There is ancther method for finding the areas of curvilincar spaces, besides what has already been shown, which is as follows : divide the sum of all the ordinates by the number of them, for a mean breadth, which is to be multiplied by the length for the content; but this rule is a very false one; it gives the area by far too great when the curve is concave, and by far too small when it is convex, and will not give the true area in any case whatever, except the curve become a straight line ; in which case, all the other rules will coincide with it. But in order to show the falsity of this rule, suppose it were required to find the contents of the same figure, as in the last example, then the sum of all the ordinates, viz. 0+7+12+15+16=50, their number is 5; and 50 divided by 5, is equal to 10 : this being multiplied by the whole length, viz. 24, gives 240 instead of 256, the exact area of the curve found by the last example ; the difference of this example being too small by 1•16th part of the true area, which is very considerable. Prob. 2. To find the superficial content of a mixed figure, partly a curve, and partly right-lined. Find the area of the curved part of the figure by the last problem, by dividing it into equidistant ordinates ; divide the right-lined parts of the figure by ordinates drawn through every angle, which will die vide the right-lined part of the figure into trapezoids and triangles ; find the area of each part, according to their respective rules ; add the areas of all the parts together, and the sum will give the area of the whole figure. Er. 1. Let ABKNORS be the figure proposed to find its area. As the end AB turns round very perpendicular to the base AS, draw the ordinate CB in such a manner, as it may cut off the most perpendicular part of the curve AB at the end, and divide it by ordinates, which are respectively 1, 2, 14, 1, 0, at the distance of 3 from each other, the part CBKL of the curvilineal space is also divided into four equal parts, between the first and last ordinates BC, KL, by the ordinates ED, GF, HI, and LK, which are respectively 12, 13, 12, 10, and 9, and their common distance 4; the other parts of the figure are divided into three trapezoids, KLMN, MNOP, OPQR, and the triangle QRS, by ordinates from the angles at K, N, O, and R; the whole figure being thus prepared, by dividing it into curvilineal spaces, trapezoids, and a triangle, each part will be measured according to their respective rules. The measures or dimensions are marked on their respective places on the figure ; the contents of each part is computed separately, as is shown in the following operation. 23 2)25 124 12 92 4 3 24 1 12 50 area of the trapezoid 9 KLMN 3)16 3)137 45.6 4 182:6 area of the part CBKL. 2014 7 7 182.6 50 14.5 131 49 area of the tri. 43:5 3 angle QRS. 27.0 49'0 43.5 area of 27 area of OPQR. MNOP. 368°1 sum of the areas, or contents of the whole figure. Prob. 3. To measure any irregular figure, bounded wholly by right ines. Divide the whole figure into trapeziums and triangles, divide each trapezium into two triangles, by means of diagonals from the other angles, let fall the perpendiculars to the diagonals; then, to find the area of any of the trapeziums, multiply the balf sum of the two perpendiculars in that trapezium by its diagonal ; find the area or content of all the trapeziums in this manner, and of the triangles, if any; add their several areas together, and the sum will give the area of the whole figure. Er. Let ABCDEFG, fig. 6, be the figure proposed, which is divided into two trapeziums ABFG, BCEF, and a triangle CDE; the diagonal BG of the trapezium ABFG is 16, and the two perpendiculars AH and IF are respectively 4 and 8; the diagonal BE of the trapezium BCEF is 10, and the perpendiculars KF and CL are respectively 14 and 6; the base CE of the triangle CDE is 8, and its perpendicular 5; required the area of the whole figure ABCDEFG. 81 a 100 100, the area of the trapezium BCDF. 20, the area of the triangle FDE. And 216, the area of the whole figure ABCDEFG. Prob. 4. To find the solidity of a solid, by means of equidistant rections or planes. Divide the length of the solid into any even number of equal parts; find the area of all the parallel sections passing through these parts perpendicular to the axis of the solid ; then, to four times the sum of the areas of all the even planes, add twice the sum of the areas of all the odd planes, except the first and last, and the areas of the two ends ; divide the sum by 3, multiply the quotient by the common distance, and the product will give the solidity. N.B. If the sections are circular, the rule may be as follows: to four times the area of the even planes, add twice the sun of the areas of the odd planes, excepting the first and last, and the sum of the areas of the ends, or the first and last planes ; then multiply the sum by 8618 equal .??, and that pro Juct by the common distance, and it will give the solidity sought. Scholium. This Problem is accurately true for parabolic curves, or solids generated from the revolution of conic sections, or right lines. For all kinde of pyramids, or frustums of pyramids, or any orber kind