v = {32806.6 r i + 0.023751-0.154113 These formulæ have been simplified, and are tolerably correct. v=50 d nearly the velocity in feet per second. {2} nearly the velocity Let D be the discharge per minute in cubic feet, then D=2356 d 2 ›(dd) $ (13.) (14.) (15.) To find the fall in a river caused by obstruction, such as the piers of a bridge, &c. Let v be the velocity of the stream in feet per second, b the whole breadth of the channel in feet, c the contracted breadth between the obstacles, and ƒ the fall, then 21 v2 1.42 62-c2 Let, as is nearly the case with the old London Bridge, v=31, b=926, c=200, Hencef = 1.42 62-c2 64 c2 Xv2=0.46 × 10 -4.73 feet, or 4 feet 8 inches by the formula, while that by experiment was 4 feet 9 inches. For a more extensive collection of these formulæ, see Dr Gregory's Mathematics for Practical Men. TO FIND THE TONNAGE OF A SHIP BY LOGARITHMS, ACCORDING TO THE COMMON METHOD. Rule. If the vessel is a ship of war, let fall a perpendicular from the fore-side of the stem, at the height of the hause-holes; but if a merchantman, the perpendicular is to be let fall from that part of the fore-side of the stem which is at the same height above the keel as the wing-transom: also let fall another perpendicular from the back of the main post, at the height of the wing-transom. Find the distance between these two perpendiculars, from which subtract three-fifths of the extreme breadth; and also the product of the height of the wing-transom above the upper edge of the keel, by 21 inches, and the remainder is the length of the keel for tonnage. To the logarithm of which add the logarithm of the breadth, and that of the half-breadth, and the constant logarithm 8.02687;* the sum, rejecting 10 from the index, will be the logarithm of the tonnage required. Ex-Let the length between the perpendicular at the fore-part of the stem, and the back of the post, be 100 feet: the extreme breadth 27 feet, and the height of the wing-transom 15 feet. Required the tonnage ?-Ans. 321 tons. The arithmetical complement of the logarithm of 94 being the common divisor for finding the tonnage. This method is far from being correct. See papers on Naval Architecture, published by Morgan and Creuze. G. B. Whittaker, London, 1826. 1. Since a cubic foot of water, at the temperature of 40° Fahrenheit, weighs 1000 ounces avoirdupois, or 62 pounds, the numbers in the preceding tables, omitting the decimal points, exhibit very nearly the respective weights of a cubic foot of the several substances in avoirdupois ounces. 2. If the weight of a body be known in avoirdupois ounces, its weight in Troy ounces will be found in multiplying it into 91145, And, if the weight be given in Troy ounces, it will be found in avoirdupois by multiplying it into 1.0971. GASES. Atmospheric air* 1.0000 Muriatic acid-gas Vapour of hydriotic ether 5-4749 Sulphuretted hydrogen oil of turpentine 5.0130 Oxygen-gas Hydriotic acid-gas 4-4430 Nitrous-gas 1.0288 Fluo-silicic acid-gas 3-5735 Olefiant-gas 0-9784 Chlorine Vapour of sulph. of carbon 2.6447 Azote, or nitrogen-gas sulphuric ether 2.5860 Oxide of carbon 2-4700 Hydro-cyanic vapour 0.9691 0.9569 0.9476 2-3709 Phosphoretted hydrogen 0.8700 0.6235 2.1920 Ammoniacal-gas 0.5967 1.8064 Carburetted hydrogen 0.5550 Vapour of absolute alcohol 1.6133 Arseniated hydrogen 0.5290 Nitrous oxide 1-5204 Hydrogen-gas 0.0732 1.5196 0.00122 water being Carbonic acid Specific Gravity of Distilled Water at different Temperatures, that at 62° being taken as Unity. 70° 0.99913 | 62° | 1·00000 | 54° 1.00064 | 26° | 46° | 1·00102 | 34° 68 0.99936 60 1·00018 52 1.00076 28 44′ 1·00107 36 66 0-99958 58 1.00035 50 1.00087 30 42 64 0.99980 56 1.00050 48 1.00095 32 40 1.00111 | 38 100113 40 MISCELLANEOUS COMPUTATIONS AND EXPERIMENTS. The pendulum vibrating seconds of mean solar time at London in a vacuum, and reduced to the level of the sea, is 39·1393 inches; consequently the descent of a heavy body from rest in one second of time, in a vacuum, will be 193.145 inches. The logarithm 2-2858835. A platina metre at the temperature of 32°, supposed to be the ten millionth part of the quadrant of the meridian, 39.3708 inches. The ratio to the imperial measure of three feet, as 1·09363 to 1, the logarithm 0.0388717. The following standards, accurately measured, give these results :General Lambton's scale, used in the Trigonometrical Survey of India, 35-99934 inches. Sir G. Shuckburgh's scale (which, for all purposes, may be considered as identical with the impe-35.99998 rial standard) Gen. Roy's scale Royal Society's standard Ramsden's bar 36.00088 36.00135 36.00249 Weight of a cubic inch of distilled water in a vacuum at the temp. 62° Fahrenheit, bar. 30 inches, as opposed to brass weights in a vacuum also, 252.766 grains Consequently a cubic foot 62.3971 pounds avoirdupois in vacuo Weight of a cubic inch of distilled water in air at -log 2.4027186 -log 1.7951644 62° of temperature, height of the barometer 30 log 2.4021892 inches, 252.458 grains Consequently a cubic foot 62.3211 pounds avoir- And an ounce of water 1.732966 cubic inches log 1.7946351 log 0.2387886 log 2.4429092 -log 1.2739096 SPECIFIC GRAVITY OF DRY AND SATURATED AIR. That at 30 in Bar., and 32° Fahr. being 1. On this subject see Biot's Traité de Physique, vol. I., ch. xix. |