{"id":1042,"date":"2018-11-12T21:27:25","date_gmt":"2018-11-13T03:27:25","guid":{"rendered":"http:\/\/jonvoisey.net\/blog\/?p=1042"},"modified":"2018-11-12T21:31:34","modified_gmt":"2018-11-13T03:31:34","slug":"almagest-book-ii-calculation-of-rising-times-at-sphaera-obliqua-for-remaining-arcs","status":"publish","type":"post","link":"https:\/\/jonvoisey.net\/blog\/2018\/11\/almagest-book-ii-calculation-of-rising-times-at-sphaera-obliqua-for-remaining-arcs\/","title":{"rendered":"Almagest Book II: Calculation of Rising Times at Sphaera Obliqua for Remaining Arcs"},"content":{"rendered":"

In the last post, we explored the rising time for one zodiacal sign which would comprise 30\u00ba of the ecliptic. Since the preliminary math is now out of the way, we can quickly do 60\u00ba of the ecliptic which constitutes Aries and<\/em> Taurus. From there, we’ll be able to more quickly compute the remaining constellations as well.<\/p>\n

As noted above, we’ve already done the math in the last post<\/a> setting up the problem so let’s go straight to the pieces.<\/p>\n

First, $arc \\; DK$ doesn’t change. It’s the latitude of Rhodes: 36\u00ba, which means\u00a0$arc \\; 2DK = 72\u00ba$ and\u00a0$Crd \\; arc \\; 2DK = 70;32,3$.<\/p>\n

Next, since $arc \\; GK$ is still 90\u00ba and\u00a0$arc \\; DK$ is still 36\u00ba, that means $arc \\; DG$ is still 54\u00ba,\u00a0$arc \\; 2DG = 108\u00ba$, and\u00a0$Crd \\; arc \\; 2DG = 97;4,55$.<\/p>\n

We can get $arc \\; LM$ from the Table of Inclinations<\/a> for 60\u00ba which tells us\u00a0$arc \\; LM = 20;30,9\u00ba$ so\u00a0$arc \\; 2LM = 41;0,18\u00ba$ and\u00a0$Crd \\; arc \\; 2LM = 42;1,48$.<\/p>\n

Since $arc \\; KL$ and $arc \\; LM$ are part of $arc \\; KM$ and $arc \\; KM = 90\u00ba$ we can subtract $arc \\; LM$ out\u00a0to get\u00a0$arc \\; KL = 69;30,51\u00ba$,\u00a0$arc \\; 2KL = 138;59,42\u00ba$, and\u00a0$Crd \\; arc \\; 2KL = 112;23,56$.<\/p>\n

Again, $arc \\; EG = 90\u00ba$ so\u00a0 $arc \\; 2EG = 180\u00ba$, and\u00a0 $Crd \\; arc \\; 2EG = 120$.<\/p>\n

Laying out Menelaus’ theorem again:<\/p>\n

$$\\frac{70;32,3} {97;4,55} =\u00a0\\frac{112;23,56}{41;1,48} \\cdot\u00a0\\frac{Crd \\; arc \\; 2EM}{120}$$<\/p>\n

Solving for the unknown:<\/p>\n

$$Crd \\; arc \\; 2EM = 120 \\cdot \\frac{70;32,3} {97;4,55}\u00a0 \\cdot \\frac{42;1,48}{112;23,56}$$<\/p>\n

$$Crd \\; arc \\; 2EM = 32;36,6$$<\/p>\n

Looking that up in our\u00a0chord table<\/a>\u00a0we get that $arc \\; 2EM = 31;32\u00ba$ and thus $arc \\; EM = 15;46\u00ba$.<\/p>\n

As with before, we need to subtract that from $arc \\; HM$ (which was 57;44\u00ba when the arc of the ecliptic was 60\u00ba<\/a>) to get $arc \\; EH = 41;58\u00ba$.<\/p>\n

