{"id":4873,"date":"2024-09-04T10:00:29","date_gmt":"2024-09-04T15:00:29","guid":{"rendered":"https:\/\/jonvoisey.net\/blog\/?p=4873"},"modified":"2024-09-04T10:00:29","modified_gmt":"2024-09-04T15:00:29","slug":"almagest-book-xi-determining-the-size-of-saturns-epicycle","status":"publish","type":"post","link":"https:\/\/jonvoisey.net\/blog\/2024\/09\/almagest-book-xi-determining-the-size-of-saturns-epicycle\/","title":{"rendered":"Almagest Book XI: Determining the Size of Saturn’s Epicycle"},"content":{"rendered":"

Next, we’ll turn our attention to determining the size of Saturn’s epicycle. As he did for Jupiter, Ptolemy introduces a new observation.<\/p>\n

[W]e took an observation which we made in the second year of Antoninus, Mecchir [VI] $6\/7$ in the Egyptian calendar [$138$ CE, December $22\/23$]. It was $4$ equinoctial hours before midnight, for according to the astrolabe, the last degree of Aries was culminating, while the longitude of the mean sun was $28;41\u00ba$ into Sagittarius. At that moment, the planet Saturn, sighted with respect to the bright star in the Hyades<\/span>1<\/sup><\/a><\/span>, was seen to have a longitude $9 \\frac{1}{15}\u00ba$ into Aquarius, and was about $\\frac{1}{2}\u00ba$ to the rear of the center of the moon (for that was its distance from the noom’s northern horn).<\/p><\/blockquote>\n

Ptolemy then calculations the moon’s positions as a reference point:<\/p>\n

Mean longitude: $8;55\u00ba$ into Aquarius
\nAnomaly: $174;15\u00ba$ from the apogee of the epicycle
\nhence, its true longitude must have been $9;40\u00ba$ into Aquarius
\nand its apparent longitude at Alexandria, $8;34\u00ba$ into Aquarius<\/span>2<\/sup><\/a><\/span>.<\/p><\/blockquote>\n

This is used as a check for Saturn’s position against the astrolabe position.<\/p>\n

Thus, from these considerations too, the planet Saturn must have had a longitude of $9 \\frac{1}{15}\u00ba$ into Aquarius (since it was about $\\frac{1}{2}\u00ba$ to the rear of the moon’s center.<\/p><\/blockquote>\n

He then give the mean position with respect to the apogee:<\/p>\n

And its distance from the apogee of the eccentre (which was [in] the same [position as at the third opposition], since its shift over so short an interval is negligible), was $76;04\u00ba.$<\/p><\/blockquote>\n

The calculation is not shown here but is the mean ecliptic longitude of Saturn ($9;04\u00ba$ into Aquarius or $309;04\u00ba$) minus the position of apogee ($23;00\u00ba$ into Scorpio or $233\u00ba$).<\/p>\n

Next, Ptolemy gives the interval of time between the observation of the third opposition and this new observation as<\/p>\n

$2$ Egyptian\u00a0 years, $167$ days, $8$ hours.<\/p><\/blockquote>\n

Using that interval, the mean motions are calculated from the mean motions tables. The increases are<\/p>\n

in longitude: $30;03\u00ba$
\nin anomaly: $134;24\u00ba.$<\/p><\/blockquote>\n

Ptolemy describes these as being “calculated roughly” but the values he gives agree to the nearest minute with my calculations.<\/p>\n

From there, we can add these to our positions from the third anomaly calculated in the last post<\/a>, to determine that the positions of Saturn are:<\/p>\n

in [mean] longitude: $86;33\u00ba$
\nin anomaly: $309;08\u00ba$<\/p><\/blockquote>\n

This gives us what we need to get started with the diagram to determine the size of the epicycle:<\/p>\n

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

In this diagram we have the line of apsides, $\\overline{AG}$ on which sit our three centers: the center of mean motion, $Z$, the center of mean distance, $D$, and the center of the ecliptic. $E$.<\/p>\n

The epicycle is centered on $B$ with Saturn at $K$. We connect $\\overline{BK}$ and $\\overline{EK}$ to it.<\/p>\n

We also connect $\\overline{DB}$ and $\\overline{ZB}$, extending it so that it his the other side at $H$. Additionally, it is extended in the other direction until it meets a perpendicular dropped from $E$ at $L$. Another perpendicular is dropped from $D$ onto that line at $M$. We’ll also draw $\\overline{EB}$ extending it as well, until it hits the opposite side of the epicycle at $\\Theta.$<\/p>\n

Lastly, a perpendicular is dropped from $B$ onto $\\overline{KE}$.<\/p>\n

To begin, we know that $\\angle AZB = 86;33\u00ba$ as we established above. Thus, its vertical angle, $\\angle DZM$ is as well.<\/p>\n

This allows us to create a demi-degrees circle about $\\triangle DZM$ in which the hypotenuse, $arc \\; DM = 173;06\u00ba$. The supplement, $arc \\; ZM = 6;54\u00ba.$ The corresponding chords are $\\overline{DM} = 119;47^p$ and $\\overline{ZM} = 7;13^p.$<\/p>\n

