Machinery''''s Handbook 27th Episode 1 Part 2 ppsx

VOLUMES OF SOLIDS 75 Volumes of Solids Cube: Square Prism: Prism: Pyramid: Example: The side of a cube equals 9.5 centimeters. Find its volume. Example: The volume of a cube is 231 cubic centimeters. What is the length of the side? Example: In a square prism, a = 6, b = 5, c = 4. Find the volume. Example: How high should a box be made to contain 25 cubic feet, if it is 4 feet long and 2 1 ⁄ 2 feet wide? Here, a = 4, c = 2.5, and V = 25. Then, V=volume A=area of end surface V=h × A The area A of the end surface is found by the formulas for areas of plane figures on the preceding pages. Height h must be mea- sured perpendicular to the end surface. Example: A prism, having for its base a regular hexagon with a side s of 7.5 centimeters, is 25 centi- meters high. Find the volume. If the base is a regular polygon with n sides, and s = length of side, r = radius of inscribed circle, and R = radius of circumscribed circle, then: Example: A pyramid, having a height of 9 feet, has a base formed by a rectangle, the sides of which are 2 and 3 feet, respectively. Find the volume. Diagonal of cube face ds2== Diagonal of cube D 3d 2 2 s 3 1.732s== = = Volume Vs 3 == sV 3 = Volume Vs 3 9.5 3 9.5 9.5× 9.5× 857.375 cubic centimeters== = = = sV 3 231 3 6.136 centimeters== = Volume Vabc== a V bc = b V ac = c V ab = Vab× c× 65× 4× 120 cubic inches=== b depth V ac 25 4 2.5× 25 10 2.5 feet=== == Area of hexagon A 2.598s 2 2.598 56.25× 146.14 square centimeters== = = Volume of prism hA× 25 146.14× 3653.5 cubic centimeters== = Volume V 1 ⁄ 3 h area of base×== V nsrh 6 nsh 6 R 2 s 2 4 –== Area of base 2 3× 6 square feet; h 9 feet== = Volume V 1 ⁄ 3 h area of base× 1 ⁄ 3 9× 6× 18 cubic feet== = = Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY 76 VOLUMES OF SOLIDS Frustum of Pyramid: Wedge: Cylinder: Portion of Cylinder: Example: The pyramid in the previous example is cut off 4 1 ⁄ 2 feet from the base, the upper part being removed. The sides of the rectangle forming the top surface of the frustum are, then, 1 and 1 1 ⁄ 2 feet long, respectively. Find the volume of the frustum. Example: Let a = 4 inches, b = 3 inches, and c = 5 inches. The height h = 4.5 inches. Find the volume. Total area A of cylindrical surface and end surfaces: Example: The diameter of a cylinder is 2.5 inches. The length or height is 20 inches. Find the volume and the area of the cylindrical surface S. Example: A cylinder 125 millimeters in diameter is cut off at an angle, as shown in the illustration. Dimension h 1 = 150, and h 2 = 100 mm. Find the volume and the area S of the cylindrical surface. Volume V h 3 A 1 A 2 A 1 A 2 ×++()== Area of top A 1 11 1 ⁄ 2 × 1 1 ⁄ 2 sq. ft.= = = Area of base A 2 23× 6 sq. ft.== = V 45⋅ 3 1 .5 6 1. 