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 shown 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 toth of an inch. 3. But if the air be of a different temperature, the 87 mụst be increased or diminished by 21 of a foot for every degree of difference of the temperature from 32o. . 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 32o and the mean temperature of the air, D the difference between the barometric heights in tenths of an inch, 8 the difference between the mercurial temperatures, m the mean barometric height, and E the correct elevation. Then E= 30D (87+210) +8 x 2.833 This formula may be given in words, thus : (X) 1. Multiply the difference between 32o and the mean temperature of the air by 21, and add the product to 87, 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. m 3. Multiply the difference between the mercurial temperatures by 2.833 feet, and add this product to the approximated elevation, if the upper barometer has been the warmer of the two, otherwise subtract it, the result will be the corrected elevation in feet. EXAMPLE LIV. Suppose that the mercury in the barometer at the lower station was at 29•4 inches, that its temperature was 50° of Fahrenheit's thermometer, and the temperature of the air 45°; and let the height of the mercury at the upper station be 25:19 inches, its temperature 46°, and the temperature of the air 39°, what would be the elevation of the higher barometer above the lower ? 45° +39° 1. =42° mean temperature of the air. (.21 x 10)+87=89•1 (=87+.21d). 2. (294-25°19) x 10=42:1 diff. of barometric heights in tenths of inches (=D) 29.4 + 25.19 =27.295 mean barometric height (=m) 2 42:1 X 30 X 89•1 -=4122•854 feet, the approximated elevation 27.295 30D (87+210) . m 3. 500 — 46o=4o diff. of mercurial temperatures (8) 2.833 x 4=11.332 (=8 x 2.833). Then 4122-854-11•332-4111•522 feet, the correct elevation. (Y) If several sets of observations be made at each station, after short intervals of time, and the mean of the separate results be taken, the conclusion will probably be more accurate than that derived from a single observation. CHAP. V. OF THE SIGNS OF TRIGONOMETRICAL QUANTITIES, &c. (Z) In the investigation of some theorems in spherical trigonometry, and in the application of trigonometry to several astronomical and analytical problems, the changes of the signs of the tangents, sines, &c. from affirmative or +, to negative or-, ought to be particularly attended to. They likewise serve to illustrate the use of the negative sign in the application of algebra to geometry: (A) The tangents, sines, and versed sines are counted from, or have their origin at B, the beginning of the arc (Plate I. fig. 1.) The cosines, secants, and co-secants, originate at the centre c. (B) The co-tangents and co-versed sines originate at the end of the first quadrant, as at A. (C) The sine increases from nothing at B till it becomes equal to the radius, at the end of the first quadrant AB; from hence it decreases along the second quadrant from a to b, and vanishes at the point b, shewing the sine of a semicircle to be nothing. After this, the sine will lie on the contrary side of the diameter: therefore being reckoned affirmative before, must now be counted negative. During the 3d quadrant bd the sine ih increases negatively, till it becomes equal to the radius; after which it decreases negatively, till it arrives at the point B, where it is nothing as before. (D) The cosine is equal to the radius when the arc is nothing; but decreases through the first quadrant Ba, at the end of which it is nothing: during the second quadrant it is negative; for the cosine ci will lie in an opposite direction to the cosine CG.. Through the third quadrant bd the cosine decreases negatively, at the negation is destroyed; and in the fourth quadrant Do it again becomes affirmative. (E) The tangent at the beginning of the arc is nothing, and increases to infinity during the first quadrant BA, at the point a there is no tangent, for ca produced can never meet BT (L. 31.) In the second quadrant ab the tangent Bt is negative; for the tangents being reckoned from the point B, the tangent it will lie in an opposite direction from Bt. During this second quadrant the tangent decreases negatively, from infinity to nothing. In the third quadrant the tangent is again affirmative, and increases from nothing to infinity. In