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Ans.-By Nautical Almanac or Table of Mean Places of Stars in Norie's Tables.

The stars with less right ascension, and visible, are Pollux, Procyon, Castor, Sirius, a Orionis, Rigel, Capella, and Aldebaran, W. of the meridian.

The stars with greater right ascension, and visible, are a Hydræ, Regulus, a Ursæ Maj., Spica, n Ursæ Maj., and Arcturus, E. of the meridian.

The bright stars Canopus, n Argus, a Crucis, B Centauri, and a Centauri are never visible in lat. 49° N.; the declination of each

being greater than co-lat. 41°. N.B.-It is often convenient to know what bright stars will pass the meridian

between any given hours. RULE.—To each of the given times add the sidereal time (Nautical Almanac, p. II.) for Greenwich noon of the day; thus you have two approximate sidereal times at place corresponding to the given hours. Now, turn to Nautical Almanac for “Mean Places of Stars,” or Table of Mean Places of Starsin Norie's Tables and allstars whose right ascension lies between the two sidereal times at place will pass the meridian in the given interval.

Example.What bright stars will pass the meridian between 6h. p.m. and ith. p.m. mean time, on November 7th ?

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21

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6 33

Ship time

II O O (N.A.) Sid. T. 15 6 33

15 6 33 Approximate R.A. mer.

0 33 Ans.-Turn to Nautical Almanac or Table of Mean Places of Stars in Norie's Tables and find right ascension 21h.; all stars with less right ascension will have already passed the meridian; the stars with greater right ascension are a Gruis, Fomalhaut, and Markab; but the stars with less right ascension, at the beginning of the Catalogue, will follow Markab; these are the stars in Andromeda, Pegasus, Cassiopeia, Eridanus, and Aries; but which of all these will be seen depends on the latitude of the place, and the declination of the star. XI.—TO FIND THE HOUR ANGLE (H. A.) OF A STAR AT A GIVEN TIME

AND PLACE RULE.—To the mean time at place add the mean sun's right ascension, and from the sum (which is the sidereal time or right ascension of the meridian) subtract the star's right ascension (borrowing 24 hours if necessary); the remainder will be the westerly hour angle or meridian distance; if the W. hour angle exceeds 12h. subtract it from 24h. for the less or E. hour angle, if required.

R. A. M. Formula.- H. A.

*'s R. A. M. T.S. + M. o's R. A. If the apparent time at place is given, find the sidereal time or right ascension of the meridian by VI., p. 265 and subtract the star's right ascension.

R. A. M. Formula.- H. A.

*'s R. A. A. T. S. + App. O's R. A. N.B.—“Star ” may here denote any heavenly body, except the sun.

Example.— January 18th, at 8h. 4om. 245. p.m. mean time at place ; the Greenwich date being Jan. 18d. Ioh. 5om. 38s. Find the hour angle of Sirius.

N.B.—The elements of the problem are brought forward from p. 265.

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The circle is the equinoctial, P Q the meridian of the place, A the first point of Aries, M the mean sun, S the star, Q AMB is the W. H. A., and Q B the E. H. A., Q M the M. T. S., A QM the M. S. R. A., A Q the Sid. T., or R. A. mer., and A B the *'s R. A.

XII.—TO FIND THE HOUR ANGLE (H.A.) OF A STAR, THROUGH THE HOUR

ANGLE OF ANOTHER STAR WHICH HAS BEEN PREVIOUSLY FOUND

RULE.—To the given star's westerly hour angle add its right ascension ; the sum will be the sidereal time at place, from which subtract the right ascension of the star for which the hour angle is required; the remainder will be the westerly hour angle which, if more than 12h., can be subtracted from 24h. for the easterly hour angle.

Formula

Sid. T. (given) *'s W. H. A. + R. A.
W. H. A. (required *) Sid. T. - R. A.

N.B.-All the necessary data must be corrected for the Greenwich date, astronomical time.

