Ptolemaic planetary models and Kepler’s laws

Abstract

In this article, we aim at presenting a thorough and comprehensive explanation of the mathematical and theoretical relation between all the aspects of Ptolemaic planetary models and their counterparts which are built according to Kepler’s first two laws (with optimized parameters). Our article also analyzes the predictive differences which arise from comparing Ptolemaic and these ideal Keplerian models, making clear distinctions between those differences which must be attributed to the structural variations between the models, and those which are due to the specific parameters Ptolemy determined in the Almagest. We expect that our work will be a contribution for a better understanding not only of the Ptolemaic theories for planetary longitudes through a clearer perception of the way in which Keplerian features are present—or absent—in Ptolemy’s models, but also for a more balanced judgement of different aspects of the contribution of the first two laws of Kepler to the modern astronomical revolution.

Appendixes

1.1 Appendix 1: Problems regarding the calculation of the position of Venus’ apsidal line and its center of uniform motion

To understand how we arrived at an apsidal line EUZ and an equant Q in Sect. 2.2.3, we must first recall the way in which Ptolemy himself calculated the position of the apsidal line of Venus. He explicitly mentions two different methods which, according to him, must have the same outcome, that is, both will result in the same final value for the longitude of the apogee of the deferent.

The first method relies on the assumption that all planetary phenomena will be exactly symmetrical with respect to the apsidal line. If that is so, he says (1984: 445–446), then two maximum mean elongations of Venus—i.e., maximum elongations with respect to the mean Sun—located on opposite sides of the epicycle, and which have equal value, will be equidistant from the apsidal line. Thus, using two such observations, he can bisect the angle between the two positions of Venus and find the position of the apsidal line. This method will be called the method of symmetries (refer to Fig. 29).

Fig. 29

Method of symmetries to find the direction of the apsidal line. According to Ptolemy, if both maximum mean elongations have the same value α, then the line bisecting the two positions of Venus is the apsidal line

The second method is only mentioned in the Almagest (1984: 471) as a means of checking the value for the longitude of the apogee obtained via the method of symmetries and consists in finding the place where the mean Sun is located when the sum of two maximum mean oppositions which happen on opposite sides of the epicycles is the greatest—or smallest—in the ecliptic. Ultimately, this method measures the apparent size of the epicycle. Given that—if the epicycle has a fixed radius—the value of that sum depends exclusively on the distance of the epicycle to the observer, then the greatest result will be obtained when the epicycle is at its closest to the Earth, that is, at its perigee. The smallest one will in turn be obtained when it is at the apogee. This method will be called the method of sizes (refer to Fig. 30).

Fig. 30

Method of sizes to find the direction of the apsidal line. When the sum α of two opposite maximum mean elongations with the mean Sun at the same longitude reaches its maximum value, then the line bisecting both positions of the planet will not only pass through the center of the epicycle D_V—this will always be the case in such sums—but most importantly, it will coincide with the apsidal line

It can be proven that not only does the method of symmetries not yield the same result as the method of sizes (Wilson 1973: 212), but that it doesn’t even yield a unique result: the direction of the line bisecting both positions of Venus changes depending on the maximum mean elongations that are chosen (see Appendix 3). Ptolemy, nevertheless, says he used this method to obtain the apsidal line, but he in fact obtains a result very close to the one which can be obtained through the method of sizes. This can hardly be a coincidence, and it is reasonable to assume that he in fact obtained the position of the apsidal line with this last method and then tuned the observations of mean maximum elongations in order to arrive at that result with the other (cf. Swerdlow 1989, for a thorough study of these interconnections in the case of Venus). Thus, because it is the only method that yields an unequivocal result, and because it is presumably the one that Ptolemy actually used, we will follow the method of sizes in our transformation.

Refer to Fig. 31. As we said, the apparent size of the epicycle—which is what is ultimately measured in the method of sizes—will have its greatest value when the center of the epicycle is at its closest position to the Earth E. This position can be easily found as the point A, the position where a line EU which passes through both the Earth and the center of the deferent intersects the deferent. Thus, the Ptolemaic apsidal line is determined by points ZUEA.

Fig. 31

Ptolemaic apsidal line, determined by the Earth and the perigee A. The line will thus also pass through U and Z

After determining the direction of the apsidal line and its perigee and apogee, Ptolemy calculates the position of the center of uniform motion of D_V. To do this, he uses a method which we will call the method of perpendicularity. Regarding this method, it is important to know that Ptolemy is aware that the period of D_V around its center of uniform motion is equal to the period of the Sun, that is, that D_V moves around its center of uniform motion just as the mean Sun moves around the Earth.

