How can a day be exactly 24 hours long? [on hold]Hours of light per day based on latitude/longitude formulaHow long was a day at the creation of Earth?What accounts for the discrepancies in my calculations of year lengths?Why a day is divided by 12/24 hours? Why the number 12?Calculating the length of day at any time of the year?How long did the first rainshowers on Earth last?How many hours will there be in a day 5,000,000,000 years from now?How have the duration of the martian day changed in the past?What determines how long a day is on a planet?Why was the fraction 1/31,556,925.9747, in the 1956-1968 definition of the second in terms of ephemeris time, chosen?
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How can a day be exactly 24 hours long? [on hold]
Hours of light per day based on latitude/longitude formulaHow long was a day at the creation of Earth?What accounts for the discrepancies in my calculations of year lengths?Why a day is divided by 12/24 hours? Why the number 12?Calculating the length of day at any time of the year?How long did the first rainshowers on Earth last?How many hours will there be in a day 5,000,000,000 years from now?How have the duration of the martian day changed in the past?What determines how long a day is on a planet?Why was the fraction 1/31,556,925.9747, in the 1956-1968 definition of the second in terms of ephemeris time, chosen?
$begingroup$
The longest solar day of year is approximately 24 hours 0 min 30 seconds (occurs at mid to late December) while the shortest solar day of year is approximately 23 hour 59min 38 seconds. If I average out both of these I come up with average solar day of 24 hour +4seconds. Why then it is said it is 24 hours 0min 0 seconds exactly??
Wouldn't using a 24 hour solar day as a definition of day cause the offset of 4 seconds each solar year
time astronomy earth solar-system metrology
edited 2 days ago
Pureferret
3061718
asked Apr 1 at 11:48
ObsessionWithElectricityObsessionWithElectricity
4532312
$endgroup$
put on hold as unclear what you're asking by Kyle Kanos, Jon Custer, Cosmas Zachos, Dvij Mankad, Emilio Pisanty 2 days ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 11 more comments
$begingroup$
The longest solar day of year is approximately 24 hours 0 min 30 seconds (occurs at mid to late December) while the shortest solar day of year is approximately 23 hour 59min 38 seconds. If I average out both of these I come up with average solar day of 24 hour +4seconds. Why then it is said it is 24 hours 0min 0 seconds exactly??
Wouldn't using a 24 hour solar day as a definition of day cause the offset of 4 seconds each solar year
time astronomy earth solar-system metrology
edited 2 days ago
Pureferret
3061718
asked Apr 1 at 11:48
ObsessionWithElectricityObsessionWithElectricity
4532312
$endgroup$
put on hold as unclear what you're asking by Kyle Kanos, Jon Custer, Cosmas Zachos, Dvij Mankad, Emilio Pisanty 2 days ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
10
$begingroup$
Related: "Leap second", Wikipedia.
$endgroup$
– Nat
Apr 1 at 12:09
5
$begingroup$
Also see leapsecond.com for a plethora of articles about timekeeping, especially the articles by Steve Allen of Lick Observatory. That's a rather old page, and some of its links are now dead, but there's still plenty of good info there.
$endgroup$
– PM 2Ring
Apr 1 at 13:09
6
$begingroup$
@Nat - This has almost nothing to do with leap seconds. It has much more to do with that thinking that the length of day averaged over the course of a few years is equal to the average of the minimum and maximum length of day is invalid. There are many pseudocyclical functions where the mean is not the average of the min and max, the equation of time being a prime example.
$endgroup$
– David Hammen
Apr 2 at 15:38
2
$begingroup$
@Nat - I disagree. The four second deviation from 86400 seconds is over three orders of magnitude larger than the tiny 2.5 millisecond deviation that results in the need for leap seconds. If the mean solar day was 86404 seconds long, we would need 4 leap seconds every day, or two leap minutes per month. But it's not. The mean solar day remains very close to being 86400 seconds long.
$endgroup$
– David Hammen
Apr 2 at 15:55
3
$begingroup$
Dear @ObsessionWithElectricity, The existence of a well-received answer indicates nothing about whether the question belongs to this site or not. Hope you appreciate the distinction. Cheers! :)
$endgroup$
– Dvij Mankad
2 days ago
|
show 11 more comments
$begingroup$
The longest solar day of year is approximately 24 hours 0 min 30 seconds (occurs at mid to late December) while the shortest solar day of year is approximately 23 hour 59min 38 seconds. If I average out both of these I come up with average solar day of 24 hour +4seconds. Why then it is said it is 24 hours 0min 0 seconds exactly??
Wouldn't using a 24 hour solar day as a definition of day cause the offset of 4 seconds each solar year
time astronomy earth solar-system metrology
edited 2 days ago
Pureferret
3061718
asked Apr 1 at 11:48
ObsessionWithElectricityObsessionWithElectricity
4532312
$endgroup$
The longest solar day of year is approximately 24 hours 0 min 30 seconds (occurs at mid to late December) while the shortest solar day of year is approximately 23 hour 59min 38 seconds. If I average out both of these I come up with average solar day of 24 hour +4seconds. Why then it is said it is 24 hours 0min 0 seconds exactly??
Wouldn't using a 24 hour solar day as a definition of day cause the offset of 4 seconds each solar year
time astronomy earth solar-system metrology
time astronomy earth solar-system metrology
edited 2 days ago
Pureferret
3061718
asked Apr 1 at 11:48
ObsessionWithElectricityObsessionWithElectricity
4532312
edited 2 days ago
Pureferret
3061718
asked Apr 1 at 11:48
ObsessionWithElectricityObsessionWithElectricity
4532312
edited 2 days ago
Pureferret
3061718
edited 2 days ago
Pureferret
3061718
edited 2 days ago
Pureferret
3061718
3061718
asked Apr 1 at 11:48
ObsessionWithElectricityObsessionWithElectricity
4532312
asked Apr 1 at 11:48
ObsessionWithElectricityObsessionWithElectricity
4532312
asked Apr 1 at 11:48
ObsessionWithElectricityObsessionWithElectricity
4532312
4532312
put on hold as unclear what you're asking by Kyle Kanos, Jon Custer, Cosmas Zachos, Dvij Mankad, Emilio Pisanty 2 days ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by Kyle Kanos, Jon Custer, Cosmas Zachos, Dvij Mankad, Emilio Pisanty 2 days ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
10
$begingroup$
Related: "Leap second", Wikipedia.
$endgroup$
– Nat
Apr 1 at 12:09
5
$begingroup$
Also see leapsecond.com for a plethora of articles about timekeeping, especially the articles by Steve Allen of Lick Observatory. That's a rather old page, and some of its links are now dead, but there's still plenty of good info there.
$endgroup$
– PM 2Ring
Apr 1 at 13:09
6
$begingroup$
@Nat - This has almost nothing to do with leap seconds. It has much more to do with that thinking that the length of day averaged over the course of a few years is equal to the average of the minimum and maximum length of day is invalid. There are many pseudocyclical functions where the mean is not the average of the min and max, the equation of time being a prime example.
$endgroup$
– David Hammen
Apr 2 at 15:38
2
$begingroup$
@Nat - I disagree. The four second deviation from 86400 seconds is over three orders of magnitude larger than the tiny 2.5 millisecond deviation that results in the need for leap seconds. If the mean solar day was 86404 seconds long, we would need 4 leap seconds every day, or two leap minutes per month. But it's not. The mean solar day remains very close to being 86400 seconds long.
$endgroup$
– David Hammen
Apr 2 at 15:55
3
$begingroup$
Dear @ObsessionWithElectricity, The existence of a well-received answer indicates nothing about whether the question belongs to this site or not. Hope you appreciate the distinction. Cheers! :)
$endgroup$
– Dvij Mankad
2 days ago
|
show 11 more comments
10
$begingroup$
Related: "Leap second", Wikipedia.
