water earlier in the former, and later in the latter case. This explanation will be received with a slight modification, arising from the consideration that the solar and lunar tides are eastward of the places of the sun and moon, and consequently this acceleration and retardation, like the times of the highest spring and the lowest neap tides, will take place a little after the moon enters the quarters mentioned. When the moon is at her least distance from the earth, the tides of course are greater than usual ; and when the full or change of the moon happens about the beginning of January, when the sun also is nearest the earth, the tides are the greatest of all. The nearer the moon passes the zenith of any place the greater are the tides which she produces at that place. In small seas which are much enclosed by land, such as the North Sea, the observed tides are supplied from those raised in the adjoining ocean ; and the Baltic, the Mediterranean, and such other seas as communicate with the ocean by very small mouths, cannot receive a sufficient supply of water in a tide to produce a material elevation in their surfaces ; in these seas, in consequence, the tides are found to be very trifling. The times of high water at any individual place are greatly influenced by its local situation; but there is at every place a mean relation between the time of high water and that of the moon's passing the meridian, which relation is subject to periodical variations, depending on the distances and relative positions of the sun and moon. The time of high water too is often materially affected by the wind; but it may be found, with sufficient exactness for any practical purpose in navigation, by means of the following problems. PROBLEM I. To find the moon's age on any day between the years 1800 and 1900. The time of a mean lunation is nearly 29d 12h 44m, twelve of which make about 354d 9h, or about 10d 21h less than a solar year of 365d 6h. Hence the age of the moon, at the beginning of any year, will be 10d 21h greater than at the beginning of the preceding one, till it reach a complete lunation, when its excess above that quantity will be her age. And if the days elapsed from the beginning of the year were divided by 29d 12h 44m, the remainder would be the age of the moon, if the new moon happened at the commencement of the year; consequently if the moon's age, at the beginning of the year, were added to that remainder, the sum would be the moon's age on the proposed day. the following table, both for leap years and common years ; every year divisible by 4, without a remainder, being a leap year. Jan. Feb. March April May June July August Sept. Oct. Nov. Dec. Common Years 0 1 0 1 2 3 4 5 7 7 9 9 Leap Years ... 0 2 1 2 3 4 5 6 8 8 10 10 The moon's age, at the beginning of the year 1800, was five days, therefore to find it for any subsequent time, till 1900, reject 1800 from the given year, multiply 10:9 by the remainder, and divide the product by 29.5; and to the integers in the remainder of the division, add 5, the day of the month, and the number for the month from the above table ; and the sum, if it does not exceed 30, will be the moon's age; but if it exceed 30, reject 30 from the sum, and the remainder will be her age. PROBLEM II. To find the time at which the moon passes the meridian on a given day. If the new moon happen at noon, the sun and moon will be on the meridian together ; and as the moon, at a mean rate, is about 49 minutes later in coming to the meridian every day, if we multiply 49 by her age in days, and divide the product by 60, the quotient will be the hours, and the remainder the minutes past noon, when she passes the meridian on the given day, on the supposition that her motion is uniform, and that she changes at noon. But as her motion is not uniform, and she seldom changes precisely at noon, this method of finding the time of her transit must be considered as only an approximate one. The method of finding it correctly, with the aid of the Nautical Almanac, has been given at Problem 5, p. 220, and those who possess the Nautical Almanac, will of course have recourse to that method. EXAMPLE. It is required to find, by the two preceding problems, the moon's age nearly, and the approximate time of her passing the meridian on September 10, 1823 ? 10:9 x 23 = 250•7, which, divided by 29.5, gives 8 for the quotient, and an integral remainder of 14. Hence, Remainder ...... 14 5 6 x 49 6, and = 4h 54m, the time 60 of her passing the meridian. 10 Moon's age EXAMPLES FOR EXERCISE. It is required from the two preceding problems, without the aid of the Nautical Almanac, to determine approximately the moon's age, and the time of her passing the meridian on each of the following days? To find the time of high water on a given day at any place where the time of high water at full and change is known. Let the time of the moon's passing the meridian of the given place be found by Prob. 5, p. 220, and to this time apply the correction, from the following table, corresponding to her meridian passage and semidiameter, and to the result add the time of high water at full and change at the given place, as given in Table 33, or 34, and the sum will be the time of high water on the given day. If this sum exceed 12h 24m, or 24h 49m, subtract those times from it, and the remainder will be nearly the time of high water on the afternoon of the given day. If the computer have not a Nautical Almanac, he may find the approximate time of the moon's passing the meridian by the two preceding problems, and applying to it the correction from the middle column in the following table, add the result to the time of high water at full and change, from Table 33, or 34, and the sum will give the time of high water, generally within less than an hour of the truth. Corrections to be applied to the time of the moon's meridian passage in finding the time of high water. Required the time of high water September 10, 1923, at Quebec ? D's Pass. meridian, Nautical Almanac, (see Table 31,) 4h 32m + 5lın, semid. 15' Correction for longitude, Table 17.... + 10 4 42 Correction from the above Table -1 3 3 39 High water at full and change, Tab. 33, 7 30 Answer 11 9 The same computed by the other method. In the example to Problem 2, the time of the moon's passing the meridian is computed to be ... 4h 54m Cor. from the middle column of the above Table.... -1 3 47 Time from Table 33 ny 30 Answer. 11 17 EXAMPLES FOR EXERCISE. The time of high water is required in the afternoon at each of the places mentioned in the following examples, on the respective days mentioned, the requisite data from the Nautical Almanac being given in Table 31 ? X The time of high water in the afternoon is required in each of the following examples, computing the moon's age, and the time of her meridian passage, by the two preceding problems? From the observed time of high water at any place on a given day, to find the time of high water at full and change. To the time at which the moon passes the meridian of the given place, found by Prob. 5, p. 220, or Prob. 2, p. 295, apply the correction from the preceding table, and the result, subtracted from the observed time of high water, will leave the time of high water on the afternoon of the days of full and change. If the time to be subtracted exceed the observed time of high water, let 12h 24m, or 24h 49m, (whichever is necessary to make it greater) be added to the observed time before the subtraction is made. EXAMPLE. It was observed to be high water at Quebec, on September 10, 1823, at 11h 9m, required the time of high water at full and change? In the last problem, the time of the moon's passing the meridian on the given day, corrected by the equation from the table, was found to be 3h 39m, which, subtracted from ilh 9m, leaves h 30m for |