41!!. To find the Altitude of one Station above another, from the Temperature of the Boiling of Water. This method is not so reliable as that by barometrical observations, although Colonel Sykes, in Australia, has found altitudes above the sea agree with those found by triangulation closer than he had anticipated. There are very valuable tables in the Smithsonian Institute's Meteorological and Physical Tables Tables XXIV, XXV, and XXVI — for finding the altitudes by this method. Take any tin pot and lay a piece of board across the top, having groove to receive the thermometer, and a button or slide to keep it steady, at about two inches from the bottom. Take several observations, carefully noting them, and at the same time the temperature of the surrounding air. Use Fahrenheit's thermometer. Example.—Boiling point, upper station, 209°, lower, 202°; temperature of the air at upper station, 72°, lower, 84°, mean temperature, 78°. From Table A, 209°. Alt., 1534 ft. 202. 5185 Approximate height, 3651 Mean temperature, 78°. Multiplier from Table B, 1096 Product, 4001 ft. Where the degrees are taken to tenths, then we interpolate. 419a.—Continued from Sec. 410. Having one side, A B, and the adjacent angles,—to find the area--Let the triangle ABC (Fig. 68,) be the triangle; the side A B s, and the angles A and B are given, also the angle C. 420. From a point, P, within a given figure, to draw a line cutting off any part of it by the line F G.-Let the figure IGBAE the required area. (See Fig. 69.) Let the ABCDEF the tract be plotted on a scale of ten feet to an inch, from which we can find the position of the required line very nearly, with reference to the points B and E. Run the assumed line, A S, through P, finding the distances A P 11 and PS n, also the angles P T A, PSG, and that the tract AS BAT is too great, by the area d. Hence the true line, T P G, must be such that the triangle PSG - PAT = d. Assume the angle SPG P, then we find the angles T and G, and by Sec. 409 we find the areas of the triangles PSG and PA F. the difference is not d, again, calculate the sides PG and P T. 420a. From the triangle A B C to cut off a given area (say one-third,) by a line drawn through the given point, D. (Fig. 69a.) Through D draw the line D G parallel to A C. If Now all the angles at A, B, and C are given, and the line DG is given to find the point I or H, through which, and the given point D, the line I DH will cut off the triangle A HI to one-third the area (Fig. 69a.) Make A F one-third of A C, then one-third of the triangle A B C, which is to be of the triangle ABC. HG: GD :: I A2 : A B × A F, and by Euclid, 6-16, G D x H A2 HGA BAF (II A ས་ AG). AF. AB Now we have P and A G given, to find A H or A I, A H = 1⁄2 P + (1⁄44 P 2 AG × P') 1⁄2, when I) is inside the triangle. A H 1⁄2 P + ( P2 + AG x P), when dis outside. × 421. Through the point D to draw the line G DE so that the triangle BGE will be the least possible. Through D draw HDI parallel to BC, make B H H G, and draw G D E, which is the required line. Geodedical Jurisprudence, p. 72, B. Fig. 69a. Chief Justice Caton's opinion adds the following in support of established lines and monuments:- Dreer v. Carskaddan, 48 Penn. State, 28. Bartlett 7. Hubert, 21 Texas, S. Thomas 7. Patten, 13 Maine, 329. To Divide pro rata. After Bailey 2. Chamblin, 20 Indiana, 33, add Francoise v. Maloney, Illinois, April Term, 1871. 309Me. After English Reports, 42, p. 307, add Knowlton v. Smith, 36 Missouri, 520. Jordan v. Deaton, 23 Arkansas, 704. United States Digest, Vol. 27-where an owner points out a boundary, and allows improvements to be made according to it, cannot be altered when found incorrect by a survey. For Laying Out Curves. Example after p. 72. Let radius 2000 feet; chord, 200; then tangential angle 2° 51′57′′; versed sine at the middle, 2,503 feet. If the ground does not admit of laying off long chord of 200 feet, make 200= 200 half feet 100, then for radius 4000 find the versed sine 1,251 and the tang. angle = 1° 25′ 57′′. 1° 25′ 57′′. If we use the chord of 200 feet, half feet, or links, then we are to take the ordinates in Table C as feet, half feet, or links. Canals. The Illinois and Michigan locks are 128 feet long, 18 feet wide, and 6 feet deep, bottom 36, surface 60, tow-path 15, berm 7, tow-path above water, 3 feet. The New York Canals.-Erie Canal, 363 miles long, when first built, 40 feet at top, 28 at bottom, 4 feet deep, 84 locks, each 90x15, lockage 6SS, 8 large feeders, 18 acqueducts. The acqueduct across the Mohawk is 1188 feet in length. The Pennsylvania Canal-top 40, bottom 28, depth 4, locks 90x15, and some, 90x17. The Ohio and Erie Canal-40 feet at top, 4 feet deep. Rideau Canal, in Canada-1291⁄2 miles long, 53 locks, each 134x33. Welland Canal, in Canada-locks, large enough to admit large vessels. It is now in progress of widening and deepening, so as to admit of the largest vessels that may sail on the lakes, and to correspond with the canals and lakes at Lachine, and on the River St. Lawrence. CORRECTIONS. Page 43, example 2, read the polygon a b c d e f g h, Fig. 38. Page 72153, soda No O read soda Na O. 72B55, 4th line, read felspathic. 72B111, after the Sth line insert Sir William Bland makes it as 17 to 13, egg-shaped. 72s, begin at 8th line from bottom and put mean base = 50 + 40 = 90 72H*, at 16th line from bottom, for r S-b A, read r S+bQ. 72н*10, at 14th from bottom, for product of the adjacent parts, read product of tan of adjacent parts. 72H*24, Sec. 388, for apparent, read mean. 72н*30, by the Heliostat, insert after Heliostat Fig. H. 72R*, under 82°, opposite 48, for 2921 put 9921. 104, under 2, opposite 12, make it 1.9300, and opposite 13 make it |