6121

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« on: April 25, 2018, 08:53:22 AM »

I thought your problem was the 12 hour offset at the halfway mark of the earth’s orbit around the sun. That’s explained by the difference between sidereal and solar day references.

That's a problem as well, but the discussion has progressed to showing that the Solar Day does not fit into the number of Solar Days in a Solar Year.

If sidereal has nothing to do with it, then what’s the problem again? “The solar day doesn’t fit into the solar year?” What does that mean? Are the “extra hours” you’re now talking about the 0.24 day tacked onto the 356 day solar year? That’s a different issue. A different problem.]If sidereal has nothing to do with it, then what’s the problem again? “The solar day doesn’t fit into the solar year?” What does that mean? Are the “extra hours” you’re now talking about the 0.24 day tacked onto the 356 day solar year? That’s a different issue. A different problem.

Yes. That is the problem we are talking about now. The .24 come out of nowhere and does not match up with the Solar Year where the sun needs to return to the point of the Equinox.

Following your link to The Problem, I see you state “The sun needs to return back to the same position every year in a Solar Year.” It doesn’t “need to.” I know for the sake of tidiness it would be nice if it did. There’d be no need for leap days or leap years. But it doesn’t line up that neatly because it doesn’t “need to” just to make it easy for us. It’s close enough that we barely notice it at first, but the mismatch between solar day and solar year can add up over time, thus the need to “catch up” with leaps.

The Solar Year is defined by the time it takes for the sun to return to the Equinox. The number of Solar Days in a Solar Year needs to match up.

If those are the “extra hours” then yes. That IS different from what I’ve been trying to explain about the NYC half year “problem” you started with. Those ARE extra hours, needed BECAUSE the solar days don’t “fit” the solar year in a nice, whole number of 365. The earth, on that 365th solar day is coming up just a bit short from where it began, relative to the sun.

The Sun needs to get back to the point of the Equinox under the definition of a Solar Year. It has to match up with the Solar Day.

6122

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« on: April 25, 2018, 08:15:21 AM »

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.

And so the number of degrees rotated in a solar day is what? Not 360. Which is what you’re using. But that’s a sidereal measure. Not solar because of the earth’s orbit means it needs to rotate just a bit more,

That number is 360.986 degrees. Use that per 24 hrs. Not 360.

The earth rotates 360 degrees in a Solar Day. In a Solar Day the sun makes one rotation around the earth, per its definition. It rotates 360.986 degrees when you compare the Solar Day to the Sidereal Day.

That link does not explain where the extra hours comes from. See The Problem post.

The “extra hours” aren’t “extra.” They’re just the difference between sidereal and solar references.

The Solar Day does not fit into the Solar Year. The Sidreal Day and the Sidreal Year have nothing to do with it.

The Solar Day is 24 hours in relation to the sun. That does not line up with the Solar Year.

The Sidreal Day is the rotation of the earth in reference to the stars. The difference between Sidreal Year and the Solar Year is only 20 minutes anyway, not 5+ hours. There is a source for the 20 minutes figure at the bottom of The Problem post.

6123

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« on: April 25, 2018, 08:03:34 AM »

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.

Of course it won't 365.24 days is not a whole number of days. It matters not whether there are 24 hours in the day or not, if you define a year as 365.24 days, then a whole number of days does not 'fit'.

The clue is in the 0.24.

Over four years, the 0.24s are accumulated into an extra 0.96 or so (4*0.24), and this forms the extra day of the leap year.

I gave the definition for Leap Year in The Problem post. The Leap Year was made to try and account for the .24. The .24 was not made to account for the Leap Year...

Leap Year

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

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.

What is your definition of the sun being 'in the same spot'? The same spot relative to what?

You keep saying the sun should return to the same position but are not really defining what that position is.

As far as I know the only requirement is that the sun should cross the celestial equator... not that it be over a certain longitude.

The same spot relative to the equinox. See The Problem post.

6124

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« on: April 25, 2018, 07:33:09 AM »

See https://www.timeanddate.com/astronomy/tropical-year.html

That link does not explain where the extra hours comes from. See The Problem post.

6125

Flat Earth Theory / Re: Areas of daylight are impossible on a bipolar map

« on: April 25, 2018, 06:53:30 AM »

Those are just calculators. You know that timeanddate.com doesn't have people at every location on earth reporting the sunrise and sunset times, right?

But YOU (or ME, or anyone) do have friends online everywhere you want.
Or find online web cameras and test day/night cycles and, where possible, shadow angles.
Publishers of TimeAndDate.com, SunCalc.org and others know very well about public scrutiny.