$endgroup$
– Nat
Apr 1 at 12:09
5
$begingroup$
Also see leapsecond.com for a plethora of articles about timekeeping, especially the articles by Steve Allen of Lick Observatory. That's a rather old page, and some of its links are now dead, but there's still plenty of good info there.
$endgroup$
– PM 2Ring
Apr 1 at 13:09
6
$begingroup$
@Nat - This has almost nothing to do with leap seconds. It has much more to do with that thinking that the length of day averaged over the course of a few years is equal to the average of the minimum and maximum length of day is invalid. There are many pseudocyclical functions where the mean is not the average of the min and max, the equation of time being a prime example.
$endgroup$
– David Hammen
Apr 2 at 15:38
2
$begingroup$
@Nat - I disagree. The four second deviation from 86400 seconds is over three orders of magnitude larger than the tiny 2.5 millisecond deviation that results in the need for leap seconds. If the mean solar day was 86404 seconds long, we would need 4 leap seconds every day, or two leap minutes per month. But it's not. The mean solar day remains very close to being 86400 seconds long.
$endgroup$
– David Hammen
Apr 2 at 15:55
3
$begingroup$
Dear @ObsessionWithElectricity, The existence of a well-received answer indicates nothing about whether the question belongs to this site or not. Hope you appreciate the distinction. Cheers! :)
$endgroup$
– Dvij Mankad
2 days ago
10
10
$begingroup$
Related: "Leap second", Wikipedia.
$endgroup$
– Nat
Apr 1 at 12:09
$begingroup$
Related: "Leap second", Wikipedia.
$endgroup$
– Nat
Apr 1 at 12:09
5
5
$begingroup$
Also see leapsecond.com for a plethora of articles about timekeeping, especially the articles by Steve Allen of Lick Observatory. That's a rather old page, and some of its links are now dead, but there's still plenty of good info there.
$endgroup$
– PM 2Ring
Apr 1 at 13:09
$begingroup$
Also see leapsecond.com for a plethora of articles about timekeeping, especially the articles by Steve Allen of Lick Observatory. That's a rather old page, and some of its links are now dead, but there's still plenty of good info there.
$endgroup$
– PM 2Ring
Apr 1 at 13:09
6
6
$begingroup$
@Nat - This has almost nothing to do with leap seconds. It has much more to do with that thinking that the length of day averaged over the course of a few years is equal to the average of the minimum and maximum length of day is invalid. There are many pseudocyclical functions where the mean is not the average of the min and max, the equation of time being a prime example.
$endgroup$
– David Hammen
Apr 2 at 15:38
$begingroup$
@Nat - This has almost nothing to do with leap seconds. It has much more to do with that thinking that the length of day averaged over the course of a few years is equal to the average of the minimum and maximum length of day is invalid. There are many pseudocyclical functions where the mean is not the average of the min and max, the equation of time being a prime example.
$endgroup$
– David Hammen
Apr 2 at 15:38
2
2
$begingroup$
@Nat - I disagree. The four second deviation from 86400 seconds is over three orders of magnitude larger than the tiny 2.5 millisecond deviation that results in the need for leap seconds. If the mean solar day was 86404 seconds long, we would need 4 leap seconds every day, or two leap minutes per month. But it's not. The mean solar day remains very close to being 86400 seconds long.
$endgroup$
– David Hammen
Apr 2 at 15:55
$begingroup$
@Nat - I disagree. The four second deviation from 86400 seconds is over three orders of magnitude larger than the tiny 2.5 millisecond deviation that results in the need for leap seconds. If the mean solar day was 86404 seconds long, we would need 4 leap seconds every day, or two leap minutes per month. But it's not. The mean solar day remains very close to being 86400 seconds long.
$endgroup$
– David Hammen
Apr 2 at 15:55
3
3
$begingroup$
Dear @ObsessionWithElectricity, The existence of a well-received answer indicates nothing about whether the question belongs to this site or not. Hope you appreciate the distinction. Cheers! :)
$endgroup$
– Dvij Mankad
2 days ago
$begingroup$
Dear @ObsessionWithElectricity, The existence of a well-received answer indicates nothing about whether the question belongs to this site or not. Hope you appreciate the distinction. Cheers! :)
$endgroup$
– Dvij Mankad
2 days ago
|
show 11 more comments
4 Answers
4
active
oldest
votes
$begingroup$
You can't calculate the length of a mean solar day just by taking the mean of the shortest & longest apparent solar days. That would work if the apparent day lengths varied in a simple linear fashion, but that's not the case.
From Wikipedia's article on the Equation of Time,
The equation of time describes the discrepancy between two kinds
of solar time. [...] 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 uniform
motion.
This graph shows the cumulative differences between mean & apparent solar time:
The equation of time — above the axis a sundial will
appear fast relative to a clock showing local mean time, and below the
axis a sundial will appear slow.
To correctly calculate the mean solar day length you need to integrate the apparent day lengths over the whole year. (And you need to decide exactly how to define the length of the year, which is a whole complicated story in its own right).
There are two primary causes of the Equation of Time.
1. The obliquity of the plane of Earth's orbit (the ecliptic plane), which is tilted approximately 23° relative to the equatorial plane. This tilt is also responsible for the seasons.
2. The eccentricity of Earth's orbit, which causes the Earth's orbital speed to vary over the year. The following graph shows how these two components combine to create the Equation of Time.
Equation of time (red solid line) and its two main components plotted
separately, the part due to the obliquity of the ecliptic (mauve
dashed line) and the part due to the Sun's varying apparent speed
along the ecliptic due to eccentricity of the Earth's orbit (dark blue
dash & dot line)
Please see the linked Wikipedia article for further details.
answered Apr 1 at 12:57
PM 2RingPM 2Ring
3,64821123
$endgroup$
4
$begingroup$
And, then, once in a while, we need leap seconds.
$endgroup$
– Draco18s
Apr 1 at 14:33
3
$begingroup$
@Draco18s Indeed! Accurate timekeeping is a complex & subtle business. See leapsecond.com, which I also linked in my comment on the question.
$endgroup$
– PM 2Ring
Apr 1 at 14:37
2
$begingroup$
It's worth noting that in terms of the actual length of a day (above or below 24 hours), the mauve line in that diagram makes a substantially bigger contribution than the dark line - more so than this diagram would suggest. The reason is that the length of a day is effectively the time-derivative of the equation of time; and the mauve line is rather steeper than the dark line (as well as having a slightly greater amplitude).
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– Dawood ibn Kareem
Apr 2 at 5:18
1
$begingroup$
@DawoodibnKareem - I wouldn't call ~20% smaller significantly less. I would call them almost the same magnitude. What's more important is the phase difference. The two curves contributors are closer to being in phase with one another in early November than they are in late February.
$endgroup$
– David Hammen
Apr 2 at 16:36
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@DavidHammen You have failed to understand my comment. It's not the magnitude that affects the length of the day, it's the slope. The mauve line has half the period of the dark line, so if you were to plot its slope, you'd find that it makes more than double the contribution that the dark line makes.
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– Dawood ibn Kareem
Apr 2 at 18:01
add a comment |
$begingroup$
The average value of a distribution is not the average of its minimum and maximum. For example, the average value of (0,0,0,4) is 1, not 2. Earth's orbit eccentricity is not 0, nor the Moon's one, so your distribution of day duration is probably slightly asymetric, hence the 4 seconds discrepancy. Sum all day duration of a year, divide by the number of days, you will get a better value.
answered Apr 1 at 12:35
MattMatt
27114
New contributor
Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The original definition of an hour was 1/24th of a day, no matter how long the day was at the time. The durations you're quoting are only possible with the help of extremely modern definitions made possible by redefining the second (keeping the minute and hour fixed as 60 seconds, and 3600 seconds, respectively).
answered Apr 1 at 11:51
Sean E. LakeSean E. Lake
14.7k12351
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$begingroup$
I don't disagree, but do you have 1) a date when that became the official definition of an hour, and 2) when it was conventionally determined to use 1/24 and not 1/10 or 1/20 of a day? I've read undocumented sources that mention the daylight being divided into tenths historically prior to twelfths.
