How can a day be of 24 hours? The Next CEO of Stack OverflowHours of light per day based on latitude/longitude formulaWhat 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 is UT1 being computed?Day/night cycle in GreenlandHow 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?Why was the fraction 1/31,556,925.9747, in the 1956-1968 definition of the second in terms of ephemeris time, chosen?Is zero shadow day the longest day
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How can a day be of 24 hours?
The Next CEO of Stack OverflowHours of light per day based on latitude/longitude formulaWhat 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 is UT1 being computed?Day/night cycle in GreenlandHow 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?Why was the fraction 1/31,556,925.9747, in the 1956-1968 definition of the second in terms of ephemeris time, chosen?Is zero shadow day the longest day
$begingroup$
The longest solar day of year is approximately 24 hours 0 min 30 seconds (occurs at mid to late December) while shortest solar day of year is approximately 23 hour 59min 38 seconds. If average out both of these I come up with average solar day of 24 hour +4seconds. Why then it is said it is 24 h 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
$endgroup$
add a comment |
$begingroup$
The longest solar day of year is approximately 24 hours 0 min 30 seconds (occurs at mid to late December) while shortest solar day of year is approximately 23 hour 59min 38 seconds. If average out both of these I come up with average solar day of 24 hour +4seconds. Why then it is said it is 24 h 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
$endgroup$
3
$begingroup$
Related: "Leap second", Wikipedia.
$endgroup$
– Nat
18 hours ago
2
$begingroup$
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
17 hours ago
add a comment |
$begingroup$
The longest solar day of year is approximately 24 hours 0 min 30 seconds (occurs at mid to late December) while shortest solar day of year is approximately 23 hour 59min 38 seconds. If average out both of these I come up with average solar day of 24 hour +4seconds. Why then it is said it is 24 h 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
$endgroup$
The longest solar day of year is approximately 24 hours 0 min 30 seconds (occurs at mid to late December) while shortest solar day of year is approximately 23 hour 59min 38 seconds. If average out both of these I come up with average solar day of 24 hour +4seconds. Why then it is said it is 24 h 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 18 hours ago
Qmechanic♦
107k121991231
107k121991231
asked 18 hours ago
ObsessionWithElectricityObsessionWithElectricity
4151312
4151312
3
$begingroup$
Related: "Leap second", Wikipedia.
$endgroup$
– Nat
18 hours ago
2
$begingroup$
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
17 hours ago
add a comment |
3
$begingroup$
Related: "Leap second", Wikipedia.
$endgroup$
– Nat
18 hours ago
2
$begingroup$
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
17 hours ago
3
3
$begingroup$
Related: "Leap second", Wikipedia.
$endgroup$
– Nat
18 hours ago
$begingroup$
Related: "Leap second", Wikipedia.
$endgroup$
– Nat
18 hours ago
2
2
$begingroup$
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
17 hours ago
$begingroup$
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
17 hours ago
add a comment |
3 Answers
3
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.
$endgroup$
2
$begingroup$
And, then, once in a while, we need leap seconds.
$endgroup$
– Draco18s
16 hours ago
2
$begingroup$
@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
16 hours ago
8
$begingroup$
Can we take a moment to acknowledge that 'The Equation of Time' is one of the most badass names for an equation that I have ever encountered?
$endgroup$
– Arcanist Lupus
11 hours ago
1
$begingroup$
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
1 hour ago
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 excentricity 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.
New contributor
$endgroup$
1
$begingroup$
Where can I get solar day length information of the whole year??
$endgroup$
– ObsessionWithElectricity
18 hours ago
$begingroup$
I could find data here: hpiers.obspm.fr/eoppc/eop/eopc04/eopc04.62-now - I think that UT1-UTC is the discrepancy to 86400 of day duration, in seconds, from 1962 to now.
$endgroup$
– Matt
17 hours ago
$begingroup$
@Matt I don't think that data is what you think it is. The longest solar days are always in late December (around Christmas); and the shortest ones are in mid September. That doesn't fit with the data you've given.
$endgroup$
– Dawood ibn Kareem
1 hour ago
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).
