twice the sum of all intermediate odd-numbered ordinates. accurate than the trapezoidal rule. This rule is n EXAMPLE.-What is the area ABCD of the polygon, in the example of Third Method, according to Simpson's rule? SOLUTION. A[19+23+4×(18+12+17)+2×(14+13)] = 4,733 sc Fifth Method.-Prepare two drawings upon paper of the same weight a quality, one of the tract of irregular outline and another of about the sa size but of a tract whose area is known, both drawings being to the same sca Cut each out carefully along the boundary and weigh in a chemist's bala sensitive to milligrams (1 mg.=.015 gr., about). The area of the irregu figure may be calculated from the proportion: The unknown area: the kno area the weight of the drawing of the unknown area the weight of the draw of the known area. Sixth Method.-Draw the figure upon cross-section paper. If each of t small squares represents a certain area to the scale of the map, the numb of whole squares and fractions thereof within the boundary, when multipli by the relative area of a single square, will give the area of the figure. Seventh Method. If the area of many irregular figures (such as a series indicator diagrams) is to be determined, a planimeter should be secure This instrument, of which there are several types, and which may be secur through any dealer in engineer's supplies, affords the most rapid and accura means for determining the areas of irregular figures. NOTE. The accuracy of the results obtained by the use of these appro mate methods for determining the areas of irregular figures depends almo entirely on the skill and judgment of the engineer. Other things being equa the smaller the subdivisions (in the method involving ordinates or small square the more accurate the results. CIRCLES A circle is a plane figure bounded by a curved line every point of which equidistant from an interior point called the center. In the formulas relating to circles the letters have the following meaning capital letters refer to the larger and lower-case letters to the smaller of tw circles concerned in the same formula. To Find Diameter of a Circle Equal in Area to a Given Square.-Multiply one side of the square by 1.12838. To Find Radius of a Circle to Circumscribe a Given Square.-Multiply one side by .7071; or take one-half the diagonal. To Find Side of a Square Equal in Area to a Given Circle.-Multiply the diameter by .88623. To Find Side of Greatest Square in a Given Circle.-Multiply the diameter by .7071. To Find Area of Greatest Square in a Given Circle.-Square the radius and multiply by 2. To Find Side of an Equilateral Triangle Equal in Area to a Given Circle. Multiply the diameter by 1.3468. Circumferences and areas of circles from 1 to 1,000 units in diameter will be found in connection with the table of squares, cubes, etc. A similar table, but for diameters from to 100, increasing by, is also given at the end of the volume. Circumferences and areas of circles whose diameters are not exactly given in the tables, but are within its limits, may be found by interpolation. If the diameters are greater than those given in the table, the circumferences and areas may be found by recalling that the former are proportional to the diameters and the latter to the squares of the diameters. Thus, the circumference and area of a circle 9,380 units in diameter are respectively 10 and 100 times those of a circle 938 units in diameter. F SECTORS E B A sector is a portion of the surface of a circle included between an arc AEB and two radii OA and OB. The central angle C must be reduced to degrees and decimal parts thereof. CIRCULAR SEGMENTS A segment is a portion of the surface of a circle included between an arc NEM and its chord NM. Area segment NEM area sector ONEM - area triangle NOM. The area of the sector is found from some one of the formulas already given and that of the triangle is found from k A' = ~(r− h) = r2 sin C = k cot Other formulas are k=2√h(2r-h) = 2r sin = (k-h)2 tan To determine the area of a heading, the upper part of which is the arc of a circle, the width thereof k and the rise of the arc h being given. Find r from the formula r= Ck Then find C from sin = 27* k2+4h2 The angle found is C and 8h must be multiplied by 2. C is to be expressed in degrees and decimals thereof. The area of the sector ONEM is now found from the formula A = .00872664r2C. k The area of the triangle NMO is found from the formula A'=(r-h). Then A -A'area segment NEM, which is to be added to that of the lower rectangular portion of the heading. ELLIPSE If A and B are the axes of an ellipse and a and b are the semi-, or half, að the area A = Tab 3.141593 ab= ==AB= No simple formula has been developed for finding the perimeter of ellipse. In terms of the semi-axes, the formula for the perimeter involves t summation of an expanding series of an infinite number of terms: (a 1 P= x(a+b) [1+1 ((a+b))2 + 1 ((@ + b)) * +; In an ellipse whose semi-axes are 4 and 3, the values within the bracke become 1+.005102041+.000006507+.000000033. It is apparent, then, exce in cases involving great accuracy, that the terms beyond the second m be dropped and the formula for the perimeter of an ellipse may be writte 1/(a P = = x(a+b) [1+1 ((a+b))3]. This, the correct formula, is fully as simp and is, naturally, far more accurate than any of the so-called