The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Investigations => Topic started by: Tom Bishop on April 22, 2018, 10:31:35 AM
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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:
(https://www.e-education.psu.edu/meteo300/sites/www.e-education.psu.edu.meteo300/files/images/lesson6/EarthOrbit.png)
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.
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?
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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?
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Napkin Calculation 3
Some sources (https://www.google.com/search?client=firefox-b-1&ei=73HcWrTtBou6tgX2yoywCg&q=earth+rotates+%22360.98+degrees%22&oq=earth+rotates+%22360.98+degrees%22&gs_l=psy-ab.3...18960.20962.0.21596.2.2.0.0.0.0.184.336.0j2.2.0....0...1c.1.64.psy-ab..0.1.152...33i160k1.0.RIj9UZSX37E) 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?
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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.
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I may be going about this entirely wrong. Can I have some help with this seemingly glaring problem?
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Are you dividing the number of hours in six months by the number of degrees the sun rotates in one day?
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Are you dividing the number of hours in six months by the number of degrees the sun rotates in one day?
Yes, that is the method I am using. I get the number of hours in 6 months and divide by 360 in the first two calculations. I am also dividing by 360.98, which is what some sources allege that the earth rotates in degrees in one day. The offset in both cases seems to suggest that NYC will be in night during its noontime.
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Not sure why you divide the number of hours in 6 months by no of degrees in one day?
If you multiply 182.5 by 360 it gives 65700 so now we have total hours in 6 months: 4380 ÷ total degrees in6 months:65700 which gives us 0.0666.
This makes more sense to me and fits with us observed or an I just being thick? (I'm totally open to that being the case here :-)
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Not sure why you divide the number of hours in 6 months by no of degrees in one day?
If you multiply 182.5 by 360 it gives 65700 so now we have total hours in 6 months: 4380 ÷ total degrees in6 months:65700 which gives us 0.0666.
This makes more sense to me and fits with us observed or an I just being thick? (I'm totally open to that being the case here :-)
I'm dividing by 360 or 360.98 because I am dividing to get the offset in a day.
Since you are dividing by the hour, wouldn't that give us the degree offset per hour?
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How about 182.5÷360 which is total days divided by degrees which gives us 0.5?
I think your sum is mixing variables too much?
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Dear Mr. Bishop,
Could you, please, help me understand why you divide the number by 360°?
Thank you.
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I don't understand. Why would we take six months in days and divide by 360?
Look at Napkin Calculation 3. I am finding out how many degrees the earth spins over 6 months (according to a Sidreal Year), and then I am dividing by 360.98. I am of the opinion that it clearly shows the issue, and I do not mix variables in this calculation:
Napkin Calculation 3
Some sources (https://www.google.com/search?client=firefox-b-1&ei=73HcWrTtBou6tgX2yoywCg&q=earth+rotates+%22360.98+degrees%22&oq=earth+rotates+%22360.98+degrees%22&gs_l=psy-ab.3...18960.20962.0.21596.2.2.0.0.0.0.184.336.0j2.2.0....0...1c.1.64.psy-ab..0.1.152...33i160k1.0.RIj9UZSX37E) 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?
Right got it, my bad I think I'm on the same page as you now
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I was editing and deleted a post. Anyone interested can read it above in Devils Advocate's post.
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I don't understand. Why would we take six months in days and divide by 360?
Look at Napkin Calculation 3. I am finding out how many degrees the earth spins over 6 months (according to a Sidreal Year), and then I am dividing by 360.98. I am of the opinion that it clearly shows the issue, and I do not mix variables in this calculation:
Napkin Calculation 3
Some sources (https://www.google.com/search?client=firefox-b-1&ei=73HcWrTtBou6tgX2yoywCg&q=earth+rotates+%22360.98+degrees%22&oq=earth+rotates+%22360.98+degrees%22&gs_l=psy-ab.3...18960.20962.0.21596.2.2.0.0.0.0.184.336.0j2.2.0....0...1c.1.64.psy-ab..0.1.152...33i160k1.0.RIj9UZSX37E) 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?
I'd like to point out that you multiply 182.62818 [days] by 360.98 [°] and then you divide the answer by 360.98 [°]. So you get 182.62818 [days]. Not very useful, is it? Or did I misunderstand that calculation?
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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.
Why should I/we indulge you, and explain this in great detail, when it has already been detailed?
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. Surely you won't conclude that you're right, and they're all wrong?
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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?
If a day is defined as the period of time in which NY, at solar noon on day 1, returns to solar noon on day 2, and on day 3, then the whole calculation is moot.
6 months later, NY will still be returning to solar noon once per day
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Dear Mr. Bishop,
Could you, please, help me understand why you divide the number by 360°?
Thank you.
I divided by 360 and 360.98 in Napkin Calculation #1 and #2 because this is a calculation involving hourly rotation (It started with 24 hours or 23.933333 hours in one rotation) and 360 represents one day. I didn't introduce the concept of a rotation when I introduced the 360 number.
I don't understand. Why would we take six months in days and divide by 360?
Look at Napkin Calculation 3. I am finding out how many degrees the earth spins over 6 months (according to a Sidreal Year), and then I am dividing by 360.98. I am of the opinion that it clearly shows the issue, and I do not mix variables in this calculation:
Napkin Calculation 3
Some sources (https://www.google.com/search?client=firefox-b-1&ei=73HcWrTtBou6tgX2yoywCg&q=earth+rotates+%22360.98+degrees%22&oq=earth+rotates+%22360.98+degrees%22&gs_l=psy-ab.3...18960.20962.0.21596.2.2.0.0.0.0.184.336.0j2.2.0....0...1c.1.64.psy-ab..0.1.152...33i160k1.0.RIj9UZSX37E) 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 182.62818 degrees.
Earth should be offset by 182.62818 degrees? NYC should be in night?
I'd like to point out that you multiply 182.62818 [days] by 360.98 [°] and then you divide the answer by 360.98 [°]. So you get 182.62818 [days]. Not very useful, is it? Or did I misunderstand that calculation?
What I was going for was to show that when you divide you are facing the opposite way, which is why I stayed with degrees.
The offset is 182.62818 days from the starting point, as you can interpret that result in hours, but isn't the offset also 182.62818 degrees if you interpret the result in degrees since all of this takes place on a big circle?
However, since I was just reversing the calculation I did, I see the apparent weakness in that argument, but still maintain the above explanation.
How would you calculate this?
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"On a prograde planet like the Earth, the sidereal day is shorter than the solar day. At time 1, the Sun and a certain distant star are both overhead. At time 2, the planet has rotated 360° and the distant star is overhead again (1→2 = one sidereal day). But it is not until a little later, at time 3, that the Sun is overhead again (1→3 = one solar day). More simply, 1-2 is a complete rotation of the Earth, but because the revolution around the Sun affects the angle at which the Sun is seen from the Earth, 1-3 is how long it takes noon to return."
https://en.wikipedia.org/wiki/Solar_time
The core issue is with how you define a "day", and you didn't define it prior to commencing your napkin calcs.
At solar noon on day 1, and referring to the first image in the wiki above, solar noon has a line from centre of Earth, through observer AT Noon, to the centre of the Sun. Next day, Earth completes its 360 degree rotation, but the same line from centre of Earth to observer to sun is now NOT aligned with the centre of the sun. It has to rotate further to achieve this, because the Earth has moved in its orbit as well as rotating.
So calculations based on 360 degree rotation may well be misplaced.
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Tumeni, thank you for participating. I need to go at the moment and can't read through your post and links, but will make a reply later.
Here is another version I came up with a little earlier, which even easier to visualize:
Napkin Calculation #4
We can agree that, according to the image in the OP, after one full route of the earth around the sun, Solar Noon will be in the same place.
Using the simple numbers --
Premise:
- The earth rotates once in a day (a day defined as once every 24 hours).
- One revolution of the earth around the sun is 365 days
Therefore, a day is a whole number in the above scheme.
If we divide 365 by 2 we do not get a whole number of days. We get 182.5 days. Therefore noon will be midnight.
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Tumeni, thank you for participating. I need to go at the moment and can't read through your post and links to give an in depth response, but will make a reply later. I think that I did address sidereal day, sidereal year, and 365.98 degree rotation though, and showed that they do not make much difference in the calculations.
Here is another version I came up with a little earlier, which even easier to visualize:
Napkin Calculation #4
We can agree that, according to the image in the OP, after one full route of the earth around the sun, Solar Noon will be in the same place.
Using the simple numbers --
Premise:
- The earth rotates once in a day (a day defined as once every 24 hours).
- One revolution of the earth around the sun is 365 days
Therefore, a day is a whole number in the above scheme.
If we divide 365 by 2 we do not get a whole number of days. We get 182.5 days. Therefore noon will be midnight.
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I don't understand. Why would we take six months in days and divide by 360?
Look at Napkin Calculation 3. I am finding out how many degrees the earth spins over 6 months (according to a Sidreal Year), and then I am dividing by 360.98. I am of the opinion that it clearly shows the issue, and I do not mix variables in this calculation:
Napkin Calculation 3
Some sources (https://www.google.com/search?client=firefox-b-1&ei=73HcWrTtBou6tgX2yoywCg&q=earth+rotates+%22360.98+degrees%22&oq=earth+rotates+%22360.98+degrees%22&gs_l=psy-ab.3...18960.20962.0.21596.2.2.0.0.0.0.184.336.0j2.2.0....0...1c.1.64.psy-ab..0.1.152...33i160k1.0.RIj9UZSX37E) 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 182.62818 degrees.
Earth should be offset by 182.62818 degrees? NYC should be in night?
I'd like to point out that you multiply 182.62818 [days] by 360.98 [°] and then you divide the answer by 360.98 [°]. So you get 182.62818 [days]. Not very useful, is it? Or did I misunderstand that calculation?
What I was going for was to show that when you divide you are facing the opposite way, which is why I stayed with degrees.
The offset is 182.62818 days from the starting point, as you can interpret that result in hours, but isn't the offset also 182.62818 degrees if you interpret the result in degrees since all of this takes place on a big circle?
(...)
No, it isn't, because 1° does NOT equal 1 day. You can't just interpret things as you please. [day*°] divided by [°] is always [day] and nothing else. You'll have to come up with a different formula.
Tumeni, thank you for participating. I need to go at the moment and can't read through your post and links, but will make a reply later.
Here is another version I came up with a little earlier, which even easier to visualize:
Napkin Calculation #4
We can agree that, according to the image in the OP, after one full route of the earth around the sun, Solar Noon will be in the same place.
Using the simple numbers --
Premise:
- The earth rotates once in a day (a day defined as once every 24 hours).
- One revolution of the earth around the sun is 365 days
Therefore, a day is a whole number in the above scheme.
If we divide 365 by 2 we do not get a whole number of days. We get 182.5 days. Therefore noon will be midnight.
So, you've come to this cońclusion: in New York, 182 and a half days after solar noon it is midnight. Bravo.
EDIT: The problem with your calculations is that you assume the Earth's orbit to be a perfect circle, its speed to be constant and the Sun to be in the exact middle. None of which is claimed by the GE model.
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tumeni is correct. sidereal days and solar days have different lengths.
a sidereal day is 4 minutes shorter than a solar day.
4 minutes per solar day * 180 solar days = 720 minutes = 12 hours.
"sidereal noon" will lag behind solar noon by 12 hours after six months. but we define time according to solar days, not sidereal days.
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Napkin Calculation #4
Premise:
- The earth rotates once in a day (a day defined as once every 24 hours).
- One revolution of the earth around the sun is 365 days
Therefore, a day is a whole number in the above scheme.
If we divide 365 by 2 we do not get a whole number of days. We get 182.5 days. Therefore noon will be midnight.
Premise;
https://en.wikipedia.org/wiki/Day
"A day, understood as the span of time it takes for the Earth to make one entire rotation with respect to the celestial background or a distant star (assumed to be fixed), is called a stellar day. This period of rotation is about 4 minutes less than 24 hours (23 hours 56 minutes and 4.1 seconds)..."
Using this premise, if we take 4 mins off each 24 hour day, for each of your 182.5 days, that adds up to (4*182.5)/60 = 12.2 hours. Roughly half a day.
So, where you've taken a day as exactly 24 hours, and concluded this places noon in the wrong place by around 12 hours half a year hence, if you take the actual day length, that places solar noon roughly twelve hours before this, which would be on the side of the Earth facing the sun, not away from it
Napkin calc #1 is therefore rendered moot.
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Napkin Calculation 2
"4370.90104712394 / 360 (since the sun rotates around the earth 360 degrees in one day) = 12.14139179 hours offset"
I fail to see the logic in this. Number of hours divided by number of degrees in a circle yields number of hours taken to cover one degree of that circle, not an "offset" number of hours....
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I see that the concept of computing offset with division of degrees and circles, and the relation of such is lost on some.
You are right. I did assume that the earth orbit around the sun was a perfect circle. But we can do this without degrees, and with a method that does not assume that the orbit of the earth around the sun is a perfect circle. With easy and simple math that will make all very clear.
Napkin Calculations # 5 & 6
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Premise:
- A Solar Day happens every 24 hours
- A Year happens every 365 days
365 days / 2 = 185.5 days in six months
Now, we need to fit 24 into 185.5
185.5 days divided by 24 = 7.729
Conclusion: 24 does not fit cleanly into 185.5. If it did, it would be a whole number.
Conclusion: The day does not fit neatly into half of the year. Solar Noon will not be at the same place at that point.
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Lets now define that a Solar Day happens every 23.933333 hours, and correct the year as well.
Premise:
- A Sidrael Day/Solar Day happens every 23.933333 hours
- A Sidrael Year happens every 365.25636 days
365.25636 / 2 = 182.62818 days in six months
182.62818 / 23.933333 = 7.730.
Conclusion: 23.93333 hours does not fit neatly into 182.62818 days. If it did it would be a whole number.*
Conclusion 2: The Sidrael Day does not fit neatly into half of the Sidrael Year. Solar Noon will not be in the same place at that point.
An aside point: When we compare the ratios we got of 7.729 and 7.730 we can see that trying to use the slightly different numbers of Sidrael Day and Sidrael Year makes hardly a difference.
* For arguments that the numbers actually do fit, and but we just aren't "accurate enough" in the measured values of Sidrael Day and Year, and there were a few more numbers after the decimal numbers, the ratio result should still be very near to a whole number. It should be 8.00001 or something of that nature. But we are way off.
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I see that the concept of computing offset with division of degrees and circles, and the relation of such is lost on some.
I see that you can be exceedingly condescending to someone who has been nothing but polite toward you. What is this 'offset concept', and where has it been discussed or mentioned, prior to your post?
You are right. I did assume that the earth orbit around the sun was a perfect circle.
I never said you did that. Once again, you took a number of hours divided by the number of degrees in a circle; all this yields is the number of hours taken to cover one degree of that circle
But we can do this without degrees, and with a method that does not assume that the orbit of the earth around the sun is a perfect circle. With easy and simple math that will make all very clear.
Napkin Calculations # 5 & 6
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Premise:
- A Solar Day happens every 24 hours
- A Year happens every 365 days
365 days / 2 = 185.5 days in six months
Now, we need to fit 24 into 185.5
185.5 days divided by 24 = 7.729
24 does not fit cleanly into 185.5. If it did, it would be a whole number.
Conclusion: The day does not fit neatly into half of the year. Solar Noon will not be at the same place at that point.
Dividing the number of days in a half year by the number of hours in one day produces .... what? A random number.
Lets now define that a Solar Day happens every 23.933333 hours, and correct the year as well.
Premise:
- A Sidrael Day/Solar Day happens every 23.933333 hours
- A Sidrael Year happens every 365.25636 days
365.25636 / 2 = 182.62818 days in six months
182.62818 / 23.933333 = 7.730.
23.93333 hours does not fit neatly into 182.62818 days. If it did it would be a whole number.
Conclusion: The Sidrael Day does not fit neatly into half of the Sidrael Year. Solar Noon will not be in the same place at that point.
Again, all you're doing is deriving a random number by dividing the number of days in a half year by the number of hours in one day ....
365/24 = 15.21. Days fit equally as badly into a year as they do into a half-year. You appear to have 'proved' that our days don't fit in with our years....
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Dividing the number of days in a half year by the number of hours in one day produces .... what? A random number.
It's not a random number. Division gives a ratio. If the number of days fit into half the year, it should be a whole number.
An example:
24 / 6 = 4.
The number 4 is whole. The number 6 fits neatly into 24 four times.
This is a very, very clear concept.
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This is a very, very clear concept.
indeed it is.
a sidereal day is 4 minutes shorter than a solar day.
4 minutes per solar day * 180 solar days = 720 minutes = 12 hours.
tbh i don't understand what you think the problem is.
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This is a very, very clear concept.
indeed it is.
a sidereal day is 4 minutes shorter than a solar day.
I accounted for that in my last example. Did you not see where I stated:
Premise:
- A Sidrael Day/Solar Day happens every 23.933333 hours
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It's not a random number. If the number of days (24 hour cycles) fit into half the year, it should be a whole number
An example:
24 / 6 = 4.
The number 4 is whole. The number 6 fits neatly into 24 four times.
This is a very, very clear concept.
.... then you need to divide the number of hours in a half year by the number of hours in a day to try and get an exact fit of days into the half year ....
Not divide the number of days in a half year by the number of hours in a day.
7 (days) divided by 24 (hours) = 0.29 - if the number of days fit into the week, it should be a whole number, shouldn't it?
24 (hours) divided by 60 (minutes) = 0.4 - if the number of hours fit into the day, it should be a whole number, shouldn't it?
1 (hour) divided by 60 (seconds) = 0.01667 - if the number of minutes fits into the hour, it should be a whole number, shouldn't it?
Are you any kin to Mr Thrive and Survive, or The Potter's Clay, both of whom are active on YouTube?
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I accounted for that in my last example. Did you not see where I stated:
Premise:
- A Sidrael Day/Solar Day happens every 23.933333 hours
then i don't see what the issue is. it's a very simple calculation. i even stuck to integers to make it clearer to see.
4 minutes per solar day * 180 solar days = 720 minutes = 12 hours.
over the course of 6 months, sidereal noon will be ~12 hours behind solar noon. what's the problem?
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It's not a random number. If the number of days (24 hour cycles) fit into half the year, it should be a whole number
An example:
24 / 6 = 4.
The number 4 is whole. The number 6 fits neatly into 24 four times.
This is a very, very clear concept.
.... then you need to divide the number of hours in a half year by the number of hours in a day to try and get an exact fit of days into the half year ....
We don't know the number of hours in half a year in this problem. That is an number or ratio we are trying to compute based on the length of day and the length of year, to see if the Solar Time cycle matches up.
If you assume that half of the year is composed of full 24/23.933.. hour cycles, then you just assumed your result. You need to do the math without that assumption.
And even if you do use that assumption for half of the year, that assumption will not fit into the full year if you do the calculations.
This last example I gave should have been very clear.
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This last example I gave should have been very clear.
And my examples make it very clear that your method generates other random numbers when applied to other periods.
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It's not a random number. If the number of days (24 hour cycles) fit into half the year, it should be a whole number
An example:
24 / 6 = 4.
The number 4 is whole. The number 6 fits neatly into 24 four times.
This is a very, very clear concept.
.... then you need to divide the number of hours in a half year by the number of hours in a day to try and get an exact fit of days into the half year ....
We don't know the number of hours in half a year in this problem. (...)
Since [hour], [day] and [year] are all simply time units which we can quite easily convert to each other if we know the time in [days] (or [years]) we automatically know it in [hours] too.
We don't need to assume anything.
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We don't know the number of hours in half a year in this problem.
.. but your premise to start with was
Premise:
- A Solar Day happens every 24 hours
- A Year happens every 365 days
365 days / 2 = 185.5 days in six months
So ....185.5 * 24 = 4452 hours.
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I accounted for that in my last example. Did you not see where I stated:
Premise:
- A Sidrael Day/Solar Day happens every 23.933333 hours
then i don't see what the issue is. it's a very simple calculation. i even stuck to integers to make it clearer to see.
4 minutes per solar day * 180 solar days = 720 minutes = 12 hours.
over the course of 6 months, sidereal noon will be ~12 hours behind solar noon. what's the problem?
I think the definitions we are using for these terms may be slightly mixed up.
Lets look at this:
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.
Sidereal time is time measured with respect to the apparent motion of the 'fixed' stars in the sky due to the Earth's rotation. While the Earth is rotating on its axis it is also moving along its orbit around the Sun. Over the course of a day the Earth moves about one degree along its orbit (360 degrees in a full orbit divided by 365.25 days in a year is about one degree). Therefore, from our perspective, the Sun moves about one degree from west to east with respect to the 'fixed' stars.
And this one:
http://astronomy.swin.edu.au/cosmos/S/Sidereal+Day
A solar day is the time it takes for the Earth to rotate about its axis so that the Sun appears in the same position in the sky. The sidereal day is ~4 minutes shorter than the solar day. The sidereal day is the time it takes for the Earth to complete one rotation about its axis with respect to the 'fixed' stars.
It says that Solar time is measured with respect to the Sun's apparent motion in the sky, and is 24 hours.
Sidrael time is measured with respect to the Stars, and is the 23.933333 hours per rotation value. Why would you use the time the stars rotate around the earth in this when we are concerned with the Sun?
Those links say that Solar Time has a 24 hour cycle. Lets use the value according to this link.
Lets now define that a Solar Day happens every 24 hours, and the year every 365.25636 days.
Premise:
- A Solar Day happens every 24 hours
- A year happens every 365.25636 days
365.25636 / 2 = 182.62818 days in six months
182.62818 / 24 = 7.609
A 24 hour cycle does not fit neatly into 182.62818 days.
Sidrael Time is star time, not the day and night time, and if that is the definition then I don't see why we should use it in this at all.
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(...)
365.25636 / 2 = 182.62818 days in six months
(...)
Which six months? The year isn't divided evenly... (and there's reason for it)
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Sidrael Time is star time, not the day and night time, and if that is the definition then I don't see why we should use it in this at all.
it's clear you simply don't fully understand the model you're criticizing.
(https://i.imgur.com/qaAGmmu.png)
i added a star to show you how these are related. at the top, solar noon and "sidereal noon" are the same. at the bottom, solar noon and sidereal noon are 12 hours apart.
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(...)
365.25636 / 2 = 182.62818 days in six months
(...)
Which six months? The year isn't divided evenly... (and there's reason for it)
But it doesn't matter if the movement of the earth around the sun is not a perfect circle. We are concerned solely with with ratios. The illustration is of Vernal and Autumnal Equinox.
Sidrael Time is star time, not the day and night time, and if that is the definition then I don't see why we should use it in this at all.
it's clear you simply don't fully understand the model you're criticizing.
(https://i.imgur.com/qaAGmmu.png)
i added a star to show you how these are related. at the top, solar noon and "sidereal noon" are the same. at the bottom, solar noon and sidereal noon are 12 hours apart.
Sideral Noon is the fixed star's position. Yes, both images are facing us (the star). Congratulations. Sideral Noon (star time) changed.
But look at the picture! Solar Noon for NYC is different between both positions. NYC is in midnight in the bottom illustration. Uh oh.
It appears that Star Time is not what we want to talk about here, and the rate the stars move in respect to the sun are not directly relevant. Why should we be concerned with the movement of the stars when we are talking about Solar Noon?
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At which point, I refer you back to reply #13 and the wikipedia link.
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We don't know the number of hours in half a year in this problem.
.. but your premise to start with was
Premise:
- A Solar Day happens every 24 hours
- A Year happens every 365 days
365 days / 2 = 185.5 days in six months
So ....185.5 * 24 = 4452 hours.
The length of the earth's day doesn't fit into the year, whether whether we use 24 hours or 23.933333 hours (which is just star time, as we have read).
It doesn't work if we assume that the year is 365 days long or 365.25636 days long.
Please show how the number of days fits perfectly into the year.
This is very easy math.
We know that Solar Time has a 24 hour cycle. We know that the year is 365.25636 days long. How does it work?
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Sideral Noon is the fixed star's position. Yes, both images are facing us (the star). Congratulations. Sideral Noon changed.
But look at the picture! Solar Noon for NYC is different between both positions. NYC is in midnight in the bottom illustration. Uh oh. Solar Noon needs to be on the opposite side.
you're still not getting this. just look at the image. for the top bit, solar noon and sidereal noon are the same. the star and the sun cross the local meridian at the same time.
for the bottom bit, sidereal noon happens at "solar midnight." they're out of phase by twelve hours. i genuinely do not understand what's troubling you about this geometry.
i think the problem is that you're assuming that over the course of an orbital period, a sidereal day and a solar day must have an integer ratio. why must that be the case?
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Sideral Noon is the fixed star's position. Yes, both images are facing us (the star). Congratulations. Sideral Noon changed.
But look at the picture! Solar Noon for NYC is different between both positions. NYC is in midnight in the bottom illustration. Uh oh. Solar Noon needs to be on the opposite side.
you're still not getting this. just look at the image. for the top bit, solar noon and sidereal noon are the same. the star and the sun cross the local meridian at the same time.
for the bottom bit, sidereal noon happens at "solar midnight." they're out of phase by twelve hours. i genuinely do not understand what's troubling you about this geometry.
i think the problem is that you're assuming that over the course of an orbital period, a sidereal day and a solar day must have an integer ratio. why must that be the case?
This is very easy to understand, Gary. I don't know why you are talking about the movement of the stars in relation to the sun. We are not talking about Star Time.
Lets break it down even simpler. Let not talk about half of the year. Lets talk about the full year. Very basic.
The Solar Time has a 24 hour cycle. That cycle does not fit into the length of the year of 365.25636 days.
Easy one, right? Just divide and we will get a whole number. But that is not what happens. The days don't fit.
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The length of the earth's day doesn't fit into the year, whether whether we use 24 hours or 23.933333 hours.
Why not?
Please show how the number of days fits perfectly into the year.
1 year = 365 days for three years.
1 year = 366 days for one year. Leap year
Repeat ad infinitum
What's the difficulty?
How does it work?
Have you considered consulting some wikipedia pages, or decent astronomical textbooks?
https://en.wikipedia.org/wiki/Day#Apparent_and_mean_solar_day
https://en.wikipedia.org/wiki/Leap_year
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Your Solar Day link says that the Solar Day is 24 hours long
The leap year happens because the year is 365.25636 days long. Why are you trying to use that as an explanation when that is already accounted for in the number.
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The link Tumeni provided does say that Solar Day is technically "24.0000006 hours"
https://en.wikipedia.org/wiki/Day#Apparent_and_mean_solar_day
In recent decades, the average length of a solar day on Earth has been about 86 400.002 seconds (24.0000006 hours) and there are about 365.242 2 solar days in one mean tropical year.
Lets divide those numbers:
Solar Day: 24.0000006 hours
Year: 365.2422 Solar Days in a Year
Year / Day = 15.21842461953938
Oh no. This is not a whole number. Even with the more accurate numbers this doesn't work.
They put their contradiction right there in the same sentence, and even specified that it is Solar Days in both cases.
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I don't know why you are talking about the movement of the stars in relation to the sun.
He's not. The star is merely a reference point. Look again at the first wikipedia page I linked for you. (Illustration in wiki referred to in reply #13)
Solar Time has a 24 hour cycle. That cycle does not fit into the length of the year of 365.25636 days.
... and the 0.25 is carried forward to the leap year.
Easy one, right? Just divide and we will get a whole number. But that is not what happens. The days don't fit.
Because it's a meaningless calculation. Dividing any number of days by the number of hours in a day won't produce anything meaningful.
Divide 24 (days) by 24 (hours) and the answer is 1, divide 48 (days) by 24 (hours) and the answer is 2.
72 days = 3
96 days = 4
What does that mean to you?
See my examples above based on weeks and days.
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@Tom Bishop
Premises: - 1 day = 24 hours
- 1 hours = 3600 seconds
1/2 day = 12 hours
12/3600 = 0.0033333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333(and so on) = not an integer
Oh no! Conclusion: seconds do not fit into 1/2 day :o
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@Tom Bishop
Premises: - 1 day = 24 hours
- 1 hours = 3600 seconds
1/2 day = 12 hours
12/3600 = 0.0033333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333(and so on) = not an integer
Oh no! Conclusion: seconds do not fit into 1/2 day :o
Exactly. Tom, please see reply #25 again.
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This is very easy to understand, Gary. I don't know why you are talking about the movement of the stars in relation to the sun. We are not talking about Star Time.
because comparing sidereal days to solar days demonstrates that the image you posted makes perfect sense.
Lets break it down even simpler. Let not talk about half of the year. Lets talk about the full year. Very basic.
The Solar Time has a 24 hour cycle. That cycle does not fit into the length of the year of 365.25636 days.
Easy one, right? Just divide and we will get a whole number. But that is not what happens. The days don't fit.
i get it. you believe that a tropical year and a sidereal year must have an integer ratio.
sorry, but that isn't part of the model. it's just bogus requirement you made up.
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@Tom Bishop
Premises: - 1 day = 24 hours
- 1 hours = 3600 seconds
1/2 day = 12 hours
12/3600 = 0.0033333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333(and so on) = not an integer
Oh no! Conclusion: seconds do not fit into 1/2 day :o
Why are you placing the hours integer first when the goal is to see if seconds fits into hours?
You need to put in the numerically bigger number first if you are looking to see if the small number fits into it.
3600 / 12 = 300 = The ratio works. Whole number.
Conclusion: A second fits into an hour.
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We have 8 mysterious pieces of a pizza pie that are the same size and a pizza tray that holds 8 pieces.
We see that the two pieces make up 2/8th's of the pizza tray.
Can we calculate whether, if put in all 8 of the pizza slices (that are of the same size, if you recall above) of the pizza pie, would they fit?
8 / 2 = 4 -- Yes, these are whole slices that will all equally fit into the pizza tray. Tray divided by pizza slices. The result is a whole number. All 8 will fit.
2 / 8 = 0.25 -- This is not a valid equation for the purpose of this. Pizza slices divided by Tray? You need to put the larger numerical integer first to see if the smaller integer fits into it by producing a whole number.
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this is a weird argument. there is nothing in physics that says planets have to complete integer numbers of rotations in a single orbital period. you're just manufacturing this criticism.
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We learned earlier that a half year later, a sidereal day and a sun day are 12 hours apart. On one side of the sun they coincide, on the other side they are opposite. Since the only relationship that has changed is Earth to sun, the sidereal noon occurs at the same relative time that it did half a year ago, and solar noon now occurs 12 hours different. Oh wow, that makes solar noon continue to be during the day doesn't it? You're bringing in a requirement that doesn't exist and proclaiming it proves everything wrong. It's already been shown the difference between the two is 12 hours over the course of half of a year. Since the relative location of the stars to Earth doesn't change, that means the sun has done exactly what we experience every year. QED.
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We learned earlier that a half year later, a sidereal day and a sun day are 12 hours apart. On one side of the sun they coincide, on the other side they are opposite. Since the only relationship that has changed is Earth to sun, the sidereal noon occurs at the same relative time that it did half a year ago, and solar noon now occurs 12 hours different. Oh wow, that makes solar noon continue to be during the day doesn't it? You're bringing in a requirement that doesn't exist and proclaiming it proves everything wrong. It's already been shown the difference between the two is 12 hours over the course of half of a year. Since the relative location of the stars to Earth doesn't change, that means the sun has done exactly what we experience every year. QED.
Sidrael Day is Star Time, and has nothing to do with this. What does it matter how fast the stars are moving in this?
Tumni has already settled the matter, by providing this link, right here:
The link Tumeni provided does say that Solar Day is technically "24.0000006 hours"
https://en.wikipedia.org/wiki/Day#Apparent_and_mean_solar_day
In recent decades, the average length of a solar day on Earth has been about 86 400.002 seconds (24.0000006 hours) and there are about 365.242 2 solar days in one mean tropical year.
Lets divide those numbers:
Solar Day: 24.0000006 hours
Year: 365.2422 Solar Days in a Year
Year / Day = 15.21842461953938
Oh no. This is not a whole number. Even with the more accurate numbers this doesn't work.
They put their contradiction right there in the same sentence, and even specified that it is Solar Days in both cases.