$endgroup$
$begingroup$
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
18 hours ago
$begingroup$
More details in en.wikipedia.org/wiki/Hour
$endgroup$
– Stéphane Rollandin
18 hours ago
$begingroup$
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
14 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
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.
$endgroup$
2
$begingroup$
And, then, once in a while, we need leap seconds.
$endgroup$
– Draco18s
16 hours ago
2
$begingroup$
@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
16 hours ago
8
$begingroup$
Can we take a moment to acknowledge that 'The Equation of Time' is one of the most badass names for an equation that I have ever encountered?
$endgroup$
– Arcanist Lupus
11 hours ago
1
$begingroup$
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
1 hour ago
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.
$endgroup$
2
$begingroup$
And, then, once in a while, we need leap seconds.
$endgroup$
– Draco18s
16 hours ago
2
$begingroup$
@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
16 hours ago
8
$begingroup$
Can we take a moment to acknowledge that 'The Equation of Time' is one of the most badass names for an equation that I have ever encountered?
$endgroup$
– Arcanist Lupus
11 hours ago
1
$begingroup$
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
1 hour ago
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.
$endgroup$
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 17 hours ago
PM 2RingPM 2Ring
3,43321023
3,43321023
2
$begingroup$
And, then, once in a while, we need leap seconds.
$endgroup$
– Draco18s
16 hours ago
2
$begingroup$
@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
16 hours ago
8
$begingroup$
Can we take a moment to acknowledge that 'The Equation of Time' is one of the most badass names for an equation that I have ever encountered?
$endgroup$
– Arcanist Lupus
11 hours ago
1
$begingroup$
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
1 hour ago
add a comment |
2
$begingroup$
And, then, once in a while, we need leap seconds.
$endgroup$
– Draco18s
16 hours ago
2
$begingroup$
@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
16 hours ago
8
$begingroup$
Can we take a moment to acknowledge that 'The Equation of Time' is one of the most badass names for an equation that I have ever encountered?
$endgroup$
– Arcanist Lupus
11 hours ago
1
$begingroup$
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
1 hour ago
2
2
$begingroup$
And, then, once in a while, we need leap seconds.
$endgroup$
– Draco18s
16 hours ago
$begingroup$
And, then, once in a while, we need leap seconds.
$endgroup$
– Draco18s
16 hours ago
2
2
$begingroup$
@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
16 hours ago
$begingroup$
@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
16 hours ago
8
8
$begingroup$
Can we take a moment to acknowledge that 'The Equation of Time' is one of the most badass names for an equation that I have ever encountered?
$endgroup$
– Arcanist Lupus
11 hours ago
$begingroup$
Can we take a moment to acknowledge that 'The Equation of Time' is one of the most badass names for an equation that I have ever encountered?
$endgroup$
– Arcanist Lupus
11 hours ago
1
1
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
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 excentricity 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.
New contributor
$endgroup$
1
$begingroup$
Where can I get solar day length information of the whole year??
$endgroup$
– ObsessionWithElectricity
18 hours ago
$begingroup$
I could find data here: hpiers.obspm.fr/eoppc/eop/eopc04/eopc04.62-now - I think that UT1-UTC is the discrepancy to 86400 of day duration, in seconds, from 1962 to now.
$endgroup$
– Matt
17 hours ago
$begingroup$
@Matt I don't think that data is what you think it is. The longest solar days are always in late December (around Christmas); and the shortest ones are in mid September. That doesn't fit with the data you've given.
$endgroup$
– Dawood ibn Kareem
1 hour ago
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 excentricity 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.
New contributor
$endgroup$
1
$begingroup$
Where can I get solar day length information of the whole year??
$endgroup$
– ObsessionWithElectricity
18 hours ago
$begingroup$
I could find data here: hpiers.obspm.fr/eoppc/eop/eopc04/eopc04.62-now - I think that UT1-UTC is the discrepancy to 86400 of day duration, in seconds, from 1962 to now.
$endgroup$
– Matt
17 hours ago
$begingroup$
@Matt I don't think that data is what you think it is. The longest solar days are always in late December (around Christmas); and the shortest ones are in mid September. That doesn't fit with the data you've given.
$endgroup$
– Dawood ibn Kareem
1 hour ago
add a comment |
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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 excentricity 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.
New contributor
$endgroup$
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 excentricity 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.
