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Following my other question about a specific “hidden” formula in Ptolemy’s model for Mercury, I am now looking for yet another “hidden” formula, this time the one used to find $\bar\kappa_0$ so that $OC=60$ in the figure below. We are given that $DC=60$ (in arbitrary units), that angles $AFD$ and $AEC$ are equal but of opposite direction (so $\angle AFD = -\angle AEC$), and that $OE=EF=FD$ (which is equal to 3, but I’d like to keep 3 out of the equation and just symbolize it by e (for “eccentricity”).

($A$ is the apogee of Mercury’s deferent centered on $D$. $D$ rides on the small circle centered on $F$. The Earth is at point $O$ [for “observer”]. Points $O$, $E$, and $F$ are always in a straight line with $A$. Mercury itself rides on the epicycle centered on $C$.)

While Ptolemy knew of absolutely no trigonometry and other modern mathematics, I don’t mind the answer to be expressed in modern terms, although simple trigonometry would be nice since math is not my forte.

enter image description here

P.S. I’ll also ask this question on MathSE.

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    $\begingroup$ Please check whether the conditions $OC=60$ and $DC=60$ are formulated correctly. The fact is that if they are true, then the $\Delta OEC$ cannot be right-angled. By the way, in my calculation I simply substituted $OE = e$, and instead of $R = R_D$ (that is, I replaced numbers with symbols). And I got the following equation: $-6e^2\cos(\tilde{\kappa}_0)^3+(R_D^2-4e^2)\cos(\tilde{\kappa}_0)^2+2e^2\cos(\tilde{\kappa}_0)-e^2=0$ $\endgroup$
    – ayr
    Commented Jan 4 at 15:16
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    $\begingroup$ @dtn: Yes, this is a different problem. Both $OC$ and $DC$ are here $60$ units and indeed, $\triangle OEC$ is not a right triangle. And awesome for the formula (which I guess applies to my first problem). $\endgroup$ Commented Jan 4 at 18:04
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    $\begingroup$ According to the conditions of this problem, I got this equation: $16 \left(2 R^2-e^2\right) \cos ^4(\tilde{\kappa}_0)+8 \left(7 e^2-4 R^2\right)\cos ^3(\tilde{\kappa}_0)+4 e^2 \cos ^2(\tilde{\kappa}_0)-16 \left(e^2-R^2\right) \cos (\tilde{\kappa}_0)+(5e^2-4 R^2)=0$ $\endgroup$
    – ayr
    Commented Jan 5 at 13:30
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    $\begingroup$ $\Delta OEC$ is not rectangular, but now $OC=DC$ :)) However, please note that the equation for such conditions will no longer have the 3rd, but the 4th degree. $\endgroup$
    – ayr
    Commented Jan 5 at 13:30
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    $\begingroup$ Oh my...It seems I made a mistake. The first coefficient is written as $16(2e^2-R^2)$ I mixed up $e$ and $R$ :)) Yes, your version shows exactly that, you are right. $\endgroup$
    – ayr
    Commented Jan 6 at 5:58

1 Answer 1

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I’ll write how I got the 4th degree polynomial, which is mentioned in the comments. I used the same formulas from the answer to the previous formulation of the problem. According to the conditions of the new problem, $OC = DC = R$.

That is:

$OC=\sqrt{\text{CE}^2-2 \text{CE} \cdot\text{OE} \cos(\angle\text{OEC})+\text{OE}^2}=R$

Hence:

$\text{CE}^2-2 \text{CE} \cdot\text{OE} \cos(\angle\text{OEC})+\text{OE}^2=R^2$

We also know that:

$\text{CE}=\sqrt{R^2-\text{DE}^2 \sin ^2(\text{CED})}+\text{DE} \cos (\text{CED})$

Let's substitute $CE$ into the formula for $OC$, open all the brackets and get:

$-2 \text{OE} \cos (\angle \text{OEC}) \sqrt{R^2-\text{DE}^2 \sin^2(\angle \text{CED})}+2 \text{DE} \cos (\angle \text{CED})\sqrt{R^2-\text{DE}^2 \sin ^2(\angle \text{CED})}-\text{DE}^2\sin ^2(\angle \text{CED})+\text{DE}^2 \cos ^2(\angle \text{CED})-2\text{DE} \cdot \text{OE} \cos (\angle \text{CED}) \cos(\angle \text{OEC})+\text{OE}^2+R^2=R^2$

Let's make a replacement $\sqrt{R^2-\text{DE}^2 \sin^2(\angle \text{CED})}=S$ in the above formula:

$-\text{DE}^2 \sin ^2(\angle \text{CED})+\text{DE}^2 \cos^2(\angle \text{CED})-2 \text{DE} \cdot \text{OE} \cos (\angle \text{CED})\cos (\angle \text{OEC})+2 \text{DE} \cdot S \cos(\angle \text{CED})+\text{OE}^2-2 \text{OE} \cdot S \cos(\angle \text{OEC})+R^2=R^2$