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We learned earlier that a half year later, a sidereal day and a sun day are 12 hours apart. On one side of the sun they coincide, on the other side they are opposite. Since the only relationship that has changed is Earth to sun, the sidereal noon occurs at the same relative time that it did half a year ago, and solar noon now occurs 12 hours different. Oh wow, that makes solar noon continue to be during the day doesn't it? You're bringing in a requirement that doesn't exist and proclaiming it proves everything wrong. It's already been shown the difference between the two is 12 hours over the course of half of a year. Since the relative location of the stars to Earth doesn't change, that means the sun has done exactly what we experience every year. QED.
Sidrael Day is Star Time, and has nothing to do with this. What does it matter how fast the stars are moving in this?
Tumni has already settled the matter, by providing this link, right here:
The link Tumeni provided does say that Solar Day is technically "24.0000006 hours"
https://en.wikipedia.org/wiki/Day#Apparent_and_mean_solar_day
In recent decades, the average length of a solar day on Earth has been about 86 400.002 seconds (24.0000006 hours) and there are about 365.242 2 solar days in one mean tropical year.
Lets divide those numbers:
Solar Day: 24.0000006 hours
Year: 365.2422 Solar Days in a Year
Year / Day = 15.21842461953938
Oh no. This is not a whole number. Even with the more accurate numbers this doesn't work.
They put their contradiction right there in the same sentence, and even specified that it is Solar Days in both cases.
Ah yes, your made up requirement doesn't work. Pity that, I suppose it's a good thing it doesn't mean anything. In fact, since we don't have a whole number of days in a year, why would you ever expect this division to result in a whole number? It's almost as though you're just making something up and claiming it disproves something else.
You originally asked about solar noon changing the side of the planet it's facing. This is readily answered in the difference between solar noon and sidereal noon. Over half of a year they end up 12 hours, or half a rotation apart. Since we know the stars don't move in relation to Earth by 180 degrees, obviously the sun does. Which leaves solar noon still happening on the day side of the Earth. Which shows your original objection is groundless.
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I believe that I have shown a significant error in Solar Time, which is the heart of this discussion.
Lets define Tropical Year
https://en.wikipedia.org/wiki/Tropical_year
A tropical year (also known as a solar year) is the time that the Sun takes to return to the same position in the cycle of seasons, as seen from Earth; for example, the time from vernal equinox to vernal equinox, or from summer solstice to summer solstice.
And the Topical Year varies on solar return points (which is why they called it the 'mean' tropical year):
http://calendars.wikia.com/wiki/Tropical_year
Current values and their annual change of the time of return to the cardinal ecliptic points[2] are:
vernal equinox: 365.24237404 + 0.00000010338×a days
northern solstice: 365.24162603 + 0.00000000650×a days
autumn equinox: 365.24201767 − 0.00000023150×a days
southern solstice: 365.24274049 − 0.00000012446×a days
Lets compare the smallest value to the largest value.
365.24162603 / 24.0000006 = 15.218400704
365.24274049 / 24.0000006 = 15.218447139
The difference isn't anywhere close to a whole number. This is way off.
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Sidrael Day is Star Time, and has nothing to do with this. What does it matter how fast the stars are moving in this?
Nobody is saying that it does matter. You seem to be manufacturing something to argue against.
Again, see https://en.wikipedia.org/wiki/Solar_time
References to a 'distant star' are merely as a reference point, a presumed stationary reference point ...
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The difference isn't anywhere close to a whole number.
So what?
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Watcha Tom,
I'm wondering if your issue is more of a problem with the calendar as opposed to astronomy?
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Lets compare the smallest value to the largest value.
365.24162603 / 24.0000006 = 15.218400704
365.24274049 / 24.0000006 = 15.218447139
The difference isn't anywhere close to a whole number. This is way off.
Once again, you're a dividing an arbitrary number OF days by the number of hours WITHIN a day and expecting some correlation, when none is expected by anyone. ever.
As an aside;
The currency system in the UK used to be pounds, shillings and pence
1 pound = 20 shillings
1 shilling = 12 pence
You're doing the equivalent of
1 (pound) divided by 12 (pence) = 0.08333 .... oh, no, the currency system doesn't work (unless you divide 12 pounds by 12 pence, and get 1; or 24 pounds by 12 pence to get 2, 36 by 12 = 3, 48 by 12 = 4 ......) !
...
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I believe that I have shown a significant error in Solar Time, which is the heart of this discussion.
Lets define Tropical Year
https://en.wikipedia.org/wiki/Tropical_year
A tropical year (also known as a solar year) is the time that the Sun takes to return to the same position in the cycle of seasons, as seen from Earth; for example, the time from vernal equinox to vernal equinox, or from summer solstice to summer solstice.
And the Topical Year varies on solar return points (which is why they called it the 'mean' tropical year):
http://calendars.wikia.com/wiki/Tropical_year
Current values and their annual change of the time of return to the cardinal ecliptic points[2] are:
vernal equinox: 365.24237404 + 0.00000010338×a days
northern solstice: 365.24162603 + 0.00000000650×a days
autumn equinox: 365.24201767 − 0.00000023150×a days
southern solstice: 365.24274049 − 0.00000012446×a days
Lets compare the smallest value to the largest value.
365.24162603 / 24.0000006 = 15.218400704
365.24274049 / 24.0000006 = 15.218447139
The difference isn't anywhere close to a whole number. This is way off.
So you have discovered something nobody else has?
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Earth takes the same position relative to orbital event in 31 556 925 seconds.
It is called Tropical Year.
During that time it turns 365.2421875 times.
Which means, it completes 365 turns about 0.242 days earlier than making full circle.
Our 24 hours day was designed to have 86400 seconds in one Solar day.
(The 86400 is on average, plus or minus 18 to 29 seconds due to speed changes in accordance with Kepler's Second Law.)
One solar day is time until the same meridian points towards Sun again.
It is not 360 degrees, it is 360.9856 degrees (almost 361).
It is because Earth moved forward in orbit and has to "look a bit back" to point the Sun again.
Time required to spin for 360 degrees is Sidereal day, and is measured towards distant, apparently unmovable star.
It is not 24 hours, but a bit shorter, 23 h, 56 min, 4.1 sec.
For our calendar it is irrelevant.
When people designed calendar they didn't know about it.
They designed day based on Sun, nor on stars, and year based on whole number of such days.
Astronomers at the time of Julius Caesar noticed that there is difference between solstices and full set of 365 days.
To correct that error they added one more day every four years to compensate for those 6 hours.
That's how Julian calendar was created.
It measures 31 557 600 seconds per year and still doesn't hit the exact number of 31 556 925 seconds.
The time between the full 365 days and the full orbital circle wasnt exactly 6 hours, it was 5 hours 48 minutes and 45 seconds.
The difference accumulated by adding one unnecessary day every 127 years.
At the time of Pope Gregory XIII the error accumulated to 10 days.
Calendar was late.
Instead of showing 21st of March for vernal equinox, it was still showing March 11th.
Solstices weren't at the same dates as in the time of Julius Caesar.
So, calendar was corrected again, and extra days were removed by skipping them.
Thursday, 4 October 1582 was followed by Friday, 15 October 1582.
Some non-catholic countries didn't accept the change yet.
Few haven't until 19th or even 20th century.
Meanwhile their error grew by one more day every 127 years.
Now you understand how "October Revolution" in Russia started in 1917 on October 25, while it was November 7 in the rest of the world.
(NASA didn't exist until 1958.)
Additionally, to prevent further accumulation of error, changes were made:
Julian calendar simply adds one leap day every 4 years.
Gregorian calendar does the same, except for years divisible by 100 but not by 400.
So, years 1700, 1800 and 1900 weren't leap years, but 2000 was.
Years 2100, 2200, 2300 won't be leap years, but 2400 will be.
Old Julian calendar makes 1 day error every 127 years.
Gregorian calendar measures 31 556 952 seconds and is in error for 26.7 seconds.
Gregorian calendar is off by 1 day every 3236 years.
What you say would happen with Sun (6 months later on the other side of Earth) actually happens with stars and constellations.
It is because of difference between Solar day and Sideral day.
In different times of the year the night side of Earth points to different stars and constellations.
Ever heard of Summer Triangle, or Winter Hexagon?
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I haven't been following the maths in this closely but I suspect this argument is as spurious as saying:
"Wait, round earthers, so you're claiming there are 365 days in a year and 52 weeks in a year BUT you're also claiming that there are 7 days in a week.
365/52 = 52.142857...
So weeks don't fit into a year. Checkmate!"
A day (on any planet) is how long it takes to rotate on its access.
A year (on any planet) is how long it takes to orbit its star.
Length of a year / length of a day doesn't have to be an integer, it would be weird if it was.
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Sidrael Day is Star Time, and has nothing to do with this. What does it matter how fast the stars are moving in this?
Nobody is saying that it does matter. You seem to be manufacturing something to argue against.
Again, see https://en.wikipedia.org/wiki/Solar_time
References to a 'distant star' are merely as a reference point, a presumed stationary reference point ...
The stars move in the sky slightly slower than the sun. But this has nothing to do with the stars. Whatever you want to interpret about the stars and how the earth moves and where it is assumed to be fixed doesn't matter. In Solar Time the earth rotation is tied to the sun. If the stars disappeared from existence, Solar Time is still wrong.
Lets compare the smallest value to the largest value.
365.24162603 / 24.0000006 = 15.218400704
365.24274049 / 24.0000006 = 15.218447139
The difference isn't anywhere close to a whole number. This is way off.
Once again, you're a dividing an arbitrary number OF days by the number of hours WITHIN a day and expecting some correlation, when none is expected by anyone. ever.
As an aside;
The currency system in the UK used to be pounds, shillings and pence
1 pound = 20 shillings
1 shilling = 12 pence
You're doing the equivalent of
1 (pound) divided by 12 (pence) = 0.08333 .... oh, no, the currency system doesn't work (unless you divide 12 pounds by 12 pence, and get 1; or 24 pounds by 12 pence to get 2, 36 by 12 = 3, 48 by 12 = 4 ......) !
...
What are you talking about? Both of the values I am using involves days. 365.24 days in a Solar Year. 24 hours in a Solar Day. Each of those days in a Solar Year should have 24 hours in it.
We can see just by looking at the numbers that the 24 hour Solar Day won't fit into a 365.24 day year. The year has a decimal point at the end of it! But it's more than that. The closest kind of year a 24 hour Solar Day could fit into is if the earth had a 360 day year, or a 384 day year. A 365.24 day year does not make any sense.
From https://en.wikipedia.org/wiki/Tropical_year --
A tropical year (also known as a solar year) is the time that the Sun takes to return to the same position in the cycle of seasons, as seen from Earth; for example, the time from vernal equinox to vernal equinox, or from summer solstice to summer solstice.
The sun needs to return to the same position in the sky after one Solar Year.
This means that if Solar Noon starts over New York City at the beginning of the circuit, it needs to end with Solar Noon over New York City at the end of the circuit. As one example in the above quote shows, the Solar Time on Vernal Equinox needs to be the same as Vernal Equinox 1 and Vernal Equinox 2.
If a Solar Day can't fit into a Solar Year, that is a huge problem. Where are those extra hours coming from?
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If a Solar Day can't fit into a Solar Year, that is a huge problem. Where are those extra hours coming from?
If a Solar Day is how long it takes the earth to rotate on its axis.
And a Solar Year is how long it takes the earth to orbit the sun.
Why must the ratio of those be an integer? That implies that for every orbit the earth makes of the sun it rotates exactly 'x' times, and x is an integer.
Why should it be? The rotation speed (which, by the way, does change over time, very slowly) and orbit speed are determined by all kinds of things, there is no reason the ratio should be an integer.
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From Mar 1 to Feb 28 Earth turns towards Sun 365 times and in that time travels 359.7612885 degrees around the orbit.
After whole number of 365 turns (not on that date) we reset days/months counter and increase number of years by 1.
After another 365 turns towards Sun it travels another 359.7612885 degrees.
Third time again, fourht time again.
After four sets of 365 days (turns towards Sun) Earth travelled 4 * 359.7612885 = 1439.045154 degrees.
Almost one degree is missing to 1440 (4 * 360 = 1440) and that's where Julian calendar adds Feb 29 to adjust number of own counted years with number of orbits.
So, during full set of four Julian years (one is leap) Earth travels 1440.045154 degrees.
Closer to full set of orbital events than the calendar would be without the Feb 29.
That almost (those 0.045154 degrees) makes Julian calendar still miss for 1 day every 127 years, leading to Gregorian calendar.
People couldn't speed up or slow down the Eaerth in orbit.
They also couldn't speed up or slow down the Earth's rotation to adjust length of solar day to fit in length of tropical year.
All they could do is to adjust the way of counting days in years.
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(...)
If a Solar Day can't fit into a Solar Year, that is a huge problem. (...)
Maybe for you but not for the GET.
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What are you talking about? Both of the values I am using involves days. 365.24 days in a Solar Year. 24 hours in a Solar Day. Each of those days in a Solar Year should have 24 hours in it.
Yes, 365 * 24 = 8760 hours, plus 0.24 of a day (6 hours approx).
Every four years, those 0.24s are rolled into (4 * 0.24, or 4 * 6 hours) = 1 extra day of 24 hours in a leap year.
You agree that humankind as a whole does this every four years, don't you?
We can see just by looking at the numbers that the 24 hour Solar Day won't fit into a 365.24 day year. The year has a decimal point at the end of it! But it's more than that. The closest kind of year a 24 hour Solar Day could fit into is if the earth had a 360 day year, or a 384 day year. A 365.24 day year does not make any sense.
Dividing 365.24 (days) BY 24 (hours) yields a meaningless figure which nobody except you expects to be an integer. The mixture of units makes it a nonsense.
Again, see my currency example and hours/seconds examples above. What am I talking about? I'm drawing an analogy intended to help you understand why expecting an integer from this calculation makes no sense....
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The stars move in the sky slightly slower than the sun. But this has nothing to do with the stars. Whatever you want to interpret about the stars and how the earth moves and where it is assumed to be fixed doesn't matter.
Nobody is saying it has anything to do with the stars, other than them being a reference point outwith the Earth/Sun system. It's nothing to do with any interpretation.
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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:
(https://www.e-education.psu.edu/meteo300/sites/www.e-education.psu.edu.meteo300/files/images/lesson6/EarthOrbit.png)
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.
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?
--- --- ---
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?
--- --- ---
Napkin Calculation 3
Some sources (https://www.google.com/search?client=firefox-b-1&ei=73HcWrTtBou6tgX2yoywCg&q=earth+rotates+%22360.98+degrees%22&oq=earth+rotates+%22360.98+degrees%22&gs_l=psy-ab.3...18960.20962.0.21596.2.2.0.0.0.0.184.336.0j2.2.0....0...1c.1.64.psy-ab..0.1.152...33i160k1.0.RIj9UZSX37E) 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?
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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.
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I may be going about this entirely wrong. Can I have some help with this seemingly glaring problem?
Why should the time it takes for the earth to spin on its axis fit in to how long it takes the Earth to orbit the Sun?
I must be misunderstanding something here otherwise you've gotten yourself into a muddle.
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Why should the time it takes for the earth to spin on its axis fit in to how long it takes the Earth to orbit the Sun?
I must be misunderstanding something here otherwise you've gotten yourself into a muddle.
Yes, I think this is another fine mess he has got himself into.
There is no reason to think that the ratio between the time it takes the planet to orbit the sun and the time it takes the planet to rotate on its access should be an integer. The fact we need leap days and even leap seconds to keep things synced is indication that is is not exact.
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Turns out that the other planets days and years don't throw up integers either, what is going on?
https://www.universetoday.com/37481/days-of-the-planets/
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If a Solar Day can't fit into a Solar Year, that is a huge problem. Where are those extra hours coming from?
If a Solar Day is how long it takes the earth to rotate on its axis.
And a Solar Year is how long it takes the earth to orbit the sun.
Why must the ratio of those be an integer? That implies that for every orbit the earth makes of the sun it rotates exactly 'x' times, and x is an integer.
Why should it be? The rotation speed (which, by the way, does change over time, very slowly) and orbit speed are determined by all kinds of things, there is no reason the ratio should be an integer.
Division of the number of days in a year and the number of hours in a day ( Days in a Year / Hrs in a Day ) must be a whole number because each Day in the Year is a representative of 24 Hours. If this is true then the ratios must relate.
Imagine that we had a planet with a Solar Day that was a 10 hour day.
Imagine that that the Solar Year year of this planet was 100 days.
Lets define Solar Year as the time it takes for the Sun to return back to the same place in the sky in Solar Time. This means that each of the days in the year must be full rotations.
Does a Solar Day fit into a Solar Year?
100 / 10 = 10. Yes. A solar day fits into a Solar Year.
If we mess around those numbers, a solar day no longer fits into a Solar Year.
142 / 10 = 14.2. A solar day does not fit into a Solar Year.
100 / 7 = 16.6. A solar day does not fit into a Solar Year.
A result of a whole number shows that the second value fits into the first value. The only types of Years a 10 hour day would return whole numbers in such a division.
A 10 hour day can fit into a 10 day year (10 / 10 = 1), a twenty day year ( 20 / 10 = 2 ), but not a 25 day year ( 25 / 10 = 2.5 ).
Each 10 doesn't fit nicely into the 25, and the result is 2.5. The Solar Days have not been completed by the time the planet makes its way to the end.
It doesn't make a difference if we call them planets and years or dimes and pennies. The relationship is defined and must be maintained.
Relationship: 100 pennies is 10 dimes. There are 10 pennies in a dime.
We have 142 pennies and 10 dimes. Does the relationship work?
142 pennies / 10 dimes = 14.2. This is not a whole number. 10 dimes does not fit into a value that is 142 pennies.
100 pennies / 7 dimes = 16.2. This is not a whole number. 7 dimes does not fit into a value that is 100 pennies.
10 dimes can fit into 10 pennies (10 / 10 = 1), 20 pennies (20 / 10 = 2), but it cannot fit into 25 pennies (25 / 10 = 2.5).
10 dimes doesn't fit nicely into the 25, and the result is 2.5.
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Turns out that the other planets days and years don't throw up integers either, what is going on?
https://www.universetoday.com/37481/days-of-the-planets/
So a day on Mercury is longer than a year on Mercury.......Tom get your calculator out, I think you can prove Mercury doesn't exist ;)
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If a Solar Day can't fit into a Solar Year, that is a huge problem. Where are those extra hours coming from?
If a Solar Day is how long it takes the earth to rotate on its axis.
And a Solar Year is how long it takes the earth to orbit the sun.
You're answering your own question Tom. The day is the time it takes for earth to rotate and the year is how long it takes for the earth to orbit the sun. These are two independent events with no logical reason why they should fit snugly into an equation together. Humans have jammed and manipulated them to fit in our calendar as best as we can but its not perfect because the universe isn't mathematically perfect. If it was then Pi would be a whole number.
;D
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If a Solar Day can't fit into a Solar Year, that is a huge problem. Where are those extra hours coming from?
If a Solar Day is how long it takes the earth to rotate on its axis.
And a Solar Year is how long it takes the earth to orbit the sun.
Why must the ratio of those be an integer? That implies that for every orbit the earth makes of the sun it rotates exactly 'x' times, and x is an integer.
Why should it be? The rotation speed (which, by the way, does change over time, very slowly) and orbit speed are determined by all kinds of things, there is no reason the ratio should be an integer.
Division of the number of days in a year and the number of hours in a day ( Days in a Year / Hrs in a Day ) must be a whole number because each Day in the Year is a representative of 24 Hours. If this is true then the ratios must relate.
Imagine that we had a planet with a Solar Day that was a 10 hour day.
Imagine that that the Solar Year year of this planet was 100 days.
Lets define Solar Year as the time it takes for the Sun to return back to the same place in the sky in Solar Time. This means that each of the days in the year must be full rotations.
Does a Solar Day fit into a Solar Year?
100 / 10 = 10. Yes. A solar day fits into a Solar Year.
If we mess around those numbers, a solar day no longer fits into a Solar Year.
142 / 10 = 14.2. A solar day does not fit into a Solar Year.
100 / 7 = 16.6. A solar day does not fit into a Solar Year.
A result of a whole number shows that the second value fits into the first value. The only types of Years a 10 hour day would return whole numbers in such a division.
A 10 hour day can fit into a 10 day year (10 / 10 = 1), a twenty day year ( 20 / 10 = 2 ), but not a 25 day year ( 25 / 10 = 2.5 ).
Each 10 doesn't fit nicely into the 25, and the result is 2.5. The Solar Days have not been completed by the time the planet makes its way to the end.
It doesn't make a difference if we call them planets and years or dimes and pennies. The relationship is defined and must be maintained.
Relationship: 100 pennies is 10 dimes. There are 10 pennies in a dime.
We have 142 pennies and 10 dimes. Does the relationship work?
142 pennies / 10 dimes = 14.2. This is not a whole number. 10 dimes does not fit into a value that is 142 pennies.
100 pennies / 7 dimes = 16.2. This is not a whole number. 7 dimes does not fit into a value that is 100 pennies.
10 dimes can fit into 10 pennies (10 / 10 = 1), 20 pennies (20 / 10 = 2), but it cannot fit into 25 pennies (25 / 10 = 2.5).
10 dimes doesn't fit nicely into the 25, and the result is 2.5.
So you're saying there should be some fixed relationship between a planets orbit around the Sun and a planets rotation on its own axis, is that right?
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Division of the number of days in a year and the number of hours in a day ( Days in a Year / Hrs in a Day ) must be a whole number because each Day in the Year is a representative of 24 Hours.
No.
Divide 7 (days in a week) by 24 (hours) and you get 0.29166. So what? Does that mean weeks are wrong?
If there are 20 shillings in one pound, and 12 pence in one shilling, there's (20*12) 240 pence in one pound.
Divide 20 by 12 and you get 1.6666. What does that MEAN to you, in this context, if anything?
If you divide 240 by 12, you get 20, but you have to divide similar units to do this. You're dividing different units, and coming up with something meaningless.
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Division of the number of days in a year and the number of hours in a day ( Days in a Year / Hrs in a Day ) must be a whole number because each Day in the Year is a representative of 24 Hours. If this is true then the ratios must relate.
OK...firstly, the number of hours in a day being 24 is completely arbitrary. It's a definition. I actually don't know where that definition came from but there is nothing magic about the number 24. I guess it was defined as such because it's so divisible, you can divide it by 2,3,4,6,8 and 12 exactly, I suspect this is also why there were 240 pennies in an old British pound, but I digress.
The length of a year and the length of a day are a defined by
a) How long a planet takes to orbit its star
b) How long a planet takes to revolve on its axis
respectively. That applies to any planet.
These are two independent motions, the ratio of them does not need to be exact. We need leap days and leap seconds because it is not exact.
Imagine that we had a planet with a Solar Day that was a 10 hour day.
Imagine that that the Solar Year year of this planet was 100 days.
Lets define Solar Year as the time it takes for the Sun to return back to the same place in the sky in Solar Time. This means that each of the days in the year must be full rotations.
Does a Solar Day fit into a Solar Year?
100 / 10 = 10. Yes. A solar day fits into a Solar Year.
Well, again, the definition of hours in a day is arbitrary. You could still have 24 hours in a day. The important part of your scenario is that the year (how long the planet takes to go around the sun) is an exact number of days (how long the planet takes to rotate).
In that scenario then because 100 is an integer, if we imagine I'm in space above that planet and I'm completely stationary (let's ignore any movement of stars and galaxies). I take a photo of the planet as it is right in front of me and then exactly 100 of that planet's days later I take another photo. In that time the planet has gone around its star once and rotated 100 times while I have stayed exactly where I was. Those two photos should exactly match because 100 is an integer, whatever side of the planet was facing towards me the first time should be facing towards me the second.
If the number of days in a year is not an integer though, as it isn't on earth, then if I took the second photo an exact solar year after the first of earth then the photos would not match, the earth would not be in exactly the same orientation as it was before. What I'm confused about is why you think that's an issue? Who says it has to be? As I said, if that was the case on earth we wouldn't need leap days or seconds, the ratio between the length of earth's year and day would be exact. But it isn't.
The speed the earth orbits the sun and the speed it rotates on its access are completely independent of one another, as someone else has pointed out one of the planets in our solar system actually has a longer day than it does a year, it spins so slowly that it has gone round the sun more than once before it makes a complete revolution around its access.
I don't understand why you think that's an issue.
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They aren't independent variables.
1 Year = 365.24 Days
24 hours = 1 Day
The related variable is Days.
The relationship isn't really any different than 100 pennies = 10 dimes, as demonstrated in my previous post.
The math doesn't need to know about the concept of money or rotations or years. The equation is a simple relationship ratio.
IF there are 24 hours in a day then it only fits into certain kinds of years. Visualize it.
Can a 24 hour day fit into a year that is 0.5 days long? NO
Can a 24 hour day fit into a year that is 2.5 days long? NO
The above examples should be easy to visualize. Very easy. The earth has not finished rotating by the time it reaches those points.
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Division of the number of days in a year and the number of hours in a day ( Days in a Year / Hrs in a Day ) must be a whole number because each Day in the Year is a representative of 24 Hours.
No.
Divide 7 (days in a week) by 24 (hours) and you get 0.29166. So what? Does that mean weeks are wrong?
There is a rule in division that the larger group on the left side needs to be a bigger number than the smaller group on the right side for this to work.
7 / 24 = 0.29166 < --- This is not a valid equation for the purpose
If we change that around to "1 Day is 3 hours. Does a 3 hour day fit into a 6 day week?"
6 / 3 = 2. Whole Number. Works.
Does a 3 hour day fit into a 7 day week?
7 / 3 = 2.33. Not a whole number. A 3 hour day does not fit neatly into a week that is 7 days long. The earth has not finished rotating by the time it reaches that point.
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Again, there being 24 hours in a day is a definition. An hour is a completely made up concept.
A day and a year are not, they have definitions which relate to the movement of the planet, an hour is not it is just an arbitrary sub-division of a day.
Why you think days / hours should be a whole number is beyond me.
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They aren't independent variables.
1 Year = 365.24 Days
24 hours = 1 Day
The related variable is Days.
The relationship isn't really any different than 100 pennies = 10 dimes, as demonstrated in my previous post.
The math doesn't need to know about the concept of money or rotations or years. The equation is a simple relationship ratio.
IF there are 24 hours in a day then it only fits into certain kinds of years.
Can a 24 hour day fit into a year that is 0.5 days long? NO
Can a 24 hour day fit into a year that is 2.5 days long? NO
The above examples should be easy to visualize. Very easy.
A year an a day, only have as much relation to one another as humans have decided they do. Your two examples are incorrect. The length of a day doesn't care how long the length of a year is, and vice versa. There's no mathematical relationship/ratio required between them.
Earth takes 365.24 days to go around the sun. It takes 1 day to go fully rotate around it's axis. All that means is the rate of rotation and the orbital rate are not a perfect match. Which completely expected. As mentioned previously, this is why we have leap days and leap seconds. To keep our calendar roughly on track.
If I have $4.21 in my pocket right now, I can't divide that equally into dimes. Looks like our money system is broken. I can't divide it equally into quarters either! Dear god help us, the economy is ruined! /s
Division of the number of days in a year and the number of hours in a day ( Days in a Year / Hrs in a Day ) must be a whole number because each Day in the Year is a representative of 24 Hours.
No.
Divide 7 (days in a week) by 24 (hours) and you get 0.29166. So what? Does that mean weeks are wrong?
There is a rule in division that the larger group on the left side needs to be a bigger number than the smaller group on the right side for this to work.
7 / 24 = 0.29166 < --- This is not a valid equation for the purpose
If we change that around to "Does a 3 hour day fit into a 6 day week?"
6 / 3 = 2. Whole Number. Works.
Does a 3 hour day fit into a 7 day week?
7 / 3 = 2.33. Not a whole number. A 3 hour day does not fit neatly into a week that is 7 days long.
Does a 24 hour day fit into a 7 day week?
24/7 = 3.429 Not a whole number, a 24 hour day doesn't fit into a 7 day week. Clearly our whole concept of time is broken Tom. Help us figure it out, will you?
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Does a 3 hour day fit into a 7 day week?
7 / 3 = 2.33. Not a whole number. A 3 hour day does not fit neatly into a week that is 7 days long.
Does a 24 hour day fit into a 7 day week?
Does a 52 week year fit into a 365 day year?
IF I said "there are 168 hours in a week and 24 hours in a day" and 168 didn't divide by 24 THEN you would be able to say
"But that means there aren't an exact number of hours in a day" and you'd be right.
But expecting arbitrary definitions like 7 days in a week and 24 hours in a day to correlate to number of days in a year and it have to be a whole number is meaningless.
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Dimensional analysis would really help here.
When multiplying or dividing quantities with units, keep the units in the equation to see what you end up with.
For example, the units of speed are distance/time, for example km/h or miles/h or meters/second. If you do math and you end up with meters^2/second or meters/dollar or something, you didn't calculate a speed.
Let me look at Napkin #5 to illustate this:
Premise:
- A Solar Day happens every 24 hours
- A Year happens every 365 days
365 days / 2 = 185.5 days in six months
Now, we need to fit 24 into 185.5
185.5 days divided by 24 = 7.729
Conclusion: 24 does not fit cleanly into 185.5. If it did, it would be a whole number.
Conclusion: The day does not fit neatly into half of the year. Solar Noon will not be at the same place at that point.
Let me rephrase that with units.
Premise:
- A Solar Day happens every 24 hours. 1 solar day = 24 hours. To get a unitless 1, which is always legal to multiply by, take the ratio. 1 = 1 solar day/24 hours
- A Year happens every 365 days 1 year = 365 days. To get a unitless 1, 1 = 365 days / 1 year
365 days / 2 = 185.5 days in six months
0.5 years is what you're looking for, so you can take 0.5 years * 365 days/year. Years cancel, and you get 185.5 days.
THIS CALCULATION IS CORRECT
Now, we need to fit 24 into 185.5
185.5 days divided by 24 = 7.729
185.5 days / (24 hours/day) = 7.729 days^2/hour
THIS CALCULATION IS MEANINGLESS - your units don't make any sense.
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There is a rule in division that the larger group on the left side needs to be a bigger number than the smaller group on the right side for this to work.
Which 'rule' is that?
7 / 24 = 0.29166 < --- This is not a valid equation for the purpose
No sh*t. The whole point of me quoting it to you is that it doesn't mean anything.... you appear to agree
If we change that around to " Day - 3 hours. Does a 3 hour day fit into a 6 day week?"
6 / 3 = 2. Whole Number. Works.
Six days, if each day is three hours, is a total of (6*3) 18 hours.
That's six separate sets of three hours. A total of 18 hours. Divide the total 18 hours by 6, and you get the length of a day. 3 hours.
Divide the total number of days by 3, and all you get three sets of 2 days.
You've calculated that one-third of a week of six days is two days.
Again, you're dividing Days by Hours to get something meaningless.
Does a 3 hour day fit into a 7 day week?
7 / 3 = 2.33. Not a whole number. A 3 hour day does not fit neatly into a week that is 7 days long. The earth has not finished rotating by the time it reaches that point.
Again, you've calculated that one-third of a 7-day week is 2.33 days.
7 days of 3 hours each is 21 hours. Divide 21 by 7 and you get the length of one day. Divide 21 by 3 and you get 7 days of three hours each.
Dividing 7 by 3 gives you one-third of seven.
Likewise, dividing any number of days by 24 gives you an answer equal to one twenty-fourth of the total number of days.
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days in a year x hours in a day = number of hours in a year
so
number of hours in a year / hours in a day = days in a year
That sort of calculation yields a result which means something.
days in a year / hours in a day = a random number
Especially given the fact that the second of those numbers is a fairly arbitrary definition which apparently comes from the ancient Egyptians
http://www.abc.net.au/science/articles/2011/11/15/3364432.htm
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Not a random number.
The result isn't a "random number". The result is a ratio of how many times the second number fits into the first number.
Premise: We have a pie with really big slices. We have a Pie that is giant slice that takes up 50% of a possible 100%.
If we divide those, the equation 100 / 50 = 2.
