Tychos link with Egypt?

Hi Simon, Patrick,

I am really impressed with all the work you have put in creating the Tychos model. Also supported you financially

I have reviewed the model and did some calculations myself. Hopefully it is of any further help to bring it even further because I found some nice additions.

You started all calculations based upon the numbers: 9256896 days/ 25344 years. Which is 365,25 day per year. I believe the correct calculations should all be based upon a year length of 365.2421875 days.

25344 year times 365.2421875 (=real average Equinox year) = 9256698 days
I checked and all equinoxes are matching if you change the code to 365.2421875 in the time constants:

Via https://codepen.io/pholmq/pen/XGPrPd
//DEFINE TIME CONSTANTS
const yearLength = 365.2421875
const earthRotations = 366.2421875

This was an eye opener for me:
Reason is the sidereal day being 23 hours, 56 minutes and 4.0905309414 seconds being 86164,0905309414 seconds in a day which is ((365+(31/128))/(366+(31/128)))*86400
and 365+(31/128) = 31 leap years in 128 years = 365.2421875

The earth rotates exactly 1 time in a year fully extra to match up the earth rotation itself. You already knew, but in my view this is the exact calculation.

So if you fill in Juliandates:
-117877139 to 104283613 contains 24 cycles of 9256698 days
You will meet juliandate 2459935 as 21-12-2022 on one of those cycles but you can also use other equinox dates as well as long as it fits in the 9256698 days cycle.

The pyramids of Egypt are researched a lot on the internet and a common understanding is there is a ratio of 1:43200 in it compared to the size of the earth. Additionally the “stadia” is used which is 300 non-royal cubits and therefore 253440 stadia will fit in the circumference of the earth. The number of times the non-royal cubit fits in the circumference of the earth is 76032000.

Actually researching the number 253440 let me to the tychos model:-)

Egypt Great Pyramid ratio is 1:43200 (half a day of seconds)

  • Great Pyramid circumference *43200 = Earth circumference

If you take the 9256698 days cycle as base for the surrounding of the pyramid:

  • 0.9256698 * 43200 = 39988.93536 km (=24847.9724409449 miles)
  • So Great Pyramid base length = 0.23141745 km (0.9256698/4).
  • This creates a non-royal cubit of 52.594875 cm (39988.93536 km divided by 76032000)
  • Which creates a stadia length of 157.784625 meter

So the earth turns 39988.93536 km / 432000 = 0,9256698 meter per (almost) 2 seconds in a day (=0.92566983060)/ 86164.0905309414 * 86400 = 1670.76756 km/hour

Which is the same the other way around as (366.2421875 rotations in a year * distance 39988.93536)/(365.2421875*24) = 1670.76756 km/hour

Another calculation is: Earth rotation engine rpm is 1/1436.06817551569 (minutes per day) = 0.00069634577 rpm.

To get the speed of the earth you have to multiple the diameter of the earth (12728.87346 km)* earth rotation engine (0.000696346)* 60 (to get to hours) * PI = 1670.767560
So earth’s rotational speed (at the equator) is 1670.767560 km/h

The speed decreases by the cosine of your latitude so that at a latitude of 45 degrees, cos(45) = .707 and the speed is .707 x 1670.767560 = 1180 km/h. You can use this formula to find the speed of rotation at any latitude.

Another link with the pyramids is the alignment of the pyramids on the date -4003-10-24. Which is an important date for the freemasons as it is the date the world was created. Anno Mundi/ Anno Lucis. We currently live around year 6026 AL. The great pyramid was said to be built by Hemiunu which also has a connection to those guys.

The date 24th of October -4003 juliandate 259259, which was exactly during the autumnal equinox, aligns with earth+moon (Great Pyramid), Sun (Khafre) and Mars (Menkaure) on that date. As above, so below. Kind of weird.


Note: just like the order in Teotihuacan with the pyramid of the sun being the middle Pyramid! And there is also a connection between the temple of the Feathered Serpent (first pyramid) and the calendar. Temple of the Feathered Serpent, Teotihuacan - Wikipedia.

Back to gizeh. The sphinx looks at Jupiter on that date. The sun is aligned with the autumnal equinox. The black sun (= the centre on which the earth revolves) is horizontally aligned with earth on that day which on the ground points to the black kabaa of mekka in east-south direction. The kabaa being 799.77870720 * 462834.9 times away from the great pyramid which coincidentally exactly is 799.77870720 miles between the Kabaa and the Black sun (=corner of the great Pyramid). Which is the circumference of the earth (39988.935360)*9256.698. The great pyramid points to Saturn in north-east direction.

There are some more strange connections to old sites when you draw a big circle with the great pyramid as centre which crosses the Kabaa. The circle crosses - besides the Kabaa - Greece places like mount athos, tower of Babylon, Nimrud, hagia sophia, temple of Amun, Old Dongola, etc. If you hold the kabaa at the top, to the left you align with baalbek, Lebanon and the birthplace of all: mount ararat/ Noah’s ark. Look it up for yourself.

So the great pyramid might hold more secrets useful for the Tychos model.

