EXAMPLE I. Let Fig. 1, Plate 54, be the curve proposed, whose equidistant ordinates, AB, CD, EF, GH, IK, LM, and NO, are respectively 5ft. 5ft. 6in. 6ft. 7ft. 9ft. and 8ft. and the distance of AC, CE, EG, or CI, is 3ft. required the area of the Plate 54, Fig. 2, let ABCD be a circle, whose diameter AC, or BD, is 10ft. it is required to find the area by means of equidistant ordinates, marked 3ft. 4ft. 4'5ft. 4'9ft. and 5ft. being at the distance of 1 foot from each other. 0 5 25 - 2.5 half sum of the outside ordinates. 3 4 4:5 4.9 189 area of one quarter. 756 feet, area of the whole. If the diameter, which is 10 feet, be multiplied by 7854, the product, 78-54, will be the area. From hence it appears, that this mode of operation, by means of equidistant ordinates, is exceedingly near the truth in measuring irregular planes; for it will produce the area of a circle, which is one of the most oblique curves possible, as the ends raise quite perpendicular to the axis, from only 10 equidistant spaces within the part of the truth; and would be still nearer when applied to measuring any plane surface, where it is bounded partly by concave and partly by convex curves: because, if wholly bounded by a convex curve, or curves, the area will be something less than the truth; but if bounded by a concave curve, or curves, the area will be something greater than the truth; and if the extremities of the ordinates are joined by straight lines, the area so 'found will be exactly true: but the following is a method of approximation still nearer to the truth, whether the curve be concave or convex to the axis. METHOD II. Divide the given curve, by ordinates, into any even number of equal parts; then add into one sum four times the sum of all the even ordinates; twice the sum of all the odd ordinates, except the first and last, and also the first and last ordinates; and if one third of that sum is multiplied by the common distance between any two ordinates, the product will be the answer, EXAMPLE I. Let Fig. 1, Plate 54, be a curve of any kind, whose equidistant ordinates, AB, CD, EF, GH, IK, LM, and NO, are respectively 5ft. 5ft. 6in. 6ft. 7ft. 9ft. 10ft. and 8ft. and the distance between the ordinates is 3ft. required the area of the curte. CD, GH, and LM, will be the even ordinates; that is, the second, fourth, and sixth; EF, and IK, the odd ordinates, that is, the third and fifth; AB, and NO, the first and last. 4 four times the sum of the even ordinates. O twice the sum of the odd ordinates. 90 30 5 0 first ordinate. Now by comparing this area, viz. 133 feet, with the area found in Method I. Example I. viz. 132 feet, there appears to be a difference of 1 foot; but this last method is the most correct. EXAMPLE II. Plate 54, Fig. 3, let ACEGILN be a concave curve, whose equidistant ordinates, A, BC, DE, FG, HI, KL, and MN, are respectively 0, 1, 3, 6, 10, 15, 21, and the common distance 2; required the area. 88 four times the sum of the even ordinates. EXAMPLE III. Plate 54, Fig. 4, let AHI be a parabola, whose ordinates, A, BC, DE, FG, and HI, are respectively 0, 7, 12, 15, and 16, and their common distance 6; required the area of the $8 four times the sum of the even ordinates. 3)128 424 6 256 the true area. There is another method for finding the areas of curvi lineal 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 |