Refer to Fig. 32. Therefore, he decides to look for two maximum mean elongations that happen on opposite sides of the epicycle, and when the mean Sun S_m is 90° away from the apsidal line he just found.

Fig. 32

Method of perpendicularity, where the position of D_V on the deferent is located via two maximum mean elongations, to calculate the position of its equant. Ptolemy’s requirement for the equant to lie on the apsidal line he found makes him assume that the equant is not W, but its projection on the apsidal line Q

Once he has the necessary observations, he bisects the angle between the two positions of Venus, thus obtaining the position of D_V on the deferent at those moments. Because S_m is 90° away from the apsidal line, line S_mE is perpendicular to the apsidal line. Thus, because of the equivalence of velocities we just mentioned, he knows that also the line determined by D_V and its center of uniform motion is perpendicular to the apsidal line. Given that the center of uniform motion of D_V is in fact W, we know that this is correct. But Ptolemy requires the center of uniform motion to lie on the apsidal line, so he assumes that the equant is not W, but the projection of line D_VW onto the apsidal line, that is, point Q. In fact, Ptolemy’s observations are slightly tuned in such a way that the position of the equant he finds is a relation between EQ and EU which is exactly 2 to 1.

In this way we have arrived at the full Ptolemaic model for Venus (refer to Fig. 33), where Venus V₁ moves uniformly around D_V, the center of its epicycle. The epicycle in turn moves uniformly with respect to Q—which Ptolemy incorrectly found to be at the same distance from U than from E—on a deferent with center U.

Fig. 33

Final Ptolemaic model for Venus, with the apsidal line EUQ, and Venus V₁ moving on an epicycle with center D_V

1.2 Appendix 2: The path of Mercury’s epicycle and the introduction of a movable center of the deferent

In Sect. 2.3.1 we presented a rather curious step 8 to achieve the final Ptolemaic model for Mercury. To understand the reason we did this, we must see in detail how Mercury’s great eccentricity lies at the heart of Ptolemy’s incorporation of a movable deferent. In Almagest IX, 8 Ptolemy checks the accuracy of his preliminary model for Mercury, one that is equivalent to the one resulting from step 7 (although remembering that in Mercury’s case there is a peculiar relation where EU = 2EQ). The accuracy of this model has been analyzed at the end of Sect. 4.5. He does this in a very specific way: he measures the size of the epicycle at different positions on the deferent to see if the size it shows is in accordance with the model he arrived at. When he does this, he finds that there is an anomaly: in at least two symmetrical positions on the ecliptic, the epicycle appears larger than it should. As we mentioned earlier, this problem is due to the fact that Mercury’s orbit has a big eccentricity.

Refer to Fig. 20. Here we can see an elliptic epicycle that, when located on the apogee of the deferent, has its minor axis perpendicular to the center of the deferent U. When the epicycle moves along the deferent, this is not the case anymore. When the center of the epicycle is located on D_V1 and D_V2, for example, the elliptic epicycle is viewed from U with its major axis perpendicular to the observer. So, although the epicycle center is at the same distance from U, the elliptical epicycle itself will be seen with a greater size.

Ptolemy’s procedure faces this problem caused by the elliptic shape of the epicycle of Mercury. Ptolemy calculates the radius of the epicycle with an observation of maximum elongation at apogee. When he checks the apparent size of the epicycle with two more maximum elongations at different longitudes, he finds that it is bigger than expected—in fact, he incorrectly thought that it was even bigger than at the opposite longitude of the apogee (see below). The solution he finds is to make the epicycle to come closer to the observer at those longitudes, thus making the trajectory of its center to lose its circular shape in favor of an oval one.

Ptolemy’s solution, although flawed in its details, is an attempt to solve the following problem: what shape must the trajectory of an elliptic epicycle have in order to make the apparent size of that epicycle constant as seen from the center of that trajectory? The answer is a non-circular, oval, shape. Thus, this step corresponds to the fact that, though in a crude manner, Ptolemy transfers the elliptic shape of Mercury’s orbit to the Earth’s orbit by making it oval.

In order to arrive at the final Ptolemaic model for Mercury we must, then, make the trajectory of the epicycle to be oval. So, we will finish step 8 by making the circular epicycle obtained in Fig. 34 move on a circular deferent, which in turn moves on a little epicycle with the appropriate angular motion in order to make the center of the epicycle to determine the oval trajectory needed.

