Perhaps the errors in Chapter 47 came from wrongly using the area as a measure of time. After all, Kepler has not yet considered area as a true measure of time, but only as a convenient way of adding an infinite number of distances. The true principle in his mind is that the distance from the sun determines the rate of motion. In this chapter, Kepler uses a sum-of-distances approach, rather than an area approach for measuring time. This has several advantages:

Using the true distance principle, rather than the approximate area-time hypothesis.

The inverse relationship does not hold if you sum up distances and then take the inverse, which is what happens if you use area. See the aside from chapter 46 for an example.

It is no trouble to add up any particular number of distances.

Since no sector areas are involved, the problem of the size of ovoid sectors does not concern us.

“And so with renewed preparations, I settled down again upon the eggs…”

A Difficult Calculation

Kepler’s path to implementing this idea is quite difficult, and will probably require several readings-through, and perhaps the use of a spreadsheet program to try to replicate Kepler’s method. (The way to be sure that you know what Kepler is doing, is to do it yourself!)

Again using the idea of the fictitious semi-circle, move angle GBD through each degree of mean anomaly, and determine the distances AG. Add up all of these distances to get a sum for the entire path of the planet. Kepler reports a sum of 36,075,562 for the entire oval when the eccentricity is 9165.

Now, by looking at each degree of mean anomaly, compare the distance at that mean anomaly to the average distance from the sun. The greater the distance, the less the planet will have moved in that amount of time. Create point C, which has moved the appropriate distance DC corresponding to the distance from the sun over that period. DC is inversely proportional to length AG. Also, to agree with chapter 45, C must be at distance AG from the sun. Lines are drawn through C from A and B, creating points E and F, respectively:

But, how can the correct angle DBC be determined, when the determined motion is along non-circular arc DC, rather than along circular arc DF?

“It was therefore not enough to know the length DC. The angle CBD ought to have been investigated as well. For because CD is shorter than FD, CD does not measure the angle FBD, or CBD.

Kepler adopts a few approximations: that CE and CF are equal, and that arc EF is imperceptible. Then, assuming CD to be the same as DF (or DE), length CD is the measure of angle EBD.

Now, calculate the length of AE and compare it to AC (equal to AG) as calculated in step 1. This difference is CE, which is assumed to be equal to CF, the measure of the ovoid’s approach to the center B. The ratio AC : AE is a measure for how much greater arc DC must appear to B -- i.e. the optical magnification.

Correct the angles CBD, according to this optical parallax. Then, with known length AC (equal to AG, remember) and eccentricity AB, the law of tangents can be used to consruct a table of the equated anomaly CAD for all the degrees of mean anomaly.

Whew! That’s a lot of work! And no one anomaly can be known without working through all the others leading up to it. Kepler:

“I can’t imagine anyone reading this not being overcome by the tedium of it even in the reading. So the reader may well judge how much vexation we (my calculator and I) derived hence, as we thrice followed this method through the 180° of anomaly, changing the eccentricity each time.”

There is a problem in all of this…

Determining the ovoid distance

To properly account for the optical magnification of each portion of the oval orbit, we must first know the total length of the orbit. Otherwise, the magnification would be incorrect and the equated anomaly for mean anomaly of 180° would not be 180°.

Just as Kepler sought the quadrature of the area of the ovoid in chapter 47, he must now determine its circumference. To this end, he uses the diagram on the left. The semi-circles DR and HK are centered on B, with radii BD and BH corresponding to the semi-major and and semi-minor axes, respectively, circumscribed outside and inscribed within, the green dotted ellipse DR. The circumference of the ovoid must be greater than HK and less than DR, but where in the middle between the two is the correct circumference located?

Kepler now draws two “mean” semi-circles: DK (in blue, centered on I) and OP (in red, centered on B). Circle DK has a radius (and thus a circumference) which is the arithmetic mean between the two bounding circles, while circle OP has a radius (and circumference) which is the geometric mean between the two. (Do you know how to construct OP?)

Kepler argues that it is correct to use the arithmetic mean circle DK. He reasons thus: the geometric mean circle has the same area as the ellipse, while the arithmetic mean circle has a larger radius (and area). But among shapes covering a given area, it is the circle which has the least circumference, and thus the geometric mean circle, having an area equal to the ellipse, must have too short a circumference. This leaves him with the arithmetic-mean semicircle, whose length is 179°23′40″ where the semicircle KR is 180°.

But…

This arithmetic length is only an estimate. Kepler must try out hypothesized lengths of the oval and work out the entire 180° to determine whether he has chosen the right length for the oval. If he has, he will get an equated anomaly of 180° for a mean anomaly of 180°.

“As for me, I… by a most laborious and dogged calculation found… that where the perfect semicircle is 180°, the oval is 179°14′15″.”

With a length for the oval, Kepler can now implement his method. (For a degree-by-degree working-through, click here for a Microsoft Excel spreadsheet.) This is the resulting motion:

The red point is the planet, while the blue point is the position of point F before the optical magnification of the oval path has taken place. In which regions of the orbit does the red point move more quickly than the blue? Do you understand why?

Results

Now Kepler can implement his tedious method, and compare his results with the “veracious vicarious,” his “index of truth.”

Mean Anomaly

Equated Anomaly

Equated (Vicarious)

Difference

45°

38°5'33"

38°4'54"

+ 39"

90°

79°31'31"

79°27'41"

+ 3'50"

135°

127°0'1"

126°51'9"

+ 8'52"

These errors, especially the fact that the 90° value is so far off, indicate to Kepler that the eccentricity (9165) is too big. Redoing his entire lengthy calculation with an eccentricity of 9230, he arrives at:

Mean Anomaly

Equated Anomaly

Equated (Vicarious)

Difference

45°

38°2'24"

38°4'54"

− 2'30"

90°

79°26'49"

79°27'41"

− 0'52"

135°

127°56'25"

126°52'0"

+ 4'25"

As in chapter 47, the planet moves too quickly in the middle longitudes. But note the reasonably good results -- this is the closest that Kepler has come to matching the accuracy of the vicarious hypothesis since he began trying to implement his physical causes in the current Part IV.

“So, when I saw that as the physical causes introduced in chapter 45 were the more skillfully, and the more fittingly advocated, for the implementation of the theoretical foundations of the calculation, I came ever nearer to the true equations furnished by the vicarious hypothesis of ch. 16, I greatly congratulated myself, and was confirmed in the opinion of chapter 45.
“On the other hand, since I disliked the many contrivances with which I contended in this chapter, I did not rest until I had established a more certain and direct way, and at the same time I began to suspect that the calculation had not accomplished everything which the opinion of chapter 45 had required.”