We’re now ready to make the correction for the third opposition in this second iteration. And the good news is that this is the last time we’ll need to calculate a correction.
Ptolemy still does a third iteration which relies on these corrections, but since he won’t be doing a fourth iteration, further corrections aren’t necessary. So let’s get to it.
We’ll again reuse the diagram from the last time around.
This time, $arc \; HP = 45;28º$ as we’d found it in this post in which we’d called it $arc \; GM$.
This means that $\angle H \Theta P$ has the same measure.
In this iteration, we also showed that the eccentricity, $\overline{D \Theta} = \overline{DN} = 5;55^p$.
Then, using a bit of trig we can state:
$$\overline{DF} = 5;55^p \cdot sin(45;28)$$
$$\overline{DF} = 4;13^p.$$
Additionally,
$$\overline{F \Theta} = 5;55^p \cdot cos(45;28)$$
$$\overline{F \Theta} = 4;09^p.$$
Recalling that $\overline{DG} = 60^p$ since it’s a radius, we then have two sides in $\triangle FGD$ allowing us to use the Pythagorean theorem to state:
$$\overline{FG} = \sqrt{60^2 – 4;13^2}$$
$$\overline{FG} = 59;51^p.$$
As with before, $\overline{QF} = \overline{F \Theta} = 4;09^p$.
We can then write:
$$\overline{GQ} = \overline{FG} – \overline{QF}$$
$$\overline{GQ} = 59;51^p-4;09^p = 55;42^p.$$
Additionally, $\overline{QN} = 8;26^p$ since it’s twice $\overline{DF}$.
That gives us two sides of $\triangle NGQ$, so we can use a bit more trig to determine,
$$\angle NGQ = tan^{-1}(\frac{\overline{QN}}{\overline{GQ}})$$
$$\angle NGQ = tan^{-1}(\frac{8;26^p}{55;42^p})$$
$$\angle NGQ = 8;37º.$$
Next, we’ll recall that $\overline{H \Theta} = 60^p$, again being a radius. We can subtract off $\overline{Q \Theta}$ to find that,
$$\overline{QH} = 51;42^p.$$
That now gives us two of the sides in $\triangle NHQ$ allowing us to write:
$$\angle NHQ = tan^{-1}(\frac{\overline{QN}}{\overline{QH}})$$
$$\angle NHQ = tan^{-1}(\frac{8;26}{51;42^p})$$
$$\angle NHQ = 9;16º.$$
Lastly, we can subtract to find $\angle GNH$:
$$\angle GNH = \angle NHQ – \angle NGQ$$
$$\angle GNH = 9;16º – 8;37º$$
$$\angle GNH = 0;39º.$$
This is also $arc \; MY$ which is our correction.
Ptolemy comes up with a slightly higher value of $0;40º$ which is in agreement with the value Toomer gives of $0;39;31º$. If I carry my calculation out to the second sexagesimal place I find it to be $0;39,22º$. So again, slightly different but within reason.
That’s the end of the corrections.
We now have everything we need to go through our third iteration which we’ll do in the next couple posts.
