gently round in contact with the paper will mark out the boundary of an ellipsis, as p A B C D. It is evident that the sides of the triangle formed by the thread will continually vary in length, one side, as pf, increasing by as much as the other diminishes. There are several other methods of constructing an ellipse, as by means of "the elliptic compasses" the «trammels," "three rulers with groves," &c. But the method above described is quite accurate, and may be very readily practised. Having given two right lines intersecting at right angles, as the diameters of an ellipse, to find the foci and construct the figure.-Let A C and B D be the given diameters. Take on the compasses the distance A O or O C, and from D or B as a centre, describe arcs cutting A Cat f and f. These points will be the foci of the ellipse; for, by the fundamental property of the figure, the position of the foci is such that the sum of the distances from them to any point in the curve is always the same; and as the distances from f and f to D are together equal to twice A C, or the transverse diameter, which is a fixed quantity, it is evident that these points fulfil the condition in question. Having thus ascertained the foci, the outline of the ellipse may be traced by means of a string, in the manner above explained. To determine whether a given oval figure is greater or less than a true ellipse. Draw the diameters at right angles and find the points which should correspond to the foci. Then take a string equal in length to the transverse diameter and fasten its ends at the focal points with two pins. Stretch the loop with a sharp, pointed instrument so as to form a triangle, as fpf, the vertex of which should lie in the periphery, and carry the instrument round towards C, keeping both parts of the string fully stretched. If now the moveable point coincide everywhere with the curve, the given figure is a true ellipse; if it extend beyond the curve, as at r, the figure is less than an ellipse, and if it fall short of the curve, as at s, the figure is greater than an ellipse. The reason of this must be obvious from what has preceded. To find the area of an elliptical segment, the base of which is parallel to either of the diameters of the ellipse.-In order to resolve a problem of this kind it is necessary to know both diameters of the ellipse and the height of the proposed segment. RULE.-Divide the height of the segment by that diameter of which it is a part, and find the area seg. answering to the quotient in the table of Areas of Segments in the Appendix. Multiply together the tabular area thus found and the two diameters of the ellipse. The last product will be the area required. If the proposed segment be greater than the semi-ellipse, find the area of the smaller segment, which subtract from the area of the entire ellipse. The remainder will be the area required. Example (1.) What is the area of an elliptical segment cut off by a chord parallel to the shorter axis, the height of the segment being 10, and the two diameters respectively 35 and 25? Let G B H figures, represent the given segment. The height of this, B P, is part of the longer diameter A B. Then, by the rule, 10 35 = 2857... which may be written •286* Opposite to the height 286 in the table is "area seg." .185425 -185425 x 35 x 25 = 162.25 Answer. When the division does not terminate at the third place of decimals, and great accuracy is required, a correction must be made for the remaining figures in the mnner directed on page 193. Example (2). What is the area of the elliptical segment G AH (same fig.) the height of which, A P, is 25, and the two diameters as before? Here, as the ratio of the height to the transverse diameter is evidently greater than, the proposed segment must exceed a semi-ellipse. Consequently, we first compute the area of the smaller segment, as in the last example, and deduct the result from the area of the whole ellipse. 35 x 25 x 7854 687.22 arca of ellipse. 162.25 area of smaller segment. With regard to the reason of the rule in this case, it will be sufficient to observe, that as in finding the area of any given circle we multiply the square of the diameter by 7854, the area of a circle whose diameter is 1; and as in finding the area of an ellipsis we multiply the product of the diameters by the same number 7854, so from any circular segment the chord of which is parallel to the diameter, we may deduce on a similar principle the corresponding elliptic segment; for when a circle is described on either of the diameters of an ellipse, parallel ordinates will intercept portions of the two surfaces, which are always in a constant ratio to one another. (See the remarks on page 195 relative to the method of determining the area of an ellipse.) To find the area of a parabola—that is, the space included by the arc of the curve, as C A D, and a chord or double ordinate, C D. RULE.-Multiply the double ordinate or base by the E axis or height of the figure; two thirds of the product will be the area. This rule is derived from the remarkable property, as demonstrated in the higher analysis (arithmetic of infinites) that the area of a parabola, whether or not its base be perpendicular to the axis, is always equal to two 2 thirds of the area of the circumscribing parallelogram, that is, to two thirds of a rectangle having the same base and altitude. Example. What is the area of the segment of a parabolic curve, CA D, the base of which, C D, is 130 and the height A B, 160 inches? 130 x 160 = 20800 and 20800 × B D Sq. inches. To describe a parabolic curve by means of a series of points, having given the axis or height and the base or double ordinate.