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The circle is the Equinoctial, P the Pole, P Q the Meridian of Greenwich, A the first point of Aries, and M the Mean Sun. AMO is the G. Sid. T., Q M the G. M. T., and A M the M. S. R. A. The right ascension is always reckoned in the direction of the arrow from A.
V.-TO CONVERT MEAN TIME AT A GIVEN PLACE INTO SIDEREAL TIME AT
The Greenwich mean time must be known, or found through the mean time at place and the longitude.
RULE. Find the mean sun's right ascension as in the previous case, or by p. 247. To the mean time at place add the mean sun's right ascension, and the sum will be the sidereal time at place, or right ascension of the meridian.
Formula- Sid. T. obs. or R. A. Mer. M. T. S. + M. O's R. A.
Example. - January 18th, at Sh. 4om. 245. p.m. mean time at place of observation, the Greenwich date being January 18d. Ioh. 50m. 38s. Find the sidereal time at place of observation.
Sid. T. (N.A. p. II.) 19 51 22071
M.T. at place 8 40 24
19 53 9:58 5om.
Sid. T. at place 38s.
4 33 33.58 Mean sun's R.A. 1953
9:58 NOTE.—The diagram is similar to that in IV., P Q being the meridian of the place.
VI.-TO CONVERT APPARENT TIME AT A GIVEN PLACE INTO SIDEREAL
TIME AT THAT PLACE
RULE.—Find the Greenwich date (apparent time); from the Nautical Almanac, p. I., take out the sun's right ascension for the Greenwich date, and correct it by the “ Var. in 1 hour” for the Greenwich Apparent time.
Then, to the apparent time at place add the sun's right ascension (corrected); the result will be the right ascension of the meridian, or sidereal time of observation at place.
Formula- R. A. Mer. or Sid. T. obs. = A. T. S. + App. o's R. A.
Example.—December 3d., at 7h. 53m. 46s. p.m. apparent time at place, in long. 30° W. Find the right ascension of the meridian, or sidereal time at place.
The circle is the Equinoctial, P Q the Meridian of the place, A the first point of Aries, and the sun. Q A is the R.A. mer., Q B the A.T.S, and A Q B the sun's R.A.
If preferred, the apparent time at place might be converted into mean time at place by II., and then into sidereal time at place by V.
VII.-TO CONVERT SIDEREAL TIME AT PLACE (OR THE RIGHT ASCENSION
OF THE MERIDIAN) INTO MEAN TIME AT PLACE
You will require a Greenwich date by which to get the mean sun's right ascension, for which see Case IV. p. 264, or p. 247. Then,
RULE.—From the sidereal time at place (i.e. the right ascension of the meridian), subtract the mean sun's right ascension, borrowing 24h. if required, before which write the date at place.
N.B.— This case is the reverse of the preceding case V.
With regard to all problems in which the right ascension of any of the heavenly bodies is introduced, attend to the following remark:
Caution.-Remember that whenever we speak of one right ascension as being subtracted from another, for the purpose of obtaining a third right
ascension, it is always tacitly supposed that 24 hours is added (if necessary) : to the one from which the subtraction is to be made ; and whenever one right ascension is to be added to another, to get a third right ascension, 24 hours is always suppressed from the sum if it exceeds that quantity. There is no displacement of a celestial object by increasing its right ascension by 24 hours, or by 360°, if the right ascension be expressed in angular measure. Also
Hour Angle of any Heavenly Body.—Remember that whenever time, or an object's hour angle, is considered astronomically, it is reckoned from oh. om. os. when on the meridian, thence westerly through 24 hours, to the meridian again. Hence when an object is east of the meridian we can convert the easterly into a westerly hour angle, thus
Westerly H. A. = 24h. Easterly H. A. i.e., by subtracting the easterly hour angle from 24 hours, we get the westerly hour angle.
VIII.---MERIDIAN PASSAGE OF THE SUN
The apparent time of the sun's meridian passage is oh. om. os., except below the pole, when it is 12h. om. os.
For the mean time of the meridian passage :- Take the equation of time from Nautical Almanac, p. I., and reduce it for the longitude as a Greenwich date (see p. 239); then apply equation of time according to precept on Nautical Almanac, p. I.
The approximate apparent time of a star's meridian passage is given in Table of Star's Mer. Passage, and for the correction see “Explanation."
For the mean time of a star's meridian passage.
RULE.-Enter Table of Star's Mer. Pass, and note whether the time there given is more or less than 12 hours. From the star's right ascension taken from the “ Apparent Places of the Stars” in the Nautical Almanac (increased by 24 hours if necessary) subtract the sidereal time (Nautical Almanac, p. II.) at mean noon of the astronomical day (this is the same as the civil day if the time in Table of Star's Mer. Pass. is less than 12h. or p.m., but is one less than the civil day if the time in the Table is more than 12h. or a.m.); the remainder is the approximate mean time of transit, civil time. From the
Retardation Table take out the retardation corresponding to this remainder and subtract it from the approximate mean time of transit. Also take from the same Table the retardation corresponding to the longitude in time, and add for E. longitude, but subtract for W. longitude. The result is the mean time of the star's meridian passage.
M. T. S.
*'s R.A. — M. O's R. A.
Example.-- January with; long. 45° W. Find the mean time of the meridian passage of Aldebaran.
The time in Table of Star's Mer. Pass. is less than 12h. Astronomical date same as Civil date.
The circle is the equinoctial, * the star on the meridian PQ, A the first point of Aries, and M the mean sun. QM is the M. T. S., A Q the Star's R. A., and also the R. A. Mer., and A QV the M. o's R. A.
Example.— January 21st, civil time; long. 120° E. Find the mean time of the meridian passage of Regulus.
The time in Table of Star's Mer. Pass. is more than 12h. Astronomical date goes back one day.
The explanation of the diagram is the same as in the previous example.
For the apparent time of a star's meridian passage.
RULE.—Proceed as in finding the mean time of meridian passage, using apparent sun's right ascension (Nautical Almanac, p. I.) instead of sidereal time, and the two corrections will be found by using " Var. in 1 hour” of sun's right ascension. Formula.
A. T. S. = *'s R: A. — App. o's R. A. Example.— January with ; long. 45° W. Find the apparent time of meridian passage of Aldebaran.
The circle is the equinoctial, * the star on the meridian PQ, A the first point of Aries, and S the sun. Q B is the A. T. S., A Q the star's R. A., and also the R. A. of the meridian, and A & B the sun's R. A.
X.-GIVEN THE MEAN TIME AT PLACE TO FIND WHAT PRINCIPAL STARS
ARE EAST AND WEST OF THE MERIDIAN RULE. Find the sidereal time at place, or right ascension of the meridian by Case V., p. 265; then, since the stars pass the meridian in the order of their right ascension, and they are catalogued in the Nautical Almanac in that order, turn to Nautical Almanac "Mean Places of Stars," or to Table of Mean Places of Stars in Norie's Tables, and find the right ascension corresponding nearest to the sidereal time at place. All the stars having less right ascension than the sidereal time are west of the meridian; all with greater right ascension are east of the meridian.
For the possible visibility of the stars, remember that if declination of star and latitude of place are of different names, one N. and the other S., and the declination be greater than the co-latitude, such stars are never above the horizon of the observer, or if declination and latitude exceed 90° the star will be below the horizon.
Also, with declination and latitude of the same name, when the polar distance is less than the latitude the star is circumpolar.
Example.— January 19th, at ih. 2om. a.m. mean time in lat. 49° N., the Greenwich date being Jan. 18d. 16h. 7m.; what principal stars are visible E. and W. of the meridian ?