Ex. 16. 1880, October 28th: find A.M. and P.M. times of high water at Ballycottin. The Standard Port for Reference is Waterford, and the Constant oh 20m. In this example there is a blank in the morning column at Standard Port, we therefore take the afternoon tide, and the Constant being of the same value precisely as the time of high water, the result is a noon tide at Ballycottin. There is only one high water on the 28th October, and this occurs at Noon. EXAMPLES FOR PRACTICE. In each of the following examples it is required to find the time of high water A.M. and P.M. 315. In pages 151 to 232 of Admiralty Tide Tables for 1880, are given the times of high water at full and change of a great number of ports, by which we are enabled to calculate approximately the time of high water on each day. The Constant is found by taking Brest as the Standard Port, at which place the time of high water, full and change, is 3" 47". The difference between the full and change at the given port and Brest will be the Constant to be employed (as in the preceding Rules), except there be a great difference of longitude, in which case the correction for the moon's meridian passage must be employed, since for the greatest longitude this correction may amount to half an hour. Should the longitude, however, not exceed 5o, it may be neglected, as doing so will scarcely make more than a difference of one minute. It must also be observed that the longitude of Brest is 44° W. of Greenwich, and in strictness, therefore, in determining this correction 4° should be subtracted, if the longitude of the place be East, or added if it be West. The correction is found in Table XVI, NORIE, or TABLE XXVIII, RAPER. Hence : 316. To find the time of high water at Foreign Ports whose Constants are not given in the Tide Tables. RULE LXXXIX. 1o. To find the Constant.—In the Alphabetical List of Ports at the end of the Admiralty Tide Tables for 188c, page 189—232, find the time of high water Full and Change, at Brest, and also that corresponding to the given port; subtract the less from the greater of these two times, and the remainder will be the CONSTANT, additive if the full and change (F. & C.) at the given port is greater than that of Brest, but subtractive if less. MM 2o. Take out the times of high water at Brest for the given day, and apply the Constant as directed in the preceding Rule LXXXVIII, pages 259-260; the result is the time of high water (nearly) at the given place. 3°. Take out the longitude of Brest and also of the given place; take the sum if the names are alike, but take the difference if the names are unlike. 4°. Take out (from the column to the left of those containing the times of high water at Brest) the moon's transit for the proposed day and the following one, if the long. is West; but for the given day and the preceding one if the long. is East. Their difference, in either case, is the Daily Variation, or Retardation. 5°. Take from Table 28, RAPER, or Table 16, NORIE, the correction corresponding to the daily variation and longitude. 6°. Apply this correction by addition in West longitude, but by subtraction in East longitude, to the approximate times of high water already found, the result is the times of high water on the proposed day at the given place. EXAMPLES. Ex. 1. 1880, March 19th: required the time of high water at Victoria River, Point (N.W. coast of Australia), longitude 130° E. Time of H.W. full and change, Victoria River Brest 7h 15m (p. 231) Constant +3 28 Turtle D's transit, March 19th, 6h 43m P.M. 18th, 5 53 Long. Victoria River 130° E. 4 W. 126 Under 50m and against 126° longitude, in RAPER, Table 28, or NORIE, 16, we find 17m to be subtracted, because the longitude is E. Ex. 2. 1882, October 9th: find the times of high water at Sandy Hook, long. 74° W. Time of H.W. full and change, Sandy Hook 7h 29m (p. 225) Constant +3 42 74° W. + 4 W. 78 Under 60m and against 78° in longitude, in RAPER, Table 28, or NORIE 16, we find 13m to be added, because the longitude is West. Ex. 3. 1882, May 25th: required the times of high water at Nelson, New Zealand, longitude 173° E. Under 65 and opposite 169° (173° — 4°) in Table 16, NORIE, or 28, RAPER, stands the correction 30o to be subtracted. Ex. 4. 