-Let A B be the axis and CD the double ordinate. Through A draw E F parallel to C D, and through C and D draw CE and D F parallel to A B, and meeting E F at E and F. Divide B C and B D into any number of equal parts-the more the better-and divide CE and D F into the same number. From the points of section in C E, D F, draw straight lines, 1 A, 2 A, &c., to the vertex A; and from the divisions in C B, B D, draw straight lines parallel to A B, cutting the lines 1 A, 2 A, &c. The points at which these lines intersect will be in the curve of the parabola. It is only necessary, therefore, to draw carefully a curved line through the several intersections thus made, and a parabola will be produced. To find the area of any plane figure bounded by an irregular curve line. Principle of the method of equidistant ordinates.—The Mensuration of irregular right-lined figures, however numerous their sides, or however various the angles at which these sides are joined, may be effected with great simplicity and exactness, as has been shown on page 179, by dividing the given space into triangles or quadrilaterals, and computing the area of each separately. But when the boundary of a surface consists wholly or in part of curved or crooked lines which are not governed by any ascertainable geometric law, and are subject to frequent and abrupt changes of direction of greater or less amount, it is impossible to determine the area otherwise than approximately; and in this case all that can be done is to devise some general method, which by judicious compensations and corrections, shall reduce to a minimum the errors that are unavoidably committed. The system of equidistant ordinates, now to be explained, possesses these advantages in a high degree, and enables us by a very ready process to obtain so close an approach to the true area of the most irregu larly shaped figure, that the difference can never be of the least practical moment. P C m D n E H In the annexed diagram, let BCDEFGHIJ be part of any curvilinear boundary, and AR, a straight line subtending it, to which perpendiculars, BA and JR, are drawn from the extremities of the B curve. These perpendiculars are called ordinates to the curve at the points B and J respectively. Let the line AR be divided into any even number of A K LM NOP Q equal parts, A K, K L, L M, &c., and let perpendiculars, K C, L D, M E, &c., be drawn from the points of division to the curve. Such perpendiculars are also termed ordinates. The space A RJ B is thus divided into an even number of parallel slices or sections, by "equi-distant ordinates." R From B and C draw Bm, Cn, parallel to A R, and complete the parallelograms B C, C D. Areas of rectangles A C and K D = KC × A K+LD × KL (AB+ KC) × A K + (KC+L D) ×_K L: or, since K L = A K, the sum is equal to (AB+2KC + L D) × A K. (1.) ................... (2.) But it is manifest from inspection of the figure, that the area numbered (1) above is less than the curvilinear area, A B C D L, and that the expression (2) is greater than it. The curvilinear area is thus between (1) and (2), so that taking the average of the two we shall obtain (AB+ 2 KC + L D) × AK (3.) as a tolerably close approximation to the area of the intermediate curvilinear surface ABCDL. If the small parallelograms D E, E F, be formed and the same reasoning be applied to the next section L D E F N, we shall have, in like manner, for an approximation to the area of that space, the expression (LD+2 ME+NF) × A K. Similarly, the area of the portion NFGHP will be denoted by (NF + 2 0 G + PH) AK, and of P HIJR by (P H + 2 Q I + RJ) × A K. Hence, taking the sum of these several sections, the area of the entire figure A B JR will be expressed as (AB × 2 KC+2 LD+2 ME + 2 NF +20 G2PH + 2 RI+ RJ) × AR: That is,-Add together the extreme ordinates and twice all the other ordinates; and multiply the sum by the common distance between the ordinates, and divide the product by 2. The result will be a good first approximation to the surface included by the curve B J and the straight lines B A, A R, and R J. A still better approximation may, however, be had as follows: On looking first at that part of the surface included between the ordinates A B and L D, it will be perceived that the rectangle A L n p is greater than the expression marked (1) above, and less than that marked (2), the difference in the former case being the rectangle B C in excess, and in the latter case, the rectangle C D in defect. Thus it appears that not only the surface bounded by the curve B D, but also the parallelogram AL np is intermediate between (1) and (2), and is, evidently, as close an approximation to the curvilinear surface as is the expression (3). Consequently, if the area of the parallelogram A Ln p be added to the two areas (1) and (2), and one-third, or the average of the amount be taken, the result will be an approximation always as close as that given by the formula (3), and generally much closer. The parallelogram A L n p is, like the areas (1) and (2), made up of a pair of parallelograms, namely, A C and Kn; so that putting A K for its equal K L in the former pair, we have АВХАК + КСХАК, = Area of AL n p.) And collecting similar terms, one-third of the sum of these rectangles, viz., (AB+4 KC + L D) × A K = area of A B C D L, very nearly. In like manner, (LD+4 ME + NF) X AK = area of L DEFN, (NF40G + PH) × AK = area of N F G H P, (PH + 4 QI + R J) × A K = area of P HIJR Adding all these portions together, the expression for the entire area will be, (AB+ 4 KC+2 LD + 4 ME + 2 NF + 4 0 G + 2 PH + 4 QI+RJ) × A K. We have, accordingly, the following general method for determining the areas of such figures as the present, by means of equi-distant ordinates :— RULE.