Conclusion: Two is not a "random number". 50 fits into 100 two times perfectly. The pie can be filled with two of those equally sized slices.
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Premise: We have a pie with really big slices. We have a Pie that is giant slice that takes up 30% of a possible 100%.
If the pie is 30% filled, the equation is 100 / 30 = 3.333...
Conclusion: 3.333... means that 30% fills pie 3.333... times.
Conclusion: Since 3.333... times isn't a whole number, 30% isn't an even ratio if 100%
Conclusion: 30% does not work. The pie slice would actually need to fill the pie by 33.333...% to fit into the pie in equal slices.
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Not a random number.
The result isn't a "random number". The result is a ratio of how many times the second number fits into the first number.
Premise: We have a pie with really big slices.
We have a Pie that is 50% filled of a possible 100%.
If we divide those, the equation 100 / 50 = 2.
Conclusion: Two is not a "random number". 50 fits into 100 two times perfectly. The pie can be filled with two of those equally sized slices.
If the pie is 30% filled, the equation is 100 / 30 = 3.333...
Conclusion: 3.3333 means that 30 fills 100 3.333... times.
Conclusion: Since 3.333 times isn't a whole number, 30% isn't an even ratio if 100%
Conclusion: The pie would need to be 33.333...% filled to fit into the 100% pie in equal slices. 30% does not work.
Great job Tom. Do you at last understand why a year is 365.24 days and not just 365 days if nothing else?
As for 'not a random number' you are correct. However it is a meaningless number in this circumstance. It tells you nothing of value. Even if we assume there were 365 days in a year exactly, what do we get when we divide that by 24? Our answer of 15.2 is equally meaningless.
Let's try another.
1 day is 24 hours.
1 hour is 60 minutes.
60/24 = 2.5
Huh, guess minutes don't fit equally into a day either. It's like our whole time system and the concept of it is made up to fit some arbitrary duration.... oh wait.
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Continuing my last scenario:
1 Slice = 1 Day
Each Slice is divided into 24 smaller slices (lets call them MiniSlices)
We know that 1 Slice goes, allegedly, into a pie 365.24 times (Year)
We can do 365.24 / 1 to see if that works. The result is 365.24. Not a whole number. 1 Slice cannot fit a pie evenly.
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We can also get rid of the concept of the single slice entirely and just focus on the MiniSlices.
We have 24 MiniSlices in the pie of equal length. Will they fit into the pie?
365.24 / 24 = 15.21. The answer is No. They will not fit.
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Not a random number. The result isn't a "random number". The result is a ratio of how many times the second number fits into the first number.
Again, you're deriving a meaningless ratio from two separate, distinct units. First unit is days, second is hours.
365 divided by 24 = 15.2083
One twenty-fourth of 365 is 15.2083
But you have 365 or 365.24 days, and each of those days is 24 hours, so you have 365 or 365.24 sets of 24. Dividing 365 or 365.24 by 24 merely tells you that one twenty-fourth of 365 or 365.24 is 15.2083 or 15.2183 of those sets.
You appear to be trying to convince yourself with this that 365.24 days is not a whole number of days. But we know that.
365 days is a whole number of days. The clue that 365.24 days is not a whole number is that it has 0.24 in it.
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Continuing my last scenario:
1 Slice = 1 Day
Each Slice is divided into 24 smaller slices (lets call them MiniSlices)
We know that 1 Slice goes, allegedly, a pie 365.24 times (Year)
We can do 365.24 / 1 to see if that works. The result is 365.24. Not a whole number. 1 Slice cannot fit a pie evenly.
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We can also get rid of the concept of the single slice entirely and just focus on the MiniSlices.
We have 24 MiniSlices in the pie of equal length. Will they fit a pie that has 365.24 slots?
365.24 / 24 = 15.21. The answer is No. They will not fit.
Which proves...what exactly? Also your second idea doesn't work anyway. Again, even if we had an exactly even 365 days in a year, you would get an integer with a remainder. Despite 1 day fitting evenly into 365.
You're not proving anything here. You're showing 1 orbit =/= an even number of days. Which is already well known and understood. Hence leap year to keep the calendar approx. correct.
So what exactly do you think this is showing? Other than the fact you either don't understand basic math, don't understand the difference between an orbital period and a rotational period, or don't understand those two have no need to be related in any fashion. You started out wondering how the sun is always up at noon throughout the year (which I felt was shown rather well) and have moved on to...what exactly?
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1 Slice = 1 Day
Each Slice is divided into 24 smaller slices (lets call them MiniSlices)
We know that 1 Slice goes, allegedly, a pie 365.24 times (Year)
So you have 365.24 sets of 24 minislices
We can do 365.24 / 1 to see if that works. The result is 365.24. Not a whole number. 1 Slice cannot fit a pie evenly.
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We can also get rid of the concept of the single slice entirely and just focus on the MiniSlices.
We have 24 MiniSlices in the pie of equal length. Will they fit into the pie?
You have space for 365.24 sets of 24 minislices, not 24 minislices
365.24 / 24 = 15.21. The answer is No. They will not fit.
Now all you've done is divide the space up differently.
First, you divided it into 365.24 sets of 24 minislices. You've divided it into 8765.76 minislices.
Now you've divided it up into 24 equal parts. 15.21 'units' is one twenty-fourth of your available space, which you said was 365.24. This is unrelated to any part of your slices or minislices.
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The problem you seem to have with the above account with the MiniSlices example, is why am I using the 365.24 / 24 at the end? Years and hours?
Well, that is the same ratio as if we multiplied both of the numbers by 24 and found the lowest common denominator.
1 Slice = 24 Minislices
365.24 x 24 = 8765.76 hours in the year
24 x 24 = 576 hours in 24 days (24 days picked because we need equivalence to the ratio)
A ratio of 8765.76 / 576 is equivalent to 365.24 / 24.
8765.76 / 576 = 15.21
365.24 / 24 = 15.21
Equivelent
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The problem one might have with the above account with the MiniSlices, is why am I using the 365.24 / 24 at the end?
Well, that is the same ratio as if we multiplies both of those numbers by 24 and found the lowest common denominator.
1 Slice = 24 Minislices
365.24 x 24 = 8765.76
24 x 24 = 576
A ratio of 8765.76 / 576 is equivalent to 365.24 / 24.
8765.76 / 576 = 15.21
365.24 / 24 = 15.21
Tom, do you need to go lie down for a bit? You're honestly just spouting nonsense that's unrelated to any of the issues people are bringing up, and you still have yet to explain what you're attempting to prove/show here in any way. Your numbers are meaningless in context, and multiplying everything by 24 (or any number) doesn't show or solve anything.
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The result is 365.24. Not a whole number. 1 Slice cannot fit a pie evenly.
Seriously what?? You are aware that non whole numbers are just as valid as whole numbers yes? Have you ever split the last slice of pizza in half? In your pie example so what if the slices don't fit exactly as whole numbers?!
This is completely irrelevant to the two separate factors of how long a day and year are.
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The problem one might have with the above account with the MiniSlices, is why am I using the 365.24 / 24 at the end?
Well, that is the same ratio as if we multiplies both of those numbers by 24 and found the lowest common denominator.
1 Slice = 24 Minislices
365.24 x 24 = 8765.76
24 x 24 = 576
A ratio of 8765.76 / 576 is equivalent to 365.24 / 24.
8765.76 / 576 = 15.21
365.24 / 24 = 15.21
Tom, do you need to go lie down for a bit? You're honestly just spouting nonsense that's unrelated to any of the issues people are bringing up, and you still have yet to explain what you're attempting to prove/show here in any way. Your numbers are meaningless in context, and multiplying everything by 24 (or any number) doesn't show or solve anything.
This is simple math.
A grouping of Twenty Four 24 hour days has 576 hours.
24 x 24 = 576
A year has 8765.76 hours in it.
365.24 x 24 = 8765.76
The ratio is 8765.76 hours / 576 hours
I am no longer using the year / day calculation.
Can we see if a grouping of Tewenty Four 24 hour days (576 hours) fits into a year that is 8765.76 hours long?
8765.76 hours / 576 hours = 15.21. No. It does not fit.
This is the same ratio as 365.24 / 24, and gives the same answer.
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The problem you seem to have with the above account with the MiniSlices example, is why am I using the 365.24 / 24 at the end? Years and hours?
Well, that is the same ratio as if we multiplies both of the numbers by 24 and found the lowest common denominator.
1 Slice = 24 Minislices
365.24 x 24 = 8765.76 hours in the year
24 x 24 = 576 hours in 24 days (24 days picked because we need equivalence to the ratio)
A ratio of 8765.76 / 576 is equivalent to 365.24 / 24.
8765.76 / 576 = 15.21
365.24 / 24 = 15.21
Equivelent
So are you saying that you believe the number of times the planet spins on its axis should fit into a single orbit the planet takes around the sun nice and neatly?
If so, then why do you feel this should be the case?
I genuinely don't get it.
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The problem one might have with the above account with the MiniSlices, is why am I using the 365.24 / 24 at the end?
Well, that is the same ratio as if we multiplies both of those numbers by 24 and found the lowest common denominator.
1 Slice = 24 Minislices
365.24 x 24 = 8765.76
24 x 24 = 576
A ratio of 8765.76 / 576 is equivalent to 365.24 / 24.
8765.76 / 576 = 15.21
365.24 / 24 = 15.21
Tom, do you need to go lie down for a bit? You're honestly just spouting nonsense that's unrelated to any of the issues people are bringing up, and you still have yet to explain what you're attempting to prove/show here in any way. Your numbers are meaningless in context, and multiplying everything by 24 (or any number) doesn't show or solve anything.
This is simple math.
A grouping of Twenty Four 24 hour days has 576 hours.
24 x 24 = 576
A year has 8765.76 hours in it.
365.24 x 24 = 8765.76
The ratio is 8765.76 hours / 576 hours
I am no longer using the year / day calculation.
Can we see if a grouping of Tewenty Four 24 hour days (576 hours) fits into a year that is 8765.76 hours long?
8765.76 hours / 576 hours = 15.21. No. It does not fit.
This is the same ratio as 365.24 / 24, and gives the same answer.
What are you trying to prove here Tom? That there isn't a whole number of days in one orbital year/period? We know this, hence leap year. That this somehow invalidates...something? How? Why? What? Why can't the rate of rotation NOT be an even ratio to the rate of orbit? What prevents this?
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So are you saying that you believe the number of times the planet spins on its axis should fit into a single orbit the planet takes around the sun nice and neatly?
If so, then why do you feel this should be the case?
I genuinely don't get it.
Look into definitions of Solar Time and Solar Year and Solar Day. We have quoted them in this thread.
From https://en.wikipedia.org/wiki/Tropical_year --
A tropical year (also known as a solar year) is the time that the Sun takes to return to the same position in the cycle of seasons, as seen from Earth; for example, the time from vernal equinox to vernal equinox, or from summer solstice to summer solstice.
We looked at the real numbers earlier in this thread.
The rest of this thread seems to be just clarification about how division and ratios work, to check if the math of dividing years by days was correct.
What are you trying to prove here Tom? That there isn't a whole number of days in one orbital year/period? We know this, hence leap year. That this somehow invalidates...something? How? Why? What? Why can't the rate of rotation NOT be an even ratio to the rate of orbit? What prevents this?
Look up the definition of a Solar Year. The sun needs to return to its same position. Solar Noon needs to be the same after a Solar Year.
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So are you saying that you believe the number of times the planet spins on its axis should fit into a single orbit the planet takes around the sun nice and neatly?
If so, then why do you feel this should be the case?
I genuinely don't get it.
Look into definitions of Solar Time and Solar Year and Solar Day. We have quoted them in this thread.
From https://en.wikipedia.org/wiki/Tropical_year --
A tropical year (also known as a solar year) is the time that the Sun takes to return to the same position in the cycle of seasons, as seen from Earth; for example, the time from vernal equinox to vernal equinox, or from summer solstice to summer solstice.
We looked at the real numbers earlier in this thread.
The rest of this thread seems to be just clarification about how division and ratios work, to check if the math of dividing years by days was correct.
What are you trying to prove here Tom? That there isn't a whole number of days in one orbital year/period? We know this, hence leap year. That this somehow invalidates...something? How? Why? What? Why can't the rate of rotation NOT be an even ratio to the rate of orbit? What prevents this?
Look up the definition of a Solar Year. The sun needs to return to its same position. Solar Noon needs to be the same after a Solar Year.
Surely the point is that the sun is in the same position, but the earth is at a different rotation because the earths rotation is not linked to its orbit.
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This is simple math.
This is condescension
A grouping of Twenty Four 24 hour days has 576 hours.
24 x 24 = 576
A year has 8765.76 hours in it.
365.24 x 24 = 8765.76
The ratio is 8765.76 hours / 576 hours
What ratio? All you've done is divide the number of hours in a year by the number of hours in 24 days. Why would you do this?
Can we see if a grouping of Twenty Four 24 hour days (576 hours) fits into a year that is 8765.76 hours long?
8765.76 hours / 576 hours = 15.21. No. It does not fit.
So what? What does that prove?
This is the same ratio as 365.24 / 24, and gives the same answer.
Again, all you're doing with this calculation is deriving one twenty-fourth of a year. You're not deriving any 'ratio' between anything that needs to be ratioed. You're calculating that 1/24th of a year is 15.21 days.
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This is simple math
It is simple, it's just a bit meaningless and doesn't demonstrate anything useful.
So you have a pie with 365 slices (let's just go with 365, for simplicity).
That year pie can be divided into 365 day-length slices.
And each of those slices can be divided into 24 hour mini-slices.
That means our pie has 365 x 24 = 8760 hour mini-slices.
But dividing the number 365 by the number 24 does not give you any meaningful result.
Multiplying them does because each of the 365 slices has 24 mini slices, so multiplying them gives you the total number of mini-slices in the pie.
But dividing them doesn't give you anything useful.
All it tells you is if you divided the whole pie into 24 slices, instead of 365, then in order to get 365 slices you'd have to divide each of the 24 slices into 15.2083333... slices. Meaningless. (If you want to try this at home you'd actually split 23 of the slices into 15 and the 24th into 20).
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The rest of this thread seems to be just clarification about how division and ratios work, to check if the math of dividing years by days was correct.
... but you didn't divide years by days.
You've persistently divided 365.24 by 24 (days by hours) to arrive at a number of days that represents one twenty-fourth of a year (15.21).
We already know that a day of 24 hours doesn't fit exactly into a year of 365.24 days. The clue is in the 0.24. Otherwise the year would be 365 days (365 sets of 24 hours), an exact number. Dividing this by 24 still gives one twenty-fourth of a year. This is why we have leap years.
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I predict #95 will be Tom's "claim of victory" post, and nothing further will be forthcoming.
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What are you trying to prove here Tom? That there isn't a whole number of days in one orbital year/period? We know this, hence leap year. That this somehow invalidates...something? How? Why? What? Why can't the rate of rotation NOT be an even ratio to the rate of orbit? What prevents this?
Look up the definition of a Solar Year. The sun needs to return to its same position. Solar Noon needs to be the same after a Solar Year.
Oh, awesome. We can /thread this then, because if you haven't noticed the equinoxes don't occur on the same day every year. Because our calendar system isn't defined by solar noons. In fact, solar noon for any location isn't even regular in time between them, due to variances in the Earths orbit and rotation mostly. See: https://www.timeanddate.com/astronomy/solar-noon.html Elastic Solar Noon.
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This is simple math.
This is condescension
A grouping of Twenty Four 24 hour days has 576 hours.
24 x 24 = 576
A year has 8765.76 hours in it.
365.24 x 24 = 8765.76
The ratio is 8765.76 hours / 576 hours
What ratio? All you've done is divide the number of hours in a year by the number of hours in 24 days. Why would you do this?
Can we see if a grouping of Twenty Four 24 hour days (576 hours) fits into a year that is 8765.76 hours long?
8765.76 hours / 576 hours = 15.21. No. It does not fit.
So what? What does that prove?
This is the same ratio as 365.24 / 24, and gives the same answer.
Again, all you're doing with this calculation is deriving one twenty-fourth of a year. You're not deriving any 'ratio' between anything that needs to be ratioed. You're calculating that 1/24th of a year is 15.21 days.
You are right. It also represents 15.21 days. 1/24th of a year is 15.21. It can also be interpreted like that. But division of this manner can also be interpreted on how many times the smaller number on the right fits into the bigger number on the left. The result should be a whole number.
The rest of this thread seems to be just clarification about how division and ratios work, to check if the math of dividing years by days was correct.
... but you didn't divide years by days.
You've persistently divided 365.24 by 24 (days by hours) to arrive at a number of days that represents one twenty-fourth of a year (15.21).
We already know that a day of 24 hours doesn't fit exactly into a year of 365.24 days. The clue is in the 0.24. Otherwise the year would be 365 days (365 sets of 24 hours), an exact number. Dividing this by 24 still gives one twenty-fourth of a year. This is why we have leap years.
362.24 / 24 is equivalent in ratio to as if we broke this out by hours, as I wrote about on the previous page.
A Solar Day needs to fit exactly into a Solar Year of 365.24 days because the sun needs to be in the same place in the sky at that point. 12 PM Solar Noon at the start needs to end up at 12 PM Solar Noon at the end.
What are you trying to prove here Tom? That there isn't a whole number of days in one orbital year/period? We know this, hence leap year. That this somehow invalidates...something? How? Why? What? Why can't the rate of rotation NOT be an even ratio to the rate of orbit? What prevents this?
Look up the definition of a Solar Year. The sun needs to return to its same position. Solar Noon needs to be the same after a Solar Year.
Oh, awesome. We can /thread this then, because if you haven't noticed the equinoxes don't occur on the same day every year. Because our calendar system isn't defined by solar noons. In fact, solar noon for any location isn't even regular in time between them, due to variances in the Earths orbit and rotation mostly. See: https://www.timeanddate.com/astronomy/solar-noon.html Elastic Solar Noon.
If you try to use the amount of time the earth "actually rotates at" in relation to the stars, the Sidreal Day, it doesn't work either.
The Solar Year isn't based on the 365 day calendar. The Solar Year is ~365.24 days. Leap year is incorporated into the 365 day calendar to try and account for that.
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You are right. It also represents 15.21 days. 1/24th of a year is 15.21. It can also be interpreted like that. But division of this manner can also be interpreted on how many times the smaller number on the right fits into the bigger number on the left. The result should be a whole number.
WHY?! Why should it?
Why should the number of DAYS in a year divided by the number of HOURS in a day be a whole number?! They are completely different units.
The number of HOURS in a year divided by the number if HOURS in a day (if we are simplifying and assuming exactly 365 days and using that to calculate the number of hours in a year) will be a whole number because the unit is the same. But if you mix up the units you get a completely meaningless number.
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No-one but you is saying that a solar day should fit exactly into a solar year though.
This is extraordinary.
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You are right. It also represents 15.21 days. 1/24th of a year is 15.21. It can also be interpreted like that. But division of this manner can also be interpreted on how many times the smaller number on the right fits into the bigger number on the left. The result should be a whole number.
Only if the units used on each side of the divisor MATCH. They do not.
At least two others have stated your calculation to be meaningless as well as I, but you persist.
What do you claim to have 'proved' here?
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No-one but you is saying that a solar day should fit exactly into a solar year though.
This is extraordinary.
Thing is, even if it did all that would mean would be a year is exactly 'x' days where 'x' is an integer.
How 'x' is divided into 'y' hours is irrelevant and 'x/y' would still be a meaningless number whether it happened to be an integer or not.
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You are right. It also represents 15.21 days. 1/24th of a year is 15.21. It can also be interpreted like that. But division of this manner can also be interpreted on how many times the smaller number on the right fits into the bigger number on the left. The result should be a whole number.
WHY?! Why should it?
Why should the number of DAYS in a year divided by the number of HOURS in a day be a whole number?! They are completely different units.
The number of HOURS in a year divided by the number if HOURS in a day (if we are simplifying and assuming exactly 365 days and using that to calculate the number of hours in a year) will be a whole number because the unit is the same. But if you mix up the units you get a completely meaningless number.
I have already broken it out by hours. If you take a group of 24 days compared to the 365.24 day year, the number of hours in both is equivalent in ratio to 365.24 days / 24 hours.
This is simple math.
A grouping of Twenty Four 24 hour days has 576 hours.
24 x 24 = 576
A year has 8765.76 hours in it.
365.24 x 24 = 8765.76
The ratio is 8765.76 hours / 576 hours
I am no longer using the year / day calculation.
Can we see if a grouping of Tewenty Four 24 hour days (576 hours) fits into a year that is 8765.76 hours long?
8765.76 hours / 576 hours = 15.21. No. It does not fit.
This is the same ratio as 365.24 / 24, and gives the same answer.
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If you take a group of 24 days compared to the 365.24 day year, the number of hours in both is equivalent in ratio to 365.24 days / 24 hours.
Of course it is. 365.24 / 24 gives the same result as (365.24*24) / (24*24)
So what, though? Why take a group of 24 days in the first place?
A week is seven days. It's not a multiple of that.
It's not the number of days in any month of the year.
It seems to be an arbitrary number you've picked at random.
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Let me have another go. Let's assume there are:
'd' days in a year,
'h' hours in a day,
'm' minutes in an hour and
's' seconds in a minute.
Out of those only 'd' is a number dictated by the globe earth, it's the number of days it takes the planet to orbit its star. And let's pretend it's a whole number.
h, m and s are all arbitrary.
You can divide a day how you like
You can divide an hour how you like
You can divide a minute how you like.
We as humans have agreed on 24, 60 and 60 but we could have equally chosen other numbers and it would make no difference.
Hours in a year = d x h
Minutes in a year = d x h x m
Seconds in a year = d x h x m x s
It's not particularly useful to know the number of seconds in a year but it does at least mean something.
And you can divide that by d, h, m or s and get a whole number, because it is the product of the 4 numbers.
Divide by s and you get back to minutes in a year.
But if you divide by h, what do you get? It would be an integer but is it a meaningful one?
I'm not sure it would be, it doesn't represent anything. It would be a 24th of a year, in seconds.
I guess it's the number of seconds between, say, midnight and 1am each day in a year.
So what? It doesn't mean anything.
The point is there is a hierarchy in time, year, day, hour, minute, second.
You can't just arbitrarily divide one of these by another and expect to get a result which means anything.
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OK, so what informs the assumption that you must have a whole number of days in a year. Put another way, why must the orbital period of a planet, and the rotational period of a planet, be evenly divisible? I have a planet that takes 12 hours to complete a full rotation. It takes 117 hours to complete an orbital cycle. Can this planet exist? Why or why not?
Earth happens to have a relatively close relationship between the two. But it's not exact. This is why the equinoxes vary year to year, why we need leap years, and why local noon doesn't always correspond to solar noon. These are conventions and similar to adjust for the fact that, as you've pointed out, there isn't an even ratio between a rotational period, and an orbital period.
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i see what's happening here. tom is assuming too much about this diagram:
(https://www.e-education.psu.edu/meteo300/sites/www.e-education.psu.edu.meteo300/files/images/lesson6/EarthOrbit.png)
tom, this diagram is schematic. you're assuming that the earth returns precisely to position 1 (the top position, let's say) at the end of a full period. it doesn't.
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To be fair that diagram even says that each point isn't the same day each year so there is a clue there.
Without leap days it would drift over time.
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Division of the number of days in a year and the number of hours in a day ( Days in a Year / Hrs in a Day ) must be a whole number because each Day in the Year is a representative of 24 Hours. If this is true then the ratios must relate.
"Year in calendar" is specific number of days, not specific number of degrees that Earth travelled around Sun.
We are tryig to make them fit, that's why our calendar year sometimes has additional day.
"Tropical year" is not in calendar, it is in astronomy, and it is not based on whole number of solar days.
It is based on location in orbit, regardless of the direction in which the same meridian looks.
Calendar year and Tropical year are two different things.
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If you take a group of 24 days compared to the 365.24 day year, the number of hours in both is equivalent in ratio to 365.24 days / 24 hours.
Of course it is. 365.24 / 24 gives the same result as (365.24*24) / (24*24)
So what, though? Why take a group of 24 days in the first place?
A week is seven days. It's not a multiple of that.
It's not the number of days in any month of the year.
It seems to be an arbitrary number you've picked at random.
It is not a random number. When you are multiplying the Year by 24 then you are breaking out the units that are contained within the Day. You need to multiply the day by 24 as well to maintain the equivalence.
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You need to multiply the day by 24 as well to maintain the equivalence.
Tom there is NO equivalence! You're trying to balance two completely separate events! If this thread was an animal any decent vet would put it out of its misery... :-[
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If you take a group of 24 days compared to the 365.24 day year, the number of hours in both is equivalent in ratio to 365.24 days / 24 hours.
Of course it is. 365.24 / 24 gives the same result as (365.24*24) / (24*24)
So what, though? Why take a group of 24 days in the first place?
A week is seven days. It's not a multiple of that.
It's not the number of days in any month of the year.
It seems to be an arbitrary number you've picked at random.
It is not a random number. When you are multiplying the Year by 24 then you are breaking out the units that are contained within the Day. You need to multiply the day by 24 as well to maintain the equivalence.
Then you're not doing anything to the equation. So of course it gives you the same answer back.
24/8=3
(24*8)/(8*8)=3
The issue is the units you're using. Currently you're dividing days by hours. Which doesn't give a meaningful answer. f.e. 7 days/24 hours=0.2916 days/hours. What is that? Nothing. It has no inherent meaning.
When you multiply you get the same issue. (7 days*24 hours)=168 hours/wk. (24 hours*24 hours)=576 hours. Sure they're both ostensibly in hours, but you've still go that 'wk' bit technically. You can't do anything with the pair of them that means anything.
Again, your claim is that x solar rotations, must go evenly into 1 orbital cycle. Unfortunately, you're just incorrect. They have no required correlation.
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your claim is that x solar rotations, must go evenly into 1 orbital cycle.
Tom, is this what you are asserting or do you wish to retract this ideal?
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It is not a random number. When you are multiplying the Year by 24 then you are breaking out the units that are contained within the Day. You need to multiply the day by 24 as well to maintain the equivalence.
At least three others assert this to be meaningless. Why are you persisting with it?
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Again, I see that you guys are trying to divide smaller numbers by bigger numbers to disprove this. You can't do that in division. The first number needs to be the biggest hierarchical group in the situation and the largest number.
You need to multiply the day by 24 as well to maintain the equivalence.
Tom there is NO equivalence! You're trying to balance two completely separate events! If this thread was an animal any decent vet would put it out of its misery... :-[
1 Year = 365.24 Days
1 Days = 24 Hours
365.24 Days / 24 Hours <--- You say this is invalid
But look at these:
--- --- ---
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 Mile = 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.
--- --- ---
I just divided unlike units and got a right answer :o
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Again, I see that you guys are trying to divide smaller numbers by bigger numbers to disprove this. You can't do that in division.
Why not? Says who, apart from you?
Are you telling me I can't divide 1 by 4 to yield 0.25, or one quarter anymore?
1 Year = 365.24 Days
1 Days = 24 Hours
365.24 Days / 24 Hours <--- You say this is invalid
It means nothing to everyone else. What does it mean to you? 365.24/24 = 1/24th of 365.24. Nothing else.
1 Dollar = 10 Dimes
1 Dime = 10 Pennies
10 Dimes / 10 Pennies = 1 <--- But this is correct. 10 Dimes fits into 10 Pennies 1 time
Once again,
1 Pound = 20 shillings
1 Shilling = 12 Pence
20 / 12 = 1.66666 - Your example only comes out with a whole number because of the 10/10 relationship of cents to dimes AND dimes to dollars
1 Yard = 5280 Feet
1 Foot = 12 inches
5280 Feet / 12 inches = 440 <--- This is correct as well. 12 Inches can fit neatly into 5280 feet 440 times
Total hogwash. 1 yard is 3 feet. 440 yards is a quarter mile, 5280 feet is 1 mile
This one only works out because
1 foot = 12 inches
3 feet = 1 yard
440 yards = quarter mile
1760 yards = 1 mile, so (1760*3) 5280 feet = 1 mile
Divide 5280 feet by 12 and you get 440 feet. Which is the same result as dividing 5280 by 4 to get one quarter mile (= 440 yards), then dividing that by 3 to get 440 feet (4*3 = 12).
You've picked another where the relationship between the units is built upon the figure that you're dividing into them, so it will naturally work out as a whole number.
Try it on anything where the relationship is not built this way, and it falls apart. See my example above.
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My mistake. You are correct. I meant to write "12 fits neatly into 5280 440 times"
----
1 Pound = 20 shillings
1 Shilling = 12 Pence
20 / 12 = 1.66666
This is actually correct. If you have 12 groups of 12 units, it doesn't fit into 20 x 12. The labels are arbitrary and do not represent real world things of real units.
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Again, I see that you guys are trying to divide smaller numbers by bigger numbers to disprove this. You can't do that in division. The first number needs to be the biggest hierarchical group and the largest number.
You need to multiply the day by 24 as well to maintain the equivalence.
Tom there is NO equivalence! You're trying to balance two completely separate events! If this thread was an animal any decent vet would put it out of its misery... :-[
1 Year = 365.24 Days
1 Days = 24 Hours
365.24 Days / 24 Hours <--- You say this is invalid
But look at these:
--- --- ---
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.
--- --- ---
1 Yard = 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.
The 1 Foot = 12 inches is implicit in the above equation.
--- --- ---
I just divided unlike units and got a right answer :o
I just want to point out this dimes/pennies one. Because your answer is total hogwash, and I can't believe you don't see it. 10 dimes doesn't fit into 10 pennies 1 time. 10 pennies doesn't fit into 10 dimes 1 time either. You've provided the exact example to help show your final statement false, within your attempt to prove it true. :O indeed. Not to mention none of these have any sort of units attached to them.
I like your last one too. Feet/inches. Just what is your unit here? 12 fits nicely into 5280, and? You've created a nonsense number again. It doesn't actually have any meaning beyond the number of times 12 fits into 5280. It's not the number of inches in 5280 feet, its not the number of feet in 5280 feet. It's a relatively meaningless number.
My mistake. I meant to write "12 fits neatly into 5280 440 times"
----
1 Pound = 20 shillings
1 Shilling = 12 Pence
20 / 12 = 1.66666
This is actually correct. If you have 12 groups of 12 units, it doesn't fit into 20. The values are arbitrary and do not represent real world things of real units.
Oh Tom, this is the best yet! Pound, pence, and shilling aren't real world things of real units? According to who? Do they not exist in your world? They're just a 'fake' as dimes, pennies, and dollars!
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This applies to measuring systems that are constant as ratios.
The dollar is divided by dimes and pennies, so it will work.
The mile is divided by foot and by inches, so it will work.
British currency values are not based on constants like that.
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This applies to measuring systems that are constant.
The dollar is divided by dimes and pennies, so it will work.
The mile is divided by foot and by inches, so it will work.
British currency values are not based on constants like that.
The pound was divided by shillings and pennies, in the same way that your dollar is divided by dimes and cents/pennies.
However;
1 Dollar = 4 Quarters
1 Quarter = 25 Cents
So .... 4/25 = ... what?
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This applies to measuring systems that are constant.
The dollar is divided by dimes and pennies, so it will work.
The mile is divided by foot and by inches, so it will work.
British currency isn't based on constants like that.
Ah, so if it doesn't support your arbitrary hypothesis, we can just toss it out. Got it.
I guess we're also going to ignore how none of your answers are 'right' in so far as actually describing any sort of unit then, as I also pointed out? That they're just largely meaningless numbers?
Or maybe we should discuss the fact that the reason for your dismissal of British currency applies equally to rotational duration and orbital duration?
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Can someone give me a TLDR of the thread? I'm really not getting the argument that Tom is trying to make.
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tom's argument (correct me if i'm wrong, tom) is that there must be an integer number of rotations in a single orbital period.
i still don't understand where he's getting that from.
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Can someone give me a TLDR of the thread? I'm really not getting the argument that Tom is trying to make.
TL;DR (I think) Because the number of hours in a day don't go neatly into the number of days in a year, the RE model can't be correct.
