Having computed the lunar parallax,
the sun’s parallax for a similar situation [i.e., as measured along an altitude circle] is immediately determined, in a simple fashion (for solar eclipses) from the number in the second column corresponding to the size of the arc from the zenith [to the sun].
Well that sure sounds easy. Let’s look at a quick example.
First, the quote I used above refers to an altitude circle. Being that this portion is in brackets, this is a note from Toomer, but as a reminder, an altitude circle is just a great circle through the zenith which will necessarily be perpendicular to the horizon. It’s along this circle that we measure altitude, starting from the horizon. However, in this case, we’re not wanting to measure altitude, but again will want to measure angular distance from the zenith, so along the same circle.
This process is ultimately no different than what we did for the moon. So let’s take an example solar position of $10º$ into Leo with the sun being an 2 hours past the meridian1, being observed from my home location in the Barony of Three Rivers or St. Louis, MO. We’ll refer to our table of zenith distances, using the one for Rhodes which has a similar latitude to St. Louis.
So I will look up the value for 2 hours in Leo which is $30;28º$. But again, this is just the first point in Leo. So I’ll do the same for the next constellation, Virgo and get $36;24º$. We were $\frac{1}{3}$ of the way through Leo so doing a bit of interpolation, this would give an arc from the zenith of $32;27º$.
This then gets looked up using the parallax table, inserting this value into column $1$ and getting the result from column $2$. Again, the value isn’t here, so we’ll need to interpolate a bit for which I get a result of $0;01;31º$.
And that’s all there is too it.
Wow… knocked off a whole two sentences in this post…