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– Bill N
Apr 1 at 11:56
1
$begingroup$
More details in en.wikipedia.org/wiki/Hour
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– Stéphane Rollandin
Apr 1 at 12:14
1
$begingroup$
The mean solar day is 86400.0025 seconds, so while the reason in this answer has some effect, it only accounts for 2.5 ms of the 4 second discrepancy in the question.
$endgroup$
– JiK
Apr 1 at 16:05
add a comment |
$begingroup$
The 24 hours (exactly) definition of the mean solar day only applies if you use the UT1 time scale. As others have mentioned, the mean solar day is not the average of the shortest and longest apparent solar day, and you have to consider the Equation of Time to calculate the mean solar day by averaging all apparent solar days in one year.
If you use the definition of a second based on the metric system, which is used by the atomic time (TAI), UTC and the terrestrial time (TT) time scales, the length of the mean solar day is not exactly 24 hours. The definition of the SI second, apart from being an atomic scale, is based on the mean solar day in 1900 as determined by Newcomb (in reality it corresponds to the mean solar day in about mid 19th century), when the Earth's rotation was faster than today. Today the mean solar day is slightly longer than in the past (when measured by an atomic clock), and the length of the mean solar day in TAI seconds is greater than in UT1. To correct for the difference, leap seconds are introduced 0-2 times per year in the UTC time scale based on observations of Earth's rotation, to keep UTC in sync with UT1 to less than 0.9 s. These corrections are publised in advance by IERS in Bulletin C. There have been 27 leap seconds inserted in the last 46 years, which equals to about 0.6 s per year difference between the length of the mean solar day between TAI and UT1, or about 1.6 ms per day (see Figure 1 and 2).
The length of the SI second depends on the location, and TAI is based on observations done by atomic clocks in various laboratories around the world ("UTC(k)") and corrected for the geopotential on the sea level. TT is a theoretical length of the SI second on the geoid, and as such is never known perfectly. Approximation of past TT is revised annualy by BIPM. In contrast, TAI and UTC are determined and kept fixed after about 1 month past the fact (published regularly in Circular-T by BIPM). Laboratory-specific UTC(k) time scales and the GPS time (based on the US Naval Observatory Master Clock) are known in real time. Other time scales include the geocentric time, barycentric time and the ephemeris time. The length of 1 s is slightly different between all of them, some of the differece is due to time passing differently based on location in the gravitation potential. For example, time passes faster in the solar system than on the Earth's surface by about 0.5 second a year.
Historically, time measured by the rotation of the Earth was the most accurate, and the mean solar day was assumed to be constant equal to 86400 s. This changed with the introduction of the ephemeris time, later superseded by the quartz clock and the atomic clock, which lead to re-definition of the second. They will deviate further as the rotation of the Earth slows down.
References:
- Urban and Seidelmann (2012): Explanatory Supplement to the Astronomical Almanac
- McCarthy and Seidelmann (2018): Time: From Earth Rotation to Atomic Physics
- BIPM time publications: https://www.bipm.org/en/bipm/tai/publications.html
- IERS Bulletins: https://www.iers.org/IERS/EN/Publications/Bulletins/bulletins.html
- USNO - The Equation of Time: https://aa.usno.navy.mil/faq/docs/eqtime.php
Figure 1. Excess to 86400s of the duration of the days, combined GPS solution, 1995-1997. From https://www.iers.org/IERS/EN/Science/EarthRotation/LODgps.html
Figure 2. TAI-UT1 and TAI-UTC. From McCarthy and Seidelmann (2018).
answered 2 days ago
peterpeter
1495
$endgroup$
$begingroup$
Thanks for mentioning Terrestrial Time, etc. I didn't want to go down the time scales rabbit-hole in my answer (although I did supply a link to articles on these topics in a couple of comments), but I guess it's good to give people a brief glimpse of this material. ;)
$endgroup$
– PM 2Ring
2 days ago
$begingroup$
Leap seconds have absolutely nothing to do with the OP's error of 4 seconds in his/her attempt to calculate the length of the mean solar day. The information in this answer is all true, but all completely unrelated to the question.
$endgroup$
– Dawood ibn Kareem
2 days ago
$begingroup$
@DawoodibnKareem I think it is related because the answer to the question how long the mean solar day is is not "24 hours 0min 0 seconds exactly" - not when you use the conventional definition of 1 s. The other answers do not explain this (but answer the main point, which is the wrong averaging).
$endgroup$
– peter
2 days ago
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can't calculate the length of a mean solar day just by taking the mean of the shortest & longest apparent solar days. That would work if the apparent day lengths varied in a simple linear fashion, but that's not the case.
From Wikipedia's article on the Equation of Time,
The equation of time describes the discrepancy between two kinds
of solar time. [...] 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 uniform
motion.
This graph shows the cumulative differences between mean & apparent solar time:
The equation of time — above the axis a sundial will
appear fast relative to a clock showing local mean time, and below the
axis a sundial will appear slow.
To correctly calculate the mean solar day length you need to integrate the apparent day lengths over the whole year. (And you need to decide exactly how to define the length of the year, which is a whole complicated story in its own right).
There are two primary causes of the Equation of Time.
1. The obliquity of the plane of Earth's orbit (the ecliptic plane), which is tilted approximately 23° relative to the equatorial plane. This tilt is also responsible for the seasons.
2. The eccentricity of Earth's orbit, which causes the Earth's orbital speed to vary over the year. The following graph shows how these two components combine to create the Equation of Time.
Equation of time (red solid line) and its two main components plotted
separately, the part due to the obliquity of the ecliptic (mauve
dashed line) and the part due to the Sun's varying apparent speed
along the ecliptic due to eccentricity of the Earth's orbit (dark blue
dash & dot line)
Please see the linked Wikipedia article for further details.
answered Apr 1 at 12:57
PM 2RingPM 2Ring
3,64821123
$endgroup$
4
$begingroup$
And, then, once in a while, we need leap seconds.
$endgroup$
– Draco18s
Apr 1 at 14:33
3
$begingroup$
@Draco18s Indeed! Accurate timekeeping is a complex & subtle business. See leapsecond.com, which I also linked in my comment on the question.
$endgroup$
– PM 2Ring
Apr 1 at 14:37
2
$begingroup$
It's worth noting that in terms of the actual length of a day (above or below 24 hours), the mauve line in that diagram makes a substantially bigger contribution than the dark line - more so than this diagram would suggest. The reason is that the length of a day is effectively the time-derivative of the equation of time; and the mauve line is rather steeper than the dark line (as well as having a slightly greater amplitude).
$endgroup$
– Dawood ibn Kareem
Apr 2 at 5:18
1
$begingroup$
@DawoodibnKareem - I wouldn't call ~20% smaller significantly less. I would call them almost the same magnitude. What's more important is the phase difference. The two curves contributors are closer to being in phase with one another in early November than they are in late February.
$endgroup$
– David Hammen
Apr 2 at 16:36
$begingroup$
@DavidHammen You have failed to understand my comment. It's not the magnitude that affects the length of the day, it's the slope. The mauve line has half the period of the dark line, so if you were to plot its slope, you'd find that it makes more than double the contribution that the dark line makes.
