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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #180 on: April 24, 2018, 10:05:43 PM »
Are you talking here about calendar year, tropical year, or sidereal year.

If you are talking about calendar year, then you must specify which one.
Some have 365 days, some have 366 days.

Calendar year is based on number of days and it has the whole number of days in it.

Tropical and sidereal years aren't.

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Offline Tumeni

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #181 on: April 24, 2018, 10:07:08 PM »
It is a good measuring system which gives consistent results when you manipulate it in this fashion.

So you're manipulating the numbers to get the result you want?
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Offline Tumeni

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #182 on: April 24, 2018, 10:21:10 PM »
Hundreds upon thousands of astronomers and others in related disciplines have looked at the relative motions of Sun, Earth and Moon over hundreds, or thousands of years.

Their work has been distilled into hundreds, possibly thousands of textbooks, and many of them have used optical instruments and high-level maths in preference to napkins and simple arithmetic.

Rather than using a school-level diagram as your starting point, why not start with a trip to your local library, and peruse some of these textbooks? Rather than using a napkin, look at what astronomers have used, and still use, for their empirical observations.

Your go-to response is to refer globe-earthers to one book, and one book only - ENaG.   I refer you to hundreds, possibly thousands, which deal with this matter, in libraries all over the world.

Surely you won't conclude that you're right, and they're all wrong?
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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #183 on: April 24, 2018, 10:30:00 PM »
It is a good measuring system which gives consistent results when you manipulate it in this fashion.

So you're manipulating the numbers to get the result you want?

I'm not manipulating anything. You will get whole numbers from units in measuring systems that are based on multiples.

1 Barrel = 36 Gallons
1 Gallon = 4 Quarts

In this measuring system 4 and 36 share a common factor.

36 Gallons / 4 Quarts = 9. Whole Number. 4 will fit into 36. This fluid measuring system is constant (at least between these entities).

Now:

1 Year = 360 Days
1 Day = 24 Hours

360 Days / 24 Hours = 15. Whole Number. 24 will fit into 360. This measurement system is constant.

For the explanation on why we can also do this if the year is 365.24 Days, refer to the explanation on the previous page.
« Last Edit: April 24, 2018, 10:40:29 PM by Tom Bishop »

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Offline Tumeni

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #184 on: April 24, 2018, 10:43:07 PM »
I'm not manipulating anything. You will get whole numbers from units in measuring systems that are based on multiples.

... if you pick the multiples to fit the result you want

1 Barrel = 36 Gallons
1 Gallon = 4 Quarts

In this measuring system 4 and 36 share a common factor.

36 Gallons / 4 Quarts = 9. Whole Number. 4 will fit into 36. This fluid measuring system is constant (at least between these two entities).

Now:

1 Year = 360 Days
1 Day = 24 Hours

360 Days / 24 Hours = 15. Whole Number. 24 will fit into 360. This measurement system is constant.

Congratulations. You have calculated that one twenty-fourth of 360 = 15. Nothing else.


1 pound = 20 shillings
1 shilling = 12 pence

20/12 = 1.66666

What does this tell you? That British currency is/was not a 'constant' system? What does that even mean, outwith your own head?

1 mile = 8 furlongs
1 furlong = 10 chains
1 chain = 66 feet
1 foot = 12 inches

Take your pick of which one you would divide by which other (8/12? 10/12? 66/12?), for I think everyone except you has lost track of what point you think you're making....
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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #185 on: April 24, 2018, 10:53:08 PM »
Quote
Congratulations. You have calculated that one twenty-fourth of 360 = 15. Nothing else.

There is more meaning than that. I have been attempting to explain it to you.

1 pound = 20 shillings
1 shilling = 12 pence

20/12 = 1.66666

What does this tell you? That British currency is/was not a 'constant' system? What does that even mean, outwith your own head?

1 mile = 8 furlongs
1 furlong = 10 chains
1 chain = 66 feet
1 foot = 12 inches

Take your pick of which one you would divide by which other (8/12? 10/12? 66/12?), for I think everyone except you has lost track of what point you think you're making....

You are showing me systems and numbers that are not based on multiples. Of course it won't work to get a whole number when you divide those things. We are trying to get rid of the Imperial System in the US because it is inferior to the Metric System, as more of those units are consistent and based on multiples.

I can only imagine that the British peoples don't like that system and would prefer units of currency that is more consistent and based on multiples as well.