I therefore introduce 1.66620564 km per hour over here which is 1.03533218503937 miles per hour and most importantly: 0.4628349 meter per second (the half of 9256698). The number is related to 365.2421875 + 9256698 and I will explain the number below.

1.66620564 km per hour = 0.4628349 meter per second. 4628349 is the half of 9256698.
Most probably also the 0.9256698 number or it’s half 0.4628349 has been used in several megalithic sites. You can check for yourself.

A little known fact is, the meter (and therefore also indirectly the mile) is related to day length.
Earth travels the same speed in ratio as the earth turns: 1/1000 PVP distance of its circumference per day

Why 1/1000 ratio? To keep the 9256698 ratio between the earth PVP and earth circumference, the earth speed needs to be exactly 1000 times lesser than the circumference of the earth. So in 9256698 days, the earth travels at a pace of 0.4628349 meters per second.

24847.972440945 miles/1000 * 23.93446959 (= sidereal day) = 24.84797244 miles per day = 1.03533218503937 miles per hour = 1.66620564 km per hour
9256698 days * 1.03533218503937 * 24 miles per day = 230010176.79815 miles PVP
Also to be calculated as = 0.4628349 meter per second * 86400 seconds a day* 365.2421875 days * 25344 years) /1000 (convert to km) = 370165497.969041 km PVP

So radius PVP earth = 36607256.5988663 miles

Earth PVP is 230010176.79815 miles/ Earth circumference 24847.9724409449 miles = 9256.698 times. Again the number 9256698.
Earth PVP radius is 36607256.5988663 miles/ Earth radius in miles 3954.67763978757 = 9256.698 times. Again the number 9256698.

Sun orbit is 2.5344 times PVP earth = 582937792.07723 miles
Sun orbit radius is 92777431.1241667 miles

Mars orbit and radius are twice the sun.

PVP constant =

  • Earth PVP is travelled in 25344 Average Equinox years: 230010176.79815 miles /25344 / 365.2421875 / 24.847972440945 = 1
  • Sun orbit is travelled in 1 Solar year: 582937792.07723 miles / 1 / 365.2421875 / 24.847972440945 (speed 1.03533218503937*24)= 64231.8336
    1/ 64231.8336 = 0.0000155686042878278

Sun travels with 92777431.1241667/ 365.2421875 / 24 = 66501.28463 miles per hour (= 107023.4434 km/hour) which is 64231.8336 times faster than earth.

So to summaries:

  • The great pyramid 4 sides adds up to 0.9256698 km
  • The great pyramid 1 side is 231.41745 meter
  • The great pyramid cubit fits 440 times in the 1 side of the pyramid (so 1760 times in the 4 sides)
  • The great pyramid is a scaled representation of the earth in ratio 1:43,200
  • The great pyramid cubit fits 76,032,000 times in earth circumference
  • The great pyramid height is related to PI and 43,200 smaller than earth radius
  • The great pyramid casing stones are missing but might have represented one year per side in which case they would have been 0.6336 meters wide
  • Earth rotates 366.2421875 times in 1 year
  • Sidereal day is 365.2421875/366.2421875 shorter than the 24 hours/86400 seconds - day
  • Earth moves 0.4628349 meters per second on PVP path
  • Earth PVP orbit is 9,256,698 days of 365.2421875 days/year
  • Earth PVP orbit is 9,256,698 times earth circumference/1000
  • Earth circumference in km is 0.9256698 * 43200
  • Earth experienced rotation at the equator is 0.4628349 km per second (but actually a little faster because of the 1/(365.2421875/366.2421875) experience)
  • Sun orbit is 2.5344 earth PVP

Might all just be some juggling with numbers, but maybe of value.

My conclusions:

  1. History is written by the victors. We cannot trust history books but need to recreate the picture with the latest information at hand. We do not know the exact function of the pyramids. My assumption it was a big calendar
  2. Old civilisations might have been aware of the structure of our galaxy because the numbers match
  3. NASA should know about the true structure of our galaxy. Unclear why they lie?
  4. There is a group of “humans” actively trying to hide certain knowledge from the rest of us. Look up the sabbatean movement.
  5. If we would have this knowledge, the earth + the human will once again be put in the centre of the galaxy again.
  6. Knowledge: Due to the missing parallax, and the earth isn’t traveling around the sun, but the sun around the earth, all stars are much closer than expected. Therefore the big bang happened about 400.000 years ago and not 13,8 billion years ago

Keep it going
Sean Sellors

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Thank you Sean for the support and this very interesting post. It certainly seems like some powerful group actively has tried to keep public science in the dark for quite some time. As Simon’s book exposes, the Copernican/Heliocentric model has been challenged previously. The most problematic event was probably the Michelson-Morley experiment and Dayton Miller’s research in the early 20th century that clearly demonstrated that Earth does not hurtle around the Sun at 107 000 km/h. Just as clearly as a Foucault pendulum or Sagnac interferometer demonstrate that Earth rotates diurnally. This was tackled with Einstein and Quantum physics that declared that the experimentally confirmable Aether-Light theory is wrong. “There is no Aether and the speed of light is constant regardless if an object moves or not. In fact light is made up of magical particles that can be at two places at once and change their behavior depending on if they are being watched or not”… If you laugh at this, then no Nobel prize for you. :slight_smile:

Sean, I’m sure you meant to type “9,282.042” - so I took the liberty to edit / change that comma to a dot.