$endgroup$
– Dawood ibn Kareem
Apr 2 at 18:01
add a comment |
$begingroup$
You can't calculate the length of a mean solar day just by taking the mean of the shortest & longest apparent solar days. That would work if the apparent day lengths varied in a simple linear fashion, but that's not the case.
From Wikipedia's article on the Equation of Time,
The equation of time describes the discrepancy between two kinds
of solar time. [...] 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 uniform
motion.
This graph shows the cumulative differences between mean & apparent solar time:
The equation of time — above the axis a sundial will
appear fast relative to a clock showing local mean time, and below the
axis a sundial will appear slow.
To correctly calculate the mean solar day length you need to integrate the apparent day lengths over the whole year. (And you need to decide exactly how to define the length of the year, which is a whole complicated story in its own right).
There are two primary causes of the Equation of Time.
1. The obliquity of the plane of Earth's orbit (the ecliptic plane), which is tilted approximately 23° relative to the equatorial plane. This tilt is also responsible for the seasons.
2. The eccentricity of Earth's orbit, which causes the Earth's orbital speed to vary over the year. The following graph shows how these two components combine to create the Equation of Time.
Equation of time (red solid line) and its two main components plotted
separately, the part due to the obliquity of the ecliptic (mauve
dashed line) and the part due to the Sun's varying apparent speed
along the ecliptic due to eccentricity of the Earth's orbit (dark blue
dash & dot line)
Please see the linked Wikipedia article for further details.
answered Apr 1 at 12:57
PM 2RingPM 2Ring
3,64821123
$endgroup$
4
$begingroup$
And, then, once in a while, we need leap seconds.
$endgroup$
– Draco18s
Apr 1 at 14:33
3
$begingroup$
@Draco18s Indeed! Accurate timekeeping is a complex & subtle business. See leapsecond.com, which I also linked in my comment on the question.
$endgroup$
– PM 2Ring
Apr 1 at 14:37
2
$begingroup$
It's worth noting that in terms of the actual length of a day (above or below 24 hours), the mauve line in that diagram makes a substantially bigger contribution than the dark line - more so than this diagram would suggest. The reason is that the length of a day is effectively the time-derivative of the equation of time; and the mauve line is rather steeper than the dark line (as well as having a slightly greater amplitude).
$endgroup$
– Dawood ibn Kareem
Apr 2 at 5:18
1
$begingroup$
@DawoodibnKareem - I wouldn't call ~20% smaller significantly less. I would call them almost the same magnitude. What's more important is the phase difference. The two curves contributors are closer to being in phase with one another in early November than they are in late February.
$endgroup$
– David Hammen
Apr 2 at 16:36
$begingroup$
@DavidHammen You have failed to understand my comment. It's not the magnitude that affects the length of the day, it's the slope. The mauve line has half the period of the dark line, so if you were to plot its slope, you'd find that it makes more than double the contribution that the dark line makes.
$endgroup$
– Dawood ibn Kareem
Apr 2 at 18:01
add a comment |
$begingroup$
You can't calculate the length of a mean solar day just by taking the mean of the shortest & longest apparent solar days. That would work if the apparent day lengths varied in a simple linear fashion, but that's not the case.
From Wikipedia's article on the Equation of Time,
The equation of time describes the discrepancy between two kinds
of solar time. [...] 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 uniform
motion.
This graph shows the cumulative differences between mean & apparent solar time:
The equation of time — above the axis a sundial will
appear fast relative to a clock showing local mean time, and below the
axis a sundial will appear slow.
To correctly calculate the mean solar day length you need to integrate the apparent day lengths over the whole year. (And you need to decide exactly how to define the length of the year, which is a whole complicated story in its own right).
There are two primary causes of the Equation of Time.
1. The obliquity of the plane of Earth's orbit (the ecliptic plane), which is tilted approximately 23° relative to the equatorial plane. This tilt is also responsible for the seasons.
2. The eccentricity of Earth's orbit, which causes the Earth's orbital speed to vary over the year. The following graph shows how these two components combine to create the Equation of Time.
Equation of time (red solid line) and its two main components plotted
separately, the part due to the obliquity of the ecliptic (mauve
dashed line) and the part due to the Sun's varying apparent speed
along the ecliptic due to eccentricity of the Earth's orbit (dark blue
dash & dot line)
Please see the linked Wikipedia article for further details.
answered Apr 1 at 12:57
PM 2RingPM 2Ring
3,64821123
$endgroup$
You can't calculate the length of a mean solar day just by taking the mean of the shortest & longest apparent solar days. That would work if the apparent day lengths varied in a simple linear fashion, but that's not the case.
From Wikipedia's article on the Equation of Time,
The equation of time describes the discrepancy between two kinds
of solar time. [...] 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 uniform
motion.
This graph shows the cumulative differences between mean & apparent solar time:
The equation of time — above the axis a sundial will
appear fast relative to a clock showing local mean time, and below the
axis a sundial will appear slow.
To correctly calculate the mean solar day length you need to integrate the apparent day lengths over the whole year. (And you need to decide exactly how to define the length of the year, which is a whole complicated story in its own right).
There are two primary causes of the Equation of Time.
1. The obliquity of the plane of Earth's orbit (the ecliptic plane), which is tilted approximately 23° relative to the equatorial plane. This tilt is also responsible for the seasons.
2. The eccentricity of Earth's orbit, which causes the Earth's orbital speed to vary over the year. The following graph shows how these two components combine to create the Equation of Time.
Equation of time (red solid line) and its two main components plotted
separately, the part due to the obliquity of the ecliptic (mauve
dashed line) and the part due to the Sun's varying apparent speed
along the ecliptic due to eccentricity of the Earth's orbit (dark blue
dash & dot line)
Please see the linked Wikipedia article for further details.
answered Apr 1 at 12:57
PM 2RingPM 2Ring
3,64821123
answered Apr 1 at 12:57
PM 2RingPM 2Ring
3,64821123
answered Apr 1 at 12:57
PM 2RingPM 2Ring
3,64821123
answered Apr 1 at 12:57
PM 2RingPM 2Ring
3,64821123
3,64821123
4
$begingroup$
And, then, once in a while, we need leap seconds.
$endgroup$
– Draco18s
Apr 1 at 14:33
3
$begingroup$
@Draco18s Indeed! Accurate timekeeping is a complex & subtle business. See leapsecond.com, which I also linked in my comment on the question.
$endgroup$
– PM 2Ring
Apr 1 at 14:37
2
$begingroup$
It's worth noting that in terms of the actual length of a day (above or below 24 hours), the mauve line in that diagram makes a substantially bigger contribution than the dark line - more so than this diagram would suggest. The reason is that the length of a day is effectively the time-derivative of the equation of time; and the mauve line is rather steeper than the dark line (as well as having a slightly greater amplitude).
$endgroup$
– Dawood ibn Kareem
Apr 2 at 5:18
1
$begingroup$
@DawoodibnKareem - I wouldn't call ~20% smaller significantly less. I would call them almost the same magnitude. What's more important is the phase difference. The two curves contributors are closer to being in phase with one another in early November than they are in late February.
$endgroup$
– David Hammen
Apr 2 at 16:36
$begingroup$
@DavidHammen You have failed to understand my comment. It's not the magnitude that affects the length of the day, it's the slope. The mauve line has half the period of the dark line, so if you were to plot its slope, you'd find that it makes more than double the contribution that the dark line makes.
$endgroup$
– Dawood ibn Kareem
Apr 2 at 18:01
add a comment |
4
$begingroup$
And, then, once in a while, we need leap seconds.
$endgroup$
– Draco18s
Apr 1 at 14:33
3
$begingroup$
@Draco18s Indeed! Accurate timekeeping is a complex & subtle business. See leapsecond.com, which I also linked in my comment on the question.
