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 8.3 The SiderealTime and HourAngle Functions SiderealTime determines which right ascension line on a star chart is crossing the local meridian at a given date and time. Similarly, HourAngle determines the number of hours before an object will cross the local meridian. You can also use this function to determine when a star will cross the meridian. Culmination determines at what time an object will cross the meridian. Timing information. This shows that the 5.7 hour right ascension line, which roughly corresponds to the constellation of Taurus, is crossing the local meridian at the given date and time. In[22]:=SiderealTime[{1993,11,17,3,20,0}] Out[22]= The star Sirius crosses the meridian in just over one hour's time from the given date. In[23]:=HourAngle[Sirius, {1993,11,17,3,20,0}] Out[23]= On 1993 November 17, Sirius crosses the local meridian at 04:22. At this time the star is at its highest point in the sky. In[24]:=Culmination[Sirius, {1993,11,17}] Out[24]= Time Conversion ModifiedJulianDay returns the modified Julian day number of a local date. To get the true Julian day number, add 2415019.5 to the output. Converting dates to and from Julian days. This is the modified Julian day number of 1993 November 17, 03:20 local time. In[25]:=ModifiedJulianDay[{1993,11,17,3,20,0}] Out[25]= Add 2415019.5 to get the true Julian day number. In[26]:=% + 2415019.5 Out[26]= Here is the true Julian day number expressed in nonexponential form. In[27]:=AccountingForm[%, 20] Out[27]//AccountingForm= This converts a modified Julian day number into a local date and time. In[28]:=LocalDate[34288.68055555556] Out[28]= All dates returned by Scientific Astronomer are in local time; that is, your time zone is always taken into account. To get Universal Time (UT) or Greenwich Mean Time (GMT), subtract your time zone value from any local date. For instance, in the examples used throughout this user's guide, where TimeZone -> 11, the local date {1993,11,17,3,20,0} corresponds to {1993,11,16,16,20,0} Universal Time. In addition, all dates returned by Scientific Astronomer are based on the Gregorian calendar. To get the date according to the Julian calendar, which was in use prior to 1752 in most British colonies, add 2-Floor[y/100]+Floor[y/400] days, where y is the year. Remember also that Scientific Astronomer includes a year zero, although it was not historically used. Hence, the year 0 is the same as 1 B.C., and the year -1 is 2 B.C., and so on. Thus, the date {-584,5,22} corresponds to May 22, 585 B.C. In the Julian calendar this date would be called May 28, 585 B.C. because 2-Floor[-584/100]+Floor[-584/400] is 6 days.