The arch corresponding to a given log. tangent may also be found in the following manner, which, it is presumed, will prove both interesting and instructive to the student in this department of science. Find the natural tangent, that is, the natural number corresponding to the given log. tangent, to the square of which add the square of the radius; extract the square root of the sum, and it will be the natural secant corresponding to the required arch; then, say, as the natural secant, thus found, is to the natural tangent, so is the radius to the natural sine; now, the degrees, &c. answering to this in the Table of Natural Sines, will be the arch required, or that corresponding to the given log, tangent, Example. Let the given log. tangent be 10: 084153; it is required to find the arch corresponding thereto by a Table of Natural Sines? Solution.-Given log. tangent .084153; the natural number corresponding to this is 1.213816; which, therefore, is the natural tangent answering to the given log. tangent. In the annexed diagram, let B C represent the natural tangent = 1.213816, and AB the radius F = 1.000000. Now, since the base and perpendicular of the right-angled triangle ABC are known, the hypothenuse or secant AC may be determined by Euclid, Book I., Prop. 47. Hence ✔ BC2 = 1.213816 + A B2 = 1.0000002 = AC 1.572689, the natural secant corresponding to the given log. tangent. Having thus found the natural secant A C, the natural sine DE may be found agreeably to the principles of similar triangles, as demonstrated in Euclid, Book VI., Prop. 4; for, as the natural secant A C is to the natural tangent B C, so is the radius AD = AB to the natural sine DE: hence, As A C 1.572689 BC 1.213816 :: AD 1. 000000; DE= 771810, the corresponding natural sine; now, the arch answering to this, in the Table of Natural Sines, is 50:31; which, therefore, is the arch corresponding to the given log. tangent, as required. Note.-The Table of log. tangents may be very readily deduced from Tables XXXV. and XXXVI., as thus :-To the log. secant of any given arch, add its log. sine; and the sum, abating 10 in the index, will be the log. tangent of that arch; the difference between which and twice the radius, will be its co-tangent. Example. Required the log. tangent, and co-tangent, of 25:27:35?? Log. secant of the given arch 25:27:35% = 10.044366 9.633344 Log. tangent corres. to the given arch= 9.677710 The Table of log. tangents may also be computed in the following manner; viz., From the log. sine of the given arch, the index being increased by 10, subtract its log. co-sine, and the remainder will be the log. tangent of that arch; the difference between which and twice the radius, will be its log. co-tangent. Example. Required the log. tangent, and co-tangent, of 32:39:40"? To reduce the Time of the Moon's Passage over the Meridian of Greenwich, to the Time of her Passage over any other Meridian. The daily retardation of the moon's passage over the meridian, given at the top of the Table, signifies the difference between two successive transits of that object over the same meridian, diminished by 24 hours; as thus: the moon's passage over the meridian of Greenwich, July 22d, 1824, is 21:7, and that on the following day 229"; the interval of time between these two transits is 252", in which interval it is evident that the moon is 1:2 later in coming to the meridian; and which, therefore, is the daily retardation of her passage over the meridian. This Table contains the proportional part corresponding to that retardation and any given interval of time or longitude; in computing which, it is easy to perceive that the proportion was, As 24 hours, augmented by the daily retardation of the moon's transit over the meridian, are to the said daily retardation of transit, so is any given interval of time, or longitude, to the corresponding proportional part of such retardation. The operation was performed by proportional logs., as in the following Example. Let the daily retardation of the moon's transit over the meridian be 12"; required the proportional part corresponding thereto, and 9:40" of time, or 145 degrees of longitude? To corresponding proportional part = 2357 Pro. log.= 0.8761; and in this manner were all the numbers in the Table obtained. The corrections or proportional parts contained in this Table are expressed in minutes and seconds, and are extended to every twentieth minute of time, or fifth degree of longitude: these are to be taken out and applied to the time of the moon's transit, as given in the Nautical Almanac, in the following manner : Find, in page VI. of the month in the Nautical Almanac, the difference between the moon's transit on the given day (reckoned astronomically) and that on the day following, if the longitude be west; but on the day preceding, if it be east. With this difference enter the Table at the top, and the given time in the left-hand, or the longitude in the right-hand column; in the angle of meeting will be found a correction, which, being applied by addition to the time of transit on the given day, if the longitude be west, but by subtraction, if east, the sum, or difference, will be the reduced time of transit. Example 1. Required the apparent time of the moon's passage over a meridian SO degrees west of Greenwich, July 22d, 1824 ? Mn's pas. over mer. of Greenw. on giv. day is 21:7" on the day following = 22.9 Ditto Apparent time of the moon's transit over the given meridian 21:20:13: Example 2. Required the apparent time of the moon's passage over a meridian 120 degrees east of Greenwich, August 20th, 1824? Mn's pas. over mer. of Greenw. on giv. day is 20:54" on the day preceding=19.54 20:54" 0: Ditto Apparent time of the moon's transit over the given meridian 20:34:48: TABLE XXXIX. Correction to be applied to the Time of the Moon's Transit in finding the Time of High Water. Since the moon is the principal agent in raising the tides, it might be expected that the time of high water would take place at the moment of her passage over the meridian; but observation has shown that this is not the case, and that the tide does not cease flowing for some time after : for, since the attractive influence of the moon is only diminished, and not entirely destroyed, in passing the meridian of any place, the ascending impulse previously communicated to the waters at that place must, therefore, continue to act for some time after the moon's meridional passage. The ascending impulse, thus imparted to the waters, ought to cause the time of the highest tide to be about 80 minutes after the moon's passage over the meridian; but owing to the disturbing force of the sun, the actual time of high water differs, at times, very considerably from that period. The effect of the moon in raising the tides exceeds that of the sun in the ratio of about 2 to 1; but this effect is far from being uniform: for, since the moon's distance from the earth bears a very sensible proportion to the diameter of this planet, and since she is constantly changing that distance, (being sometimes nearer, and at other times more remote in every lunation,) it is evident that she must attract the waters of the ocean with very unequal forces: but the sun's distance from the earth being so very immense, that, compared with it, the diameter of this planet is rendered nearly insensible, his attraction is consequently more uniform, and therefore it affects the different parts of the ocean with nearly an equal force. By the combined effect of these two forces, the tides come on sooner when the moon is in her first and third quarters, and later when in the second and fourth quarters, than they would do if raised by the sole lunar agency: it is, therefore, the mean quantity of this acceleration and retardation that is contained in the present Table, the arguments of which are, the apparent times of the moon's reduced transit; answering to which, in the adjoining column, stands a correction, which, being applied to the apparent time of the moon's passage over the meridian of any given place by addition or subtraction, according to its title, the sum, or difference, will be the corrected time of transit. Now, to the corrected time of transit, thus found, let the time of high water on full and change days, at any given place in Table LVI., be applied by addition, and the sum will be the time of high water at that place, reckoning from the noon of the given day: should the sum exceed 1224", or 2448", subtract one of those quantities from it, and the remainder will be the time of high water very near the truth. Example 1. Required the time of high water at Cape Florida, America, March 7th, 1824; the longitude being 80:5' west, and the time of high water on full and change days 7:30"? Moon's transit over the meridian of Greenwich, per Nautical west= 1. 5. 0 Moon's transit reduced to the meridian of Cape Florida Corrected time of transit Time of high water at Cape Florida on full and change days Time of high water at Cape Florida on the given day = Example 2. 4 9 23: 7.30. 0 11:39 23: Required the time of high water in Queen Charlotte's Sound, New Zealand, April 13th, 1824; the longitude being 174:56 east, and the time of high water on full and change days 9:0"? |