{"id":4373,"date":"2023-12-23T18:14:29","date_gmt":"2023-12-24T00:14:29","guid":{"rendered":"https:\/\/jonvoisey.net\/blog\/?p=4373"},"modified":"2023-12-23T18:14:29","modified_gmt":"2023-12-24T00:14:29","slug":"almagest-book-ix-position-of-venus-about-epicycle-dec-156-138-ce","status":"publish","type":"post","link":"https:\/\/jonvoisey.net\/blog\/2023\/12\/almagest-book-ix-position-of-venus-about-epicycle-dec-156-138-ce\/","title":{"rendered":"Almagest Book X: Position of Venus About the Epicycle on Dec 15\/16 138 CE"},"content":{"rendered":"

Now that we’ve worked out the basic parameters of Venus’ epicycle and eccentre, we’ll need to determine the motion about the epicycle<\/span>1<\/sup><\/a><\/span>.<\/p>\n

As we did for Mercury, we’ll determine the position at two<\/a> points<\/a> in time and then use that to calculate the speed. In this post, we’ll concentrate on determining the position for the first of these observations.<\/p>\n

[$1$] In the second year of Antoninus, Tybi [V] $29\/30$ in the Egyptian calendar, [December $15\/16$ $138$ CE], we observed the planet Venus, after its greatest elongation as a morning star, using the astrolabe and sighting it with respect to Spica: its apparent longitude was $6 \\frac{1}{2}\u00ba$ into Scorpio. At that moment, it was also between and on a straight line with the northernmost of the stars in the forehead of Scorpius and the apparent centre of the moon, and was in advance of the moon’s centre $1 \\frac{1}{2}$ times the amount it was to the rear of the northernmost of the stars in the forehead. Now, the [latter] fixed star had, at that time, according to our coordinates, a longitude of $6;20$ in Scorpio, and is $1;20\u00ba$ north of the ecliptic. The time was $4 \\frac{3}{4}$ equinoctial hours after midnight, since the sun was about $23\u00ba$ into Sagittarius, and the second degree of Virgo [i.e. $1\u00ba-2\u00ba$ into Virgo] was culminating according to the astrolabe.<\/p>\n

At that moment, the positions were as follows<\/span>2<\/sup><\/a><\/span>:<\/p>\n

– mean longitude of the sun: $22;09\u00ba$ into Sagittarius
\n– mean longitude of the moon: $11;24\u00ba$ into Scorpio
\n– [argument of] latitude of the moon, from the northern limit: $87;30\u00ba$
\n– hence, true position of the moon’s centre: $5;45\u00ba$ into Scorpio
\n– [moon’s latitude] $5\u00ba$ North of the ecliptic
\n– apparent position [of the moon] as Alexandria in longitude: $6;45\u00ba$ into Scorpio
\n– [apparent position of the moon in latitude]: $4;40\u00ba$ North of the ecliptic<\/p>\n

From these considerations too, then, Venus’ longitude was $6;30\u00ba$ into Scorpio, and it was $2;40\u00ba$ north of the ecliptic.<\/p><\/blockquote>\n

With all of the preliminary configurations out of the way, let’s go ahead and jump into determining the position of Venus at that time.<\/p>\n

To do so, Ptolemy lays out the following diagram:<\/p>\n

\"\"<\/a><\/p>\n

This diagram is one that looks a lot scarier than it really is in my opinion, largely due to how scrunched up many of the points are. However, it’s pretty simple.<\/p>\n

As usual, we have $\\overline{AE}$ as the line of apsides, again with $A$ as the point $25\u00ba$ into Taurus (apogee) and $E$, the point $25\u00ba$ into Scorpio (perigee).<\/p>\n

The circle about $Z$ is the epicycle with $Z$ at its center. We’ve taken points $B$ (the equant about which the center of the epicycle rotates uniformly) and $D$ (the observer on Earth), and extended lines through the center of the epicycle producing points $\\Theta$ and $H$, on the far side, respectively. A line is also produced from $G$ (the point about which the eccentre rotates) but terminates at $Z$.<\/p>\n

We’ll let Venus be at point $K$, which we can see it just past greatest elongation, and connect that to both $D$ and $Z$.<\/p>\n

All that remains from there is making a few right triangles. We’ll make one by producing lines from $G$ and $D$ such that they fall on $\\overline{BH}$ perpendicularly, as well as producing a line from $Z$ such that it meets $\\overline{DK}$ at a right angle.<\/p>\n