Fig. 34

The influence of Mercury’s eccentricity in the Ptolemaic detection of a problem with the size of the epicycle. D_V, D_V1 and D_V2 are the three positions of the center of the epicycle. Ptolemy obtains the size of the epicycle when it is at D_V. To maintain the same radius of the epicycle that at the last two longitudes the center of the epicycle is not on D_V1 and D_V2, but on B₁ and B₂, respectively. This diagram, of course, does not represent either the real Keplerian or the Ptolemaic model, where motion is not measured from the center. When such a change is introduced—an equant, for example—some approximations have to be made, though with no significant consequences

In Fig. 35 we can see a center O of the deferent, which moves on a little epicycle around point U. The motion of O around U is equal to the motion of the center of the epicycle D_V around the equant Q, but in the opposite direction. This combination, coupled with the appropriate radii for both circles, assures that the trajectory of D_V will be the one needed to account for the observations.

Fig. 35

Model with moving deferent. The center of the deferent O moves around U with a uniform motion equal to the motion of D_V around Q, but in the opposite direction. This causes an oval trajectory of D_V. In the diagrams we can see two random positions for D_V. In the true Ptolemaic model, the Hellenistic astronomer finds EU is exactly bisected by Q, and the radius of the little epicycle is equal to UQ. This is not, however, a necessary feature of the model, as we show in these examples

In this way we have arrived at Ptolemy’s final model for Mercury (refer to Fig. 36). It should be added that Ptolemy’s final result is not extremely precise, and this is because of his empirical input. The most striking example of this is his conclusion that, in two places of the ecliptic 120° apart from the apogee, the distance from the Earth to the center of the epicycle is less than at the longitude opposite to the apogee. This is in fact not true: although it is really closer than expected, it is never closer than at the opposite longitude of the apogee. This, along with some other notable errors in Ptolemy’s model (see Sect. 3), makes Mercury’s model the worst in the Almagest, and thus the most difficult case to deal with when researching the relation between Ptolemy’s and Kepler’s models.

Fig. 36

Final Ptolemaic model. The observer is at Earth E, and the equant is at Q, bisecting the eccentricity of U. This feature, as we said, while present in Ptolemy’s actual model for Mercury, is not a correct parameter of the model (see Sect. 3). Point U is the center of a little epicycle around which the center of the deferent O moves. The motion of O around U is determined by speed γ, as is the motion of the center of the deferent D_V and D_V1 around Q. Step 8 makes the center of the deferent to move from D_V—the one which corresponds to a fixed deferent with center U—to D_V1—the one which corresponds to a moving deferent with center O. This change produces a dotted trajectory of the center of deferent which is oval. It also moves the planet Mercury from M to M₁

In Fig. 36 we can clearly appreciate the modification introduced by step 8: in order to produce the oval trajectory of the center of the epicycle, we have introduced a motion in the center of the deferent, which is now O. This motion causes the center of the epicycle to change from D_V to D_V1. These two centers of the epicycle will now move following different paths, coinciding only on the apsidal line. The change of the position of the center of the epicycle, finally, produces in turn a change in the position of the planet itself, from M to M₁. This, of course, introduces a new difference in the predicted longitude.

Thus, we have concluded our analysis of the geometrical and theoretical relation between Keplerian and Ptolemaic planetary models. In the next section we are going to focus on the errors due to the structural differences between those models.

1.3 Appendix 3: The ambiguity of the method of symmetries for finding the apsidal line of the inferior planets

The method of symmetries says that if two maximum elongations with respect to the mean Sun that are opposite (i.e., one morning maximum elongation and the other evening) are equal, then the apsidal line will pass through the Earth and the midpoint between the two positions of the mean Sun at both maximum elongations. Figure 37 is similar to Fig. 29. The center of the orbit of the center of the epicycle is U, but the center of uniform motion of the center of the epicycle is W. The Earth is at E; the line joining the Earth and the mean Sun is always parallel to the line joining W and the center of the epicycle. Let us assume that, when the center of the epicycle is at C and when it is at C^′ (and the mean Sun at S_m and S ^′_m , respectively), the opposite maximum elongations involving V₂ and V ^′₂ are exactly the same. This defines an apsidal line bisecting the two positions of the mean Sun, S_m and S ^′_m . But, because the center of the deferent is at U and C and C^′ are not symmetrical with respect line EU, C and C^′ are not equidistant from E. Actually, C is closer to the Earth than C^′. This means that the apparent sizes of the epicycle are not equal, i.e., that V₂EV₁ ≠ V ^′₂ EV ^′₁ . Therefore, if two opposite maximum elongations are equal (V₂ES_m = V ^′₂ ES ^′_m ), then the other two opposite maximum elongations cannot be equal (S_mEV₁ ≠ S ^′_m EV ^′₁ ). Consequently, the two different pairs of opposite maximum elongations will define two different apsidal lines.