People can communicate nowadays with whomever wants to respond.
ANYONE can test those "calculators" any time with people all over the planet.
Instead of making database with separate data for every minut of every day, it was much more efficient and ellegant to organize it by common function.
You DO know that in many cases table CAN be represented by function.

"Anyone can do it" and "People have done it" are not acceptable levels of proof on this forum. Did you think anyone was going to be swayed by those arguments when you came up with it in your head?

6126

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« 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

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.

6127

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« 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

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/

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

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

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

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:

	days	hours	min	s
1985–1986	365	5	48	58
1986–1987	365	5	49	15
1987–1988	365	5	46	38
1988–1989	365	5	49	42
1989–1990	365	5	51	06

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

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

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

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

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

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

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.

6128

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« 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

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

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:

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.

6129

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« on: April 24, 2018, 11:29:52 PM »

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.

6130

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« on: April 24, 2018, 10:53:08 PM »

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

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.

6131

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« 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.

6132

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« on: April 24, 2018, 09:49:29 PM »

It doesn't matter what you call the units, or the history on how it was defined. The numbers are multiples and therefore can be divided to get whole numbers.

If we had a system that was composed of multiples of 8 it would work as well.

1 SuperLegoBlock = 64 MegaLegoBlocks
1 MegaLegoBlock = 16 SmallLegoBlocks

64 MegaLegoBlocks / 16 SmallLegoBlocks = 4. Whole number. 16 SmallLegoBlocks will fit into 64 MegaLegoBlocks.

It doesn't matter what you call it. 64 and 16 share a common multiple. Different units of this measuring system can be divided in this manner.

360 and 24 share a common multiple, and so these units, whatever you call them, can be divided. This measuring system is based on multiples. You can call it the Spinning Ball Earth Measuring System or the Piles of Popcorn Measuring System. It is a good measuring system which gives consistent results when you manipulate it in this fashion.

6133

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« on: April 24, 2018, 09:40:26 PM »

Step 1.

If you take a circle that is 360 degrees around and imagine that each of those degrees had 24 sub units in it (lets call them Sub-Degrees), the 24 sub-degrees should fit neatly into the 360 degree circle.

No! 

Are you doing this deliberately?!

If you take a circle that is 360 degrees around and imagine that each of those degrees had 24 sub units in it then all that means is that there are
360 x 24 = 8640 sub-degrees in the circle. That is literally all that means.

360 DOES happen to exactly divide into 24 but that's only because you happen to have picked two numbers where one is the multiple of the other.

360 degrees in a circle is a definition - it may be related to the days in a year, there is some debate about that, other theories are that some ancient civilisations used a base 60 numbering system and it's related to that.

24 hours in a day is also a definition - it seems to be something to do with the ancient Egyptians.

Both are highly divisible numbers which may also be a reason for those numbers being defined that way, so it's not such a reach that 360 happens to divide exactly by 24. But they are defined in different ways for different historic reasons, there is no particular reason one should divide neatly into the other.

You are saying that it's a "coincidence" that 360 is a multiple of 24. But the reason for defining it that way doesn't matter. It's a multiple.

We can divide by the two for the exact same reason we can divided dollars by dimes and cents, despite being units of different names. There is a common factor. Since there is a common factor, the measuring system is constant. It is not just meaningless numbers being divided together.

Addressing my previous example:

1 Miles = 5280 Feet
1 Foot = 12 inches
5280 Feet / 12 inches = 440

This works because there is a common multiple between miles and feet and inches.

Now you're doing units wrong again.
5280 feet / 12 (inches/foot) = 440 foot^2/inches

This is only an integer coincidentally. There is no reason the mile could not have been defined as 5380 feet or 5281 feet.

Sure, there is no hard reason it could not have been defined differently; but people who defined it wanted some kind of ratio or common factor to other smaller units of measurements. It is not the greatest idea to define your units willy nilly. The Imperial System isn't entirely constant with some of the unit types either, which is why there is a (failed) push in the US to change to the Metric System which is constant all throughout.

1 kilometer = 1000 meters
1 meter = 100 centimerers

There is a common factor between the two numbers which is good for a measuring system.

1000 meters / 100 centimeters = 10. Whole number. 100 fits into 1000 10 times. We can compare "unlike" units of measurements because there is a common factor.

6134

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« on: April 24, 2018, 06:15:15 PM »

I will attempt an explanation.

--- --- ---

Step 1.

