Relations of Arc, Chord, Chord of Half the Arc, = 2 √ R2 — (R − V )2; = 2√(D − V) × Chord of half the arc, ch = 1 Cd2+4V2; =√ D x V V. 2 = 8ch-3 Arc. ; (or (D+√D2 – Ca2), if V is greater than radius Half the chord of the arc is a mean proportional between the versed sine and diameter minus versed sine: %Cd = √V × (D V) Length of the Chord subtending an angle at the centre = twice the sine of half the angle. (see Table of Sines, p. 157.) Length of a Circular Arc.-Huyghens's Approximation. Let Crepresent the length of the chord of the arc and c the length of the chord of half the arc; the length of the arc Professor Williamson shows that when the arc subtends an angle of 30°, the radius being 100,000 feet (nearly 19 miles), the error by this formula is about two inches, or 1/600000 part of the radius. When the length of the arc is equal to the radius, i.e., when it subtends an angle of 570.3, the error is less than 1/7680 part of the radius. Therefore, if the radius is 100,000 feet, the 100000 error is less than 13 feet. The error increases rapidly with the 7680 increase of the angle subtended. In the measurement of an are which is described with a short radius the error is so small that it may be neglected. Describing an arc with a radius of 12 inches subtending an angle of 30°, the error is 1/50000 of an inch. For 57°.3 the error is less than 0.0015. In order to measure an arc when it subtends a large angle, bisect it and measure each half as before-in this case making B = length of the chord of half the arc, and b = length of the chord of one fourth the arc; then Relation of the Circle to its Equal, Inscribed, and Circumscribed Squares. x .88623 Diameter of circle = side of equal square. = perimeter of equal square. Sectors and Segments.-To find the area of a sector of a circle. RULE 1. Multiply the arc of the sector by half its radius. RULE 2. As 360 is to the number of degrees in the arc, so is the area of the circle to the area of the sector. RULE 3. Multiply the number of degrees in the arc by the square of the radius and by .008727. To find the area of a segment of a circle: Find the area of the sector which has the same arc, and also the area of the triangle formed by the chord of the segment and the radii of the sector. Then take the sum of these areas, if the segment is greater than a semicircle, but take their difference if it is less. Another Method: Area of segment = 1⁄2R(arc sin 4), in which A is the central angle, R the radíus, and are the length of arc to radius 1. To find the area of a segment of a circle when its chord and height only are given. First find radius, as follows: 2. Find the angle subtended by the arc, as follows: half chord ÷ radius sine of half the angle. Take the corresponding angle from a table of sines, and double it to get the angle of the arc. 3. Find area of the sector of which the segment is a part; area of sector = area of circle x degrees of arc + 360. 4. Subtract area of triangle under the segment]: Area of triangle = half chord × (radius – height of segment). The remainder is the area of the segment. From the When the chord, arc, and diameter are given, to find the area. length of the are subtract the length of the chord. Multiply the remainder by the radius or one-half diameter; to the product add the chord multiplied by the height, and divide the sum by 2. Given diameter, d, and height of segment, h. When h is from 0 to 44d, area = 66 66 66 66 h V1.766dh-h2; (approx.). Greatest error 0.23%, when h = 4d. To find the chord: From the diameter subtract the height; multiply the remainder by four times the height and extract the square root. When the chords of the arc and of half the arc and the rise are given: To the chord of the arc add four thirds of the chord of half the arc, multiply the sum by the rise and the product by .40426 (approximate). Circular Ring. To find the area of a ring included between the circumferences of two concentric circles: Take the difference between the areas of the two circles; or, subtract the square of the less radius from the square of the greater, and multiply their difference by 3.14159. The area of the greater circle is equal to R2; 2). 1.2 Their difference, or the area of the ring. is π(R2 The Ellipse.-Area of an ellipse product of its semi-axes x 3.14159 product of its axes x .785398. = The Ellipse.