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Offline Tumeni

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #200 on: April 25, 2018, 07:19:15 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.
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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #201 on: April 25, 2018, 07:22:31 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.

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.

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Offline Bobby Shafto

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #202 on: April 25, 2018, 07:30:45 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.

Half way round the sun takes 182.62 days.
At an extra 0.986 degs of rotation a day, that’s 180 extra degrees. A half turn. So, if it was noon in NYC at the start of the solar year, it’ll be midnight when reaching the halfway point.

You don’t even need to get into the weeds for it to make sense. Rounding to 365 days in a year, half is 182.5.  0.5 of a 24 hour day is 12 hours. The “offset” makes sense. NYC will mark the halfway point in darkness if the clock started at noon NYC time.

In sidereal time, that’s when you use 360 degs because the sun is not the reference. You get the same result (NYC facing away from the sun at 1/2 year) but the clock isn’t synced to 360 degs per 24 hrs. It’s 360 degs per 23.934 hrs. And instead of 365.24 solar days, use 366.24 sidereal days. Halfway is 183.12 sidereal days. How much shorter are sidereal days? 3.93 minutes. So in 183.12 sidereal days, you lose about 12 hours compared to the solar day.

It adds up. The “offset” you were curious about and considered a problem is just the difference between sidereal and solar days, depending on what your reference point. If talking solar, count 24 hrs in a day, but account for the added degree of rotation in a solar day. If talking sidereal, use 360 degrees of rotation, but shorten the “day” parameter by 3 mins 56 secs. Figuring with 360 degrees (a sidereal measure) with 24 hours (a solar measure) is mixing terms.


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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #203 on: April 25, 2018, 07:33:09 AM »
« Last Edit: April 25, 2018, 07:59:07 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 #204 on: April 25, 2018, 08:03:07 AM »
There is something that seems wrong with the way the earth rotates around the Sun. Consider the following image that we are taught in school:




Assume that New York City is in its Solar Noon (look at where New York City is in the top and bottom September and March figures in the above illustration). After 6 months the motions suggests that New York City will be in darkness during its noontime.

I’d like to point out that, like the flat earth animation graphic, that picture you’ve chosen of earth’s orbit about the sun is just a visualization. It’s illustrating a point about tilt and equinox/solstice and isn’t meant to be a scale, accurate representation. The globe earth is a copy and paste icon massively larger and close to a tiny cartoon sun.

But, your question is understandable nonetheless...

Some Rough Calculations

Napkin Calculation 1

Day = 24 hours
Year = 365 days

365 days / 2 = 182.5 days in 6 months
24 hours x 182.5 days = 4380 hours in 6 months
4380 hours / 360 (since the sun rotates around the earth 360 degrees in one day) = 12.16666 hour offset

Earth should be offset by 12.16666 hours (similar to the above image)? NYC should be in night?


Yes. The approach you took is confusing, but the solar day and sidereal day will be offset by about 12 hours, and the side of the world facing the sun at the starting gun will be facing away from the sun at the halfway point.
Napkin Calculation 2

According to RET particulars, the earth doesn't rotate at exactly 24 hours a day, and the earth doesn't have an exactly 365 day year, which is why we have to change times and add a leap year every 4 years.

Sidreal Day = 23.933333 hours
Sidreal Year = 365.25636 days

365.25636 days per year / 2 = 182.62818 days in 6 months
23.933333 hours per day x 182.62818 days = 4370.90104712394 hours in 6 months
4370.90104712394  / 360 (since the sun rotates around the earth 360 degrees in one day) = 12.14139179 hours offset

Earth should be offset by 12.14139179 hours? NYC should be in night?


Again, don’t agree with the method, but same result: an “offset” as should be expected between a star field frame of reference and a solar frame of reference at the half yearly mark.


Napkin Calculation 3

Some sources say that the earth "actually" rotates 360.98 degrees per day.

360.98 degrees in a day x 182.62818 days in 6 months
= 65925.1204164 degrees in 6 months
ans / 360.98 = offset is 183.62818 degrees.

Earth should be offset by 183.62818 degrees? NYC should be in night?

--- --- ---

Corrections with the 360.98 figure

Using 360.98 degrees per day in the second calculation, replacing 360 with 360.98, gives an offset of 12.1082 hours. The offset still says that NYC should be in night.

