Now, by comparing this computed distance with that directly deduced from the Table, as in the preceding example, it will be seen that the difference amounts to very little more than the fifth part of a second in twenty years; which evidently demonstrates that the tabular distances may be reduced to any subsequent period, for a considerable series of years, with all the accuracy that may be necessary for the common purposes of navigation.

Note. The tabular distances will be found particularly useful in determining the latitude, at sea, by the altitudes of two stars, as will be shown hereafter.

TABLE XLV.

Acceleration of the Fixed Stars; or to reduce Sidereal to Mean Solar Time.

Observation has shown that the interval between any two consecutive transits of a fixed star over the same meridian is only 23 564.09, whilst that of the sun is 24 hours:-the former is called a sidereal day, and the latter a solar day; the difference between those intervals is 355'.91, and which difference is called the acceleration of the fixed stars.

This acceleration is occasioned by the earth's annual motion round its orbit: and since that motion is from west to east at the mean rate of 59.8". 3 of a degree each day; if, therefore, the sun and a fixed star be observed on any day to pass the meridian of a given place at the same instant, it will be found the next day when the star returns to the same meridian, that the sun will be nearly a degree short of it; that is, the star will have gained 3 56.55 sidereal time, on the sun, or 3" 55′ .91 in mean solar time; and which amounts to one sidereal day in the course of a year :-for 3:55′.91 × 365548" 48: 2356" 4::-hence in 365 days as measured by the transits of the sun over the same meridian, there are 366 days as measured by those of a fixed star.

=

Now, because of the earth's equable or uniform motion on its axis, any given meridian will revolve from any particular star to the same star again in every diurnal revolution of the earth, without the least perceptible difference of time shewn by a watch, or clock, that goes well :-and this presents us with an easy and infallible method of ascertaining the error and the rate of a watch or clock :-to do which we have only to observe the instant of the disappearance of any bright star, during several successive nights, behind some fixed object, as a chimney or corner of a house at a

little distance, the position of the eye being fixed at some particular spot, such as at a small hole in a window-shutter nearly in the plane of the meridian; then if the observed times of disappearance correspond with the acceleration contained in the second column of the first compartment of the present Table, it will be an undoubted proof that the watch is well regulated: hence, if the watch be exactly true, the disappearance of the same star will be 356′. earlier every night; that is, it will disappear 356! sooner the first night; 752: sooner the second night; 1148: sooner the third night, and so on, as in the Table.-Should the watch, or clock deviate from those times, it must be corrected accordingly; and since the disappearance of a star is instantaneous, we may thus determine the rate of a watch to at least half a second.

The first compartment of this Table consists of two columns; the first of which contains the sidereal days, or the interval between two successive transits of a fixed star over the same meridian, and the second the acceleration of the stars expressed in mean solar time; which is extended to 30 days, so as to afford ample opportunities for the due regulation of clocks or watches. The five following compartments consist of two columns each, and are particularly adapted to the reduction of sidereal time into mean solar time-the correction expressed in the column marked acceleration, &c. being subtracted from its corresponding sidereal time, will reduce it to mean solar time; as thus.

Required the mean solar time corresponding to 14:40 55% sidereal time?

Remark. This Table was computed in the following manner; viz.,

Since the earth performs its revolution round its orbit, that is, round the sun, in a solar year; therefore as 36554848:; 360°::1; 59′8′′. 3; which, therefore, is the earth's daily advance in its orbit: but while the earth is going through this daily portion of its orbit, it turns once round on its axis, from west to east, and thereby describes an arc of 360:59:8".3 in a mean solar day, and an arc of 360: in a sidereal day.

Hence, as 360:59:8".3; 24::360: 23:56" 4'.09, the length of a sidereal day in mean solar time; and which, therefore, evidently anticipates 355.91 upon the solar day as before-mentioned. Now,

As one sidereal day, is to 355.91, so is any given portion of sidereal time to its corresponding portion of mean solar time :-and hence, the method by which the Table was computed.