Let express $S$ from it, obtaining:

$S = \frac{(\text{DE}-\text{OE})(\text{DE}+\text{OE})}{2 \text{DE} \cos (\angle \text{CED})-2\text{OE} \cos (\angle \text{OEC})}-\text{DE} \cos(\angle \text{CED})$

Hence:

$\sqrt{R^2-\text{DE}^2 \sin^2(\angle \text{CED})} = \frac{(\text{DE}-\text{OE})(\text{DE}+\text{OE})}{2 \text{DE} \cos (\angle \text{CED})-2\text{OE} \cos (\angle \text{OEC})}-\text{DE} \cos(\angle \text{CED})$

Let's square both sides, rewriting them in the following form:

$R^2-\text{DE}^2 \sin^2(\angle \text{CED})=\frac{\left(2 \text{DE}^2 \cos ^2(\angle \text{CED})-2 \text{DE} \cdot \text{OE} \cos (\angle \text{CED}) \cos(\angle \text{OEC})-\text{DE}^2+\text{OE}^2\right)^2}{4(\text{DE} \cos (\angle \text{CED})-\text{OE} \cos(\angle \text{OEC}))^2}$

Well, therefore:

$R^2-\text{DE}^2 \sin ^2(\angle \text{CED})-\frac{\left(2 \text{DE}^2 \cos ^2(\angle \text{CED})-2 \text{DE} \cdot \text{OE} \cos (\angle \text{CED}) \cos(\angle \text{OEC})-\text{DE}^2+\text{OE}^2\right)^2}{4(\text{DE} \cos (\angle \text{CED})-\text{OE} \cos(\angle \text{OEC}))^2}=0$

From the previous problem (and taking into account the new conditions) we know that:

$\angle CED=\frac{3}{2}\tilde{\kappa}_0; \angle OEC=\pi-\tilde{\kappa}_0; DE=2e\cos(\frac{\tilde{\kappa}_0}{2}); OE=FE=FD=e$

Let’s substitute everything into the previous formula, open the brackets and take out the common factor:

$\frac{\left(26 e^2-8 R^2\right) \cos (\tilde{\kappa}_0)+2 \left(9 e^2-4R^2\right) \cos (2 \tilde{\kappa}_0)+14 e^2 \cos (3 \tilde{\kappa}_0)+4 e^2 \cos (4\tilde{\kappa}_0)+19 e^2-8 R^2 \cos (3 \tilde{\kappa}_0)-2 R^2 \cos (4 \tilde{\kappa}_0)-10 R^2}{2\cos (\tilde{\kappa}_0)+\cos (2 \tilde{\kappa}_0)}=0$

From the expression below we will find the polynomial we need: $pol=\left(26 e^2-8 R^2\right) \cos (\tilde{\kappa}_0)+2 \left(9 e^2-4R^2\right) \cos (2 \tilde{\kappa}_0)+14 e^2 \cos (3 \tilde{\kappa}_0)+4 e^2 \cos (4\tilde{\kappa}_0)+19 e^2-8 R^2 \cos (3 \tilde{\kappa}_0)-2 R^2 \cos (4 \tilde{\kappa}_0)-10 R^2$

Using formulas for multiple angles of trigonometric functions and substitution $\sin(\tilde{\kappa}_0)=\sqrt{1-\cos(\tilde{\kappa}_0)^2}$, we obtain the following expression:

$16 \left(2 e^2-R^2\right) \cos ^4(\tilde{\kappa}_0)+8 \left(7 e^2-4R^2\right) \cos ^3(\tilde{\kappa}_0)+4 e^2 \cos ^2(\tilde{\kappa}_0)-16 \left(e^2-R^2\right) \cos(\tilde{\kappa}_0)+(5 e^2-4 R^2)=0$

This is the required equation for the angle $\tilde{\kappa}_0$ at which $OC=DC=R$.

For $e=3$ and $R=60$, polynomial coefficients will be respectively: $16 \left(2 e^2-R^2\right)=-57312; 8 \left(7 e^2-4 R^2\right)=-114696;4 e^2=36; 16 \left(e^2-R^2\right)=-57456;(5e^2-4 R^2)=-14355$.

Then, polynomial equation will look like this:$-57312X^4-114696X^3+36X^2+57456X-14355=0$, where $X=cos(\tilde{\kappa}_0)$. One of the solutions to this polynomial will be $X=cos(\tilde{\kappa}_0)=0.37887$, and $acos(0.37887)=67.736^{\circ}$

Which coincides with the result of measuring the angle in a sketch constructed according to given geometric conditions in SolidWorks.

enter image description here

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