TABLE XXXVII. Logarithmic Tangents. This Table is arranged in a manner so very nearly similar to that of the log. sines, that it is not deemed necessary to enter into its description any farther than by observing, that it is computed to every second in the two first and two last degrees of the quadrant, or semicircle, and to every fifth second in the intermediate degrees. The log. tangent, or co-tangent, of a given arch, and conversely, is to be found by the rules for the log. sines in pages 94 and 95. Example 1. Required the log. tangent, and co-tangent, corresponding to 31:10:47?? Given arch= 31:10:47" Corresponding log. tang.=9.781855 Odd seconds To find the Log. Co-tangent: 31:10:45, ans. to which is . . 10.218155 Given arch 31:10:47" Corres. log. co-tang. = 10.218145 Example 2. Required the log. tangent, and co-tangent, corresponding to 139:11:53?? Given arch 139:11:53 Corres, log. co-tang. 10.063871 H Example 3. Required the arch corresponding to the given log. tang. 10. 155436? Solution. The next less log. tangent in the Table is 10. 155423, corresponding to which is 55:2:25; the difference between this log. tangent and that given, is 13; answering to which, in the column of proportional parts abreast of the tabular log., is 3"; which, being added to the abovefound arch, gives 55:2:28" for that required. Example 4. Required the arch corresponding to the given log. co-tang. 9. 792048? Solution. The next greater log. co-tangent in the Table is 9.792057, corresponding to which is 58:13:15; the difference between this log. cotangent and that given, is 9; answering to which, in the column of proportional parts abreast of the tabular log., is 2"; which, being added to the above-found arch, gives 58:13:17" for that required. Remark. The arch corresponding to a given log. tangent may be found by means of a Table of log. sines, in the following manner; viz., Find the natural number corresponding to twice the given log. tangent, rejecting the index, to which add the radius, and find the common log. of the sum; now, half this log. will be the log. secant, less radius, of the required arch; and which, being subtracted from the given log. tangent, will leave the log. sine corresponding to that arch. Example. Let the given log. tangent be 10. 084153; required the arch corresponding thereto by a table of log. sines? Given log. tang.. 084153 × 2.168306, Nat. num. = 1.473349 to which add the radius = 1.000000 common log. of which is 0.393286; the half of this is 0. 196643, secant, less radius of the required arch. the the answering to which is 50:31; and which, therefore, is the required arch corresponding to the given log, tangent. 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 AC is to the natural tangent B C, so is the radius AD AB to the natural sine DE: hence, = B As AC 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 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?? Log. sine of the given arch 32:39:40% = 9.732128 9.925249 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 12" 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 = 23 57: 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 ? |