Night 2
As the light breaks forth from the dark-blue and black, the chasms appear to possess a certain depth.
Meteorology I.5
Jake retweeted
This is fun. Cut the whole Great Books survey. Cut modern science. Cut the attempt at the look at modern mathematics. Move formal theology to the very summit. Use as much recovered time as possible to teach the seven liberal arts as actual arts. And do this through their historically authoritative texts (and I assume TAC is only interested in the Latin West tradition and not the Greek East). Only then continue on to natural and moral philosophy, metaphysics, and finally theology in that order. Four years isn’t nearly enough time for all this, but it’s the best that can be done if freshman aren’t entering with any arts.
Something like this:
Year 1: Trivium focus plus Arithmetic. Donatus and Priscian for Grammar (along with the usual Latin authors as matter—Virgil, Salish, Terence). In Dialectic start with Porphyry and try to get through the Organon through Book 2 of Prior Analytics. Rhetoric, Cicero’s De inventione, Ad Herennium, and selections of Quintillian. Arithmetic is Boethius’s Arithmetic obviously—every word of it.
Year 2: Complete the Organon—scrap Sophistical Refutatjons if need to. Advanced Rhetoric: invention, style, imitation, arrangement, declamation. Geometry: the one art TAC probably successfully already does well, so stick with getting as deep into Elements as possible without skipping any propositions. Music: every word of Boethius’s De Musica. Astronomy: ideally the Almagest but there’s no way they’ll be prepared adequately. It’ll have to be a deep study of Sacrobosco’s De sphaera mundi. Continue reading the Latin authors for Grammar.
Year 3: Natural philosophy. Physics, On the Heavens, On Generation and Corruption, On the Soul, and as much of the biological works as possible. Plow through Nic Ethics and Poltics for Moral and political philosophy. Aquinas’s commentaries after Aristotle, but only if there’s time.
Year 4: Metaphysics and Aquinas’s De ente et essentia. Finish with sacred theology. Scripture, Lombard, Augustine, Aquinas.
So even with that massive butcher job we are still desperate for time.
It’s just embarrassing. As soon as the sophists and fakers of the classical education movement feel exposed, they plug their ears and flee.
Was it bringing to light that one of the biggest names in the modern movement doesn’t know what classical education is—or that it continues to use AI to write its posts? What is known is that pretenders hate being exposed.
Classical Education - Quadrivium: Solve this Almagest Problem
In Book I.13 of the Almagest, Ptolemy is sludging through the proofs for the geometrical theorem known as Menelaus’ theorem. This theorem is part of the groundwork necessary for spherical astronomy.
As is typical for Ptolemy, in his proofs he does not spell out every little step. He expects the student to be able to fill in a lot of the geometry. Sometimes it is manageable and sometimes it is quite difficult. In this proof he breezes over one step that he apparently finds obvious that I was only able to figure out a year after initially giving up on and moving on from: how a certain angle in a triangle is “given.”
So I want to see who here can show exactly how angle EDZ is given using the same classical geometry Ptolemy is using.
The attached images show Fig. I.13 with the angle in question. Also is the same figure marked pointing out the angle. And lastly is the whole lemma as it is found in the Almagest.
Neither Google, nor LLMs, nor even the ancient commentaries will be able to do this for you. You gotta know the geometry.
***Helpful Notes***
- By “given,” think of it as "determinate" based on prior givens or geometrical propositions.
For example: if two sides of a right triangle are given, the third side is also given (Pythagorean theorem). Or if two angles in a triangle are given, the remaining angle is also given, since the angles of a triangle are equal to two right angles.
- A "chord" is the straight line joining the endpoints of an arc.
So the chord of arc GB is the straight line GB.
“Crd arc 2GA” means the chord belonging to an arc twice GA. It does not mean “twice the line GA.”
In the Almagest the diameter of a circle is assumed as 120 parts, where “parts” is the standard unit of measurement we use. The circle itself is assumed as 360 degrees.
So, since the diameter bisects the circle and joins the two endpoints of either semicircular arc, each arc is 180 degrees.
That is:
Chord arc 180° = 120P.
The following values were determined previously and will be needed.