If I've erred, a correction would be greatly appreciated however.
tom's argument (correct me if i'm wrong, tom) is that there must be an integer number of rotations in a single orbital period.
i still don't understand where he's getting that from.
I think it has to do with solar noon year to year. He doesn't seem to get that solar noon on important days (like the equinoxes) don't happen at the same time on the calendar every year. That, in fact, the only reason they even stay close is because of adjustments such as leap year.
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tom's argument (correct me if i'm wrong, tom) is that there must be an integer number of rotations in a single orbital period.
i still don't understand where he's getting that from.
I don't either. Why does he think this should be if he thinks this? Does he not know leap years exist to act as a correction?
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This applies to measuring systems that are constant.
The dollar is divided by dimes and pennies, so it will work.
The mile is divided by foot and by inches, so it will work.
British currency isn't based on constants like that.
Ah, so if it doesn't support your arbitrary hypothesis, we can just toss it out. Got it.
I guess we're also going to ignore how none of your answers are 'right' in so far as actually describing any sort of unit then, as I also pointed out? That they're just largely meaningless numbers?
Or maybe we should discuss the fact that the reason for your dismissal of British currency applies equally to rotational duration and orbital duration?
The rotational and orbital duration should have constant ratios. It is not merely arbitrary that the number of hours in a day should fit into the number of days in a year.
This applies to measuring systems that are constant.
The dollar is divided by dimes and pennies, so it will work.
The mile is divided by foot and by inches, so it will work.
British currency values are not based on constants like that.
The pound was divided by shillings and pennies, in the same way that your dollar is divided by dimes and cents/pennies.
However;
1 Dollar = 4 Quarters
1 Quarter = 25 Cents
So .... 4/25 = ... what?
As I described earlier, the first number needs to be the larger one and the greatest hierarchical unit in the scenario.
Can someone give me a TLDR of the thread? I'm really not getting the argument that Tom is trying to make.
TL;DR (I think) Because the number of hours in a day don't go neatly into the number of days in a year, the RE model can't be correct.
If I've erred, a correction would be greatly appreciated however.
tom's argument (correct me if i'm wrong, tom) is that there must be an integer number of rotations in a single orbital period.
i still don't understand where he's getting that from.
I think it has to do with solar noon year to year. He doesn't seem to get that solar noon on important days (like the equinoxes) don't happen at the same time on the calendar every year. That, in fact, the only reason they even stay close is because of adjustments such as leap year.
The Solar Year says that the sun should be in the same place in the sky after 1 Solar Year.
A visualization.
The sun travels across the earth's surface each day.
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.
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The Solar Year says that the sun should be in the same place in the sky after 1 year.
this is the discrepancy. that's not really true. you're conflating two different quantities.
what i mean to say is: the solar year is not precisely one period long. the sun is not in exactly the same place in the sky after one period of the earth's orbit. there is not an integer number of rotations in one period.
this article does a pretty good job of describing all of the different ways to define a year: http://www.orlandosentinel.com/news/space/go-for-launch/os-think-you-know-how-many-days-are-in-a-year-think-again-20170228-story.html
none of them are an integer numbers of days. that's not part of the model.
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The Solar Year says that the sun should be in the same place in the sky after 1 year.
this is the discrepancy. that's not really true. you're conflating two different quantities.
what i mean to say is: the solar year is not precisely one period long. the sun is not in exactly the same place in the sky after one period of the earth's orbit. there is not an integer number of rotations in one period.
What do you mean? The sun isn't in the same place in the sky after a Solar Year? It says that in the definition:
https://en.wikipedia.org/wiki/Tropical_year
A tropical year (also known as a solar year) is the time that the Sun takes to return to the same position in the cycle of seasons, as seen from Earth; for example, the time from vernal equinox to vernal equinox, or from summer solstice to summer solstice.]A tropical year (also known as a solar year) is the time that the Sun takes to return to the same position in the cycle of seasons, as seen from Earth; for example, the time from vernal equinox to vernal equinox, or from summer solstice to summer solstice.
this article does a pretty good job of describing all of the different ways to define a year: http://www.orlandosentinel.com/news/space/go-for-launch/os-think-you-know-how-many-days-are-in-a-year-think-again-20170228-story.html
none of them are an integer numbers of days. that's not part of the model.
The Solar Day should fit into the Solar Year. The Solar Day is 24 hours long.
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What do you mean? The sun isn't in the same place in the sky after a Solar Year? It says that in the definition:
https://en.wikipedia.org/wiki/Tropical_year
yes, that's the definition of a tropical year; but, i mean that one tropical year is not precisely the same duration as the time it takes the earth to go around the earth one time.
one tropical year is not equal to one period (one orbit of the earth around the sun).
the earth does not rotate on its axis an integer number of times in one orbit of the sun.
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What do you mean? The sun isn't in the same place in the sky after a Solar Year? It says that in the definition:
https://en.wikipedia.org/wiki/Tropical_year
yes, that's the definition of a tropical year; but, i mean that one tropical year is not precisely the same duration as the time it takes the earth to go around the earth one time.
one tropical year is not equal to one period (one orbit of the earth around the sun).
the earth does not rotate on its axis an integer number of times in one orbit of the sun.
It says right in the link that a Solar Year is 354.24 days. Why isn't it equal to one sun rotation around the earth if the sun needs to get back into the same place in the sky according to the definition?
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It says right in the link that a Solar Year is 354.24 days. Why isn't it equal to one sun rotation around the earth if the sun needs to get back into the same place in the sky according to the definition?
because the earth moves a little bit along its orbit during one rotation. i think someone already posted this, but here is it again:
(https://upload.wikimedia.org/wikipedia/commons/8/8b/Sidereal_day_%28prograde%29.svg)
also, a solar day is not equal to one rotation around the earth's axis. that's a sidereal day.
the earth does not rotate on its axis an integer number of times in one orbit, so neither of these (solar or sidereal days) are evenly divisible into one period.
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It says right in the link that a Solar Year is 354.24 days. Why isn't it equal to one sun rotation around the earth if the sun needs to get back into the same place in the sky according to the definition?
because the earth moves a little bit along its orbit during one rotation. i think someone already posted this, but here is it again:
(https://upload.wikimedia.org/wikipedia/commons/8/8b/Sidereal_day_%28prograde%29.svg)
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.
The orbital path is elongated, however. This is true.
On Page 3 I posted a quote showing how little the Solar Year varies from where you measure it:
And the Topical Year varies on solar return points (which is why they called it the 'mean' tropical year):
http://calendars.wikia.com/wiki/Tropical_year
Current values and their annual change of the time of return to the cardinal ecliptic points[2] are:
vernal equinox: 365.24237404 + 0.00000010338×a days
northern solstice: 365.24162603 + 0.00000000650×a days
autumn equinox: 365.24201767 − 0.00000023150×a days
southern solstice: 365.24274049 − 0.00000012446×a days
It varies very little.
also, a solar day is not equal to one rotation around the earth's axis. that's a sidereal day.
the earth does not rotate on its axis an integer number of times in one orbit, so neither of these (solar or sidereal days) are evenly divisible into one period.
The Sidrael Day is a rotation calculated based how much the stars move in relation to the sun. The stars move slightly slower than the sun. Lets assume that the stars don't exist and only focus on Solar Day. According to Solar Day the sun moves around the earth once in 24 hours. Form the last exercise I did with miles around the earth's circumference, which showed the sun traveling over all of the earth's circumference, we can see that the sun is offset at the end.
The sun should have returned to the same position over the earth. It did not.
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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.
The sun should have returned to the same position over the earth. It did not.
there is no such requirement in ret.
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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.
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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.
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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.
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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.
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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.
which is longer than the time required for the earth to rotate once with respect to its axis. a solar day is longer than a sidereal day.
sigh. last try: http://astro.unl.edu/naap/motion3/sidereal_synodic.html
A synodic year is the time it takes for a planet-sun alignment to reoccur. For the case of the sun, it is the time it takes the sun to come to the same place on the ecliptic (equinox to equinox) and is called a Tropical Year. A tropical year is 365.242 mean solar days (366.242 sidereal days). It is just over 20 minutes shorter than a sidereal year (again, the effect of precession).
everyone in astronomy is already very well aware that the earth does not rotate an integer number of times in one orbit. there's no reason it should have to. it would be a noteworthy coincidence if it did.
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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.
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Tom, a solar year and a solar day (despite having similar names) have no required correlation. Presuming they do is getting you into trouble. A solar year is defined by the sun returning to a specific ecliptic, or arc through the sky. NOT a specific point IN the sky. You've been missing this important point this whole time it seems, it looks like it hasn't been included when you copy the definitions over. The equinoxes are pretty normally spaced apart. Yet they can vary (according to the calendar) between some point in two different days. Do you wonder why that is if they're so consistent? It's because of the difference between Earth's orbital period, and it's rotational period. Which, once again, have no requirement to be correlated in any way outside of your own mind.
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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 is still wrong.
Lets try to do the math on this hunch you have.
The Sidrael Day is about 4 seconds less than the Solar Day.
4 x 364.24 / 60 / 60 = 0.4072 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.
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.
Why do you need to discuss this here when there are explanations online. Do you think you have discovered something?
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The rotational and orbital duration should have constant ratios.
This bit is actually true. The ratio between how long the earth takes to rotate on its axis - 1 day, that is actually the definition of a day - and how long the earth takes to orbit the sun should be constant. And it is, it’s 365.24. That IS the ratio, and it is constant but it is not an integer, nor does it need to be. There is no law of physics which says it does. If you put a spinning top on a roundabout and spin the top and the roundabout at certain speeds the number of times the top spins while the roundabout revolves once does not need to be an integer. It would be a pretty amazing coincidence if it was.
It is not merely arbitrary that the number of hours in a day should fit into the number of days in a year.
Yes it is.
They are completely different units.
The number of HOURS in a day should fit into the number of HOURS in a year but that’s about it.
The number of days in a year is defined by how many times the earth rotates as it orbits the sun - can we agree that this number is fairly random? All the planets rotate at different speeds depending on their distance from the sun.
The number of hours in a day is a definition which seems to come from the ancient Egyptians. It isn’t random but it is fairly arbitrary and comes from the mists of time. It could have been defined differently.
There is no reason that two numbers like this should have an integer ratio. It could be on some planets the number of days in a year is a prime number and nothing will divide by it evenly. We have 12 months and they aren’t even the same number of days each because 365/12 isn’t an integer.
What if we introduce a new coin into the American system, let’s call it the Tom. And a Tom is 6 cents. So if I want to pay for something which costs 30 cents I could pay using 5 Toms. If something costs a dollar then I can’t pay for it exactly in Toms because 100/6 is not an integer so you could argue it’s not that useful a coin, but it could exist and you could use it. It doesn’t break the whole concept of American money or mean dollars can’t exist.
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Tom is messing with us, he knows full well what he's saying is nonsense but he obviously gets a kick out of the confusion he causes.
I'm not feeding this particular troll anymore.
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Tom is messing with us, he knows full well what he's saying is nonsense but he obviously gets a kick out of the confusion he causes.
I'm not feeding this particular troll anymore.
I have wondered this. He's either pretty dumb or pretty clever and playing us. I honestly can't work out which.
But his lack of understanding of lots of areas of maths and science is pretty poor - and, strangely, coupled with an unshakeable confidence in his understanding of them.
Odd. Or, as you say, possibly he's just trolling. Hard to tell.
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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?
.. because they form (essentially) a stationary reference point outwith our solar system
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There is precession in Earth's orbit, hence the difference between Sidereal year and Tropical year.
Sidereal year is time for 360 degrees around Sun.
It is 31 558 149.5 seconds.
You can measure it, for example, when Earth takes the same position between Sun and, say, Bode's Galaxy (M81), at 12 million light years.
Solar year is between consequent repetition of the same orbital event, like equinox, or solstice.
On average it is 31 556 925 secons.
As you can see, the difference is 1224.5 seconds (20 min 24.5 sec).
Sidereal day is time between consequent facing of the same meridian towards the same distant star, or, for example, the same Bode's Galaxy.
It is 86 164.1 second.
Solar day is the time between consequent facing of the same meridian towards Sun.
We defined our day and hour by that.
On average it is 86400 seconds.
Additionally:
Tropical year varies due to gravitational influence of other planets.
Planets and Moon wobble Earth in orbit, Jupiter even wobbles Sun.
Here is the variation of Tropical year from 1900 to 2099:
https://www.timeanddate.com/astronomy/tropicalyearlength.html (https://www.timeanddate.com/astronomy/tropicalyearlength.html)
Solar day varies plus or minus 29 seconds, depending on Earth's orbital speed variations, explained by Kepler's Second Law.
We still use Solar day to count our calendars, not Sidereal day.
That's because what synchronizes our lives is Sun, not distant stars.
(Unless our superstition makes us believe in astrology.)
So, what would be the most convenient for us:
1. Whole number of sidereal days in sidereal year?
2. Whole number of sidereal days in tropical year?
3. Whole number of solar days in sidereal year?
4. Whole number of solar days in tropical year?
The number 4, ofcourse, but we don't have either.
All we can do is count our days and organize them into our calendar years the best we can,
to follow seasonal markers (solstices and equinoxes) as close as possible.
The closest way was Gregorian calendar.
For now it is close enough.
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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?
The Earth doesn't rotate slower towards the stars, it faces same distant star 4 minutes earlier than it faces Sun, because of one day movement along the orbit.
They matter to show that "360 degrees" and "one solar day" are not the same thing.
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Tom is messing with us, he knows full well what he's saying is nonsense but he obviously gets a kick out of the confusion he causes.
I'm not feeding this particular troll anymore.
Our personalities are not important here.
Only facts and figures.
And it doesn't depend on Tom's or anyone else's "trolling" or "lack of understanding".
If he's trolling then it's easy: no reason to worry about his mental abilities.
If he doesn't understand, try to simplify and help him.
If you can.
In both cases, it is important for other readers to understand as well.
And that doesn't depend on WHO said it, but on WHAT has been said.
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It is not merely arbitrary that the number of hours in a day should fit into the number of days in a year.
No, the number of HOURS in a day should fit the number of HOURS in a year. There's no requirement for the number of days to fit, for, as you keep saying, a year is 365.24 days. The clue that this is not a multiple of 24 is in that 0.24 that's additional to the 365.
(1 Dollar = 4 Quarters
1 Quarter = 25 Cents
So .... 4/25 = ... what?)
As I described earlier, the first number needs to be the larger one and the greatest hierarchical unit in the scenario.
so ...
9 dollars = 36 quarters
1 quarter = 25 cents
36/25 = 1.44
10 dollars = 40 quarters
1 quarter = 25 cents
40/25 = 1.60
A visualization.
The sun travels across the earth's surface each day.
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.
This visualisation fails because it takes no account of axial tilt and oblateness of the Earth, and because it takes no account of the movement of Earth in its orbit. The sun travels a greater distance in the sky than the circumference, because once the Earth has moved on in its orbit, it needs to rotate more than one circumference to place the sun in the same position. Back to that diagram above showing how you need a reference point outwith the solar system to show this (reply #135, and the wikipedia that I quoted around page 2 or 3 ....)
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Tie motorized model of airplane to a rope and let it fly around some pole.
Why the propeller doesn't turn the whole number of times during one revolution around the pole?
----------------------------------------------------
What we count is not Tropical year, nor Sidereal year.
We count Calendar year.
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Lets try to do the math on this hunch you have.
By all means, let us do that. Then we can explore how condescending this comment is, when your basis for this whole thread is a big 'hunch' ....
Seriously, there is not ONE comment in this thread that I can find that agrees with you. Nobody supports your position. Nobody agrees with your calculations. Every comment from everyone who is not you either questions your premise, or explicitly disagrees with your figures.
What does this tell you?
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Nobody supports your position.
To be honest it's not even clear what Tom's position is anymore, he's been digging himself into a hole so deep it's all gone dark and the lack of Oxygen is doing strange things to him...
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To be honest it's not even clear what Tom's position is anymore, he's been digging himself into a hole so deep it's all gone dark and the lack of Oxygen is doing strange things to him...
Honestly no idea why he things the number of days in a year (which is defined as the number of times the earth rotates as it makes one orbit of the sun, and isn't actually an integer a value itself) and the number of hours in a day (which is defined arbitrarily based on a system the ancient Egyptians made up and could easily be defined a different way) should have an integer ratio.
It's this which makes me think he might be trolling, but agree with Macarios that even if he is other people will see and understand the arguments for what they are.
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In sort of related news:
According to Rubeus Hagrid...
A Galleon is made up of 17 Sickles and
A Sickle is made up of 29 Knuts.
http://harrypotter.wikia.com/wiki/Wizarding_currency
But wait...29/17 isn't an integer, nor is 17/29. Help!
It's almost like JK made up some numbers. BUT...that currency system would work. It would be a bit inconvenient to work with but you could use it.
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When you multiply you get the same issue. (7 days*24 hours)=168 hours/wk. (24 hours*24 hours)=576 hours. Sure they're both ostensibly in hours, but you've still go that 'wk' bit technically. You can't do anything with the pair of them that means anything.
Again, your claim is that x solar rotations, must go evenly into 1 orbital cycle. Unfortunately, you're just incorrect. They have no required correlation.
24 hours * 24 hours = 576 hours^2.
24 days * 24 hours/day = 576 hours.
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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 Yard = 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 :o
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?
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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 :o
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.
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A large number of feet should divide into a whole number of inches. Diving by different variables does work.
Right so...2939 feet divided by 12 inches = 244.916666
Help! I broke distances!
Per the earth hours and days in the year, those should be multiples too.
OK, once again: Why would you think that the number of days in a year (which is the number of times the earth rotates in one orbit and an accurate value of which is not itself an integer) should be a multiple of 24, an arbitrary division of a day made up by the ancient Egyptians.
They are two completely unrelated numbers.
The whole cycle, seasons and all, should be divisible by the hour.
If you measure the year in hours, sure. Then that should divide by 24 (if you've simplified and used an integer number of days to do the calculation).
But if you don't, or if you mix up days and hours you'll get a meaningless answer.
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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.
...and our system of seconds, minutes, hours, days, etc isn't?
One works on multiples of 10 and the other system works on multiples of 8 (there is also the furlong between the gap of feet and miles, but no one uses that anymore).
Actually, someone does. Horse races are measured in furlongs.
My examples make sense.
There are many here who disagree
A large number of feet should divide into a whole number of inches.
etc etc
The whole cycle, seasons and all, should be divisible by the hour.
Why? Says who, apart from you?
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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.
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Again, it doesn't work between all units of measurement. There must be a common multiple. Not all measurement systems are constants.
Right!
And here's the point. There is no common multiple between days in a year and hours in a day.
They are defined completely separately in different ways for different reasons.
The first of them, if we're being accurate, isn't even an integer.
This is why your divisions aren't working.
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Dear Mr. Bishop,
Could you, please, help me understand why you think a single orbit around the Sun should consist of an integer of rotations (of the Earth, of course)? Thank you.
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Dear Mr. Bishop,
Could you, please, help me understand why you think a single orbit around the Sun should consist of an integer of rotations (of the Earth, of course)? Thank you.
I second that request.
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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?
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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), the 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 higher hierarchical entity of the sub-degree, similar to how 100% of a pie is the higher hierarchical entity of the pie slices that fit within it.
--- --- ---
Step 2.
Not lets cut the circle in Step 1. and lay the line out flat on a circle. Lets also rename Degrees to Mega-Lengths and Sub-Degrees to Sub-Lengths now for less confusion.
We have 360 Mega-Legths. Each Mega Units has 24 Sub-Lengths.
360 Mega-Lengths / 24 Sub-Lengths = 15
The Sub-Length needs to fit into the 360 Mega-Length. Same thing, we are just 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 is maintained.
If we consider the 365.24 days year as the length of the oval. We can call the days Mega-Lengths and the hours Sub-Lengths.
Now lay the lines down on a flat surface. We are working with lengths now, like the above analogy.
We have 365.24 Mega-Units.
We want to see if 24 Sub-Units fit into it..
365.24 / 24 = 15.21
This is not a while number. The Sub-Units does not fit into the whole Mega-Unit length.
--- --- ---
Does that make a little more sense?
You've misunderstood what I've written. I understand your calculations (possible flaws aside). I'm asking you why this would pose a problem to the GET.
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2939 inches divided by 12 inches in a foot = 244.916666 feet. The foot does not fit into 2939 inches.
Look! You just did units correctly!
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.
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.
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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.
You have that backwards. The 365.24 days in the year came first, and the degrees in the circle came as an approximation to that. Nobody knows for sure who did it, but it's pretty clear that it was chosen because 360 is a nice number that is divisible by lots of factors. It's even possible that some civilizations thought that the year was 360 days long. If so, it is because they either hadn't perfected astronomy or they let dogma drive their decisions.
That doesn't change the fact that the number of days in a year is 365.24
It's an oval. But ovals still have 360 degrees in them, so the analogy with Step 1 is maintained.
If we consider the 365.24 days year as the length of the oval. We can call the days Mega-Lengths and the hours Sub-Lengths.
Now lay the lines down on a flat surface. We are working with lengths now, like the above analogy.
We have 365.24 Mega-Units.
We want to see if 24 Sub-Units fit into it..
365.24 / 24 = 15.21
This is not a while number. The Sub-Units does not fit into the whole Mega-Unit length.
--- --- ---
Does that make a little more sense?
Consider the nautical mile. This is defined as the distance between 1 minute of latitude lines on the earth. I know that you believe that latitude is a real thing that can be measured by looking at the angle to the north star, independent of whether there's a globe or not.
So, people went out and measured and determined that the nautical mile is 6072.12 feet. Would you insist that this can't be true because the nautical mile must be an integer number of feet?
People went out and measured and determined that the year is 365.24 days long. Why would you insist this can't be true because it can't be evenly divided by 360 or any factors thereof?
Neither number is nice and divisible, but the universe is under no obligation to use nice round numbers or to be easily understandable by humans.
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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! :D
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.
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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 higher hierarchical entity of the sub-degree, similar to how 100% of a pie is the higher 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 needs to fit into the 360 Mega-Length. Same thing, we are just 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.
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 365.24 Mega-Lengths.
We want to see if 24 Sub-Lengths fit into it..
365.24 / 24 = 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?
The earth doesn't orbit the sun at a constant speed, so how would that ever work unless the earths spin varied its speed to match?
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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! :D
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.
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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.
Yes, (24*360) 8640 of them will. 24 sub-degrees still only fills one degree, though.
360 degrees / 24 sub-degrees = 15
Only an integer because you've picked two numbers which neatly fit. 36/24 = 1.5, so 360/24 = 15.
One twenty-fourth of 360 is 15. 24 times 15 = 360. So what?
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You are saying that it's a "coincidence" that 360 is a multiple of 24.
No, I think you picked those figures to fit your calculation. No coincidence involved, other than in your head
But the reason for defining it that way doesn't matter. It's a multiple.
It's not just meaningless numbers being divided together.
That's exactly what it is
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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.
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I've been trying to follow the discussion, but I've lost the thread. Seems there's some meta-debate.
But as for solar vs. sidereal days, I sketched this up to try to illustrate the relationship between the rotation of the earth vs its orbit about the sun.
(http://oi64.tinypic.com/vrv5nt.jpg)
Might help resolve the meta issue. Might not. Not sure.
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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 MediumLegoBlock = 16 SmallLegoBlocks
64 MegaLegoBlocks / 16 SmallLegoBlocks = 4. Whole number.
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 Lego Block 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.
Great, you've discovered multiplication tables and such.
Tell us what bearing this has upon the rotational duration of the Earth and its orbit around the Sun.
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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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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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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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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 (https://forum.tfes.org/index.php?topic=9478.msg148374#msg148374).
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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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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.
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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.
(http://oi67.tinypic.com/29wn8u1.jpg)
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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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.
(http://oi67.tinypic.com/29wn8u1.jpg)
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 (https://forum.tfes.org/index.php?topic=9478.msg148392#msg148392).
And this explanation for why we can also divide 365.24 by 24 (https://forum.tfes.org/index.php?topic=9478.msg148374#msg148374) 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 (https://forum.tfes.org/index.php?topic=9478.msg148280#msg148280).
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What IS the problem?
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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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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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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.)
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Per the Sidrael Day comment I refer you to this post which shows that the Sidrael Day is not the solution to this problem (https://forum.tfes.org/index.php?topic=9478.msg148280#msg148280).
"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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Per the Sidrael Day comment I refer you to this post which shows that the Sidrael Day is not the solution to this problem (https://forum.tfes.org/index.php?topic=9478.msg148280#msg148280).
"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.
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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.
I haven't checked your math. I'm just noting confusion of terms/variables.
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Dear Mr. Bishop,
Please, do not avoid my question. Why would this "issue" pose a problem to the GET?
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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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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
(https://i.imgur.com/aUWxCjg.jpg)
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).
...
(https://upload.wikimedia.org/wikipedia/commons/thumb/7/7a/Equation_of_time.svg/500px-Equation_of_time.svg.png)
...
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.
(https://i.imgur.com/WqIgmjd.png)
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.
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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.
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See https://www.timeanddate.com/astronomy/tropical-year.html
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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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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.
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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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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See https://www.timeanddate.com/astronomy/tropical-year.html
That link does not explain where the extra hours comes from. See The Problem (https://forum.tfes.org/index.php?topic=9478.msg148424#msg148424) post.
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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:
(https://www.e-education.psu.edu/meteo300/sites/www.e-education.psu.edu.meteo300/files/images/lesson6/EarthOrbit.png)
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 (https://www.google.com/search?client=firefox-b-1&ei=73HcWrTtBou6tgX2yoywCg&q=earth+rotates+%22360.98+degrees%22&oq=earth+rotates+%22360.98+degrees%22&gs_l=psy-ab.3...18960.20962.0.21596.2.2.0.0.0.0.184.336.0j2.2.0....0...1c.1.64.psy-ab..0.1.152...33i160k1.0.RIj9UZSX37E) 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.
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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.
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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 (https://forum.tfes.org/index.php?topic=9478.msg148424#msg148424) 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 (https://forum.tfes.org/index.php?topic=9478.msg148424#msg148424) post.
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That link does not explain where the extra hours comes from. See The Problem (https://forum.tfes.org/index.php?topic=9478.msg148424#msg148424) post.
The “extra hours” aren’t “extra.” They’re just the difference between sidereal and solar references.
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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 (https://forum.tfes.org/index.php?topic=9478.msg148424#msg148424) 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 (https://forum.tfes.org/index.php?topic=9478.msg148424#msg148424) post.
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That link does not explain where the extra hours comes from. See The Problem (https://forum.tfes.org/index.php?topic=9478.msg148424#msg148424) 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.)
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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?
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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.
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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. 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.
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.
Equinox is just the sun passing the equatorial plane, its not a fixed point.
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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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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:
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.
(https://i.imgur.com/WqIgmjd.png)
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
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).
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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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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.
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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
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
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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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.
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.
(...)
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?
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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.
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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.
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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.
The equinox shifts 6 hours a year?
The equinox's variation occurs with a rotation of about once every 25,772 years. The shift takes a very long time. At the moment The Equinox is aligned with the constellation of Pisces, and we are moving into the "Age of Aquarius." The time between Zodiac points is about 2,150 years.
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.
The Equinox shifts with a rotation of about once every 25,772 years, due to the movement of the Galaxy.
Why is it shifting 6 hours a year? What kind of Galaxy do we live in?
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)
There are 24 hours in a Solar Day. There are 354.24 Solar Days in a Solar Year.
From https://en.wikipedia.org/wiki/Tropical_year we read
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).
The Solar Year is defined at the moment where sun intersects the Equinox, which only occurs twice a year (March Equinox and September Equinox). The Time between one March Equinox to the next March Equinox is 354.24 Solar Days, with incredibly slight variation at decimal points beyond that.
Where do the extra hours come from?
The Equinox moves 6 hours in a year? The Equinox moves due to the rotation of the Galaxy, and a quarter rotation of the Equinox takes many eons.
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Tom, we're up to 12 pages now, and as far as I can see, there's NOBODY who even slightly agrees with you.
Don't you think this might be telling you .... something?
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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)
There are 24 hours in a Solar Day. There are 354.24 Solar Days in a Solar Year.
From https://en.wikipedia.org/wiki/Tropical_year we read
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).
The Solar Year is defined where the celestial equator of the sun intersects with the equinox, which only occurs twice a year (March Equinox and September Equinox). The Time between one March Equinox to the next March Equinox is 354.24 Solar Days, with incredibly slight variation at decimal points beyond that.
Where do the extra hours come from?
The Equinox moves 6 hours in a year? The Equinox moves due to the rotation of the Galaxy, and a quarter rotation of the Equinox takes many eons.
That's not an answer to my question. Nothing there says a solar year is an integer multiple of solar days. It in fact explicitly defines it as NOT being such. Still not seeing the problem here. A solar year and a solar day have nothing to do with one another. One is in no way dependent on the other. A solar year happens. A solar day happens. We can subdivide a solar year by the number of solar days, but there's no guarantee of an integer value.
The Equinox is 6 hours different from the last one, according to our clocks, which are based around the timing of the solar day. They care nothing about the duration of a solar year. Looking at the background stars, the equinox is far more regular.
Once again. Show me anywhere that states a solar year should be evenly divisible by a solar day, or indeed that they have any mathematical correlation that must result in an integer. They are based upon two separate occurrences, that have no need, nor even reason, to be related. You appear to be the only one claiming they have to be, but have yet to present any evidence that indicates they should be, beyond things you apparently cannot be bothered to fully read.
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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.
The equinox shifts 6 hours a year?
The equinox's variation occurs with a rotation of about once every 25,772 years. The shift takes a very long time. At the moment The Equinox is aligned with the constellation of Pisces, and we are moving into the "Age of Aquarius." The time between Zodiac points is about 2,150 years.
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.
The Equinox shifts with a rotation of about once every 25,772 years, due to the movement of the Galaxy.
Why is it shifting 6 hours a year? What kind of Galaxy do we live in?
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)
There are 24 hours in a Solar Day. There are 354.24 Solar Days in a Solar Year.
From https://en.wikipedia.org/wiki/Tropical_year we read
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).
The Solar Year is defined at the moment where sun intersects the Equinox, which only occurs twice a year (March Equinox and September Equinox). The Time between one March Equinox to the next March Equinox is 354.24 Solar Days, with incredibly slight variation at decimal points beyond that.
Where do the extra hours come from?
The Equinox moves 6 hours in a year? The Equinox moves due to the rotation of the Galaxy, and a quarter rotation of the Equinox takes many eons.
Yes, equinoxes do, in fact, "get delayed" by 6 hours every year. (https://www.timeanddate.com/calendar/seasons.html)
Where do the extra hours come from?
Didn't you start this thread to prove that days don't fit into one year? That's the reason exactly why equinoxes shift (see the image below).
(https://www.scienceabc.com/wp-content/uploads/2016/10/Leap-year-earth-orbit-sun-calendar-29-february-space-year-end.jpg) (https://www.scienceabc.com/wp-content/uploads/2016/10/Leap-year-earth-orbit-sun-calendar-29-february-space-year-end.jpg)
(Not mine, click to visit)
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Tom, we're up to 12 pages now, and as far as I can see, there's NOBODY who even slightly agrees with you.
Don't you think this might be telling you .... something?
Not quite true. I agreed with him when he said that 360 divides exactly by 24. I didn’t agree with him if he thought there was any significance in that.
I think the trouble here is Tom seems to think that the number of rotations the earth makes as it orbits the sun should be an integer. But it isn’t and there is no reason why it should.
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The equinox shifts 6 hours a year?
The equinox's variation occurs with a rotation of about once every 25,772 years. The shift takes a very long time. At the moment The Equinox is aligned with the constellation of Pisces, and we are moving into the "Age of Aquarius." The time between Zodiac points is about 2,150 years.
You are still trying to decipher where the "extra hours" in the solar year come from, right? So you need to talk apples and apples. The variation you're speaking of there has to do with movement of equinox along the plane of earth's orbit, aka the ecliptic. That's oranges. The apples are the yearly shift in time when the equinox occurs from an earth perspective, and that is just under 6 hours each year. As in about 1/4 of a 24 hour day. That's where you're "extra hours" are coming from.