However, there’s one more step in this post than in the last post, because this time, we did the rising time of\u00a0two<\/em> zodiacal signs. Thus, we need to subtract the result from the last post where we showed $arc \\; EH$ for 30\u00ba was 19;12\u00ba. Thus $41;58\u00ba – 19;12\u00ba = 22;46\u00ba$ which is the rising time for $arc \\; EH$ which is the arc of the celestial equator that rises with Taurus.<\/p>\n

Again using the rule for arcs equidistant from the same equinox<\/a>\u00a0this means that Aquarius also rises in 22;46\u00ba. We can then get the signs of Leo and Scorpio using the rule about arcs equidistant from the same solstice<\/a>. For that, we need to know the rising time of Taurus (along the ecliptic) at\u00a0sphaera recta<\/em>, which is 29;54\u00ba (as determined here<\/a>). As the rule requires, we need to add Leo’s rising time (at\u00a0sphaera recta)\u00a0<\/em>which is the same so we’re effectively doubling, that which is 59;48\u00ba. We then subtract the rising time for Taurus’ section of the celestial equator at the latitude in question. Thus, $59;48\u00ba –\u00a022;46\u00ba = 37;2\u00ba$ as the arcs of the celestial equator that rise with Leo and Scorpio.<\/p>\n

So that nets us another four of the zodiacal constellations. After the four from the last post, that leaves us with four to go.<\/p>\n

But before moving on, Ptolemy drops a few more pieces. The first is this statement:<\/p>\n

Now since the longest day is $14 \\frac{1}{2}$ equinoctial hours, and the shortest $9 \\frac{1}{2}$ equinoctial hours, it is obvious that the semi-circle [of the ecliptic] from Cancer to Sagittarius will rise with 217;30\u00ba of the equator, and the semi-circle from Capricorn to Gemini with 142;30\u00ba.<\/p><\/blockquote>\n

True fact: I\u00a0hate<\/em> it when Ptolemy says “it is obvious”. It never is<\/span>1<\/sup><\/a><\/span>. So let’s break this down. To start, I want to rearrange what he’s said into two parts:<\/p>\n

1. On the longest day, which is\u00a0$14 \\frac{1}{2}$ hours, the portion of the ecliptic that contains the constellations Cancer to Sagittarius rises with 217;30\u00ba of the equator.<\/p>\n

2. On the shortest day, which is\u00a0$9 \\frac{1}{2}$ hours, the portion of the ecliptic that contains the constellations Capricorn to Gemini rises with 142;30\u00ba of the equator.<\/p>\n

I want to tackle the second of these first. The reason being, that the value of 142;30\u00ba shows up in this post<\/a> which is usually a good hint that there’s something of an explanation there. Specifically, I want to use a variation of one of the diagrams there. In that post, I had one showing the position of the Sun on the winter solstice. I’ll again show the Sun on the winter solstice, but this time, things are slightly rotated so you can see the ecliptic while the Sun is on the horizon.<\/p>\n

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

I’ve labelled everything here so I don’t feel like the general setup needs more explanation so let’s look at a few details. First off, the motion for the Sun on that single day is the semi-transparent small circle. The path we’re concerned about, I’ve highlighted in red. Why is that the important one? Take a look at my zodiac diagram I sketched awhile back.<\/p>\n

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

Capricorn to Gemini is from the winter solstice to the summer solstice going counter clockwise. And what we want to know is how long that would take to rise? What we notice is that,\u00a0if the Sun is at the winter solstice, directly opposite it on the horizon (180\u00ba difference in azimuth) is the summer solstice.<\/p>\n

So now let’s allow the Sun to set, still on the winter solstice and see what happens.<\/p>\n

