where the apparent distance of the body from apogee [at one moment] equals its apparent distance from perigee [at another], the equation of anomaly will be the same at both positions.<\/p><\/blockquote>\n
Fortunately, the proofs for this are quite simple.<\/p>\n
We’ll start off with the eccentric hypothesis with eccentre ABG having center E and an observer at Z.<\/p>\n
We’ll consider the angular distance from apogee at point B, which is $\\angle{AZB}$ and note that this is equal to $\\angle{DZG}$ because they are vertical angles. The two points, B and D, are the ones Ptolemy is referring to. At B the angular distance from apogee is the same as the angular distance point B is from perigee. And B doesn’t have to be at that specific location. It could be anywhere and there will always be a point D opposite B through Z that meets this criteria.<\/p>\n According to Ptolemy, at these two points, the equation of anomaly<\/span>1<\/sup><\/a><\/span> is equal. To show this, we’ll draw in lines connecting B and D to E.<\/p>\n Here, look at $\\triangle{EBD}$. Since $\\overline{EB}$ and $\\overline{ED}$ are equal (since they’re both radii), this means that $\\triangle{EBD}$ is isosceles and its base angles, $\\angle EBZ$ and $\\angle EDZ$, must be equal. And since we showed back in this post<\/a> that $\\angle EBZ$ is the equation of anomaly at B, we can use the same logic to state $\\angle EDZ$ is the equation of anomaly at D. Their being equal proves Ptolemy’s claim.<\/p>\n Next up, we’ll look at the same argument for the epicyclic model. I’m not a big fan of the diagram in the book as it’s hard to understand where it came from so I’m going to deviate and then bring things back to try to explain what Ptolemy has done.<\/p>\n First, let’s start with the object at Z’ while it’s at apogee. We’ll let some interval of time pass recalling that, because we’re exploring the case where the epicycle rotates clockwise at the same angular speed as the epicycle moves counter-clockwise on the deferent, it will retain it’s upwards positioning.<\/p>\n Now, the point that the object was on the epicycle (Z’) is now called E and the object has moved to point Z. We’ll draw in $\\overline{DZ}$ which represents the line of sight for the observer. From the observer’s point of view, the angular distance of the object from apogee is $\\angle Z’ D Z$. However, in this post<\/a>, we showed that this angle is equal to $\\angle{AZD}$<\/span>2<\/sup><\/a><\/span>.<\/p>\n Now, let’s continue letting time pass until the object is the same angular distance before the perigee as the previous image was from apogee:<\/p>\n Here, I’ve renamed the object from Z to H. You’ll see why in a moment.<\/p>\n As with before, $\\overline{DH}$ is the line of sight to the object and so it is apparent distance from perigee of $\\angle{HDG}$, which is the same as $\\angle{BHA}$ and, as required by the hypothesis is also equal to $\\angle{A’ D A}$ which in turn was equal to $\\angle{AZD}$ as discussed above. Therefore these two angles ($\\angle{BHA}$ and $\\angle{AZD}$) are equal.<\/p>\n But they’re not the equations of anomaly, so we’ll still need to do a bit more work. To get us to the equation of anomaly, I’ll now switch back to Ptolemy’s proof <\/span>3<\/sup><\/a><\/span> in which he stacked these three images on top of one another:<\/p>\n![\"\"](\"https:\/\/i0.wp.com\/jonvoisey.net\/blog\/wp-content\/uploads\/2019\/11\/AlmagestFig3-7a.jpg?ssl=1\")
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