Replacing 360 with 360.98 in the third calculation gives an offset answer of 182.625 degrees. The offset still says that NYC should be in night.

---

I may be going about this entirely wrong. Can I have some help with this seemingly glaring problem?

Even when cross mixing sidereal and synodic parameters, you’re still getting a affirmation that there’s about a half a day/half  a rotation delta between sidereal and solar. And yes, if NYC was at high noon at the start, it’ll be in darkness at half way around the orbit.

The fact that the earth orbits while it rotates is why.
Solar day is 24.000 hours per 360.986 degrees.
Sidereal day is 23.934 hrs per 360.000 degrees.
After halfway thru 365.24 solar days (which is also 366.24 sidereal days) the two measures will be offset by about 12 hours, which means it will be oriented the same as it was to the stars when it started the year but 180 degrees out in relation to the sun.
« Last Edit: April 25, 2018, 08:53:12 AM by Bobby Shafto »

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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #205 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...

Quote
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.

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.
« Last Edit: April 25, 2018, 08:08:36 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 #206 on: April 25, 2018, 08:07:43 AM »
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.

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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #207 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.
« Last Edit: April 25, 2018, 08:38:15 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 #208 on: April 25, 2018, 08:37:51 AM »
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.
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.

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.

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.

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.

(Frankly, I wouldn’t worry about the sidereal year. That’s yet a whole ‘nuther ball of wax that, if having trouble with sidereal days versus solar days, and solar days fitting into solar years, it’ll only compound the confusion, and it isn’t necessary for resolving what you’ve been asking.)
« Last Edit: April 25, 2018, 08:41:18 AM by Bobby Shafto »

Macarios

Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #209 on: April 25, 2018, 08:41:06 AM »
The Problem

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

Mean Solar Day fits in Calendar Year.
Some Calendar Years have 365 some 366 Solar Days, and in both cases it is the whole number.
Calendar Year doesn't fit in Tropical nor Sidereal years because of the way we count it.
Our calendar simply counts days for each year.

Solar (Tropical) year or Sidereal year are not determined by number of Earth's rotations, but by orbital positions.

Were you expecting to synchronize Earth's rotation with Earth's revolution?
« Last Edit: April 25, 2018, 08:43:58 AM by Macarios »

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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #210 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.

Quote
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.

Quote
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.

Quote
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.
« Last Edit: April 25, 2018, 08:56:48 AM by Tom Bishop »

Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #211 on: April 25, 2018, 09:01:48 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.

Quote
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. The .24 come out of nowhere and don't match up with the Solar Year where the sun needs to return to the point of the Equinox.

Quote
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.

Quote
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.

Equinox is just the sun passing the equatorial plane, its not a fixed point.

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Offline Tumeni

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #212 on: April 25, 2018, 09:20:04 AM »
Equinox is just the sun passing the equatorial plane, its not a fixed point.

... as shown by the equinoxes shown over two days in the illustration which started this thread.
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Offline Tom Bishop

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #213 on: April 25, 2018, 09:31:44 AM »
Equinox is just the sun passing the equatorial plane, its not a fixed point.

... as shown by the equinoxes shown over two days in the illustration which started this thread.

The March and September Equinoxes are  single points:

Quote from: Tom Bishop
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.

A random quote:

http://www.slate.com/blogs/bad_astronomy/2014/01/01/new_year_2014_how_astronomers_define_the_year.html

Quote
You could, for example, measure it from the exact moment of the vernal equinox—a specific time of the year when the Sun crosses directly over the Earth’s equator in March—in one year to the vernal equinox in the next. That’s called a tropical year (which is 31,556,941 seconds long).
« Last Edit: April 25, 2018, 09:35:55 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 #214 on: April 25, 2018, 09:59:18 AM »
Equinox is just the sun passing the equatorial plane, its not a fixed point.
Right. And there’s nothing in that definition requiring it to occur after a whole number of 365 solar day rotations of the earth. The extra 0.24 hours don’t “come out of nowhere.” That event doesn’t occur at the same moment every 365 solar says. It happens about 6 hours later every year.

