About leap year.

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16 years 9 months ago #12785 by tvanflandern
Time is simply a measure of change. As such, it has no need to be measured in any particular way. Think of International Atomic Time (as used widely in astronomy) as simply a count of nanoseconds since atomic clocks were first used in timekeeping (1955).

Time for civilian purposes has many oddities. Leap seconds are inserted here and there to compensate for changes in Earth's rotation rate. Months have variable lengths. Leap years have an extra day to keep the calendar in phase with the seasons. Some century years that should be leap years are not leap years to make a further correction for drift with respect to seasons. There was a discontinuity in calendar dates at the change from the Julian calendar to the Gregorian calendar. Etc.

None of this affects astronomical time, which is based on continuous counts of time units, whether nanoseconds, seconds, days, or centuries. Astronomers simply have a number of different time scales, depending on the application.

Back in the 1960s, when programming languages were first being applied to the bulk of astronomical applications, a small group of us developed simple, one-code-line versions of "continuous day count" algorithms using the "Julian day number" system (a day count with an origin about 4000 BC). Here is one example:
G=367*Y-7*(Y+(M+9)/12)/4-3*((Y+(M-9)/7)/100+1)/4+275*M/9+D+1721029

Here, Y = 4-digit year, M = month number, D = day of month. G is the continuous day count. This works around all calendar irregularities. However, it utilizes integer arithmetic, as implemented in the old computer languages such as FORTRAN. That means the remainder in every division must always be discarded. -|Tom|-

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16 years 9 months ago #12814 by thebobgy
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
<br />Back in the 1960s, when programming languages were first being applied to the bulk of astronomical applications, a small group of us developed simple, one-code-line versions of "continuous day count" algorithms using the "Julian day number" system (a day count with an origin about 4000 BC). Here is one example:
G=367*Y-7*(Y+(M+9)/12)/4-3*((Y+(M-9)/7)/100+1)/4+275*M/9+D+1721029

Here, Y = 4-digit year, M = month number, D = day of month. G is the continuous day count. This works around all calendar irregularities. However, it utilizes integer arithmetic, as implemented in the old computer languages such as FORTRAN. That means the remainder in every division must always be discarded. -|Tom|-<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Excellent Tom, thank you. With a little work I will make a four year daily chart.

thebobguy

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