5 6×++()1.5 7.5 9+()1.5 10.5× 15.75 cubic feet==== Volume V 2ac+()bh 6 == V 2ac+()bh 6 245+×()3 4.5×× 6 85+()13.5× 6 == = 175.5 6 29.25 cubic inches== Volume V 3.1416r 2 h 0.7854d 2 h== = Area of cylindrical surface S 6.2832rh 3.1416dh== = A 6.2832rr h+()3.1416d 1 ⁄ 2 dh+()== V 0.7854d 2 h 0.7854 2.5 2 × 20× 0.7854 6.25× 20× 98.17 cubic inches== = = S 3.1416dh 3.1416 2.5× 20× 157.08 square inches== = Volume V 1.5708r 2 h 1 h 2 +()== 0.3927d 2 h 1 h 2 +()= Cylindrical surface area S 3.1416rh 1 h 2 +()== 1.5708dh 1 h 2 +()= V 0.3927d 2 h 1 h 2 +()0.3927 125 2 × 150 100+()×== 0.3927 15 625,× 250× 1 533 984 cubic millimeters,, 1534 cm 3 === S 1.5708dh 1 h 2 +()1.5708 125× 250×== 49 087.5 square millimeters, 490.9 square centimeters== Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY VOLUMES OF SOLIDS 77 Portion of Cylinder: Hollow Cylinder: Cone: Frustum of Cone: Use + when base area is larger, and − when base area is less than one-half the base circle. Example: Find the volume of a cylinder so cut off that line AC passes through the center of the base circle — that is, the base area is a half-circle. The diameter of the cylinder = 5 inches, and the height h = 2 inches. In this case, a = 2.5; b = 0; area ABC = 0.5 × 0.7854 × 5 2 = 9.82; r = 2.5. Example: A cylindrical shell, 28 centimeters high, is 36 centi- meters in outside diameter, and 4 centimeters thick. Find its vol- ume. Example: Find the volume and area of the conical surface of a cone, the base of which is a circle of 6 inches diameter, and the height of which is 4 inches. Example: Find the volume of a frustum of a cone of the follow- ing dimensions: D = 8 centimeters; d = 4 centimeters; h = 5 centi- meters. Volume V 2 ⁄ 3 a 3 b area ABC×±() h rb± == Cylindrical surface area Sadblength of arc ABC×±() h rb± == V 2 3 2.5 3 × 0 9.82×+ ⎝⎠ ⎛⎞ 2 2.5 0+ 2 3 15.625× 0.8× 8.33 cubic inches=== Volume V 3.1416hR 2 r 2 –()0.7854hD 2 d 2 –()== = 3.1416ht 2Rt–()3.1416ht D t–()== 3.1416ht 2rt+()3.1416ht d t+()== 3.1416ht R r+()1.5708ht D d+()== V 3.1416ht D t–()3.1416 28× 436 4–()× 3.1416 28× 4× 32×== = 11 259.5 cubic centimeters,= Volume V 3.1416r 2 h 3 1.0472r 2 h 0.2618d 2 h== = = Conical surface area A 3.1416rr 2 h 2 + 3.1416rs== = 1.5708ds= sr 2 h 2 + d 2 4 h 2 +== V 0.2618d 2 h 0.2618 6 2 × 4× 0.2618 36× 4× 37.7 cubic inches== = = A 3.1416rr 2 h 2 + 3.1416 3× 3 2 4 2 +× 9.4248 25×== = 47.124 square inches= V 0.2618 5 8 2 844 2 +×+()× 0.2618 5 64 32 16++()×== 0.2618 5× 112× 146.61 cubic centimeters== Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY V volume= A area of conical surface= V 1.0472hR 2 Rr r 2 ++()0.2618hD 2 Dd d 2 ++()== A 3.1416sR r+()1.5708sD d+()== aRr–= sa 2 h 2 + Rr–() 2 h 2 +== 78 VOLUMES OF SOLIDS Sphere: Spherical Sector: Spherical Segment: Ellipsoid: Example: Find the volume and the surface of a sphere 6.5 centimeters diameter. Example: The volume of a sphere is 64 cubic centimeters. Find its radius. Example: Find the volume of a sector of a sphere 6 inches in diameter, the height h of the sector being 1.5 inch. Also find the length of chord c. Here r = 3 and h = 1.5. Example: A segment of a sphere has the following dimensions: h = 50 millimeters; c = 125 millime- ters. Find the volume V and the radius of the sphere of which the segment is a part. In an ellipsoid of revolution, or spheroid, where c = b: Example: Find the volume of a spheroid in which a = 5, and b = c = 1.5 inches. Volume V 4πr 3 3 πd 3 6 4.1888r 3 0.5236d 3 == = = = Surface area A 4πr 2 πd 2 12.5664r 2 3.1416d 2 == = = = r 3V 4π 3 0.6024 V 3 == V 0.5236d 3 0.5236 6.5 3 × 0.5236 6.5× 6.5 6.5×× 143.79 cm 3 == = = A 3.1416d 2 3.1416 6.5 2 × 3.1416 6.5× 6.5× 132.73 cm 2 == = = r 0.6204 64 3 0.6204 4× 2.4816 centimeters=== V 