julian-date

How to add JulianDate objects or offsets in Skyfield

由 让人想犯罪 __ 提交于 2019-12-01 12:30:55
问题 The JulianDate object in Skyfield is a handy way to quickly produce and hold a set of time values in Julian Days, and pass them to Skyfield's at() method to calculate astronomical positions in various coordinates. (see an example script) However, I can't seem to find an add or offset method so that I can add a time offset or an iterable of offsets to a JulianDate object. I always seem to struggle with dates and times. Here is a very simple, abstracted example. I generate jd60 which is offset

SQL Server : convert Julian Date to YYYY-MM-DD

由 空扰寡人 提交于 2019-12-01 00:13:34
I have searched far and wide, but I can't seem find a way to convert julian to yyyy-mm-dd . Here is the format of my julian: The Julian format consists of the year, the first two digits, and the day within the year, the last three digits. For example, 95076 is March 17, 1995 . The 95 indicates the year and the 076 indicates it is the 76th day of the year. 15260 I have tried this but it isn't working: dateadd(d,(convert(int,LAST_CHANGED_DATE) % 1000)-1, convert(date,(convert(varchar,convert(int,LAST_CHANGED_DATE) /1000 + 1900) + '/1/1'))) as GrgDate You can select each part of the date using

Convert a Julian Date to an Instant

由 点点圈 提交于 2019-11-30 20:43:57
I'm running into a situation where I would like to convert from a Julian date to an java.time.Instant (if that makes sense), or some Java time that can be more easily understood. My understanding of what a Julian date is comes from reading the Wikipedia page . There are bunch of different variants , and the date I am trying to read uses a different epoch than any of these. For example, let's say the epoch is the beginning of the Calendar (New Style) Act 1750 , and the Julian date is 95906.27600694445 which in this case I believe is CE 2015 April 15 06:37:26.9 UT , how do I get an instant from

Convert julian day into date

由 非 Y 不嫁゛ 提交于 2019-11-30 19:28:57
问题 I have files named day00000.nc, day00001.nc, day00002.nc, ...day00364.nc for several years. They represent the 365 or 366 days. I want to rename my files like this day20070101.nc, day20070102.nc , ...day20071231.nc How can I do that ? Thank you 回答1: Use the datetime module to get date from day of the year. I am assuming the year is 2007 as in your examples, since your filenames do not seem to have an year value. Feel free to replace the hardcoded 2007 in the code with a variable if required.

Converting a Gregorian date to Julian Day Count in Objective C

由 那年仲夏 提交于 2019-11-30 14:13:29
I need Objective C method for converting Gregorian date to Julian days same as this PHP method (GregorianToJD). According to http://en.wikipedia.org/wiki/Julian_day , the Julian day number for January 1, 2000, was 2,451,545. So you can compute the number of days between your date and this reference date. For example (Jan 1, 2014): NSUInteger julianDayFor01012000 = 2451545; NSCalendar *cal = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar]; [cal setTimeZone:[NSTimeZone timeZoneForSecondsFromGMT:0]]; NSDateComponents *comp = [[NSDateComponents alloc] init]; comp.year = 2014;

Convert unix timestamp to julian

由 风格不统一 提交于 2019-11-30 08:18:54
How can I convert from a unix timestamp (say 1232559922) to a fractional julian date (2454853.03150). I found a website ( http://aa.usno.navy.mil/data/docs/JulianDate.php ) that performs a similar calculation but I need to do it programatically. Solutions can be in C/C++, python, perl, bash, etc... Jason Cohen The Unix epoch (zero-point) is January 1, 1970 GMT. That corresponds to the Julian day of 2440587.5 So, in pseudo-code: function float getJulianFromUnix( int unixSecs ) { return ( unixSecs / 86400.0 ) + 2440587.5; } I know that this is an old post, but I'll just say ... The answer given

Convert a Julian Date to an Instant

由 百般思念 提交于 2019-11-30 05:15:11
问题 I'm running into a situation where I would like to convert from a Julian date to an java.time.Instant (if that makes sense), or some Java time that can be more easily understood. My understanding of what a Julian date is comes from reading the Wikipedia page. There are bunch of different variants, and the date I am trying to read uses a different epoch than any of these. For example, let's say the epoch is the beginning of the Calendar (New Style) Act 1750, and the Julian date is 95906

Converting a Gregorian date to Julian Day Count in Objective C

由 做~自己de王妃 提交于 2019-11-29 20:04:09
问题 I need Objective C method for converting Gregorian date to Julian days same as this PHP method (GregorianToJD). 回答1: According to http://en.wikipedia.org/wiki/Julian_day, the Julian day number for January 1, 2000, was 2,451,545. So you can compute the number of days between your date and this reference date. For example (Jan 1, 2014): NSUInteger julianDayFor01012000 = 2451545; NSCalendar *cal = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar]; [cal setTimeZone:[NSTimeZone

Convert unix timestamp to julian

由 放肆的年华 提交于 2019-11-29 11:17:01
问题 How can I convert from a unix timestamp (say 1232559922) to a fractional julian date (2454853.03150). I found a website ( http://aa.usno.navy.mil/data/docs/JulianDate.php ) that performs a similar calculation but I need to do it programatically. Solutions can be in C/C++, python, perl, bash, etc... 回答1: The Unix epoch (zero-point) is January 1, 1970 GMT. That corresponds to the Julian day of 2440587.5 So, in pseudo-code: function float getJulianFromUnix( int unixSecs ) { return ( unixSecs /

Convert a Julian Date to Regular Calendar Date

由 三世轮回 提交于 2019-11-29 10:32:25
How do I convert a 7-digit julian date into a format like MM/dd/yyy? scott Found a useful site: http://www.rgagnon.com/javadetails/java-0506.html This should do the trick: public static int[] fromJulian(double injulian) { int jalpha,ja,jb,jc,jd,je,year,month,day; double julian = julian + HALFSECOND / 86400.0; ja = (int) julian; if (ja>= JGREG) { jalpha = (int) (((ja - 1867216) - 0.25) / 36524.25); ja = ja + 1 + jalpha - jalpha / 4; } jb = ja + 1524; jc = (int) (6680.0 + ((jb - 2439870) - 122.1) / 365.25); jd = 365 * jc + jc / 4; je = (int) ((jb - jd) / 30.6001); day = jb - jd - (int) (30.6001
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