Ex. 1. 1875, Aug. 12th: find the times of high water at Caracus River, Ecuador, longitude 67° W. Ex. 2. 1875, September 22nd: find the times of high water at Auckland, New Zealand, longitude 175° E. Ex. 3. 1875, May 15th: find the times of high water at Point de Galle, Ceylon, longitude 80° E. Ex. 4. 1875, February 22nd: find the times of high water at San Francisco Bay, longitude 122° W. Ex. 5. 1875, September 23rd: find the times of high water at Malacca Fort, longitude 102° 15' E. Ex. 6. 1875, July 22nd: find times of high water at Port Jackson, North Head, longitude 151° 16' E. Ex. 7. 1875, July 27th: find times of high water at St. Julian, longitude 67° 38′ W. GREENWICH DATE BY CHRONOMETER. 290. 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. 291. 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. 292. 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 3o 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 35 a day gaining. Hence the amount of the daily rate (supposed uniform) is found by dividing the change of the error by the number of days in the interval between observations. 293. 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; if 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. 1. A chronometer was 25m 20s slow for mean time at Greenwich on Nov. 20th, and on November 30th, was 24m 45 slow on Greenwich mean time: required the daily rate. November 20th, chronometer slow 25m 20% November 30th, slow 24 45 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 28m 5s on mean time at Greenwich, Feb. 27th, 1876, and on March 11th was slow 29m 36 on mean time at Greenwich: find daily rate. 1876, February 27th, chronometer slow 28m 58 slow 29 36 Feb. 29 (leap year). 27 The error of chronometer, which is slow, is increasing, it is therefore losing 70. Ex. 3. A chronometer was fast 1m 23s on mean time at Greenwich, June 2nd, and on July 1st, was fast 1m 375 on mean time at Greenwich: find daily rate. The error of chronometer is fast and increasing, hence the daily rate is os 5 gaining. Ex. 4. A chronometer was fast 1m 51 on mean time at Greenwich, May 1st, and on May 15th was 418 fast on mean time at Greenwich: find daily rate. May 1st, chronometer fast Im 518 fast o 41 In this example the chronometer is fast and the error decreasing, the rate therefore is losing. 294. When the error is found to have changed from fast to slow, or from slow to fast, the rate is the sum of the errors divided by the number of days elapsed. EXAMPLES. Ex. 1. July 28th, at 3h P.M., the chronometer was om 60 fast, and on August 4th at same time, it was om 178°1 slow: required the daily rate. In this example the error has changed from fast to slow, the chronometer therefore is losing. Ex. 2. A chronometer was slow 1m 4 on mean time at Greenwich, March 1st, and on March 23rd, was om 1986 fast on mean time at Greenwich: required the rate of chronometer. March 1st, chronometer slow Im 48 fast o 19.6 March 23 In this example the error of chronometer has changed from slow to fast, it is evident, therefore, that the chronometer is gaining. EXAMPLES FOR PRACTICE. Ex. 1. A chronometer was slow 2m 14 on mean time at Greenwich, March 3rd, and on March 25th was slow 50°4 on mean time at Greenwich: find the daily rate. Ex. 2. A chronometer was slow 5m 19 on mean time at Greenwich, January 30th, and on February 17th was slow 6m 13s on mean time at Greenwich: find the daily rate. Ex. 3. A chronometer was 2m 25 fast on mean time at Greenwich, January 24th, and on February 10th was fast 3m 185 for mean time at Greenwich: find the daily rate. Ex. 4. A chronometer was slow 493 on mean time at Greenwich, March 17th, and on April 1st was 1m 587 fast for mean time at Greenwich: find daily rate of chronometer. Ex. 5. A chronometer was fast 1m 4 on mean time at Greenwich, January 10th, and on February 10th was 1m 62 slow for mean time at Greenwich: required the daily rate. Ex. 6. A chronometer was fast im 29" on mean time at Greenwich, July 1st, and on July 23rd was fast 1. 59 on mean time at Greenwich: find daily rate. Ex. 7. A chronometer was fast 48 on mean time at Greenwich, February 28th, and on March 15th was slow 48* on mean time at Greenwich: find daily rate. Ex. 8. A chronometer was slow 20° on mean time at Greenwich, September 1st, and on September 15th was fust 1m 18" on mean time at Greenwich: find daily rate. Explanation. Multiply the decimal 26 by the number of whole days, namely, 32. Next consider that 12 hours is the of 1 day, and 4 hours is the of 12 hours. 12 and 4 make up the whole number of hours, namely, 16. Divide 26 by 2 and the quotient 13 by 3 (see example). Add the products and quotients together; its sum is 849 = 1m 249; and observe that the decimal is rejected, and since it is above 5, therefore 1 is added to the seconds. 30m 6,0 11,6'4 Im56*4 or, 1m56 Explanation.Multiply by 12; then 6 hours is t of 1 day, and 3 hours of 6 hours, and 30 minutes is 1-6th of 3 hours. Divide the daily rate, 94, by 4, which will give 2.3, the proportional part of rate in 6 hours (of a day); next, divide 23 by 2, which gives the rate for 3 hours ( of 6 hours); again, divide 11 by 6, which gives the change for 30 minutes (1-6th of 3 hours); then, add the product and several quotients together, the result is the accumulated rate for the interval. 296. The accumulated rate may also be found by decimals; thus: RULE XCIII. 1o. Affix two cyphers to the hours, and divide the result by 6 and the quotient by 4, (i.e., divide by 24); the last quotient is the hours expressed as decimals of a day. (See Rule XVIII, page 44). 2o. Multiply the days and decimals of a day by the seconds and decimals of a second (if any) for the daily rate; the product is the accumulated rate. 297. Before going to sea, the error of the chronometer on Greenwich mean time, and its daily rate, are supposed to have been accurately determined, either at an observatory by means of daily comparison with an astronomical clock, or by observations taken by a sextant at a place whose longitude is known. 298. When the error of a chronometer on Greenwich mean time, and also its daily rate, are known, we may determine Greenwich mean time at some other instant, as when an observation is taken, by the following: RULE XCIV. 1. To the time by chronometer apply the original error, adding it if the chronometer was slow, rejecting 24" if greater than 24", and putting the day one forward; but if the chronometer is fast, subtract original error, increasing time shown by chronometer by 24", if necessary, and putting the day one back. Find the number of days and parts of a day, to the nearest hour, elapsed since the original error was ascertained. 3o. Multiply the daily rate of chronometer by the elapsed time, and add thereto the proportionate part for the fraction of a day, found by proportion or otherwise; the result is the accumulated rate in the interval. 4°. To the result found by 1°, add the accumulated rate, if chronometer is losing ; but subtract if gaining; the result will be mean time at Greenwich, at the instant of observation. Ex. I. EXAMPLES. 1876, Jan. 30th, P.M. at ship, time by a chronometer, Jan. 29d 15h 47m 488-3, which was 9m 19′′6 slow for Greenwich mean time Dec. 1st, 1875, and on January 1st, 1876, was 10m 2487 slow on Greenwich mean time. The chronometer being slow and the error increasing, the rate must be marked losing. |