TABLE XLVI.

To reduce Mean Solar Time into Sidereal Time.

Since this Table is merely the converse of the preceding, it is presumed that it does not require any explanation farther than by observing, that the correction is to be applied by addition to the corresponding mean solar time, in order to reduce it into sidereal time ; as thus.

Required the sidereal time corresponding to 20:1533: mean solar time?

Time from Noon when the Sun's Centre is in the Prime Vertical; being the instant at which the Altitude of that Object should be observed in order to ascertain the apparent Time with the greatest Accuracy.

Since the change of altitude of a celestial object is quickest when that object is in the prime vertical, the most proper time for observing an altitude from which the apparent time is to be inferred, is therefore when the object is due east or west; because then the apparent time is not likely to be affected by the unavoidable errors of observation, nor by the inaccuracy of the assumed latitude.-This Table contains the apparent time when a celestial object is in the above position.-The declination is marked at top and bottom, and the latitude in the left and right hand marginal columns: hence, if the latitude be 50 degrees, and the declination 10 degrees, both being of the same name, the object will be due east or west at 5:26" from its time of transit or meridional passage.

Remark. This Table was computed by the following rule; viz.,

To the log. co-tangent of the latitude, add the log. tangent of the declination; and the sum, abating 10 in the index, will be the log. co-sine of the hour angle, or the object's distance from the meridian when its true bearing is either east or west.

Example.

Let the latitude be 50 degrees, north or south, and the sun's declination 10 degrees, north or south; required the apparent time when that object will bear due east or west?

Note.-During one half of the year, or while the sun is on the other side of the equator, with respect to the observer, that object is not due east or west while above the horizon; in this case, therefore, the observations for determining the apparent time must be made while the sun is near to the horizon; the altitude, however, should not be under 3 or 4 degrees, on account of the uncertainty of the effects of the atmospheric refraction on low altitudes.

TABLE XLVIII.

Altitude of a Celestial Object (when its centre is in the Prime Vertical,) most proper for determining the apparent Time with the greatest Accuracy.

This Table is nearly similar to the preceding; the only difference being that that Table shows the apparent time when a celestial object bears due east or west, and this Table the true altitude of the object when in that position; being the altitude most proper to be observed in order to ascertain the apparent time with the greatest accuracy :-thus, if the latitude be 50 degrees, and the declination 10 degrees, both being of the same name, the altitude of the object will be 13:6, when it bears due east or west from the observer; which, therefore, is the altitude most proper to be observed, for the reasons assigned in the explanation to Table XLVII.

Note. This Table was computed by the following rule; viz.,

If the declination be less than the latitude; from the log. sine of the former (the index being increased by 10), subtract the log. sine of the latter, and the remainder will be the log. sine of the altitude of the object when its centre is in the prime vertical:-But, if the latitude be less than the declination, a contrary operation is to be used ; viz., from the log. sine of the latitude, the index being increased by 10, subtract the log. sine of the declination, and the remainder will be the log. sine of the altitude of the object when its centre is in the prime vertical, or when it bears due east or

west.

Example 1.

Let the latitude be 50%, and the declination of a celestial object 10, both being of the same name; required the altitude of that object when its centre is in the prime vertical.

Let the latitude be 3°, and the declination of a celestial object 14:, both being of the same name; required the altitude of that object when its in the prime vertical.

Note.-Altitudes under 3 or 4 degrees should not be made use of in computing the apparent time, on account of the uncertainty of the atmospheric refraction near the horizon.

And since the Table only shows the altitude of a celestial object most. favourable for observation when the latitude and declination are of the same name; therefore during that half of the year in which the sun is on the other side of the equator, with respect to the observer, and in which he does not come to the prime vertical while above the horizon, the altitude is to be taken whenever it appears to have exceeded the limits ascribed to the uncertainty of the atmospheric refraction in page 120.