Fig. 37

The references are those of Fig. 29. The Earth is at E, the center of the deferent is at U, but the motion of the center of the epicycle is uniform from W. When the mean Sun is at S_m, the center of the epicycle is at C and V₂ and V₁ are the maximum elongations; when the mean Sun is at S ^′_m , the center of the epicycle is at C^′ and V ^′₂ and V ^′₁ are the maximum elongations. While V₂ES_m = V ^′₂ ES ^′_m , S_mEV₁ ≠ S ^′_m EV ^′₁

1.4 Appendix 4: Formulae for calculating the errors

Here we offer the formulae that govern each step explained in Sect. 2 and allow to calculate the error analyzed in Sect. 3, as well as the way we arrive at those formulae.

1.4.1 Calculation of the heliocentric longitude of the planet in a Keplerian orbit

In our analysis, we calculate the heliocentric longitude of the planet using the iterative method proposed by Meeus (1998, p. 196), called by him “First Method,” that works well for small eccentricities.

In Fig. 38, the planet K moves in the elliptical orbit PKA, F and S are the foci of the ellipse, S is also the Sun, C is the center of the ellipse, and the dotted circle is centered at C and has radius PC. A is the aphelion, and P, the perihelion. Angle M (PSK^′) is the mean anomaly and moves uniformly with period equal to the planet. It shows the position that the planet would have if there were no eccentricities with respect to perihelion. Angle v (PSK) is the true anomaly; it shows where the planet actually is (with respect to the perihelion). Angle E (PCQ) is the angle called the eccentric anomaly, and it is defined in the figure.

Fig. 38

Reproduction of Figures 1 and 2 of chapter 30 of Meeus 1998, 193, 195)

Kepler’s equation is

$$ E = M + e\sin E $$

where e is the eccentricity of the ellipse. It is a transcendental equation which cannot be solved directly. The iterative method allows to find E as a function of M (and the eccentricity e).

Having E it is possible to find r and v according to the following equations:

$$ v = 2 \cdot \tan^{ - 1} \left( {\sqrt {\frac{1 + e}{1 - e}} \cdot \tan \frac{E}{2}} \right) $$

$$ r = a\left( {1 - e\cos E} \right) $$

where a is the semi-major axis.

For obtaining the values of a, e and M, we use Meeus (1998, chapter 31, pp. 209–216). M is obtained as the difference between the mean heliocentric longitude of the planet and the longitude of the perihelion. Once v is obtained, the heliocentric longitude of the planet is obtained adding the longitude of the perihelion to v.

1.4.2 From heliocentric longitudes to geocentric longitudes

Having calculated the heliocentric longitude of the planet and the Earth, we can obtain the geocentric longitude of the planet. See Fig. 39:

Fig. 39

S is the Sun; E, the Earth; and M, the planet. EA and SA correspond to the direction of Aries 0°. ASM is the heliocentric longitude of the planet and ASE the heliocentric longitude of the Earth. AEM is the geocentric longitude of the planet

A is in the direction of Aries 0. S is the Sun; E, the Earth; and M, the planet, in this case Mars. Angle Λ_e (ASE) is the heliocentric longitude of the Earth, and angle Λ_m (ASM) is the heliocentric longitude of Mars, while angle λ_m (AEM) is the desired geocentric longitude of Mars. Distance SE is the distance between the Sun and the Earth, and distance SM is the distance between the Sun and Mars. Both distances are given by the application of the formula of Kepler. (It is the value of r of the corresponding planet, Earth or Mars).

We first obtain the distance from the Earth and Mars, EM, applying the cosine rule:

$$ EM = \sqrt {SE^{2} + SM^{2} - 2 \cdot SE \cdot SM \cdot \cos \left( {\varLambda_{m} - \varLambda_{e} } \right)} $$

Now, ρ can be obtained using the sine rule.

$$ \rho = \sin^{ - 1} \left( {\frac{{SE. \sin \left( {\varLambda_{m} - \varLambda_{e} } \right)}}{EM}} \right) $$

Finally, we know that:

$$ \lambda_{m} = \varLambda_{m} + \rho $$

In the analysis of all the steps from now on, we will arrive at the heliocentric longitude implied by the change; in order to obtain the geocentric longitude, the method here explained must be applied.