Also, I’d be most grateful if we could stick - here at this forum - to the metric system (i.e. km rather than miles), since that’s what I’ve used in my TYCHOS book.

For now, I’ll just wish you a merry christmas - for I would need more time to look into your interesting calculi… :slight_smile:

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EDIT1: Reworded my previous post to metric system.

You are right Simon. I am from Europe, but tried to put all in american format with . = , and vice versa. I again looked at all and tried to stick with dots as separator (american) and tried to stick to meters instead of miles.

But let’s get the nobel price for both of you!

I looked at the numbers a bit more and made a mistake.

Originally I used the number 1.66620564 km/hour as the earth PVP speed, but that’s not right. 1.67076739764795 km/hour is the right number of earth PVP speed based upon the difference stellar day/ sidereal day.

I stumbled on the stellar day time compared to sidereal daytime.

  • Sidereal day time - which is the exact time between equinox - is 86,164.0905309414 (as explained above calculated + measured in real life)
  • Stellar day time - which is the exact time between star alignment - is 8.4 ms longer: 86,164.098903691 (as measured in real life)

If you look at the little bit of time difference of the stellar day (alignment stars) regards to the sidereal day (alignment equinox) you can make formulas based upon the size of the earth resulting in the speed of the PVP orbit, etc.

With the small difference of 0.00837274963851087 seconds in 1 day, in 9256698 days = 0.00837274963851087 * 9256698 days / 86,164.098903691 = 0.899493127873751 seconds difference after 1 PVP orbit.

EDIT2: In my previous deleted post I compared this number to the sidereal year. This doesn’t seem to be correct, so made some changes

So the expected distance to have travelled in 1 day, with an earth circumference of 39,988.93536 km = (39,988.93536 / 9256698 days) * the 0.899493127873751 = the expected distance to have been passed = 0.0038858103124146 meter

So the expected distance to have travelled in 1 day, in meters per second is 0.0038858103124146/ 0,00837274963851087 = 0.4641020549022 meters per second = 1.67076739764795 km per hour = 1.03816672982777 miles per hour

Now we can calculate the rest. Start with earth PVP:
9256698 days * 1.03816672982777 * 24 miles per day = 230,639,901.399919 miles PVP
Also to be calculated as = 0.4641020549022 meter per second * 86400 seconds a day* 365.2421875 days * 25344 years) /1000 (convert to km) = 371,178,941.478551 km PVP

So radius PVP earth = 36,707,480.3820244 miles = 59,074,963.3079287 km

Sun orbit is 2.5344 times PVP earth = 584,533,766.107954 miles = 940,715,909.283239 km
Sun orbit radius is 93,031,438.2802026 miles = 149,719,587.007614 km

Mars orbit and radius are twice the sun.

PVP constant =

  • Earth PVP is travelled in 25,344 Average Equinox years: 230,639,901.399919 miles /25,344 / 365.2421875 / 1.03816672982777 * 24 (24.9160015158665) = 1
  • Sun orbit is travelled in 1 Solar year: 584,533,766.107954 miles / 1 / 365.2421875 / 24.9160015158665 (speed 1.03816672982777*24)= 64,231.8336
    1/ 64,231.8336 = 0.00155686042878278

Sun travels with 584,533,766.107954/ 365.2421875 / 24 = 66,683.35264 miles per hour (= 107,316.4535 km/hour) which is 64,231.8336 times faster than earth.

1/ PVP constant = 1/ 0.00155686042878278 = 642.318336

And to see the relation with 25,344, again to my surprise: 25,344^2 = 642318336

So to summaries:

  • The great pyramid 4 sides adds up to 0.9256698 km
  • The great pyramid 1 side is 231.41745 meter
  • The great pyramid cubit fits 440 times in the 1 side of the pyramid (so 1,760 times in the 4 sides)
  • The great pyramid is a scaled representation of the earth in ratio 1:43,200
  • The great pyramid cubit fits 76,032,000 times in earth circumference
  • The great pyramid height is related to PI and 43,200 smaller than earth radius
  • The great pyramid casing stones are missing but might have represented one year per side in which case they would have been 0.6336 meters wide
  • Earth rotates 366.2421875 times in 1 year
  • Sidereal day is 365.2421875/366.2421875 shorter than the 24 hours/86,400 seconds - day
  • Earth moves 0.46410205490221 meters per second on PVP path
  • Earth PVP orbit is 9,256,698 days of 365.2421875 days/year
  • Earth PVP orbit is 9,282.04109804414 times earth circumference
  • Earth circumference in km is 0.9256698 * 43,200
  • Earth rotation speed at the equator is 0.4641021 km per second
  • Sun PVP is 2.5344 earth PVP

I added these as well, although I can’t explain it but there seem to be a link between all those numbers

  • Ratio Sun - Earth = Earth circumference 39,988.93536/ 365.2421875 = 109.48608

  • Ratio Sun - Moon = Earth circumference in km/100 = 399.8893536:1

  • Sun radius = Earth radius * 109.48608 = 432,982.152443993 miles = 696,817.229142825 km