If you've moved on from that new "problem" too and are talking now about something else dealing with the cyclic precession of the earth's rotational axis, I apologize. But it seems to me you're getting yourself wound around the axle, trying to work in all the complex motions of earth's rotation and orbit to understand the simple notion that the earth doesn't rotate a whole number of days during the course of one orbit around the sun. That's all the extra hours are is that those two parameters (solar day and solar year) aren't integrated as a whole number.
Forget the 25K year precession cycle. It has nothing (very little) to do with the so-called "problem" you think you've identified.
The Salon article you linked was pretty good, I thought. Cheeky explanation of what can be a confusing subject, I know. Your first problem (which apparently wasn't resolved to your satisfaction) was why at the exact halfway point of a solar year, NYC isn't oriented toward the sun as you presumed it should be. Your second was the "mystery" of where the "extra hours" came from in a solar year. Both of those questions are addressed in the Salon article pretty well. You may not like that it "doesn't fit" in an orderly way, but the universe doesn't conform to how we wish we could measure and quantify. We choose from nature what we want to use as a standard reference, and if we find that there's variation or deviation from what we thought, we either have to adjust our methods or pick a new standard. We can't expect the universe to conform to what we think would make calculations easier or more aesthetically ordered.
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In 2016, the March equinox occurred at 04:30 (UTC) on March 20th.
Start the clock. Start the calendar.
The earth keeps rotating and begins its orbital trek around the sun. 365 days after we've started our clock, we approached 04:30 UTC on March 20th, 2017. The earth had rotated 365 times in relation to the sun (366 times in relation to the stars) during that trek. But the earth hasn't reached the same spot in its orbit as it was when we started the clock. It still has a little ways to go.
So the earth keeps spinning and we don't hit the equinox point (plane) until 10:28 UTC on March 20th, 2017. It doesn't change anything as far as our clocks are concerned. It still feels like 10:28 (or whatever time zone you're in). But the time of day of reaching that orbital spot has gotten later by almost 6 hours. We note it, but the earth is still spinning on its axis and we keep the clock rolling and keep ticking off the calendar as we start a new trip around the sun.
Another year passes and we confront the same situation as before. It's now March 20th, 2018, but the time of equinox is later in the day, again. Now it's 16:15 UTC. Same reason as before. The earth hasn't completed a full orbit in exactly 365 calendar days and still needs to finish the last bit of that orbit. But the earth keeps rotating and a little under 6 hours, we reach the equinox.
That's two trips around the sun and we've seen the equinox slip 11 hrs and 30 minutes. It isn't a set time difference each time. It varies, but it's within a range of 5 1/2 to 6 hours. But still we roll with it, start the trip around the sun again. On March 20th, 2019, the equinox arrives 21:58 UTC. Same reason. Same approximate time slippage.
A fourth trip around the sun produces yet another slippage in the time of equinox. It's now occurring at 03:50 UTC. But what day? If we didn't input a "leap day" sometime prior, the equinox would fall on 03:50 21 March (UTC). But we do, adding a day to February 2020 and catch the calendar up all in one fell swoop. We basically add 24 "extra hours" to make up for the fact that our calendar had slipped 23 hrs and 40 mins over the previous 4 orbits around the sun, using the equinox as a marker.
It's not perfect because we haven't restored the equinox to the exact minute that it was back in 2016, but that variance takes much longer to matter to our calendar date keeping. But those ~6 "extra hours" per year are simply due to the fact that the solar day DOESN'T "fit" into the solar year as a whole integer. The earth's rotation isn't perfectly matched with the earth's orbit about the sun. We complete the 365 solar rotations (or 366 sidereal rotations) just a little short of 1 full orbit about the sun.
And yes, there are other factors that affect the calculus, but those are less impacting on our annual calendars as is the mismatch of solar day with solar year. Precession, wobble, drag...some of those effects take centuries, millenia or eons to matter, phenomenological. They do matter to astronomers who must be precise, but not to our daily calendar life. Those "extra hours" do matter to us regular folks because we'd notice that seasons would appear to gradually shift during the calendar if we didn't make those "leaping" adjustments.
It would be nice and tidy if we didn't have to worry about those adjustments and the time piece of the sun/earth was tuned for our convenience. But it's not, and it doesn't have to be. We get "extra hours" in a solar year because of that and they don't appear out of nowhere.
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This is how our calendar based on counting solar days misses the orbital events like solstices and equinoxes,
and how we adjust our calendar every 4 years to make it as close as possible to those events:
http://www.astropixels.com/ephemeris/soleq2001.html (http://www.astropixels.com/ephemeris/soleq2001.html)
Pay attention to 2004, 2008, 2012...
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What I am hearing is that the Solar Day does not match up to a Solar year that starts and stops on the Equinox. You are calling the two are "arbitrarily" in relation to each other. But the March and September Equinoxes are at a certain point on the earth's path around the sun. The earth will be in the same physical position looking at the sun.
If the Solar day at the start of the day is over New York City, it must end on New York City.
The Equinox only varies over eons, as the Equinox and Solstice lines in the below diagram rotates according to the Precession of the Equinoxes at a period of 25,772 years.
(https://s3.amazonaws.com/ai2-vision-textbook-dataset/dataset_releases/rc2/train/question_images/seasons_6284.png)
See the diagram above. The Solar Day has to be the same when the earth returns back to the same Equinox point in relation to the sun. It's looking at the sun at the same position.
Why is the Solar Day not the same? What is causing the variance?
The diagram is not an exact representation of everything Tom. Don't get caught up in it please. If that's where this entire issue arises we might have bigger problems. There's a reason the equinoxes and the solstices vary from year to year. They are not exactly on the same time every year. Do you know why that is? The answer to that question is the same as yours, and it's because the Earth's rotation around it's axis, and the Earth's orbit around the sun are not related. One is not dependent on the other. They happen to match up quite well, but it's pure coincidence.
You're right though, the March and September equinoxes happen at about the same time every year. Because of leap year. Without leap year, as mentioned a number of times now, seasons would eventually flip based on the calendar month. This is why we have leap year, so that doesn't happen. The Earth isn't changing it's orbit, or it's rotation. We're changing how we track both of these things in order to keep things similar to what everyone expects.
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I was editing my post and lost one. Here it is again. I added on to it at the bottom.
What I am hearing is that the Solar Day does not match up to a Solar year that starts and stops on the Equinox. You are calling the two are "arbitrarily" in relation to each other. But the March and September Equinoxes are at a certain point on the earth's path around the sun. The earth will be in the same physical position looking at the sun.
If the Solar day at the start of the day is over New York City, it must end on New York City.
The Equinox only varies over eons, as the Equinox and Solstice lines in the below diagram rotates according to the Precession of the Equinoxes at a period of 25,772 years.
(https://s3.amazonaws.com/ai2-vision-textbook-dataset/dataset_releases/rc2/train/question_images/seasons_6284.png)
See the diagram above. The Solar Day has to be the same when the earth returns back to the same Equinox point in relation to the sun. It's looking at the sun at the same position.
Why is the Solar Day not the same? What is causing the variance?
Go here:
http://astro.unl.edu/classaction/animations/coordsmotion/siderealSolarTime.html
(https://cdn.pbrd.co/images/HijFQAC.gif)
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The diagram is not an exact representation of everything Tom. Don't get caught up in it please. If that's where this entire issue arises we might have bigger problems. There's a reason the equinoxes and the solstices vary from year to year. They are not exactly on the same time every year. Do you know why that is? The answer to that question is the same as yours, and it's because the Earth's rotation around it's axis, and the Earth's orbit around the sun are not related. One is not dependent on the other. They happen to match up quite well, but it's pure coincidence.
You're right though, the March and September equinoxes happen at about the same time every year. Because of leap year. Without leap year, as mentioned a number of times now, seasons would eventually flip based on the calendar month. This is why we have leap year, so that doesn't happen. The Earth isn't changing it's orbit, or it's rotation. We're changing how we track both of these things in order to keep things similar to what everyone expects.
Look at the second animated image I added above. Solar Time returns to the same position after one year when it returns to the Equinox it started on. The Solar Time is opposite on the opposite Equinox. The Solstices are opposites too.
If there is a difference over the year and the Solar Day is just arbitrarily related to the year, why are there exact opposites?
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The diagram is not an exact representation of everything Tom. Don't get caught up in it please. If that's where this entire issue arises we might have bigger problems. There's a reason the equinoxes and the solstices vary from year to year. They are not exactly on the same time every year. Do you know why that is? The answer to that question is the same as yours, and it's because the Earth's rotation around it's axis, and the Earth's orbit around the sun are not related. One is not dependent on the other. They happen to match up quite well, but it's pure coincidence.
You're right though, the March and September equinoxes happen at about the same time every year. Because of leap year. Without leap year, as mentioned a number of times now, seasons would eventually flip based on the calendar month. This is why we have leap year, so that doesn't happen. The Earth isn't changing it's orbit, or it's rotation. We're changing how we track both of these things in order to keep things similar to what everyone expects.
Look at the second animated image I added above. Solar Time returns to the same position after one year when it returns to the Equinox it started on. The Solar Time is opposite on the opposite Equinox. The Solstices are opposites too.
If there is a difference over the year and the Solar Day is just arbitrarily related to the year, why are there exact opposites?
Because the total difference over the entire year, is just 6 hours? I would note, even this simulation isn't exact. Look at the times of the equionoxes here if you wish. https://www.timeanddate.com/calendar/seasons.html
12 hour difference from sidreal, - 3 hour difference from half the year rotation offset. Bam, the time of the opposite equinox or solstice. Actually doesn't look quite right for the solstice, just the equinox. Interesting. Not sure why, would have to look a bit more.
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The diagram is not an exact representation of everything Tom. Don't get caught up in it please. If that's where this entire issue arises we might have bigger problems. There's a reason the equinoxes and the solstices vary from year to year. They are not exactly on the same time every year. Do you know why that is? The answer to that question is the same as yours, and it's because the Earth's rotation around it's axis, and the Earth's orbit around the sun are not related. One is not dependent on the other. They happen to match up quite well, but it's pure coincidence.
You're right though, the March and September equinoxes happen at about the same time every year. Because of leap year. Without leap year, as mentioned a number of times now, seasons would eventually flip based on the calendar month. This is why we have leap year, so that doesn't happen. The Earth isn't changing it's orbit, or it's rotation. We're changing how we track both of these things in order to keep things similar to what everyone expects.
Look at the second animated image I added above. Solar Time returns to the same position after one year when it returns to the Equinox it started on. The Solar Time is opposite on the opposite Equinox. The Solstices are opposites too.
If there is a difference over the year and the Solar Day is just arbitrarily related to the year, why are there exact opposites?
Because the total difference over the entire year, is just 6 hours? I would note, even this simulation isn't exact. Look at the times of the equionoxes here if you wish. https://www.timeanddate.com/calendar/seasons.html
12 hour difference from sidreal, - 3 hour difference from half the year rotation offset. Bam, the time of the opposite equinox or solstice. Actually doesn't look quite right for the solstice, just the equinox. Interesting. Not sure why, would have to look a bit more.
Are you saying that the Astronomy Department at the University of Nebraska-Lincoln got Round Earth Astronomy wrong, and that you know better than them?
Their simulation is clearly showing that Solar Time is related to the Equinoxes and Solstices of the year. It is not arbitrary. The sun needs to return to its same position above the earth after 1 year.
Your TimeandDate.com link isn't showing Solar Time, and isn't based on the Solar Year. It's showing in PDT for me and it says the Year is based on the The Gregorian Calendar. This is unlike the Solar Day and the Solar Year. There does not seem to be a way on that page to change it to Solar Day and Solar Year.
As an aside point, TimeandDate has been discredited on this forum because we emailed them and they refused to answer on whether they were using prediction models that were based on Round Earth Theory or on observed patterns and trends, citing proprietary data.
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Look at the second animated image I added above. Solar Time returns to the same position after one year when it returns to the Equinox it started on. The Solar Time is opposite on the opposite Equinox. The Solstices are opposites too.
If there is a difference over the year and the Solar Day is just arbitrarily related to the year, why are there exact opposites?
That’s a teaching tool to demonstrate the relationship between solar and sidereal days. It’s using generic times and isn’t accounting for the extra .24 of a day. Hard to manipulate on mobile, but it resets at the vernal equinox at exactly 356 solar days and 366 sidereal days.
It’s not a calculator for actual equinox and solstice times. It’s for illustrating the concept to students. It ought to help you resolve the question you had had the opening of this topic. But I can understand why it might be confusing in trying to understand the relationship of solar days to solar years. It wasn’t coded for that.
Look at it for what it is intended: a simplified interactive illustration of the solar/sidereal relationship concept.
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I would note, even this simulation isn't exact. Look at the times of the equionoxes here if you wish. https://www.timeanddate.com/calendar/seasons.html
It’s not exact because that’s not what it’s meant to be. Actual equinox and solstice times vary year to year. This is generic.
It’s a nice aid for demonstrating how solar days and sidereal days are related. But the times are purely representative and not actual or predictive. And unfortunately for explaining to Tom how solar days don’t fit as whole integers into a solar year, or show how the time of vernal equinox varies from year to year, it wasn’t coded for that. It shows a single solar year divided crisply into 365 solar days and 366 sidereal days with the earth returning to the same spot in its orbit around the sun. That’s what Tom thinks should be true of a globe earth/sun model if it were reality. but the solar year, equinox and earth rotations aren’t synchronized that way. The animation is simplified for the introductory student.
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The diagram is not an exact representation of everything Tom. Don't get caught up in it please. If that's where this entire issue arises we might have bigger problems. There's a reason the equinoxes and the solstices vary from year to year. They are not exactly on the same time every year. Do you know why that is? The answer to that question is the same as yours, and it's because the Earth's rotation around it's axis, and the Earth's orbit around the sun are not related. One is not dependent on the other. They happen to match up quite well, but it's pure coincidence.
You're right though, the March and September equinoxes happen at about the same time every year. Because of leap year. Without leap year, as mentioned a number of times now, seasons would eventually flip based on the calendar month. This is why we have leap year, so that doesn't happen. The Earth isn't changing it's orbit, or it's rotation. We're changing how we track both of these things in order to keep things similar to what everyone expects.
Look at the second animated image I added above. Solar Time returns to the same position after one year when it returns to the Equinox it started on. The Solar Time is opposite on the opposite Equinox. The Solstices are opposites too.
If there is a difference over the year and the Solar Day is just arbitrarily related to the year, why are there exact opposites?
Because the total difference over the entire year, is just 6 hours? I would note, even this simulation isn't exact. Look at the times of the equionoxes here if you wish. https://www.timeanddate.com/calendar/seasons.html
12 hour difference from sidreal, - 3 hour difference from half the year rotation offset. Bam, the time of the opposite equinox or solstice. Actually doesn't look quite right for the solstice, just the equinox. Interesting. Not sure why, would have to look a bit more.
Are you saying that the Astronomy Department at the University of Nebraska-Lincoln got Round Earth Astronomy wrong, and that you know better than them?
Their simulation is clearly showing that Solar Time is related to the Equinoxes and Solstices of the year. It is not arbitrary. The sun needs to return to its same position above the earth after 1 year.
Your TimeandDate.com link isn't showing Solar Time, and isn't based on the Solar Year. It's showing in PDT for me and it says the Year is based on the The Gregorian Calendar. This is unlike the Solar Day and the Solar Year. There does not seem to be a way on that page to change it to Solar Day and Solar Year.
As an aside point, TimeandDate has been discredited on this forum because we emailed them and they refused to answer on whether they were using prediction models that were based on Round Earth Theory or on observed patterns and trends, citing proprietary data.
I'm saying, as is Bobby, that this tool is not designed with what you are expecting of it in mind. It's a simple tool to show the difference between solar days and sidereal days. Nothing more. You really need to actually look at the tools you're going to use and make sure they fit the purpose you want out of them Tom. It's a failing of yours.
I'm not pointing out solar time here. I'm showing you how the equinoxes vary from year to year. Solar day = Calendar day. Do you see how the timing of the equinoxes and solstices vary each year by about 6 hours? Due to the fact a solar year is not an exact number of solar days?
To your aside, I was part of that discussion. You decided it was discredited for the purposes of it's sunrise/sunset times being based on a RE model, NOT that the site was 100% false. Stop attempting to pull the wool over peoples eyes with dishonesty. It's unbecoming. This information can be found many other places if you wish to check it against what is listed there as well. The information the site presents is accurate, but it's not for sure if it's based on observational data, a round Earth model, or something else entirely. You may claim the sunrise/set times can't be trusted all you want as well, but this isn't one of them. Feel free to fact check if you feel so inclined, but until you can provide evidence the information is incorrect (and not just that we don't know how they get it, that's not showing the information is incorrect) my own experience is their information is spot on. Confirmed by friends and family around the globe, as well as our resident seafarer as mentioned in another thread.
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Dear Mr. Bishop,
Why are you surprised and deny that equinoxes delay? Didn't you start this thread with the intention to prove that they delay in the GE model? Didn't your own calculations confirm that? Have you forgotten your own opinions or do you disregard them intentionally?
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The diagram is not an exact representation of everything Tom. Don't get caught up in it please. If that's where this entire issue arises we might have bigger problems. There's a reason the equinoxes and the solstices vary from year to year. They are not exactly on the same time every year. Do you know why that is? The answer to that question is the same as yours, and it's because the Earth's rotation around it's axis, and the Earth's orbit around the sun are not related. One is not dependent on the other. They happen to match up quite well, but it's pure coincidence.
You're right though, the March and September equinoxes happen at about the same time every year. Because of leap year. Without leap year, as mentioned a number of times now, seasons would eventually flip based on the calendar month. This is why we have leap year, so that doesn't happen. The Earth isn't changing it's orbit, or it's rotation. We're changing how we track both of these things in order to keep things similar to what everyone expects.
Look at the second animated image I added above. Solar Time returns to the same position after one year when it returns to the Equinox it started on. The Solar Time is opposite on the opposite Equinox. The Solstices are opposites too.
If there is a difference over the year and the Solar Day is just arbitrarily related to the year, why are there exact opposites?
Because the total difference over the entire year, is just 6 hours? I would note, even this simulation isn't exact. Look at the times of the equionoxes here if you wish. https://www.timeanddate.com/calendar/seasons.html
12 hour difference from sidreal, - 3 hour difference from half the year rotation offset. Bam, the time of the opposite equinox or solstice. Actually doesn't look quite right for the solstice, just the equinox. Interesting. Not sure why, would have to look a bit more.
Are you saying that the Astronomy Department at the University of Nebraska-Lincoln got Round Earth Astronomy wrong, and that you know better than them?
Their simulation is clearly showing that Solar Time is related to the Equinoxes and Solstices of the year. (...)
Or could it be that, just like the FE "map", that simulation is meant to be only illustrative?
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I was editing my post and lost one. Here it is again. I added on to it at the bottom.
What I am hearing is that the Solar Day does not match up to a Solar year that starts and stops on the Equinox. You are calling the two are "arbitrarily" in relation to each other. But the March and September Equinoxes are at a certain point on the earth's path around the sun. The earth will be in the same physical position looking at the sun.
If the Solar day at the start of the day is over New York City, it must end on New York City.
The Equinox only varies over eons, as the Equinox and Solstice lines in the below diagram rotates according to the Precession of the Equinoxes at a period of 25,772 years.
(https://s3.amazonaws.com/ai2-vision-textbook-dataset/dataset_releases/rc2/train/question_images/seasons_6284.png)
See the diagram above. The Solar Day has to be the same when the earth returns back to the same Equinox point in relation to the sun. It's looking at the sun at the same position.
Why is the Solar Day not the same? What is causing the variance?
Go here:
http://astro.unl.edu/classaction/animations/coordsmotion/siderealSolarTime.html
(https://cdn.pbrd.co/images/HijFQAC.gif)
Dear Tom
You owe me about 30 mins of my life.
That's the roughly 30 mins I have wasted on this childish joke of yours over the last few days.
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See the diagram above. The Solar Day has to be the same when the earth returns back to the same Equinox point in relation to the sun. It's looking at the sun at the same position.
The diagrams you cite here, and at the beginning of the thread, are both approximations, not exactnesses (!).
They're intended to show a high-level, general principle of how things work. They have no data, they are merely illustrative. I'll say that again - they have no low-level data (numbers 'n' stuff).
For actual data, you need to look at astronomical textbooks, papers, journals, etc. and see how astronomers of the past figured this stuff out from first principles and their observations.
I'm sure the data, from year upon year upon year of actual observation, is out there to be found, but you won't find it from skimming webpages and lifting diagrams from those pages intended for student or general public audiences.
It will be in the work done by astronomers of university level or beyond. You may need to do some groundwork to find it. It will most likely be in real books, in libraries, and you may need to expend some effort to find it, including asking librarians to retrieve little-used tomes from dusty, remote archives.
Are you willing to do this?
Or is it truly beyond the scope of the 30 mins per day that you devote to this field? If so, just say so.
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Lets Recap
I have several pages giving a rebuttal is that the Solar Day and Solar Time is unconnected to the Solar Year. There are 24 hours in a Solar Day and 356.24 Solar Days in a Solar Year. Where does the .24 come from? The argument I am hearing is that the two are "unconnected". There are posts with arguments that the two are "arbitrary" and the names Solar Day and Solar Year coincidental. Lets work with that.
The Solar Year is measured starting at the Equinox. It can be measured from elsewhere to get very slight variance, but starting from the Equinox is the standard definition. The Equinox is two points on the Earth's route around the Sun. The March Equinox and the September Equinox. The Equinox is defined as the point where the Equatorial Plane intersects the Ecliptic Plane.
(https://i.imgur.com/VLsfYWz.gif)
The Equinox is the line that runs through the Earth-Sun System. Its an exact point on the earth's path around the sun.
Starting at point at the September Equinox and after one year when the Earth is again at the September Equinox the Solar Year will have accumulated 356.24 Solar Days.
But I am hearing that the Solar Day and the Solar Year are "unconnected". The 24 Hour Solar Day will have accumulated almost 6 hours, according to that argument.
Consider if the September Equinox in the above illustration was at Solar Noon. Solar Noon is when the Sun is directly overhead.
After 1 Solar Year the new Solar Noon can't be at + ~6 Solar Hours. The earth has returned to its location on the orbit. The sun can't be pointing in a different direction. How does that work?
Solar Noon is when the sun is directly overhead of the local celestial meridian:
https://en.wikipedia.org/wiki/Noon
Solar noon is the time when the Sun appears to contact the local celestial meridian. This is when the Sun apparently reaches its highest point in the sky, at 12 noon apparent solar time. The local or clock time of solar noon depends on the longitude and date.
Meridian Illustration
(https://i.imgur.com/aUWxCjg.jpg)
The 24 hour Solar Time is, therefore, related to the Solar Year. They cannot be "arbitrary". This question is not answered by that argument.
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Lets Recap
I have several pages giving a rebuttal is that the Solar Day and Solar Time is unconnected to the Solar Year. There are 24 hours in a Solar Day and 356.24 Solar Days in a Solar Year. Where does the .24 come from? The argument I am hearing is that the two are "unconnected". There are posts with arguments that the two are "arbitrary" and the names Solar Day and Solar Year coincidental. Lets work with that.
The Solar Year is measured starting at the Equinox. It can be measured from elsewhere to get very slight variance, but from the Equinox is the standard definition. The Equinox is two points on the Earth's route around the Sun. The March Equinox and the September Equinox. The Equinox is defined as the point where the Sun's Equalatorial Plane intersects the Orbital Plane.
(http://www.crbond.com/images/eclip2.gif)
The Equinox is the line that runs through the Solar System. Its an exact point on the earth's path.
Starting at point at the September Equinox and after one year when the Earth is again at the September Equinox the Solar Year will have accumulated 356.24 Solar Days.
But I am hearing that the Solar Day and the Solar Year are "unconnected". The 24 Hour Solar Day will have accumulated almost 6 hours, according to that argument.
Consider if the September Equinox in the above illustration was at Solar Noon. Solar Noon is when the Sun is directly overhead.
After 1 Solar Year the Solar Noon can't be Solar Noon plus 5+ Solar Hours. The sun would be pointing off into space. How does that work?
Solar Noon is when the sun is directly overhead of the local celestial meridian:
https://en.wikipedia.org/wiki/Noon
Solar noon is the time when the Sun appears to contact the local celestial meridian. This is when the Sun apparently reaches its highest point in the sky, at 12 noon apparent solar time. The local or clock time of solar noon depends on the longitude and date.
Meridian Illustration
(https://i.imgur.com/aUWxCjg.jpg)
The 24 hour Solar Time is, therefore, related to the Solar Year. They cannot be "arbitrary".
The equinox doesn't happen at the same time every year. Boom, mystery solved. That's what we've been telling you this whole time. If the September equinox of 2023 happens on the 20th, at solar noon local time, the September equinox of 2024 happens on the 20th at around 18:00 local time. It does not occur at solar noon year to year at any location on Earth, and it would be an extraordinary coincidence if it did.
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I'm really hoping that Tom isn't thinking that the fact its 365.24 is in some way related to there being 24 hours a day :D
For a start, 365.24 is rounding, according to:
https://www.timeanddate.com/astronomy/tropical-year.html
A tropical year, also known as a solar year, an astronomical year, or an equinoctial year, is, on average, approximately 365 days, 5 hours, 48 minutes and 45 seconds long (365.24219 days)
And secondly, the 24 hour day apparently comes from the ancient Egyptians who would not have had the ability to measure the length of a year that accurately.
I have no idea why anyone would think there would be any correlation between the time it takes a planet to orbit its star and the time it takes a planet to rotate on its axis. They are two completely separate motions.
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The equinox doesn't happen at the same time every year. Boom, mystery solved.
That's not the answer to the question. That is the question I am asking. The Solar Day and the Solar Year that is based on the Equinox do not match up.
That's what we've been telling you this whole time. If the September equinox of 2023 happens on the 20th, at solar noon local time, the September equinox of 2024 happens on the 20th at around 18:00 local time. It does not occur at solar noon year to year at any location on Earth, and it would be an extraordinary coincidence if it did.
Furthermore, the equinox only moves at a rate measured in eons.
https://www.britannica.com/science/precession-of-the-equinoxes
Precession of the equinoxes, motion of the equinoxes along the ecliptic (the plane of Earth's orbit) caused by the cyclic precession of Earth's axis of rotation. ... Such a motion is called precession and consists of a cyclic wobbling in the orientation of Earth's axis of rotation with a period of 25,772 years.
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I have no idea why anyone would think there would be any correlation between the time it takes a planet to orbit its star and the time it takes a planet to rotate on its axis. They are two completely separate motions.
The geometry of the scene shows you wrong. The time of Solar Noon when the earth returns to its position of the September Equinox can't be Solar Noon + 5+ Solar Hours. The sun and earth is in the same geometric position on the orbit. Solar Noon can't change.
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Those diagrams are only representative. They are not exactly what happens.
The solstice and equinox dates aren't even the same each year
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(...) Where does the .24 come from? (...)
You've calculated it yourself - it's because days don't fit exactly into one year.
(...) Furthermore, the equinox only moves at a rate measured in eons. (..)
Moreover, they also move by 6 hours each year. (https://www.timeanddate.com/calendar/seasons.html)
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I have no idea why anyone would think there would be any correlation between the time it takes a planet to orbit its star and the time it takes a planet to rotate on its axis. They are two completely separate motions.
The geometry of the scene shows you wrong. The time of Solar Noon when the earth returns to its position of the September Equinox can't be Solar Noon + 5+ Solar Hours. The sun and earth is in the same geometric position on the orbit. Solar Noon can't change.
Incorrect. That's why the time of the equinox changes every year. I've lost what your problem is with this again. You keep repeating the same thing over and over and declaring it can't be true for some reason. Why? The Earth moves around the sun at X speed. The Earth rotates about it's own axis at Y speed. Why do you believe X and Y have any correlation to one another, such that an integer amount of Y's should happen in every X? There's none. Zero. Zilch. One is not dependent upon the other in any way. I don't understand how this concept is so difficult for you to grasp.
"The sun and earth is in the same geometric position on the orbit." This, is correct. Their spatial relationship to one another is the same each time the equinox happens.
"Solar Noon can't change." This is incorrect, and in fact a non-sequitur. Solar Noon has nothing whatsoever to do with the equinox.
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Lets Recap
I have several pages giving a rebuttal is that the Solar Day and Solar Time is unconnected to the Solar Year. There are 24 hours in a Solar Day and 356.24 Solar Days in a Solar Year. Where does the .24 come from? The argument I am hearing is that the two are "unconnected". There are posts with arguments that the two are "arbitrary" and the names Solar Day and Solar Year coincidental. Lets work with that.
I stopped reading here, and I haven't scrolled down to see if anyone else has responded.
Tom,
I can appreciate if you feel kind of like you're being ganged up on, and I don't want to contribute to what might feel like a mob. I think we're all independently trying to explain to you the same thing, though maybe in different ways. And there's obviously history with some of you that is leading others to respond more caustically.
I, for one, don't like to discuss things that way. But I do wish I could help you work through the question, but your recap is not right, and you can't use that as a place from which to start to "work with."
You seem to expect something to be true that isn't. These solar days, solar years, etc. are not "arbitrary." They aren't "unconnected." They're just not connected in a manner that you consider whole.
1 rotation of the earth with respect to the sun is a solar day. That's not arbitrary. It's 1. 1 day. 1 solar day.
1 orbit around the sun is a solar year. That's not arbitrary. It's 1. 1 year. 1 solar year.
The solar day happens 365 times in 1 solar year, but then there's a fractional period of rotation that must occur before the orbit of 1 solar year is complete. That's not arbitrary. It's measurable. It's "arbitrary" I suppose, but with good reason, that you might want to choose a starting point for measuring the year at an equinox or solstice. In this case, via convention (call it arbitrary) we use the vernal equinox. But the number of earth rotations (relative to the sun) during the length of that orbit to the next vernal equinox is not arbitrary. They happen naturally, are measurable, and the solar day doesn't line up with the solar year in a whole number. The earth reaches a point in its orbit around the sun a little before the vernal equinox when it completes its 365th solar day rotation.
That's where the extra time is coming from. The earth rotates a little more while it has to complete that orbit, returning to the point of the vernal equinox.
Calling things "arbitrary," "coincidental" and "unconnected" indicates you aren't absorbing what's being explained. Some things, like words or units of measurement might be "arbitrary" or based on something that is a matter of convenience or convention, but what we are measuring aren't. And you can't consider what's being measured as "unconnected" just because the way we measure doesn't use units that divide evenly as integers. Fractions are a way of life in the world.
I hope that helps. (If I'm redundant and someone else has already made that point, I apologize, but I hope you aren't being resistant just because you feel defensive or ganged up on.)
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The equinox doesn't happen at the same time every year. Boom, mystery solved. That's what we've been telling you this whole time. If the September equinox of 2023 happens on the 20th, at solar noon local time, the September equinox of 2024 happens on the 20th at around 18:00 local time. It does not occur at solar noon year to year at any location on Earth, and it would be an extraordinary coincidence if it did.
AND, Tom, to go back to your diagram on the first page, it shows the equinoxes and solstices on two possible days in each instance.
Please consider why.
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Consider if the September Equinox in the above illustration was at Solar Noon. Solar Noon is when the Sun is directly overhead.
After 1 Solar Year the Solar Noon can't be Solar Noon plus 5+ Solar Hours. The sun would be pointing off into space. How does that work?
You're right. It can't. That's why the last solar noon (on the 365th day) happens before the end of the solar year. Those two moments don't coincide. If starting the measure of the solar year is at the equinox (you've chosen September instead of March, but whatever), that isn't at solar noon either (necessary; I mean it could be on rare occasions.)
Solar day doesn't have to start at solar noon either. You could pick any measurable moment of sun elevation and use that as a starting point. But the day is the amount of time to rotate and see the sun back at that point.
Similar with solar year. Doesn't have to start count at an equinox. It's just a good marking point. But the amount of time for the earth to complete its orbit to get back to that point, relative to the sun, is a solar year.