-Measure an odd number of equi-distant ordinates between the perpendiculars drawn to the extremities of the curve boundary. Then To the sum of the first and last or extreme ordinates add four times the sum of the even ordinates, that is, the second, fourth, sixth, &c., and twice the sum of all the others, not including the extreme ones. Multiply the result by the common distance between the ordinates, and one-third of the product will be the area of the figure bounded by the curve, the extreme ordinates, and the straight line joining these ordinates. Example (1.) Suppose the lengths of the nine equi-distant ordinates, in the preceding figure to be respectively, A B = 3.5; K C 32 ME 17; NF 21; OG 24; PH inches; and the common distance between them on the line A K to be 3.5 inches. The spirit of this illustration is derived from a treatise on Mensuration by J. R. Young. If this question be worked also by the formula marked (3) the area will prove to be 551-78, showing a disparity of 0.87 square inches. But the result obtained by the second method, or the rule above laid down, is to be regarded as the more accurate, inasmuch as it represents the mean of three rectangular areas, each of which affords a different approximation to the given space. If the entire curve,-or, what will be sufficient for the purpose of an illustration-that part of it which lies between B and D, were replaced by a straight line, it is evident that the formula marked (3) would exactly express the areas of the trapezoids A B C K, and KC DL, thus formed, since, according to the Rule on page 178, the area of the trapezoid A B C K is equal to (A B+ KC) × × A K, that is, half the sum of the parallel sides multiplied by the perpendicular distance between them. Similarly, area of KCDL = (KC + L D) X X KL. But KC AK; hence, the sum of the two trapezoidal areas, A B C K, KCDL = (A B+ 2 K C + L D) × A K, a result identical with (3). Now these trapezoids are together less than the figure with the curvilinear boundary B D, by the small segments intercepted between that curve and the straight line joining B and D; while the rectangle AL np, it may be seen, somewhat exceeds this curvilinear space in consequence of the triangular figure Bpc being rather greater than the similar figure DCn. The two errors, one of defect and one of excess, tend therefore to compensate one another, when the areas of the trapezoids and the rectangle are added together and the mean of their sum is taken as in the preceding Rule. What is here shown to be true of the space A B C D L, is of course, also true of each of the remaining sections included by the ordinates L D, N F; N F, P H, &c., and holds good accordingly as respects the entire figure. It should be observed that the Rule furnishes precisely the same result as that given by formula (3), when instead of a curve, as B J, the upper extremities of the ordinates are joined by a straight line. In that case, the figure ABCDL becomes a trapezoid equal in area to the rectangle A L n p; for the triangle Bp C, by which the latter how exceeds the trapezoid in the direction n p, is of the same area as the triangle D C n, by which it falls short of the trapezoid in the direction n D. It is manifest also, that if BD were a straight line, it would intersect KC at a lower point than C, making the length of K C equal to the mean of the lengths of A B and L D. The area of the rectangle A L np might then be found by multiplying the new value of K C by the base A L, and would prove to be equal to the area of the trapezoid A B C D L. It thus appears that when BJ assumes the form of a straight line it is immaterial whether the area of the figure be determined by the method prescribed in the Rule, or the more simple process indicated by formula (3) as the results will be identical, and also strictly correct. The Rule, moreover, furnishes the exact area when B J consists of a true parabolic curve; and in all other cases, it gives an approximation which will differ less and less from the truth, according as the curvilinear boundary approaches more and more nearly to a straight line or to a parabola. As every possible degree of divergence may exist between the sides of a parabola, it is considered that no curve can arise or be produced which will not, for a short distance at least, coincide with a part of some parabolic curve; and hence it is obvious that the greater the number of the ordinates taken in any instance, or the nearer these are to one another, the smaller will be the amount of the error introduced by treating the curved boundary between any pair of ordinates, as if it were a straight line or a parabolic arc, and, consequently, the less will be the sum of all the errors for the entire surface. The facility with which a parabolic area may be determined, from its relation to the investing parallelogram, supplies an additional reason for assuming generally that the small successive portions of every curvilinear outline correspond to ares of parabolas. It is essential that the number of ordinates should always be odd, as this mode of finding the area is based on the division of the given space into pairs of parallel sections, each of which must be included by three ordinates, every odd ordinate, except the first and last, serving as a boundary both of the section preceding and that following. Thus, four sections, or two pairs, require five ordinates; six sections or three pairs, require seven ordinates, and so on. Let the case now be considered of a surface enclosed wholly by a line of indefinite or irregular curvature, and let it be inquired in what manner the system of equi-distant ordinates can properly be applied to the computation of the area of such a surface. It is chiefly with respect to figures having an oval shape or contour, and yet, in reality, differing to a greater or less extent from a true ellipsis, that the approximative method in question is resorted to in the practice of gauging. |