EZ = 130P(arts)
DZ = 48P
ED = 138.5784976P
In the post under this I will even provide the demonstration of the entire lemma up to the point where I want you to jump in.
How exactly does angle EDZ become given? Show all steps of your proof.
Classical education experts?
@athenaeumbc
@JeremyTate41
@0Beanie05923291
@LordIxabert
@MartinCothran
@zenahitz
@JamesWHankins1
Looking at you especially, St. John's and TAC folks.
SOLUTION
To solve this one, we go--as we often do--to the Elements. All we do is circumscribe a circle around the right triangle EDZ, and the rest falls into place.
Circumscribe a circle about the right triangle EZD. (Eucl. IV.5)
Since EZD is a right triangle, the hypotenuse ED is the diameter, and the circle passes through all three points of the triangle EZD. (Eucl. IV.5 Cor.)
Since ED is the diameter, let it now be 120P, and thus arc EZD is 180°.
And therefore, angle EZD = 90°. (Eucl. III.20)
Therefore, with respect to circle ABG, the ratio of 120:ED can be used to determine the arc of chords EZ and DZ.
So:
angle DEZ = arc (DZ * 120/ED) / 2;
angle EDZ = arc (EZ * 120/ED) / 2.
For example, since it was found originally that:
EZ = 130
DZ = 48;
ED = 138.5784976;
Then:
angle EZD = arc (ED * 120/ED) /2 = arc (138.5784976 * 120/138.5784976) /2 = 180 / 2 = 90°
angle DEZ = arc (DZ * 120/ED) / 2 = arc (48* 120/138.5784976) / 2 = 40°31'53" / 2 = 20°15'57"
angle EDZ = arc (EZ * 120/ED) / 2 = arc (130* 120/138.5784976) / 2 = 139°28'5" / 2 = 69°44'3"
Therefore, angle EDZ is given.
SOLUTION
To solve this one, we go--as we often do--to the Elements. All we do is circumscribe a circle around the right triangle EDZ, and the rest falls into place.
Circumscribe a circle about the right triangle EZD. (Eucl. IV.5)
Since EZD is a right triangle, the hypotenuse ED is the diameter, and the circle passes through all three points of the triangle EZD. (Eucl. IV.5 Cor.)
Since ED is the diameter, let it now be 120P, and thus arc EZD is 180°.
And therefore, angle EZD = 90°. (Eucl. III.20)
Therefore, with respect to circle ABG, the ratio of 120:ED can be used to determine the arc of chords EZ and DZ.
So:
angle DEZ = arc (DZ * 120/ED) / 2;
angle EDZ = arc (EZ * 120/ED) / 2.
For example, since it was found originally that:
EZ = 130
DZ = 48;
ED = 138.5784976;
Then:
angle EZD = arc (ED * 120/ED) /2 = arc (138.5784976 * 120/138.5784976) /2 = 180 / 2 = 90°
angle DEZ = arc (DZ * 120/ED) / 2 = arc (48* 120/138.5784976) / 2 = 40°31'53" / 2 = 20°15'57"
angle EDZ = arc (EZ * 120/ED) / 2 = arc (130* 120/138.5784976) / 2 = 139°28'5" / 2 = 69°44'3"
Therefore, angle EDZ is given.
Classical Education - Quadrivium: Solve this Almagest Problem
In Book I.13 of the Almagest, Ptolemy is sludging through the proofs for the geometrical theorem known as Menelaus’ theorem. This theorem is part of the groundwork necessary for spherical astronomy.
As is typical for Ptolemy, in his proofs he does not spell out every little step. He expects the student to be able to fill in a lot of the geometry. Sometimes it is manageable and sometimes it is quite difficult. In this proof he breezes over one step that he apparently finds obvious that I was only able to figure out a year after initially giving up on and moving on from: how a certain angle in a triangle is “given.”
So I want to see who here can show exactly how angle EDZ is given using the same classical geometry Ptolemy is using.
The attached images show Fig. I.13 with the angle in question. Also is the same figure marked pointing out the angle. And lastly is the whole lemma as it is found in the Almagest.
Neither Google, nor LLMs, nor even the ancient commentaries will be able to do this for you. You gotta know the geometry.
