Ex. 3. To find the content of a conic frustum, the altitude being 18, the greatest diameter 8, and the least diameter 4. Ans. 527.7888. Ex. 4. What is the solidity of the frustum of a cone, the altitude being 25, also the circumference at the greater end being 20, and at the less end 10? Ans. 464216. Ex. 5. If a cask, which is two equal conic frustums joined together at the bases, have its bung diameter 23 inches, the head diameter 20 inches, and length 40 inches; how many gallons of wine will it hold? Ans. 79.0613. PROBLEM VII. To find the Surface of a Sphere, or any Segment. Rule I. MULTIPLY the circumference of the sphere by its diameter, and the product will be the whole surface of it *. RULE II. * These rules come from the following theorems for the surface of a sphere, viz. That the said surface is equal to the curve surface of its circumscribing cylinder; or that it is equal to 4 great circles of the same sphere, or of the same diameter : which are thus proved. Let ABCD be a cylinder, circumscribing the sphere EFGH ; the former generated by the rotation of the rectangle FBCH - 培 M about the axis or diameter FH; and the E latter by the rotation of the semicircle FGH about the same diameter Fh. Draw two lines KL, MN, perpendicular to the axis, intercepting the parts LN, OP, of the D) cylinder and sphere; then will the ring or cylindric surface generated by the rotation of Ln, be equal to the ring or spherical surface generated by the arc op. For first, suppose the parallels ki and MN to be indefinitely near together; drawing lo, and also og parallel to LN, Then, the two triangles iko, 09P, being equiangular, it is, as OP: 0:9 or UN: : 10 or Kl: KO :: circumference described by Kl: circumf. described by KO; therefore the rectangle op X circumf, of ko is equal to the rectangle LN X circumf. of KL; that is, the ring desc. ibed by Op on the sphere, is equal to the ring described by Ln on the cylinder. VOL. II. E And Rule II. Square the diameter and multiply that square by 3.1416, for the surface. Rule III. Square the circumference; then either multiply that square by the decimal •3183, or divide it by 3.1416, for the surface. Note. For the surface of a segment or frustum, multiply the whole circumference of the sphere by the height of the part required. Ex. 1. Required the convex superficies of a sphere, whose diameter is 7, and circumference 22. Ans. 154. Ex. 2. Required the superficies of a globe, whose diameter is 24 inches. Ans. 1809:5616. Ex. 3. Required the area of the whole surface of the earth, its diameter being 7957 miles, and its circumference 25000 miles. Ans. 198943750 sq. miles. Ex. 4. The axis of a sphere being 42 inches, what is the convex superficies of the segment whose height is 9 inches? Ans. 1187:52 18 inches. Ex. 5. Required the convex surface of a spherical zone, whose breadth or height is 2 feet, and cut from a sphere of 12, feet diameter. Ans. 78.54 feet, And as this is every where the case, therefore the sums of any corresponding number of these are also equal; that is, the whole surface of the sphere, described by the whole semicircle FGH, is equal to the whole curve surface of the cylinder, described by the height BC; as well as the surface of any segment described by Fo, equal to the surface of the corresponding segment described by BL. Corol. 1. Hence the surface of the sphere is equal to 4 of its great circles, or equal to the circumference EFGH, or of DC, multiplied by the height bc, or by the diameter fH. Corol. 