And as Macharios says, (I think) the calendar we use for convention’s sake ignores those “extra hours” for a few years even though celestially the equinox is shifted. Rather than adjust yearly, incrementally, we let it go, accounting for those extra hours with a whole day adjustment every 4 years to realign. But the sun and earth don’t wait for man’s calendar. If we didn’t adjust, our calendars would fall behind because of those “extra hours” and after awhile, we’d notice the seasons weren’t right.

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Offline Bobby Shafto

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #215 on: April 25, 2018, 10:02:01 AM »
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.
First sentence, I agree. It “needs to” for the definition to be true.

But what I’m not getting is why you think the second sentence is a “has to” situation and, if it doesn’t, it’s a problem.

It clearly doesn’t meet your “has to” expectation since Equinox occurrence slides later by about 5.8 “extra hours” each year, and would keep sliding forward into the calendar year if we didn’t add a day to the calendar every four years.

I’m not picking up on why this is a problem. Solar days don’t “have to” fit neatly  and non-fractionally into the solar year. Not in reality, and not by definition.

Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #216 on: April 25, 2018, 12:48:34 PM »
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.
Hmm, I can't find the one that explicitly says it right now like I did last night (it was one of the links you provided) but the wikipedia page says it too. 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.

As well it's brought up in the definition on timeanddate.com https://www.timeanddate.com/astronomy/tropicalyearlength.html

Quote from: solar year
The tropical year can be measured from any equinox or solstice to the next. timeanddate.com calculates a tropical year from the March equinox one year to March equinox the next year.

timeanddate.com unfortunately doesn't itself define equinox/solstice in there though. But it matches what wikipedia says in this regard.

Now, would you care to point out where *anywhere* uses a whole number of solar days to define a solar year? Or where *anywhere* even suggests this should be the case? Other than your misintepretation of something, or not reading something fully, nothing suggests this but you. You are completely alone in suggestion there must be a whole number of solar days in a solar year. (remember, a calendar year =/= a solar year)

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Offline Stagiri

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Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #217 on: April 25, 2018, 02:14:16 PM »
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.

Indeed, we are well aware of that fact. That's why we have leap days and leap seconds.

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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.

(...)

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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.

Are you aware that equinox is, indeed, 6 hours (~0.25 days; exactly as you've calculated) later each year?
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 #218 on: April 25, 2018, 04:40:15 PM »
I think some of the confusion here comes from there being two senses of "the sun returning to a point".

1) The equinoxes happen when the sun moves across a line in the sky with reference to the "fixed" stars. It is absolutely true that this happens at an exact moment every year, and that that moment is gradually precessing over a period about 26000 years.
So, the sun returning to a point in the sky might mean with respect to the stars, and as observed by any observer at any location on earth at the same time (if you can see the sun, of course)
2) The sun also returns to the same meridian every day at solar noon for any observer. This is a different time for every different meridian on earth.
So, the sun returning to a point in the sky might mean with respect to local meridian, and this varies for observers with their longitude (which we do not need to mean a spherical earth, just with respect to angles of the sun and which Tom has agreed is an observable measurement.)

After one solar day, the sun returns to the same meridian.

After a solar year, the sun returns to the same location in the sky, but observers on the ground may not see it at the same time as they did on other years. In other words, the meridian the sun is over at the end of the solar year will be different from the meridian it was at at the beginning of the year.

There is no requirement that a particular observer see the sun at solar noon at both ends of a solar year - there is only the requirement that as measured against the stars the sun is in the same position.

Re: Possible Issue with Solar Noon in Round Earth Theory
« Reply #219 on: April 25, 2018, 04:44:53 PM »
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.


Is your contention that humans cooked up the number of days in the year and thus picked 360? This is not the case.

Humans also did not pick pi - the ratio of a circumference of a circle to its diameter. The number of radians in a circle is 2 * pi, which is not a round number. No matter how hard you try, you can't change the number of radians in a circle. If you unroll a circle into a line, it will be of length 2 * pi * the radius. You could treat it as 360 degrees * some constant * the radius, but you can never make that constant a nice round number.
The number of solar days in a solar year is 365.24, which is not a round number. The same problem happens here - you can define some other subdivision of the year, but you don't get to change how many rotations the earth makes (or the sun makes in the sky if you think the earth does not rotate) per year.
Humans can invent other measurement systems, but nature has no obligation to fit into those measurement systems.