We can then convert this into the context in which the diameter of the eccentres is $120^p$:<\/p>\n

$$\\frac{3;25^p}{120^p} = \\frac{\\overline{DM}}{119;47^p}$$<\/p>\n

$$\\overline{DM} = 3;25^p$$<\/p>\n

and $$\\frac{3;25^p}{120^p} = \\frac{\\overline{ZM}}{7;13^p}$$<\/p>\n

$$\\overline{ZM} = 0;12^p.$$<\/p>\n

We then find $\\overline{BM}$ using the Pythagorean theorem:<\/p>\n

$$\\overline{BM} = \\sqrt{60^2 – 3;25^2} = 59;54^p.$$<\/p>\n

We also know that $\\overline{ZM} = \\overline{ZL}$ and $\\overline{EL} = 2 \\cdot \\overline{DM}.$ This allows us to determine that $\\overline{BL} = 60;06^p$ and $\\overline{EL} = 6;50^p.$<\/p>\n

That’s two sides of $\\triangle BEL,$ so we can use the Pythagorean theorem to find the remaining side:<\/p>\n

$$\\overline{BE} = \\sqrt{60;06^2 + 6;50^2} = 60;29^p.$$<\/p>\n

We’ll now create a demi-degrees circle about this triangle in which the hypotenuse, $\\overline{BE} = 120^p.$<\/p>\n

$$\\frac{120^p}{60;29^p} = \\frac{\\overline{EL}}{6;50^p}$$<\/p>\n

$$\\overline{EL} = 13;33^p.$$<\/p>\n

We can then look up the corresponding arc, for which I find $arc \\; EL = 12;58\u00ba.$ This means that the angle this arc subtends on the opposite side of the demi-degrees circle, $\\angle EBL = 6;29\u00ba.$<\/p>\n

Recalling that $\\angle AZB = 86;33\u00ba,$ we can subtract $\\angle EBL$ from this to determine $\\angle AEB = 80;04\u00ba.$<\/p>\n

Ptolemy then tells us to recall that we determined $\\angle AEK,$ the apparent distance of the planet from apogee, to be $76;04\u00ba$ above. We can subtract to determine:<\/p>\n

$$\\angle KEB = \\angle AEB – \\angle AEK = 4;00\u00ba.$$<\/p>\n

We then focus on $\\triangle BEN$ which contains this angle, drawing a demi-degrees circle about it in which the hypotenuse, $\\overline{BE} = 120^p.$ In that context, $arc \\; BN = 8;00\u00ba$ and the corresponding chord, $\\overline{BN} = 8;22^p.$<\/p>\n

We’ll convert this back into our context in which the diameters of the eccentre are $120^p$:<\/p>\n

$$\\frac{60;29^p}{120^p} = \\frac{\\overline{BN}}{8;22^p}$$<\/p>\n

$$\\overline{BN} = 4;13^p.$$<\/p>\n

We’ll now recall that we determined the planet, at $K$ was $309;08\u00ba$ about its epicycle from its apogee, at $H$. Thus, $\\angle HBK = 50;52\u00ba.$<\/p>\n

We’ll also recall that we determined $\\angle EBL$ to be $6;29\u00ba$ above. This is a vertical angle with $\\angle HB \\Theta$ so we can subtract to determined:<\/p>\n

$$\\angle \\Theta BK = \\angle HBK – \\angle HB \\Theta = 44;23\u00ba.$$<\/p>\n

Next, let’s look at $\\triangle KEB$. In it, we know $\\angle KEB$ which we found above to be $4;00\u00ba.$ Additionally, we can quickly find $\\angle KBE$ as it’s the supplement of $\\angle \\Theta BK,$ so $135;37\u00ba.$ Thus, the remaining angle,<\/p>\n

$$\\angle BKE = 180\u00ba – 135;37\u00ba – 4;00\u00ba = 40;23\u00ba.$$<\/p>\n

Now, we’ll focus on $\\triangle BKN$ which contains this angle, creating a demi-degrees circle about in in which the hypotenuse, $\\overline{BK} = 120^p.$ In that context, $arc \\; BN = 80;46\u00ba$ and the corresponding chord, $\\overline{BN} = 77;45^p.$<\/p>\n

However, we also know $\\overline{BN} = 4;13^p$ in our context in which the diameters of the eccentre is $120^p,$ so we can use that to convert.<\/p>\n

$$\\frac{4;13^p}{77;45^p} = \\frac{\\overline{BK}}{120^p}$$<\/p>\n

$$\\overline{BK} = 6;30^p.$$<\/p>\n

This is the radius of the epicycle.<\/p>\n

Ptolemy quickly sums up what we’ve covered:<\/p>\n

Thus, we have computed the following:<\/p>\n

[R]ound about the beginning of the reign of Antoninus, the longitude of Saturn’s apogee was $23\u00ba$ into Scorpio;
\nwhere the radius of the eccentre carrying the epicycle is $60^p,$
\nthe distance between the centres of the ecliptic and the eccentre which produces the uniform motion is $6;50^p,$
\nand the radius of the epicycle $6;30^p.$<\/p>\n

\n

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

\n

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

Next, we’ll turn our attention to determining the size of Saturn’s epicycle. As he did for Jupiter, Ptolemy introduces a new observation.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"jetpack_post_was_ever_published":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":true,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2}},"categories":[24],"tags":[25,16,14,51],"class_list":["post-4873","post","type-post","status-publish","format-standard","hentry","category-almagest","tag-almagest","tag-epicycle","tag-ptolemy","tag-saturn"],"acf":[],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p9ZpvC-1gB","_links":{"self":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/4873","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/comments?post=4873"}],"version-history":[{"count":4,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/4873\/revisions"}],"predecessor-version":[{"id":4880,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/4873\/revisions\/4880"}],"wp:attachment":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/media?parent=4873"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/categories?post=4873"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/tags?post=4873"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}