Hi everyone,

I've double checked my notes today about the above topics, and found myself in a confusion of sorts.


- I know that the MC point is where the ecliptic and meridian intersect above the horizon (towards the south in the north hemisphere, and towards the north in the south hemisphere).


- I also have the following note, which I'm less certain of:

The MC is the high point in time (what is culminating half way between the times of its rise and set) while the nonagesimal (cusp of 10th Equal House) is the high point in space (closest to Zenith above).

It seems non exact to me, as the nonagesimal is just the highest point on the ecliptic plane, but not necessarily the highest point in space.


- Another point of uncertainty is under what conditions, in the northern hemisphere, the MC point falls in the 9th house vs in the 11th house. That is, in equal or whole sign houses.
I have the following note, which I also doubt:

In north hemisphere in summer, MC would be closer to 11th (Sun reaches mid-sky in less time) and in winter closer to 9th (Sun reaches mid-sky in more time).

Also not sure about this one. Tried to test some chart in my Solar Fire, but wasn't able to notice a pattern.


Anyone who has a good understanding of celestial dynamics, please come forth

--

Moderators: If another forum is a better fit for this post, please feel free to move the thread, or let me know and I'll repost.
General Astrology seemed fine.

The confusion that many astrologers seem to experience with regard to this topic stems largely, I think, from starting at the wrong end. The stable point in these dynamics is the MC, which (as you say) is always found due south for observers north of the tropics and is defined by the meridian, which marks the culmination (highest point reached in the course of a day) of any star, planet or degree of the ecliptic (zodiacal degree).

What varies is the location of the ascendant, which is only found due east twice a day (at the rising of 0 tropical Aries and Libra, respectively). The rest of the time, the ascendant degree (the intersection of the ecliptic and the horizon) is found either to the north-east or to the south-east, which is what causes the varying dynamics between rising and culminating signs. This in turn is due to the obliquity of the ecliptic: the equator always rises due east, but as the ecliptic is skewed relative to the equator, it is 'dragged along' with it in a lopsided manner. (All this is, of course, expressed from a purely observational point of view.)

The MC is the high point in time (what is culminating half way between the times of its rise and set) while the nonagesimal (cusp of 10th Equal House) is the high point in space (closest to Zenith above).

No, the MC is the highest point reached by a given degree in its daily motion, as well as the point midway (in time and space) between rising and setting. The so-called nonagesimal (which is located 90 degrees before the ascendant = 270 degrees after it) is the point of the ecliptic that is highest above the horizon at a given moment in time. Unlike the meridian, it cannot be located visually.

In north hemisphere in summer, MC would be closer to 11th (Sun reaches mid-sky in less time) and in winter closer to 9th (Sun reaches mid-sky in more time).

It's the other way round: in summer, the days are longer, meaning that the Sun takes longer both from rising to culminating and from culminating to setting. If you put the Sun on the ascendant in those months, and for a place far enough from the equator, you will see the ascendant and MC degrees forming a trine, more or less (MC in the 9th sign from the ascendant). If you do so six months later, you will see them forming a sextile (MC in the 11th sign).

Or you could visit northern Sweden and see the ascendant and MC in the same or opposing signs.

Martin hi,
First of all, a big thanks!

Martin Gansten wrote: Tue Feb 04, 2025 9:09 am What varies is the location of the ascendant, which is only found due east twice a day (at the rising of 0 tropical Aries and Libra, respectively). The rest of the time, the ascendant degree (the intersection of the ecliptic and the horizon) is found either to the north-east or to the south-east, which is what causes the varying dynamics between rising and culminating signs.

- How come the ASC is only found due east twice a day, when 0 Ari or 0 Lib is on top of that degree?
- Isn't the ASC degree the intersection between the ecliptic and the equator?

the MC is the highest point reached by a given degree in its daily motion, as well as the point midway (in time and space) between rising and setting. The so-called nonagesimal (which is located 90 degrees before the ascendant = 270 degrees after it) is the point of the ecliptic that is highest above the horizon at a given moment in time. Unlike the meridian, it cannot be located visually.