$endgroup$
– PM 2Ring
Apr 1 at 14:37
2
$begingroup$
It's worth noting that in terms of the actual length of a day (above or below 24 hours), the mauve line in that diagram makes a substantially bigger contribution than the dark line - more so than this diagram would suggest. The reason is that the length of a day is effectively the time-derivative of the equation of time; and the mauve line is rather steeper than the dark line (as well as having a slightly greater amplitude).
$endgroup$
– Dawood ibn Kareem
Apr 2 at 5:18
1
$begingroup$
@DawoodibnKareem - I wouldn't call ~20% smaller significantly less. I would call them almost the same magnitude. What's more important is the phase difference. The two curves contributors are closer to being in phase with one another in early November than they are in late February.
$endgroup$
– David Hammen
Apr 2 at 16:36
$begingroup$
@DavidHammen You have failed to understand my comment. It's not the magnitude that affects the length of the day, it's the slope. The mauve line has half the period of the dark line, so if you were to plot its slope, you'd find that it makes more than double the contribution that the dark line makes.
$endgroup$
– Dawood ibn Kareem
Apr 2 at 18:01
4
4
$begingroup$
And, then, once in a while, we need leap seconds.
$endgroup$
– Draco18s
Apr 1 at 14:33
$begingroup$
And, then, once in a while, we need leap seconds.
$endgroup$
– Draco18s
Apr 1 at 14:33
3
3
$begingroup$
@Draco18s Indeed! Accurate timekeeping is a complex & subtle business. See leapsecond.com, which I also linked in my comment on the question.
$endgroup$
– PM 2Ring
Apr 1 at 14:37
$begingroup$
@Draco18s Indeed! Accurate timekeeping is a complex & subtle business. See leapsecond.com, which I also linked in my comment on the question.
$endgroup$
– PM 2Ring
Apr 1 at 14:37
2
2
$begingroup$
It's worth noting that in terms of the actual length of a day (above or below 24 hours), the mauve line in that diagram makes a substantially bigger contribution than the dark line - more so than this diagram would suggest. The reason is that the length of a day is effectively the time-derivative of the equation of time; and the mauve line is rather steeper than the dark line (as well as having a slightly greater amplitude).
$endgroup$
– Dawood ibn Kareem
Apr 2 at 5:18
$begingroup$
It's worth noting that in terms of the actual length of a day (above or below 24 hours), the mauve line in that diagram makes a substantially bigger contribution than the dark line - more so than this diagram would suggest. The reason is that the length of a day is effectively the time-derivative of the equation of time; and the mauve line is rather steeper than the dark line (as well as having a slightly greater amplitude).
$endgroup$
– Dawood ibn Kareem
Apr 2 at 5:18
1
1
$begingroup$
@DawoodibnKareem - I wouldn't call ~20% smaller significantly less. I would call them almost the same magnitude. What's more important is the phase difference. The two curves contributors are closer to being in phase with one another in early November than they are in late February.
$endgroup$
– David Hammen
Apr 2 at 16:36
$begingroup$
@DawoodibnKareem - I wouldn't call ~20% smaller significantly less. I would call them almost the same magnitude. What's more important is the phase difference. The two curves contributors are closer to being in phase with one another in early November than they are in late February.
$endgroup$
– David Hammen
Apr 2 at 16:36
$begingroup$
@DavidHammen You have failed to understand my comment. It's not the magnitude that affects the length of the day, it's the slope. The mauve line has half the period of the dark line, so if you were to plot its slope, you'd find that it makes more than double the contribution that the dark line makes.
$endgroup$
– Dawood ibn Kareem
Apr 2 at 18:01
$begingroup$
@DavidHammen You have failed to understand my comment. It's not the magnitude that affects the length of the day, it's the slope. The mauve line has half the period of the dark line, so if you were to plot its slope, you'd find that it makes more than double the contribution that the dark line makes.
$endgroup$
– Dawood ibn Kareem
Apr 2 at 18:01
add a comment |
$begingroup$
The average value of a distribution is not the average of its minimum and maximum. For example, the average value of (0,0,0,4) is 1, not 2. Earth's orbit eccentricity is not 0, nor the Moon's one, so your distribution of day duration is probably slightly asymetric, hence the 4 seconds discrepancy. Sum all day duration of a year, divide by the number of days, you will get a better value.
answered Apr 1 at 12:35
MattMatt
27114
New contributor
Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The average value of a distribution is not the average of its minimum and maximum. For example, the average value of (0,0,0,4) is 1, not 2. Earth's orbit eccentricity is not 0, nor the Moon's one, so your distribution of day duration is probably slightly asymetric, hence the 4 seconds discrepancy. Sum all day duration of a year, divide by the number of days, you will get a better value.
answered Apr 1 at 12:35
MattMatt
27114
New contributor
Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The average value of a distribution is not the average of its minimum and maximum. For example, the average value of (0,0,0,4) is 1, not 2. Earth's orbit eccentricity is not 0, nor the Moon's one, so your distribution of day duration is probably slightly asymetric, hence the 4 seconds discrepancy. Sum all day duration of a year, divide by the number of days, you will get a better value.
answered Apr 1 at 12:35
MattMatt
27114
New contributor
Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The average value of a distribution is not the average of its minimum and maximum. For example, the average value of (0,0,0,4) is 1, not 2. Earth's orbit eccentricity is not 0, nor the Moon's one, so your distribution of day duration is probably slightly asymetric, hence the 4 seconds discrepancy. Sum all day duration of a year, divide by the number of days, you will get a better value.
answered Apr 1 at 12:35
MattMatt
27114
New contributor
Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
answered Apr 1 at 12:35
MattMatt
27114
New contributor
Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 1 at 12:35
MattMatt
27114
answered Apr 1 at 12:35
MattMatt
27114
27114
New contributor
Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
The original definition of an hour was 1/24th of a day, no matter how long the day was at the time. The durations you're quoting are only possible with the help of extremely modern definitions made possible by redefining the second (keeping the minute and hour fixed as 60 seconds, and 3600 seconds, respectively).
answered Apr 1 at 11:51
Sean E. LakeSean E. Lake
14.7k12351
$endgroup$
$begingroup$
I don't disagree, but do you have 1) a date when that became the official definition of an hour, and 2) when it was conventionally determined to use 1/24 and not 1/10 or 1/20 of a day? I've read undocumented sources that mention the daylight being divided into tenths historically prior to twelfths.
$endgroup$
– Bill N
Apr 1 at 11:56
1
$begingroup$
More details in en.wikipedia.org/wiki/Hour
$endgroup$
– Stéphane Rollandin
Apr 1 at 12:14
1
$begingroup$
The mean solar day is 86400.0025 seconds, so while the reason in this answer has some effect, it only accounts for 2.5 ms of the 4 second discrepancy in the question.
$endgroup$
– JiK
Apr 1 at 16:05
add a comment |
$begingroup$
The original definition of an hour was 1/24th of a day, no matter how long the day was at the time. The durations you're quoting are only possible with the help of extremely modern definitions made possible by redefining the second (keeping the minute and hour fixed as 60 seconds, and 3600 seconds, respectively).
answered Apr 1 at 11:51
Sean E. LakeSean E. Lake
14.7k12351
$endgroup$
$begingroup$
I don't disagree, but do you have 1) a date when that became the official definition of an hour, and 2) when it was conventionally determined to use 1/24 and not 1/10 or 1/20 of a day? I've read undocumented sources that mention the daylight being divided into tenths historically prior to twelfths.