2πr 2 h 3 2.0944r 2 h Volume== = A 3.1416r 2h 1 ⁄ 2 c+()= total area of conical and spherical surface= c 2 h 2rh–()= V 2.0944r 2 h 2.0944 3 2 × 1.5× 2.0944 9× 1.5× 28.27 cubic inches== = = c 2 h 2rh–()2 1.5 2 3× 1.5–()2 6.75 2 2.598× 5.196 inches== === V volume= A area of spherical surface= V 3.1416h 2 r h 3 – ⎝⎠ ⎛⎞ 3.1416h c 2 8 h 2 6 + ⎝⎠ ⎛⎞ == A 2πrh 6.2832rh 3.1416 c 2 4 h 2 + ⎝⎠ ⎛⎞ == = c 2 h 2rh–()=;r c 2 4h 2 + 8h = V 3.1416 50× 125 2 8 50 2 6 + ⎝⎠ ⎛⎞ × 157.08 15 625, 8 2500 6 + ⎝⎠ ⎛⎞ × 372 247 mm, 3 372 cm 3 ==== r 125 2 450 2 ×+ 850× 15 625, 10 000,+ 400 25 625, 400 64 mi ll im et er s== == Volume V 4π 3 abc 4.1888abc== = V 4.1888ab 2 = V 4.1888 5× 1.5 2 × 47.124 cubic inches== Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY VOLUMES OF SOLIDS 79 Spherical Zone: Spherical Wedge: Hollow Sphere: Paraboloid: Example: In a spherical zone, let c 1 = 3; c 2 = 4; and h = 1.5 inch. Find the volume. Example: Find the area of the spherical surface and the volume of a wedge of a sphere. The diameter of the sphere is 100 millimeters, and the center angle α is 45 degrees. Example: Find the volume of a hollow sphere, 8 inches in outside diameter, with a thickness of mate- rial of 1.5 inch. Here R = 4; r = 4 − 1.5 = 2.5. Example: Find the volume of a paraboloid in which h = 300 millimeters and d = 125 millimeters. Volume V 0.5236h 3c 1 2 4 3c 2 2 4 h 2 ++ ⎝⎠ ⎛⎞ == A 2πrh 6.2832rh area of spherical surface== = r c 2 2 4 c 2 2 c 1 2 –4h 2 – 8h ⎝⎠ ⎛⎞ 2 += V 0.5236 1.5× 33 2 × 4 34 2 × 4 1. 5 2 ++ ⎝⎠ ⎛⎞ × 0.5236 1.5× 27 4 48 4 2.25++ ⎝⎠ ⎛⎞ × 16.493 in 3 === V volume= A area of spherical surface= α center angle in degrees= V α 360 4πr 3 3 × 0.0116αr 3 == A α 360 4πr 2 × 0.0349αr 2 == V 0.0116 45× 50 3 × 0.0116 45× 125 000,× 65 250 mm 3 , 65.25 cm 3 == == A 0.0349 45× 50 2 × 3926.25 square millimeters 39.26 cm 2 == = V volume of material used= to make a hollow sphere V 4π 3 R 3 r 3 –()4.1888 R 3 r 3 –()== π 6 D 3 d 3 –()0.5236 D 3 d 3 –()== V 4.1888 4 3 2.5 3 –()4.1888 64 15.625–()4.1888 48.375× 202.63 cubic inches== == Volume V 1 ⁄ 2 πr 2 h 0.3927d 2 h== = Area A 2π 3p d 2 4 p 2 + ⎝⎠ ⎛⎞ 3 p 3 –== in which p d 2 8h = V 0.3927d 2 h 0.3927 125 2 × 300× 1 840 781,, mm 3 1 840.8, cm 3 == = = Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY 80 VOLUMES OF SOLIDS Paraboloidal Segment: Torus: Barrel: Ratio of Volumes: Example: Find the volume of a segment of a paraboloid in which D = 5 inches, d = 3 inches, and h = 6 inches. Example: Find the volume and area of surface of a torus in which d = 1.5 and D = 5 inches. V = approximate volume. If the sides are bent to the arc of a circle: If the sides are bent to the arc of a parabola: Example: Find the approximate contents of a barrel, the inside dimensions of which are D = 60 centi- meters, d = 50 centimeters; h = 120 centimeters. If d = base diameter and height of a cone, a paraboloid and a cyl- inder, and the diameter of a sphere, then the volumes of these bod- ies are to each other as follows: Example: Assume, as an example, that the diameter of the base of a cone, paraboloid, and cylinder is 2 inches, that the height is 2 inches, and that the diameter of a sphere is 2 inches. Then the volumes, written in formula form, are as follows: Volume V π 2 hR 2 r 2 +()1.5708hR 2 r 2 +()== = π 8 hD 2 d 2 +()0.3927hD 2 d 2 +()== V 0.3927hD 2 d 2 +()0.3927 6× 5 2 3 2 +()×== 0.3927 6× 34× 80.11 cubic inches== Volume V 2π 2 Rr 2 19.739Rr 2 == = π 2 4 Dd 2 2.4674Dd 2 == Area of surface A 4π 2 Rr 39.478Rr== = π 2 Dd 9.8696Dd== V 2.4674 5× 1.5 2 × 2.4674 5× 2.25× 27.76 cubic inches=== A 9.8696 5× 1.5× 74.022 square inches== V 1 12 πh 2D 2 d 2 +()0.262h 2D 2 d 2 +()== V 0.209h 2D 2 Dd 3 ⁄ 4 d 2 ++()= V 0.262h 2D 2 d 2 +()0.262 120× 260 2 50 2 +×()×== 0.262 120× 7200 2500+()× 0.262 120× 9700×== 304 968 cubic centimeters, 0.305 cubic meter== Cone:paraboloid:sphere:cylinder 1 ⁄ 3 : 1 ⁄ 2 : 2 ⁄ 3 :1= Cone Paraboloid Sphere Cylinder 3.1416 2 2 × 2× 12 : 3.1416 2p() 2 × 2× 8 : 3.1416 2 3 × 6 : 3.1416 2 2 × 2× 4 1 ⁄ 3 : 1 ⁄ 2 : 2 ⁄ 3 :1= Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY CIRCLES IN A CIRCLE 81 Packing Circles in Circles and Rectangles Diameter of Circle Enclosing a Given Number of Smaller Circles.—Four of many possible compact arrangements of circles within a circle are shown at A, B, C, and D in Fig. 1. To determine the diameter of the smallest enclosing circle for a particular number of enclosed circles all of the same size, three factors that influence the size of the enclosing circle should be considered. These are discussed in the paragraphs that follow, which are based on the article “How Many Wires Can Be Packed into a Circular Conduit,” by Jacques Dutka, Machinery, October 1956. 1) Arrangement of Center or Core Circles: The four most common arrangements of cen- ter or core circles are shown cross-sectioned in Fig. 1. It may seem, offhand, that the “A” pattern would require the smallest enclosing circle for a given number of enclosed circles but this is not always the case since the most compact arrangement will, in part, depend on the number of circles to be enclosed. Fig. 1. Arrangements of Circles within a Circle 2) Diameter of Enclosing Circle When Outer Layer of Circles Is Complete: Successive, complete “layers” of circles may be placed around each of the central cores, Fig. 1, of 1, 2, 3, or 4 circles as the case may be. The number of circles contained in arrangements of com- plete “layers” around a central core of circles, as well as the diameter of the enclosing cir- cle, may be obtained using the data in Table 1. Thus, for example, the “A” pattern in Fig. 1 shows, by actual count, a total of 19 circles arranged in two complete “layers” around a central core consisting of one circle; this agrees with the data shown in the left half of Table 1 for n = 2. To determine the diameter of the enclosing circle, the data in the right half of Table 1 is used. Thus, for n = 2 and an “A” pattern, the diameter D is 5 times the diameter d of the enclosed circles. 3) Diameter of Enclosing Circle When Outer Layer of Circles Is Not Complete: In most cases, it is possible to reduce the size of the enclosing circle from that required if the outer layer were complete. Thus, for example, the “B” pattern in Fig. 1 shows that the central core consisting of 2 circles is surrounded by 1 complete layer of 8 circles and 1 partial, outer layer of 4 circles, so that the total number of circles enclosed is 14. If the outer layer were complete, then (from Table 1) the total