- Sorry, not sure I understand the difference between the MC and the nonagesimal in your explanation. I mean, if the MC is the highest point a degree can reach, and the nonagesimal the highest point above the horizon in a given moment, what's the difference?
Is it about the MC point constantly moving and shifting? If 'yes', I also wondered about that. Why such a thing happens anyways? Meaning, when I animate a chart in my Solar Fire, the MC degree shifts left (to 11th) and right (to 9th), every 24H or so (by whole sign).
Edit: Is the difference about the fact, that a chart point can't reach the nonagesimal, while it can reach the MC?

In north hemisphere in summer, MC would be closer to 11th (Sun reaches mid-sky in less time) and in winter closer to 9th (Sun reaches mid-sky in more time).

It's the other way round: in summer, the days are longer, meaning that the Sun takes longer both from rising to culminating and from culminating to setting. If you put the Sun on the ascendant in those months, and for a place far enough from the equator, you will see the ascendant and MC degrees forming a trine, more or less (MC in the 9th sign from the ascendant). If you do so six months later, you will see them forming a sextile (MC in the 11th sign).

- Clear, thanks.
So to summarize, the MC would be closer to the 9th in the summer months, in the north hemisphere (days longer so more time till Sun reaches mid sky). The same MC would be closer to the 11th in the winter months, in the north hemisphere (the reverse of what's in the brackets, above).

Sorry for taking long to reply, busy time and these topics require concentration (on my behalf, at least ).

Hi Dima Gur

Isn't the ASC degree the intersection between the ecliptic and the equator?

No - the asc is the degree on the ecliptic that lies on the horizon at the time for which the chart is drawn.

How come the ASC is only found due east twice a day, when 0 Ari or 0 Lib is on top of that degree?

Because the ecliptic is inclined at around 23.46 degrees to the equator, the ascendant moves along the eastern half of the horizon in a back and forth motion throughout the course of twenty-four hours. At the equator, the movement is limited to 23.46 degrees either side of the east point of the horizon, but at northern and polar regions the range of movement can be extreme, to the extend that the ascendant can conjoin the MC as one approaches the poles.

The equator always intersects the horizon at the east and west points of the horizon. However the ascendant may be found either north or south (in horizontal terms) of the east point of the horizon. It is only at 6H sidereal time and 18H sidereal time that the ascendant momentarily is perfectly aligned east and west. At 6H 00 Libra rises and 00 Aries sets; at 18H 00 Aries rises and 00 Libra sets.

Is it about the MC point constantly moving and shifting?

It is only that astrological chart that gives this impression. If you are in the northern hemisphere, the MC in non-polar regions will always be above the horizon and due south. This is a constant point. On your left the ascendant would be moving back and forth across the eastern horizon, sometimes towards the south point of the horizon and sometimes towards the north point of the horizon, throughout the course of the day. The descendant would be doing the same on the western horizon, at 180 degrees from the ascendant measured along the horizon.

So to summarize, the MC would be closer to the 9th in the summer months, in the north hemisphere (days longer so more time till Sun reaches mid sky).

The angles of the chart and house cusps are phenomenon of diurnal motion - the daily rotation of the earth on its polar axis in the course of 24 hours. So at points in the day the MC will appear (on an astrological chart) to be inclined towards the west and the 9th house, and at other times of the day, the MC will appear (again in an astrological chart) to be inclined towards the east and the 11th house. This is a phenomenon of diurnal motion not seasonal changes.

The image attached below will give you a guide to the movement of the angles and house cusps for around 51N for Porphyry cusps. The line across the graph from top left to bottom right is the 10th house cusp (MC). If you track the first house cusp (ascendant) you will see that it alters its course over the course of the day. This represents it's movement along the horizon over the course of the day relative to the constant orientation of the MC.