$endgroup$
– Bill N
Apr 1 at 11:56
1
$begingroup$
More details in en.wikipedia.org/wiki/Hour
$endgroup$
– Stéphane Rollandin
Apr 1 at 12:14
1
$begingroup$
The mean solar day is 86400.0025 seconds, so while the reason in this answer has some effect, it only accounts for 2.5 ms of the 4 second discrepancy in the question.
$endgroup$
– JiK
Apr 1 at 16:05
add a comment |
$begingroup$
The original definition of an hour was 1/24th of a day, no matter how long the day was at the time. The durations you're quoting are only possible with the help of extremely modern definitions made possible by redefining the second (keeping the minute and hour fixed as 60 seconds, and 3600 seconds, respectively).
answered Apr 1 at 11:51
Sean E. LakeSean E. Lake
14.7k12351
$endgroup$
The original definition of an hour was 1/24th of a day, no matter how long the day was at the time. The durations you're quoting are only possible with the help of extremely modern definitions made possible by redefining the second (keeping the minute and hour fixed as 60 seconds, and 3600 seconds, respectively).
answered Apr 1 at 11:51
Sean E. LakeSean E. Lake
14.7k12351
answered Apr 1 at 11:51
Sean E. LakeSean E. Lake
14.7k12351
answered Apr 1 at 11:51
Sean E. LakeSean E. Lake
14.7k12351
answered Apr 1 at 11:51
Sean E. LakeSean E. Lake
14.7k12351
14.7k12351
$begingroup$
I don't disagree, but do you have 1) a date when that became the official definition of an hour, and 2) when it was conventionally determined to use 1/24 and not 1/10 or 1/20 of a day? I've read undocumented sources that mention the daylight being divided into tenths historically prior to twelfths.
$endgroup$
– Bill N
Apr 1 at 11:56
1
$begingroup$
More details in en.wikipedia.org/wiki/Hour
$endgroup$
– Stéphane Rollandin
Apr 1 at 12:14
1
$begingroup$
The mean solar day is 86400.0025 seconds, so while the reason in this answer has some effect, it only accounts for 2.5 ms of the 4 second discrepancy in the question.
$endgroup$
– JiK
Apr 1 at 16:05
add a comment |
$begingroup$
I don't disagree, but do you have 1) a date when that became the official definition of an hour, and 2) when it was conventionally determined to use 1/24 and not 1/10 or 1/20 of a day? I've read undocumented sources that mention the daylight being divided into tenths historically prior to twelfths.
$endgroup$
– Bill N
Apr 1 at 11:56
1
$begingroup$
More details in en.wikipedia.org/wiki/Hour
$endgroup$
– Stéphane Rollandin
Apr 1 at 12:14
1
$begingroup$
The mean solar day is 86400.0025 seconds, so while the reason in this answer has some effect, it only accounts for 2.5 ms of the 4 second discrepancy in the question.
$endgroup$
– JiK
Apr 1 at 16:05
$begingroup$
I don't disagree, but do you have 1) a date when that became the official definition of an hour, and 2) when it was conventionally determined to use 1/24 and not 1/10 or 1/20 of a day? I've read undocumented sources that mention the daylight being divided into tenths historically prior to twelfths.
$endgroup$
– Bill N
Apr 1 at 11:56
$begingroup$
I don't disagree, but do you have 1) a date when that became the official definition of an hour, and 2) when it was conventionally determined to use 1/24 and not 1/10 or 1/20 of a day? I've read undocumented sources that mention the daylight being divided into tenths historically prior to twelfths.
$endgroup$
– Bill N
Apr 1 at 11:56
1
1
$begingroup$
More details in en.wikipedia.org/wiki/Hour
$endgroup$
– Stéphane Rollandin
Apr 1 at 12:14
$begingroup$
More details in en.wikipedia.org/wiki/Hour
$endgroup$
– Stéphane Rollandin
Apr 1 at 12:14
1
1
$begingroup$
The mean solar day is 86400.0025 seconds, so while the reason in this answer has some effect, it only accounts for 2.5 ms of the 4 second discrepancy in the question.
$endgroup$
– JiK
Apr 1 at 16:05
$begingroup$
The mean solar day is 86400.0025 seconds, so while the reason in this answer has some effect, it only accounts for 2.5 ms of the 4 second discrepancy in the question.
$endgroup$
– JiK
Apr 1 at 16:05
add a comment |
$begingroup$
The 24 hours (exactly) definition of the mean solar day only applies if you use the UT1 time scale. As others have mentioned, the mean solar day is not the average of the shortest and longest apparent solar day, and you have to consider the Equation of Time to calculate the mean solar day by averaging all apparent solar days in one year.
If you use the definition of a second based on the metric system, which is used by the atomic time (TAI), UTC and the terrestrial time (TT) time scales, the length of the mean solar day is not exactly 24 hours. The definition of the SI second, apart from being an atomic scale, is based on the mean solar day in 1900 as determined by Newcomb (in reality it corresponds to the mean solar day in about mid 19th century), when the Earth's rotation was faster than today. Today the mean solar day is slightly longer than in the past (when measured by an atomic clock), and the length of the mean solar day in TAI seconds is greater than in UT1. To correct for the difference, leap seconds are introduced 0-2 times per year in the UTC time scale based on observations of Earth's rotation, to keep UTC in sync with UT1 to less than 0.9 s. These corrections are publised in advance by IERS in Bulletin C. There have been 27 leap seconds inserted in the last 46 years, which equals to about 0.6 s per year difference between the length of the mean solar day between TAI and UT1, or about 1.6 ms per day (see Figure 1 and 2).
The length of the SI second depends on the location, and TAI is based on observations done by atomic clocks in various laboratories around the world ("UTC(k)") and corrected for the geopotential on the sea level. TT is a theoretical length of the SI second on the geoid, and as such is never known perfectly. Approximation of past TT is revised annualy by BIPM. In contrast, TAI and UTC are determined and kept fixed after about 1 month past the fact (published regularly in Circular-T by BIPM). Laboratory-specific UTC(k) time scales and the GPS time (based on the US Naval Observatory Master Clock) are known in real time. Other time scales include the geocentric time, barycentric time and the ephemeris time. The length of 1 s is slightly different between all of them, some of the differece is due to time passing differently based on location in the gravitation potential. For example, time passes faster in the solar system than on the Earth's surface by about 0.5 second a year.
Historically, time measured by the rotation of the Earth was the most accurate, and the mean solar day was assumed to be constant equal to 86400 s. This changed with the introduction of the ephemeris time, later superseded by the quartz clock and the atomic clock, which lead to re-definition of the second. They will deviate further as the rotation of the Earth slows down.
References:
- Urban and Seidelmann (2012): Explanatory Supplement to the Astronomical Almanac
- McCarthy and Seidelmann (2018): Time: From Earth Rotation to Atomic Physics
- BIPM time publications: https://www.bipm.org/en/bipm/tai/publications.html
- IERS Bulletins: https://www.iers.org/IERS/EN/Publications/Bulletins/bulletins.html
- USNO - The Equation of Time: https://aa.usno.navy.mil/faq/docs/eqtime.php
Figure 1. Excess to 86400s of the duration of the days, combined GPS solution, 1995-1997. From https://www.iers.org/IERS/EN/Science/EarthRotation/LODgps.html
Figure 2. TAI-UT1 and TAI-UTC. From McCarthy and Seidelmann (2018).
answered 2 days ago
peterpeter
1495
$endgroup$
$begingroup$
Thanks for mentioning Terrestrial Time, etc. I didn't want to go down the time scales rabbit-hole in my answer (although I did supply a link to articles on these topics in a couple of comments), but I guess it's good to give people a brief glimpse of this material. ;)
$endgroup$
– PM 2Ring
2 days ago
$begingroup$
Leap seconds have absolutely nothing to do with the OP's error of 4 seconds in his/her attempt to calculate the length of the mean solar day. The information in this answer is all true, but all completely unrelated to the question.
