or end is a, and radius r, moving in the direction of its axis; because then s= 1, and a = pr?, where pe 3.1416 ; then png?r2 will be the resisting force R, and Prava v2 the retarding 45 4gw force f. 402. Corol. 5. This is the value of the resistance when the end of the cylinder is a plane perpendicular to its axis, or to the direction of motion. But were its face an elliptic section, or a conical surface, or any other figure everywhere equally inclined to the axis, or direction of motion, the sine or inclination being s: then, the number of particles of the fluid striking the face being still the same, but the force of each, opposed to the direction of motion, diminished in the duplicate ratio of radius to the sine of inclination, the resist para q2 m2 ing force R would be 4g PROPOSITION LXXIX. 403. The Resistance to a Sphere moving through a Fluid, is but Half the Resistance to its Great Circle, or 10 the End of a Cylinder of the same Diameter, moving with an Equal Velocity. Let any LET AFEB be half the sphere, moving А. in the direction Ceg. Describe the para- H boloid AIEKB on the same base. particle of the medium meet the semicir C cle in F to which draw the tangent FG, the radius Fc, and the ordinate fi! Then the force of any particle on the surface at F, is to its force on the base at n, as the B! square of the sine of the angle G, or its equal the angle Fch, to the square of radius, that is, as HFto cr? Therefore the force of all the particles, or the whole fluid, on the whole surface, is to its force on the circle of the base, as all the hi? to as many times cf?. But cr? is CA? = CB, and AF2 X AH HB by the nature of the circle : also, al . HB : AC. CB:: H1: ce by the nature of the parabola : consequently the force on the spherical surface, is to the force on its circular base, as all the hi's to as many cE's, that is, as the content of the paraboloid to the content of its circumscribed cylinder, namely, as 1 to 2 404. Corol. Hence, the resistance to the sphere is a = pns2 2 , being the half of that of a cylinder of the same diameter AC. . diameter. For example, a 9lb iron ball, whose diameter is 4 inches, when moving through the air with a velocity of 1600 feet per second, would meet a resistance which is equal to a weight of 1323ib, over and above the pressure of the at-, mosphere, for want of the counterpoise behind the wall. PRACTICAL EXERCISES CONCERNING SPECIFIC GRAVITY. The Specific Gravities of Bodies are their relative weights contained under the same given magnitude ; as a cubic foot, or a cubic inch, &c. The specific gravities of several sorts of matter, are expressed by the numbers annexed to their names in the Table of Specific Gravities, at page 211 ; from which the numbers are to be taken, when wanted. Note. The several sorts of wood are supposed to be dry. Also, as a cubic foot of water weighs just 1000 ounces avoirdupois, the numbers in the table express, not only the specific gravities of the several bodies, but also the weight of a cubic foot of each in avoirdupois ounces ; and hence, by proportion, the weight of any other quantity, or the quantity of any other weight, may be known, as in the following problems. PROBLEM I. To find the Magnitude of any Body, from its Weight. As the tabular specific gravity of the body, EXAMPLES, Exam. 1. Required the content of an irregular block of common stone, which weighs Icwt. or 1121b, Ans. 12284 cubic inches. Exam. 2. How many cubic inches of gunpowder are there in 1lb weight? Ans 29 cubic inches nearly. Exam 3. How many cubic feet are there in a ton weight of dry oak? Ans. 3813 cubic feet. PROBLEM PROBLEM II. To find the Weight of a Body from its Magnitude. As one cubic foot, or 1728 cubic inches, EXAMPLES. Exam. 1. Required the weight of a block of marble, whose length is 63 feet, and breadth and thickness each 12 feel ; quing the dimensions of one of the stones in the walls of Balbeck ? Ans. 68375 ton, which is nearly equal to the burden of an East-India ship. Exam 2 What is the weigh: of 1 pint, ale measure, of gunpowder ? Ans. 19 oz. nearly. Exam. 3 What is the weight of a block of dry oak, which measures 10 feet in length, 3 feet broad, and 24 feet deep : Ans. 483541b. PROBLEM III. To find the Specific Gravity of a Body. Case 1. When the body is heavier than water, weigh it both in water and out of water, and take the difference, which will be the weight lost in water. Then say, As the weight lost in water, EXAMPLE. A piece of stone weighed 10lb, but in water only 6 lb, required its specific gravity ? Aus. 2609. Case 2 When the body is lighter than water, so that it will not quite sink, affix to it a piece of another body, heavier than water, so that the mass compounded of the iwo may sink together. Weigh the denser body and the compound mass separately, both in water and out of it; then find how much each Joses in water, by subtracting its weight in water from its weight in air ; and subtract the less of these remainders from the greater. Then say, As As the last remainder, EXAMPLE. Suppose a piece of elm weighs 15lb in air ; and that a piece of copper which weighs 18lb in air, and 161b in water, is affixed to it, and that the compound weighs 6lb in water ; required the specific gravity of the elm ? Ans, 600. PROBLEM IV. To find the Quantities of Two Ingredients, in a Given Compound. Take the three differences of every pair of the three specific gravities, namely, the specific gravities of the compound and each ingredient; and multiply the difference of every two specific gravities by the third. Then say, as the greatest product. is to the whole weight of the compound, so is each of the other products, to the two weights of the ingredients. EXAMPLE. A composition of 112lb being made of tin and copper, whose specific gravity if found to be 8784 ; required the quantity of each ingredient, the specific gravity of tin being 7320, and of copper 9000 ? Ans. there is 100lb of copper in the composition. and consequently 121b of tins OF THE WEIGHT AND DIMENSIONS OF BALLS AND SHELLS. The weight and dimensions of Balls and Shells might be found from the problems last given, concerning specific gravity But they may be found still easier by means of the experimented weight of a ball of a given size, from the known proportion of similar figures, namely, as the cubes of their diametery. PROBLEM I. To find the Weight of an Iron Ball, from its Diameter. An iron ball of 4 inches diameter weighs 9ib, and the weights being as the cubes of the diameters, it will be, as 64 (which I i VOL. II. (which is the cube of 4) is to 9 its weight, so is the cube of the diwmeter, of any other ball, to its weight. Or, take the cube of the diameter, for the weight. Or, take f of the ' cube of the diameter, and } of that again, and add the two together, for the weight. EXAMPLES. Exam. 1. The diameter of an iron shot being 6.7 inches, required its weight? Ans. 42-2941b. Èxam 2. What is the weight of an iron ball, whose diameter is 5.54 inches ? Ans 241b nearly. PROBLEM. II. To find the Weight of a Leaden Ball. A leaden ball of one inch diameter weighs is, of a Ib; therefore as the cube of 1 is to or as 14 is to 3, so is the cube of the diameter of a leaden ball, to its weight. Or, take is of the cube of the diameter, for the weight, nearly. EXAMPLES. Exam. 1. Required the weight of a leaden ball of 6.6 inches diameter ? Ans 61.6061b. Exam. 2. What is the weight of a leaden ball of 5.30inches diameter? Ans. 321b nearly PROBLEM II. To find the Diameter of an Iron Ball . MULTIPLY the weight by 7, and the cube root of the product will be the diameter. EXAMPLES. Exam. 1. Required the diameter of a 421b iron ball? Ans. 6.685 inches: Exam. 2. What is the diameter of a 241b iron ball ? Ans, 5.54 inches. PROBLEM IV. To find the Diameter of a Leaden Ball. Multiply the weight by 14, and divide the product by 3; then the cube root of the quotient will be the di ameter. EXAMPLES |