways of doing this so here’s my version. Essentially what we want to know is what percentage this value is between the 10º and 20º marks from our table, much like we’ve done previously. Except this time we wont’ be applying it to the total rising time; we’ll apply it to the degrees. Since we know it’s between the 10º and 20º marks, we’ll subtract the total rising time at 10º from it:
$$168;18º – 156;53º = 11;25º$$
Divide this by the change in degees between the two:
$$\frac{11;25º}{11:35º} = 0.9856$$
So we have that we’re 98.56% the way from 10º to 20º degrees. So let’s multiply that by the interval (10º) to determine what that is in degrees:
$$10º \cdot .9856 = 9;51º$$
And add that back to the 10º we’re already past to get the point just rising at that time is not just in Virgo, but 19;51º into Virgo.
The concept of upper culmination is one that I just realized I haven’t defined yet despite making extensive use of it. In general, a culmination is when an object is on the observer’s meridian. This happens twice each day, although one is often beneath the ground. The upper culmination is the one is the higher of the two. To find when this occurs:
we take in every case [i.e. for both day and night] the total seasonal hours from the last midday to the given time, multiply it by the appropriate length(s) of the hour(s) in time-degrees, and add the product to the rising-time at sphaera recta of the sun’s degree: the degree [of the ecliptic] with rising time at sphaera recta equal to the total will be at upper culmination at that moment.
Continuing the above example at Alexandria, we first find the total seasonal hours from the last midday. The previous problem was $8 \frac{1}{4}$ seasonal hours past sunset, which in turn was (by definition) 6 seasonal hours past the previous mid-day. But bear in mind that these the night hours and day hours aren’t the same as these are seasonal. So we’ll need to figure them differently.
We’d already found the length of a night hour above as 16;38º. From that, we can quickly find the length of a seasonal day hour for that date by subtracting that from 30. The reason is that:
$$12 \; day \; hours + 12 \; night \; hours = 360º$$
Dividing both sides by 12 we get:
$$day \; hours + night \; hours = 30º$$
Thus, we find that the day hours are 13;22º each.
Now we multiply the length of each hour by the number of each and sum them. In other words:
$$6 \; day \; hours \cdot 13;22º + 8\frac{1}{4} \; night \;hours \cdot 16;38º = 217;26º$$
This then gets added to the rising time at sphaera recta of the location of the sun. So we hop back to our rising time table and do a bit of calculation to discover that the accumulated rising time of the sun 13;17º into Scorpio is 220;46º. Adding that to the previous value we get:
$$217;26º + 220;46º = 438;12º = 78;12º$$
We then head back to our rising time table for Lower Egypt to see where that falls, which is in Gemini. Again, we can calculate exactly where as I outlined in the last section to see it is 19;11º into Gemini specifically.
Upper Culmination (Alternative Method using Rising Point)
Ptolemy also provides another method to find the upper culmination.
[F]ind from the table of rising-times for the relevant latitude the cumulative rising-times corresponding to the degree which is rising. Subtract from it, in every case, the 90º of the quadrant [of the equator between horizon and meridian]. The degree corresponding to the result in the column for rising-times at sphaera recta will be at upper culmination at that moment.
There’s a bit of word salad here, but the first thing Ptolemy instructs us to find is the “cumulative rising-times corresponding to the degree which is rising.” This is another way of saying what point is just rising which is what we found in the section on horoscopes above. So to continue that example, the point that is on the horizon at that moment is the one at 168;18º. If that’s on the horizon, 90º from it (along the ecliptic) would be on the meridian, which would be the point of upper culmination. However, we must remember that constellations rise in their zodiac order. Thus, we need to go 90º backwards since we’re referring to constellations that have already risen. Thus, we need to subtract:
$$168;18º – 90º = 78;18º$$
Using the table for sphaera recta, we find this point is 19;16º into Gemini9.
Rising Point (from Upper Culmination)
As with any good math, we should be able to go backwards as well. Thus, if we have the point at upper culmination we should be able to derive the point currently on the horizon,
by taking the degree corresponding to the culminating point in the column for rising-times at sphaera recta, adding to it, in every case, the above 90º, and finding the degree corresponding to the result in the column for rising-times for the latitude in question: This degree will be rising at that moment.
No example is provided here as this one is simply the exact opposite of the previous one: Start with a point at upper culmination, add 90º, and look that up on the sphaera recta table.
Position of Sun
Ptolemy closes the chapter with one final statement that doesn’t really contain much in the way of any math:
[F]or those living beneath the same meridian the sun is the same distance from noon or midnight, counted in equinoctial hours, while for those living beneath different meridians the sun’s distance from noon or midnight differs by an amount, counted in time-degrees, equal to the distance of one meridian from the other in degrees.
There’s no example provided for this as there’s really little to no math involved. It’s more of a statement of fact that’s largely self obvious simply due to the fact we’re using a spherical coordinate system. When you look down the central axis, it behaves as a polar coordinate system and the projection of the meridian falls on all locations beneath it.
- And thank goodness they did. I tried doing one by myself and was utterly confused by Ptolemy’s explanation!
- This math is not demonstrated in the example. While it certainly produces the correct answer, I am uncertain what method Ptolemy may have used although it certainly did not involve modern decimals and percentages as I have done. However, this method assumes a linearity between the rising times which is likely fundamental to Ptolemy’s methods as well given that we saw a similar application in the table of chords when Ptolemy assumed that all sixtieths were equal, or at least close enough that it didn’t make a significant difference.
- Notice that the day (142;37º) plus the night (217;23º) equals a full 360º. This makes sense since it’s a full day. However, this actually assumes that the Sun has not moved during the course of the day. Obviously it does, buy about 1º per day, which apparently Ptolemy did not consider sufficiently important to take into consideration. I don’t blame him as it would make the problem much more difficult.
- If you’re interested in learning more about this topic, here is an interesting essay here about the transition to equal hours that we use today.
- These equal hours are also known as “equinoctial hours” as during the equinox, the day and night are equal, and thus the length of their hours is also equal.
- The text does not say what to do when it is on the equator, but given we already know the answer (that the unequal hours are in fact equal on that date, being 15º each), we need not concern ourselves with it.
- As a reminder, this method no longer gives the correct answer as the vernal equinox is no longer the first point in Aries due to precession of the equinoxes.
- This is a reason to use the first method over the second which would have directly given us the rising time only in Scorpio. The first went through both without the extraneous steps of looking things up at sphaera recta.
- For those paying close attention, you’ll notice that this value is 7 minutes of arc off of the value we previously found. The reason for this is that we have repeatedly engaged in rounding to the nearest minute in both this method and the others. As such, the remainders add up and have caused them to diverge. Should we have kept things more accurate, they would be in agreement.