902 



THE IRRIGATION AGE. 



THE PRIMER OF HYDRAULICS* 



By FREDERICK A. SMITH, C. E. 



Article VI. Mensuration of Solids. 



1. Prisms. 



A prism is a solid bounded by 2 parallel polygons as 

 end surfaces and parallelograms as sides; if the end sur- 

 faces are parallelograms the solid is called a parallelepiped,, 

 and if it is bounded by 6 rectangles then it is called a right 

 parallelepiped, and if in such a solid 

 all faces are squares it is called a 

 cube. Thus Fig. 46 represents a 

 cube and if AB = 1 ft. the volume 

 of the solid is called 1 cubic ft. ; if 

 AB = 1 inch the solid is called one 

 cubic inch, etc. Such a cube is used 

 as an unit to measure the volume 

 of the other solids with. 



The volume of a prism is found 

 by multiplying the area of the base 



I 

 J 



s 



Fig. 46 



by the perpendicular height; the height is to be taken per- 

 pendicular between the two end surfaces. Thus in Fig. 47 



the two prisms 

 I and II con- 

 tain the same 

 volume if the 

 bases have the 

 same area and 

 the heights are 

 equal ; observe 

 the height in 

 Prism II to be 

 the perpendicu- 

 lar distance be- 



J-4-- 



Fig. 47 



tween the end surfaces. 



Problem. Find the volume of a rectangular prism with 

 a base 3' X *' and a height of 5 ft. (Fig. 47-1.) 



Solatia)!. Let V = volume, then V = 3 X 4 X 5 = 60 

 cub. ft. 



The surface consists of 6 rectangles, 2 end rectangles, 

 each 3' X 4', or 12 sq. ft. each = 24 sq. ft. 



4 sides, each 4' X 5', or 20 sq. ft. each = 80 sq. ft. 



Total area in surface, 104 sq. ft. 



2. Pyramids. 



A pyramid is a solid having a plane polygon for its base 

 and is bounded by as many triangles as the base has sides, 

 all having a common vertex P; thus in Fig. 48 let A, B, C, 

 D, E, be the base and P the vertex of the pyramid, then the 

 base is a pentagon, and the other 

 surfaces are 5 triangles with the 

 point P in common ; if a perpen- 

 dicular is drawn from P to the base 

 like P H, then P H is called the 

 height of the pyramid. 



Rule. The volume of a pyramid 

 is found by multiplying the area of 

 the base by the height and divide by 

 3. The area of the surfaces bound- 

 ing a pyramid consists of the sum of 

 the area of the base plus the areas 

 of the triangles ; thus in Fig. 48 the 



surface of the pyramid is equal to area of the pentagon A B 

 CUE plus the 5 triangles, A P B + 

 BPC+CP, D+DPE + 

 EPA 



If the top is cut away from a pyra- 

 mid like in Fig. 49, it is called a frus- 

 tum of a pyramid if the area of the 

 base = B and the area of the top = C 

 and the height h (distance between top 

 and bottom), then the volume of the 

 h 



Fig. 49 frustum V = ( B + C + V B O- 



3 

 "Copyright by D. H. Anderson. 



Fig. 48 



3. Prismoids. 



In Fig. 50 is shown a solid known as a prismoid ; it is 

 a solid bounded by two parallel end surfaces and the sides 

 are formed by trapezoids. The perpen- 

 dicular distance h between the top 

 surface B and the bottom surface 

 A, Fig. 50, is called the height of 

 the prismoid ; if a section M is taken 

 through the solid, midway between 

 the two end surfaces, then the vol- 

 ume V of the solid is found by the 

 following formula : 



h 

 V (A 



Fig. 50 6 



expressed in words reads : The volume 

 found by adding 4 times the area of the 



4 M + B), which 



of a prismoid is 

 middle section to 



the sum of the top and bottom surface and multiply by 1/6 of 

 the height. 



Application. In Fig. 50, let A be a rectangle 3X5 ft., 

 B a rectangle 2 X 4 ft., and let h = 9 ft. ; find volume of 

 prismoid. 



Solution. The area of A = 3 X 5 = 15 ; the area of B 



3+2 5+4 

 = 2X4 = 8; the sides of the rectangle M are -- and 



2 2 



= 2.5 and 4.5 respectively; hence M = 11.25 and 4 M = 45; 



' h 

 apply now formula V = (A + 4 M + B.) 



6 

 9 

 V = (15 + 45 + 8) 1.5 X 68 = 102 cub. ft. The 



6 



surface of such a solid is equal to the sum of surfaces of the 

 bounding side; thus let S = the surface of the prismoid just 

 considered, then S =- A + B + 2C+ 2>; where C and D 

 respectively are the trapezoidal sides; care must be taken in 

 figuring the surfaces not to confound the slant height of 

 the trapezoids with the vertical height of the body. 



4. Cylinders. 



To find the volume of a cylinder multiply the area of 

 the base by the height, because a cylinder can be considered 



a prism of a very large num- 

 ber of sides and therefore the 

 volume of the cylinder is 

 found by multiplying the area 

 of base by height, just like the 

 prism. When the two bases 

 of a cylinder are circles with 

 the centers perpendicular above 

 each other, as shown in Fig. 

 51. it is called a right cylinder, 

 and the line OO' is called the 

 axis or height of the cylinder ; 

 if the radius of the cylinder is r and the height h, then V = 

 Tt r" h. 



The surface of the cylinder consists of the two end cir- 

 cles = TT r' plus the shell, which is a rectangle h ft. high 

 and the width of which is 2 ff r, hence S = 2 TT r" + 2 v r h, 

 which can be simplified to : S = 2 v r (r + h) ; this can be 

 expressed in words : The surface of a right cylinder is equal 

 to the sum of radius and height multiplied by the circum- 

 ference. To apply to a problem let the radius of a cylinder 

 be 3 ft. and the height 5 ft.; find volume and surface. 

 V = 3.1416 X 3 X 3 X 5 = 141.372 cub. ft. 

 S 2 X 3.1416 X 3 X 8 = 6.2832 X 24 = 150.81 sq. ft. 

 Cylinders may be oblique, as shown in Fig. 52; then the 

 height must be perpendicular to the two end surfaces for the 

 figuring of the volume, but parallel to axis when figuring the 

 area of surface. 



5. Cones. 



A cone, shown in Fig. 53, may be considered as a pyra- 

 mid with a large number of sides; it has a base A which is 

 a plane surface from which innumerable tri- 

 angles reach to a common point V , called the 

 vertex. A line VV , drawn perpendicular 

 from the vertex to base is called the height of 

 the cone; if the base A is a circle and VV 

 strikes its center it is called a right circular 

 cone. The volume of such a cone is found as 



Fig. 51 



follows: V = 



and the surface S 



Fig. 53 