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

Now the Sun is on the horizon in the west. But if you look, the summer solstice is\u00a0still<\/em> opposite it on the horizon, 180\u00ba in azimuth. Which means that Capricorn to Gemini rose during the length of that day. Which was\u00a0$9 \\frac{1}{2}$ hours. In the post linked above, we went through the math to demonstrate that the arc of the celestial equator that rose in\u00a0$9 \\frac{1}{2}$ hours was 142;30\u00ba.<\/p>\n

The same reasoning applies to the other half of the statement. The proportion of the ecliptic again goes from sunrise to sunset, but this time it’s in\u00a0$14 \\frac{1}{2}$ hours which, just as\u00a0$14 \\frac{1}{2} + 9 \\frac{1}{2}$ adds up to 24 hours, so too must the\u00a0142;30\u00ba + 217;30\u00ba = 360\u00ba.<\/p>\n

Next, Ptolemy chops those up:<\/p>\n

Therefore each of the quadrants on either side of the spring equinox will rise in 71;15 time-degrees, and each of the quadrants on either side of the autumnal equinox will rise in 108;45 time degrees.<\/p><\/blockquote>\n

This follows from the symmetries of rising times equidistant from the same equinox<\/a>. In short, we’re breaking the from Capricorn to Gemini into two equal parts: Capricorn to Pisces and Aries to Gemini. Because those two arcs are equidistant (by which we mean immediately adjacent) to the spring equinox, their rising times must be the same. If they’re the same and add up to 142;30\u00ba, each must be half: 71;15\u00ba. The same is true for the other two.<\/p>\n

Now that we’ve taken this huge detour, let’s remember that in the last post we got the rising times for Aires, Pisces, Virgo and Libra. In the beginning of this post we figured out Taurus, Aquarius, Leo, and Scorpio. That’s $\\frac{2}{3}$ of each quadrant we already had figured out, and now we have the rising times for the full quadrants, so we can subtract out the previously known $\\frac{2}{3}$ to determine the rising times for the last ones.<\/p>\n

Specifically, we showed above that Aries + Taurus =\u00a041;58\u00ba. Just now we’ve shown that Aries + Taurus + Gemini = 71;15\u00ba, so Gemini’s rising time is $71;15\u00ba – 41;58\u00ba = 29;17\u00ba$. Which means that Capricorn is the same.<\/p>\n

By using the symmetries of rising times equidistant from the same solstice<\/a>,\u00a0 we can also show that Cancer and Sagittarius rise in 35;15\u00ba.<\/p>\n

Next up, Ptolemy isn’t just satisfied breaking the ecliptic up into 30\u00ba chunks. No. He wants to break it into 10\u00ba intervals. But he promises there’s an easier way to do it. Given that we’ve already developed a formula, I feel like that’s probably the easier way, but then again, we have the advantage of a calculator to plug all the numbers in. Ptolemy didn’t. So I expect “easier” to him means less computation and more geometry. But we’ll cover that next time!<\/p>\n

\n","protected":false},"excerpt":{"rendered":"

In the last post, we explored the rising time for one zodiacal sign which would comprise 30\u00ba of the ecliptic. Since the preliminary math is now out of the way, we can quickly do 60\u00ba of the ecliptic which constitutes Aries and Taurus. From there, we’ll be able to more quickly compute the remaining constellations … <\/p>\n

Continue reading “Almagest Book II: Calculation of Rising Times at Sphaera Obliqua for Remaining Arcs”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","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":[25,36,37,39,28,26,14,34],"class_list":["post-1042","post","type-post","status-publish","format-standard","hentry","category-almagest","tag-almagest","tag-celestial-equator","tag-celestial-sphere","tag-equinox","tag-geometry","tag-mathematics","tag-ptolemy","tag-solstice"],"acf":[],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p9ZpvC-gO","_links":{"self":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/1042","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=1042"}],"version-history":[{"count":16,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/1042\/revisions"}],"predecessor-version":[{"id":1088,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/1042\/revisions\/1088"}],"wp:attachment":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/media?parent=1042"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/categories?post=1042"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/tags?post=1042"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}