of areas and solidities, it is a very near approximation. It is evident, that the greater the number of ordinates or sections used, the more accurate will the area or solidity be determined : but in praca tice a few sections will be found sufficieut to answer the purpose. This is the best method that possibly van be devised for the practice of gauging, or for measuring curved timber trees, or the like unequally thick ; or any kind of carved solids whatever, generated about an axis ; and when all other methods fail, this is the only one that can be depended upon for its accuracy. OF MEASURING TIMBER. Prob. 1. To measure timber scantling. Find the area at either end, and multiply it by the lengtu, will give the solidity. Er. I. Suppose a joist is fin. by gin. and 8ft. long, what is the solidity ? in. 4 9 3i Oii 8 2 O feet, the solidity. Ex. 2. What is the solidity of a joist 3 in. by gin, and 10f. long? i. ii. 3 6 gi. 2i. 7ii. iif. 10 feet 2 2 3 Qüi. Prob. 2. To measure timber-trees, or unsquared timber, equally unck. Method 1. Multiply the square of of the tree's compass for the side of a square equal in area to the end of the tree, by the length of the tree, and the product will give the solidity. This method, though easy in practice, is very erroneous in principle, as the content by this rule is too small by above one-fourth of itself. The true rules for measuring round timber, have been already given for measuring a cylinder : but if this rule should be thought trouble. some, the following is a method which will come very near the truth, and nearly as expeditious in practice as the above method, and therefore may be esteemed true. Melhod 2. Multiply the square of one-fifth of the tree's compass by the length of the tree, and double the product will be the content. MISCELLANEOUS QUESTIONS. 1. What difference is there between a floor 48 feet long, and 30 teet broad, and two others each of balf the dimensions ? Ans. 720 feet. 2. From a mabogany plank 26 inches broad, a yard and a half is to be sawed off; at what distance from the end must the line be struck ? Ans 6 23 feet. 3. The sides of three squares being 4, 5, and 6 feet, respectively, it is required to find the side of a square that shall be equal in area io all the three. Ans. 807749 feet. 4. What quantity of canvas will be necessary for forming a conical tent, whose height is 8 feet, and the diameter at bottom 13 feet? Ans. 210 square yards. per cwt. 3. How many square feet of board are required to make a rectaogulur box, whose length is to be 31 feet, breadth 2 feet, and depth 20 inches ? Ans. 32 feet. 6. A joist is 8 inches deep, and 3 broad; what will be the dimensions of a scantling just as big again as the joist, that is 4* inches broad? Ans. 12:52 inches deep. 7. A roof which is 24 feet 8 inches by 14 feet 6 inches, is to be covered with lead at 8 lbs to the foot ; what will it come to at 185. Ans. 221. 198. 10:d. 8. If the side of an equilateral triangle be 10 chains, what will be the side of another equilateral triangle, whose area is one-fourth of the former? Ans. 5 chains. 9. What is the side of that equilateral triangle whose area cost as much paving at 8d. per foot, as the pallisading the three sides did at a guinea a yard ? Ans. 72-746 feet. 10. What would a circular reservoir, hose diameter at top is 40 yards, at bottom 389 yards, and its side, or slant depth 11 feet, cost lining with brick-work, at 3s. 10d. the square yard? Ans. 3111. 18s. 2d. 11. The four sides of a field, whose diagonals are equal to each other, are 25, 35, 31, and 19 poles, respectively; what is the area? Ans. 4 acres, 1 rood, 39 poles. 12. What is the length of a cord that will cut off one-third of the area from a circle whose diameter is 289 ? Ans. 278 6716. 13. A cable which is 3 feet long, and 9 inches in compass, weighs 22 lbs. what will a fathom of that cable weigh whose diameter is 9 inches : Ans. 434.26 lbs. 14. A circular fish-pond is to be dug in a garden, that shall take up just half an acre; what must the length of the chord be that strikes the circle ? Aus. 27 75 yards. 15. A carpenter is to put an oaken curb to a round well, at 8d. per square foot; the breadih of the curb is to be 7 inches, and the diameter within 34 feet; what will be the expense : Ans. 5s. 2 d. 16. Suppose the expense of paving a semicircular plot, at 2s. 4d. per foot, amounted to 10l. what is the diameter of it ? Aus. 14.7737. 17. Seven men bought a grinding-stone of 60 inches in diameter, each paying one-seventh part of the expense; what part of the diameter must each grind down for his share ? Ans. first, 4:4509; second, 4 8400; third, 5.3535; fourth, 6 0765 ; fitth, 7-2079; sixth, 9:3935; and the seventh, 22:6778. 18. A garden 100 feet long, and 80 feet broad, is to have a gravel walk of an equal width half round it; what must the width of the walk be so as to take up just half the ground ? Ans. 25.968 feet. 19. How many gallons, wine measure, will a cistern hold, supposing its length and breadth at top to be 5 and 4 feet respectively, and at bottom 4 and 3 feet; the perpendicular depth being 3 feet? Ans. 414 4 gallons. 20. A malster has a kiln that is 16 feet 6 inches square, which he |