The first, however, who ascertained this ratio to any great degree of ex. actness, was Ludolph Van Ceulen, a Dutchman. He found that if the diameter of a circle be 1, the circumference will be 3.141592653589793 238462643. 383279502884 nearly, which is true to 36 places of decimals. This was thought so extraordinary a performance, that the numbers were cut on his tomb-stone, in St. Peter's church-yard, at Leyden. Since the invention of fluxions, by the illustrious Sir Isaac Newton, the squaring of the circle has become more easy; and the late ingenious Mr. Abraham Sharp, of Little Horton, near Bradford, in Yorkshire, has not only confirmed Ceulen's ratio, but extended it to 72 places of decimals. Mr. John Machin, professor of Astronomy in Gresham College, London, has also given us a quadrature of the circle, which is true to 100 places of figures; and even this has been extended, by the French mathematicians, to 128. 2. The first Rule is the proportion of Archimedes; the second that of Vieta and Metius; and the third is an abridgement of Van Ceulen's ratio. This Rule is not quite so accurate as the second; but is most commonly used, as being more convenient, and in most cases, correct enough for prac tice. EXAMPLES. 1. If the diameter AB of a circle be 12; what is the circumference ? Here 3.1416 x 12 = 37.6992, the circumference required. 2. If the circumference of a circle be 45; what is the diameter ? By Rule I. As 22 7 45 14.318181, the diameter required. By Rule II. As 355 113: 45: 14.323943, the diameter required. By Rule III. Here 45 3.1416 14.323911, the diameter required. 3. If the diameter of a well be 3 feet 9 inches; what is its circumference ? Ans. 11 ft. 9 in. 4 pa. 4. The diameter of a circular plantation is 100 yards; what did it cost fencing round, at 6s. 9d. per rood? Ans. £15. 2s. 111d. 5. What is the diameter of a stone column whose circumference measures 9 feet 6 inches? Ans. 3 ft. 0 in. 3 pa. 6. The circumference of the earth is 25000 miles; what is its diameter, supposing it a perfect sphere? Ans. 7957.72854 miles. 7. The diameter of the sun is 883220 miles; what is his circumference? Ans. 2774723.952 miles. 8. The circumference of the moon is 6850 miles; what is her diameter ? Ans. 2180.41762 miles. 9. The diameter of Venus is 7680 miles, what is her circumference? Ans. 24127.488 miles. Note, Those who are desirous of making themselves acquainted with the method of finding the distances of the sun, moon, and planets from the earth; and also their diameters, are referred to Martin's Trigonometry, Vol. I, page 203; Ferguson's Astronomy, page 100; and Bonnycastle's Astronomy, page 277. PROBLEM XIV. To find the length of any arc of a circle. RULE I. From 8 times the chord of half the arc subtract the chord of the whole arc, and of the remainder. will be the length of the arc, nearly. Note. Half the chord of the whole arc, the chord of half the arc, and the versed sine are sides of a right-angled triangle; any two of which being given, the third may be found by Problem VI. EXAMPLES. 1. The chord AB of the whole arc is 24, and the versed sine CD 9; what is the length of the arc ACB? Here 122 +92 = 144 + 81 = 225; and ✅225 = 15 = AC, the chord of half the arc; then 120-24 = 96 15 X 8-24 = 32, the length of the arc required. 3 2. The chord of the whole arc is 45, and the chord of half the arc is 25.5; what is the length of the arc? Ans. 53. 3. The chord of half the arc is 21.25, and the versed sine 10; what is the length of the arc? 4. The chord of the whole arc is 30, sine 8; what is the length of the arc ? RULE II. Ans. 44.1666. and the versed Ans. 35. Divide the square of half the chord by the versed sine; to the quotient add the versed sine, and the sum will be the diameter. Subtract divide 2 3 41 50 of the versed sine from the diameter; of the versed sine by the remainder; and to to the quotient last found, add 1; then this sum being multiplied by the chord of the whole arc, will give the length of the arc, nearly. Note. When great accuracy is required, the second Rule should be used, as the first gives the length of the arc too little; though near enough for most cases in practice. EXAMPLES. 