TIME

A.  Hour Angle (HA)

Hour Angle = The angular distance of an object E/W from its upper transit point on the LCM.
The arc representing HA may be imagined to extend from the upper transit point on the LCM
toward an object, along the diurnal circle of that object.
Hence, when an object is at upper transit, it's hour angle is exactly 0h 00m 00s.

HA may be expressed in either arc or time units, where 15o=1h00m and 4m = 1o.

Eastern HAs are always negative. Western HAs are always positive.

An example: HA = -3h20m = 50oE.

All objects with the same RA have the same HA for a given observer at a given time.

All objects with the same RA make UT at the same time.

B. Local Apparent Solar Time (LAT)

Local Apparent Solar Time is time determined by the Sun's position in the sky relative
to one's local celestial meridian.  One's local apparent solar time may be computed from the
following relation:

 LAT = HAS+ 12h00m,  where S stands for the Sun.

In the above equation, which defines LAT, use only a 24 hour time system, e.g.
5:30 p.m. = 17:30

Do not confuse LAT with latitude, which we usually abbreviate as Lat.

Example: If the sun is observed to be 35oW of the LCM, what is the LAT? First convert arcmeasure
to time units by dividing the HA in degrees by the rate of rotation of the Earth (15o per hour). The
result is 35oW=+2h20m, then:

LAT = 2h20m + 12h00m = 14h20m

Conversely, if LAT = 14h20m, then:
HA¤= LAT - 12h00m = 14h20m - 12 h00m = +2h20m

C.  SIDEREAL TIME (LST)

One's local sidereal time (time by the stars) is defined by the following relation:

LST = Western HA of VE.

Conceptually, this relation is saying that LST is the amount of time elapsed (at the sidereal rate)
since the last UT of VE.

As the Earth rotates from W to E, an observer sees successive hour circles (HCs) of RA
coinciding with their LCM.

Hence, the time on the sidereal clock is the same as the RA of all objects now making UT.

Because the Sun moves eastward along the ecliptic by approximately 1o per day, time by the
vernal equinox and time by the Sun do not pass at the same rate. The sidereal clock runs faster
than the solar clock, gaining 4 minutes per day relative to the solar clock. The two clocks agree
only on one day of the year, this is the day of the autumnal equinox. One of the consequences
of this is that the stars rise, make upper transit, or set 4 minutes earlier each day when using
solar time.

The algebraic sum of the RA and HA for any object is equal to the sidereal time for any observer,
that is,

LST = RA + HA

Also see Ex. 8.0 in the Course Manual and the diagram at the end of Exercise 9.0

D.  Computing Time Intervals:

There are several occasions, in doing the lab exercises, that require the computation of how much
time has passed between two events.  This is the case  for  Ex. 13.0 and 15.1.

An example would be to find how many years and decimal parts thereof have passed between two
dates and times.  To do this,you need to convert both dates and times to a decimal equivalent of
the largest unit of time, be it years, months, days, etc.

A specific example would be to find the number of decimal years between June 22, 2022 at 14:00
and Feb. 5, 2055, at 9:00, with a precision of hours, such as in Exercise 15.1

Let us do this for the first date and time:
First divide the number of hours by 24.  E. G.  14/24= 0.58 days.  Add this to the day minus 1:
(22-1) + 0.58 = 21.58 days.   We subtract 1 since 22 days have not passed in June until
midnight, that is, the end of the 22nd.

Now take the number of days and divide by the average number of days per month, 30.5;
21.58/30.5 = 0.708 months.  Don't do this step if the time interval you are computing is over
only a few months or less.  Then just work totally in days and decimal parts thereof.
See below.

Add the decimal part of a month to the month number minus 1.  Again we subtract 1, since 6
months have not passed in the year until the end of June 30th. June is month 6, so we get
5.708.

Now take the number of decimal months and divide by the number of months in a year:
5.708/12 = 0.476
Take this decimal part of  a year and add to the year number minus 1, getting:   2021.476.

Now do the same thing for the other date to get 2054.???.  You should calculate what the
decimal numbers are that replace the question marks

Now simply subtract the two years to get how many years and decimall parts thereof have
passed between the two dates and times;   2054.??? - 2021.478 = ??.???

Don't convert to years for Ex. 13.0   There you  need only work in days by calculating
the amount of days and decimal parts of a day that have passed between two times and dates
of different months of the same year. For example, if the first date is in June and the second
date is in July, you must allow for the fact that there are 30 days in June.  You cannot simply
subtract the date in June from the date in July.  Instead add 30 to the date in July and then
subtract.  Also, it may be necessary to work with a precision of minutes instead of hours.  In the
latter case, you need to find what decimal part of an hour a certain number of minutes is before
you divide the number of hours by 24..

Copyright 2003, by R. J. Pfeiffer