Edit: They changed it in 1971. See:

https://www.milesfaster.co.uk/information/uk-currency.htm

Quote
February 15th 1971 the UK moved to a new system called decimalisation and brought the currency into line with the metric systems used in Europe which are based on a logical system of 10 or factors of 10's. So with decimalisation came a system of pounds and pence doing away with shillings altogether. UK currency is known UK currency is known as BRITISH STERLING.
« Last Edit: April 24, 2018, 11:20:20 PM by Tom Bishop »

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Offline Bobby Shafto

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #186 on: April 24, 2018, 11:22:15 PM »
Quote
Congratulations. You have calculated that one twenty-fourth of 360 = 15. Nothing else.

There is a deeper meaning than that. I have been attempting to explain it to you.
Don't divide 360° by 24 hours if trying to figure out earth rotation vs orbit. That's mixing terms.


If using 24 hours, that's a solar day figure and you need to apply it to 360.986°.

If wishing to use 360° of rotation, that's a sidereal day, and you need to use 23 hrs, 56 mins, 4.09 secs.

You need to pick one or the other to work out how many of which type of rotations has happened along an arc of orbit around the sun. You can't cross the terms.

I'm not entirely sure what the debate has evolved into, but I have a sense this is the source of confusion or miscommunication.

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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #187 on: April 24, 2018, 11:29:52 PM »
Quote
Congratulations. You have calculated that one twenty-fourth of 360 = 15. Nothing else.

There is a deeper meaning than that. I have been attempting to explain it to you.
Don't divide 360° by 24 hours if trying to figure out earth rotation vs orbit. That's mixing terms.


If using 24 hours, that's a solar day figure and you need to apply it to 360.986°.

If wishing to use 360° of rotation, that's a sidereal day, and you need to use 23 hrs, 56 mins, 4.09 secs.

You need to pick one or the other to work out how many of which type of rotations has happened along an arc of orbit around the sun. You can't cross the terms.

I'm not entirely sure what the debate has evolved into, but I have a sense this is the source of confusion or miscommunication.

You can mix terms. You can call the variables anything you want to call it. Just make sure the measuring system you have created is constant and based on multiples. You will get whole numbers from units in measuring systems that are based on multiples.

I refer you to this explanation for why we can divide 360 and 24.

And this explanation for why we can also divide 365.24 by 24 in this scenario.

Per the Sidrael Day comment I refer you to this post which shows that the Sidrael Day is not the solution to this problem.
« Last Edit: April 24, 2018, 11:36:43 PM by Tom Bishop »

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Offline Tumeni

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #188 on: April 24, 2018, 11:39:10 PM »
What IS the problem?
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Offline Tumeni

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #189 on: April 24, 2018, 11:43:10 PM »
You are showing me systems and numbers that are not based on multiples.

You're cherry-picking your systems to only select those which fit your needs.
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Offline Bobby Shafto

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #190 on: April 24, 2018, 11:59:31 PM »

You can mix terms. You can call the variables anything you want to call it...
"Terms" are not what you call them. "Terms" are the variables you are applying.

Sure, you can make up any standards or units of measurement you want for those terms, but whatever you use, you have to be clear on what it is you're working with. The ratio you are working with is either the degrees of earth's rotation to bring the same line back toward the sun in 24 hours, OR 360° of earth rotation per the time it takes to complete that rotation.

But you aren't working with the right terms if applying a ratio of numerator of one (360°) with the denominator of the other (24 hrs). Doing that results in the question you posed in the opening post. Keep the right degrees of rotation with the correct, corresponding time period. You can use whatever units of measurement for those terms you like.


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Offline Bobby Shafto

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #191 on: April 25, 2018, 12:03:40 AM »
You're cherry-picking your systems to only select those which fit your needs.
I'm sorry for butting in. I can't figure out what the argument is about. I thought I had my finger on the disconnect, but maybe not. Maybe you are arguing about numbers/units and not terms/variables.

I'll go back to lurk mode (after responding to Tom's suggestion I read his earlier post.)
« Last Edit: April 25, 2018, 12:19:37 AM by Bobby Shafto »

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Offline Bobby Shafto

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #192 on: April 25, 2018, 12:18:25 AM »
Per the Sidrael Day comment I refer you to this post which shows that the Sidrael Day is not the solution to this problem.