1. The chord AB is 24, and the versed sine CD 9; what is the length of the arc ACB? 122 Here +9== +9=16+9= 25 = the di 9 144 9 2 ameter CE; and 9 x 3 -(25 - 9 x 41) =6÷ 25 — 7.38 = 6 ÷ 17.62 = .34052; then 1 + .34052 × 24 = 1.34052 × 24 = 32.17248, the length of the arc required. 2. The upper part or head of a window is the segment of a circle, whose chord line measures 6 feet 9 inches, and versed sine 2 feet 6 inches; what is the length of the arch? Ans. 8 ft. 11 in. 11 pa. 3. What is the length of the circular arch of a bridge, the span of which is 15 feet 6 inches, and height above the top of the piers, 6 feet 9 inches? Ans. 22 ft. 4 in. 9 pa. 4. If the span of the circular roof of a cellar be 21 feet 9 inches; what is the length of the arch, its height being 6 feet 6 inches? Ans. 26 ft. 7 in. 5 pa. 5. If the span of a circular pediment is 18 feet 6 inches; what is the length of the arch, its height above the top of the entablature being 4 feet 6 inches? Ans. 21 ft. 3 in. 7 pa. RULE III. As 180 is to the number of degrees in the arc, so is 3.1416 times the radius to its length. Or, multiply together the number of degrees in the arc, the radius, and the number .01745329; and the product will be the length of the arc. Note. The length of the arc of a semicircle, a quadrant, &c. may also be found by taking one half, one fourth, &c. of the whole circumference, EXAMPLES. 1. Required the length of an arc of 46 degrees 35 minutes, the radius being 12 feet. As 180° 46° 35′:: 3.1416 x 12: 9.75641, the length of the arc required. 2. What is the length of the semi-circular arch of a bridge, the span of which is 18 feet 6 inches? Ans. 29 ft. 0 in. 8 pa. 3. The radius of a cart wheel, from the centre of the nave to the outside of the felloe, is 2 feet 3 inches; what is the length of one-sixth of the rim or circumference? Ans. 2 ft. 4 in. 3 pa. Note. Those who wish to understand the principles upon which carriage wheels should be made, so that the carriage may be drawn by the least power, may consult Marrat's Mechanics, Article 788. PROBLEM XV. To find the area of a circle. 1. Multiply half the circumference by half the diameter, and the product will be the area. Or, divide the product of the whole circumference and diameter by 4, and the quotient will be the area. 2. Multiply the square of the diameter by .7854, and the product will be the area. 3. Multiply the square of the circumference by .07958, and the product will be the area. Note 1. If the area of a circle be divided by .7854, the quotient will be the square of the diameter. 2. A circle may be considered as a regular polygon of an infinite number of sides, the perimeter of which being equal to the circumference, and the perpendicular equal to the radius; consequently, by Problem XI. the area of a circle is equal to balf the circumference multiplied by half the diameter. Also, the area of a circle whose diameter is 1, is .7854 nearly; and by Theo. 18, circles are to each other as the squares of their diameters; hence we derive the second rule. The area of a circle whose circumference is 1, is .07958; and the areas of circles are to each other as the squares of their circumferences hence our third rule. The following Rules will solve most of the useful Problems relating to the circle and its equal or inscribed square, &c. Rule 1. The diameter of a circle multiplied by .8862269, will give the side of a square equal in area. 2. The circumference of a circle multiplied by .2820948, will give the side of a square equal in area. 3. The diameter of a circle multiplied by .7071068, will give the side of the inscribed square. 4. The circumference of a circle multiplied by .2250791, will give the side of the inscribed square. 5. The area of a circle multiplied by .6366197, and the square root of the product extracted, will give the side of the inscribed square. 6. The side of a square multiplied by 1.414214, will give the diameter of its circumscribing circle. G |