the fourth quadrant it decreases from an infinite negative to nothing, just the same as in the second quadrant. (F) The co-tangent at the beginning B, of the arc, that is when the arc is very small, is infinite affirmative, but in the second quadrant it will fall on the contrary side of ad, and consequently will be negative. In the third quadrant it will again be affirmative, and in the fourth negative, exactly the same as the tangent. (G) The secant at the commencement of the first quadrant, when the arc is nothing, is equal to the radius, and increases affirmatively to the end of the first quadrant Ba, where it ceases to exist (M. 32.) In the second quadrant it is negative, for the revolving radius falls on the contrary side of ad, and continues . . so through the third quadrant; but in the fourth quadrant it is again affirmative, or falls on the same side of ad as in the first quadrant. (H) The co-secant will agree with the sine, for the same reason that the secant agrees with the cosine. (I) The versed sine increases from nothing during the first two quadrants, till it becomes equal to the diameter its utmost limit. It decreases for the last two quadrants till it becomes nothing, but being always counted in the same direction from B to b is always affirmative. From 0° From 90° From 180° From 270° (K) Hence, to 90° to 180° to 270° to 360° + + (L) In analytical enquiries, arcs of all magnitudes", whether greater or less than 180°, are frequently used; but in trigonometry, every arc or angle made use of must be less than 180°. It has already been shewn that equal arcs have equal sines, tangents, &c.; but if any arc be considered as affirmatives its equal arc in a contrary direction will be negative. Let BF and Bh be two equal arcs (Plate I. Fig. 1.); now if the arc BF be considered as positive, the arc By in a contrary direction will be negative. The arcs BF and Bh are equal, and have the same cosine gc, and the same versed sine BG: but the sine hg, lying in a contrary direction to the sine FG, becomes negative, or contrary to the sine FG of the positive arc BF. It is shown in the xıth proposition following, that if a=any arc, tang. A = x rad; cot. A= x rad; and co sine A rad.2 These quantities will consequently have a different sign for a negative arc, to that which they will have +1 +1 1+1 sine A COS. A COS. A sec. A = sine A * For if to any arc BP, there be added one or more circumferences of the circle, they will terminate exactly in the point F, and the augmented arc will have the same positive or negative sine, cosine, &c. with the arc BF. Thus if c denote an entire circumference, or 360°, sine x=sine (c + x)=sine (2c +x)=sine (3c+x), &c. and the same will take place with respect to the cosine, tangent, &c. Legendre's Geometry, 6th edit. page 536. 2 9 COS. A A A rad. for a positive arc; but as sec. A= the secant will have the same sign for a negative arc as it will have for an affirmative arc. (M) If the side, tangent, cotangent, or cosecant of A-B be found at the conclusion of a problem, and the arc a be less than the arc B, the signs of the sine, tangent, cotangent, or cosecant of A-B, will be contrary to those which are given in the foregoing table. It may also be remarked, that in trigonometrical calculations, when a required quantity comes out a sine, the case is frequently ambiguous; because the sine of an arc and the sine of its supplement are equal, and have the same sign. But when the required quantity is expressed by a cosine, tangent, or a cotangent, there is no ambiguity, for a positive cosine, tangent, or cotangent, shows the arc to be less than 90°; and a negative cosine, tangent, or cotangent, denotes an arc between 90° and 180°. (N) Since the sine of an arc=sine of its supplement, tangent of an arc=tangent of its supplement, &c.; if a represent any arc less than 90°, we shall have Sine (90° + A) = sine (90° A) = cos. A) = cosec. A A) = tang. A (180° - A) A) Cot. A = - cot. (180° A) (O) The cosine of half an arc is equal to the sine of half the supplement of that arc, or the sine of half an arc is equal to the cosine of half the supplement of that arc, &c. for the tangent and secant. For if La be substituted for a (N. 101.) we shall have, 180°Cos. A = sine (90° – JA) = sine 2 180° - A (90° 2 1849 Cot. A = tang. (90° JA) = tang 2 180° Tang. ŽA = cot. (90° - Ja) LA) = cot. 2 (90° A (90° A A = COS. -A A |