—Circumference (approximate) = 3.1416 √ D2 + d2 D and d being the two axes. B Trautwine gives the following as more accurate: When the longer axis D is not more than five times the length of the shorter axis, d, When D is more than 5d, the divisor 8.8 is to be replaced by the following: For D/d = 6 7 8 9 10 12 14 16 18 20 30 40 50 Divisor 9 9.2 9.3 9.35 9.4 9.5 9.6 9.68 9.75 9.8 9.92 9.98 19 A2 A4 A® An accurate formula is C = ☛(a+b) (1 which A= a-b a+b + 4 + 16 --Ingenieurs Taschenbuch, 1896. + 250 2548 + +.......), in 16381 Carl G. Barth (Machinery, Sept., 1900) gives as a very close approximation to this formula Area of a segment of an ellipse the base of which is parallel to one of the axes of the ellipse. Divide the height of the segment by the axis of which it is part, and find the area of a circular segment, in a table of circular segments, of which the height is equal to the quotient; multiply the area thus found by the product of the two axes of the ellipse. Cycloid.-A curve generated by the rolling of a circle on a plane. Length of a cycloidal curve = 4 × diameter of the generating circle. Length of the base = circumference of the generating circle. Area of a cycloid = 3 × area of generating circle. Helix (Screw).-A line generated by the progressive rotation of a point around an axis and equidistant from its centre. Length of a helix.-To the square of the circumference described by the generating-point add the square of the distance advanced in one revolution, and take the square root of their sum multiplied by the number of revolu tions of the generating point. Or, (c2 + h2)n = length, n being number of revolutions. Spirals.-Lines generated by the progressive rotation of a point around a fixed axis, with a constantly increasing distance from the axis. A plane spiral is when the point rotates in one plane. A conical spiral is when the point rotates around an axis at a progressing distance from its centre, and advancing in the direction of the axis, as around a cone. Length of a plane spiral line.-When the distance between the coils is uniform. RULE. Add together the greater and less diameters; divide their sum by 2; multiply the quotient by 3.1416, and again by the number of revolutions. Or, take the mean of the length of the greater and less circumferences and multiply it by the number of revolutions. Or, length d+d' d and d' being the inner and outer diameters. Length of a conical spiral line.-Add together the greater and less diameters; divide their suni by 2 and multiply the quotient by 3.1416. To the square of the product of this circumference and the number of revolutions of the spiral add the square of the height of its axis and take the square root of the sum, The Prism.-To find the surface of a right prism: Multiply the perimeter of the base by the altitude for the convex surface. To this add the areas of the two ends when the entire surface is required. Volume of a prism = area of its base X its altitude. The pyramid.-Convex surface of a regular pyramid = perimeter of its base half the slant height. To this add area of the base if the whole surface is required. Volume of a pyramid = area of base X one third of the altitude. To find the surface of a frustum of a regular pyramid: Multiply half the slant height by the sum of the perimeters of the two bases for the convex surface. To this add the areas of the two bases when the entire surface is required. To find the volume of a frustum of a pyramid: Add together the areas of the two bases and a mean proportional between them, and multiply the sum by one third of the altitude. (Mean proportional between two numbers = square root of their product.) Wedge.-A wedge is a solid bounded by five planes, viz.: a rectangular base, two trapezoids, or two rectangles, meeting in an edge, and two triangular ends. The altitude is the perpendicular drawn from any point in the edge to the plane of the base. To find the volume of a wedge: Add the length of the edge to twice the length of the base, and multiply the sum by one sixth of the product of the height of the wedge and the breadth of the base. Rectangular prismoid.-A rectangular prismoid is a solid bounded by six planes, of which the two bases are rectangles, having their corresponding sides parallel, and the four upright sides of the solids are trapezoids. To find the volume of a rectangular prismoid: Add together the areas of the two bases and four times the area of a parallel section equally distant from the bases, and multiply the sum by one sixth of the altitude. Cylinder.-Convex surface of a cylinder = perimeter of base altitude. To this add the areas of the two ends when the entire surface is required. Volume of a cylinder = area of base X altitude. Cone.-Convex surface of a cone = circumference of base X half the slant side. To this add the area of the base when the entire surface is required. Volume of a cone area of base X one third of the altitude. To find the surface of a frustum of a cone: Multiply half the side by the sum of the circumferences of the two bases for the convex surface; to this add the areas of the two bases when the entire surface is required. To find the volume of a frustum of a cone: Add together the areas of the two bases and a mean proportional between them, and multiply the sum by one third of the altitude. Or, Vol. = 0.2618a(b2+ c2 + bc); a = altitude; b and c, diams, of the two bases. Sphere.