1882, August 8th: find the times of high water at Cape Virgin, Straits of Magellan, longitude 68° W. On the dates given, find the times of high water at the undermentioned places. 1. 1882, January 12th: Rio Janeiro, longitude 43° 9′ W. " 2. 3. 4. 5. 6. 7. 8. 9. IO. II. 12. August 1st: Caracus River, Ecuador, longitude 67° W. ,, July 12th: Port Jackson, North Head, longitude 151° 16′ E. February 4th: Awatska Bay, longitude 158° 47' E. GREENWICH DATE BY CHRONOMETER. 317. The Error of Chronometer on Mean Time at any place is the difference between the time indicated by the chronometer and the mean time at that place. The error of chronometer on Mean Time at Greenwich is the difference between the time indicated by the chronometer and the mean time at Greenwich. The error is said to be fast or slow as the chronometer is in advance of or behind the mean time at Greenwich. 318. Rate of Chronometer is the daily change in its error, or the interval it shows more or less than twenty-four hours in a mean solar day. If the instrument is going too fast, the rate is called gaining; if too slow, losing. 319. To find the Rate.-The rate of a chronometer is determined by comparing its errors for mean time, as found by observation at a given place, on different days. Thus, if by observation a chronometer is found 20° slow, and at the end of ten days is found to be 50' slow for mean time at the same place, it has evidently lost 30° in ten days, whence its mean daily rate is 3' losing. If on a given day, chronometer be 12* fast, and at the end of thirteen days 57' fast for mean time at any place, it must have gained 45" in thirteen days, or its rate is about 3o5 a day gaining. Hence the amount of the daily rate (supposed uniform) is found by the following RULE XC. Write one error under the other, then Both errors fast, or both slow, take their difference. One error fast and the other slow, take their sum. Bring the sum or remainder into seconds, and divide by the number of days between the dates of the two errors; the result will be the daily rate in seconds and tenths, or (perhaps) tenths only. 320. To name the Rate.-When the chronometer is fast either on Greenwich mean time, or on the time at place, if the error is increasing, the rate is gaining; ff decreasing, the rate is losing. When the chronometer is slow, if the error is increasing, it is losing; if decreasing, it is gaining. When the chronometer is fast, and the error changes to slow, the rate is losing; if the error changes from slow to fast, the rate is gaining. EXAMPLES. Ex I. A chronometer was 25m 20 slow for mean time at Greenwich on Nov. 20th, and on November 30th was 24TM 45o slow on Greenwich mean time: required the daily rate. In this example the chronometer is slow on November 20th, and the error is decreasing, therefore the chronometer is gaining. Ex. 2. A chronometer was slow 18m 5a on mean time at Greenwich, Feb. 27th, 1880, and on March 11th was slow 29m 360 on mean time at Greenwich: find daily rate. 1880, February 27th, chronometer slow 28m 58 Change of error in 13 days slow 29 36 99 1 31 13)91(70 Feb. 29 (leap year) 27 2 March II The error of chronometer, which is slow, is increasing, it is therefore losing 70. Ex. 3. A chronometer was fast 1m 23" on mean time at Greenwich, June 2nd, and on July 1st was fast im 375 on mean time at Greenwich: find daily rate. The error of chronometer is fast and increasing, hence the daily rate is 0o5 gaining. Ex. 4. A chronometer was fast 1m 51° on mean time at Greenwich, May 1st, and on May 15th was 41o fast on mean time at Greenwich: find daily rate. In this example the chronometer is fast and the error decreasing, the rate therefore is losing. Ex. 5. July 28th, at 3h P.M., the chronometer was om 60 fast, and on August 4th at same time, it was om 17′′ 1 slow; required the daily rate. In this example the error has changed from fast to slow, the chronometer therefore is losing. Ex. 6. A chronometer was slow 1m 4o on mean time at Greenwich, March 1st, and on March 23rd, was om 19o6 fast on mean time at Greenwich: required the rate of chronometer. March 1st, chronometer slow Im 4o March 23rd, Change of error in 22 days March 23 fast o 19.6 23 I 22d Rate 38 gains. In this example the error of chronometer has changed from slow to fast, it is evident, therefore, that the chronometer is gaining. |