<p>You misunderstood Ptolemy’s model: The epicycle doesn’t “roll on” the deferent; rather, the epicycle’s <em>center</em> is on deferent. Mind you, it’s only a minor detail in this situation.</p>
<p>In the diagram you drew up, it is impossible to know φ* in terms of γ alone, as φ₀ could take any value. Also, in Ptolemy’s model, φ₀ does <em>not</em> depend on φ*, so φ₀ ≠ δφ*. However, both vary uniformly, by quantities that we will here call ωₐ and ωₜ after Pedersen.¹ For example, in the case of Mars, ωₜ = 0°;31,26,36,53,51,33/day and ωₐ = 0°;27,41,40,19,20,58/day.</p>
<p>You are correct to presume that c, rₐ, and r<sub>A</sub> were known and specified in Ptolemy’s model—a good part of the <em>Almagest</em> is spent determining such values for rₐ and r<sub>A</sub> supposing that c = 60 units.</p>
<p>Finally, nowhere in the <em>Almagest</em> are φ* or φ₀ expressed in terms of γ, as both of them, as hinted two paragraphs above, are only time-dependent. Because none of the angular speeds given for ωₐ and ωₜ is a factor of 360°, this means that for a given γ, <em>any</em> value of φ* and φ₀ is possible, depending on how long has elapsed since the reference date (the “epoch”), which Ptolemy chooses, for historical reasons, to be February 26, −746 (aka 747 BCE), Noon Alexandria true solar time (Julian Day 1448637.91666…).</p>
<p>If you understand the language, I have translated the <em>Almagest</em> to French, with some diagrams animated or interactive. It’s freely available at <a href="https://ecliptiqc.ca" rel="nofollow noreferrer">https://ecliptiqc.ca</a>.</p>
<hr />
<p>¹Pedersen, O. <em>A Survey of the Almagest.</em> With Annotation and New Commentary by Alexander Jones. New York: Springer, 2011, 480 p., ISBN 978-0-387-84825-9.</p>

CC BY-SA 4.0
2023-06-16T05:19:38.370Z
2023-06-16T05:19:38.370Z
Pierre Paquette
<p>I am about to work with Ptolemy's Model for the motion of an upper Planet (Mars). I use a epicycle rolling on the deferent. As a first step, I am just interested in the shape of the trajectory, which is why I can ignore the equant, because the uniform motion arount the equant does not change the fact, that the epicycle rolls on the deferent, which is why <span class="math-container">$\phi_0^*$</span> can be expressed in terms of <span class="math-container">$\phi^*$</span>, <span class="math-container">$r_A$</span> and <span class="math-container">$r_a$</span>. I want to express the angle <span class="math-container">$\phi^*$</span> or <span class="math-container">$\phi_0^*$</span> in terms of <span class="math-container">$\gamma$</span>. I assume <span class="math-container">$r_A$</span>, <span class="math-container">$r_a$</span> and <span class="math-container">$c$</span> (distance from Earth A to the center of the deferent <span class="math-container">$B$</span> and distance from this center to the equant as well). How can I express <span class="math-container">$\phi^*$</span> geometrically?
<a href="https://i.sstatic.net/sQ0fE.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/sQ0fE.png" alt="enter image description here" /></a></p>

<observational-astronomy><history><mathematics><modeling><geocentrism>
Angles in Ptolemaic Model
Joe_base
# Angles in Ptolemaic Model

(Asked by: username_0 on 2020-11-24)

I am about to work with Ptolemy's Model for the motion of an upper Planet (Mars). I use a epicycle rolling on the deferent. As a first step, I am just interested in the shape of the trajectory, which is why I can ignore the equant, because the uniform motion arount the equant does not change the fact, that the epicycle rolls on the deferent, which is why $\phi\_0^\*$ can be expressed in terms of $\phi^\*$, $r\_A$ and $r\_a$. I want to express the angle $\phi^\*$ or $\phi\_0^\*$ in terms of $\gamma$. I assume $r\_A$, $r\_a$ and $c$ (distance from Earth A to the center of the deferent $B$ and distance from this center to the equant as well). How can I express $\phi^\*$ geometrically?
[![enter image description here](https://i.sstatic.net/sQ0fE.png)](https://i.sstatic.net/sQ0fE.png)

**Tags:** observational-astronomy, history, mathematics, modeling, geocentrism

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## 1 answers

### Answer by username_1 on 2023-06-16. Score: 3

You misunderstood Ptolemy’s model: The epicycle doesn’t “roll on” the deferent; rather, the epicycle’s *center* is on deferent. Mind you, it’s only a minor detail in this situation.

In the diagram you drew up, it is impossible to know φ\* in terms of γ alone, as φ₀ could take any value. Also, in Ptolemy’s model, φ₀ does *not* depend on φ\*, so φ₀ ≠ δφ\*. However, both vary uniformly, by quantities that we will here call ωₐ and ωₜ after Pedersen.¹ For example, in the case of Mars, ωₜ = 0°;31,26,36,53,51,33/day and ωₐ = 0°;27,41,40,19,20,58/day.

You are correct to presume that c, rₐ, and rA were known and specified in Ptolemy’s model—a good part of the *Almagest* is spent determining such values for rₐ and rA supposing that c = 60 units.

Finally, nowhere in the *Almagest* are φ\* or φ₀ expressed in terms of γ, as both of them, as hinted two paragraphs above, are only time-dependent. Because none of the angular speeds given for ωₐ and ωₜ is a factor of 360°, this means that for a given γ, *any* value of φ\* and φ₀ is possible, depending on how long has elapsed since the reference date (the “epoch”), which Ptolemy chooses, for historical reasons, to be February 26, −746 (aka 747 BCE), Noon Alexandria true solar time (Julian Day 1448637.91666…).

If you understand the language, I have translated the *Almagest* to French, with some diagrams animated or interactive. It’s freely available at <https://ecliptiqc.ca>.

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¹Pedersen, O. *A Survey of the Almagest.* With Annotation and New Commentary by Alexander Jones. New York: Springer, 2011, 480 p., ISBN 978-0-387-84825-9.


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