If you look along the 6H line (find in the left hand column) and work along to the right you will see that 1st house cusp (ascendant) intersects in the vertical axis with 00 Libra. If you continue to work along the 6H line you will see that the 7th house cusp (descendant) intersects in the vertical axis with the Pisces/Aries cusp (effectively wrapping around to 00 Aries). For clarity, in this graph sidereal time (24 hours) is the y-axis (vertical) and longitude (zodiacal degrees) is the x-axis (horizontal).

I won't put this inline as it is rather large but you should be able to open it and see it at full-size. There is a second diagram showing the celestial sphere with the relative positions of the ascendant, descendant, MC, and nonagesimal.

I hope that helps.

Best wishes. Ed

Dima Gur wrote: Wed Feb 19, 2025 6:25 pm - How come the ASC is only found due east twice a day, when 0 Ari or 0 Lib is on top of that degree?
- Isn't the ASC degree the intersection between the ecliptic and the equator?

No, the Asc degree is the intersection of the ecliptic and the horizon. So we have four great circles here. The horizon and meridian, fixed at right angles, are immovable. The equator, intersecting the horizon due east and west and rotating at a constant rate of 360° in 23h 56m, is tilted with respect to it. (How great a tilt depends on your location on earth: 90 minus your terrestrial latitude will tell you how high above the horizon the celestial equator culminates.) Finally, the ecliptic may for practical purposes be considered fixed in relation to the equator, at an angle of around 23°27', and is dragged along with the equator as it rotates, making for a wobbling motion. This is perhaps more easily understood with the aid of an armillary sphere such as this one:
https://youtu.be/M0chCdFEaP0

So to answer your first question, as the equator always rises and sets due east and west, the ecliptic only does so when the intersection points between equator and ecliptic are on the horizon. Those points define 0 tropical Aries and Libra.

- Sorry, not sure I understand the difference between the MC and the nonagesimal in your explanation. I mean, if the MC is the highest point a degree can reach, and the nonagesimal the highest point above the horizon in a given moment, what's the difference?

Let's say you've got 15° tropical Taurus rising with 15° tropical Capricorn on the MC. That means 15° Cap is due south and at the highest point it will ever reach in the sky at the latitude where you are. At the same time, 15° Taurus is slightly off to the northwest (behind your left shoulder if you're facing south), and 15° Scorpio slightly to the southeast (in front of your right shoulder). The 'nonagesimal' of 15° Aquarius (really the anti-nonagesimal!) is a little to the left (east) of the MC/meridian. It is not yet at its highest point (it's still climbing towards the meridian); it just happens to be the highest point on the ecliptic at that point in time, relative to the horizon. About two hours later, 15° Aquarius will have reached its highest (culminating) point, but at that time it will no longer be the highest point on the ecliptic.

In one case, you follow the actual daily motion of a given degree across the sky, and it will always culminate due south (for an observer north of the tropics). In the other case, you freeze-frame the ecliptic, which is almost certainly rising and setting at a wonky angle to the cardinal directions, and find the point exactly halfway between the rising and setting degrees, a point which will of necessity be equally wonky.

Is it about the MC point constantly moving and shifting? If 'yes', I also wondered about that. Why such a thing happens anyways? Meaning, when I animate a chart in my Solar Fire, the MC degree shifts left (to 11th) and right (to 9th), every 24H or so (by whole sign).

You need to change your Solar Fire settings! No, really: if you want a reasonably accurate depiction of what actually happens in the sky, you need to keep the meridian and horizon at right angles. The meridian (and hence the MC) never moves and shifts. Only the ascendant (= the intersection of ecliptic and horizon) does that.

I see you beat me to it, Ed. Thank you, particularly for that first illustration.

No problem Martin - Dima Gur's post is timely. I've been working on a house comparison module for Planetdance for the last few days. The plot of Porphyry houses is a sample of the output.

I've attached a text-based table of houses for Porphyry for the same latitude. It's essentially the same data but presented in the traditional tabular format. Due to the amount of data, it's only possible to tabulate the positions for every twelve minutes of the day. The graphical representation is calculated for every minute during the course of 24 hours.


If you right click on this image and go to open in new tab you should be able to see it.

Best wishes. Ed

astralwanderer wrote: Wed Feb 19, 2025 7:14 pm If you right click on this image and go to open in new tab you should be able to see it.