number of enclosed circles would be 24 and the diameter of the enclosing circle would be 6d; however, since the outer layer is com- posed of only 4 circles out of a possible 14 for a complete second layer, a smaller diameter of enclosing circle may be used. Table 2 shows that for a total of 14 enclosed circles arranged in a “B” pattern with the outer layer of circles incomplete, the diameter for the enclosing circle is 4.606d. Table 2 can be used to determine the smallest enclosing circle for a given number of cir- cles to be enclosed by direct comparison of the “A,” “B,” and “C” columns. For data out- side the range of Table 2, use the formulas in Dr. Dutka's article. Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY CIRCLES IN A CIRCLE 83 98 11.584 11.441 11.264 153 14.115 14 14.013 208 16.100 16 16.144 99 11.584 11.441 11.264 154 14.115 14 14.013 209 16.100 16.133 16.144 100 11.584 11.441 11.264 155 14.115 14.077 14.013 210 16.100 16.133 16.144 101 11.584 11.536 11.264 156 14.115 14.077 14.013 211 16.100 16.133 16.144 102 11.584 11.536 11.264 157 14.115 14.077 14.317 212 16.621 16.133 16.144 103 11.584 11.536 12.016 158 14.115 14.077 14.317 213 16.621 16.395 16.144 104 11.584 11.536 12.016 159 14.115 14.229 14.317 214 16.621 16.395 16.276 105 11.584 11.817 12.016 160 14.115 14.229 14.317 215 16.621 16.395 16.276 106 11.584 11.817 12.016 161 14.115 14.229 14.317 216 16.621 16.395 16.276 107 11.584 11.817 12.016 162 14.115 14.229 14.317 217 16.621 16.525 16.276 108 11.584 11.817 12.016 163 14.115 14.454 14.317 218 16.621 16.525 16.276 109 11.584 12 12.016 164 14.857 14.454 14.317 219 16.621 16.525 16.276 110 12.136 12 12.016 165 14.857 14.454 14.317 220 16.621 16.525 16.535 111 12.136 12.270 12.016 166 14.857 14.454 14.317 221 16.621 16.589 16.535 112 12.136 12.270 12.016 167 14.857 14.528 14.317 222 16.621 16.589 16.535 113 12.136 12.270 12.016 168 14.857 14.528 14.317 223 16.621 16.716 16.535 114 12.136 12.270 12.016 169 14.857 14.528 14.614 224 16.875 16.716 16.535 115 12.136 12.358 12.373 170 15 14.528 14.614 225 16.875 16.716 16.535 116 12.136 12.358 12.373 171 15 14.748 14.614 226 16.875 16.716 17.042 117 12.136 12.358 12.373 172 15 14.748 14.614 227 16.875 16.716 17.042 118 12.136 12.358 12.373 173 15 14.748 14.614 228 16.875 16.716 17.042 119 12.136 12.533 12.373 174 15 14.748 14.614 229 16.875 16.716 17.042 120 12.136 12.533 12.373 175 15 14.893 15.048 230 16.875 16.716 17.042 121 12.136 12.533 12.548 176 15 14.893 15.048 231 16.875 17.094 17.042 122 13 12.533 12.548 177 15 14.893 15.048 232 16.875 17.094 17.166 123 13 12.533 12.548 178 15 14.893 15.048 233 16.875 17.094 17.166 124 13 12.533 12.719 179 15 15.107 15.048 234 16.875 17.094 17.166 125 13 12.533 12.719 180 15 15.107 15.048 235 16.875 17.094 17.166 126 13 12.533 12.719 181 15 15.107 15.190 236 17 17.094 17.166 127 13 12.790 12.719 182 15 15.107 15.190 237 17 17.094 17.166 128 13.166 12.790 12.719 183 15 15.178 15.190 238 17 17.094 17.166 129 13.166 12.790 12.719 184 15 15.178 15.190 239 17 17.463 17.166 130 13.166 12.790 13.056 185 15 15.178 15.190 240 17 17.463 17.166 131 13.166 13.125 13.056 186 15 15.178 15.190 241 17 17.463 17.290 132 13.166 13.125 13.056 187 15 15.526 15.469 242 17.371 17.463 17.290 133 13.166 13.125 