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I refer here to Ptolemy’s epicycle-and-deferent model of the Solar System, specifically that of Mercury (see drawing).

enter image description here

In this model, Mercury (not shown) revolves on an epicycle of center C, which itself turns on an eccentric circle (later called “deferent”) of center D, which in turn moves on a small circle of center F. The size of this small circle is such that DF = FE = EO = 3 arbitrary units where DC = 60 of the same units.

Point A is the direction of the apogee, the deferent’s point which is furthest from the Earth. Angle AFD increases uniformly and is called the mean centrum $ \bar{\kappa} $. The true centrum, $ \kappa $, is angle AOC—obviously, this one does not increase uniformly, and a good part of the Almagest consists of the explanation of the procedure to find it.

The calculation of latitudes involves finding out distance OC for when $ \kappa = 90° $. Its calculation is not mentioned in Ptolemy’s Almagest, but its value in the model for Mercury is specified as being about 56.7° (Ptolemy actually says it was found previously, but such is not the case). For other planets, for which the deferent is fixed and centered where point E is for Mercury, the calculation is easy: $ \displaystyle OC = \sqrt{R^2 - e^2} $, but the calculation is made more complex in Mercury’s case, because the deferent moves, with angle AFD equal to (but in opposite direction from) angle AEC, in both cases being $ \bar{\kappa} $.

In A History of Ancient Mathematical Astronomy, Otto Neugebauer states that, in order to find this value, “One finds a cubic equation for the sine of the angle under which the eccentricity e = 3 is seen from C” (p. 221, n. 1).

I have looked in other books and articles commenting the Almagest, but I have never been able to find the said “cubic equation for the sine of the angle.”

Can someone please help me find this equation? EDIT: What I’m looking for is the equation to find angle ∠AEC = ∠AFD = κ̄₀. My image mentions “OC = ???,” but only because we’re not supposed to know it in advance, not because I want to know how to calculate it.

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  • $\begingroup$ I have also asked this question in MathSE. $\endgroup$ Commented Dec 24, 2023 at 1:26
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    $\begingroup$ Hello, Pierre! Please look carefully at the edit of my answer. I've added new information there. $\endgroup$
    – ayr
    Commented Dec 27, 2023 at 17:13
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    $\begingroup$ Pierre, I prepared a new answer and posted it here on Astronomy. Please read carefully. $\endgroup$
    – ayr
    Commented Jan 3 at 10:51

2 Answers 2

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My answer:

$-54\cos(\tilde{\kappa}_0)^3+3564\cos(\tilde{\kappa}_0)^2+18\cos(\tilde{\kappa}_0)-9=-6\cos(\tilde{\kappa}_0)^3+396\cos(\tilde{\kappa}_0)^2+2\cos(\tilde{\kappa}_0)-1=0$

The new explanations consist of three blocks. With the most detailed explanations and demonstration of calculations.

Block 1 ## This block contains basic symbolic calculations based on the methodology outlined in my other answer.

enter image description here

Block 2 ## This block shows what expression was used to fulfill one of the important conditions of the problem (the rectangularity of the $OEC$ triangle) and some of its transformations.

enter image description here

Block 3 ## The third block shows the required equation, its solution and verification calculation.

enter image description here

enter image description here

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  • $\begingroup$ Excellent! You also gave me the solution on MathSE, but you deserve the bounty nonetheless. :) $\endgroup$ Commented Jan 3 at 13:56
  • $\begingroup$ Have you seen my latest comment on MathSE? I’d love to know a formula for generic values of e = OE. $\endgroup$ Commented Jan 4 at 2:29
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    $\begingroup$ @PierrePaquette Thank you! Since the result turned out, I had to publish it, sooner or later. And I'm glad that the answer was useful. I read the answer to MathSE. It was prepared by another person, not me, he outlined his approach to the solution, but our results are the same. $\endgroup$
    – ayr
    Commented Jan 4 at 5:05
  • $\begingroup$ @PierrePaquette As for the general value of $e$...do you mean that symbols must now be substituted for numbers? Should the condition of the rectangularity of the $\Delta OEC$ triangle and the ratio $OE/R = 3/60$ be preserved? $\endgroup$
    – ayr
    Commented Jan 4 at 5:08
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    $\begingroup$ @PierrePaquette Well, I'll see what happens as a result in my free time. $\endgroup$
    – ayr
    Commented Jan 4 at 5:16
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The links below may contain hints:

  1. Trigonometry According to Ptolemy - Ptolemy's theorem is presented in trigonometric form.
  2. Almagest Book IX: Exploring Mercury’s Double Perigee - dedicated to the analysis of book 9 of the Almagest in more modern mathematical terms.