New contributor
New contributor
answered 18 hours ago
MattMatt
1812
1812
New contributor
New contributor
1
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Where can I get solar day length information of the whole year??
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– ObsessionWithElectricity
18 hours ago
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I could find data here: hpiers.obspm.fr/eoppc/eop/eopc04/eopc04.62-now - I think that UT1-UTC is the discrepancy to 86400 of day duration, in seconds, from 1962 to now.
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– Matt
17 hours ago
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@Matt I don't think that data is what you think it is. The longest solar days are always in late December (around Christmas); and the shortest ones are in mid September. That doesn't fit with the data you've given.
$endgroup$
– Dawood ibn Kareem
1 hour ago
add a comment |
1
$begingroup$
Where can I get solar day length information of the whole year??
$endgroup$
– ObsessionWithElectricity
18 hours ago
$begingroup$
I could find data here: hpiers.obspm.fr/eoppc/eop/eopc04/eopc04.62-now - I think that UT1-UTC is the discrepancy to 86400 of day duration, in seconds, from 1962 to now.
$endgroup$
– Matt
17 hours ago
$begingroup$
@Matt I don't think that data is what you think it is. The longest solar days are always in late December (around Christmas); and the shortest ones are in mid September. That doesn't fit with the data you've given.
$endgroup$
– Dawood ibn Kareem
1 hour ago
1
1
$begingroup$
Where can I get solar day length information of the whole year??
$endgroup$
– ObsessionWithElectricity
18 hours ago
$begingroup$
Where can I get solar day length information of the whole year??
$endgroup$
– ObsessionWithElectricity
18 hours ago
$begingroup$
I could find data here: hpiers.obspm.fr/eoppc/eop/eopc04/eopc04.62-now - I think that UT1-UTC is the discrepancy to 86400 of day duration, in seconds, from 1962 to now.
$endgroup$
– Matt
17 hours ago
$begingroup$
I could find data here: hpiers.obspm.fr/eoppc/eop/eopc04/eopc04.62-now - I think that UT1-UTC is the discrepancy to 86400 of day duration, in seconds, from 1962 to now.
$endgroup$
– Matt
17 hours ago
$begingroup$
@Matt I don't think that data is what you think it is. The longest solar days are always in late December (around Christmas); and the shortest ones are in mid September. That doesn't fit with the data you've given.
$endgroup$
– Dawood ibn Kareem
1 hour ago
$begingroup$
@Matt I don't think that data is what you think it is. The longest solar days are always in late December (around Christmas); and the shortest ones are in mid September. That doesn't fit with the data you've given.
$endgroup$
– Dawood ibn Kareem
1 hour ago
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).
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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
18 hours ago
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More details in en.wikipedia.org/wiki/Hour
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– Stéphane Rollandin
18 hours ago
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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.
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– JiK
14 hours ago
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).
$endgroup$
$begingroup$
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
18 hours ago
$begingroup$
More details in en.wikipedia.org/wiki/Hour
$endgroup$
– Stéphane Rollandin
18 hours ago
$begingroup$
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
14 hours ago
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).
$endgroup$
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 18 hours ago
Sean E. LakeSean E. Lake
14.6k12350
14.6k12350
$begingroup$
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
18 hours ago
$begingroup$
More details in en.wikipedia.org/wiki/Hour
$endgroup$
– Stéphane Rollandin
18 hours ago
$begingroup$
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
14 hours ago
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
18 hours ago
$begingroup$
More details in en.wikipedia.org/wiki/Hour
$endgroup$
– Stéphane Rollandin
18 hours ago
$begingroup$
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
14 hours ago
$begingroup$
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
18 hours ago
$begingroup$
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
18 hours ago
$begingroup$
More details in en.wikipedia.org/wiki/Hour
$endgroup$
– Stéphane Rollandin
18 hours ago
$begingroup$
More details in en.wikipedia.org/wiki/Hour
$endgroup$
– Stéphane Rollandin
18 hours ago
$begingroup$
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
14 hours ago
$begingroup$
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
14 hours ago
add a comment |
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Related: "Leap second", Wikipedia.
$endgroup$
– Nat
18 hours ago
2
$begingroup$
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
17 hours ago