# Chapter 24 — The “Solar Year” versus the “Sidereal Year”

We will now see how Earth’s 1-mph-motion accounts for the 20.41-min. difference between the solar (a.k.a. tropical) year and the sidereal year. You may have asked yourself (rightfully) this good & fair question: “Why-oh-why is the sidereal day shorter (by 4 min.) than the solar day, whereas the sidereal year is 20.41 min. longer than the solar year?”

The Copernican model offers yet another incredibly convoluted “explanation” for this major conundrum. If you are not familiar with it, you may read all about it in Wikipedia type sources, but here is a compilation of example data from an Arizona State University professor of philosophy (Michael J. White).

In any case, here is a most pertinent question (highlighting this particular riddle) raised by the Binary Research Institute:

Sidereal vs. Solar Time: Why is the delta (time difference) between a sidereal and solar day attributed to the curvature of the Earth’s orbit (around the Sun), but the delta between a sidereal ‘year’ and solar year is attributed to precession? […] The burden of proof lies with those who support the current lunisolar precession theory which requires a different explanation for the two deltas.”

Understanding Precession of the Equinox: Evidence our Sun may be part of a long cycle binary system by Walter Cruttenden and Vince Dayes (2003)

Let us first take a look at these 20.41 min. which, in fact, represent the time difference between a solar and a sidereal year.

Average duration of a solar (or “tropical”) year:

365.24219 days or 525,948.753 minutes (by the Gregorian calendar’s year count)

Average duration of a sidereal year:

365.256363 days or 525,969.163 minutes (as of empirical observation)

Ergo, a discrepancy of 20.41 minutes – or a difference of 0.00388%

We see that 20.41 min. is 0.00388% of 525,960 min. (i.e.; 365.25 days). And in fact, the currently-observed amount of annual “equinoctial precession” (50.29 arc seconds) amounts to 0.00388% of 1,296,000 arcseconds (remember: 1,296,000″ arcseconds = a full 360° circle). Hence, those 20.41 minutes are, manifestly, a direct consequence of the so-called equinoctial precession.

Earlier on, we determined that our ACP (Annual Constant of Precession) amounts to:

51.136 arc seconds-per-year
i.e.; a value 1.68% larger than the currently-observable precession rate of 50.29″.

Note that the modern, academic estimate of the duration of one full 360° equinoctial precession (i.e.; one Great Year) is ca. 25,770 years. This is, in fact, about 1.68% longer than the TYCHOS reckoning of 25344 solar years.

Please see the Wikipedia entry on “Axial precession“.

My next graphic should help visualizing why a small portion (about 1.68% in our epoch) of the equinoctial precession will always remain unobservable from Earth. The reason for this “hidden” angle of precessional drift (which increases over time) is something that, yet again, can be demonstrably attributed to Earth’s yearly 14,035.847-km motion.

The Sun’s orbit’s “radial equivalence” of Earth’s yearly 14,035.847-km motion is 37,087 km (as we found in Chapter 22).

In 20.41 min. (or 0.3401667 hours), the Sun — traveling at 107,226 km/h — will cover about 36,475 km, i.e.; about 1.68% less than 37,087.424 km.

It thus becomes plainly evident what causes this 20.41-min. difference between the solar and sidereal year: it simply represents the extra time needed for the Sun (as viewed from Earth) to line up again with a given star, after the completion of a solar year. These 20.41 minutes will effectively reset the Earth-Sun-Stars alignment which, in actuality, has been offset by Earth’s yearly motion. Let me describe in due detail the above graphic by using the solar positions marked A, B and C:

As Joe, our earthly observer, moves from A to B (i.e.; from June 21, 2001 to June 21, 2002) he will have experienced a “Solar year”. Since Earth (traveling at 1 mph) has, in that time period, moved along by 14,035.847 km, Joe will “meet up” with the Sun at an earlier point (compared to the previous year) of the solar orbit. As the Sun’s orbit is 2.642336X larger than Earth’s PVP orbit, Joe’s lateral displacement will be proportionally equivalent to a 37,087.424-km-section of the solar orbit (14,035.847 km X 2.642336 ≈ 37,087 km). This is the distance between A and B, as of the above graphic.

This small angular offset (with respect to the Sun-Stars alignment) that Earth’s motion causes will then quickly be “regained” by the Sun’s speedy motion. In only 20.41 minutes, the Sun will once again line up (at point “C” ) with the same star that it faced one year earlier. Traveling at 107,226 km/h, in 20.41 minutes the Sun covers 36,474 km (the distance between B and C) which is 1.68% less than 37,087.424 km.

Note that our earthbound Joe will not realize the full extent of the true annual stellar precession; a small portion of it (currently ca. 1.68%) will remain unobservable to him. This, because Joe is unaware of Earth’s 1-mph-motion and he therefore (wrongly) believes that Earth has returned at the same physical place as the previous year. Joe will thus conclude that the annual stellar precession rate amounts to 50.29” instead of the true constant rate of 51.136” per year (our ACP).