But I tried to derive formulas and carry out my own calculations. As a basis, I took the second model from your drawing, which shows $OC=???$. Therefore, the length of this segment needs to be calculated? The figure below shows the complete model, which I rebuilt in another software (Mercury rotates along an epicycle with center $C$, which rotates along a deferent with center $D$, which rotates in a circle with center $F$). Relationships with some points have been preserved, and auxiliary vectors, angles and lines have been added. True, I slightly violated the relationship between the distance scales, but this was done for clarity.

enter image description here

Next will be a demonstration of calculations and corresponding geometric constructions.


Let's choose the length of the segments $OE=EF=FD=5u$, $R=25u$ and the angles $K_0=\tilde{\kappa}_0=125^{\circ}$ and $\measuredangle DF0=35^{\circ}$, $u=1$.

enter image description here


Auxiliary vectors $v_1,v_2,v_3$, $V_1$ connecting a point on the internal sphere along which the center of deferent rotates and $V_2$ connecting the origin of coordinates and a moving deferent center $D$, or .

enter image description here


Work with the triangle $\bigtriangleup DEF$. It is equilateral and you can calculate the angle of the $\measuredangle DFE$ at the vertex through the scalar product of vectors $V_2$ and $v_3$, and then find the length of the base of the $DE$ (it will be one of the sides of the next triangle) and the remaining angles $\measuredangle BEF$. Don’t pay attention to the “pale” area in Mathcad Prime, this is the same formula, only in the active one, instead of the vector norms, the constant lengths of the corresponding segments are written.

enter image description here


Next, we determine the coordinates of the auxiliary vectors $c_t ("tail") = v_3$, $c_h ("head")$ and we get $C_n || ED$, which are necessary to calculate the angle at the side of $\bigtriangleup CED$.

enter image description here


Using the formulas from here (Solution of triangles) we find the length of the segment $CE$. I note that according to this theory, two solutions are possible, but I chose the one that corresponds to the Ptolemaic model.

enter image description here


From the angle $K_0=\tilde{\kappa}_0$ we find the $\measuredangle OEC$ of the triangle $OCE$, and then, using the cosine theorem, we calculate the length of the required side $OC$.

enter image description here

Below is the same model, but taking into account the relationships between distances, that is $OE=EF=FD=3u$, $\tilde{\kappa}_0=-\tilde{\kappa}_0$, and $DC=60u$, $u=1$.

enter image description here

The length of the segment $OC$ was calculated using the same formulas as above. Also note that it is equal to $56.72$. This is the number you entered on your original model. Is this not what you were looking for?

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    $\begingroup$ Hi @dtn! Thanks for this extra work, it is great indeed! However, I don’t see how you found ∠OEC = 86.97° or ∠AEC = 93.03°, and—most importantly—there’s no “cubic equation for the sine of the angle” as mentioned by Neugebauer. I should have specified (I’ll edit the OP in a second) that I’m looking for the formula needed to find ∠AEC = κ̄₀, not for OC or the formula to find OC. $\endgroup$ Commented Dec 27, 2023 at 17:34
  • $\begingroup$ @PierrePaquette I created the drawing in SolidWorks and specified the required restrictions on distances and angular relationships. And then SolidWorks completed the remaining geometry on its own. In any case, you can carry out some “inversion” of the mathematical apparatus in question. Those. swap known and unknown quantities, expressing the latter of the resulting formulas. $\endgroup$
    – ayr
    Commented Dec 27, 2023 at 17:40
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    $\begingroup$ @PierrePaquette As for this mysterious cubic equation...I tried to get it, but I haven't succeeded yet. It may appear if we try to reformulate the problem in terms of systems of equations, but I haven’t done that yet. I mean such a system, by solving which we could get both the $OC$ (and not use it in advance) and the angle $\tilde{\kappa}_0$ $\endgroup$
    – ayr
    Commented Dec 27, 2023 at 17:42
  • $\begingroup$ I’m also looking for a solution that doesn’t involve SolidWorks, Mathematica, and the likes. I want to be able to do it with pen-and-paper like the ancients did. (OK, maybe with a calculator, since it’s a third-degree trig function.) $\endgroup$ Commented Dec 27, 2023 at 20:07

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