Almagest Book X: Second Iteration Correction for Equant – Second Opposition

Next, we’ll work on the correction for the second opposition in our second iteration.

As with before, we’ll make use of our previous diagram:

In this, $arcXZ=39;04º$ as this was the distance we determined the second observation was from the line of apsides.

This is also equal to $\mathrm{\angle }Z\mathrm{\Theta }X$ and its vertical angle $\mathrm{\angle }D\mathrm{\Theta }F$.

As a reminder, $\overline{D\mathrm{\Theta }}=\overline{DN}=5;{55}^p$ in this second iteration.

With that, we can use some trig to determine $\overline{DF}$:

$$\overline{DF}=\overline{D\mathrm{\Theta }}\cdot sin(\mathrm{\angle }D\mathrm{\Theta }F)=5;{55}^p\cdot sin(39;04º)$$

$$\overline{DF}=3;{44}^p.$$

Next, $\overline{DB}={60}^p$ since it’s a radius of one of the eccentres. So we now know two sides in $\mathrm{△}DBF$ allowing us to state via the Pythagorean theorem:

$$\overline{BF}=\sqrt{{60}^2–3;{44}^2}=59;{53}^p.$$

And using a bit more trig or the Pythagorean theorem, we can determine $\overline{F\mathrm{\Theta }}=4;{36}^p$ which has the same measure as $\overline{QF}$.

We can now add $\overline{BF}+\overline{FQ}$ to find,

$$\bar{B}Q=59;{53}^p+4;{36}^{=}46;{29}^p.$$

We can also state that

$$\overline{NQ}=2\cdot \overline{DF}=2\cdot 3;{44}^p=7;{28}^p.$$

That gives us two sides in $\mathrm{△}NBQ$ so we can use a bit of trig to determine:

$$\mathrm{\angle }NBQ=ta{n}^{-1}(\frac{\overline{QN}}{\overline{QB}})$$

$$\mathrm{\angle }NBQ=ta{n}^{-1}(\frac{7;{28}^p}{64;{29}^p})=6;36º.$$

We’ll now add a few pieces together to find $\overline{QZ}$:

$$\overline{QZ}=\overline{Z\mathrm{\Theta }}+\overline{\mathrm{\Theta }F}+\overline{FQ}={60}^p+4;{36}^p+4;{36}^p=69;{12}^p.$$

That gives us two of the sides in $\mathrm{△}NZQ$ allowing us to use a bit more trig to determine,

$$\mathrm{\angle }NZQ=ta{n}^{-1}(\frac{\overline{QN}}{\overline{QZ}})$$

$$\mathrm{\angle }NZQ=ta{n}^{-1}(\frac{7;{28}^p}{69;{12}^p})=6;09º.$$

Lastly, we can subtract to find

$$\mathrm{\angle }BNZ=\mathrm{\angle }NBQ–\mathrm{\angle }NZQ=6;36º–6;09º=0;27º.$$

This is also $arcLT$ which is what we were looking for in this post.

Ptolemy gives a value of $0;28º$ whereas Toomer gives a value of $0;26,51º$ in good agreement with mine¹.

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1. If I carry my value to more sexagesimal places, I get $0;26;46º$, so slightly different from Toomer, but easily within reason given rounding discrepancies.