For illustration purposes, you can start the measure at solar noon and, for the sake of understanding, assume that is also the time of the vernal (Spring) equinox. Let the motions occur and at the next vernal equinox, it won't be solar noon. That's because that 365th solar noon happened earlier, before the equinox was reached. When the equinox is finally reached (completing the solar year), about 5.5 to 5.8 more hours will have elapsed because the earth kept rotating until vernal equinox.
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That's not the answer to the question. That is the question I am asking. The Solar Day and the Solar Year that is based on the Equinox do not match up.
They don't match up as a whole number. So what? I don't understand why you expect them to, especially after having it repeatedly explained why they don't.
1 solar year = 365.24 solar days.
The equinox is about 0.24 solar days later each year because they "don't match up" without the fraction. They match up well with the fraction.
We're all telling you why that is, but you keep claiming that's a problem or a question.
Furthermore, the equinox only moves at a rate measured in eons.
https://www.britannica.com/science/precession-of-the-equinoxes
Precession of the equinoxes, motion of the equinoxes along the ecliptic (the plane of Earth's orbit) caused by the cyclic precession of Earth's axis of rotation. ... Such a motion is called precession and consists of a cyclic wobbling in the orientation of Earth's axis of rotation with a period of 25,772 years.
I tried to tell you that that's a different influence on the motion, and it's much more marginal, so don't worry about it for trying to comprehend this other, much larger (relatively) variance that is the focus of your question.
Until the solar year/solar day relationship is understood, the precession affect will just add to the confusion. And it's way down in the weeds, astronomically, for the short time reference periods we've been focusing on.
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The geometry of the scene shows you wrong. The time of Solar Noon when the earth returns to its position of the September Equinox can't be Solar Noon + 5+ Solar Hours. The sun and earth is in the same geometric position on the orbit. Solar Noon can't change.
Until you can discard yourself of that incorrect premise, attempting to work through it with you is bound to be fruitless.
It's disappointing that that UNL animated illustration seems to have cemented your misunderstanding. That same Astronomy Department also has published this page on the Web. Maybe it will help you get over the rigid insistence that a starting marker to the solar day must line up with the starting marker to the solar year and the end of 1 solar year.
http://astro.unl.edu/naap/motion3/sidereal_synodic.html (http://astro.unl.edu/naap/motion3/sidereal_synodic.html)
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Those diagrams are only representative. They are not exactly what happens.
The solstice and equinox dates aren't even the same each year
Solar Noon can't change. The question has not been answered.
(...) Where does the .24 come from? (...)
You've calculated it yourself - it's because days don't fit exactly into one year.
(...) Furthermore, the equinox only moves at a rate measured in eons. (..)
Moreover, they also move by 6 hours each year. (https://www.timeanddate.com/calendar/seasons.html)
That is not the answer to the question. That was the question.
From your link (ignoring that is it from TimeandDate that has been discredited as a source on RET models, and also ignoring that it is displaying as PDT for me and on a 365 Day Gregorian Calendar, not Solar Time):
Year March Equinox
2013 Mar 20 4:01 am PDT
2014 Mar 20 9:57 am PDT -- ~6 hr difference
2015 Mar 20 3:45 pm PDT -- ~6 hr difference
2016 Mar 19 9:30 pm PDT -- ~ -1 day + 6 hr difference
2017 Mar 20 3:28 am PDT -- ~6 hr difference
2018 Mar 20 9:15 am PDT -- ~6 hr difference
2019 Mar 20 2:58 pm PDT -- ~6 hr difference
We already know that the Solar Year that is based on the equinox adds .24 hours to our local calendars and clocks which we must create innovative adjustment methods for (ie. Leap Year). That is what needs to be solved and reconciled with Solar Time that is based on a 24 hour solar clock.
Why doesn't Solar Noon move?
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That's not the answer to the question. That is the question I am asking. The Solar Day and the Solar Year that is based on the Equinox do not match up.
And I answered that question few times in this thread already.
Solar day is based on Earth's rotation towards Sun.
Calendar year is based on solar days.
We, humans count it for our needs.
Tropical year is not based on solar days.
Tropical year is based on Earth's orbital events and it is not calendar year.
EDIT:
I already placed analogy with propeller of the plane model revolving pole.
We count our calendar year by counting "rotations of the propeller" (days), and trying to adjust "counted sets of rotations" (calendar year) as close as possible to "number of revolutions" (tropical year).
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Dear Mr. Bishop,
Just to make this situation clear, what exactly is your question?
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Are you saying that the Astronomy Department at the University of Nebraska-Lincoln got Round Earth Astronomy wrong, and that you know better than them?
Their simulation is clearly showing that Solar Time is related to the Equinoxes and Solstices of the year. It is not arbitrary. The sun needs to return to its same position above the earth after 1 year.
Think of that animation like this animation of a flat earth model.
https://imgur.com/r/educationalgifs/D7aOXVA (https://imgur.com/r/educationalgifs/D7aOXVA)
Should I use that animation to work out things like azimuth of the sun at sunset from San Diego at the equinox?
http://oi66.tinypic.com/213iz6f.jpg (http://oi66.tinypic.com/213iz6f.jpg)
And if the azimuth of the sun is not what that animation predicts, should I reject the flat earth model?
I read you elsewhere tell someone that animation (or others like it) are visualizations. They convey the concept, but aren't intended to be accurate.
Same with that simple animation you showed that isn't synchronized to any actual time, can't be used for predicting or comparing to actual truth data (time/date) and doesn't even bother to illustrate the very aspect of the "extra time" that you are interested in. But that's because it's just a visualization. A fancy cartoon to help explain the relationship of sidereal days with solar days.
Unfortunately, it's reinforced a notion that a solar day (solar noon) must line up always with the solar year (equinox) or else something's wrong with the globe/solar model. If you are sincere in wanting to find resolution to what you see as a problem with solar noon in round earth theory, you need to break through that incorrect concept that "The sun needs to return to its same position above the earth after 1 year."
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Why doesn't Solar Noon move?
Why doesn't what move? Your question is a nonsense question here Tom. If this is what you're attempting to ask, you will need to explain what exactly it is that you mean.
I had more to say about Timeanddate, but it's not conducive to the discussion, just as your dishonest ramblings aren't conducive. Stick to the topic, and see if you can be clear with what exactly your issue is, because after this sentence it seems your issue is complete nonsense.
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Solar Noon can't change...Why doesn't Solar Noon move?
You started with "solar noon can't change" and then ended asking "why doesn't solar noon move?"
Solar noon is one moment out of a solar day. Solar noon doesn't care about the equinox or where the earth is in its orbit or how the ecliptic plane is oriented. As long as the earth is spinning, solar noon will come around again and again, as will any other demarcation point in the solar day.
Even if the earth didn't orbit the sun at all but somehow just hung above the sun in one spot, you'd still have solar days if the earth is rotating making the sun appear to rise to a solar noon.
Solar noon can't change? Sure. As long as your clock is tied to the sun and the rotation of earth is constant, solar noon will be solar noon always. When the sun is at its highest elevation.
But if you tie your clock to, say, the stars? Then solar noon isn't always at the same time. It's still solar noon, but it's happening at a different sidereal time each day.
Right? But we're talking about a solar day clock, so yeah. Solar noon is constant.
So, why doesn't it move then if the solar year doesn't "match?" Why should it? Why would you expect it to? The solar year is not based on the rotation of the earth. It's based on the orbit of the earth around the sun. It's a different cyclical motion: one "inside" of the other.
They match up and are relateable fractionally, but you can have orbit without rotation and rotation without orbit (well, at least temporarily). In the sun/earth case, we have 365 and a fraction rotations for every orbit. This fraction won't change the timing of solar noon. But it will change how we relate the marking of the new solar year to the timing of a solar day.
You've got to remember your reference frames and understand the relationships between them, else you will create conundrums that don't really exist.
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Calling things "arbitrary," "coincidental" and "unconnected" indicates you aren't absorbing what's being explained.
I said that it can't be arbitrarily connected. Read the post again. This indicates that you are not really absorbing this content.
Why doesn't Solar Noon move?
Why doesn't what move? Your question is a nonsense question here Tom. If this is what you're attempting to ask, you will need to explain what exactly it is that you mean.
Solar Noon is based on a 24 hour Solar Day. After 364.24 days Solar Noon should have moved ~6 Hours. The geometry of the scene when the earth returns to the September Equinox point shows that Solar Noon is not in that place.
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That's not the answer to the question. That is the question I am asking. The Solar Day and the Solar Year that is based on the Equinox do not match up.
And I answered that question few times in this thread already.
Solar day is based on Earth's rotation towards Sun.
Calendar year is based on solar days.
We, humans count it for our needs.
Tropical year is not based on solar days.
Tropical year is based on Earth's orbital events and it is not calendar year.
EDIT:
I already placed analogy with propeller of the plane model revolving pole.
We count our calendar year by counting "rotations of the propeller" (days), and trying to adjust "counted sets of rotations" (calendar year) as close as possible to "number of revolutions" (tropical year).
I have shown several quotes in this thread which say that there are 365.24 Solar Days in a Solar Year.
I have also shown with the Recap post that the Solar Day must be connected to the Solar Year.
You have not yet provided the + ~6 hour solution.
A Sidrael Year is only different by the Solar Year by 20 minutes, and is not a solution. Why doesn't the Solar Day fit into the Solar Year?
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Why doesn't Solar Noon move?
Why doesn't what move? Your question is a nonsense question here Tom. If this is what you're attempting to ask, you will need to explain what exactly it is that you mean.
Solar Noon is based on a 24 hour Solar Day. After 364.24 days Solar Noon should have moved ~6 Hours. The geometry of the scene when the earth returns to the September Equinox point shows that Solar Noon is not in that place.
Solar noon is based on Earth's rotation, not its orbit (in contrary to equinoxes).
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(...)
You have not yet provided the +6 hour solution.
(...)
It takes 365.24 solar days (i. e. time units are not defined by the tropical year and that do not fit exactly to one tropical year) to complete a tropical year.
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I said that it can't be arbitrarily connected. Read the post again. This indicates that you are not really absorbing this content.
I wasn't addressing what you said about your own thoughts. I was addressing how you were mis-characterizing the explanations provided to you in your recap.
(http://oi68.tinypic.com/69dz5w.jpg)
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Why doesn't Solar Noon move?
Why doesn't what move? Your question is a nonsense question here Tom. If this is what you're attempting to ask, you will need to explain what exactly it is that you mean.
Solar Noon is based on a 24 hour Solar Day. After 364.24 days Solar Noon should have moved ~6 Hours. The geometry of the scene when the earth returns to the September Equinox point shows that Solar Noon is not in that place.
Wrong. Once again. Solar Noon for a place on Earth happens after a set period of 24 hours (roughly). The movement of the Earth around the sun has no relation to a solar day (beyond the difference to a sidereal day). Why is Solar Noon moving? What is making Solar Noon move? Solar Noon is happening 24/7. Every second of every day somewhere is experiencing Solar Noon. From Equinox A to Equinox B, if location X was experiencing solar noon during A, it will not be in solar noon during B. Location Y will. What's wrong with that? Nothing says location X should be experiencing solar noon at B, if it was at A. If you think it is, explain exactly what you think is saying this should be the case.
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Solar Noon is based on a 24 hour Solar Day. After 364.24 days Solar Noon should have moved ~6 Hours. The geometry of the scene when the earth returns to the September Equinox point shows that Solar Noon is not in that place.
Assuming solar noon was the start of the solar year, after 364.24 solar days, the last solar noon would have happened ~6 hours before the equinox. Thus, at equinox, solar noon is past. Solar noon didn't change. It still came and went as predicted. The return to equinox after one solar year will not line up with solar noon, as you apparently expect it to. That's not the "geometry of the scene." The earth has rotated past solar noon by the time equinox arrives.
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I have also shown with the Recap post at the top that the Solar Day must be connected to the Solar Year.
This is the crux of the issue, right here. There's no hope of resolution if we don't understand why the "solar day must be connected to the solar year" in the whole integer way you expect, or if you don't understand that the relationship is fractional.
Without that being resolved, this is a merry-go-round.
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I have also shown with the Recap post at the top that the Solar Day must be connected to the Solar Year.
This is the crux of the issue, right here. There's no hope of resolution if we don't understand why the "solar day must be connected to the solar year" in the whole integer way you expect, or if you don't understand that the relationship is fractional.
Without that being resolved, this is a merry-go-round.
I'm getting off, unless I see an explanation from Tom why the earth's rotation relative to the sun (solar day) must be synchronized to the orbit around the sun (solar year), without any fractional variance.
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This thread is the smoking gun that proves beyond reasonable doubt that Tom is just here for the banter, to wind up folks, to troll. The guys one of the finest examples of modern troll, he's led 14 pages here and it will still grow, his technique is sublime. I salute you Tom, master of the mass debate!
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Solar noon is based on Earth's rotation, not its orbit (in contrary to equinoxes).
Solar Noon is part of the Solar Day and Solar Time, which is period of the rotation of the sun around the earth.
The number of Solar Days and Solar Hours must fit into the Solar Year. See the Recap post (https://forum.tfes.org/index.php?topic=9478.msg148664#msg148664). After 365.24 Solar Days the earth will be back at the September Equinox point. Solar Noon will be at a position that is + ~6 Hours. Solar Noon can't change.
(...)
You have not yet provided the +6 hour solution.
(...)
It takes 365.24 solar days (i. e. time units are not defined by the tropical year and that do not fit exactly to one tropical year) to complete a tropical year.
See https://encyclopedia2.thefreedictionary.com/tropical+year
Tropical Year
the time interval between two successive passages of the sun through the vernal equinox. The tropical year contains 365.242196 mean solar days.
It says right there that the Tropical Year, also known as the Solar Year, contains 365.24 Solar Days.
The variation between years is very small.
Solar/Tropical Year Variation
See 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 variation between the years does not account for those hours.
I have also shown with the Recap post at the top that the Solar Day must be connected to the Solar Year.
This is the crux of the issue, right here. There's no hope of resolution if we don't understand why the "solar day must be connected to the solar year" in the whole integer way you expect, or if you don't understand that the relationship is fractional.
Without that being resolved, this is a merry-go-round.
I'm getting off, unless I see an explanation from Tom why the earth's rotation relative to the sun (solar day) must be synchronized to the orbit around the sun (solar year), without any fractional variance.
This was explained. See the Recap Post (https://forum.tfes.org/index.php?topic=9478.240). The Sun is back in its position in its path around the sun. Solar Noon can't change.
The Sidrael Year only has a difference of 20 minutes from the Solar Year.
See http://astro.unl.edu/naap/motion3/sidereal_synodic.html
the sidereal year is about 20 minutes longer than the tropical year.
Not a solution to the number of Solar Days in a Solar Year.
Solar Noon is based on a 24 hour Solar Day. After 364.24 days Solar Noon should have moved ~6 Hours. The geometry of the scene when the earth returns to the September Equinox point shows that Solar Noon is not in that place.
Assuming solar noon was the start of the solar year, after 364.24 solar days, the last solar noon would have happened ~6 hours before the equinox. Thus, at equinox, solar noon is past. Solar noon didn't change. It still came and went as predicted. The return to equinox after one solar year will not line up with solar noon, as you apparently expect it to. That's not the "geometry of the scene." The earth has rotated past solar noon by the time equinox arrives.
You say "The earth has rotated past solar noon by the time equinox arrives."
Solar Time is based on the sun moving over the earth along a local celestial meridian. If the earth has rotated past Solar Noon after a Solar Year, then the Solar Noon has as well. The Solar Day does not fit into the Solar Year.
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This topic is getting a little stale now. There is still no answer. Nothing new.
Lets go back to the Equation of Time. Maybe this is an answer for this.
http://nfo.edu/solarday.htm
Timekeeping on Earth is loosely based on the position of the Sun in the sky. Apparent solar time is the time like a Sundial would tell, where local noon is when the Sun would be directly to the South. However the length of the apparent solar day is not constant throughout the year because of two contributing factors. To account for these varitions the solar day is averaged so all the days of the year will be of equal length. Mean solar time is the time that clocks display. The difference between apparent solar time and mean solar time is called the equation of time.
From https://en.wikipedia.org/wiki/Equation_of_time
The equation of time describes the discrepancy between two kinds of solar time. The word equation is used in the medieval sense of "reconcile a difference". The two times that differ are the apparent solar time, which directly tracks the diurnal motion of the Sun, and mean solar time, which tracks a theoretical mean Sun with noons 24 hours apart. Apparent solar time can be obtained by measurement of the current position (hour angle) of the Sun, as indicated (with limited accuracy) by a sundial. Mean solar time, for the same place, would be the time indicated by a steady clock set so that over the year its differences from apparent solar time would resolve to zero
(https://upload.wikimedia.org/wikipedia/commons/thumb/0/02/Tijdvereffening-equation_of_time-en.jpg/465px-Tijdvereffening-equation_of_time-en.jpg)
The equation of time is this difference, which is cyclical and does not accumulate from year to year.
It seems that this is not the answer to this problem either. Starting from a point in the year, after one year, any seconds added will be subtracted, giving no extra time.
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You say "The earth has rotated past solar noon by the time equinox arrives."
Solar Time is based on the sun moving over the earth along a local celestial meridian. If the earth has rotated past Solar Noon after a Solar Year, then the Solar Noon has as well. The Solar Day does not fit into the Solar Year.
If by "doesn't fit" you mean there can't be a fractional solar day in a solar year, you're absolutely right. But if you do accept that a fraction of a solar day is possible, then it does "fit." 365.24 solar days fit into a solar year. The "extra time" is what makes it "fit."
Now, can you provide some "explanation...why the earth's rotation relative to the sun (solar day) must be synchronized to the orbit around the sun (solar year), without any fractional variance," else it "doesn't fit?"
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You say "The earth has rotated past solar noon by the time equinox arrives."
Solar Time is based on the sun moving over the earth along a local celestial meridian. If the earth has rotated past Solar Noon after a Solar Year, then the Solar Noon has as well. The Solar Day does not fit into the Solar Year.
If by "doesn't fit" you mean there can't be a fractional solar day in a solar year, you're absolutely right. But if you do accept that a fraction of a solar day is possible, then it does "fit." 365.24 solar days fit into a solar year. The "extra time" is what makes it "fit."
Now, can you provide some "explanation...why the earth's rotation relative to the sun (solar day) must be synchronized to the orbit around the sun (solar year), without any fractional variance," else it "doesn't fit?"
Read through the Recap post (https://forum.tfes.org/index.php?topic=9478.240) please.
After 1 Solar Year starting on the September Equinox the earth will return to the same place on the Earth-Sun orbit. The 24 Hour Solar Clock will not be in sync. This is a problem. The Sun did not change position in the Earth-Sun system. If the start point were Solar Noon, we can't have Solar Noon now + ~6 hours at the end point.
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Read through the Recap post (https://forum.tfes.org/index.php?topic=9478.240) please.
After 1 Solar Year starting on the September Equinox the earth will return to the same place on the Earth-Sun orbit. The 24 Hour Solar Clock will not be in sync. This is a problem. The Sun did not change position in the Earth-Sun system. If the start point were Solar Noon, we can't have Solar Noon now + ~6 hours at the end point.
I can read it a million times and it still restates the error.
Solar year and the solar day clock not being in sync is not a problem.
If the start point of the solar year is solar noon, we CAN (and will) have the end point of that solar year ~6 hours after the last solar noon. The new solar year start point is then ~6 hours after solar noon and after another solar year, THAT endp point will be ~12 hours after the last solar noon, and so on. What you seem to want is for each new solar year to be able to start at solar noon. But that's not possible. The timing of equinox (which is how we mark the start and end of the solar year for reasons of ease of measurement) is not synchronized with the timing of solar noon (which is also an indexing convention related to ease of measurement.)
Solar days and solar years not being in sync is not the problem you think it is.
Solar year is vernal equinox to vernal equinox (or autumnal if you prefer), measuring 1 orbital cycle.
Solar day is solar noon to solar noon (or some other indexing point of the solar day if you prefer, measuring 1 rotational cycle.
There is no reason for them to be synchronized. There's no reason to believe it's a problem if they are not.
I would actually be quite fascinated by the situation if they were.
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I'm pretty sure that's it, especially since you keep referring us back to your recap. You seem to believe that the solar year can only start at solar noon. If we haven't reached the equinox by the time of the 365th solar noon, then there must be a problem because the subsequent solar year can't start except at solar noon. In other words, if you start the clock with both orbit and rotation synchronized, you think he's new cycle of the orbit must start synchronized with the rotation.
That's a fallacy.
If I'm restating your position incorrectly, tell me; in which case my question surfaces again as to why you think solar year and solar day must be synchronized in a whole number ratio. If the earth's rotation and the earth's orbit of the sun were toothed gears with a ratio of 365:1, then yeah. I'd consider it a problem if they weren't in sync after 1 cycle of the orbital gear.
But there's not. They are "synced" to a ratio of ~365.24:1 though. There's no reason I know of why that ratio exists, but they are. So at the end of 1 orbital cycle, the rotation gear will have turned a fraction more than it would have in a 365:1 ratio system.
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This appears to be the misunderstanding on your recap post:
After 1 Solar Year the new Solar Noon can't be at + ~6 Solar Hours. The earth has returned to its location on the orbit. The sun can't be pointing in a different direction. How does that work?
Solar Noon is when the sun is directly overhead of the local celestial meridian:
Yes, solar noon can be any amount of time later, just for a different meridian.
If you stand even 100 feet to the West of new, your solar noon is slightly later than mine.
It is always solar noon somewhere.
The sun isn't pointing anywhere different, the Earth is.
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Solar noon is a local phenomenon. The length of the solar day is global.
It is always solar noon somewhere, so the equinox happens at solar noon for someone, just not the same person/meridian every year.
Does this help?
It might help Tom to distinguish it that way.
Looking up times for the next two autumnal equinoxes, 2018 occurs 01:53 UTC on Sep 23. A year later, in 2019, it occurs 07:41 UTC.
Estimating using 15° of longitude per (solar) hour, that means that in 2018, solar noon at the time of autumnal equinox will be somewhere near 150°E meridian. But in 2019, solar noon at the time of autumnal equinox will be somewhere near the 70°E meridian.
The 365th solar noon will have passed those on the 150° meridian because the earth hasn't reached the autumnal equinox yet. It will take another 5 hrs and 48 mins to get there. Meanwhile, the earth will keep rotating with respect to the sun and the solar noon line will keep moving west for that amount of time until...equinox.
(My meridians might be off. I'm sure there's a calculator that can figure what longitude the sun is at given UTC, but I don't have a link handy. The point ought to be made, though.)
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This appears to be the misunderstanding on your recap post:
After 1 Solar Year the new Solar Noon can't be at + ~6 Solar Hours. The earth has returned to its location on the orbit. The sun can't be pointing in a different direction. How does that work?
Solar Noon is when the sun is directly overhead of the local celestial meridian:
Yes, solar noon can be any amount of time later, just for a different meridian.
If you stand even 100 feet to the West of new, your solar noon is slightly later than mine.
It is always solar noon somewhere.
The sun isn't pointing anywhere different, the Earth is.
I am comparing the 24 Hour Clock in Solar Time to the 365.24 Solar Days per Solar Year.
Sidereal Time, the time in comparison to the stars, which is the time of rotation the earth "really" moves at isn't even being discussed in this.
The difference between the Sidrael Year and the Solar Year is only 20 minutes a year anyway, and does not account for this.
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I'm pretty sure that's it, especially since you keep referring us back to your recap. You seem to believe that the solar year can only start at solar noon. If we haven't reached the equinox by the time of the 365th solar noon, then there must be a problem because the subsequent solar year can't start except at solar noon. In other words, if you start the clock with both orbit and rotation synchronized, you think he's new cycle of the orbit must start synchronized with the rotation.
That's a fallacy.
If I'm restating your position incorrectly, tell me; in which case my question surfaces again as to why you think solar year and solar day must be synchronized in a whole number ratio. If the earth's rotation and the earth's orbit of the sun were toothed gears with a ratio of 365:1, then yeah. I'd consider it a problem if they weren't in sync after 1 cycle of the orbital gear.
But there's not. They are "synced" to a ratio of ~365.24:1 though. There's no reason I know of why that ratio exists, but they are. So at the end of 1 orbital cycle, the rotation gear will have turned a fraction more than it would have in a 365:1 ratio system.
You, like douglips, seem to think that the "true" rotation of the earth has something to do with this. We are not even talking about the "true" rotation of the earth. The "true" rotation of the earth, Sidrael Time, which is gauged in relation to the movement of the stars, could be rotating at 902.76 rotations in a Solar Day, and the definitions of Sidrael changed to reflect as much. The stars can be moving at any rate, and if you want to consider them to be fixed stars and that it is really the earth that is moving that rate of 902.76 rotations in a Solar Day, then sure.
The Solar Clock would still be defined by the rate the Sun (Sol) is moving around the earth, and, will still maintain, no matter how fast the stars moved or if they even existed at all, one rotation around the earth in one Solar Day.
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What?
I've long since moved past sidereal day reference frame.
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What?
I've long since moved past sidereal day reference frame.
I don't know what you are trying to convey. Please restate your criticism.
I don't think that the equinox could only start at Solar Noon. It was an example. Any other value offset from Solar Noon would have the same issue since Solar Noon is always a certain number of hours away from it. An offset changes the place of Solar Noon. The 24 Hour Solar Day and the Solar Year must be in sync.
Correction: The equinox does have to start at Solar Noon at some point on earth, but not necessarily your point on earth.
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They are in sync. Just not by a whole number. Every 356.24 solar days per 1 solar year. That "extra time" of 0.24 solar day is the ~6 hours that is the sync.
What's the problem?
My read of your id of "the problem" is the ~6 hours as if that means solar day and solar year are not in sync. But there's no reason for synchronization to be exactly 365 solar days to 1 solar year.
Equinox to same equinox takes more time than 365 solar days. The same aspect of the earth doesn't face the sun from vernal equinox to next vernal equinox.
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They are in sync. Just not by a whole number. Every 356.24 solar days per 1 solar year. That "extra time" of 0.24 solar day is the ~6 hours that is the sync.
What's the problem?
An hour in Mean Solar Time is the same as an hour in a stop watch. These aren't hours of different lengths from what we know.
See this Google Dictionary Definition (https://www.google.com/search?q=Mean+solar+time)
mean so·lar time
nounAstronomy
noun: mean solar time
time as calculated by the motion of the mean sun. The time shown by an ordinary clock corresponds to mean solar time.
And https://www.britannica.com/science/solar-time#ref144523
Mean solar time, kept by most clocks and watches, is the solar time that would be measured by observation if the Sun traveled at a uniform apparent speed throughout the year rather than, as it actually does, at a slightly varying apparent speed that depends on the seasons.
Refer to my Equation of Time (https://forum.tfes.org/index.php?topic=9478.msg148704#msg148704) post for more on the meaning of "mean".
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Refer to my Equation of Time (https://forum.tfes.org/index.php?topic=9478.msg148704#msg148704) post for more on the meaning of "mean".
Red herring, Tom. Or, if not intentionally a distraction, then misunderstanding of these concepts. That's a different variance aspect (and minor compared to the 0.24 solar day shift in equinox) that you can ignore while searching for understanding the why of that 0.24 day or ~6 hr "extra time." You're off on tangents with that and precession influences.
Do you honestly want to understand? Or are you vested in debating because you are sure you've found a flaw? I'm trying to avoid debate. You act like you're trying to educate me so that I can come to your conclusion that what you're educating me about is flawed.
If you don't know why what you're referring me to is a red herring, then don't lecture me as if we're debating. I chose to engage to help you understand and resolve the conundrum you think you've identified. If you're committed to that and can't fathom that you're conflating concepts, I'm wasting my time.
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It's not a red herring. The Solar Time hour is identical to the time in our clocks and watches, and a Solar Day is 24 Solar Time hours. It relates to the sun over one day:
https://www.britannica.com/science/solar-time#ref144523
Solar time, time measured by Earth’s rotation relative to the Sun. Apparent solar time is that measured by direct observation of the Sun or by a sundial.
A Tropical Year is 365.24 Solar Days, which is 24 Solar Hours x 365.24. Per the earlier discussion and the Recap post (https://forum.tfes.org/index.php?topic=9478.msg148664#msg148664), this does not match up to put Solar Noon into the spot it should be at for Solar Noon.
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It's not a red herring. The Solar Time hour is identical to the time in our clocks and watches, and a Solar Day is 24 Solar Time hours. It relates to the sun over one day:
When "mean solar time" and "equation of time" have zero to do with understanding the "extra" 0.24 hours that you say are confounding, invoking them is a red herring.
Those concepts matter insofar as why 0.24 hours isn't exact or why the exact value differs from year to year.
Forget that because it's not germane to comprehending the presence of the "extra time." You're either tying yourself in knots trying conflate concepts, or you're doing it intentionally to preserve a conclusion you've become invested in. I can't tell which. I want to believe you're making a sincere effort.
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What?
I've long since moved past sidereal day reference frame.
I don't know what you are trying to convey. Please restate your criticism.
I don't think that the equinox could only start at Solar Noon. It was an example. Any other value offset from Solar Noon would have the same issue since Solar Noon is always a certain number of hours away from it. An offset changes the place of Solar Noon. The 24 Hour Solar Day and the Solar Year must be in sync.
IT IS ALWAYS SOLAR NOON SOMEWHERE. This has nothing to do with sidereal anything.
Consider the number of radians in a circle. It’s 2*pi. I can assert that the number of times the radius goes into the circumference must be an integer, but I would be wrong.
You cannot just assert that solar day and solar year are related without any justification. The only justification I see from you implies that you think solar noon is a certain time (it’s not) or that you think the sun is “pointing” somewhere. Neither of these things are features of RET - they are either fundamental misunderstandings or straw men you have constructed.
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IT IS ALWAYS SOLAR NOON SOMEWHERE. This has nothing to do with sidereal anything.
It is always Solar Noon somewhere. You are correct. I corrected that post on how I worded that.
Start the 24 hour clock on the September Equinox point where it is 12PM Solar Noon at the point of the Equinox and it will end on ~6PM after one Solar Year. Why would it not stop at 12PM if you are in the same position on the earth's orbit around the sun, as per the diagrams?
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IT IS ALWAYS SOLAR NOON SOMEWHERE. This has nothing to do with sidereal anything.
It is always Solar Noon somewhere. Correct. But start the clock on the September Equinox point where it is 12PM Solar Noon and it will end on ~6PM after one year. Why doesn't it stop at 12PM if you are in the same position on the earth's orbit around the sun?
Because at that same location (not a point, but a meridian) at the 365th solar noon, the equinox is still ~6 hours away. The earth isn't going to stop its rotating to wait. The earth keeps turning as that meridian where the solar year began (almost) a year ago sees its local solar noon pass without the solar year ending. A little under 90° of longitude to the west, the equinox occurs and it's at that meridian's solar noon that the solar year is complete.
A little less than 90° of latitude equates a little less than a quarter of a rotation, a little less than a quarter of a day (0.24), or a little less than 6 hours.
I posted this a few hours ago.
It might help Tom to distinguish it that way.
Looking up times for the next two autumnal equinoxes, 2018 occurs 01:53 UTC on Sep 23. A year later, in 2019, it occurs 07:41 UTC.
Estimating using 15° of longitude per (solar) hour, that means that in 2018, solar noon at the time of autumnal equinox will be somewhere near 150°E meridian. But in 2019, solar noon at the time of autumnal equinox will be somewhere near the 70°E meridian.
The 365th solar noon will have passed those on the 150° meridian because the earth hasn't reached the autumnal equinox yet. It will take another 5 hrs and 48 mins to get there. Meanwhile, the earth will keep rotating with respect to the sun and the solar noon line will keep moving west for that amount of time until...equinox.
(My meridians might be off. I'm sure there's a calculator that can figure what longitude the sun is at given UTC, but I don't have a link handy. The point ought to be made, though.)
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IT IS ALWAYS SOLAR NOON SOMEWHERE. This has nothing to do with sidereal anything.
It is always Solar Noon somewhere. You are correct. I corrected that post on how I worded that.
