(cos. BDP X rad.2)-(sine DBS X sine DPS X rad.) (0) M. de Lambre gives the following formula for finding the angle BSP, which he demonstrates by spherical trigonometry, viz. Sine BSP rad. sine (BDP+ DBS DPS) X sine (BDP + DPS—DBS) COS. DBS X COS. DPS The sides BS and Ps of the triangle BSP are easily found, from the height Ds and the angles DBS and DPS being given. EXAMPLE L. Wanting to know the height of an object SD (standing on the horizontal plane PSB), and also the horizontal distances sp and SB. I measured a base PB=300 yards, at p the of elevation DPS was 8°.15', and the DPB, inclined to the horizon, was 57°. 15': at B the of elevation DBS was 8°. 45'. 50", and the 4 DBP, inclined to the horizon, was 63°. 15'. Required the horizontal BSP, the horizontal distances SP, and SB, and the height of the object above the plane. Answer. LBPS=56°.51′. 50′′; 4PBS=62°. 54′. 30′′; (L. 85.) BS 289.41; Ps=307.69; and DS 44.614 yards. The height Ds, and the distances BS and Ps may be found without reducing the angles. For, in the triangle DBP all the angles are given, and the side BP, whence BD and PD may be found; then in the triangle DPS, the DPS and the side PD are given to find Ps and SD; also, in the triangle DBS, DB and the DBS are given, from which вs may be found. EXAMPLE LI. From the top of a tower 44.614 yards high, the angle BDF, subtended by two distant objects B and P, was 59°. 30′; and the angles of depression m DB = 8°. 45', 50", and " DP = 8°. 15'. Hence it is required to find the horizontal distances PB, BS, and Ps. Answer. The <m DB = DBS = 8°. 45'. 50′′; and ▲n DP = DPS=8°. 15′, and ▲ BDP=59°.30'. By the formula (N. 87.) Cos. / BSP =*49655, hence / BSP=60°.13'.40"-4. and, (0.87.) Bs= 289.41, ps=307·69, and PB=300. The same answer may be obtained without reducing the angles, and without the formula. (P) The formulæ at the conclusion of N. 86, and O. 87, may be applied to any two triangles, whether their bases be in the same horizontal plane and their vertices be elevated, as in examples L and LI, or their vertices be in the same horizontal plane, plane, and the extremities of the base of the one triangle be elevated above the extremities of the base of the other. cosine SDO = (cos. PDB X rad.2) — (sine PDS X sine BDO X rad.) From a station at D in the horizontal plane Dso, I took the angle PDB, subtended by the tops of two towers,=37°. 53′. 20′′; and also the angles of elevation BD0=4°. 23′. 55′′, and PDs= 4°. 17.21". The height of the tower BO is known to be 49 yards, and that of PS 30 yards; from which it is required to find the horizontal distance of my station from each of the towers, and their horizontal distances from each other. Answer. By the formula (P. 88.) ZSDO=37°. 59'; Ds=400 yards; DO 520; and so=320 yards. (Q) The same answer may be found without the formula. For, with the BDO and height Bo, find DB-521-0536, and Do=520; with the PDS and height Ps, find DP=401·1234, and DS 400. Then with Dr, DB and the ZPDB, find PB= 320-1562; lastly, =320 yards. PB2-EO-PS so, the horizontal distance, D -F C B OF THE DIP, or DEPRESSION OF THE HORIZON AT SEA. (R) The dip, or depression of the horizon at sea, is the angle contained between the horizon of the observer, and the farthest visible point on the surface of the sea. For, if an observer whose eye is situated at D, takes the altitude of a celestial object by a sextant, or Hadley's quadrant, and brings that object to the surface of the water at B, instead of the horizon DF, he evidently makes the altitude too great by the FDB. E * Traité de Trigonométrie par M. Cagnoli, Appendice, page 467. Now Now the ADF is a right angle, and the ABD is likewise a right angle, therefore the ZFDB=LDAB. Hence, in the right angled triangle ABD, there is given AB the radius of the sphere, ADAC+DC, the radius of the sphere increased by the height of the eye above the surface of the sea, to find the quantity to be subtracted from the observed altitude of Hence cos. DAB COS. FDB= lestial object. DAB, or any ce rad. X AB AC + CD Where ACAB is the radius of the sphere, and DC the height of the observer's eye above the surface of the sea. OR, (EC+ CD) X DC DB2 (Euclid III. and 36.), that is (ECX DC)+DC2=DB'; or (2AC x DC)+DC=DB, and rejecting DC2 as being small √2AC X DC = DB, but ABAC : rad. :: DB=√2AC X DC: tangt. DAB = tangt. FDB, therefore quan Now the first of these terms is a constant quantity, and if the diameter of the earth be 7964 miles, the logarithin of this tity in feet will be 6:48915, hence log. tangt. 