  • Moon radius = Sun radius / 399.8893536 = 1,082.7548884362 miles = 1,742.52508317547 km

  • Moon distance = 93,031,438.2802026 / 399.8893536 = 232,642.948462338 miles = 374,402.533250174 km

  • So Earth circumference is 109.48608 times 365.2421875

  • So Sun is 109.48608 times bigger than earth

  • So Moon is 399.8893536 times smaller than the sun

  • So distance earth to moon is 399.8893536 times smaller than earth to sun

Could all be coincidence but I am hoping it is of value to you both

EDIT3: Compared to my deleted post an hour ago, I additionally played with the stellar day number (changed it) but because of the way of calculations, that doesn’t seem to have any effect on 1.670767560 km/hour. That remains the same no mather of the stellar day length. Corrected this to look at the stellar day length instead of the sidereal length. This doesn’t have much impact on the expected velocity which is now a little higher: 1.67076739764795 km/ hour

Nice christmas to all of you

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Hi Simon, Patrick,

Did some more calculations and I am even more convinced the heliocentric model is incorrect.

What helped me was this paper: https://www.researchgate.net/publication/358118027_The_Sun_and_the_Stars_are_Set_in_Motion_-_New_Model_of_Solar_System_Legitimate_Refutation_of_Heliocentric_Model

The model is not the same, but the measurements are. The model is based upon a celestial sphere which rotates. Nice reading material.

But back to Tychos. First some key numbers/terms

  • Measured sidereal day (alignment equinox) = 365.24218734080 days or 86,164.0905308388 seconds (wiki)
    This is the most important number because the tychos should take this number : The mean tropical year length 365.2421875 days a year, 25344 years = 9,256,698 days.
  • Because the earth spins 1 time extra round in 1 year: Measured sidereal day (alignment equinox) = 366.24218734080 days of 86,164.0905308388 seconds = 86,400 seconds a day.
  • A sidereal day, calculated in 1 year is therefore 31,556,924.9862451 seconds. The inputs for tychos should be 365.2421875 * 86400 = 31,556,925 seconds
  • Measured stellar day (alignment stars) is 86,164.0989036907 seconds. Because the earth spins 1 time extra resulting in a year length of 365.24215200864 days
  • Measured sidereal year (alignment sun in same place on earth) is 86,403.3533292235 seconds. Because the earth spins 1 time extra resulting in a year length of 365.256363004. A sidereal year is therefore 31,558,149.76 seconds
  • Each day we see the sun moves 3.3533292235 seconds moving back on the zodiac ring, resulting in the precession of the equinox.

The current precession of the equinox, by calculating the difference between the time it takes for the sun to align, not only on earth, but also in the same place in the sky, will take 25,765.4391323937 years.

To get to the tychos model, there is no precession and we only notice the pattern of 25,344 years.

To make the pattern work, the sun shows a bit faster because the earth moves in the opposite direction along on its PVP. So instead of 3.353329223 seconds, there will be actually a delay of an extra 0.055893998 seconds, resulting in 86,403.4092232202 for a sidereal year (alignment sun in same place on earth). The 3.4092232202 seconds a day * 9,256,698 days in 25,344 years results in 31,558,149.76 seconds. The earth rotated 1 PVP in total after these 9,256,698 days.

We have also seen however there is a difference between stellar day (alignment stars) regards to the sidereal day (alignment equinox) you can make formulas based upon the size of the earth, the speed of the orbit.

There was 1 piece missing in my previous calculations, and that was the extra 0.055893998 seconds to get to the 25344 year cycle instead of 25,765.4391323937 year cycle was not taken into account. I only showed the observed speed, orbit, etc. I think we can now proof the earth moves along its PVP path on a 25344-year cycle of 9,256,698 days with a mean tropical year length 365.2421875 days a year!

For some reason I can’t explain (yet) the PVP constant seems to be 1/ ((25,344^2)/ 10000 = 64,231.8336) = 0.00155686042878278. Meaning the speed of the sun is 64,231.8336 times faster than earth speed.

The extra 0.055893998 seconds can therefore be divided between earth PVP speed and Sun orbit speed taking this 64,231.8336 key into account.

  • earth PVP speed -0.00000087017799138 seconds
  • Sun orbit speed 0.0558948683249640 seconds

Let’s get back to comparing stellar to sidereal day:
With the small difference of 0.00837285189481918 seconds in 1 day between Sidereal day and Stellar day, in 9,256,698 days = 0.00837285189481918 * 9,256,698 days / 86,164.0989036907 *100 = 89.9493215 % on the round trip difference for 1 PVP orbit. 1 full PVP earth orbit can be calculated to be taken place if there is a difference of 0.00837285189481918/ 0.899493215 = 0.009308298 seconds a day.

We have seen the sidereal year (alignment sun in same place on earth) is 365.256363004 days. We therefore need to multiply 0.009308298 * 366.256363004 (1 extra orbit) = 3.409223190168 seconds. Almost there. The extra 0.055893998 seconds part of the not directly showing part of the earth PVP speed need to be added as well (the minus figure -0.00000087017799138) = 3.409223190168 - 0.00000087017799138 = 3.409222319990. The expected value was 3.4092232202

That’s the Tychos number which shows there is no precession!

