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TO FIND THE TIME OF HIGH WATER BY THE

MOON'S MERIDIAN PASSAGE AND THE
ESTABLISHMENT OF THE PORT.*

Given the date, the longitude of the place, and the time of high water at the full and change of the Moon, to find the a. M. and P. M. times of high water at the given place on the given date.

Take out of the Nautical Almanac the following quantities, viz., from Page IV. for each month, the time of the Moon's meridian passage at Greenwich for the given day; the difference between this time and that of the time opposite the day before the given date, if the longitude is E. ; but that of the time opposite the day following the given date, if the longitude is W.; from Page III. the Moon's semi-diameter; and from Page II. the nearest minute of the equation of time, noting whether it is to be added to, or subtracted from, mean time.

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RULE. (1.) First Correction. Enter Table XVI. of Norie, with the difference of the times of meridian passage" and the longitude. The quantity there found is + if the longitude is W., and if E. The result is the mean time at ship.

(2) Second Correction. To the mean time at ship apply the equation of time to get the apparent

time at ship, with which and the Moon's semi-diameter enter Table XVI.* The quantity there found is to be applied to the mean time at ship, according to the sign prefixed.

(3) Add the establishment of the place, that is, the time of high water at full and change.

The result is the P. M. tide. From this take { the difference between the times of passage of the given and the preceding day, and the remainder is the A. M. tide.

EXAMPLE 1. 1870, January 7th. Find the times of high water, A. M. and P. M., at Cape Recife, Algoa Bay. Longitude (Norie, Table LVI.) 26° E.; high water, full and change, (Norie, Table LVII.) 3h. 20m.

* This Formula is used at the Examinations at the Scotch Ports.

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EXAMPLE 2. 1870, February 9th. Find the times of high water, A. M. and P. M., at Bahia. Longitude 38° W. Establishment 4h. 15m. )'s meridian passage

6 45 Diff. time pass. between 9th & 10th=48m.

1st cor. + 5 Longitude

38° W. Mean time at ship.

6 50 Mean time.... 6 50 Eq. of time

14 Apparent time.. 6 36 Table XVI.* 2nd cor. —

0 45 } )'s semi-diam... 15 0

6 05 Establishment

+ 4 15

h. m.

10 20 P.M.

diff. of passage, 9th & 8th

22

9 58 A.M.

If after the establishment is added, the result is above 12 hours, the tide thus found will be the A. M. tide of the next day.

To find the P. M. tide of the given day. Take i the difference of times of meridian passage between the given day and the day before, from the mean time at ship, and proceed as before.

EXAMPLE 3. 1870, March 11th. Find the times of high water A. M. and P. M. at Busheer. Longitude 51° E. High water at full and change 7h. 30m.

h. m.

7

6

's meridian passage
Diff. of passage between 11th & 10th=352m; } 1st cor.—
Longitude.

7

6 59

Mean time at ship .

h. m. Mean time 6 59

10

Eq. time

App. time 6 49 & )'s semi-diam. 15' 20",

2nd cor.

39

6 20 Establishment..

+ 7 30

13 50

This being above 12 hours, we must take ţ difference of passage 26m. from the mean time at ship, and proceed as before, thus :

h. m.

Mean time at ship .

difference of passage

6 59

26

h. m.

6 33

Mean time 6 33
Eq. time -

10

App. time 6 23 & )'s semi-diam. 15' 20", 2nd cor.

52

Establishment

5 41
7 30

13 11
12 00

difference of passage .

1 11 P.M.

26 0 45 A.M.

If, after adding the establishment, the result is above 24 hours, the tide thus found will be the P.m. tide of the next day.

To find the P. M. tide of the given day. Subtract the whole difference of meridian passage between the given and the preceding day, and proceed as in the last Example.

EXAMPLE 4. 1870, September 20th. Find the times of high water, A. M. and P. M. at Shoreham.

h. m.

.20 38

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1st. cor. + 0 0

20 38 Note,- The Establishment of Shoreham is 11h.15m.

and the meridian passage is 20h. 38m.
therefore the sum is more than 24 hours;

so instead of working the sum over twice,
Subtract at once the whole diff. of passage, viz..... 0 54

19 44

m.

Mean time at ship

h. Mean time....19 44 Eq. of time + 0 7

Apparent time 19 51 and )'s S.D. 15' 28"=2nd cor. — - 0 3

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EXAMPLES FOR EXERCISE. 1. 1870, April 17th. Find the A. M. and P. M. tides at

Nagasaki. 2. 1870, May 19th. Find the A. M. and P. M. tides at

Rio Janeiro. 3. 1870, June 2nd. Find the A. M. and P. M. tides at

Rangoon. 4. 1870, July 7th. Find the A. M. and P. M, tides at

Goa.

SEXTANTS AND QUADRANTS.

Sextants are divided on the Arch either into 10, 15, or 20 minutes, which may be known by the number of divisions in the degree, or by the number of minutes or miles marked on the vernier at the bottom. (In reading the Sextant, either on or off the Arch, care should be taken that the reading commences from the position of the arrow or nonius.) To find the Index Error, place the arrow or nonius on the vernier to 0 on the Arch, and then if the true and reflected Suns exactly cover each other, there is no Index Error; if they do not, put the vernier so that it reads about 40 minutes on the Arch ; then make the edges of the true and reflected Suns touch each other; read it and note it down; then put 40 minutes off the arc, and make their other edges touch; read it and note it down as before; the half-difference between these two readings will be the Index Error; to be subtracted if on is greatest, and added if off is greatest.

To prove that your observations are right, add the two readings together and divide by 4, which will give the Sun's semidiameter in the Nautical Almanac for the day; if not, you must repeat the observations.

ADJUSTMENTS.

18t. To set the plane of the Index-glass perpendicular to the plane of the Instrument.

Place the arrow or nonius near to the middle of the Arch, and holding the eye close to the Index-glass, look aslant into the glass, and observe the part of the Arch towards the right, and also its reflection towards the left. If they are in one line, the glass is perpendicular; if not, it must be adjusted by the screws at the back of the Index-glass.

2nd. To set the plane of the Horizon-glass perpendicular to the plane of the Instrument.

Bring the arrow or nonius to 0 on the Arch, and, holding the Quadrant horizontally, observe whether the real and reflected horizons coincide; if so, the glass is perpendicular; if not, the horizons must be made to coincide by means of the scrows at the back of the Horizon-glass.

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