1.4.3 Steps 1 and 3: second law versus equant point

We want to obtain the angular position of the planet P, from the Sun S (angle v), assuming that angle α, from the empty focus F, grows uniformly. Angle α is the difference between the mean longitude of the planet and the longitude of the perihelion, i.e., equal to M in Fig. 38.

We can obtain distance FP by the following formula:

$$ FP = \frac{{a \cdot \left( {1 - e^{2} } \right)}}{1 - e \cdot \cos \alpha } $$

where a is the semi-major axis (Cπ in the diagram) and e is the eccentricity, FC or CS (Fig. 40).

Fig. 40

The planet P moves in an elliptical orbit, with foci at S, the Sun, and F, the empty focus. The empty focus is also the equant point, so that α increases uniformly over time. π is the perihelion of the planet. Angle v is the angular distance of the planet from its perihelion, seen from the Sun

Because it is an ellipse, we know that

$$ SP = 2a - FP $$

Applying the sine rule in triangle FSP we can obtain angle FSP.

$$ FSP = \sin^{ - 1} \left( {\frac{FP \cdot \sin \alpha }{SP}} \right) $$

And we know that v = 180 − FSP.

1.4.4 Steps 2 and 4: elliptical versus circular orbit

This step consists in replacing the elliptical orbit with a circular one. In Fig. 41, P is the position of the planet in the elliptical orbit (moving uniformly from F) and P^′ in a circular orbit with radius equal to the semi-major axis and the same center of the ellipse. The difference between these two models seen from the Sun S is expressed by angle ρ. We have to calculate the heliocentric distance from the perihelion, angle v.

Fig. 41

The planet moves in an elliptical orbit, with foci at S, the Sun, and F, the empty focus, is at P. C is the center of the ellipse. The empty focus is also the equant point, so that α increases uniformly over time. π is the perihelion of the planet. If the planet moves in a circular orbit with center in C and radius Cπ, and uniform motion from F, the planet is at P^′. Angle v is the angular distance of the planet P^′ from its perihelion, seen from the Sun

We already know distances SP and FP and FC = CS = e. We also know that CP^′ = a, i.e., the semi-major axis of the ellipse, which is also the radius of the circle.

First, we will calculate angle δ = FP^′C, applying the sine rule:

$$ \delta = \sin^{ - 1} \left( {\frac{e \cdot \sin \alpha }{a}} \right) $$

Angle ε is 180 − α − δ

Again applying the sine rule we can obtain FP^′

$$ FP '= \left( {\frac{a \cdot \sin \varepsilon }{\sin \alpha }} \right) $$

Now, we can calculate SP^′, applying the cosine rule to triangle CSP^′. Because e = CS and CP^′ = a,

$$ SP^{\prime} = \sqrt {e^{2} + a^{2} - 2ea \cdot \cos \left( {180 - \varepsilon } \right)} $$

Having SP^′, we can calculate angle γ, applying the sine rule to triangle FSP^′:

$$ \gamma = \sin^{ - 1} \left( {\frac{FP' \cdot \sin \alpha }{SP'}} \right) $$

And we know that:

$$ v = 180 - \gamma $$

1.4.5 Step 7 for outer planets: moving the center of the orbit of the (outer) planet to the equant point

Step 7 for outer planets consists of moving the center of the orbit of the Earth to the equant point. Therefore, in Fig. 42, we have to move the center of the orbit from D to Q, keeping Q as the center of uniform motion.

Fig. 42

D is the center of the circular orbit of the planet P that moves uniformly from Q. P^′ moves also uniformly from Q, but the center of its orbits is Q, too. Angle v is the angular distance of the planet P^′ from its perihelion, seen from the Sun

We already know α, and we have to calculate ρ. We know also that QS is twice the eccentricity of the earth, e, and that QP^′ is equal to R, the radius of the Earth’s orbit. Applying the cosine rule, we can obtain SP^′.

$$ SP^{\prime} = \sqrt {\left( {2e} \right)^{2} + R^{2} - 4eR \cdot \cos \alpha } $$

Now, applying the sine rule to triangle DP^′S, we obtain ρ:

$$ \rho = \sin^{ - 1} \left( {\frac{2e \cdot \sin \alpha }{SP'}} \right) $$

And we know that:

$$ v = \alpha + \rho $$

1.4.6 Step 8 for the outer planets: changing the center of the orbit of the outer planet, and so the direction of the apsidal line and the eccentricity