"If the stars did not exist the Solar Day is still wrong." - the stars are a reference point. If no reference point for 360° earth rotation, the earth would still have to rotate 360.986° to face the sun again (and we'd call that 24 hours). But without a reference point, we might be excused for thinking that we'd rotated 360° and divided those degrees into 24 hours. And we'd probably find it normal, maybe, that time of day would slip as we orbited the sun. Maybe, eventually, humans would figure it out that the 24 hours was dividing 360.986° and not 360° without the distant starfield to provide a reference. Who knows?

But you are right. If mankind continued to think 1 rotation of the earth was solar noon to solar noon, and that equated to 360°, then solar days would be "wrong," at least as we know them now. They probably wouldn't be "wrong" to a person in a no-stars world though. If they didn't adjust, they'd just adapt and accept that day/night shift during the year.


"The Sidrael Day is about 4 seconds less than the Solar Day."  - actual value is 3 mins, 56 secs. Rework your calculation using the correct delta.

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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #193 on: April 25, 2018, 12:32:16 AM »
Per the Sidrael Day comment I refer you to this post which shows that the Sidrael Day is not the solution to this problem.

"If the stars did not exist the Solar Day is still wrong." - the stars are a reference point. If no reference point for 360° earth rotation, the earth would still have to rotate 360.986° to face the sun again (and we'd call that 24 hours). But without a reference point, we might be excused for thinking that we'd rotated 360° and divided those degrees into 24 hours. And we'd probably find it normal, maybe, that time of day would slip as we orbited the sun. Maybe, eventually, humans would figure it out that the 24 hours was dividing 360.986° and not 360° without the distant starfield to provide a reference. Who knows?

But you are right. If mankind continued to think 1 rotation of the earth was solar noon to solar noon, and that equated to 360°, then solar days would be "wrong," at least as we know them now. They probably wouldn't be "wrong" to a person in a no-stars world though. If they didn't adjust, they'd just adapt and accept that day/night shift during the year.


"The Sidrael Day is about 4 seconds less than the Solar Day."  - actual value is 3 mins, 56 secs. Rework your calculation using the correct delta.

Thank you. The difference is stated in these links:

From: http://astro.unl.edu/naap/motion3/sidereal_synodic.html

Quote
A sidereal year is the time it takes for the sun to return to the same position with respect to the stars. Due to the precession of the equinoxes the sidereal year is about 20 minutes longer than the tropical year.

Another Source: https://en.wikipedia.org/wiki/Sidereal_year

Quote
The sidereal year differs from the tropical year, "the period of time required for the ecliptic longitude of the sun to increase 360 degrees",[2] due to the precession of the equinoxes. The sidereal year is 20 min 24.5 s longer than the mean tropical year

This 20 minute difference between the Sidrael Year and the Solar Year (also called the Tropical Year) is still not the solution to this problem:

Quote from: Tom Bishop
The sun travels across the earth's surface once each day. In Solar Time: There is 1 Solar Day in 24 Hours. There are 365.24 Days in a Solar Year.

Earth circumference = 24,901 mi. In 1 Day the sun travels over 24,901 mi. of earth.

24,901 / 24 = 1037.54166667 miles. Over 1 hour the sun travels over 1037.54166667 miles

After 365 days:  24,901 mi. x 365 days = 9088865 miles

After 365.24 days:  24901 x 365.24 = 9094841.24 miles

Difference = 5976.24 miles

5976.24 miles / 1037.54166667 miles = 5.76. The hours in miles fits into the difference by 5.76 times. Where are those extra hours coming from? The sun will not be in the same place over the earth.
« Last Edit: April 25, 2018, 12:58:41 AM by Tom Bishop »

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Offline Bobby Shafto

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #194 on: April 25, 2018, 01:13:16 AM »

Thank you. The difference is stated in these links:

From: http://astro.unl.edu/naap/motion3/sidereal_synodic.html

Quote
A sidereal year is the time it takes for the sun to return to the same position with respect to the stars. Due to the precession of the equinoxes the sidereal year is about 20 minutes longer than the tropical year.
That's the difference between a sidereal year and a solar (tropical year). Different from a sidereal day and solar day.

I understand the confusion, but you're mixing terms. (Not units of measurement; just terms.) There's a rotation of the earth (days) and there's orbital rotation (years). Each has a difference measurement based on whether sun is a reference point or a distant star field. The solar day is different from the sidereal day for one reason. The solar year is different from the sidereal year for others.

All I'm trying to say is make sure when doing your calculations you're using the same terms; convert if necessary, so that you're not dividing the time parameters into non-agreeing angular parameters. There are two "circles": earth's rotation and earth's orbit. If using sidereal for either, stick with the time intervals for sidereal angular displacement. If using solar, apply the solar time intervals. If relating the two, use proper conversion.