Start the 24 hour clock on the September Equinox point where it is 12PM Solar Noon at the point of the Equinox and it will end on ~6PM after one Solar Year. Why would it not stop at 12PM if you are in the same position on the earth's orbit around the sun, as per the diagrams?
Because Earth's orbit around the Sun takes 365.24 days.
There's no need for the equinox to be always at 12 am at some (single) place.
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Bobby, can you provide a source on the equinox not being a point, "but a meridian"? Everything I have seen says that it is a point.
http://stars.astro.illinois.edu/celsph.html
Twice a year, the Sun crosses the equator, on or about March 20 at a point called the Vernal Equinox, and on September 23 at the Autumnal Equinox (the terms derived from a northern hemisphere perspective).
IT IS ALWAYS SOLAR NOON SOMEWHERE. This has nothing to do with sidereal anything.
It is always Solar Noon somewhere. You are correct. I corrected that post on how I worded that.
Start the 24 hour clock on the September Equinox point where it is 12PM Solar Noon at the point of the Equinox and it will end on ~6PM after one Solar Year. Why would it not stop at 12PM if you are in the same position on the earth's orbit around the sun, as per the diagrams?
Because Earth's orbit around the Sun takes 365.24 days.
There's no need for the equinox to be always at 12 am at some (single) place.
How then, does that reconcile with the 24 Hour Solar Day?
Solar Noon is always somewhere on earth. It will be somewhere on the point of the Equinox. The Solar Day is based on the sun rotation around the earth - 24 hours per rotation.
How do you explain the incompatibility between Solar Day and the Number of Solar Days in a Solar Year? Where does the extra time come from?
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(...)
IT IS ALWAYS SOLAR NOON SOMEWHERE. This has nothing to do with sidereal anything.
It is always Solar Noon somewhere. You are correct. I corrected that post on how I worded that.
Start the 24 hour clock on the September Equinox point where it is 12PM Solar Noon at the point of the Equinox and it will end on ~6PM after one Solar Year. Why would it not stop at 12PM if you are in the same position on the earth's orbit around the sun, as per the diagrams?
Because Earth's orbit around the Sun takes 365.24 days.
There's no need for the equinox to be always at 12 am at some (single) place.
How then, does that reconcile with the 24 Hour Solar Day?
Solar Noon is always somewhere on earth. It will be somewhere on the point of the Equinox. The Solar Day is based on the sun rotation around the earth - 24 hours per rotation.
How do you explain the incompatibility between Solar Day and the Number of Solar Days in a Solar Year? Where does the extra time come from?
Why would I need to explain it? It simply is. There's no need for days to fit exactly into one year. It's just a number we've measured of units we've decided to use.
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The world record in 400m sprint is 43,03 s. Why isn't it an integer?
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There's no incompatibility, there's no "extra time" appearing from nowhere. Stop taking teaching tools as gospel, and think it through. You've been given all the information you need to sort this out. Multiple times and ways even.
I'm bowing out of this one I think. It's clear he doesn't wish to understand, is continuing to pretend he doesn't understand for some reason, or he never will because he somehow can't grasp basic concepts of motion. Best of luck to those who continue to try, and to you as well Tom.
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At the point of September Equinox the sun is illuminating half of the earth.
Solar Time is 24 Hours. One Solar Day is 24 Hours.
After 365.24 Solar Days the earth has returned to the starting point on the earth's orbit around the sun, yet illuminated differently. 24 Hour Solar Clock is misaligned. Solar Noon is no longer at 12PM in Solar Time.
Can't be explained by the 20 minute Sidreal Year difference.
Big Problem.
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Bobby, can you provide a source on the equinox not being a point, "but a meridian"? Everything I have seen says that it is a point.
He didn't say the equinox was a meridian. He said that one meridian has its solar noon about six hours before the equinox, and some other meridian has its solar noon right at the time the Earth passes the equinox point.
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IT IS ALWAYS SOLAR NOON SOMEWHERE. This has nothing to do with sidereal anything.
It is always Solar Noon somewhere. You are correct. I corrected that post on how I worded that.
Start the 24 hour clock on the September Equinox point where it is 12PM Solar Noon at the point of the Equinox and it will end on ~6PM after one Solar Year. Why would it not stop at 12PM if you are in the same position on the earth's orbit around the sun, as per the diagrams?
Thanks for the correction.
As to your question, imagine you have a music box in your car with a ballerina on top that is spinning at a constant rate. You then drive around the block at a constant speed. Which way the ballerina is facing when you pass your house is not necessarily the same each time, unless you plan your speed very carefully.
If you got close to your house and then stopped when the ballerina was facing a certain way, you might be just short or just past your house.
The ballerina in your car doesn't know how fast you are driving and just keeps spinning.
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Bobby, can you provide a source on the equinox not being a point, "but a meridian"? Everything I have seen says that it is a point.
I understand how my word choice might be confusing. I just didn't want you to think that solar noon happens on a point the way equinox does.
Equinox is a point on the equator, and it's solar noon for all points along that meridian that intersects with the equator at that point. It might have been unnecessary for me to make that parenthetical but I was trying to keep clear how solar noon and equinox are related.
The difficulty I'm having is conveying to you that solar noon and equinox are the same point over the earth from equinox to equinox, 1 solar year apart. The equinox moves westward toward a new meridian where it's solar noon when the sun crosses the ecliptic (equator). It's solar noon for all of the points along that meridian but the sun is only directly overhead on the point on the equator.
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Correction: The equinox does have to start at Solar Noon at some point on earth, but not necessarily your point on earth.
Good edit. Does this mean you understand now where the extra time comes from?
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How can you post this:
Correction: The equinox does have to start at Solar Noon at some point on earth, but not necessarily your point on earth.
And then later ask this?
How then, does that reconcile with the 24 Hour Solar Day?
Solar Noon is always somewhere on earth. It will be somewhere on the point of the Equinox. The Solar Day is based on the sun rotation around the earth - 24 hours per rotation.
How do you explain the incompatibility between Solar Day and the Number of Solar Days in a Solar Year? Where does the extra time come from?
The answer is in your correction.
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IT IS ALWAYS SOLAR NOON SOMEWHERE. This has nothing to do with sidereal anything.
It is always Solar Noon somewhere. You are correct. I corrected that post on how I worded that.
Start the 24 hour clock on the September Equinox point where it is 12PM Solar Noon at the point of the Equinox and it will end on ~6PM after one Solar Year. Why would it not stop at 12PM if you are in the same position on the earth's orbit around the sun, as per the diagrams?
Thanks for the correction.
As to your question, imagine you have a music box in your car with a ballerina on top that is spinning at a constant rate. You then drive around the block at a constant speed. Which way the ballerina is facing when you pass your house is not necessarily the same each time, unless you plan your speed very carefully.
If you got close to your house and then stopped when the ballerina was facing a certain way, you might be just short or just past your house.
The ballerina in your car doesn't know how fast you are driving and just keeps spinning.
If a laser from your house was hitting the ballerina in the face at Point A on the circular path, and you KNOW that the ballerina will turn around 365.24 times after whatever method of rotating around your house, and then if the laser hits it in the face again when it returns to point A, you have a problem.
What is so hard to understand about this?
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At the point of September Equinox the sun is illuminating half of the earth.
Solar Time is 24 Hours. One Solar Day is 24 Hours.
After 365.24 Solar Days the earth has returned to the starting point on the earth's orbit around the sun, yet illuminated differently. 24 Hour Solar Clock is misaligned. Solar Noon is no longer at 12PM in Solar Time.
Can't be explained by the 20 minute Sidreal Year difference.
Big Problem.
Put this 20-minute sidereal year difference out of your head. That's not where the answer lies.
Now look at the statement you made that I bolded.
It's no longer solar noon at the point of equinox from the previous solar year. Why? Because that came earlier than the new equinox. That solar noon on that meridian, and that point on the equator, came 365 solar days after the previous one.
But the solar year is ~6 hours longer than 365 solar days.
The new equinox is occurring at a new meridian where it's solar noon, on a point on the equator about a quarter of a turn west from the previous year's point.
The 24 hour solar clock is "misaligned" from the 365.24 solar day solar year. I guess, yes, though I think "misaligned" sounds like you think it shouldn't be this way. But it is, so instead of "misaligned," I'd call it aligned 365.24 solar days to 1 solar year. There's no reason why it has to be 365:1. It's not a problem. It's just a different ratio. One that involves a fraction of the solar day instead of a whole number.
I've yet to understand why that's an issue for you.
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Bobby. The earth is like the ballerina in the analogy above. The laser is hitting it in the face again when you know that it needs to rotate 356.24 times.
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That's not the answer to the question. That is the question I am asking. The Solar Day and the Solar Year that is based on the Equinox do not match up.
And I answered that question few times in this thread already.
Solar day is based on Earth's rotation towards Sun.
Calendar year is based on solar days.
We, humans count it for our needs.
Tropical year is not based on solar days.
Tropical year is based on Earth's orbital events and it is not calendar year.
EDIT:
I already placed analogy with propeller of the plane model revolving pole.
We count our calendar year by counting "rotations of the propeller" (days), and trying to adjust "counted sets of rotations" (calendar year) as close as possible to "number of revolutions" (tropical year).
I have shown several quotes in this thread which say that there are 365.24 Solar Days in a Solar Year.
I have also shown with the Recap post that the Solar Day must be connected to the Solar Year.
You have not yet provided the + ~6 hour solution.
A Sidrael Year is only different by the Solar Year by 20 minutes, and is not a solution. Why doesn't the Solar Day fit into the Solar Year?
And it confirms that (Solar) Tropical year is not based on number of days but on orbital events (position).
What is based on days is Calendar year.
I already asked you "Were you expecting to synchronize rotation and revolution?"
After 365 rotations towards Sun (Calendar year), Earth simply haven't yet reached exact point where Sun apparently crosses Equator (Tropical year).
When Sun crosses Equator Earth will simply have some other meridian facing Sun.
EDIT:
Do you describe difference between Calendar year and Tropical year in some other way?
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Bobby. The earth is like the ballerina in the analogy above. The laser is hitting it in the face again when you know that it needs to rotate 356.24 times.
Except that the Earth is not a ballerina. The laser wouldn't hit the same place on the Earth again, solar noon is at a different place after 365,24 rotations. Or do you have data that would prove me wrong?
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Bobby. The earth is like the ballerina in the analogy above. The laser is hitting it in the face again when you know that it needs to rotate 356.24 times.
No. The earth is like the ballerina and the laser is hitting her in the ear after an additional 0.24 turn.
You are insisting on that 365.00 solar days must = 1 solar year and if it's not, we've got a problem.
Looking at your analogy...
...imagine you have a music box in your car with a ballerina on top that is spinning at a constant rate. You then drive around the block at a constant speed. Which way the ballerina is facing when you pass your house is not necessarily the same each time, unless you plan your speed very carefully.
If you got close to your house and then stopped when the ballerina was facing a certain way, you might be just short or just past your house.
The ballerina in your car doesn't know how fast you are driving and just keeps spinning.
If a laser from your house was hitting the ballerina in the face at Point A on the circular path, and you KNOW that the ballerina will turn around 365.24 times after whatever method of rotating around your house, and then if the laser hits it in the face again when it returns to point A, you have a problem.
What is so hard to understand about this?
Yeah. That would raise a discrepancy. It would mean you have a whole number ratio of ballerina spin and orbit around the block (365:1), and the extra 0.24 amount of spin that you thought you knew was true would be wrong.
But that's not the case with the earth's solar day rotation (ballerina spinning) and earth's solar year orbiting (drive around the block). The laser from the house DOESN'T hit the ballerina's face again at Point A. It happens 0.24 of a spin later, on her ear. That's the analogy to what's happening on earth. The equinox moves west a quarter of an earth's turn, just like we'd expect the laser to hit the ballerina's side of the head and not her face if we know the ballerina will turn around 365.24 and not 365.00 times during the one trip around the block.
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I already asked you "Were you expecting to synchronize rotation and revolution?"
The earth is in the same position on the orbit from where it started and 24 Hour Solar Time needs to have the sun at 12 Noon. There does need to be synchronization.
After 365 rotations towards Sun (calendar year), Earth simply didn't yet reached exact point where Sun apparently crosses Equator (Tropical year).
When Sun crosses Equator Earth will simply have some other meridian facing Sun.
EDIT:
Do you describe difference between Calendar year and Tropical year in some other way?
Refer to the analogy from a little earlier. The laser is hitting the ballerina in the face after the circuit around the house when the ballerina has rotated 365.24 times.
Replace ballerina with the 24 hour Solar Clock and you will see the issue.
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IT IS ALWAYS SOLAR NOON SOMEWHERE. This has nothing to do with sidereal anything.
It is always Solar Noon somewhere. You are correct. I corrected that post on how I worded that.
Start the 24 hour clock on the September Equinox point where it is 12PM Solar Noon at the point of the Equinox and it will end on ~6PM after one Solar Year. Why would it not stop at 12PM if you are in the same position on the earth's orbit around the sun, as per the diagrams?
Thanks for the correction.
As to your question, imagine you have a music box in your car with a ballerina on top that is spinning at a constant rate. You then drive around the block at a constant speed. Which way the ballerina is facing when you pass your house is not necessarily the same each time, unless you plan your speed very carefully.
If you got close to your house and then stopped when the ballerina was facing a certain way, you might be just short or just past your house.
The ballerina in your car doesn't know how fast you are driving and just keeps spinning.
If a laser from your house was hitting the ballerina in the face at Point A on the circular path, and you KNOW that the ballerina will turn around 365.24 times after whatever method of rotating around your house, and then if the laser hits it in the face again when it returns to point A, you have a problem.
What is so hard to understand about this?
The laser doesn't hit the ballerina in the face after 365.24 rotations. It his the ballerina in the face .24 rotations sooner, and at point A it hits the ballerina in the ear or something.
You are right, If the laser were coming from a different direction that would be a problem. It's not, and it isn't coming from a different direction in the Earth case either.
The ballerina gets hit in the face after 365 rotations, gets hit in the ear at 365.24 rotations at point A, and gets hit in the face again after point A at 366 rotations.
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Bobby. The earth is like the ballerina in the analogy above. The laser is hitting it in the face again when you know that it needs to rotate 356.24 times.
No. The earth is like the ballerina and the laser is hitting her in the ear after an additional 0.24 turn.
You are insisting on that 365.00 solar days must = 1 solar year and if it's not, we've got a problem.
Looking at your analogy...
...imagine you have a music box in your car with a ballerina on top that is spinning at a constant rate. You then drive around the block at a constant speed. Which way the ballerina is facing when you pass your house is not necessarily the same each time, unless you plan your speed very carefully.
If you got close to your house and then stopped when the ballerina was facing a certain way, you might be just short or just past your house.
The ballerina in your car doesn't know how fast you are driving and just keeps spinning.
If a laser from your house was hitting the ballerina in the face at Point A on the circular path, and you KNOW that the ballerina will turn around 365.24 times after whatever method of rotating around your house, and then if the laser hits it in the face again when it returns to point A, you have a problem.
What is so hard to understand about this?
Yeah. That would raise a discrepancy. It would mean you have a whole number ratio of ballerina spin and orbit around the block (365:1), and the extra 0.24 amount of spin that you thought you knew was true would be wrong.
But that's not the case with the earth's solar day rotation (ballerina spinning) and earth's solar year orbiting (drive around the block). The laser from the house DOESN'T hit the ballerina's face again at Point A. It happens 0.24 of a spin later, on her ear. That's the analogy to what's happening on earth. The equinox moves west a quarter of an earth's turn, just like we'd expect the laser to hit the ballerina's side of the head and not her face if we know the ballerina will turn around 365.24 and not 365.00 times during the one trip around the block.
Provide a source for this claim that "the equator moves".
-
If you believe 1 solar year = 365 solar day, but the equinox occurs 365.24 solar days later, then you've got a problem to resolve. (your premise)
If you believe 1 solar year = 365.24 solar days, but the equinox occurs 365.00 solar days later, then you've got a problem to resolve. (your analogy)
But...
If we say 1 solar year = 356.24 solar days and the equinox does, in fact, occur 365.24 solar days later, we've got a good match.
You don't seem to like it because of that 0.24 of a solar day 'extra time,' claiming it doesn't match. Doesn't match with what? 365 solar days. Why does it have to match with that? It matches with 365.24.
-
The ballerina gets hit in the face after 365 rotations, gets hit in the ear at 365.24 rotations at point A, and gets hit in the face again after point A at 366 rotations.
I'm glad you agree that either Solar Time or the Solar Year are wrong.
-
I already asked you "Were you expecting to synchronize rotation and revolution?"
The earth is in the same position on the orbit from where it started and 24 Hour Solar Time needs to have the sun at 12 Noon. There does need to be synchronization.
Earth is NOT in the same position, that's where Calendar year and Tropical year differ.
After 365 rotations Earth is 0.24 degrees earlier than full Tropical revolution.
After 365 rotations towards Sun (calendar year), Earth simply didn't yet reached exact point where Sun apparently crosses Equator (Tropical year).
When Sun crosses Equator Earth will simply have some other meridian facing Sun.
EDIT:
Do you describe difference between Calendar year and Tropical year in some other way?
Refer to the analogy from a little earlier. The laser is hitting the ballerina in the face after the circuit around the house when the ballerina has rotated 365.24 times.
Replace ballerina with the 24 hour Solar Clock and you will see the issue.
The ballerina didn't finish the full circle around the house after whole number of 365 rotations. That's why she has additionla 0.242 rotations in the remaining 1/4 of a foot.
-
...The equinox moves...
Provide a source for this claim that "the equator moves".
I'll assume this snark was just a mistake in reading what I wrote.
-
The ballerina gets hit in the face after 365 rotations, gets hit in the ear at 365.24 rotations at point A, and gets hit in the face again after point A at 366 rotations.
I'm glad you agree that either Solar Time or the Solar Year are wrong.
I don't see how you got there from what I said.
You are the one who insists that the ballerina gets hit in the face somehow after a non integer number of rotations. That is not what happens.
-
The ballerina gets hit in the face after 365 rotations, gets hit in the ear at 365.24 rotations at point A, and gets hit in the face again after point A at 366 rotations.
I'm glad you agree that either Solar Time or the Solar Year are wrong.
Ballerina's 365 rotations : Earth's 365 solar days
Ballerina getting hit on ear by laser after 365.24 rotations : Earth's equinox after 365.24 solar days (1 solar year)
There's nothing wrong. It's only wrong if the laser DOES hit the ballerina's face (1 solar year is really an even 365 days) while we keep insisting a solar year is 365.24. But it's not. The laser hits the ballerina's ear and equinox occurs on the earth a quarter of a world away from where it did the previous equivalent equinox.
It's really only you that's stuck on the laser hitting the ballerina's face and believing that 1 solar year must be equal to 365 solar days. You see it as wrong because you have an unbending, incorrect premise that 1 solar year has to equal 365 solar days.
-
At the point of September Equinox the sun is illuminating half of the earth.
Solar Time is 24 Hours. One Solar Day is 24 Hours.
After 365.24 Solar Days the earth has returned to the starting point on the earth's orbit around the sun, yet illuminated differently.
Yes. Because of the .24. 365.24 is a count of the number of times the earth has rotated. That's what a day is.
So the earth is in the same place but it isn't in the same orientation, it has done 365 complete rotations and then .24 of another.
Which is about a quarter. And what's a quarter of a 24 hour day? 6! There's your 6 hours.
Hooray! I've solved another mystery of the univers.
-
This is getting to be genuinely funny now.
Seeing us all try dozens of ways to explain the same thing to Tom over and over again, while Tom finds ever more creative ways to sidestep those explanations and keep the debate going.
At this point it is so clear that Tom is yanking our collective chain that I'm now more interested in why people are still trying to explain this to him?
He already knows. Anyone past the 5th grade would have understood by now, and clearly Tom is not stupid.
-
He already knows. Anyone past the 5th grade would have understood by now, and clearly Tom is not stupid.
You say that...he thinks that shadows change angle because of perspective and that spectroscopy is looking at something and thinking "ooh, that's a bit red".
There are a load of things he doesn't seem to understand but thinks he understands.
But...yeah, maybe he does understand and is having fun. Which is fine, other people will see the arguments for what they are.
-
He already knows. Anyone past the 5th grade would have understood by now, and clearly Tom is not stupid.
You say that...he thinks that shadows change angle because of perspective and that spectroscopy is looking at something and thinking "ooh, that's a bit red".
There are a load of things he doesn't seem to understand but thinks he understands.
But...yeah, maybe he does understand and is having fun. Which is fine, other people will see the arguments for what they are.
Other people will also see the debate as some kind of validation for his nonsense assertion.
Isn't there a risk that debating nonsense makes everyone a little dumber.
-
I already asked you "Were you expecting to synchronize rotation and revolution?"
The earth is in the same position on the orbit from where it started and 24 Hour Solar Time needs to have the sun at 12 Noon. There does need to be synchronization.
After 365 rotations towards Sun (calendar year), Earth simply didn't yet reached exact point where Sun apparently crosses Equator (Tropical year).
When Sun crosses Equator Earth will simply have some other meridian facing Sun.
EDIT:
Do you describe difference between Calendar year and Tropical year in some other way?
Refer to the analogy from a little earlier. The laser is hitting the ballerina in the face after the circuit around the house when the ballerina has rotated 365.24 times.
Replace ballerina with the 24 hour Solar Clock and you will see the issue.
Do you have any empirical evidence for that?
-
Start the 24 hour clock on the September Equinox point where it is 12PM Solar Noon at the point of the Equinox and it will end on ~6PM after one Solar Year. Why would it not stop at 12PM if you are in the same position on the earth's orbit around the sun, as per the diagrams?
.. because the diagrams are an approximation, not an exactness.
Work through the data for each day from 1 - 182.5 and calculate the angles, as per the diagram in the first few pages (EDIT - Reply #135, and in the Wikipedia page for Solar Time (https://en.wikipedia.org/wiki/Solar_time)), which illustrated the sidereal day. See what happens when you work with actual data, not high-level illustrations.
-
He already knows. Anyone past the 5th grade would have understood by now, and clearly Tom is not stupid.
You say that...he thinks that shadows change angle because of perspective and that spectroscopy is looking at something and thinking "ooh, that's a bit red".
There are a load of things he doesn't seem to understand but thinks he understands.
But...yeah, maybe he does understand and is having fun. Which is fine, other people will see the arguments for what they are.
Other people will also see the debate as some kind of validation for his nonsense assertion.
Isn't there a risk that debating nonsense makes everyone a little dumber.
I don't think nonsense should be left unchallenged but I do think we've indulged him for too long in this thread.
Quite how you can think that a year is 365.24 days and then be confused that the number of rotations of earth in a year is not an integer is beyond me.
I do suspect at times he's only in this for a laugh and does really understand all this.
-
Okay I’m calling a halt to this, now.
This non -argument was pretty much wrapped at the end of the first page, Tumeni doing much of the lifting work, by the time GG, Macarios, the tree Rat, AATW and others had weighed in it was finished.
Since then you’ve all been running around like cats chasing a laser pointer.
Either Bish’ doesn’t understand that the day is the night and day cycle, and the year isn’t linked, and we have been trying with leap years/seconds to reconcile the two with increasingly complex tweaks/success, in which case he is an idiot and needs to look at all the other planets and point to a precedent for his premiss, or he is fucking with you.
I know it is hard to walk away as he (they) will no doubt chalk it up to a victory, but anyone else will see that you won this early on and he (they) are just trying to bury the fact in heaps of repetition.
Do a wind-up statement as concise as possible and leave him, and for your own sanity as well as mine resist all enticements to return.
Jura has spoken!
-
Okay I’m calling a halt to this, now.
This non -argument was pretty much wrapped at the end of the first page, Tumeni doing much of the lifting work, by the time GG, Macarios, the tree Rat, AATW and others had weighed in it was finished.
...
Do a wind-up statement as concise as possible and leave him, and for your own sanity as well as mine resist all enticements to return.
I tired to do that with reply #10, but ...
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.
Why should I/we indulge you, and explain this in great detail, when it has already been detailed?
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. Surely you won't conclude that you're right, and they're all wrong?
-
"But these go to eleven"
-
My last question to Tom was going to be, well, if there's something wrong with solar noon in the round earth "theory," how would flat earth explain it.
But then I read the words of Samuel Birley Rowbotham: "... the Zetetic process only permits us to say:--'The peculiar motion is visible to us, but, of the cause, at present we are ignorant.'"
I think that until a flat earth model comes up with something better, I'll be okay with Tom Bishop having a problem with where the "extra time" comes from in current, conventional Earth Is A Globe explanations and calculations.
-
At the point of September Equinox the sun is illuminating half of the earth.
Solar Time is 24 Hours. One Solar Day is 24 Hours.
After 365.24 Solar Days the earth has returned to the starting point on the earth's orbit around the sun, yet illuminated differently.
Yes. Because of the .24. 365.24 is a count of the number of times the earth has rotated. That's what a day is.
So the earth is in the same place but it isn't in the same orientation, it has done 365 complete rotations and then .24 of another.
Which is about a quarter. And what's a quarter of a 24 hour day? 6! There's your 6 hours.
Hooray! I've solved another mystery of the univers.
The problem seems to be that you still seem to misunderstand what a Solar Day is. The earth makes one full rotation in a Solar Day. In a Solar Day, and in Solar Time, Noon is always 12PM, and is facing the sun. Midnight is 12 AM. The sun can also be thought to going around the earth in a Solar Da6y.
The earth can't rotate 24 full Solar Days in a 365.24 Solar Day Year. Impossible.
The sun can't rotate around the earth 365.24 times. Impossible.
Something will be misaligned when the earth returns to the same point on the orbit around the sun.
The illuminated portion of the earth will be pointing in the wrong direction. It needs to point in the same geometric position it started in on the earth's orbit around the sun.
-
Yes, Tom. It's definitely me who is confused. Just like with the laser and the boat.
Jura is right, you've had your fun.
</thread>
-
When my daughter had "a problem with" algebra, she didn't argue as if she understood algebra and it was my failing to not recognize that algebra was flawed.
-
Yes, Tom. It's definitely me who is confused. Just like with the laser and the boat.
Jura is right, you've had your fun.
</thread>
You are still arguing that the rotation of the earth in Solar Time and the duration of the year are unconnected, when this is clearly not the case.
Solar Time is the illuminated portion of the earth turning around the earth. Think of it like that. 1 rotation in 24 hours.
Looking at the earth on the orbit around the sun in a diagram, the illuminated portion of the earth will be misaligned with the direction of the sun at the end after 365.24 days when the earth returns to the point on the orbit around the sun that it started in.
This is a very clear depiction of the situation, and a refusal to accept the problem is mere denial.
-
The problem seems to be ....
The earth can't rotate 24 Solar Days in a 365.24 Solar Day Year.
Time to call a halt, Tom, when you can't even muster the right figures.
24 days is less than a month.
365 calendar days in a year, 366 in a leap year
365.24 days in a tropical year
-
Yes, Tom. It's definitely me who is confused. Just like with the laser and the boat.
Jura is right, you've had your fun.
</thread>
You are still arguing that the rotation of the earth in Solar Time and the duration of the year are unconnected, when this is clearly not the case.
Solar Time is the illuminated portion of the earth turning around the earth. Think of it like that. 1 rotation in 24 hours.
Looking at the earth on the orbit around the sun in a diagram, the illuminated portion of the earth will be misaligned with the direction of the sun at the end after 365.24 days when the earth returns to the point on the orbit around the sun that it started in.
This is a very clear depiction of the situation, and a refusal to accept the problem is mere denial.
If you're truly still not understanding this (and your second line indicates you aren't, assuming you aren't simply trolling at this point, something I find far more likely) it's no longer an issue of us failing to teach, but you failing to learn. You've had it all spelled out for you quite plainly. I would suggest you take some courses in Astronomy and perhaps Geometry at this point. You're not going to have your disconnect solved over an internet forum if you still fail to see how your basic assumptions are incorrect. Presuming you aren't trolling, this thread has certainly gone a long way to help show how/why you can believe so firmly in the Flat Earth.
-
The problem seems to be ....
The earth can't rotate 24 Solar Days in a 365.24 Solar Day Year.
Time to call a halt, Tom, when you can't even muster the right figures.
24 days is less than a month.
365 calendar days in a year, 366 in a leap year
365.24 days in a tropical year
Typo, clearly.
It should read: There cannot be full solar days in a 365.24 solar day year. There cannot be.
The Solar Day is the illuminated portion of the earth rotating around the sun. It will be misaligned with the position of the sun on ANY diagram of the earth's oval path around the sun when the earth returns to the starting point.
It really is not a difficult concept.
-
Yes, Tom. It's definitely me who is confused. Just like with the laser and the boat.
Jura is right, you've had your fun.
</thread>
You are still arguing that the rotation of the earth in Solar Time and the duration of the year are unconnected, when this is clearly not the case.
Solar Time is the illuminated portion of the earth turning around the earth. Think of it like that. 1 rotation in 24 hours.
Looking at the earth on the orbit around the sun in a diagram, the illuminated portion of the earth will be misaligned with the direction of the sun at the end after 365.24 days when the earth returns to the point on the orbit around the sun that it started in.
This is a very clear depiction of the situation, and a refusal to accept the problem is mere denial.
If you're truly still not understanding this (and your second line indicates you aren't, assuming you aren't simply trolling at this point, something I find far more likely) it's no longer an issue of us failing to teach, but you failing to learn. You've had it all spelled out for you quite plainly. I would suggest you take some courses in Astronomy and perhaps Geometry at this point. You're not going to have your disconnect solved over an internet forum if you still fail to see how your basic assumptions are incorrect. Presuming you aren't trolling, this thread has certainly gone a long way to help show how/why you can believe so firmly in the Flat Earth.
If you do not want to participate, leave.
There is no explanation for why the illuminated portion of the earth does not line up with the sun on a diagram after 365.24 Solar Days.
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The Solar Day is the illuminated portion of the earth rotating around the sun. It will be misaligned with the position of the sun on ANY diagram of the earth's oval path around the sun when the earth returns to the starting point.
It really is not a difficult concept.
No, it isn't.
But you are the only person on this thread who thinks that the earth spinning a non-integer number of times in a solar year is a "problem".
Or, more likely, you're just stringing people along for your own amusement.
-
The Solar Day is the illuminated portion of the earth rotating around the sun. It will be misaligned with the position of the sun on ANY diagram of the earth's oval path around the sun when the earth returns to the starting point.
It really is not a difficult concept.
No, it isn't.
But you are the only person on this thread who thinks that the earth spinning a non-integer number of times in a solar year is a "problem".
Or, more likely, you're just stringing people along for your own amusement.
You CAN'T explain it. The illuminated portion of the earth spins around 365.25 times in a Solar Year.
The illuminated portion needs to be pointing at the sun when it reaches the point it started from.
On a diagram of the earth going around the sun the problem is clear. The illuminated portion needs to be pointing in the same place.
-
On a diagram of the earth going around the sun the problem is clear. The illuminated portion needs to be pointing in the same place.
Why?
-
On a diagram of the earth going around the sun the problem is clear. The illuminated portion needs to be pointing in the same place.
Why?
The illuminated portion needs to point at the sun because it comes from the sun. The illuminated portion can't turn 365.24 times on a diagram that illustrates the earth going around the sun to its same spot on the oval path.
-
On a diagram of the earth going around the sun the problem is clear. The illuminated portion needs to be pointing in the same place.
Why?
He's hung up on the illustrations that don't show the extra quarter turn of the globe believing after a full solar year. He thinks the same meridian needs to be facing the earth after 365.24 solar days and that that doesn't make sense if the earth is supposed to have rotated more.
I don't know how to explain it to him any better. I don't want to believe he's messing with us. But I don't know how to proceed without being condescending.
-
On a diagram of the earth going around the sun the problem is clear. The illuminated portion needs to be pointing in the same place.
Why?
He's hung up on the illustrations that don't show the extra quarter turn of the globe believing after a full solar year. He thinks the same meridian needs to be facing the earth after 365.24 solar days and that that doesn't make sense if the earth is supposed to have rotated more.
I don't know how to explain it to him any better. I don't want to believe he's messing with us. But I don't know how to proceed without being condescending.
