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

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

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

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

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

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

Next, $\overline{DB} =60^p$ since it’s a radius of one of the eccentres. So we now know two sides in $\triangle DBF$ 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 $\overline{F \Theta} = 4;36^p$ which has the same measure as $\overline{QF}$.

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

$$\overline BQ = 59;53^p + 4;36^ = 46;29^p.$$

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 $\triangle NBQ$ so we can use a bit of trig to determine:

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

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

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

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

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

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

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

Lastly, we can subtract to find

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

This is also $arc \; LT$ 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.


 

  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.