shorter approx mations. PRISMOID AND PRISMOIDAL FORMULA A prismoid is any solid having two parallel ends or faces of any shape similar or dissimilar, regular or irregular, provided these ends are united by M B H It surfaces, whether plane or curved, on which and through every point of which, a straight line may be drawn from one of the parallel ends to the other. embraces all polyhedrons, parallelopipeds, prisms, cylinders, cones, pyramids, etc., and even the sphere, which may be regarded as a regular polyhedron of an infinite number of faces. This is true whether the solids are regular or irregular, right or oblique, and applies to their frustums when cut parallel to the base. The formula for the volume of a prismoid is in which A and B = areas of two ends, respectively; M = area of section taken midway between ends; This is known as the prismoidal formula, and from it, by making the proper substitutions for A, M, and B, an equation for the volume of any solid may be deduced, provided its form is included in the definition of a trapezoid. Unless the parallel ends are similar polygons, the mid-section M, is not the mean of the sections A and B. This formula is extensively used in calculating excavations in railroad work. REGULAR POLYHEDRONS A polyhedron is a solid contained within any number of plane sides. A regular polyhedron is one whose bounding planes (faces) are regular polygons of the same shape and area, and whose solid (polyhedral) angles are equal each to each. Unless the sphere is considered a regular polyhedron of an infinite number of sides, only five regular polyhedrons are possible; these are shown in the accompanying figure. The surface of any regular polyhedron is equal to the number of faces multiplied by the area of a single face. It maybe found from the accompanying table by squaring the length of the edge of a face and multiplying this by the number, in the column headed Surface, opposite the name of the polyhedron. Thus, the surface area of an octahedron whose edge is 2 in. long, is 22, or 4X 3.464102=14.856408 sq. in. The volume of any polyhedron may be obtained by taking the sum of the volumes of the pyramids into which it may be divided. There will be as many pyramids as there are faces in the polyhedron and the bases of these pyramids will be the several faces of the polyhedron. The volumes of the regular polyhedrons may be obtained from the table by multiplying the cube of the length of the edge by the number, in the column headed Volume, opposite the name of the polyhedron concerned. Thus, the volume of a tetrahedron whose edge is 2 in. long, is 23, or 8X.117851.942808 cu. in. The sphere may be defined as a regular polyhedron of an infinite number of sides; or as a solid generated by the revolution of a semicircle about its diameter; or as a solid, every point of whose surface is equidistant from a fixed interior point called the center. A great circle of a sphere is the line formed by the intersection of a plane through the center with the surface of the sphere, as ABCD or AbCc. Its radius and diameter r and d are the same as those of the sphere; its area R2 may be placed equal to Z. B SPHERICAL SEGMENTS A spherical segment is that portion of a sphere that is included between its surface and a plane cutting it, as EABCD. If the plane passes through the center, the sphere is divided into two equal parts, each of which is a hemisphere. Let EF-h-height of segment, and r' and d' be the radius and diameter of the basal @plane, and R be the radius of the sphere. E -d- S=2πRh=6.2832Rh= (d'2+4h2)=.7854 (d'2+4h2) = Ph 4 To this must be added the area of the base, d'2, if the entire surface is required. SPHERICAL ZONES A spherical zone is that portion of a sphere that is included between two parallel intersecting planes. Let d be the diameter of the sphere and d' and d" be the diameters AC and ac of the two planes, whose distance apart is EF=h. h d A cylindrical ring is produced by bending a cylinder upon itself. Using the notation in the accompanying figure S = 42Rr = 39.4786Rr = T2Dd = h 9.8697 Dd. V = 2π2Rr2 Dd2 2.4674Dd2. 4 PARALLELOPIPEDS h Cube Rectangular Prism Rhombohedron Rhombic Prism A parallelopiped is a solid bounded by six faces, all of which are parallelograms; opposite faces being parallel. In the cube, there are six equal faces and eight equal solid angles; in the rectangular prism, there are three pairs of equal opposite faces and eight equal solid angles; in the rhombohedron, there are six equal faces and four pairs of equal and diagonally opposite angles; in the rhombic prism, there are three pairs of equal opposite faces and four pairs of equal and diagonally opposite angles. S sum of areas of six faces. In the cube and rhombohedron S-6A. In the two prisms (calling the equal faces A, B, and C) S=2A+2B+2C. V=Ah; that is, the volume is equal to the area of any face multiplied by the perpendicular distance to the opposite face. In the cube, V-cube of length of one edge. The diagonal joining opposite vertices = an edge X 1.732051. The radius of the inscribed sphere- the edge X.5. The radius of the circumscribed sphere -one-half the diagonal joining opposite vertices = the edge X.866026. Frustum of Prism.-If a section perpendicular to the edges is a triangle, square, parallelogram, or regular polygon, sum of lengths of edges, |