Another way to calculate this is that the sun orbit does take 86,403.3533292235 + 0.0558948683249640 (The extra 0.055893998 seconds part for the sun) = 86,403.4092240918 seconds a day - 0.00000087017799138 seconds for the opposite direction = 86,403.4092232216 which is not directly visible because the earth moves in opposite direction.

I will not fully calculate the rest of the puzzle, because there may be some assumptions and round-up mismatches. I made some rough calculations for myself and will just add them here:
Earth PVP orbit speed = 1.670611222 km/h
Sun orbit speed = 107,306.4220705 km/h
Earth PVP orbit circumference = 371,144,245.4 km
Sun orbit radius = 940,627,973.8 km

Also made a drawing that could be of help that shows the tilt. Where does the sun tilt to?

This was also a nice exploration for myself. Hopefully all this information helps you both as well. Let’s expose NASA!

Good luck!

Hello Sean, very interesting stuff. i wish i had your math abilities and understanding.

“Chi va piano va sano — e va lontano.”

Razzaq’s paper ends with this statement:
“Now no doubt is left that the sun revolves around the earth. Amazingly, orbital speed of
30 km/s could never affect any physical phenomenon on or around the earth. Thank
God, the earth has stopped running in the orbit with extremely high speed that could
cause alarming situation any time. Now you can feel easy”

However, I personally think it is crazy to think that the entire celestial sphere rotates.

Abdul Razzaq needs to hear of the PVP orbit of earth.

Why is this so fun to me, sitting around reading scientific papers?

Hi Muu, I am no expert in math, just have a nice big excel which took me a number of years to build with lots of data in it. This way you can recognize paterns which are hidden to the eye. And the only way to learn, is to dive into a topic. At first it is all blur but once you work with it, you learn. Specific to understand the Tychos I also learned Stellarium. Opened a whole new world to me:-)

The paper made it clear to me there could be other explanations for the mismatches we see in the heliocentric model. If you look at the data, the model doesn’t add up. It is better explained we are geocentric.

“the simplest answer is often the correct one”

Would indeed be nice if Abdul Razzaq and Simon/ Patrick met to share more information with eachother. What I missed in the model of Mr Razzaq was an explanation of the small difference of 0.00837285189481918 seconds in 1 day between Sidereal day and Stellar day. To me that’s the key information for the PVP orbit of earth.

“The universe is all in perfect harmony based upon number 432”

“The great pyramid was the key for discovery”

Is the actual precession of the equinox explained with phi * 2 PI?

Did a lot of discoveries and had some eureka moments last weeks when playing around with the numbers 365.2421875, 25,344 and 9,256,698. I hope you like what I am about to share:

Sun Solar year is actually 0.05580121678016 seconds a day longer as seems visible because of the movement of the earth. Therefore the actual sidereal year of the sun - if the earth didn’t move - is based upon:

  • Sidereal year = 86,403.40909090909091 seconds a day
  • Sidereal year = 31,556,945.3809585 seconds a year
  • Solar day = 86,400.0558012168 seconds a day
  • Solar year/ Solar day length would have been 365.242423390723 days/year
  • Sidereal day would have been the same as Stellar day if the earth didn’t move = 86,164.0905309414 seconds a day
  • Ratio to actual day duration if the earth didn’t move would be 1.00000064584742

But since the earth orbits in opposite direction causing the precession of the equinox:

  • Sidereal day = 86,164.0905309414 seconds a day
  • Sidereal day = 1,436.06817551569 minutes a day
  • Sidereal day = 23.9344695919282 hours a day
  • Stellar day = 86,164.098903691 seconds a day
  • Sidereal year = 86,403.3532896923 seconds per day
  • Sidereal year = 31,556,925 seconds a year (366.2421875 * 86,164.0905309414) NOTE: 1 extra spin of the earth in 1 year
  • Sidereal year = 8,765.8125 hours a year

Resulting in:

  • Solar day length = 365.256362996094 days/ year
  • Solar year length = 31,558,149.76286250 seconds
  • Solar day = 86,400 seconds a day
  • Solar year = 365.2421875 days/ year

I found the explanation of these numbers when doing some calculations on the small difference between stellar day (alignment stars) regards to the sidereal day (alignment equinox):

  • Stellar day = 86,164.09890369 seconds a day
  • Sidereal day = 86,164.0905309414 seconds a day

In a full tychos cycle that would be: ((86,164.098903691- 86,164.0905309414)* 9,256,698 days)/ 86,400 = 89.7037208718799 % of the circle.
Meaning the difference in seconds should have been: (86,164.098903691- 86,164.0905309414) / 0.897037208718799 = 0.00933378187340670 seconds a day.

Check: 0.00933378187340670 * 9,256,698 days = 86,400.

The current difference between sidereal year and solar year = 3.3532896923 a day
Sidereal year = 31,556,925 seconds a year - (3.3532896923 * 365.2421875) = 31,558,149.76286250 seconds.

