# κ₀ for Mercury—Formula (another definition)

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.

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

• 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: $-6e^2\cos(\tilde{\kappa}_0)^3+(R_D^2-4e^2)\cos(\tilde{\kappa}_0)^2+2e^2\cos(\tilde{\kappa}_0)-e^2=0$
– ayr
Commented Jan 4 at 15:16
• @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). Commented Jan 4 at 18:04
• According to the conditions of this problem, I got this equation: $16 \left(2 R^2-e^2\right) \cos ^4(\tilde{\kappa}_0)+8 \left(7 e^2-4 R^2\right)\cos ^3(\tilde{\kappa}_0)+4 e^2 \cos ^2(\tilde{\kappa}_0)-16 \left(e^2-R^2\right) \cos (\tilde{\kappa}_0)+(5e^2-4 R^2)=0$
– ayr
Commented Jan 5 at 13:30
• $\Delta OEC$ 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.
– ayr
Commented Jan 5 at 13:30
• 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.
– ayr
Commented Jan 6 at 5:58

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.