I haven't checked your math. I'm just noting confusion of terms/variables.

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Offline Stagiri

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #195 on: April 25, 2018, 03:48:58 AM »
Dear Mr. Bishop,
Please, do not avoid my question. Why would this "issue" pose a problem to the GET?
Dr Rowbotham was accurate in his experiments.
How do you know without repeating them?
Because they don't need to be repeated, they were correct.

Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #196 on: April 25, 2018, 04:59:32 AM »
The sun travels across the earth's surface once each day. In Solar Time: There is 1 Solar Day in 24 Hours. There are 365.24 Days in a Solar Year.

Earth circumference = 24,901 mi. In 1 Day the sun travels over 24,901 mi. of earth.

24,901 / 24 = 1037.54166667 miles. Over 1 hour the sun travels over 1037.54166667 miles

After 365 days:  24,901 mi. x 365 days = 9088865 miles

After 365.24 days:  24901 x 365.24 = 9094841.24 miles

Difference = 5976.24 miles

5976.24 miles / 1037.54166667 miles = 5.76. The hours in miles fits into the difference by 5.76 times. Where are those extra hours coming from? The sun will not be in the same place over the earth.
Ah, I think I found it, and wouldn't you know it's something I already said earlier, but you ignored. A Solar Year is the time between the sun being in the same 'place' in the sky to it being there again. But, as mentioned in the definition for Solar Year, 'place' is defined as the ecliptic. The ecliptic being an arc of the sky. In the case of the winter solstice, this is the arc where the sun is lowest in the sky. For the summer, it's the opposite. For the equinoxes, it's the one right in between. Solar Year carries no reference to a point above the Earth. It's the ecliptic of the sky. Solar Day carries the connotation of the sun being above a certain line/point of the Earth. I believe if you look, you'll see that the difference you've noted is about 1/4 the circumference of the Earth. Which is why a year contains about an extra 1/4 of a Solar Day. Hence why our calendar includes leap years, to keep solstices and equinoxes at about the same time of the calendar year, every year. Otherwise we would slowly drift until January was summer in the North, and then back again.

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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #197 on: April 25, 2018, 06:18:56 AM »
The Problem

The Mean Solar Day does not fit into the number of Mean Solar Days in a Mean Solar Year.

The Mean Solar Day is 24 Hours Per Day and the Mean Solar Year is 365.24217 Mean Solar Days.

The terms are "means," but the information shows very little variance over the year for the terms. The Mean Solar Day equalizes out (whatever is lost is gained) over a year, and there is very very tiny variance in the Solar Year.

There are an extra 5+ hours in a Solar Year (the .24217 at the end) which come out of nowhere, and which cannot be accounted for by these variations.

The sun needs to return back to the same position of the equinox every year in a Solar Year.

Terms

Solar Time

https://en.wikipedia.org/wiki/Solar_time

Quote
Apparent solar time

The apparent sun is the true sun as seen by an observer on Earth.[4] Apparent solar time or true solar time is based on the apparent motion of the actual Sun. It is based on the apparent solar day, the interval between two successive returns of the Sun to the local meridian.

Meridian Illustration




Mean Solar Day

https://www.universetoday.com/78107/solar-day/

Quote
The length of a solar day varies throughout the year, a result of the Earth’s elliptical orbit and axial tilt. In this model, the length of the day varies and the accumulated effect is a seasonal deviation of up to 16 minutes from the mean. The second type, Solar Mean Time, was devised as a way of resolving this conflict. Conceptually, Mean solar time is based on a fictional Sun that is considered to move at a constant rate of 360° in 24 hours along the celestial meridian. One mean day is 24 hours in length, each hour consisting of 60 minutes, and each minute consisting of 60 seconds. Though the amount of daylight varies significantly throughout the year, the length of a mean solar day is kept constant, unlike that of an apparent solar day.

Solar Day Variation

https://en.wikipedia.org/wiki/Solar_time

Quote
The length of a solar day varies through the year, and the accumulated effect produces seasonal deviations of up to 16 minutes from the mean. The effect has two main causes. First, Earth's orbit is an ellipse, not a circle, so the Earth moves faster when it is nearest the Sun (perihelion) and slower when it is farthest from the Sun (aphelion) (see Kepler's laws of planetary motion).

...



...

The equation of time is this difference, which is cyclical and does not accumulate from year to year.