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The circle is the equinoctial, P Q the meridian of the place, A the first point of Aries, S the star, and M the moon. & B is the E. H. A., and OAB the W. H. A. of the star, Q C the W. H. A. of the moon, ACQB the R. A. of the star, and AC the R. A. of the moon, ACQ the Sid. T. or R. A. meridian.

XIII.—THE HOUR ANGLE (H. A.) OF A STAR HAVING BEEN FOUND AT A

GIVEN PLACE AND DATE, TO FIND THE MEAN TIME AT PLACE

RULE.—To the star's westerly hour angle add the star's right ascension ; the sum will be the sidereal time at place or right ascension of the meridian (R. A. M.), from which subtract the mean sun's right ascension; and the remainder will be the mean time at place.

N.B.—If the star be east of the meridian, subtract its hour angle from 24h. for the westerly hour angle; and then proceed as above.

R. A. M.

Formula

M. o's R. A.

- M. T. S.

*'s W. H. A. + R. A.

Example.- January 4th a.m. at ship; when the mean time at Greenwich was Jan. 4d. 3h. 22m. 245. ; the hour angle of Regulus W. of the meridian was 2h. 14m. 49s. Find the mean time at place.

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Sid. T. (N.A. p. II.) 18 56 10.9 Regulus H.A.
Accel, for Gr. date

+ 332

R.A.

31 Mean sun's R.A. 18 56 44'I

R.A.M. or Sid. T.

17 20 Mean sun's R.A. 18 56 44 M.T. at ship, Jan. 3rd 17 20 36

12

Example.-Or suppose a star's hour angle east of the meridian, then

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SPECIFIC GRAVITY AND MISCELLANEOUS

CALCULATIONS

The specific gravity of a body may be defined as the ratio which exists between the weight, or density, of any substance when compared with another substance taken as a standard of comparison. For solids, distilled water is taken as the standard to which all other solids are referred when finding their specific gravity. The specific gravity of distilled water is taken as I, or, as it is called, unity.

Whilst specific gravity does not tell you the absolute weight of an element, it conveys to you a clear idea of its weight compared with distilled water, and when the volume of the element is known its weight can readily be found, as will be seen in the following calculations

A cubic foot of distilled water weighs 1,000 oz., but it would be incorrect to say that the specific gravity of distilled water is 1,000 oz

It is necessary to know the weight of a cubic foot of distilled water in order to find the weight of any other body when its specific gravity an1 volume are given.

A cubic foot of the Atlantic weighs about 1,025 oz., but its specific gravity is 1.025. or twenty-five thousandths heavier than distilled water If you were to evaporate a cubic foot of the Atlantic water you would find about 25 oz. of solid matter, principally salt, lying at the bottom of the receptacle in which it was boiled.

We shall now show how to find the specific gravity of a solid heavier than water. First weigh the element, whose specific gravity is required, in water ; then weigh it in air ; divide its weight in air by the loss of weight when weighed in water; the result is its specific gravity.

Example.-A piece of iron when weighed in water was found to be 24 lbs., and when weighed in air was 28 lbs. Required its specific gravity.

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To find the Specific Gravity of a Log of Square Timber of Uniform Section

throughout Put the log into the water and note the depth to which it is immersed ; divide the depth, to which it is immersed by the sectional depth of the log, and the result is its specific gravity.

Example.--A log of wood 70 feet long and 2 feet square was floating at a draft of 18 inches. Required its specific gravity.

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Answer.-Specific gravity 75.

If this log were constrained to stand in a vertical position it would float at a draft of 70 x = 52 feet.

Let it now be required to find the weight of the log and the pressure per square inch on the immersed end.

To find the Weight of the Log Multiply the number of cubic feet in the log by its specific gravity, and then by 62.5 lbs., and the result is the weight required.

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To find the Pressure per Square Inch on Lower End Divide the weight of the log by the area of end section, in inches, and the result is the pressure per square inch.

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If the spar were floating in a horizontal position the pressure per square inch would be 65 lb.

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