Problem 30. Show that the center of gravity of the segment of a circle (Fig. 36), included between the arc 2s and the chord 2d, is where F is the area of the segment. given by x = 8 d8 12 F' 25. Center of Gravity of Counterbalance of Locomotive Drive Wheel.-In Fig. 37 the drive wheel is indicated by the circle and the counterbalance by the portion inclosed by the heavy lines, the point O is the center of the wheel, and a is the angle subtended by the counterbalance. The point O' is the center of the circle forming the inner boundary of the counterbalance, and B is the angle subtended by the counterbalance at this point. Let F1 represent the area of the segment of radius r and F2 the area 2 41 of the segment of radius r1. Also let x represent the distance of the center of gravity of F1 from O, and x the distance of the center of gravity of F2 from O'. Then, from Problem 30, 2 21= = 8 a3 12F1 -and x'= 8 a3 12 F the distance of the center of gravity of F2 from 0, 8 a3 = x2' — 00' = = 12 F2 s). 2 So that x, the distance of the center of gravity of the counterbalance from O, equals F1-F. It is seen that F1-F2 F1 equals area of sector minus area of triangle equals ar2 r cos 2 2 curve is known, it is expressed as y = f(x), and the area When the algebraic equation of a between the curve and either axis is always determined by integration. In Fig. 38 the area ABCD is expressed by the integral when the curve represented by y=f(x) is continuous between A and B. In many engineering problems the curve is such that its equation is not known, so that approximate methods of obtaining the areas under the curve must be resorted to. One of these methods of approximation is known as Simpson's Rule. Suppose the curve in question is the curve AB (Fig. 39) and it is desired to find the area between the portion AB and the x-axis. Divide the length ba into an even number of equal parts n (here n=10). Consider the portion CDEF and imagine it magnified as shown in Fig. 40. Pass a parabolic arc through the points C, D, G; then the area CDEF is approximated by the area of the parabolic segment CGDI plus the area of the trapezoid CDEF, therefore area CGDEF = {(Y2+Y1)EF+ {(Y3− 1[Y2+Y1])EF, since the area of = 43 the parabolic segment is the area of the circumscribing parallelogram. Since EH= written b n α { ▲ x(Y2 + 4 y2 + Y4)• In a similar way the next two strips to the right will have an area,(+4+y), and the next two strips, an Ax 3 Ax area, (Y6+4y+yg), and so on. Adding all these so as to get the total area under the portion of the curve AB, we get total area = b 3.10 · [Yo + 4(Y1 + Y3 + Y5 + Y7 + Y9) or in general for n divisions, +2(Y2+Y4+Y6+Y8) + Y10], total area = b = α [yo + 4 (Y1 + Y 3 + Y5 + ··· Yn-1) +2(Y2+Y4+Y 6+ ··· Yn-2)+Yn], 3 n a ... ... and this is Simpson's formula for determining approxi mately the area under a curve. It is easy to see that the smaller Ax, the less the approximation will be. 27. Application of Simpson's Rule. Simpson's Rule may be made use of in determining approximately not only areas, but volumes and moments. On account of its use in adding moments Simpson's formula may be employed in finding the center of gravity of areas or volumes bounded by lines or surfaces whose equations are not known. Suppose, for example, it is desired to know the volume and position of the center of gravity of a coal bunker of a ship as shown in Fig. 41. The bunker is 80 ft. long and the areas A, A1, A2, A3, A4, are as A。= 400 sq. ft., follows: A1 = 700 sq. ft., The distance between the successive areas is 20 ft. Applying Simpson's formula for volume, volume = 80 (3) (4) [Ao +4(A1 + Ag) + 2 A2 + A4]. Summing the values А。。, А11, А22, etc., we obtain 80 (3)(4) where x = 0, x1 = 20, x1⁄2= 40, x3 = 60, x4: = 80. The po sition of the center of gravity from the fore end can now be obtained from the relation |