$endgroup$
– Dawood ibn Kareem
2 days ago
$begingroup$
@DawoodibnKareem I think it is related because the answer to the question how long the mean solar day is is not "24 hours 0min 0 seconds exactly" - not when you use the conventional definition of 1 s. The other answers do not explain this (but answer the main point, which is the wrong averaging).
$endgroup$
– peter
2 days ago
add a comment |
$begingroup$
The 24 hours (exactly) definition of the mean solar day only applies if you use the UT1 time scale. As others have mentioned, the mean solar day is not the average of the shortest and longest apparent solar day, and you have to consider the Equation of Time to calculate the mean solar day by averaging all apparent solar days in one year.
If you use the definition of a second based on the metric system, which is used by the atomic time (TAI), UTC and the terrestrial time (TT) time scales, the length of the mean solar day is not exactly 24 hours. The definition of the SI second, apart from being an atomic scale, is based on the mean solar day in 1900 as determined by Newcomb (in reality it corresponds to the mean solar day in about mid 19th century), when the Earth's rotation was faster than today. Today the mean solar day is slightly longer than in the past (when measured by an atomic clock), and the length of the mean solar day in TAI seconds is greater than in UT1. To correct for the difference, leap seconds are introduced 0-2 times per year in the UTC time scale based on observations of Earth's rotation, to keep UTC in sync with UT1 to less than 0.9 s. These corrections are publised in advance by IERS in Bulletin C. There have been 27 leap seconds inserted in the last 46 years, which equals to about 0.6 s per year difference between the length of the mean solar day between TAI and UT1, or about 1.6 ms per day (see Figure 1 and 2).
The length of the SI second depends on the location, and TAI is based on observations done by atomic clocks in various laboratories around the world ("UTC(k)") and corrected for the geopotential on the sea level. TT is a theoretical length of the SI second on the geoid, and as such is never known perfectly. Approximation of past TT is revised annualy by BIPM. In contrast, TAI and UTC are determined and kept fixed after about 1 month past the fact (published regularly in Circular-T by BIPM). Laboratory-specific UTC(k) time scales and the GPS time (based on the US Naval Observatory Master Clock) are known in real time. Other time scales include the geocentric time, barycentric time and the ephemeris time. The length of 1 s is slightly different between all of them, some of the differece is due to time passing differently based on location in the gravitation potential. For example, time passes faster in the solar system than on the Earth's surface by about 0.5 second a year.
Historically, time measured by the rotation of the Earth was the most accurate, and the mean solar day was assumed to be constant equal to 86400 s. This changed with the introduction of the ephemeris time, later superseded by the quartz clock and the atomic clock, which lead to re-definition of the second. They will deviate further as the rotation of the Earth slows down.
References:
- Urban and Seidelmann (2012): Explanatory Supplement to the Astronomical Almanac
- McCarthy and Seidelmann (2018): Time: From Earth Rotation to Atomic Physics
- BIPM time publications: https://www.bipm.org/en/bipm/tai/publications.html
- IERS Bulletins: https://www.iers.org/IERS/EN/Publications/Bulletins/bulletins.html
- USNO - The Equation of Time: https://aa.usno.navy.mil/faq/docs/eqtime.php
Figure 1. Excess to 86400s of the duration of the days, combined GPS solution, 1995-1997. From https://www.iers.org/IERS/EN/Science/EarthRotation/LODgps.html
Figure 2. TAI-UT1 and TAI-UTC. From McCarthy and Seidelmann (2018).
answered 2 days ago
peterpeter
1495
$endgroup$
$begingroup$
Thanks for mentioning Terrestrial Time, etc. I didn't want to go down the time scales rabbit-hole in my answer (although I did supply a link to articles on these topics in a couple of comments), but I guess it's good to give people a brief glimpse of this material. ;)
$endgroup$
– PM 2Ring
2 days ago
$begingroup$
Leap seconds have absolutely nothing to do with the OP's error of 4 seconds in his/her attempt to calculate the length of the mean solar day. The information in this answer is all true, but all completely unrelated to the question.
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– Dawood ibn Kareem
2 days ago
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@DawoodibnKareem I think it is related because the answer to the question how long the mean solar day is is not "24 hours 0min 0 seconds exactly" - not when you use the conventional definition of 1 s. The other answers do not explain this (but answer the main point, which is the wrong averaging).
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– peter
2 days ago
add a comment |
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The 24 hours (exactly) definition of the mean solar day only applies if you use the UT1 time scale. As others have mentioned, the mean solar day is not the average of the shortest and longest apparent solar day, and you have to consider the Equation of Time to calculate the mean solar day by averaging all apparent solar days in one year.
If you use the definition of a second based on the metric system, which is used by the atomic time (TAI), UTC and the terrestrial time (TT) time scales, the length of the mean solar day is not exactly 24 hours. The definition of the SI second, apart from being an atomic scale, is based on the mean solar day in 1900 as determined by Newcomb (in reality it corresponds to the mean solar day in about mid 19th century), when the Earth's rotation was faster than today. Today the mean solar day is slightly longer than in the past (when measured by an atomic clock), and the length of the mean solar day in TAI seconds is greater than in UT1. To correct for the difference, leap seconds are introduced 0-2 times per year in the UTC time scale based on observations of Earth's rotation, to keep UTC in sync with UT1 to less than 0.9 s. These corrections are publised in advance by IERS in Bulletin C. There have been 27 leap seconds inserted in the last 46 years, which equals to about 0.6 s per year difference between the length of the mean solar day between TAI and UT1, or about 1.6 ms per day (see Figure 1 and 2).
The length of the SI second depends on the location, and TAI is based on observations done by atomic clocks in various laboratories around the world ("UTC(k)") and corrected for the geopotential on the sea level. TT is a theoretical length of the SI second on the geoid, and as such is never known perfectly. Approximation of past TT is revised annualy by BIPM. In contrast, TAI and UTC are determined and kept fixed after about 1 month past the fact (published regularly in Circular-T by BIPM). Laboratory-specific UTC(k) time scales and the GPS time (based on the US Naval Observatory Master Clock) are known in real time. Other time scales include the geocentric time, barycentric time and the ephemeris time. The length of 1 s is slightly different between all of them, some of the differece is due to time passing differently based on location in the gravitation potential. For example, time passes faster in the solar system than on the Earth's surface by about 0.5 second a year.
Historically, time measured by the rotation of the Earth was the most accurate, and the mean solar day was assumed to be constant equal to 86400 s. This changed with the introduction of the ephemeris time, later superseded by the quartz clock and the atomic clock, which lead to re-definition of the second. They will deviate further as the rotation of the Earth slows down.
References:
- Urban and Seidelmann (2012): Explanatory Supplement to the Astronomical Almanac
- McCarthy and Seidelmann (2018): Time: From Earth Rotation to Atomic Physics
- BIPM time publications: https://www.bipm.org/en/bipm/tai/publications.html
- IERS Bulletins: https://www.iers.org/IERS/EN/Publications/Bulletins/bulletins.html
- USNO - The Equation of Time: https://aa.usno.navy.mil/faq/docs/eqtime.php
Figure 1. Excess to 86400s of the duration of the days, combined GPS solution, 1995-1997. From https://www.iers.org/IERS/EN/Science/EarthRotation/LODgps.html
Figure 2. TAI-UT1 and TAI-UTC. From McCarthy and Seidelmann (2018).
answered 2 days ago
peterpeter
1495
$endgroup$
The 24 hours (exactly) definition of the mean solar day only applies if you use the UT1 time scale. As others have mentioned, the mean solar day is not the average of the shortest and longest apparent solar day, and you have to consider the Equation of Time to calculate the mean solar day by averaging all apparent solar days in one year.