Solar Year (also called Tropical Year)

https://en.wikipedia.org/wiki/Tropical_year

Quote
Since antiquity, astronomers have progressively refined the definition of the tropical year. The entry for "year, tropical" in the Astronomical Almanac Online Glossary (2015) states:

     "the period of time for the ecliptic longitude of the Sun to increase 360 degrees. Since the Sun's ecliptic longitude is measured with respect to the equinox, the tropical year comprises a complete cycle of seasons, and its length is approximated in the long term by the civil (Gregorian) calendar. The mean tropical year is approximately 365 days, 5 hours, 48 minutes, 45 seconds."

An equivalent, more descriptive, definition is "The natural basis for computing passing tropical years is the mean longitude of the Sun reckoned from the precessionally moving equinox (the dynamical equinox or equinox of date). Whenever the longitude reaches a multiple of 360 degrees the mean Sun crosses the vernal equinox and a new tropical year begins" (Borkowski 1991, p. 122).

The mean tropical year in 2000 was 365.24219 ephemeris days; each ephemeris day lasting 86,400 SI seconds.[1] This is 365.24217 mean solar days (Richards 2013, p. 587).

Solar/Tropical Year Variation

https://en.wikipedia.org/wiki/Tropical_year

Quote
Mean time interval between equinoxes

As already mentioned, there is some choice in the length of the tropical year depending on the point of reference that one selects. But during the period when return of the Sun to a chosen longitude was the method in use by astronomers, one of the equinoxes was usually chosen because it was easier to detect when it occurred. When tropical year measurements from several successive years are compared, variations are found which are due to nutation, and to the planetary perturbations acting on the Sun. Meeus & Savoie (1992, p. 41) provided the following examples of intervals between northward equinoxes:

dayshoursmins
1985–198636554858
1986–198736554915
1987–198836554638
1988–198936554942
1989–199036555106

Until the beginning of the 19th century, the length of the tropical year was found by comparing equinox dates that were separated by many years; this approach yielded the mean tropical year (Meeus & Savoie 1992, p. 42).


The Equinox

http://www.schoolphysics.co.uk/age14-16/Astronomy/text/Equation_of_time/Equinoxes_/index.html

Quote
Because of the angle between the celestial equator and the ecliptic the path of the Sun through the sky varies from one time of year to another.



The equinox is a point where the ecliptic crosses the celestial equator – it does this twice a year as you can see from Figure 1. At the Spring (vernal) equinox the Sun crosses the celestial equator from the south to the north. At the autumnal equinox the Sun crosses the celestial equator from the north to south.

Precession of the Equinox

Quote
What is Precession:

The precession of the equinoxes refers to the observable phenomena of the rotation of the heavens, a cycle which spans a period of (approximately) 25,920 years, over which time the constellations appear to slowly rotate around the earth, taking turns at rising behind the rising sun on the vernal equinox.

This remarkable cycle is due to a synchronicity between the speed of the earth's rotation around the sun, and the speed of rotation of our galaxy.


Other terms that have been brought up in this discussion:

Sidreal Time / Motion

http://astro.unl.edu/naap/motion3/sidereal_synodic.html

Quote
The word sidereal derives from the Latin word for “star”. This is because sidereal motion is motion with respect to the stars. One sidereal day is the time it takes for a star in the sky to come back to the same place in the sky. Because, for all intents and purposes, the sky is “fixed”, a sidereal day is when the earth rotates 360°. A sidereal day is 23 hours 56 minutes and 4.09 seconds long.

A sidereal year is the time it takes for the sun to return to the same position with respect to the stars. Due to the precession of the equinoxes the sidereal year is about 20 minutes longer than the tropical year. The tropical year is the interval at which seasons repeat and is the basis for the calendar year.

Sidereal Time

https://www.britannica.com/science/sidereal-time#ref99274

Quote
Sidereal time, time as measured by the apparent motion about the Earth of the distant, so-called fixed, stars, as distinguished from solar time, which corresponds to the apparent motion of the Sun. The primary unit of sidereal time is the sidereal day, which is subdivided into 24 sidereal hours, 1,440 sidereal minutes, and 86,400 sidereal seconds. Astronomers rely on sidereal clocks because any given star will transit the same meridian at the same sidereal time throughout the year. The sidereal day is almost 4 minutes shorter than the mean solar day of 24 of the hours shown by ordinary timepieces.

Sidereal time may be defined for any place on the Earth, but in the international system used by astronomers each sidereal day begins at the instant the vernal equinox transits the prime meridian. The vernal equinox is the point on the celestial sphere at which the Sun crosses the plane of the Equator, moving from south to north.