In step 8 the center of the orbit of the planet is moved so that it bisects the line joining the mean Sun and the equant point. In Fig. 43, S is the Sun and line SCF is the apsidal line of Mars, with its perihelion at π_M. Mars is in two different positions, in P and P^′. Line SDQ represents the apsidal line of the Earth orbit. D was the center of the ellipse (and of the orbit until step 7). Now the center of the orbit is Q, the equant point. Therefore, Q is at the same time the center of uniform motion and of the orbit, and it is located at 2e_T from the Sun. π_T is the perihelion of the Earth. In step 5 we moved the center of the Earth’s orbit from D to Q. This move affected the distance relation between the Earth and the planet. In order to solve this problem, we will move the center of the outer planet orbit the same distance and direction, i.e., from C to B. (CB and DQ are equal and parallel.) If the equant point is still F but the center of Mars’ orbit is B, then the planet will be at P^′. Consequently, we have to calculate angle v, the heliocentric angular distance between P^′ and the perihelion of Mars.

Fig. 43

S is the Sun, and FCS is the apsidal line of the planet P. The planet P moves uniformly from F, in a circular orbit centered at C. π_M is the perihelion of the planet. SDQ is the apsidal line of the Earth T that moves uniformly from Q, the equant point in a circular orbit with center at D. π_T is the perihelion of the Earth. T^′ is the position of the Earth once the center of its orbit is moved to Q. P^′ is the position of the planet once the center of its orbit is moved from C to B. Angle v is the angular distance of the planet P^′ from its perihelion, seen from the Sun

In order to do that, we first will calculate FB applying the cosine rule to triangle FEB. We already know that FE is equal to the eccentricity of the earth (e_t), while EB is the eccentricity of Mars, e_m (because EB = FC). Of course, they have to be normalized using the proportion between the radii of the orbits. Angle (π_MSπ_T) is the difference between the longitude of the perihelion of the Earth and the longitude of the perihelion of Mars. Angle π = FEB is equal to π_MSπ_T, so it is also given.

Therefore:

$$ FB = \sqrt {e_{t}^{2} + e_{m}^{2} - 2 \cdot e_{t} \cdot e_{m} \cdot \cos \pi } $$

Now we can obtain angle γ, using the sine rule to triangle FEB.

$$ \gamma = \sin^{ - 1} \left( {\frac{{e_{t} \cdot \sin \pi }}{FB}} \right) $$

We know that BP^′ is one radius of Mars’ orbit = R_M. We also know that angle P^′FB is α + γ, so, applying the sine rule to triangle FBP^′, we can calculate angle ε.

$$ \varepsilon = \sin^{ - 1} \left( {\frac{{FB \cdot \sin \left( {\alpha + \gamma } \right)}}{{R_{M} }}} \right) $$

Looking at triangle FBP^′, we also know that

$$ \zeta = 180 - \varepsilon - \left( {\gamma + \alpha } \right) $$

So, looking again at triangle FBP^′, we can calculate side FP^′ applying the cosine rule:

$$ FP' = \sqrt {FB^{2} + R_{M}^{2} - 2 \cdot FB \cdot R_{M} \cdot \cos \zeta } $$

And, again using the cosine rule but now to triangle FSP^′, we are able to calculate SP^′.

$$ SP' = \sqrt {FP'^{2} + 2e_{m}^{2} - 4 \cdot FP' \cdot e_{m} \cdot \cos \alpha } $$

And applying the sine rule to triangle FSP^′, we can calculate angle θ.

$$ \theta = \sin^{ - 1} \left( {\frac{FP '\cdot \sin \alpha }{SP '}} \right) $$

And we know that

$$ v = 180 - \theta $$

1.4.7 Step 6 of inner planets: making the planet to move uniformly from its center (and not from the equant point)

Step 6 for the inner planets implies to change the center of uniform motion from the equant point Q to the center of the circle, D.

Refer to Fig. 44. We already know α, and we have to calculate ρ. We know also that DS is the eccentricity of the earth, e, and that DP^′ is equal to R, the radius of the Earth’s orbit. Applying the cosine rule, we can obtain SP^′.

Fig. 44

D is the center of the circular orbit of the planet P that moves uniformly from Q. P^′ moves uniformly from the center of its orbit, D. Angle v is the angular distance of the planet P^′ from its perihelion, seen from the Sun

$$ SP^{\prime} = \sqrt {e^{2} + R^{2} - 2eR \cdot \cos \alpha } $$

Now, applying the sine rule to triangle DP^′S, we obtain ρ:

$$ \rho = \sin^{ - 1} \left( {\frac{e \cdot \sin \alpha }{SP'}} \right) $$

Finally, we know that:

$$ v = \alpha + \rho $$

1.4.8 Steps 7 and 8 of the inner planets: changing the apsidal line and determining the equant point

These steps consist of keeping the center of the orbit of the Earth where it was but moving the equant point from F to H.