The TYCHOS model thus explains why the sidereal year is, in all logic, longer than the solar year.

Note that, as viewed under Earth’s rotational frame of reference, in 20.41 min. Earth will ROTATE by 18,369” arcseconds. This, because 1 min. of our daily 1440-min. rotation corresponds to 900 arcseconds (1,296,000”/ 1440 = 900”). Hence, 20.41 X 900 = 18,369”. If we now divide this value by 365.25, we obtain:

18,369”/ 365.25 = 50.29” (the currently observed annual “equinoctial precession”).

Thus, the ratio between the observed stellar precession (Sun vs. stars) and Earth’s rotational motion is, as would be expected, circa 1 : 365.25.

Later on (in Chapter 30) we shall see why the rate of increase of the equinoctial precession is observed to grow over the centuries — due to the less-than-optimal year count of our current Gregorian calendar, which lets the Sun drift too much Eastwards over time.

Astronomy describes the so-called “anomalistic year” as follows (the below is from Wikipedia).

“The anomalistic year is usually defined as the time between perihelion passages. Its average duration is 365.259636 days (or 365 d 6 h 13 min. 52.6 s – at the epoch J2011.0).”

The oddly-named “anomalistic year” (the period in which the Sun returns to its closest or furthest point from Earth) lasts on average for 365.259636 days. Incidentally, this is approx. 4.7 minutes more than the sidereal year of 365.256363 days. It is defined as “the time interval between perihelion passages”.

In the TYCHOS, a more aptly-worded description would be “the time interval between the Sun’s perigree transits”. In our current epoch, the Sun’s perigree transit occurs around January 3.

In short, the “anomalistic year” is defined from Sun’s perigee procession, and lasts for about 4.7 minutes longer than a sidereal year. We see that in the course of 4.7 minutes a given point on Earth’s equator will rotate (within the terrestrial rotational reference frame) by 4230”. This is because 1 minute of our celestial sphere of 1440 min. (24 hours) corresponds to 900 arcseconds.

Thence, 4.7 min. X 900” = 4230” (arcseconds)

Now, let us imagine two hypothetical signposts (“A” and “S”) being moved around Earth’s equator year by year with the following parameters.

Signpost “A” is kept pointing towards the celestial spot of each year’s passage of the Anomalistic year.

Signpost “S” is kept pointing towards the celestial spot of each year’s passage of the Sidereal year.

Since signpost “S” is conceptually always being kept oriented towards a given fixed star (in this thought exercise we disregard Earth’s daily rotations around its axis), it will complete 1 revolution around Earth’s equator in 25344 years.

On the other hand, signpost “A” will be moved each year by an extra 4230” (arcseconds) in relation to signpost “S”.

In the TYCHOS Great Year (25344) signpost “A” will have revolved around Earth by 4230” X 25344 = 107,205,120”

107,205,120” / 1,296,000” (i.e.; 360°) = 82.72

Therefore, “A” will complete 82.72 full revolutions in 25344 years, making the spin ratio between signpost “A” and “S”

82.72 : 1

Since we know that signpost “A” moved by 4230” annually, we can now find out by how much signpost “S” was moving annually:

4230” / 82.72 = 51.136 arcseconds-per-year

Hence, the so-called “anomalistic” year (which is observed to be 4.7 min. longer than the sidereal year) further corroborates our Annual Constant of Precession of 51.136 arc seconds (representing Earth’s 1-mph-motion as posited by the TYCHOS model).

We may further confirm our ACP by using the value of 11.75” arcseconds, which (as described in old astronomy books) is the observed annual amount of precession of the Sun’s apogee measured from Earth:

“On the anomalistic year: the year called the anomalistic year is sometimes used by astronomers, and is the time from the sun’s leaving its apogee till it returns to it. Now, the progressive motion of the apogee in a year is 11”.75, and hence the anomalistic must be longer than the sidereal year, by the time the sun takes in moving over 11”.75 of longitude at its apogee.”

— p.48, The Elements of Astronomy: Designed for the Use of Students in the University by Samuel Vince (1811)

In 25344 years, the Sun’s apogee will thus precess by:

25344 X 11.75” = 297,792” arcseconds (i.e.; less than 1/4th of a full circle).

To complete a full 360° precession around our system the Sun’s apogee will therefore need:

1,296,000” / 297,792” ≈ 4.35203094777562862669

4.35203094777562862669 X 25344 ≈ 110,297.87 years*

And once again, we can find our ACP by multiplying this factor of 4.35203094777562862669 by 11.75” :

4.35203094777562862669 X 11.75” ≈ 51.136
or
1,296,000” / 297,792” X 11.75” = 51.136