Google Dictionary

Quote
si·de·re·al year
nounAstronomy
noun: sidereal year; plural noun: sidereal years

    the orbital period of the earth around the sun, taking the stars as a reference frame, being 20 minutes longer than the tropical year because of precession.


Leap Year

http://astro.unl.edu/naap/motion3/sidereal_synodic.html

Quote
Leap Year (Optional)

Because a tropical year is 365.242 mean solar days long, the vernal equinox would be later and later every year if our calendar year were strictly 365 days long. In an attempt to keep the Vernal Equinox very near March 21st, the Leap Year was introduced. According to the Gregordian calendar a leap year occurs every 4 years except years evenly divisible by 100, unless that year is evenly divisible by 400. The year 1900 was not a leap year, but 2000 was.
« Last Edit: April 25, 2018, 08:43:23 AM by Tom Bishop »

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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #198 on: April 25, 2018, 06:22:15 AM »

Thank you. The difference is stated in these links:

From: http://astro.unl.edu/naap/motion3/sidereal_synodic.html

Quote
A sidereal year is the time it takes for the sun to return to the same position with respect to the stars. Due to the precession of the equinoxes the sidereal year is about 20 minutes longer than the tropical year.

That's the difference between a sidereal year and a solar (tropical year). Different from a sidereal day and solar day.

I understand the confusion, but you're mixing terms. (Not units of measurement; just terms.) There's a rotation of the earth (days) and there's orbital rotation (years). Each has a difference measurement based on whether sun is a reference point or a distant star field. The solar day is different from the sidereal day for one reason. The solar year is different from the sidereal year for others.

All I'm trying to say is make sure when doing your calculations you're using the same terms; convert if necessary, so that you're not dividing the time parameters into non-agreeing angular parameters. There are two "circles": earth's rotation and earth's orbit. If using sidereal for either, stick with the time intervals for sidereal angular displacement. If using solar, apply the solar time intervals. If relating the two, use proper conversion.

Yes, I am using Solar Days and Solar Years in the equation. I do not believe that I am mixing up terms. The Mean Solar Day has 24 hours and the Mean Solar Year has 365.24 years.

The sun travels across the earth's surface once each day. In Solar Time: There is 1 Solar Day in 24 Hours. There are 365.24 Days in a Solar Year.

Earth circumference = 24,901 mi. In 1 Day the sun travels over 24,901 mi. of earth.

24,901 / 24 = 1037.54166667 miles. Over 1 hour the sun travels over 1037.54166667 miles

After 365 days:  24,901 mi. x 365 days = 9088865 miles

After 365.24 days:  24901 x 365.24 = 9094841.24 miles

Difference = 5976.24 miles

5976.24 miles / 1037.54166667 miles = 5.76. The hours in miles fits into the difference by 5.76 times. Where are those extra hours coming from? The sun will not be in the same place over the earth.
Ah, I think I found it, and wouldn't you know it's something I already said earlier, but you ignored. A Solar Year is the time between the sun being in the same 'place' in the sky to it being there again. But, as mentioned in the definition for Solar Year, 'place' is defined as the ecliptic. The ecliptic being an arc of the sky. In the case of the winter solstice, this is the arc where the sun is lowest in the sky. For the summer, it's the opposite. For the equinoxes, it's the one right in between. Solar Year carries no reference to a point above the Earth. It's the ecliptic of the sky. Solar Day carries the connotation of the sun being above a certain line/point of the Earth. I believe if you look, you'll see that the difference you've noted is about 1/4 the circumference of the Earth. Which is why a year contains about an extra 1/4 of a Solar Day. Hence why our calendar includes leap years, to keep solstices and equinoxes at about the same time of the calendar year, every year. Otherwise we would slowly drift until January was summer in the North, and then back again.

Can you quote something that says what you are saying about the ecleptic? The Solar Year is a place where the ecleptic crosses the celestial equator. That is a point in space, not "on an arc". See my post above that has a quote for how the Solar Year is defined. It goes back to the same point every year and the variation of the terms in question is extremely little.

The Solar Year is 365.24. Right. Where does that .24 come from? That's over 5 hours. Almost 6. Saying "The solar year has .24 at the end, that's where it comes from" is not the answer to this. The sun won't be in the same spot at the end of the year.
« Last Edit: April 25, 2018, 07:06:16 AM by Tom Bishop »