If the earth wouldn’t move, a day in a sidereal year would last 0.00933378187340670 seconds a day * 365.2421875 = 3.40909090909091 extra on top of the 86,400 = 86,403.40909090909091 seconds a day.
Sidereal year would have been 86,403.40909090909091 * 365.2421875 = 31,558,170.143821 seconds instead of 31,558,149.76286250 seconds a year = 20.3809584751725 extra seconds. These need to be added to the 86,400: 20.3809584751725 / 365.2421875 + 86,400 = 86,400.0558012168.

You can calculate the rest of the actual sun day for yourself.

NOTE: The actual stellar day difference wouldn’t have any impact on the outcome to get this number. The outcome shows the actual duration of the orbit of the sun. So even if the earth moved really fast - resulting in a bigger difference between stellar day and sidereal day - we would have found this number. If the difference would be bigger, the PVP constant would have shifted and all numbers would have changed (also the 25,344). This is a harmonic number belonging to 25,344.

It’s also calculated simply as (9,256,698 / 86,400 =) 0.00933378187340670 * 365.2421875 = 3.40909090909091.

So the sun orbits the earth in a pace of 86,403.40909090909091 seconds a day.

If the sun solar day length would have been 3.40909090909091 a day * 9,256,698 days = 31,556,925 seconds which is a year, there was no precession.
What we notice is 3.3532896923 seconds delay: 31,556,925 seconds/ 3.3532896923 seconds = 9,410,736.29044946 days to get back to a full circle of 31,556,925 seconds.

Therefore Precession of the equinox is 9,410,736.29044946 days which is 25,765.74287561 years = 25,765 years, 271 Days and around 8 hours

This number is also the circumference of the sun orbit.
Earth PVP orbit = 2.5344 times smaller because of the 25,344 year connection. The rest follows with the figures below.

Some more numbers to keep all in harmony:

  • Earth circumference = 39,988.93536 km
  • Earth Nautical mile = 1.8513396 km = meters per 1/60 degree = 3,520 royal cubit = 1 arc minute
  • Earth Nautical mile/2 = 0.9256698 km = Great pyramid circumference (also 39,988.93536/43200)
  • Earth royal cubit = 52.5948750 cm (74,649,600 royal cubit = earth circumference)
  • Earth inch = 0.0025344 m = 2.5344 cm (15,778,462,500 tychos inch = earth circumference)
  • Earth circumference = 253,440 stadia (1 stadia = 300 royal cubits)
  • Earth rotation speed = 1,670.76756 km/h at equator
  • Ratio earth/sun (both size and distance) = 109.48608
  • Ratio moon/sun (both size and distance) = 399.8893536
  • Ratio size moon/earth = 3.652421875
  • Sun circumference = 4,378,231.77593979 km
  • Moon circumference = 10,948.608 km
  • Earth PVP orbit = 371,320,087.21786 km
  • Sun orbit = 941,073,629.044946 km
  • Sun orbit radius = 149,776,519.875932 km
  • Distance earth to moon = 374,544.9048032160 km

Sun orbit speed = 107,357.261981698 km/h km/h
Earth orbit speed = 1.67140272921772 km/h

107,357.261981698 * 24 * 365.2421875 = 941,073,629.044946 = orbit sun
1.6714027292177200 * 24 * 9,256,698 = 371,320,087.217861 = PVP orbit

PVP constant = 107,357.261981698 km/h / 1.67140272921772 km/h = 64,231.8336
PVP constant = 1/64,231.8336 = 1/25,344^2 / 1,000

Tychos completes in 25,344 years in 365.2421875 days/ year or 9,256,698 days.
Precession of the equinox is 9,410,736.29044946 days
25,765 years, 271 Days, 7 Hours, 54 minutes, 29.83326617 seconds

The relation between precession and tychos:

  • The number of days of precession / phi * number of days of tychos * 10 = Tau (2 PI)
  • Sun orbit circumference / phi * radius of sun orbit * 10 = Tau (2 PI)
  • 941,073,629.044946 / (phi * 9,256,698 * 10) = Tau (2 PI)

So there is a 1.01664073846305 relation between precession and tychos.
If you multiple phi * (2 PI=Tau) / 10 you will get that number. I don’t know the name for this number but let’s call:

  • phi * (2 PI=Tau) = phitau = 10.1664073846305
  • phi * (2 PI=Tau) / 10 = PhiTau = 1.01664073846305

The PVP orbit and sun orbit are based on this lost knowledge.

Some more numbers to keep it in harmony:

  • Radius PVP earth orbit = 36,524,218.75 * phi
  • Circumference PVP earth orbit = 365,242,187.5 * PhiTau
  • Earth distance to moon = 1/ 0.00000432 * phi
  • Earth circumference = 0.00432 * 9,256,698
  • Speed of the earth = 0.608256 * PhiTau
  • Earth distance to sun = 92,566,980 * phi
  • Radius sun = 92,566,980 * phi
  • PhiTau = phi * tau/10
  • Circumference Sun orbit = 925,669,800 * PhiTau
  • Speed of sun = 105,600 * PhiTau

And as the cherry on top of the cake:

Another way to calculate the sidereal year: 86,400 + (31,556,925 / ((925,669.8) * phitau)) = 86,403.3532896923

It’s beautiful:-)

If you see any mismacthes in my conclusions/ calculations, feel free to respond.