If you take a circle that is 360 degrees around and imagine that each of those degrees had 24 sub units in it (lets call them Sub-Degrees), 24 Sub-Degrees should fit neatly into the 360 degree circle.

360 degrees / 24 sub-degrees = 15

The result should be a whole number.

The 360 degree circle is the highest hierarchical entity of the sub-degree, similar to how 100% of a pie is the highest hierarchical entity of the pie slices that fit within it.

--- --- ---

Step 2.

Now lets cut the line of the circle in Step 1. and lay the line out flat on a flat surface. Lets also rename Degrees to Mega-Lengths and Sub-Degrees to Sub-Lengths now for less confusion.

We have a line that is 360 Mega-Legths. Each Mega-Length has 24 Sub-Lengths in it.

360 Mega-Lengths / 24 Sub-Lengths = 15

The Sub-Length fits neatly into the 360 Mega-Length. Same thing, we are just mentally visualizing it as lengths now to show that the scenario can be laid out flat.

--- --- ---

Step 3.

The 365.24 days year is like that 360 degree circle. The extra 5.24 days was added for for, I believe, the elongation of the earth's route along the sun.

It's an oval. But ovals still have 360 degrees in them, so the analogy with Step 1 is maintained.

Lets now consider the 365.24 day year as the length of the oval. We can call the days Mega-Lengths and the hours Sub-Lengths for less confusion. Each Mega-Length has 24 Sub-Lengths.

Now cut the oval and lay the lines down on a flat surface. We are working with lengths now, like the in Step 2.

We have a line that has 365.24 Mega-Lengths in it. Each Mega-Length has 24 Sub-Lengths.

365.24 Mega-Lengths / 24 Sub-Lengths = 15.21

This is not a whole number. The Sub-Length does not fit into the whole Mega-Length.

--- --- ---

Does that make a little more sense?

6135

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« on: April 24, 2018, 05:00:34 PM »

A large number of feet should divide into a whole number of inches. Diving by different variables does work.

Right so...2939 feet divded by 12 inches = 244.916666

Help! I broke distances!

You are right. I made a mistake. Let me see if I can clear that up.

I should have said "A large number of inches will divide into a small number of feet with a whole number, if those feet fit".

2939 inches divided by 12 inches in a foot = 244.916666 feet. The foot does not fit into 2939 inches.

If we try another number that is divisible by an appropriate factor, it does work.

2928 inches divided by 12 inches in a foot = 244 feet. The foot does fit into 2928 inches.

Addressing my previous example:

1 Miles = 5280 Feet
1 Foot = 12 inches
5280 Feet / 12 inches = 440

This works because there is a common multiple between miles and feet and inches.

34 Miles = 179520 Feet
34 Feet = 408 Inches

179520 Feet / 408 Inches = 440

Change both the numbers on the left to 34 (or another number), compute the number of feet and inches, and we get the same ratio.

Again, it doesn't work with all units of measurement. There must be a common multiple. Not all measurement systems are constants.

6136

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« on: April 24, 2018, 03:12:59 PM »

1 Dollar = 10 Dimes
1 Dime = 10 Pennies

10 Dimes / 10 Pennies = 1  <--- But this is correct. We got a whole number. 10 Dimes fits into 10 Pennies 1 time. 10 fits into 10 1 time.

1 = 10 dimes/dollar
1 = 10 pennies/dime

10 dimes / (10 pennies/dime) = 1 dimes^2/pennies

The only reason you think this calculation makes sense is because both factors are 10.

If you had a different money system that was still based on whole numbers, it wouldn't work.

1 pound = 20 shillings
1 shilling = 12 pence

20 shillings / (12 pence/shilling) = (5/3)(shillings^2/pence)

Does this mean the old english money system was invalid?

--- --- ---

1 Miles = 5280 Feet
1 Foot = 12 inches

5280 Feet / 12 inches = 440  <--- This is correct as well. We got a whole number. 12 inches can fit neatly into 5280 feet 440 times. 12 fits neatly into 5280 440 times.

The 1 Foot = 12 inches is implicit in the above equation.

1 nautical mile = 6076.12 feet
1 foot = 12 inches

1 nautical mile / 12 (inches/foot) = 506.34 nautical mile * feet/inches

Does this mean nautical miles don't exist?

--- --- ---

I just divided unlike units and got a right answer 

I can do that too:
The speed limit on I-5: 70 miles/hour
Number of days in a week: 7 days / week

70 miles/hour  /  7 days/week = 10 miles*week/hour*days

It's a whole number, it has to mean something.

