This is A stunt pure and simple, but it is an interesting stunt.
Look at any star chart or examine the table in the back of the Nautical Almanac giving the “apparent places of stars,” and you will note that the pointers of the dipper have a right ascension of 11 hours. They point directly to the North Star and revolve around Polaris like the hour hand of a clock.
Mean time clocks, with which everyone is familiar, show on their faces only 12 hours. Sidereal clocks show 24 hours. Therefore, the 1-hour division on a mean time clock occupies space equal to 2 hours on a sidereal clock.
Go on deck, or on a lawn, or on a roof—any place with a clear view to northward—and face the polar star. You will note that the pointers of the dipper (at different times during the night and on different months of the year) form various angles with your celestial meridian as they revolve around Polaris. If the two pointers are directly above Polaris (Fig. 1), they are in the meridian, and the right ascension of the pointers equals your local sidereal time, 11 hours.
We all know that L.S.T. = H.A.+R.A. Therefore, if the hour angle equals zero (the star in the meridian) the right ascension equals the sidereal time.
Let us assume another case: Suppose when you face north the pointers are in the position of seven o’clock on the face of an ordinary mean time clock (Fig. 2). Measuring to the left from the 12-hour position you have an angle of 5 hours on mean time clock, or 10 hours sidereal time clock as the hour angle of the pointers. Adding this 10 hours (H.A.) to the 11 hours right ascension gives you 21 hours as the local sidereal time.
We all know how to convert sidereal time into mean time, but that necessitates the use of tables, paper, and pencil. Here is a method which requires none of these.
We all know that on March 23, approximately, the right ascension of the mean sun is zero hours. This occurs at the vernal equinox which is the point where the celestial equator and the ecliptic cross. Month by month the mean sun’s right ascension advances until it again becomes zero on March 23, the succeeding year. In other words, the right ascension of the mean sun advances from 0 to 24 hours in 12 months, or at the rate of 2 hours per month. Assuming a 30-day month, the 2-hour monthly rate (120 minutes) gives us a daily rate of 4 minutes, which is sufficiently accurate for all practical purposes.
Now let us assume that the night on which you face north is August 23. From March 23 to August 23 is five months. At the monthly rate of 2 hours, the right ascension of the mean sun equals 10 hours on that night. If you will look at the table on page 3 of the American Nautical Almanac for 1931, you will see that the sidereal time of 0 hours civil time at Greenwich on August 23 is 22 hours and 1 minute plus. This is equal to the right ascension of the mean sun plus 12 hours. Therefore, the right ascension of the mean sun on that date is 10 hours and 1 minute plus as calculated by the mental method. Subtracting this right ascension of the mean sun (10 hours) from the local sidereal time previously calculated as 21 hours, gives the local mean time as 11 hours.
Of course, there is a small correction to the sidereal interval for longitude which must be considered when making an absolutely accurate calculation, but for the practical purpose of telling the time at night, this correction may be disregarded.
Take another example: Suppose the pointers are in the position of 8 o’clock on the clock face. Then the hour angle, measuring from 12 to the left, equals 4 hours, or 8 hours on a sidereal clock face. The right ascension being 11, hour angle 8, gives a sidereal time of approximately 19 hours.
Let us assume that the night is August 8, 1931. From March 23 to July 23 is 4 months. At the monthly rate of 2 hours this equals 8 hours. From July 23 to August 8 is 16 days. At the daily rate of 4 minutes this equals 64 minutes, or 1 hour and 4 minutes. The right ascension of the mean sun is, therefore, 8 hours plus 1 hour and 4 minutes, or 9 hours and 4 minutes. Subtracting this right ascension of the mean sun, 9 hours and 4 minutes, from the previously determined sidereal time, 19 hours, gives us 9 hours and 56 minutes as the local mean time.
As you cannot estimate an angle so closely, you undoubtedly would estimate the mean time as 10 o’clock.
Try this scheme with all combinations and you will quickly see how simple the operation is.
There is another method of telling time which eliminates some of the calculations necessary in the foregoing method, but which necessitates your memorizing the positions of a number of stars in the heavens.
Again noting that when the star is in the meridian its right ascension is equal to the local sidereal time and that you need only subtract from the local sidereal time the right ascension of the mean sun to obtain your local mean time, you are now invited to examine any star chart or the table of apparent places of stars in the back of the Nautical Almanac.
The two stars, Alpheratz and Algenib, forming the left-hand side of the square of Pegasus, are on the line of zero hours right ascension. A line from Polaris through Beta Cassiopea (Caph) and continued on for 40° of declination will pass through Alpheratz and Algenib. Face north, and roll your eyes upward from Polaris towards your zenith. Your eyes will scribe the line of your celestial meridian. If this line passes through Caph, Alpheratz, and Algenib, your local sidereal time is zero hours. Subtract the right ascension of the mean sun, calculated by the mental method. The result is your local mean time. Note that this line continued through Polaris and extended on the other side passes between the third and fourth stars of the big dipper.
If at the time you are trying this method, the two stars Scheat and Markab forming the right-hand side of the square of Pegasus, are in your meridian, your local sidereal time is 23 hours. Note that this line, if continued through Polaris, passes through the pointers of the dipper.
A line from Polaris to the left of the third star in Cassiopea’s chair and continued on for 25° of declination will pass through Mizar (Beta Andromedae). If these stars are in your meridian, the local sidereal time is 1 hour.
If your meridian line passes through Gamma Andromedae (Almach) and Alpha Aries (Hamel) the local sidereal time will be 2 hours.
A line from Polaris slightly to the right of Capella will give you a local sidereal time of 5 hours.
A line slightly to the left of Beta Aurigae (Menkalinan) will give you a local sidereal time of 6 hours.
If your meridian line passes very slightly to the right of Regulus (Alpha Leo), your local sidereal time will be 10 hours.
The 11-hour line passes through the pointers of the dipper.
The 12-hour line passes midway between the third and fourth stars of the big dipper.
The 13-hour line passes slightly to the left of the fifth star of the dipper.
This system can be continued on throughout the heavens. It is merely a question of how many stars you wish to memorize. By dividing the heavens into quarters, you need only remember four cardinal stars, or, if you want greater refinement, memorize the positions of eight stars, which will give you a check on your local sidereal time every 3 hours.
Don’t approach this problem with the idea that it is difficult. You will be surprised how easily and quickly you will learn the positions of the principal stars after a very little practice.
To complete the problem you need only remember the mental method for calculating the right ascension of the mean sun for any day. Applying this to the local sidereal time gives you your local time. The whole operation can be done in a couple of minutes.
*
Fortunate the soldier to whom destiny assigns the role of assailant. To make war is to attack. Von der Goltz.