***Helpful Notes***
- By “given,” think of it as "determinate" based on prior givens or geometrical propositions.
For example: if two sides of a right triangle are given, the third side is also given (Pythagorean theorem). Or if two angles in a triangle are given, the remaining angle is also given, since the angles of a triangle are equal to two right angles.
- A "chord" is the straight line joining the endpoints of an arc.
So the chord of arc GB is the straight line GB.
“Crd arc 2GA” means the chord belonging to an arc twice GA. It does not mean “twice the line GA.”
In the Almagest the diameter of a circle is assumed as 120 parts, where “parts” is the standard unit of measurement we use. The circle itself is assumed as 360 degrees.
So, since the diameter bisects the circle and joins the two endpoints of either semicircular arc, each arc is 180 degrees.
That is:
Chord arc 180° = 120P.
The following values were determined previously and will be needed.
EZ = 130P(arts)
DZ = 48P
ED = 138.5784976P
In the post under this I will even provide the demonstration of the entire lemma up to the point where I want you to jump in.
How exactly does angle EDZ become given? Show all steps of your proof.
Jake retweeted
The CLT is 40 minutes of Verbal Reasoning, 35 minutes of Grammar/Writing, and 45 minutes of Quantitative Reasoning and the SAT is 64 minutes of Reading/Writing and 70 minutes of Math. You’ve said publicly the primary difference between the CLT and the SAT/ACT is “source material.” The CLT’s FAQ explains that it requires no prior knowledge of the texts. It is open to students “regardless of educational background,” and “classic” refers to the use of enduring texts. It is a timed college-readiness exams testing modern verbal, writing, and mathematical performance like the others.
The exam simply isn’t one that attempts to test the learning of a classical education. Acknowledging this isn’t insulting it or calling it irrelevant. And frankly if it did attempt that, then it wouldn’t be a college-readiness exam. And even more, maybe a dozen people in the country would be equipped to take it.
More Great Books mythology from the modern fake classical movement. Homer’s epics weren’t even accessible to students until recently in the early modern period in the Latin West. And even in the Greek East where Homer was integral, it wasn’t this Great Books silliness of sitting around reading and discussing. They used Homer as school matter very specifically for grammar and rhetoric.
unavailable
Jake retweeted
Porphyry’s Isagoge is now on the Aristotle Reader — the 3rd-century gateway to Aristotle’s logic. Greek + Owen’s 1853 English side by side, clickable word parsing (and a modern diagram of the Porphyrian Tree).
johnhboyer-sys.github.io/ari…
Metaphysics absolutely should be read through Posterior Analytics. Metaphysics is prior in object and dignity, but Posterior Analytics is prior in method. A student would do best to study Posterior Analytics before Metaphysics.
Metaphysics is not a demonstrative science with a subject genus but that does not mean it should not be read through the doctrines and arguments of Posterior Analytics. And the Organon as a whole is crucial for understanding Aristotle’s method of inquiry, which he uses masterfully in Metaphysics.
But like with much of his corpus, they can and should be read through one another, and can be read together well or poorly.
Man is doing good work. Help out the project.
🏛️ The Aristotle Reader now has a way to chip in.
If it’s useful in your work, you can now support its development. Click the link on the homepage!
👉 johnhboyer-sys.github.io/ari…
If stuck on the classical astronomy problem (xitter.cf/le_master/status/20708…), give this one a shot.
Classical Education - Quadrivium: Solve this Musical ‘Enigma’
In Boethius’s de musica (the canonical text on music/harmonics in the Latin West), chapter 29 of Book I investigates why consonances are found in some ratios and not in others.
At the end of the short chapter comes the puzzle:
“the sesquialter is twice the half;
the sesquitertian is three times the third part.”
Translator Calvin Bower mentions in a footnote that this explanation of the superparticular ratios “remains something of an enigma,” that “the sesquialter, as 3:2, could be described as three halves related to two halves, and the sesquitertian, as 4:3, as four thirds related to three thirds.“
Recall:
Sesquialter = 3:2
Sesquitertian = 4:3
I’m confident I know the solution, but I’m interested in what others come up with.
Attached is the full chapter for context.
Full English text here:
classicalliberalarts.com/res…