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Just to clarify…

The previous post was more related to the precession as we observe on earth. Since the earth moves a little it seems to be 25,765.74287561 years/ 9,410,736.29044946 days. But the actual precession of the equinox is - as mentioned by Simon and Patrik - indeed 25,344 years (in my view 25,344 years of 365.2421875 days each = 9,256,698 days).

Let me explain: These are the most acurate numbers both as observed on earth and as calculated mathematically.

There are 4 types of daylength (but sidereal day and stellar day are more or less the same; sidereal day being the real figure to use since it reflects the equinox alignment):

  1. Sidereal day (daylength relative to the moving vernal equinox) = 86,164.0905309414 seconds a day = 23.9344695919282 hours a day = 1,436.06817551569 minutes a day
  2. Stellar day (daylength relative to the fixed inertial frame of background stars) = 86,164.09890369 seconds a day which shows the small movement of the earth, so can’t be used for further calculations.
  3. Solar day = 86,400 seconds a day = 24 hours a day = 1,440 minutes a day
  4. Sidereal year calculated back to a day (The sidereal year day ) = 86,403.3532896923 seconds a day = 24.000931469359 hours a day = 1,440.05588816154 minutes a day

Resulting in 2 types of yearlength both calculated in 2 ways:

  1. Sidereal year = 365.256362996094 days/ year of 86,400 seconds per day = 31,558,149.76286250 seconds a year = 8,766.1527 hours a year
  2. Sidereal year = 365.2421875 days/ year of 86,403.3532896923 seconds per day = 31,558,149.76286250 seconds a year = 8,766.1527 hours a year
  3. Solar year = 365.2421875 days/ year of 86,400 seconds per day = 31,556,925 seconds a year = 8,765.8125 hours a year
  4. Solar year = 366.2421875 spins/ year of 86,164.0905309414 seconds per day = 31,556,925 seconds a year = 8,765.8125 hours a year

Why do I think these exact numbers are right?
There is a connection between those numbers and mostly this is not shown directly at the surface. As far as I know, nobody has mentioned this anywhere:

  • (86,400 - 86,164.0905309414)* 366.2421875 = 86,400
  • (31,558,149.76286250 - 31,556,925)/ 365.2421875 = 3.353289692331 + 86,400 = the Sidereal year length of 86,403.3532896923 seconds per day
  • (86,403.3532896923 * 365.2421875) = 31,558,149.76286250
  • (86,164.0905309414 * 366.2421875) = 31,556,925 NOTE: Exactly 1 extra spin of the earth in 1 year!
  • (31,556,925 / 365.2421875) = 86,400
  • (365.2421875/ 366.2421875)* 86,400 = 86,164.0905309414
  • (86,403.3532896923/ 86,400) * 365.2421875 = 365.256362996094
  • (9,256,698 / 25,344) = 365.2421875
  • 86,400 + (31,556,925 / ((925,669.8) * phitau)) = 86,403.3532896923
  • 365.2421875 is 31 leap years in 128 years.
  • You can find a lot of evidence this 365.2421875 is the actual correct figure for the duration of a year.

These numbers create the illusion of 25,765.74287561 years/ 9,410,736.29044946 days precession of the equinox:

  • (365.2421875*86,400/25,765.74287561) + 31,556,925 = 31,558,149.76286250
  • 25,765.74287561 years = 9,410,736.29044946 days
  • 9,256,698 * phitau = 9,410,736.29044946 days

These numbers are really what we also experience on earth.

The exactly 1 round of extra earth spin is also really amazing. It’s the reason why the earth seasons are stable. If it was only a fraction different, the equinox dates would have shifted. Before I dived into this topic I always thought we spinned a little over 365 times but it is exactly 1 extra spin: 366.2421875 times. So a year, contains exactly one sidereal day more than solar days exactly. Can’t be a coincedence.

If you start with 365.2425 or 365.25 the numbers are not in sync/ as observed on earth:
eg (365.2425/ 366.2425)* 86,400 = 86,164.0907322334 = not as observed on earth. The value must be 86,164.0905…
eg (365.25/ 366.25)* 86,400 = 86,164.0955631399 = not observed on earth
365.2421875 is the correct number.

I tried playing with the Tychos simulation where you can set the parameters yourself: https://codepen.io/pholmq/pen/XGPrPd, and might have found 3 updates to reflect this number:

In the Tychos simulation, the set Earth rotationspeed is 2301.169495. This is most probably based upon a gregorian year of 365.2425 days a year * 25,344 = 9,256,705.92 days in a tychos cycle.
That is slightly off for what we observe on earth.
The correct formula should be 0.000247916086931013 * (9,256,698* 366.2421875/ 365.2421875) = 2301.16753136931

{
“name”: “Earth”,
“rotationSpeed”: 2301.16753136931,
},

Additionally the time constraints can be set to what we observe on earth with the number above:

//DEFINE TIME CONSTANTS
const yearLength = 365.2421875
const earthRotations = 366.2421875

I tried to find out how the other numbers came to be (like the moon speed, mercury speed, etc), but could not crack the code completely. But when playing with these numbers, and comparing it to stellarium it seems to be you already have them with a high degree correct. Really impressed on what you both have achieved.

