![\"\"](\"https:\/\/i0.wp.com\/jonvoisey.net\/blog\/wp-content\/uploads\/2019\/11\/AlmagestFig3-18.jpg?ssl=1\")
<\/a><\/p>\nTo get us started, we’ll notice that the angular distance of the point of mean motion, at A, is $arc \\; AG$ away from perigee, at G on the deferent<\/span>1<\/sup><\/a><\/span>. This subtends $\\angle{ADG}$, which we’ll again be taking as 30\u00ba. From there, we can point out that this is the same angle as $\\angle{HAK}$ because $\\overline{AH} \\parallel \\overline{GD}$ making the two angles alternate interior angles. Thus, the arc that represents the angular distance of the mean from perigee can also be the arc that subtends this angle, $arc \\; H \\Theta$. So this arc too is 30\u00ba in the context of the overall diagram.<\/p>\n We’ll again jump into the demi-degrees method, drawing small circle AKH with diameter $\\overline{AH}$. Here, $\\angle{KAH}$ retains its measure, but in the context of this smaller circle, it will mean that the arc subtending at angle is twice that. So $arc \\; KH = 60\u00ba$ which has a chord length of 60.<\/p>\n We can again find $\\angle{KHA} = 60\u00ba$<\/span>2<\/sup><\/a><\/span> which means $arc \\; AK = 120\u00ba$ and its chord 103;55. As a reminder, $\\overline{AH} = 120$ in the context of the small circle as it’s the diameter.<\/p>\n Now let’s put things back in terms of the the overall diagram, staring with $\\overline{KH}$ which we’ll compare to $\\overline{AH}$:<\/p>\n $$\\frac{\\overline{KH}}{\\overline{AH}} = \\frac{60}{120} = \\frac{\\overline{KH}}{2;30}$$<\/p>\n Solving:<\/p>\n $$\\overline{KH} = 2;30 \\cdot \\frac{60}{120} = 1;15$$<\/p>\n Next up, $\\overline{AK}$:<\/p>\n $$\\frac{\\overline{AK}}{\\overline{AH}} = \\frac{103;55}{120} = \\frac{\\overline{AK}}{2;30}$$<\/p>\n Solving:<\/p>\n $$\\overline{AK} = 2;30 \\cdot \\frac{103;55}{120} = 2;10$$<\/p>\n And now that we’re back in the context of the big circle, we can subtract $\\overline{KA}$ from $\\overline{DA}$ to determine $\\overline{DK} = 57;50$<\/p>\n Now we’ll focus on triangle $\\triangle{DHK}$. Again, this is a right triangle and we know two sides, so we can determine the hypotenuse, $\\overline{DH}$ via the Pythagorean theorem:<\/p>\n $$\\overline{DH} = \\sqrt{1;15^2 + 57;50^2} = 57;51$$<\/p>\n We’ll do one more demi-degrees method, this time about circle DK\u0398 on radius $\\overline{DH}$ with the goal of determining $\\overline{KH}$.<\/p>\n $$\\frac{\\overline{KH}}{\\overline{DH}} = \\frac{1;15}{57;51} = \\frac{\\overline{KH}}{120}$$<\/p>\n Solving<\/span>3<\/sup><\/a><\/span>:<\/p>\n $$\\overline{KH} = 120 \\cdot \\frac{1;15}{57;51} = 2;34$$<\/p>\n This is the chord, so we can use the chord table to reverse lookup and determine that $arc \\; KH = 2;27\u00ba$, which would be twice the value of the subtended angle if that angle is on the circumference which means $\\angle{HDK} = 1;14\u00ba$ which was the equation of the anomaly and in agreement with our figure from the eccentric hypothesis.<\/p>\n This can be added to the 30\u00ba of actual motion to determine that the apparent position along of the sun in angular distance from perigee is 31;14\u00ba.<\/p>\n Lastly, if we know the apparent position instead of the position of the mean, we can again calculate the equation of anomaly from the following diagram.<\/p>\n<\/a><\/p>\n