-To find the surface of a sphere: Multiply the diameter by the circumference of a great circle; or, multiply the square of the diameter by 3.14159. Surface of sphere = 4 × area of its great circle. 66 66 66 = convex surface of its circumscribing cylinder. Surfaces of spheres are to each other as the squares of their diameters. To find the volume of a sphere: Multiply the surface by one third of the radius; or, multiply the cube of the diameter by π/6; that is, by 0.5236. Value of /u to 10 decimal places = .5235987756. The volume of a sphere = 2/3 the volume of its circumscribing cylinder. Volumes of spheres are to each other as the cubes of their diameters. Spherical triangle.—To find the area of a spherical triangle: Compute the surface of the quadrantal triangle, or one eighth of the surface of the sphere. From the sum of the three angles subtract two right angles; divide the remainder by 90, and multiply the quotient by the area of the quadrantal triangle. Spherical polygon.-To find the area of a spherical polygon: Com. pute the surface of the quadrantal triangle. From the sum of all the angles subtract the product of two right angles by the number of sides less two; divide the remainder by 90 and multiply the quotient by the area of the quadrantal triangle. The prismoid.-The prismoid is a solid having parallel end areas, and may be composed of any combination of prisms, cylinders, wedges, pyramids, or cones or frustums of the same, whose bases and apices lie in the end areas. Inasmuch as cylinders and cones are but special forms of prisms and pyramids, and warped surface solids may be divided into elementary forms of them, and since frustums may also be subdivided into the elementary forms, it is sufficient to say that all prismoids may be decomposed into prisms, wedges, and pyramids. If a formula can be found which is equally applicable to all of these forms, then it will apply to any combination of them. Such a formula is called Let A A1, A2, Am The Prismoidal Formula. area of the base of a prism, wedge, or pyramid;, the two end and the middle areas of a prismoid, or of any of h = altitude of the prismoid or elementary solid; h V = (41 +44m+Ag). h For a prism, A、, Am and A, are equal, = A; V = = 0 x64 = hA. The prismoidal formula is a rigid formula for all prismoids. The only approximation involved in its use is in the assumption that the given solid may be generated by a right line moving over the boundaries of the end areas. The area of the middle section is never the mean of the two end areas if the prismoid contains any pyramids or cones among its elementary forms. When the three sections are similar in form the dimensions of the middle area are always the means of the corresponding end dimensions. This fact often enables the dimensions, and hence the area of the middle section, to be computed from the end areas. Polyedrons.-A polyedron is a solid bounded by plane polygons. A regular polyedron is one whose sides are all equal regular polygons. To find the surface of a regular polyedron.-Multiply the area of one of the faces by the number of faces; or, multiply the square of one of the edges by the surface of a similar solid whose edge is unity. A TABLE OF THE REGULAR POLYEDRONS WHOSE EDGES ARE UNITY. To find the volume of a regular polyedron.- Multiply the surface by one third of the perpendicular let fall from the centre on one of the faces; or, multiply the cube of one of the edges by the solidity of a similar polyedron whose edge is unity. Solid of revolution.-The volume of any solid of revolution is equal to the product of the area of its generating surface by the length of the path of the centre of gravity of that surface. The convex surface of any solid of revolution is equal to the product of the perimeter of its generating surface by the length of path of its centre of gravity. 1 1 4 Cylindrical ring.-Let d = outer diameter; d' inner diameter; 1 (d-d') = thickness = t; π t2 = sectional area; (d+d') = mean diameter = M; t circumference of section; π M = mean circumference of ring; surface = πt × π M; = · π2 (d2 — d12); = 9.86965 t M; = 2.46741 (d2 —d′2); volume = 1 1 π t2 Mπ; = 2.4674112 M. Spherical zone.-Surface of a spherical zone or segment of a sphere = its altitude x the circumference of a great circle of the sphere. A great circle is one whose plane passes through the centre of the sphere. Volume of a zone of a sphere.-To the sum of the squares of the radii of the ends add one third of the square of the height; multiply the sum by the height and by 1.5708. Spherical segment.-Volume of a spherical segment with one base.— |