2. Hence also, the surface of any such part as a segment or frustun, or zone, is equal to the same circumference of the sphere, multiplied by the height of the said part. And consequently such spherical curve surfaces are to one another in the same proportion as their altitudes. PROBLEM PROBLEM VIII. To find the Solidity of a Sphere or Globe. Rule I. Multiply the surface by the diameter, and take of the product for the content* Or, which is the same thing, multiply the square of the diameter by the circumference, and take of the product. Rule II: Take the cube of the diameter, and multiply it by the decimal •5236, for the content. RULE III. Cube the circumference, and multiply by 01688. Ex. 1. To find the content of a sphere whose axis is 12. Ans. 904•7808. Ex. 2. To find the solid content of the globe of the earth, supposing its circumference to be 25000 miles. Ans. 263750000000 miles. PROBLEM IX, To find the Solid Content of a Spherical Segment. + Rule 1. From 3 times the diameter of the sphere take * For, put d = the diameter, c = the circumference, and s = the surface of the sphere, or of its circumscribing cylinder; also, a = the number 3.1416. Then, Is is = the base of the cylinder, or one great circle of the sphere; and d is the height of the cylinder; therefore ds is the content of the cylinder. But of the cylinder is the sphere, by th. 117, Geom. that is, of ds, or ds is the sphere; which is the first rule. Again, because the surface s is = all'; therefore ds = ad =5236d3, is the content, as in the 2d rule. Also, d being=c4, therefore ad = fcu=01688, the 3d rule for the content. † By corol. 3, of theor. 117, Geom, it appears that the spheric segment PFN, is equal to the difference between the cylinder ABLO, and the conic frustum ABMO. But, putting d = AB or FH the diameter of the sphere or cylinder, h = FK the height of the segment, r = PK the radius of its base, and a = 3.1416; then the content of the cone Abi is = Lad? * {F1 = zad> ; and by the similar cones ABI, OMI, as E 2 B OH take double the height of the segment; then multiply the remainder by the square of the height, and the product by the decimal .5236, for the content. Rule II. To 3 times the square of the radius of the segment's base, add the square of its height; then multiply the sum by the height, and the product by ·5236, for the content. Ex. 1. To find the content of a spherical segment, of 2 feet in height, cut from a sphere of 8 feet diameter. Ans, 41.888. Ex. 2. What is the solidity of the segment of a sphere, its height being 9, and the diameter of its base 20 ? Ans, 1795.4244. Note. The general rules for measuring all sorts of figures having been now delivered, we may next proceed to apply them to the several practical uses in life, as follows. 13 : KI :: 21ad: 2 ad X hd - h -)3 = the cone QMI;. d therefore the cone ABI the cone QMI = zadi 24ad3 x hd - h ( J'adol-adh + ah is the conic frustum ABMO. dd And ad-h is = the cylinder ABLO. Then the difference of these two is žadh- Lah fah2 x (30 - 2h), for the spheric segment PFN; which is the first rule. Again, because PKR = FK X KH (cor. to theor. 87, Geom.) 7.2 or m2 =h (d – h), therefore d = th, and 3d h 372 + 12 ths -; which being substituted in the former h h 3r the rule, it becomes Jah? x = sah' X (3r2 + ho), which is the 2d rule. Note. By subtracting a segment from a half sphere, or from another segment, the content of any frustum or zone may be found. 2h Зr? LAND LAND SURVEYING. SECTION I. 1. OF THE CHAIN. LAND is measured with a chain, called Gunter's Chain, from its inventor, the length of which is 4 poles, or 22 yards, or 66 feet. It consists of 100 equal links; and the length of each link is therefore 6 of a yard, or 1 of a foot, or 7.92 inches. Land is estimated in acres, roods, and perches. roods, and perches. An acre is equal to 10 square chains, or as much as 10 chains in length and I chain in breadth. Or, in yards, it is 220 x 22=4840 square yards. Or, in poles, it is 40 X 4=160 square poles. Or, in links, it is 1000 x 100 = 100000 square links: these being all the same quantity. Also, an acre is divided into 4 parts called roods, and a rood into 40 parts called perches, which are square poles, or the square of a pole of 54 yards long, or the square of of a chain, or of 25 links, which is 625 square links. So that the divisions of land measure, will be thus: 625 sq. links = 1 pole or perch 4 roods El acre. Exam. Suppose the length of a rectangular piece of ground be 792 links, and its breadth 385; to find the area in acres, Toods, and perches. 792 3.04920 385 4 1 3960 •19680 6336 40 7.87200 2. Or |