The work may be shortened by omitting to multiply by 7854 till the difference of the squares of the diameters is found, as below. Set 7854 on A, to 1 on B, then against 700 (the difference of the squares of the diameters) on B, will be found 549.78 for the area of the ring, on A. To find the Area of the Segment of an Ellipsis cut by a right Line parallel to either Axis; the Altitude of the Segment and both Diameters of the Ellipsis being given. CASE 1. When the section is by a right line parallel to the transverse diameter. RULE. Find the area of a circular segment of the same altitude, the diamèter of the circle being equal to the conjugate diameter of the ellipsis. Then multiply this area by the transverse axis, and divide the product by the conjugate, for the area of the elliptical segment. EXAMPLE. Let ABC be a segment of an ellipsis cut by a right line parallel to the longer axis DE; and let BG (the altitude of the segment) be 6 inches, the conjugate diameter BF of the ellipsis being 30 inches, and the transverse diameter DE, 50 inches; the area of the segment ABC is demanded. The altitude 6 divided by the diameter 30 of the inscribed circle, quotes 2 for the versed sine with which to enter the Table of Circular Segments; and the area corresponding is 111823 tabular number Multiply by 900 square of BF. Product 100-6407|00 area of HBI. Multiply by 50 transverse diameter Conjugate 30) 5032-0350 product Quotient 167-7345 area required. CASE II. When the section is by a right line parallel to the conjugate diameter. RULE. Find the area of a circular segment of the same altitude, the diameter of the circle being equal to the transverse diameter of the ellipsis. Then multiply this area by the conjugate, and divide the product by the transverse axis of the ellipsis, for the area of the elliptical segment. EXAMPLE. Let KLM be the segment of an ellipsis cut by a right line parallel to the shorter diameter, the altitude LQ being 10 inches, and the two diameters OP and LN, being respectively 27 inches, and 40 inches; the area of the segment KLM is required. The altitude 10 divided by the diameter 40 of the circumscribing circle, quotes 25 for the versed sine with which to enter the Table of Circular Segments; and the area corresponding is 153546 tabular number Multiply by 1600 square of LN. 92127600 153546 Product 245.6736|00 area of RLS. 17197152 4913472 Transv. 40) 6633 1872 product Quotient 165-82968 area required. PROBLEM XVII. To find, by equidistant Ordinates, the Area of an Elliptical plane Figure differing from a true Oval. Let ABCD be an elliptical plane figure differing from a true oval; the area is required by the method of equidistant ordinates. Measure the longest diameter AC, which let be 54.9 inches, and through the middle of AC draw the straight line BD at right angles to it. Then parallel to BD, at equal distances, draw the lines ef, gh, ik, lm, no, pq, intersecting AC in the points 1, 2, 3, 4, 5, 6, 7; and let the distances be such that Al, and 7C, may be comparatively small. This may easily be effected by dividing the longest diameter AC (here supposed to be 54.9 inches) by the proposed number of spaces, which will always be even, since the number of ordinates will evidently be odd. In the instance before us the number of spaces we have chosen is 6; and if 54.9 inches be divided by 6, the quotient in whole numbers will be 9 inches for the distance between any two ordinates, and of an inch will remain for the sum of A1 and 7C. The ordinates being all drawn, let them be severally measured, and let their lengths be as under: 1. no...... 16.2 inches 2. ik...... 21.8 inches |