In Fig. 45, S is the Sun and line SCF is the apsidal line of the Earth, with its perihelion at π_T. The Earth is in two different positions, in T and T^′. Line SDQ represents the apsidal line of the orbit of the inner planet. D is now the center of the orbit and of uniform motion (step 6 of inner planets). π_P is the perihelion of Venus. In step 6 of inner planets we moved the center of the orbit of the planet from D to Q, so the planet moved from P to P^′. The center of the deferent will still be C, but the center of uniform motion will be H, the intersection of a line perpendicular to QF going through F. If the center of the orbit of the Earth is still C but the equant point is not F but H, then the Earth will be at T^′. Consequently, we have to calculate angle v, the heliocentric angular distance between T^′ and the perihelion of the Earth.

Fig. 45

S is the Sun, and FCS is the apsidal line of the Earth T. The Earth T moves uniformly from F, in a circular orbit centered at C. π_T is the perihelion of the Earth. SDQ is the apsidal line of the planet P that moves uniformly from Q, the equant point in a circular orbit with center at D. π_P is the perihelion of the planet. P^′ is the position of the planet once the center of the uniform motion is moved to D. T^′ is the position of the Earth once its equant point is moved from F to H, the projection of F onto the line DC. Angle v is the angular distance of the Earth T^′ from its perihelion, seen from the Sun

In order to do that, we first will calculate FB applying the cosine rule to triangle FEB. We already know that FE is equal to the eccentricity of Venus (e_V), while EB is the eccentricity of the Earth, e_T (because EB = FC). Of course, they have to be normalized using the proportion between the radii of the orbits. Angle (π_TSπ_V) = π is the difference between the longitude of the perihelion of Venus and the longitude of the perihelion of the Earth.

Therefore:

$$ FB = \sqrt {e_{t}^{2} + e_{m}^{2} - 2 \cdot e_{t} \cdot e_{m} \cdot \cos \pi } $$

Now we can obtain angle γ, using the sine rule to triangle FEB.

$$ \gamma = \sin^{ - 1} \left( {\frac{{e_{V} \cdot \sin \pi }}{FB}} \right) $$

Now we will obtain FH. Looking at triangle FHC, we know that:

$$ \sin \gamma = \frac{HF}{FC} $$

We can obtain HC applying Pythagoras’s theorem to triangle FHC:

$$ HC = \sqrt {e_{T}^{2} - HF^{2} } $$

We know that angle T^′HC = α + γ. So, we can obtain angle ε applying the sine rule to triangle HCT^′

$$ \varepsilon = \sin^{ - 1} \left( {\frac{{HC \cdot \sin \left( {\alpha + \gamma } \right)}}{{R_{T} }}} \right) $$

Looking at triangle HCT^′ and knowing ε we can calculate ξ.

$$ \xi = 180 - \left( {\alpha + \gamma + \varepsilon } \right) $$

We also know that angle T^′CS = η is:

$$ \eta = \alpha + \varepsilon $$

And so we can calculate side ST^′, applying the cosine rule:

$$ ST '= \sqrt {e_{T}^{2} + R_{T}^{2} - 2 \cdot e_{T} \cdot R_{T} \cdot \cos \eta } $$

And, applying the sine rule, we can obtain angle θ.

$$ \theta = \sin^{ - 1} \left( {\frac{{R_{T} \cdot \sin \eta }}{ST '}} \right) $$

And we know that

$$ v = 180 - \theta $$

1.4.9 Step 9 for Mercury: making the orbit of the Earth a pseudo-ellipse

Step 9 only introduces a modification in the distance from the center of the orbit to the planet. So, we will only show how to calculate this distance. In the following figure, DHC is the apsidal line of the orbit of the Earth, H is the equant point (remember that in the case of Mercury it is closer to the Earth than the center of the deferent). So, the Earth, T, moves uniformly from H. Consequently, angle α increases uniformly. C is the previous center of the deferent and now is the center of the small epicycle that carries the moving center of the deferent, B. CB is, therefore, the radius of this small epicycle. The model implies that angle HCB = −angle DHA = −α. If the Earth moves according to the previous model, it will be at T, while step 9 implies that the Earth is at T^′. D is the center of the orbit of Mercury, which corresponds to the center of the epicycle in a geocentric frame of work. Therefore, distance DT^′ is the distance between the Earth and the center of the epicycle. Our aim, therefore, is to calculate distance DT^′. Pedersen (2010, p. 319, 320) gives the equations for calculating this distance for the Ptolemaic model for Mercury, but they are not useful in this particular case, because the development takes advantage of the fact that CB = CH = HD, which is true in the case of the Ptolemaic model, but not in the ideal case (Fig. 46).