I have however 4 questions for Patrik/Simon regarding the Tychos simulation:

  1. The book mentions a lot of 365.25, 29.22, etc. Like I mentioned, that’s not what we observe on earth/ mathematically correct. I don’t know the consequence but maybe 365.25 is actually 365.2421875? and therefore 29.22 is actually 29.219375? What’s your opinion on this point?
  2. Can you explain how to get to the numbers for speed (moon has 83.28517, mercury 26.087623, etc) in the Tychos simulation?
  3. Can you explain how to get to the numbers for the rotation speed for the sun 83.995 in the Tychos simulation? NOTE: The sun should turn 25,345 times in 25,344 years. The same extra spin as on earth.
  4. Is it correct the Size and Orbit radius are only for visible efects in the Tychos simulation?

I might have found some more clues that are key to solve this puzzle.

The eye opener for me was, the sun spins actually one extra spin around the earth in 25,344 Real-Solar-Years on it’s trip of the PVP because of the same reason the earth spins one extra time in a solar year.

The duration of the Real-Solar-Year is therefore longer than the Earth-Solar-Year with the factor 25,344 : 25,345.

In one of the post above I mentioned the actual sidereal year of the sun - if the earth didn’t move. That is actually the measurement of the extra spin! You can look at the details above. You can also find it by implying the factor 25,344 : 25,345 to 86,400 = 86,403.409090909091 seconds a day.

I created a picture to show it visual.

It is drawn exaggerated but gives a good idea of how it works:

  • At the start position the earth is in position 1
  • One year of 365.2421875 days a year later, Earth moved to position 2, and the sun is in the same position in the sky as observed from earth, completing a circle of seasons, showing the sun in position A
  • A little later (one year of 365.256362996094 days a year), Earth moved to position 3, and the sun is aligned with the fixed stars, showing the sun in position B. This is shown as the precession of the equinox.
  • A little later (one year of 365.256598886817 days a year), Earth moved to position 4, and only then the sun has made 1 full turn around earth. The sun is not at exactly the same place as with position 1 because of the movement of the earth, but only now the circle is full (360 degrees), showing the sun in position C (exaggerated)

In a great year the sun is exactly 25,345 times (9,257,063.2421875 days of 86,400 seconds a day) in position A in Earth-Solar-Years and exactly 25,344 times (9,256,698 days of 86,403.409090909091 seconds a day) in position C in Real-Solar-Years.

  • Real-Solar-Year = 31,558,170.143821 seconds a year = based upon 86,403.409090909091 seconds a day / 365.256598886817 days a year/ 9,256,698 days in a great year
  • Earth-Solar-Year = 31,556,925 seconds a year = based upon 86,400 seconds a day/ 365.2421875 days a year/ 9,257,063.2421875 days in a great year

The total duration of a great year = 799,810,264,125 seconds.

  • 365.2421875 * 86,403.409090909091 * 25,344 = 799,810,264,125 seconds.
  • 365.2421875 * 86,400 * 25,345 = 799,810,264,125 seconds.
  • 799,810,264,125 / 25,344 = Real-Solar-Year of 31,558,170.143821 seconds a year
  • 799,810,264,125 / 25,345 = Earth-Solar-Year of 31,556,925 seconds a year

So the sun spins actually 25,345 times around the earth in the pace of 86,400 seconds a day.

To reflect this in the Tychos model:

Earth speed is: the sun speed in the tychos model (2*PI()) / 25,345 = 0.000247906305274397
Earth rotation speed is: earth speed (0.000247906305274397) * (9,257,063.2421875 * 366.2421875 / 365.2421875)

{
“name”: “Earth”,
“speed”: -0.000247906305274397,
“rotationSpeed”: 2301.16753136931,

},

Additionally the time constraints can be set to what we observe on earth with the number above:

//DEFINE TIME CONSTANTS
const yearLength = 365.2421875
const earthRotations = 366.2421875

Every day we learn a bit more.

Dear Sean,

I will reply to your various questions (regarding how the Tychosium simulator data & settings were calculated) in due time - i.e. as time permits (I will have to look up the notes & memos I wrote down in the early days of the Tychosium’s creation). For now though, I will reply to this most interesting sentence / observation of yours:

“The duration of the Real-Solar-Year is therefore longer than the Earth-Solar-Year with the factor 25,344 : 25,345.”

As you can easily verify with this handy Percentage Calculator, the difference between 25344 and 25345 is 0.0039457%.

Well, here is the very first paragraph of Chapter 12 of my TYCHOS book:

" Each year, planet Earth (travelling at the tranquil speed of 1.6 km/h) covers a distance of ≈14036 km along its PVP orbit. This distance equals amounts to 0.0039457% of the PVP orbit’s circumference of 355 724 597 km. From one year to the next, the Earth and the Sun will thus meet up at a slightly ‘earlier’ point in space, by an amount corresponding to a 0.0039457% “slice” of the solar orbit’s circumference."

Just saying… :slight_smile:

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