For the points on the ecliptic with which each fixed star simultaneously culminates, rises, or sets can be derived geometrically by means of the theorems [already] established.<\/span><\/p><\/blockquote>\n In this post, we’ll explore specifically the culminations and leave the risings and settings for the next post.<\/p>\n <\/p>\n To begin, Ptolemy lays out the following diagram:<\/p>\n In this diagram, circle $AEG$ is the celestial equator with a pole at $Z$. Circle $BED$ is the ecliptic with a pole at $H$. Circle $ABGD$ passes through both of these poles.<\/p>\n Ptolemy considers a star at $\\Theta$, and draws great circles through this point and the poles, intersecting the ecliptic and celestial equator at $K$, $M$, $L$, and $N$.<\/p>\n Let’s pause a moment and remind ourselves of what we’re after here: It’s the point on the ecliptic, $M$, that culminates at the same time as the star, $\\Theta$. We need to know where that falls along the ecliptic in relation to some other known point, such as $K$ which would be the ecliptic longitude of the star or $B$ which is the winter solstice as it’s drawn here and, again, has a known ecliptic longitude.<\/p>\n [It] is obvious that the star at $\\Theta$ culminates simultaneously with points $M$ and $N$ of the ecliptic and equator [respectively].<\/p><\/blockquote>\n This statement took me a bit to convince myself of, so I’ll walk through my thought process.<\/p>\n First, I’ll simplify the drawing, removing all the labeled points, except the ones we’re interested in (although I’ll still remove the labels). Then, I’ll draw in an arbitrary horizon in brown, labeling the cardinal points<\/span>1<\/sup><\/a><\/span> as well as the zenith. The meridian is drawn in in a light blue.<\/p>\n This doesn’t quite get us there, but it gave me a nice starting point to think about this.<\/p>\n Next, we can recall that, as the earth rotates<\/span>2<\/sup><\/a><\/span>, east and west will follow the celestial equator around while, north, south, and the zenith will all trace small circles.<\/p>\n So let’s rotate things a bit and see what happens:<\/p>\n And there it is. Everything snaps into place. The three original points all fall on the meridian.<\/p>\n Thinking about it a bit more, if any points are on a great circle that connects the celestial equatorial poles, then at some point, they will all necessarily be on the meridian together.<\/p>\n Moving on,<\/p>\n But these points, and arc $\\Theta N$, are given, as will be clear from the following considerations.<\/p><\/blockquote>\n As a reminder, when Ptolemy says “are given” he’s really saying “we can figure out relatively easily.” How so?<\/p>\n We will use<\/p>\n what we proved at the beginning of our treatise.<\/p><\/blockquote>\n In other words, Menelaus’ theorem<\/a>.<\/p>\n [S]ince the [two] great circle arcs $HL$ and $NZ$ have been drawn to meet the two great circle arcs $AH$ and $AN$<\/p><\/blockquote>\n $$\\frac{crd \\; arc \\; 2HA}{crd \\; arc \\; 2AZ} = \\frac{crd \\; arc \\; 2HL}{crd \\; arc \\; 2L \\Theta} \\cdot \\frac{crd \\; arc \\; 2N \\Theta}{crd \\; arc \\; 2ZN}$$<\/p>\n Let me highlight the Menelaus configuration in question here:<\/p>\n As usual, let’s go through the parts of the Menelaus configuration that we need to know, keeping in mind that we’re wanting to solve for $arc \\; N \\Theta$.<\/p>\n First off, $arc \\; AH$ is composed of $arc \\; AZ + arc \\; ZH$ where $arc \\; AZ = 90\u00ba$ as it’s the arc between a great circle (the celestial equator) and its pole and $arc \\; ZH$ is the obliquity of the ecliptic<\/span>3<\/sup><\/a><\/span>. Thus, we know this piece.<\/p>\n For $arc \\; AZ$, this too is $90\u00ba$ as it is also the arc between a great circle (the ecliptic) and its pole.<\/p>\n Next, $arc \\; HL$ is made up of $arc \\; HK + arc \\; KL$ where $arc \\; HK$ is also $90\u00ba$ for the same reason as above. If we consider $arc \\; KL$, this is the arc between the celestial equator and ecliptic – something we looked at in I.14<\/a> and built into a table in I.15<\/a>. We’d just need to know where on the ecliptic we’re starting from, which is point $K$. Fortunately, we do know this as this is just the ecliptic longitude of the star in question, so we would be able to look up.<\/p>\n Similarly, $arc \\; L \\Theta$ is the declination<\/span>4<\/sup><\/a><\/span> of the star in question.<\/p>\n Lastly, $arc \\; ZN$ is $90\u00ba$ for the same reason as we’ve seen previously.<\/p>\n Thus, we clearly has all the pieces necessary to solve for $arc \\; N \\Theta$ even though Ptolemy doesn’t go through a specific example.<\/p>\n But that’s not what we were after.<\/p>\n So next, Ptolemy invokes other form of Menelaus’ theorem on the same configuration:<\/p>\n $$\\frac{crd \\; arc \\; 2ZH}{crd \\; arc \\; 2HA} = \\frac{crd \\; arc \\; 2Z \\Theta}{crd \\; arc \\; 2 \\Theta N} \\cdot \\frac{crd \\; arc \\; 2NL}{crd \\; arc \\; 2LA}$$<\/p>\n Let’s go through the parts.<\/p>\n Again, $arc \\; ZH$ is the obliquity of the ecliptic.<\/p>\n $Arc \\; HA$ is that plus $90\u00ba$ ($arc \\; ZA$).<\/p>\n $Arc \\; Z \\Theta$ is $90\u00ba$ minus the declination of the star.<\/p>\n $Arc \\; \\Theta N$ is the declination of the star.<\/p>\n Next, we need $arc \\; LA$. Ptolemy then states that<\/p>\n $arc \\; LA$ is given from $arc \\; KB$, by means of [the arcs of] the equator which rise together with [those of] the ecliptic at sphaera recta<\/em>…<\/p><\/blockquote>\n This statement is a bit opaque, but what Ptolemy is stating is that we can determine $arc \\; LA$ because we know $arc \\; KB$ which is related to the ecliptic longitude of the star.<\/p>\n Knowing that, we can apply the same logic as used in I.16<\/a>, which looks at what section of the equator rises with an equal part of the ecliptic.<\/p>\n Thus, we can solve for $arc \\; NL$.<\/p>\n Why is that useful?<\/p>\n Well, $arc \\; LA + arc \\; NL$, both of which we know, equals $arc \\; AN$.<\/p>\n Then,<\/p>\n [by means of rising times at\u00a0sphaera recta<\/em>], $arc \\; MB$ of the ecliptic will be given from $arc \\; NA$.<\/p><\/blockquote>\n In other words, applying the same logic of rising times of the ecliptic we just mentioned, we can use $arc \\; AN$ of the celestial equator to determine the related arc of the ecliptic, $arc \\; BM$ which is what we’re after.<\/p>\n I’ll stop this post here. In the next one, we’ll explore a similar problem of which points on the ecliptic and equator rise and set with a fixed star.<\/p>\n![\"\"](\"https:\/\/i0.wp.com\/jonvoisey.net\/blog\/wp-content\/uploads\/2023\/04\/5.8.jpg?ssl=1\")
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