If you use the definition of a second based on the metric system, which is used by the atomic time (TAI), UTC and the terrestrial time (TT) time scales, the length of the mean solar day is not exactly 24 hours. The definition of the SI second, apart from being an atomic scale, is based on the mean solar day in 1900 as determined by Newcomb (in reality it corresponds to the mean solar day in about mid 19th century), when the Earth's rotation was faster than today. Today the mean solar day is slightly longer than in the past (when measured by an atomic clock), and the length of the mean solar day in TAI seconds is greater than in UT1. To correct for the difference, leap seconds are introduced 0-2 times per year in the UTC time scale based on observations of Earth's rotation, to keep UTC in sync with UT1 to less than 0.9 s. These corrections are publised in advance by IERS in Bulletin C. There have been 27 leap seconds inserted in the last 46 years, which equals to about 0.6 s per year difference between the length of the mean solar day between TAI and UT1, or about 1.6 ms per day (see Figure 1 and 2).
The length of the SI second depends on the location, and TAI is based on observations done by atomic clocks in various laboratories around the world ("UTC(k)") and corrected for the geopotential on the sea level. TT is a theoretical length of the SI second on the geoid, and as such is never known perfectly. Approximation of past TT is revised annualy by BIPM. In contrast, TAI and UTC are determined and kept fixed after about 1 month past the fact (published regularly in Circular-T by BIPM). Laboratory-specific UTC(k) time scales and the GPS time (based on the US Naval Observatory Master Clock) are known in real time. Other time scales include the geocentric time, barycentric time and the ephemeris time. The length of 1 s is slightly different between all of them, some of the differece is due to time passing differently based on location in the gravitation potential. For example, time passes faster in the solar system than on the Earth's surface by about 0.5 second a year.
Historically, time measured by the rotation of the Earth was the most accurate, and the mean solar day was assumed to be constant equal to 86400 s. This changed with the introduction of the ephemeris time, later superseded by the quartz clock and the atomic clock, which lead to re-definition of the second. They will deviate further as the rotation of the Earth slows down.
References:
- Urban and Seidelmann (2012): Explanatory Supplement to the Astronomical Almanac
- McCarthy and Seidelmann (2018): Time: From Earth Rotation to Atomic Physics
- BIPM time publications: https://www.bipm.org/en/bipm/tai/publications.html
- IERS Bulletins: https://www.iers.org/IERS/EN/Publications/Bulletins/bulletins.html
- USNO - The Equation of Time: https://aa.usno.navy.mil/faq/docs/eqtime.php
Figure 1. Excess to 86400s of the duration of the days, combined GPS solution, 1995-1997. From https://www.iers.org/IERS/EN/Science/EarthRotation/LODgps.html
Figure 2. TAI-UT1 and TAI-UTC. From McCarthy and Seidelmann (2018).
answered 2 days ago
peterpeter
1495
edited 2 days ago
answered 2 days ago
peterpeter
1495
answered 2 days ago
peterpeter
1495
answered 2 days ago
peterpeter
1495
1495
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Thanks for mentioning Terrestrial Time, etc. I didn't want to go down the time scales rabbit-hole in my answer (although I did supply a link to articles on these topics in a couple of comments), but I guess it's good to give people a brief glimpse of this material. ;)
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– PM 2Ring
2 days ago
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Leap seconds have absolutely nothing to do with the OP's error of 4 seconds in his/her attempt to calculate the length of the mean solar day. The information in this answer is all true, but all completely unrelated to the question.
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– Dawood ibn Kareem
2 days ago
$begingroup$
@DawoodibnKareem I think it is related because the answer to the question how long the mean solar day is is not "24 hours 0min 0 seconds exactly" - not when you use the conventional definition of 1 s. The other answers do not explain this (but answer the main point, which is the wrong averaging).
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– peter
2 days ago
add a comment |
$begingroup$
Thanks for mentioning Terrestrial Time, etc. I didn't want to go down the time scales rabbit-hole in my answer (although I did supply a link to articles on these topics in a couple of comments), but I guess it's good to give people a brief glimpse of this material. ;)
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– PM 2Ring
2 days ago
$begingroup$
Leap seconds have absolutely nothing to do with the OP's error of 4 seconds in his/her attempt to calculate the length of the mean solar day. The information in this answer is all true, but all completely unrelated to the question.
$endgroup$
– Dawood ibn Kareem
2 days ago
$begingroup$
@DawoodibnKareem I think it is related because the answer to the question how long the mean solar day is is not "24 hours 0min 0 seconds exactly" - not when you use the conventional definition of 1 s. The other answers do not explain this (but answer the main point, which is the wrong averaging).
$endgroup$
– peter
2 days ago
$begingroup$
Thanks for mentioning Terrestrial Time, etc. I didn't want to go down the time scales rabbit-hole in my answer (although I did supply a link to articles on these topics in a couple of comments), but I guess it's good to give people a brief glimpse of this material. ;)
$endgroup$
– PM 2Ring
2 days ago
$begingroup$
Thanks for mentioning Terrestrial Time, etc. I didn't want to go down the time scales rabbit-hole in my answer (although I did supply a link to articles on these topics in a couple of comments), but I guess it's good to give people a brief glimpse of this material. ;)
$endgroup$
– PM 2Ring
2 days ago
$begingroup$
Leap seconds have absolutely nothing to do with the OP's error of 4 seconds in his/her attempt to calculate the length of the mean solar day. The information in this answer is all true, but all completely unrelated to the question.
$endgroup$
– Dawood ibn Kareem
2 days ago
$begingroup$
Leap seconds have absolutely nothing to do with the OP's error of 4 seconds in his/her attempt to calculate the length of the mean solar day. The information in this answer is all true, but all completely unrelated to the question.
$endgroup$
– Dawood ibn Kareem
2 days ago
$begingroup$
@DawoodibnKareem I think it is related because the answer to the question how long the mean solar day is is not "24 hours 0min 0 seconds exactly" - not when you use the conventional definition of 1 s. The other answers do not explain this (but answer the main point, which is the wrong averaging).
$endgroup$
– peter
2 days ago
$begingroup$
@DawoodibnKareem I think it is related because the answer to the question how long the mean solar day is is not "24 hours 0min 0 seconds exactly" - not when you use the conventional definition of 1 s. The other answers do not explain this (but answer the main point, which is the wrong averaging).
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– peter
2 days ago
add a comment |
10
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Related: "Leap second", Wikipedia.
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– Nat
Apr 1 at 12:09
5
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Also see leapsecond.com for a plethora of articles about timekeeping, especially the articles by Steve Allen of Lick Observatory. That's a rather old page, and some of its links are now dead, but there's still plenty of good info there.
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– PM 2Ring
Apr 1 at 13:09
6
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@Nat - This has almost nothing to do with leap seconds. It has much more to do with that thinking that the length of day averaged over the course of a few years is equal to the average of the minimum and maximum length of day is invalid. There are many pseudocyclical functions where the mean is not the average of the min and max, the equation of time being a prime example.
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– David Hammen
Apr 2 at 15:38
2
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@Nat - I disagree. The four second deviation from 86400 seconds is over three orders of magnitude larger than the tiny 2.5 millisecond deviation that results in the need for leap seconds. If the mean solar day was 86404 seconds long, we would need 4 leap seconds every day, or two leap minutes per month. But it's not. The mean solar day remains very close to being 86400 seconds long.
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– David Hammen
Apr 2 at 15:55
3
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Dear @ObsessionWithElectricity, The existence of a well-received answer indicates nothing about whether the question belongs to this site or not. Hope you appreciate the distinction. Cheers! :)
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– Dvij Mankad
2 days ago