How about

7 days in a week
24 hours in a day
60 minutes in an hour

7 days / 60 minutes/hour  = 7/60  (days*hours/minutes)

7 days / 24 hours / day = 7/24  (days^2/hours)

Does this mean weeks don't exist?

The reason for my example with dollars, dimes and the example with pennies and with miles, feet, and inches work to get whole numbers is because both are constant measuring systems (There is also the furlong between the gap of feet and miles, but no one uses that anymore).

My examples make sense. A large number of feet should divide into a whole number of inches. Diving by different variables does work.

Some measuring systems or the units in them may not work, since they are not constant. If you are going to go around testing them all you will find issues, sure. The point is that systems that are based on multiples can be divided to get whole numbers.

Per the earth hours and days in the year, those should be multiples too. There should be some multiple where the number of hours in a day fits into the days in a year that is defined by the Sun cycle returning to the same spot in the sky after one north/south seasonal cycle (although, I see that someone refuted that definition in this thread). The whole cycle, seasons and all, should be divisible by the hour.

6137

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« on: April 24, 2018, 02:59:19 AM »

Gary, the earth rotates 24 hours in relation to the sun. The earth rotates slightly slower in relation to the stars. Why do the stars matter? If the stars did not exist the Solar Day and Solar Year are still wrong.

Lets try to do the math on this.

The Sidrael Day is about 4 seconds less than the Solar Day.

4 x 365.24 / 60 / 60 = 0.4058 hours. This is nowhere close to the 5.76 hour offset that I got with Solar Time:

The sun travels across the earth's surface 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.

6138

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« on: April 24, 2018, 02:27:28 AM »

That's the definition of a Solar Year. The time it takes the sun to return to its same position. I have given you the quote.

i understand that you're talking about solar years.

a solar year is not the same the same length of time it takes for the earth to orbit the sun once.  or: a tropical year is not the same duration as the earth's period around the sun.

there is no requirement in ret that tropical years and orbital periods be the same.

Are you suggesting that a Solar Day is not 24 hours? Are you suggesting that the Solar Year varies in length greater than the numbers I've given? Look up Solar Day. It is the Day as it relates to the Sun.

https://community.dur.ac.uk/john.lucey/users/e2_solsid.html

Solar time is time measured with respect to the Sun's apparent motion in the sky. The clocks we use for civil timekeeping are based on this motion. Of course, the apparent motion of the Sun across the sky is actually caused by the rotation of the Earth. So, our clocks measure the length of time required for the Earth to rotate once with respect to the Sun. From our perspective, the Sun revolves around the Earth every 24 hours. This period is known as a solar day.

6139

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« on: April 24, 2018, 02:14:50 AM »

I don't see how it matters if the radius of the circle of the earth's path is 1 million miles or 1 mile. A year has 364.24 days.

you're missing the point.  because the earth moves along its orbit (whatever the size) over the course of a rotation, a solar day is necessarily longer than a single rotation.

It says that the Solar Year is 364.24 Solar Days. Now you are telling me that a Solar Year is a different number of Solar Days?

we can see that the sun is offset at the end.

yes, it is.  the earth does not rotate on its axis an integer number of times in one orbit of the sun.  imagine a planet that rotates 1.5 times over one orbit of its sun.  at the end of one orbit, the sun will be offset from its starting point.

I am talking about Solar Days. The Earth rotates in one rotation in relation to the Sun, and the Sun makes one rotation around the earth, over a Solar Day.

The sun should have returned to the same position over the earth. It did not.

there is no such requirement in ret.

Yes, it is the definition of a Solar Year. It is defined as the time it takes the sun to return to its same position. I have given you the quote.

6140

Flat Earth Investigations / Re: Possible Issue with Solar Noon in Round Earth Theory

« on: April 24, 2018, 02:12:17 AM »

I don't see how it matters if the radius of the circle of the earth's path is 1 million miles or 1 mile. A year has 364.24 days.

you're missing the point.  because the earth moves along its orbit (whatever the size) over the course of a rotation, a solar day is necessarily longer than a single rotation.

we can see that the sun is offset at the end.

yes, it is.  the earth does not rotate on its axis an integer number of times in one orbit of the sun.  imagine a planet that rotates 1.5 times over one orbit of its sun.  at the end of one orbit, the sun will be offset from its starting point.

I am talking about Solar Days. The Earth rotates in one rotation, and the sun makes one rotation around the earth, over a Solar Day.

The sun should have returned to the same position over the earth. It did not.

there is no such requirement in ret.

That's the definition of a Solar Year. The time it takes the sun to return to its same position. I have given you the quote.