Fig. 46

The Earth at T and T^′ moves uniformly from H, the equant point. D is the center of the orbit and uniform motion of the planet P^′ (not drawn in the figure, refer to Fig. 45 for complete the figure). The center of the orbit of T is fixed at C, while the center of the orbit of T^′ is B, that revolves around C in a circle with radius CB with the same speed but opposite direction that T revolves around H, so that DHT and HCB are equal but opposite

CH is known, because it is the distance between the center of the moving eccentric and the equant point, and also CB is known, because it is the radius of the small epicycle; therefore, BH can be obtained applying the cosine rule to triangle CBH:

$$ BH = \sqrt {CH^{2} + CB^{2} - 2 \cdot CH \cdot CB \cdot \cos \alpha } $$

Now, applying the sine rule to triangle CBH, we can obtain angle β:

$$ \beta = \sin^{ - 1} \left( {\frac{CB \cdot \sin \alpha }{{{\rm B}{\rm H}}}} \right) $$

We know that BT^′ is R and that δ = 180 − α. So, applying the sine rule to triangle BHT^′, we can obtain angle BT^′H = γ.

$$ \gamma = \sin^{ - 1} \left( {\frac{{{\rm B}{\rm H} \cdot \sin \left( {180 - \alpha + \beta } \right)}}{{{\rm B}{\rm T}^{\prime } }}} \right) $$

Looking at triangle CBH, we know that angle CBH = η is 180 − α − β. Now, applying the cosine rule we can obtain side HT^′:

$$ HT' = \sqrt {BT'^{2} + BH^{2} - 2 \cdot BT' \cdot BH \cdot \cos \eta } $$

And, finally, we can calculate DT^′, applying the cosine rule, but now to triangle HDT^′:

$$ DT' = \sqrt {HT'^{2} + HD^{2} - 2 \cdot HT' \cdot HD \cdot \cos \alpha } $$

1.5 Appendix 5: How we arrived at the correct value of the radii of the epicycle and the little circle for Mercury

In all the planets, because the eccentricity is so small, the semi-major axis expressed in UA multiplied by 60 gives us the value of the epicycle in parts. But the eccentricity of Mercury is so big that the difference it is not negligible. Therefore, we must be more careful, because the ellipse would look greater or smaller depending on the observer’s point of view. According to Ptolemy, the size of the epicycle is measured at the apsidal line; this means that the elliptical orbit will be seen as small as possible. In Fig. 47Q and S are the foci of Mercury’s orbit; therefore, line QDSO is the apsidal line. D is the mean point between the foci. The semi-major axis of Mercury (line DN) is 23.22 parts, while the semi-minor axis (line DA) is 22.73 parts. The circle with center in D and radius DO represents the Earth’s orbit (made circular for simplicity). The Earth is at O when it is in the apsidal line. Seen from the apsidal line, the tangents of Mercury’s orbit are OF and OG. Radius DF of the ellipse is close to the minimum, 22.79 parts. Therefore, the radius of the epicycle will be 22.79 parts (represented by the dotted circle)

Fig. 47

Q and S are the foci of the elliptical orbit of Mercury, S is the Sun, and D is the center of the orbit. The elliptical orbit is bold. When the Earth is at O, the apsidal line of Mercury, F and G are the tangent points; therefore, FOG is the apparent size of the ellipse seen from O. Therefore, if circular, the orbit of Mercury has the radius DF. The circular orbit is dotted. When the Earth is at H, the angular size of the ellipse is IHJ, greater than FOG. In order to see the circular orbit with this apparent size, the Earth must be at M, because JHI and LMK are equal and K and K are tangents to the circular orbit

When Mercury’s orbit is considered from the quadrature with respect to the apsidal line, the apparent size of the orbit will be angle IHJ, because HJ and HI are the tangents of the orbit seen from H. In order to keep the same absolute size of the epicycle we have to move from H to M, because KMK is equal to IHJ, but K and L are tangent to the epicycle with radius 22.79 parts. Therefore, at quadrature the Earth must be at M and not at H. Distance MH is 1 part. So, the radius of the little epicycle must be 0.5 parts and the radius of the deferent 59.5 parts, so that when the Earth is at the apsidal line, distance DO is 59.5 + 0.5 = 60 parts, while when the Earth is at M, the distance DM is 59.5 − 0.5 = 59 parts.