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<\/a><\/p>\nIn this diagram, circle $ABGD$ is now the meridian and circle $AEG$ the celestial equator with $Z$ as one of its poles.<\/p>\n
We’ll let circle $BED$ be the horizon with a star on it at point $H$.<\/p>\n
We’ll then draw a great circle through the star and the celestial poles intersecting the celestial equator at point $\\Theta$.<\/p>\n
Lastly, we’ll draw in point $K$ which is equidistant as $E$ from $\\Theta$.<\/p>\n
Again, what we’re wanting to understand here is what point on the celestial equator is on the horizon at the same time as $H$, which would be $E$.<\/p>\n
As in the previous post, we’ll use a Menelaus configuration which I’ve already highlighted:<\/p>\n
$$\\frac{Crd \\; arc \\; 2ZB}{Crd \\; arc \\; 2BA} = \\frac{Crd \\; arc \\; 2ZH}{Crd \\; arc \\; 2H \\Theta} \\cdot \\frac{Crd \\; arc \\; 2 \\Theta E}{Crd \\; arc \\; 2AE}$$<\/p>\n
Let’s go through the pieces:<\/p>\n
$Arc \\; ZB$ can be found by subtracting $90\u00ba – arc \\; AB$ since $arc \\; AZ$ is $90\u00ba$ by definition (since it’s a great circle to its pole).<\/p>\n
But what is $arc \\; AB$? It’s the latitude of the observer or, as Ptolemy puts it, the “elevation of the pole”<\/span>1<\/sup><\/a><\/span>.<\/p>\n For $arc \\; ZH$, this is again, $90\u00ba – arc \\; \\Theta H$.<\/p>\n Where does $arc \\; \\Theta H$ come from? Ptolemy tells us that we can get this from the calculations we did in the previous post in which we looked at the simultaneous culmination. While $H$ won’t be on the horizon in that case, it does maintain its position relative to the celestial equator, and thus its value will be unchanged between these two scenarios.<\/p>\n Lastly, we have $arc \\; AE$ which is $90\u00ba$ since it’s between a point on the horizon and a point on the meridian.<\/p>\n Thus, we have everything we need to solve for the missing piece, $arc \\; \\Theta E$, recalling that $\\Theta$, in this diagram, is the Right Ascension of the star in question. Since that is a known quantity<\/span>2<\/sup><\/a><\/span>, we could then figure out where $E$ lies along the celestial equator.<\/p>\n You may notice that we never actually used point $K$ in this situation. Well, that one comes in to play if we were to consider the simultaneous setting. As Ptolemy puts it,<\/p>\n For the simultaneous settings, too, it can easily be seen that if we cut off an arc, $\\Theta K$, in advance of $\\Theta$ equal to $arc \\; Theta E$, the star will set together with point $K$ of the equator. For in that situation the setting takes place on an arc [of the horizon measured from the meridian] equal to $arc \\; BH$, and cuts off an angle in advance of the meridian equal to that enclosed to the rear [of it] by $arc \\; AZ$ and $arc \\; Z \\Theta$ in the present situation.<\/p><\/blockquote>\n In other words, the rising point on the horizon will trail the Right Ascension of the star where as the setting point will precede it in a symmetrical manner. This is quite easy to see if I just let the sphere rotate and carry the star around to the other horizon:<\/p>\n Here, we can clearly see that $K$ is the point on the celestial equator that hits the horizon simultaneously with the star.<\/p>\n But what about the point of the ecliptic that rises and sets with the star?<\/p>\n For that, Ptolemy tells us that<\/p>\n from the arcs of the equator and ecliptic which rise and set together, which we have computed for each clima [II.8<\/a>], there will immediately be given that point on the ecliptic which rises together with point $E$ of the equator and the star, and that point which sets together with point $K$ and the star.<\/p><\/blockquote>\n In other words, once we know the arc of the celestial equator, we can use the rising time table to determine the related section of the ecliptic.<\/p>\n Lastly, Ptolemy states,<\/p>\n It is clear that, at the moment when the sun is exactly in those points of the ecliptic, there will come to pass the risings, culminations, and settings of the fixed star [in question] taken with respect to the sun’s centre which are called ‘true simultaneous cardinal positions.’<\/p><\/blockquote>\n This is really more of a definition than anything else, but is how Ptolemy ends Chapter $5$.<\/p>\n In the next post, we’ll follow along as Ptolemy explores the first and list visibilities of a given star.<\/p>\n<\/a><\/p>\n
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