4FDB=6°48915+ log. DC in feet. OF THE PARALLAX OF THE CELESTIAL BODIES. (S) That part of the heavens in which a planet would appear, if viewed from the surface of the earth, is called its apparent place: and the point in which it would be seen at the same instant from the centre of the earth, is called its true place, the difference between the true and apparent place is called the parallax in altitude. D. Z K Let c be the centre of the earth, A the place of an observer on its surface, whose visible horizon is AB, true horizon cd, and zenith z. Let ZIFD be a portion of a great circle in the heavens, and E the absolute place of any object in the visible horizon; join CE and produce it to F; then F is the true place of the object, and B its apparent place in the heavens; and the angle BEFAEC is the parallax. The parallax is the greatest when the object is in the horizon; for the angle AEC is greater than AGC, hence the more elevated an object is (its distance from the earth's centre continuing the same) the less is the parallax. When the object is in the zenith the parallax vanishes, for then AC and Az are in the same straight line cz. Since the apparent place (H) of a planet is more distant from the zenith (z) than the true place (1) it therefore follows, that the parallax in altitude must be added to the observed altitude, in order to obtain the place of a planet as seen from the earth's centre. The stars on account of their immense distance from the earth have no sensible parallax, and the sun's mean parallax is only 8".6. The moon's greatest horizontal parallax is 61'. 32". least 54′. 4′′. (T) The horizontal parallax being given, to find the parallax et any given altitude. In the right-angled triangle EAC Radius EC :: sine AEC : AC And in the triangle GAC GC EC sine GAC :: AC : sine AGC And by comparing these two proportions radius sine GAC :: sine AEC : sine AGC. But the sine of an arc is equal to the sine of its supplement, therefore sine GAC sine GAK, and GAK is the complement of CAE, therefore sine of GAK=cosine GAE, hence radius cosine GAE :: sine AEC sine AGC. The last two terms being small, the arcs may be substituted for their sines without sensible error. Radius Hence the following rule: : Cosine of the apparent altitude, EXAMPLE LIII. The apparent altitude of the moon's centre is 24°. 29′. 44′′. the horizontal parallax 55'. 2". Required the parallax in altitude? Radius, sine of 90° : cosine moon's altitude 24°. 29′ 44′′. 10. 9.95904 :: horizontal parallax 55′. 2′′=3302" log.= 3.51878 Hence the parallax in altitude is 3005"=50′.5". 3'47782 OF THE ADMEASUREMENT OF ALTITUDES BY THE BAROMETER AND THERMOMETER. (U) One of the most simple and easy practical rules for measuring the elevations and depressions of objects by means of the the barometer and thermometer, is that given in the Encyclopædia Brit. Article Pneumatics, or in Dr. Rees' New Cyclopædia, under the word Barometer. The directions are as follow. -Let the observers be provided with two portable barometers, each of the same construction, with a nonius properly adapted to the scale, and a thermometer attached to each, having their bulbs each of the same diameter, nearly, as the diameters of the barometric tubes. Place one of these barometers in the shade at the top of the eminence, with a detached thermometer near it; and let the other barometer be placed below in like manner, with a detached thermometer near it. When the thermometers have acquired the temperature of the air, that is, when the fluid in each becomes stationary, the observers must note down the temperatures shewn by the thermometers and the heights of the mercurial columns in the barometers. Then the elevation of the higher barometer above the lower may be determined by the formula, which is deduced from the following observations: (W) 1. The height through which we must rise in order to produce any fall of the mercury in the barometer, is inversely proportional to the density of the air; that is to the height of the mercury in the barometer. 2. When the barometer stands at 30 inches, and the air and quicksilver are of the temperature 32°, we must rise through 87 feet, in order to produce a depression of th of an inch. 3. But if the air be of a different temperature, the 87 must be increased or diminished by 21 of a foot for every degree of difference of the temperature from 32°. 4. Every degree of difference of the temperatures of the mercury at the two stations makes a change of 2.833 feet, or 2 feet 10 inches in the elevation. Hence, If d be the difference between 32° and the mean temperature of the air, D the difference between the barometric heights in tenths of an inch, the difference between the mercurial temperatures, m the mean barometric height, and E the correct 30D (87+21d) elevation. Then E= m +x2·833. This formula may be given in words, thus: (X) 1. Multiply the difference between 32° and the mean temperature of the air by 21, and add the product to $7, if the mean temperature be above 32°, but subtract it if below. 2. Multiply the sum or difference, found above, by thirty times the difference between the barometric heights, in tenths of inches, and divide the product by the mean of the barometric heights, the quotient will give the approximated elevation. 3. Multiply N 2 |