13.056 188 15.423 15.526 15.469 243 17.371 17.523 17.290 134 13.166 13.125 13.056 189 15.423 15.526 15.469 244 17.371 17.523 17.290 135 13.166 13.125 13.056 190 15.423 15.526 15.469 245 17.371 17.523 17.290 136 13.166 13.125 13.221 191 15.423 15.731 15.469 246 17.371 17.523 17.290 137 13.166 13.289 13.221 192 15.423 15.731 15.469 247 17.371 17.523 17.654 138 13.166 13.289 13.221 193 15.423 15.731 15.743 248 17.371 17.523 17.654 139 13.166 13.289 13.221 194 15.423 15.731 15.743 249 17.371 17.523 17.654 140 13.490 13.289 13.221 195 15.423 15.731 15.743 250 17.371 17.523 17.654 141 13.490 13.530 13.221 196 15.423 15.731 15.743 251 17.371 17.644 17.654 142 13.490 13.530 13.702 197 15.423 15.731 15.743 252 17.371 17.644 17.654 143 13.490 13.530 13.702 198 15.423 15.731 15.743 253 17.371 17.644 17.773 144 13.490 13.530 13.702 199 15.423 15.799 16.012 254 18.089 17.644 17.773 145 13.490 13.768 13.859 200 16.100 15.799 16.012 255 18.089 17.704 17.773 146 13.490 13.768 13.859 201 16.100 15.799 16.012 256 18.089 17.704 17.773 147 13.490 13.768 13.859 202 16.100 15.799 16.012 257 18.089 17.704 17.773 148 13.490 13.768 13.859 203 16.100 15.934 16.012 258 18.089 17.704 17.773 149 13.490 14 13.859 204 16.100 15.934 16.012 259 18.089 17.823 18.010 150 13.490 14 13.859 205 16.100 15.934 16.012 260 18.089 17.823 18.010 151 13.490 14 14.013 206 16.100 15.934 16.012 261 18.089 17.823 18.010 152 14.115 14 14.013 207 16.100 16 16.012 262 18.089 17.823 18.010 Table 2. (Continued) Factors for Determining Diameter, D, of Smallest Enclosing Circle for Various Numbers, N, of Enclosed Circles (English or metric units) No. N Center Circle Pattern No. N Center Circle Pattern No. N Center Circle Pattern “A” “B” “C” “A” “B” “C” “A” “B” “C” Diameter Factor K Diameter Factor K Diameter Factor K Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY CIRCLES IN A CIRCLE 85 Fig. 5. N = 7 Fig. 6. N = 8 Fig. 7. N = 9 Fig. 8. N = 10 Fig. 9. N = 11 Fig. 10. N = 12 Fig. 11. N = 13 Fig. 12. N = 14 Fig. 13. N = 15 Fig. 14. N = 16 Fig. 15. N = 17 Fig. 16. N = 19 Fig. 17. N = 20 Fig. 18. N = 21 Fig. 19. N = 22 Fig. 20. N = 23 Fig. 21. N = 24 Fig. 22. N = 25 Fig. 23. N = 31 Fig. 24. N = 37 Fig. 25. N = 55 Fig. 26. N = 61 Fig. 27. N = 97 Fig. 28. B C A Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY [...]... 0. 21 0 5 62 0. 21 0 849 0. 21 1 136 0. 21 1 423 0. 21 1 710 0. 21 1 998 0. 21 2 28 6 0. 21 2 574 0. 21 2 863 0. 21 3 1 52 0. 21 3 4 41 0. 21 3 7 31 0. 21 4 0 21 0. 21 4 311 0. 21 4 6 02 0. 21 4 6 02 0. 21 4 893 0. 21 5 184 0. 21 5 476 0. 21 5 768 0. 21 6 0 61 0. 21 6 353 0. 21 6 646 0. 21 6 940 0. 21 7 23 4 0. 21 7 528 0. 21 7 822 0. 21 8 117 0. 21 8 4 12 0. 21 8 708 0. 21 9 004 0. 21 9 300 0. 21 9 596 0. 21 9 893 0 .22 019 0 0 .22 0488 0 .22 0786 0 .2 21 0 84 0 .2 21 3 83 0 .2 21 6 82 0 .2 21 9 81 0 .22 22 81 0 .22 25 81 0 .22 28 81 0 .22 318 2 0 .22 3483... 0 . 12 0 929 0 . 12 111 0 0 . 12 12 91 0 . 12 1473 0 . 12 1655 0 . 12 1837 0 . 12 2 019 0 . 12 22 01 0 . 12 23 84 0 . 12 25 67 0 . 12 27 50 0 . 12 29 33 0 . 12 311 7 0 . 12 3300 0 . 12 3484 0 . 12 3668 0 . 12 3853 0 . 12 4037 0 . 12 422 2 0 . 12 4407 0 . 12 45 92 0 . 12 4778 0 . 12 4964 0 . 12 515 0 0 . 12 5336 0 . 12 5 522 0 . 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