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RULES FOR FINDING THE ERROR AND RATE OF CHRONOMETERS, BY SINGLE ALTITUDES AND BY EQUAL ALTITUDES.
To find the error and rate of chronometers.
THERE are two methods of determining the error of a chronometer on mean time, the one by a single altitude of a heavenly body observed at some distance from the meridian, the other by means of equal altitudes of a heavenly body observed on both sides of the meridian.
The mean daily rate of a chronometer is found by dividing the increase or decrease in its error by the number of days elapsed between the times when the observations were taken to determine its error; thus, suppose on April 27, at 9h 30m A.M., the error of a chronometer was found to be fast 10m 10s.5 on Greenwich mean time, and that on April 30th about the same hour its error was found to be 10m 40s.5 fast: then it appears that in the three days elapsed between the observations the chronometer has gained 30s, hence its mean daily rate is 10° gaining.
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 comparisons with an astronomical clock, or by observations taken with a sextant at a place whose longitude is known.
When the error and rate of a chronometer are given we may determine what its error will be on some future day, provided the rate of the chronometer continues uniform in the interval, by the following rule.
Given, the error of a chronometer on Greenwich mean time, and also its daily rate, to find Greenwich mean time at some other instant, as when an observation is taken, &c. 1. Get a Greenwich date.
2. Find the number of days and part of a day that have elapsed from the time when the error and rate were determined by the hour of the Greenwich date.
3. Multiply the rate of the chronometer by the number of days elapsed, and add thereto the proportionate part for the fraction of a day, found by proportion or otherwise. The result is the accumulated error in the interval.
4. If the chronometer is gaining, subtract the accumulated error from the time shown by the chronometer; if losing, add.
5. To the result apply the original error of chronometer, adding if slow, subtracting if fast (increasing the time shown by chronometer by 24h if necessary, and putting the day one back). The result, (rejecting 24h if greater than 24h and putting the day one forward), will be mean time at Greenwich at the instant of the observation.
NOTE. If this time differs from the Greenwich date by 12 hours nearly; in that case 12 hours must be added to the Greenwich time, determined as above, to get the astronomical Greenwich mean time.
1. June 13, 1851, at 10h 52m P.M., mean time nearly, in long. 60° W., an observation was taken when a chronometer showed 2h 50m 42s. On June 1, its error was known to be 3m 10.2 fast on Greenwich mean time, and its mean daily rate was 3.5 gaining, required mean time at Greenwich when the observation was taken.
2. Aug. 10, 1853, at 3h 42m A.M., mean time nearly in long. 100° 30′ W., an observation was taken when a chronometer showed 10h 30m 458.5.
On Aug. 1, its error was known to be 12m 10-5 slow on Greenwich mean time and its rate 11-2 gaining, required mean time at Greenwich when the observation was taken.
If the Greenwich time thus determined differs considerably from the Greenwich date used, the work should be repeated, using for the Greenwich date the approximate Greenwich time first found.
(162.) Nov. 20, 1851, at 6h 42m P.M., mean time nearly, in long. 32° 0′ E., an observation was taken when a chronometer showed 4h 30m 6s.
On Oct. 9, its error was known to be 5m 524 slow on Greenwich mean time, and its rate 28-7 losing: required mean time at Greenwich when the observation was taken.
Ans., 4 36m 528.3. (163.) Dec. 31, 1851, at 10h 10m A.M. mean time nearly, in long. 150° E., an observation was taken when a chronometer showed 0h 0m 22.3.
On Nov. 20, its error was known to be 3m 52.4 slow on Greenwich mean time, and its rate 27 losing: required mean time at Greenwich when the observation was taken. Ans., 12h 6m 4s.0.
(164) April 11, 1851, at 3h 14m P.M. mean time nearly, in long. 56° 42′ W., an observation was taken, when a chronometer showed 7h 2m 10s.5.
On March 15, its error was known to be 1m 32s.7 fast, on Greenwich mean time, and its daily rate 6.3 losing: required mean time at Greenwich, when the observation was taken. Ans., 7h 3m 29s.7.
To find the error of a chronometer on mean time at the place of observation, by a single altitude of the sun.
Let p be the pole, z the zenith, and x the place of the sun, bearing as nearly east or west as possible, aq the celestial equator, a the first point of Aries, and m the place of the mean Then in the triangle ZPX are given the zenith distance z x, the polar distance P X, and the colatitude of the
spectator P z, to find the hour angle z PX, which is also the
apparent solar time, if the sun is west of the meridian, or what it wants of 24 hours if the sun is east of meridian, To this apparent solar time thus found, apply the equation of time XP with its proper
sign, as given in the Nautical
Almanac, the result will be q pm, or mean time at the place of observation: the difference between which and the time shown by the chronometer at the instant of the observation will manifestly be the error of the chronometer on mean time at the place. Hence this rule.
1. Find a Greenwich date.
2. Correct the sun's declination and equation of time for this date. Take out of the Nautical Almanac the sun's semidiameter, at the same time the declination and equation of time are taken out.
3. Correct the observed altitude for index correction, dip, semidiameter, and correction in altitude, and thus get the true altitude; subtract the true altitude from 90° to obtain the zenith distance.
4. To find ship apparent time (using log. haversines*).
*If the student have no table of haversines, he may proceed as follows to find apparent solar time :—
Under the latitude put the sun's declination, and, if the names be alike, take the difference; but if unlike, take their sum. Under the result put the zenith distance, and find their sum and difference, and half-sum and half-difference.
Add together the log. secants of the two first terms in this form (rejecting the tens in index) and the log, sines of the two last, and divide the sum by 2; look out the result as a log. sine and multiply the angle taken out by 2.