Thank you, I could. I actually meant the illustration with the great circles, though. That was the one that showed up first for me.

Hi Dima Gur - I hope the input from Martin and myself has been helpful. You can actually use the Table of Houses to see the difference between the nonagesimal and MC. If you open the tabular format and look at the first line in the first column (H1-H12) you can see that at 00H 00M of sidereal time 00 AR 00 is culminating (H10 - on the meridian/MC). At the same time 26 CN 32 is rising (H1 - ascendant).

The nonagesimal is the degree formed by the midpoint between the ascendant and descendant. If 26 CN 32 is rising, 26 CP 32 must be setting (H7 - descendant). Just look along the row to confirm this. The midpoint between the two degrees is 26 AR 32. Note that this is not the culminating degree at this time - this degree is 00 Aries 00.

Now look down the H10 column and find the degree closest to 26 AR 32. Given that the table entries are only calculated every twelve minutes it's unlikely you will find an exact match. However 25 AR 53 is on the MC (H10 cusp) at 01H 36M. This is the ninth row down the table. At a guess 26 AR 32 culminates a few minutes later at around 01H 38M of sidereal time. You could work out the precise time of culmination by interpolating the data.

So the nonagesimal degree at 00H 00M of sidereal time (26 AR 32) comes to the MC around 1 hour and 38 minutes later. The two points are therefore quite distinct.

I have posted a pair of charts for comparison. Today, at the latitude 51 N 27, 00 AR 00 culminates at 13:56:07 (00H 00M of sidereal time). The Porphyry chart is on the left and the equal house chart on the right. You will note that the nonagesimal point (26 AR 32) is clearly identified by the equal H10 cusp and the MC in both charts is at 00 AR 00.


In the second pair of charts 26 AR 32 is brought to the MC. This occurs at 15:34: 16 today. If you subtract the earlier time from this later time you get 01H 38M. This exactly conforms to what we have learnt from the Table of Houses. (My guess was correct!) However you will also note that there is a new nonagesimal degree - 14 TA 06, at the midpoint of the ascendant and descendant. The MC at 26 AR 32 is identified in each chart as a separate point.


(For some reason the charts don't seem to be showing inline but hopefully you can access them.)

All the best. Ed

I find that an armillary sphere really is the most helpful device in understanding these celestial dynamics. This eight-minute video may give a better explanation than the one I linked to yesterday:
https://youtu.be/N4fK4uWE4gc

The focus of the presenter is on the daily motion of the Sun, but what he says can be applied to any degree of the ecliptic, whether the Sun happens to be in it or not. You can see clearly how the horizon and meridian are fixed while the celestial equator rotates across them, carrying the ecliptic/zodiac with it at an angle, which makes for the back-and-forth motion of the ascendant across the east point of the horizon.

Once again, this is all from an observational viewpoint. Today we would say that the earth rotates around its axis rather than that the celestial equator moves across the sky; but the latter is what we actually see (or would see, if the equator itself were visible).

PS: I see that Ed and I keep posting at the same times!

Astrologers can have quite a lot of fun with a Table of Houses! Or perhaps more precisely, a table of rising and setting times for degrees of the ecliptic. Here is a neat party trick that you can use to impress your astrologically-minded friends. Perhaps you could invite them round to your place, get some pizza in, and have a Placidus house party in a manner of speaking.

If you look at the Table of Houses again for 26 AR 32 culminating you will find that it crosses the meridian at 01H 38M of sidereal time. Now find when this degree rises by looking down the H1 columns. You will see in the third full column that a degree close to 26 AR 32 rises at 18H 48M of sidereal time. Again, at a guess, given that the precise degree rising at this time is 27 AR 34, 26 AR 32 rises a few minutes earlier 18H 46M. Now find when 26 LI 32 rises and 26 AR 32 sets. This occurs at about 08H 31M (approximately midway between 08H 24M and 08H 36M by interpolation).