First, let’s determine $\\angle EBZ$. To do that, first recall that $E$ is the point $25\u00ba$ into Scorpio. Then, at the time of the observation the sun was at $22;09\u00ba$ into Sagittarius. Recalling that the position of the sun is taken from the center about which the eccentre rotates ($B$), this means that $\\angle EBZ$ is the difference between these two values which is $27;09\u00ba$.<\/p>\n

Now we’ll focus on $\\triangle BGL$ creating a demi-degrees context about it. In it, the hypotenuse, $\\overline{BG} = 120^p$ and we just determined $\\angle GBL = 27;09\u00ba$.<\/p>\n

Thus, the corresponding arc, $arc \\; GL = 54;18\u00ba$ and its chord $\\overline{GL} = 54;46^p$.<\/p>\n

We can then use the Pythagorean theorem to determine $\\overline{BL} = 106;46^p$. Ptolemy instead does this through the arcs and chords, determining $arc \\; BL = 125;42\u00ba$ (as it’s the supplement of $arc \\; GL$) and then uses the chord table to find $\\overline{BL}$ which gets him the slightly different value of $106;47^p$.<\/p>\n

However, in previous post<\/a>, we showed that the distance between the earth and the center of the eccentre ($\\overline{DG}$ in this diagram) was $1;15^p$ which is equal to $\\overline{BG}$. And since this was half of the distance between earth and the equant ($\\overline{BD}$ in this diagram), that means that the other half, $\\overline{BG}$ is also $1;15^p$ in our larger context for this model.<\/p>\n

We can use this to switch the context of this small triangle to this context. Doing so, I find that $\\overline{GL} = 0;34^p$ and $\\overline{BL} = 1;07^p$.<\/p>\n

Now, we’ll concentrate on $\\triangle GLZ$. In this, we just determined $\\overline{GL}$, but we also know $\\overline{GZ}$ as it’s the center around which the eccentre rotates and thus has a radius of $60^p$.<\/p>\n

Thus, we can use the Pythagorean theorem to determine $\\overline{LZ} \\approx 60^p$ as well<\/span>3<\/sup><\/a><\/span>.<\/p>\n

Next, we can also state that $\\overline{BL} = \\overline{LM}$. I suspect there’s a few ways to prove this, but we can see it’s true by noting that $\\triangle BGL$ is similar to $\\triangle BDM$, just twice the size. Then, because $\\overline{BG} = \\overline{GD}$, so too must $\\overline{BL} = \\overline{LM}$. Thus, $\\overline{LM} = 1;07^p$ as well.<\/p>\n

This similarity of triangles also means that $\\overline{DM} = 2 \\cdot \\overline{GL} = 1;08^p$.<\/p>\n

We can then subtract:<\/p>\n

$$\\overline{MZ} = \\overline{LZ} – \\overline{LM} = 60^p – 1;07^p = 58;53^p.$$<\/p>\n

Next, we can use the Pythagorean theorem on $\\triangle DMZ$ to determine<\/p>\n

$$ \\overline{DZ} = \\sqrt{\\overline{MZ}^2 + \\overline{DM}^2} = \\sqrt{{58;53^p}^2 + {1;08^p}^2} = 58;54^p.$$<\/p>\n

We’ll now create a demi-degrees circle about this triangle. Within it, the hypotenuse, $\\overline{DZ} = 120^p$. In this context, $\\overline{DM} = 2;19^p$ although Ptolemy rounds down to come up with $2;18^p$.<\/p>\n

In that case, $arc \\; DM = 2;12\u00ba$ which means the opposite angle, $\\angle DZM = 1;06\u00ba$<\/span>4<\/sup><\/a><\/span>.<\/p>\n

Next, we’ll add $\\angle EBZ + \\angle DZM = 27;09\u00ba + 1;06\u00ba = 28;15\u00ba$.<\/p>\n