You are misunderstanding a Solar Day and what this is referring to. This is not referring to one point or meridian on earth.
This is referring to the illuminated portion of the earth that rotates around it at one rotation per Solar Day, and which also must work with 365.24 Solar Days in a Solar Year.
In a top down diagram of the earth traveling around the sun, the illuminated portion needs to be the same when it returns to the same point. It needs to point at the sun when it gets to the same point it started from, since obviously, that illuminated portion comes from the sun.
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I don't know how to explain it to him any better. I don't want to believe he's messing with us. But I don't know how to proceed without being condescending.
This is the part of the earth that will be facing the sun at the next autumnal equinox (2018):
(http://oi68.tinypic.com/514svo.jpg)
-----------------------------
365 solar days later (2019), the same part of the world will be facing the sun at the same time.
But it won't be autumnal equinox yet.
The sun will not have crossed the ecliptic yet.
That won't happen for another 5 hrs and 48 mins when the sun is over the Indian Ocean.
A solar year will not have yet transpired at this point when the sun is illuminating this part of the world:
(http://oi64.tinypic.com/10q9jjt.jpg)
-----------------------------
This is the autumnal equinox in 2019.
This is where the sun will be crossing the equator from north to south in 2019.
Here is where 356.24217 solar days will have passed since the 2018 autumnal equinox.
A new side of the earth will be experiencing solar noon at the equinox.
(http://oi63.tinypic.com/15hbes8.jpg)
These images are more accurate than a simplistic diagram/animation that is intended to explain a concept and meant to cause confusion.
These should help anyone having difficulty with this solar year/solar day/equinox relationship comprehend it.
-
I think this guy is really sure that he found a big hole in the whole calendar system, overlooked by generations of astronomers for hundreds, even thousands of years. At that it needs a smart ass guy as this flat earther to show it to the world. That's the real flat earth spirit :-)
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It's clear I cannot explain this in a way that will make sense to you, so I will stop trying.
This video covers this problem in detail, maybe it will help.
Be aware that the animation at 11:30 is simplified and shows the wrong rotation just like you imagine it. He goes into a more detailed animation immediately thereafter.
https://youtu.be/IJhgZBn-LHg
-
This video covers this problem in detail, maybe it will help.
I contemplated posting that because he provides a good explanation of what we're talking about in light-hearted way that is nonetheless correct (at least I didn't have any quibbles).
But I thought the last part of the video would be too much of a distraction.
-
On a diagram of the earth going around the sun the problem is clear. The illuminated portion needs to be pointing in the same place.
Why?
The illuminated portion needs to point at the sun because it comes from the sun.
No idea what that's supposed to mean.
The illuminated portion can't turn 365.24 times on a diagram that illustrates the earth going around the sun to its same spot on the oval path.
I've highlighted your problem. The diagrams you're looking at are for illustration only and don't exactly represent the reality. After a solar year the earth is in the same PLACE in its orbit but is not in the same orientation, it's a about a quarter turn different, there's your .24 days or 6 hours.
-
Typo, clearly.
It should read: There cannot be full solar days in a 365.24 solar day year. There cannot be.
(It read, originally; "The earth can't rotate 24 Solar Days in a 365.24 Solar Day Year")
Wow, that's some typo.
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The illuminated portion can't turn 365.24 times on a diagram that illustrates the earth going around the sun to its same spot on the oval path.
Once more; the diagrams are an approximation, not an exactness.
They represent a high-level indication of what is going on, not the exact data relating to the scenario.
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It should read: There cannot be full solar days in a 365.24 solar day year. There cannot be.
(It read, originally; "The earth can't rotate 24 Solar Days in a 365.24 Solar Day Year")
I don't know what the corrected edit means.
There ARE 365 full solar days in a 365.24 solar day solar year.
Red is for irony, in case it isn't obvious.
These diagrams are what Tom has cited, and they don't show an earth rotating through an extra 0.24 solar day after 1 solar year. Ergo, the 0.24 'extra time" must be wrong since these must be accurate?
(https://s3.amazonaws.com/ai2-vision-textbook-dataset/dataset_releases/rc2/train/question_images/seasons_6284.png)
(https://www.e-education.psu.edu/meteo300/sites/www.e-education.psu.edu.meteo300/files/images/lesson6/EarthOrbit.png)
This simulation is what Tom has sited, and it doesn't even show the extra 0.24 solar day (or the earth's tilt). Ergo, the 0.24 "extra time" must be wrong since this must be accurate? If we dispute it, are we smarter than the astronomy department at the University of Nebraska-Lincoln? UNL professors certainly must be supporting Tom's identification of "the Problem" since they authorized publication of this simulation showing no extra time:
(http://oi64.tinypic.com/2i07mhh.jpg)
Everyone knows that a solar year MUST be -- has to be - exactly 365 solar days because the same part of the world needs to face the sun 365 solar days later.
But then how can the solar year ALSO be defined by the crossing of the ecliptic plane by the sun if it doesn't happen on the 365th solar day?
That's "a problem."
"The Solar Day and the Solar Year that is based on the Equinox do not match up."
The diagrams show the globe earth having a solar year of 365 days. Not 365.24 days.
Solar year according to the diagrams is 365 solar days.
Solar year according to equinox is 365.24 solar days.
It doesn't match up (right Tom?)
-----------
Let's turn to the FE simulation for the right answer.
(https://i.imgur.com/D7aOXVA.gif)
Hmm. This shows the sun never leaving the equator. Is it always "equinox" according to flat earth? This diagram shows it to be so.
If not -- if, contrary to the animation, the sun actually spirals between the Tropics of Cancer and Capricorn, when does that happen. Here?
(http://oi67.tinypic.com/15nogw6.jpg)
And if the sun spirals around 365 more days after that, will it be back in the same position and it will cross the equator again on the same day, and the same time? Where's the diagram? The simulation? The Flat Earth calculator telling us the correct time and date of the sun's crossing of the equator?
If it's every 365 days with no fractions, hey! No need for leap anythings. There's no drift in the point of equinox. There's no need for an added day every four years. The sun's crossing of the equator will happen over the same point above the earth every year.
But what if there is variation in the timing and placing of the equinox? Even if we do see the solstices and equinoxes drift, and the seasons start to shift, we know it can't be because of some specious 1/4 of a day extra time arising out of nowhere. Right? But we just may never know why, because "... the Zetetic process only permits us to say:--'The peculiar motion is visible to us, but, of the cause, at present we are ignorant.'"
-
"But these go to eleven"
Heyo, keep the shit posting in CN. Warned.
-
It's clear I cannot explain this in a way that will make sense to you, so I will stop trying.
This video covers this problem in detail, maybe it will help.
Be aware that the animation at 11:30 is simplified and shows the wrong rotation just like you imagine it. He goes into a more detailed animation immediately thereafter.
https://youtu.be/IJhgZBn-LHg (https://youtu.be/IJhgZBn-LHg)
He does not address the problem. He's talking about how we use calendars to account for the .24 extra day with leap years and such.
The Leap Year/Leap Day was created to account for the extra .24. The .24 was not created to account for the Leap Year.
-
It's clear I cannot explain this in a way that will make sense to you, so I will stop trying.
This video covers this problem in detail, maybe it will help.
Be aware that the animation at 11:30 is simplified and shows the wrong rotation just like you imagine it. He goes into a more detailed animation immediately thereafter.
https://youtu.be/IJhgZBn-LHg (https://youtu.be/IJhgZBn-LHg)
He does not address the problem. He's talking about how we use calendars to account for the .24 extra day with leap years and such.
The Leap Year/Leap Day was created to account for the extra .24. The .24 was not created to account for the Leap Year.
You should be discussing this in a science forum, not here. There you would find experts.
-
Everyone knows that a solar year MUST be -- has to be - exactly 365 solar days because the same part of the world needs to face the sun 365 solar days later.
Bobby, this has nothing to do with "facing the same part of the world." You still have a misunderstanding of how Solar Time/Solar Day works.
Solar Day represents the rotation of the following illuminated part of the earth over the earth. The illuminated half is rotating around the earth once per Solar Day.
Top-Down View of Earth Day/Night:
(https://msresource.files.wordpress.com/2011/11/image4.png)
The illuminated area in the top down scene above is rotating around the static earth at a rate of 1 rotation in 1 Solar Day.
(https://www.shareicon.net/data/128x128/2015/10/30/664014_arrows_512x512.png)
Next we have the earth orbiting around the sun.
(https://my.homecampus.com.sg/images/notes/P6_speed_advanced_problem_earth_revolution.png)
Any diagram. Circle Shape. Oval Shape. It doesn't matter.
The Illuminated Day side of the earth is always pointing at the sun.
(https://i.imgur.com/WnjABbF.gif)
The Equinox intersects a point on the path of the earth around the sun.
(https://i.imgur.com/aRDIuL2.gif)
There are 364.24 Solar Days in one Solar Year. It takes one Solar Year to return to that point on the orbit.
When the earth starts on the Equinox Point the illuminated portion of the earth is looking at the sun.
When the earth gets to the Equinox Point again on the orbit, after one Solar Year/Tropical Year, the day and night cycle of Solar Time / Solar Day is misaligned, and will not point at the sun again.
Why?
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The illuminated portion can't turn 365.24 times on a diagram that illustrates the earth going around the sun to its same spot on the oval path.
Once more; the diagrams are an approximation, not an exactness.
They represent a high-level indication of what is going on, not the exact data relating to the scenario.
This has nothing to do with the accuracy of the diagram. The earth returns to its same point on its path around the sun. The illuminated part must be pointing at the sun. If there are 365.24 Solar Days in a Solar Year/Tropical Year, then it is not pointing at the sun.
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The western Pacific Ocean side of the earth will be pointed toward the sun on the next autumnal equinox.
365 solar days later, the western Pacific Ocean side of the earth will be again be pointed toward the sun.
At 2019's autumnal equinox, ~6 hours into the 366th solar day since the 2018 autumnal equinox, the Indian Ocean side of the earth will be pointed toward the sun.
365 even solar days is not 1 solar year. The autumnal equinox in 2019 will not happen while the Western Pacific is experiencing solar noon like it will in 2018. The equinox will move to the west by a quarter turn of earth. The side of the earth being illuminated will thus be different too...over the Indian Ocean instead of the western Pacific.
You want to stop the solar year clock when the sun illuminates the same side of the earth as it did the previous year, but the earth doesn't stop rotating to wait for the orbit to complete. Solar year is an orbit measurement. Not a rotation measurement. Solar year and solar day aren't going to line up in the same way at the end of each solar year. It's not a problem. It's just a ratio.
(http://oi63.tinypic.com/2elz8uc.jpg)
Here's a gear system of ratio 30:7. Line them up with the red indices and then start turning. After 1 complete turn of the large gear (call it a gear year) the small gear will have rotated 4.29 times and the red lines won't line up. So what? If 1 turn of the small gear relative to the large gear is a "gear day" and 1 turn of the large gear is a gear year, you won't get a gear year being an even multiple of a gear day unless the ratio is a whole number. You shouldn't expect to.
(http://oi63.tinypic.com/2ltng9l.jpg)
Earth & sun aren't meshed like gears. There's no connection (I'm aware of) that makes earth's rotation period contingent on earth's orbit period. But they each have measure and the ratio is 356.24217 to 1. Not 356 to 1. Solar year isn't both 356 solar days AND equinox to like equinox. It's only the latter. The former is not a solar year.
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You want to stop the solar year clock when the sun illuminates the same side of the earth as it did the previous year, but the earth doesn't stop rotating to wait for the orbit to complete.
It's not the same side of the physical world. The earth needs to be lit from the same side side as in starts in the diagram.
Refer to the last post I created for you.
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You want to stop the solar year clock when the sun illuminates the same side of the earth as it did the previous year, but the earth doesn't stop rotating to wait for the orbit to complete.
It's not the same side of the physical world. The earth needs to be lit from the same side side as in starts in the diagram.
Refer to the last post I created for you.
Okay. Reading it next, because I cannot fathom why you think the earth needs to be lit from the same side as it starts. Let's see if your last post to me explains why.
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It's not the same side of the physical world. The earth needs to be lit from the same side side as in starts in the diagram.
For why? To satisfy what rule?
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It's not the same side of the physical world. The earth needs to be lit from the same side side as in starts in the diagram.
For why? To satisfy what rule?
The Solar Day is in regards to illumination, not how fast the earth may be rotating or how fast a physical feature move or what they are aligned with on the earth's surface. It deals with Day and Night.
The Solar Day is defined by the sun moving around a static earth. It may also be spoken of as the rate the earth rotates from the earth's reference frame in regards to the sun, but the key here is "Sol," which means sun. It is sun time, and represents our day and night.
The earth is illuminated on the side facing the sun. It travels around the sun in 356.24 Solar Days. This does not make sense if it goes in a circle around the sun, back to its start point, and the Day/Night cycle is not in the same Solar Day.
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Bobby, this has nothing to do with "facing the same part of the world." You still have a misunderstanding of how Solar Time/Solar Day works.
Looking forward to being corrected.
Solar Day represents the rotation illuminated part of the earth over the earth. The illuminated half of the is rotating around the earth once per Solar Day.
Okay. You're describing it as the illumination sweeping over a static earth, but I get it. Same concept. Just a difference frame of reference. I'm with you.
Top-Down View of Earth Day/Night:
(https://msresource.files.wordpress.com/2011/11/image4.png)
The illuminated area in the top down scene above is rotating around the static earth at a rate of 1 rotation per 24 hours, or 1 rotation in 1 Solar Day.
(https://www.shareicon.net/data/128x128/2015/10/30/664014_arrows_512x512.png)
Got it. Same as before. Static earth. Motion of sunlight illumination. Now from a "polar" view. Still think I'm grasping the concept of a solar day so far. Rotation arrow is backwards if your talking static earth and rotating sun, but no biggie. I follow.
Next we have the earth orbiting around the sun.
(https://my.homecampus.com.sg/images/notes/P6_speed_advanced_problem_earth_revolution.png)
Any diagram. Circle Shape. Oval Shape. It doesn't matter.
The Illuminated Day side of the earth is always pointing at the sun.
(https://i.imgur.com/WnjABbF.gif)
Hmm. "The illuminated day side of the earth is always pointing at the sun." Isn't that self-explanatory? The side of the earth facing the sun is illuminated? Who needs a diagram to explain that? And since the earth is also rotating, the illuminated day side of the earth is always changing. I don't think we're breaking any new ground yet.
The Equinox intersects a point on the path of the earth around the sun.
(https://i.imgur.com/aRDIuL2.gif)
Not going to quibble about how you articulate that because I get the point. There a line where the equatorial plane and the ecliptic plane intersect, and an equinox occurs where the earth's path around the sun intersects that line. (Two such points per orbit).
I think I'm with you so far. Waiting to see where I'm misunderstanding.
There are 364.24 Solar Days in one Solar Year.
Yes
It takes one Solar Year to return to that point on the orbit.
Yes
When the earth starts at point A on the Equinox Point the illuminated portion of the earth is looking at the sun.
Sure. Whatever area on the earth that is facing the sun at that point is being illuminated. Not Where am I misunderstanding?
When the earth gets to point A again on the orbit, after one Solar Year/Tropical Year, the day and night cycle of Solar Time / Solar Day is misaligned, and will not point at the sun again.
Whoa! Everything was fine up to here. I don't understand what's misaligned. The earth is at point A again. The illuminated side is facing the earth.
The only thing substantially different is that the area on earth that is being illuminated is different this time at point A than it was last time at point A. But that can't be it because when I said that before you told me no.
Why?
Why what? Why is a different part of the earth pointing toward the sun and being illuminated by it compared to the start?
Overlaying your polar view image on a picture of static globe with the illumination side rotating around it...
(http://oi67.tinypic.com/24xm4jq.jpg)
...are you asking why the rotating illumination isn't over the same part of the earth as it was at the start?
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Why what? Why is a different part of the earth pointing toward the sun and being illuminated by it compared to the start?
We are just talking about the day and night cycle, that has a period of 24 hours, not a "part of the earth."
Why what? Why is a different part of the earth pointing toward the sun and being illuminated by it compared to the start?
Overlaying your polar view image on a picture of static globe with the illumination side rotating around it...[/color]
http://oi67.tinypic.com/24xm4jq.jpg
...are you asking why the rotating illumination isn't over the same part of the earth as it was at the start?
This has little to do with the "same part of the earth" that needs to be lined up. The day and night cycle of 24 hours does not match up to the Solar Year. The illuminated portion in the day and night cycle is misaligned with the sun if, when the earth returns to the Equinox point on the earth's path, there has not been a full Solar Day.
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Why what? Why is a different part of the earth pointing toward the sun and being illuminated by it compared to the start?
We are just talking about the day and night cycle, that has a period of 24 hours, not a "part of the earth."
Why what? Why is a different part of the earth pointing toward the sun and being illuminated by it compared to the start?
Overlaying your polar view image on a picture of static globe with the illumination side rotating around it...[/color]
http://oi67.tinypic.com/24xm4jq.jpg
...are you asking why the rotating illumination isn't over the same part of the earth as it was at the start?
This has little to do with the "same part of the earth" that needs to be lined up. The day and night cycle of 24 hours does not match up to the Solar Year. The illuminated portion in the day and night cycle is misaligned with the sun if, when the earth returns to the Equinox point on the earth's path, there has not been a full Solar Day.
I don't understand.
Here's your polar view of a generic earth with no identification of where or when the solar day begins in relationship to the physical earth surface. It's solar noon somewhere on the planet at 00:00. And if this is point A on the orbital plane at equinox, then the sun is illuminating the earth from a point over the equator on the right "half" of the 00:00 planet, viewing it from this angle.
Start solar day and solar year clocks.
(http://oi67.tinypic.com/zim9fk.jpg)
This is your chosen frame of reference for viewing the solar day cycle: a static earth and sun's illumination sweeping around the globe. That's fine. Each one of these cycles we agree is a solar day.
At the end of 365 of these solar days, the earth is oriented back like in the 00:00 image.
1. But is the earth back at Point A on the orbit?
2. Has a solar year elapsed yet?
3. Is the sun crossing the equator with the conclusion of this 365th solar day cycle?
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At the end of 365 of these solar days, the earth is oriented back like in the 00:00 image.
1. But is the the earth back at Point A on the orbit?
2. Has a solar year elapsed yet?
3. Is the sun crossing the equator with the conclusion of this 365th solar day cycle?
I'll go ahead and give you my answers and we'll see if yours compare.
1. No. The earth has not completed it's full orbital journey yet on that 365th solar day.
2. No. The solar year hasn't concluded yet even though the 365th solar day has.
3. No. Wherever over the earth the sun is illuminating that side of the planet, it is still N of the equator, however slightly. The Point A intersection of the ecliptic/equatorial plane and the earth's orbital path hasn't been intersected yet.
Using a N view of a static earth with sun's illumination rotating the face of the earth, the sun won't reach the equator until just under 6 hours later. While the sun's approaching the equator, it's light keeps rotating around the earth until:
(http://oi68.tinypic.com/qo8s92.jpg)
There's the equinox. There's the conclusion to the solar year. It takes fraction of the 366th solar day for the sun to finally reach a point directly overhead on the equator, which means the earth has completed one solar year orbit around the sun.
I look forward to seeing your take on these questions. Perhaps it'll help identify what you see as the problem (or why my understanding is incorrect that blinds me from the problem).
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It's not the same side of the physical world. The earth needs to be lit from the same side side as in starts in the diagram.
It is the same side, but not at the exactly same spot.
Explain yourself difference between Calendar year and Tropical year.
After full Solar day Sun is crossing Meridian.
After full Tropical year Sun is crossing Parallel.
There is no gear in the vacuum of space to synchronize these two different motions.
EDIT: If you still insist on gear, take small one with four teeth: morning, noon, evening, midnight.
Now, big one would have 1441 1461 (NOT 1440 1460) teeth around.
So, each four teeth are for one day of small gear, and the last 1441st 1461st is those "6 missing hours".
Calendar year is 365 rotations of small gear (1440 1460 teeth).
We wake up, go to work, go back home, go to sleep by rotations of small gear.
Tropical year is 365.25 rotations of small gear (full circle on 1441 1461 teeth).
We plow our land, and now also go to vacations, by seasons, not by days.
After one Calendar year "noon" will point 1 tooth less than full circle.
After one more Calendar year "noon" will point one more tooth earlier (2 teeth less than full circle).
In fourth year we have to count one more rotation (4 more teeth) to have our Calendar year point to the original first tooth of Tropical year.
That is just because we don't want July 1st to be in winter, we will not have warm sea in Belize then.
For example, Arabic calendar doesn't make those corrections. Their months circulate through the whole year and each month can be in any season.
https://en.wikipedia.org/wiki/Islamic_calendar (https://en.wikipedia.org/wiki/Islamic_calendar)
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You CAN'T explain it.
Well, not in a way you can understand, apparently. But I honestly think that says more about you than me.
You are the only person in this thread who thinks there is anything to explain here.
The illuminated portion of the earth spins around 365.25 times in a Solar Year.
Correct. Well, close enough. The whole earth spins 365.25 (roughly) times in a Solar Year.
That is 365.25 Solar Days. That's what a Solar Day IS - it's a rotation of the earth.
And a Solar Year is how long it takes the earth to orbit the sun. It takes 365.25 times as long to do that as it does to rotate on its axis.
The illuminated portion needs to be pointing at the sun when it reaches the point it started from.
No it doesn't. That would only be true if there were exactly 365 solar days in a solar year, or some other integer. But there aren't.
So if the sun is exactly overhead, say, New York at the start of one solar year it will not be at the start of the next, it will have been overhead about 6 hours previously.
You are the only person who sees this as a problem.
This is the entire reason we have leap years, to keep things in sync.
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(http://oi67.tinypic.com/zim9fk.jpg)
Above: Simple view of how the sun's illumination rotates around a globe earth during a single day, from a N pole view of the earth (northern hemisphere only in view).
Below: Simple view of how the sun's illumination circles around a flat earth during a single day, from a God's eye view of the earth (whole earth in view).
(http://oi64.tinypic.com/jac654.jpg)
Despite the obvious differences, are these two versions of a day on earth of the same duration? Of course, right?
On flat earth, is the sun directly overhead the same point on earth after 525,600 minutes ("how do you measure...measure a year?")
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Bobby and AllAround, you seem to have the same argument. This is how I have interpreted your argument. Correct me if I am wrong:
Response. The Solar Day, which is the rate the sun moves around a static earth (once per 24 hours), can also be thought of as the rate of the earth's rotation against a static sun (once per 24 hours).
At Day 365 the lit side of the earth is pointing at the sun. At Day 365.24, with a static sun, the lit side stays pointed at the sun and earth has simply rotates a quarter of the way around while the lit side remains pointing at the sun.
My Reponse. That would mean we would have to set our clocks to be 6 hours ahead every year, to keep Solar Time (which our clocks are based on) consistent.
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Bobby says "yes."
Except we don't. We wait every fourth year and our clocks a whole day ahead then (4x6 hrs)
Mark me wrong. The answer is actually "no." I explained why, later, after Tom's follow-up made me think it through; but I didn't realize until just now (after reading Macarios' response) that I was contradicting this earlier reply.
I owe someone a pint.
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Bobby says "yes."
Except we don't. We wait every fourth year and our clocks a whole day ahead then (4x6 hrs)
If we don't reset our clocks then what we thinks as 12pm at the beginning of the equinox start point is now at 6pm. Does that not seem like an issue?
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Response. The Solar Day, which is the rate the sun moves around a static earth (once per 24 hours), can also be thought of as the rate of the earth's rotation against a static sun (once per 24 hours).
Yes, a solar day is one rotation of the earth.
After 365 of those the earth is nearly back where it started in its orbit around the sun but not quite, it takes (roughly) another 6 hours to get there.
We start a new year at the end of the 365th day though which would mean, over time, the seasons would get out of sync with our calendar.
After 4 years it's a whole day out of sync - this is why the dates of the equinoxes and solstices vary from year to year. So to sort that out we add the extra leap day and we're back in sync again.
Except not quite, as you say it's 365.24, not .25, it' not exactly a quarter
So to sort that out there are some extra rules: If the year is a century then we do NOT have a leap year UNLESS the century also divides by 4.
So 1900 was NOT a leap year, 2000 was, 2100 won't be.
And more recently as we've got even more accurate measurements of earth's orbit they've started adding leap seconds to keep everything in sync.
https://en.wikipedia.org/wiki/Leap_second
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Bobby says "yes."
Except we don't. We wait every fourth year and our clocks a whole day ahead then (4x6 hrs)
If we don't reset our clocks then what we thinks as 12pm at the beginning of the equinox start point is now at 6pm. Does that not seem like an issue?
No, because if the equinox was at 12pm local time last year, it'll be at around 6pm local time this year.
The point of equinox moves. That's why the time of equinox changes.
Where are you? Bay area, right? The last March equinox for you was at 9:15AM on March 20th, 2019
The next March equinox for you will be at 2:58PM, March 20th, 2019.
For the entire globe, the March equinox will be at 21:58 UTC, and to know when that is for you, you adjust according to your longitude. The equinox (the point at which the sun is crossing the celestial equator) is on a meridian where it's solar noon at that moment -- the side of the earth that's facing the sun, thus being illuminated by it when that event happens.
The next vernal equinox will occur somewhere out over the Pacific, where on some meridian it will be solar noon. It won't be solar noon for you, so don't reset your clocks. We don't set our clocks by the solar year. We set them by the solar day. The calendar year is a different story. We do reset them: every four years to try to recage our calendar to the solar year, because of the annual drift westward of the equinox (which is due to the fact we are not completing a full orbit around the sun in an even 365 days.
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According to the flat earth model, Tom, all this ecliptic plane, equatorial plane, earth rotation and sun orbiting is nonsense. According to FET, the sun circles on its own plane above the plane of a flat earth. What dictates the motion of the sun? Who knows? Doesn't matter. What matters is that the sun follows a pattern. (Right?)
From a FET perspective, the sun spirals annually between the Tropics of Cancer and Capricorn. Spring equinox is when the sun is at the equatorial circle on it's spiral from the Tropic of Capricorn inward toward the Tropic of Cancer. The Autumnal equinox is when the sun is at the equatorial circle spiraling away from the centroid of earth, toward the Tropic of Capricorn.
It'll make that equatorial crossing over a specific point over flat earth. Is that the same point every year? Or does flat earth agree with globe earth that that point will migrate year to year, roughly about 1/4 way around the circle of the equator each year?
Globe and flat earth years are both 365 calendar days. Even on flat earth, if that equinox point keeps changing without adjustment via leap years, the "spiral" from equinox to solstice to equinox to solstice (etc) will fall out of sync with the calendar. Even if the mechanism for that equinox drift isn't known, flat earth must agree that these leap year adjustments are necessary to keep the calendar in sync with the seasons.
Yes? Or is it no? Does flat earth model say that the year is 365 and equinox is at the same time and occurs over the same place on earth every year?
If it isn't "no" and the timing of flat earth agrees with the timing of globe earth, then flat earth is experiencing this same "extra time" that you find troublesome.
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Bobby says "yes."
Except we don't. We wait every fourth year and our clocks a whole day ahead then (4x6 hrs)
If we don't reset our clocks then what we thinks as 12pm at the beginning of the equinox start point is now at 6pm. Does that not seem like an issue?
If you cannot get your questions answered then why not look elsewhere?
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Bobby and AllAround, you seem to have the same argument. This is how I have interpreted your argument. Correct me if I am wrong:
Response. The Solar Day, which is the rate the sun moves around a static earth (once per 24 hours), can also be thought of as the rate of the earth's rotation against a static sun (once per 24 hours).
At Day 365 the lit side of the earth is pointing at the sun. At Day 365.24, with a static sun, the lit side stays pointed at the sun and earth has simply rotates a quarter of the way around while the lit side remains pointing at the sun.
My Reponse. That would mean we would have to set our clocks to be 6 hours ahead every year, to keep Solar Time (which our clocks are based on) consistent.
No, it wouldn't. It means we end our Calendar year before it reaches orbital position.
We end our Calendar year before Earth reaches end of Tropical year.
We don't move our clock, we every 4 years count one more day.
EDIT: You still haven't learned the difference between Calendar year and Tropical year.
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Bobby and AllAround, you seem to have the same argument. This is how I have interpreted your argument. Correct me if I am wrong:
Response. The Solar Day, which is the rate the sun moves around a static earth (once per 24 hours), can also be thought of as the rate of the earth's rotation against a static sun (once per 24 hours).
At Day 365 the lit side of the earth is pointing at the sun. At Day 365.24, with a static sun, the lit side stays pointed at the sun and earth has simply rotates a quarter of the way around while the lit side remains pointing at the sun.
My Reponse. That would mean we would have to set our clocks to be 6 hours ahead every year, to keep Solar Time (which our clocks are based on) consistent.
No, it wouldn't. It means we end our Calendar year before it reaches orbital position.
We end our Calendar year before Earth reaches end of Tropical year.
We don't move our clock, we every 4 years count one more day.
Macarios is right. I was wrong earlier to answer "yes," though my subsequent explanation for why we don't change our clocks mates with Macarios' answer.
I change my answer to "Bobby says no"
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Response: The Solar Day, which is the rate the sun moves around a static earth (once per 24 hours), can also be thought of as the rate of the earth's rotation against a static sun (once per 24 hours).
At Day 365 the lit side of the earth is pointing at the sun. At Day 365.24, with a static sun, the lit side stays pointed at the sun and earth has simply rotates a quarter of the way around while the lit side remains pointing at the sun.
There is another problem with this response. We can't assume a static sun that is casting light in one direction on a spinning ball. In the scenario the earth is spinning at 24 hours per day and it is also going around the sun. The lit area isn't static on the earth over the year. This is another variable and messes up the assumption in that response.
The scenario doesn't work if we think of the lit portion of the earth is spinning at once per day. It doesn't line up with the sun on the diagram at the end.
Nor does it work if we think of a sun that is going around the earth. Same as above. It doesn't line up in the end on a diagram.
It might seemingly work if you think of a static sun and spinning earth; but that is not taking into account the geometrical nature of the the circular orbit and the illumination of light - that light isn't static on the spinning ball. There is more work that needs to be done there.
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Response: The Solar Day, which is the rate the sun moves around a static earth (once per 24 hours), can also be thought of as the rate of the earth's rotation against a static sun (once per 24 hours).
At Day 365 the lit side of the earth is pointing at the sun. At Day 365.24, with a static sun, the lit side stays pointed at the sun and earth has simply rotates a quarter of the way around while the lit side remains pointing at the sun.
There is another problem with this response. We can't assume a static sun that is casting light in one direction on a spinning ball. In the scenario the earth is spinning at 24 hours per day and it is also going around the sun. The lit area isn't static on the earth over the year. This is another variable and messes up the assumption in that response.
The scenario doesn't work if we think of the lit portion of the earth is spinning at once per day. It doesn't line up with the sun on the diagram at the end.
Nor does it work if we think of a sun that is going around the earth. Same as above. It doesn't line up in the end on a diagram.
It might seemingly work if you think of a static sun and spinning earth; but that is not taking into account the geometrical nature of the the circular orbit and the illumination of light - that light isn't static on the spinning ball. There is more work that needs to be done there.
Sorry, why would we be considering a static sun casting light in just one direction? I'm not sure what model that represents.
Also could you clarify that last paragraph? The light isn't static on a spinning ball?
I think one thing that is over complicating this discussion is the switching between ref points of what is moving and what is static.
As your point is attempting to show a problem with a heliocentric globe earth it may help the discussion to stick to a spinning earth orbiting a static sun.
Just a suggestion.
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We can't assume a static sun that is casting light in one direction on a spinning ball.
Why not?
In the scenario the earth is spinning at 24 hours per day and it is also going around the sun. The lit area isn't static on the earth over the year. This is another variable and messes up the assumption in that response.
No, this is taken into account in the difference between sidereal day and solar day. See that Wiki that was referred to in the early pages again.
The scenario doesn't work if we think of the lit portion of the earth is spinning at once per day. It doesn't line up with the sun on the diagram at the end.
Again - difference between calendar year and tropical year