Now find the difference between the rising and setting times of the required degree 26 AR 32. We have to be a bit careful here to work out the length of the day. We will need to add 24 hours to 08H 31M and then subtract the rising time of 26 AR 32.

32H 31M - 18H 46M = 13H 45M. We can conclude that the length of daylight hours is 13H 45M (the time when the Sun would be above the horizon if it occupied this degree) at this latitude. If we divide 13H 45M by 12 we can find the length of each diurnal hour (the basis of planetary hours). This figure is 01H 09M. So each planetary hour is 01H 09M long.

Now here's the neat trick - if we divide the length of the daylight hours by 6 we get a figure of 02H 18M. So the degree 26 AR 32 traverses one-sixth of its journey across the sky in this time. From this we can work out when the degree becomes the Placidus H12 cusp. This is because each cusp between rising and culminating in the Placidus system is demarcated by 1/3 of the arc of time between the degree crossing the horizon and then later crossing the meridian (MC). This is the meaning of trisection of the semi-arc which is the technical definition underpinning the Placidus house system.

If we add 2H 18M to 18H 46M we get 21H 04M of sidereal time. If we check a Placidus ToH for H12 we see that this degree (or one very close to it) forms the cusp at this latitude just after 21H 00M. To get the H11 cusp, add a further 2H 18M to 21H 04M to get 23H 22M and go back to the ToH. We can see that 27 AR 29 forms this cusp at around 23 24H. We can conclude that the slightly earlier degree will have been 2 /3 of its journey from rising to culminating a few minutes earlier, at around 23H 22M of sidereal time.

Just to note, if you add a further 2H 18M to 23H 22M you get get to the time of culmination of this degree, or very close to it. Of course 26 AR 32 has then completed 3 / 3 of its journey from ascendant to midheaven. The culminating time of the degree was 01H 38M. Given some rounding errors and a bit of guesswork we get 01H 40M adding 2H 18M to 23H 22M, pretty close. A more precise set of calculations using detailed interpolation of figures would have given 01H 38M.

This feature of Placidean cusps underpins the astrological lore that the reason Raphael's ToH used the Placidus system exclusively is that it is very easy to calculate every cusp using this method (addition, subtraction, division of the rising and setting times of each degree) rather than complex spherical trigonometry. (Although deriving Placidean cusps from first principles requires some extremely complex mathematics.) This is how Placidus took over the world, along with some misunderstanding of Ptolemy's method of directions.


Anyway, by the time you and your friends have demolished a couple of pizzas and spent some time bashing the keypads of a few calculators, you should have a complete set of Placidus house cusps! What a great way to spend an evening. Apologies if the maths is a little bit out, but the principles are the key thing.

Best wishes. Ed

Hi Dima Gur - these diagrams may also be helpful. The two graphs below show a view of the horoscope as if the viewer is looking due south at the MC/meridian. The culminating degree in the first graph is 0 Aries. This sidereal time is 00H 00M. To the left, the nonagesimal degree is marked with the vertical red arrow. Note that altitude (vertical height above the horizon) of the nonagesimal is higher than that of the MC. As Martin has said, this degree will always be the highest point on the ecliptic above the horizon at any particular time.

A degree of the ecliptic will always culminate at an altitude represented by the sum of the co-latitude of the place + the declination of the degree. In the case of 0 Aries (declination 0) for this location (51N30) this will be at altitude 38N30 (this figure being the co-latitude of the place). When 26 Aries culminates it will be at 39.5 degrees plus the declination of the degree which is approximately +10 degrees - around 49.5 degrees above the horizon.

Declination is the measure of a position (degree or planet) above or below the celestial equator.

To interpret the graph, the horizontal line running across the middle of the image is the horizon. The curving red line is the ecliptic. The curving green line is the equator. The pair of curving pink lines represent the limits of rising and setting. Stars above or below these lines are circumpolar and never rise (if below) or set (if above) at this place. The pair of horizontal black lines at the meridian represent the difference in height (roughly) between the altitude of the nonagesimal degree and the MC degree culminating.

00 Aries 00 culminating example


26 Aries 32 culminating example