This is equal to $\\angle EDZ$.<\/p>\n

This can, again, be proven using a system of three equations:<\/p>\n

$$\\angle EDZ + \\angle ZDM + \\angle MDB = 180\u00ba$$<\/p>\n

$$\\angle ZDM + \\angle DZM = 90\u00ba$$<\/p>\n

$$\\angle MDB + \\angle DBM = 90\u00ba$$<\/p>\n

If we solve the second and third equations for $\\angle ZDM$ and $\\angle MDB$ respectively, and then plug them into the first equation and then simplify, we get:<\/p>\n

$$\\angle EDZ = \\angle DBM + \\angle DZM.$$<\/p>\n

However, $\\angle DBM = \\angle EBZ$ which allows us to write this as:<\/p>\n

$$\\angle EDZ = \\angle DZM + \\angle EBZ.$$<\/p>\n

Next, we’ll look back at the observation and note that the Venus was actually observed at $6 \\frac{1}{2}\u00ba$ into Scorpio. This is $18;30\u00ba$ away from the perigee at $25\u00ba$ into Scorpio. Thus, $\\angle EDK = 18;30\u00ba$.<\/p>\n

We can then add:<\/p>\n

$$\\angle KDZ = \\angle EDK + \\angle EDZ = 18;30\u00ba + 28;15\u00ba = 46;45\u00ba.$$<\/span><\/p>\n

Now we’ll take a look at $\\triangle DZN$, creating a demi-degrees circle about it.<\/p>\n

In it, the hypotenuse, $\\overline{DZ} = 120^p$. Then $arc \\; NZ = 93;30\u00ba$ and its corresponding chord, $\\overline{ZN} = 87;24^p$. Ptolemy comes up with $87;25^p$ which is an interesting choice given that the value I listed is directly out of the table of chords<\/a> with no interpolation required where his is rounded up for no reason.<\/p>\n

We can now return the larger context, where we showed that $\\overline{DZ} = 58;54^p$. Doing so, we find that $\\overline{ZN} = 42;54^p$.<\/p>\n

In this same context, we’ve also shown<\/a> that the radius of the epicycle, $\\overline{ZK} = 43;10^p$.<\/p>\n

Now, we’ll create a demi-degrees circle about $\\triangle KNZ$. In this, the hypotenuse, $\\overline{ZK} = 120^p$ which allows us to also convert $\\overline{NZ}$ into this context. Doing so, I find $\\overline{NZ} = 119;16^p$. Ptolemy again has done some inexplicable rounding and come up with $119;18^p$<\/span>5<\/sup><\/a><\/span> in this context.<\/p>\n

We can then determine the corresponding arc, $arc \\; ZN = 167;37\u00ba$. Again, Ptolemy has evidently done some rounding or miscalculation and comes up with $167;38\u00ba$.<\/p>\n

Thus, the opposite arc, $arc \\; ZKD = 83;49\u00ba$.<\/p>\n

Next, we’ll need to do another little side proof. We’ll write two equations:<\/p>\n

$$\\angle KZH + \\angle KZD = 180\u00ba.$$<\/p>\n

$$\\angle KZD + \\angle KDZ + \\angle ZKD = 180\u00ba.$$<\/p>\n

We’ll solve the second of these for $\\angle KZD$ and substitute it into the first. Then, simplifying we find that:<\/p>\n

$$\\angle KZH = \\angle\u00a0 KDZ + \\angle ZKD.$$<\/p>\n

Thus,<\/p>\n

$$\\angle = 46;45\u00ba + 83;49\u00ba = 130;34\u00ba.$$<\/p>\n

Lastly, we’ll subtract $\\angle \\Theta ZH$ from this which is the vertical angle of $\\angle DZM$ which was $1;06\u00ba$.<\/p>\n

Thus, $\\angle KZ \\Theta = 130;34\u00ba – 1;06\u00ba = 129;28\u00ba$.<\/p>\n

Recalling that $\\Theta$ was the apogee of the epicycle, this means that, at the time of this observation, Venus was $129;28\u00ba$ in advance of apogee on its epicycle. Stating this another way, it is $230;32\u00ba$\u00a0after<\/em> apogee (measuring counter-clockwise) which is what we wanted to prove for this observation.<\/p>\n

In the next post, we’ll repeat this calculation for an ancient observation of Venus and then use the change in the intervening time to determine the mean motion.<\/p>\n

\n

\"\"<\/a><\/p>\n

\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

Now that we’ve worked out the basic parameters of Venus’ epicycle and eccentre, we’ll need to determine the motion about the epicycle. As we did for Mercury, we’ll determine the position at two points in time and then use that to calculate the speed. In this post, we’ll concentrate on determining the position for the … <\/p>\n

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