SUPERFICIAL MENSURATION.] 



APPLIED MECHANICS. 



909 



or 37-625 X 37-625=1415-64 sq. ins. 



i of 1415 64 = 707 82 nearly. 

 I of 707-82 = 353-91 

 fof 353-91= 50-56 



Area 



Tlie correct area is 



1112-29 nearly. 

 1111-84 



Error in the approximate method 0-45 sq. ins. 



The converse operation of finding the diameter of a 

 circle when its area is given, is the following : 



Rule. Divide the area by '7854, or multiply the area 

 by 1 -273 (the reciprocal of "7854), and extract the square 

 root of the result. 



ExampU 9. Required the diameter of a circle having 

 12-5664 sq. ft. area. 



^or 12-5064 x 1-273=16 ; and the square root of 



* I o>4 



1C is 4 feet. 



Approximate method. 



liule. To the given area add its fourth part, and ^th 

 of that fourth, or the fourth with one decimal place 

 pointed off, and extract the square root.* 



Esample 10. Required the diameter of a circle whose 

 area is 1111-84 tq. ins. 



Area = llll-84 

 Add Jof 1111 84= 277 !W 

 Vo of 277-96= 27796 



1417596 



The square root of which is 37 '65, or 37 J inches 

 nearly. 



The area or an ellipse is "7354 times the product of its 

 two axes. 



ExampU. 11. Required the area of an ellipse whose 

 axes are 2 feet 3 inches and 1 foot 8 inches respec- 

 tively. 



27 ins. X 20 ins. X 7854 = 424 -110 sq. in. 



or> ~~T4T~ = 29452so .- feet - 



The area of a sector of a circle is the name part 

 of the area of the whole circle, as the arc of the sector 

 is of the whole circumference. 



Example 12. Required the area of a sector ; radius 

 12 ins., and arc 67 30'. 



The area of the whole circle is 12 X 12 X 3-1416 

 = 452-39 sq. ins. 



C w Q,V - G7J 67* X 2 135 

 ---= 



the whole circumference ; there- 



Area of sector is Aths of 452 '39 sq. ins. = 84 '823 sq. ins. 

 In measuring the area of figures bounded by Hues of 



Fig. 270. 



parts at the points 1, 2, 3, 4, 5. If we bisect each of 

 those parts at C, D, E, F, G, H, and draw perpendiculars 

 to the line A B, meeting the curve in I, P, K, Q, <fec., 

 by drawing parallels through these points, meeting per- 

 pendiculars through the points A, C, D, <fec. , we form a 

 number of rectangles, each of which is very nearly equal, 

 in area, to the part of the curve contained between its 

 two perpendicular sides. The area of the whole curved 

 figure is, therefore, very nearly equal to the sum of the 

 areas of all those rectangles. Now, as the area of each 

 rectangle is equal to its base multiplied by its perpen- 

 dicular height, and as the bases of all are equal, the area 

 of all the rectangles is equal to the length of one of the 

 bases, such as C D multiplied by the sum of all the 

 heights I P, K Q, L R, <tc. It is to be observed, that 

 the distances A C and B H of the first and last perpen- 

 diculars from the extremities of the line, are each half 

 the distance C D or D E between any two perpen- 

 diculars. Or, we may view the question in another 

 way, thus : Having divided the line A B as before, and 

 set off the perpendiculars I P, K Q, <fec. (technically 

 called offsets), we may find a mean or average V X or 

 W Y, by adding together the lengths of all the perpen- 

 diculars, and dividing their sum by their number, and 

 then the whole rectangle V W Y Z has very nearly 

 the same area with the curved figure. Hence the fol- 

 lowing^ 



Rule I. Multiply the sum of all the offsets by the 

 distance between two adjoining offsets ; or, 



Rule II. Add together the lengths of all the offsets, 

 divide by their number, and multiply by the whole 

 length of the base line. 



Example 13. The length of A B is 24, and the offsets 

 are I P = 3, K Q = 5, L R = 7, M S = 8, N T = 6, 

 O U = 4 : required the area. 



By Rule I. There are six offsets ; therefore the dis- 

 tance between any two adjoining is -g- = 4 (note, A C 



and B H are each 2), and the area is 



(3 + 6 + 7 + 8 + 6 + 4) X4 = 132. 

 By Rale II. The mean of all the offsets is 

 3 + 5 + 7 + 8 + 6 + 4 _ and the ^ ig 



C 



5-5 X 24 = 132. 



The surface of a cone may be developed, or unrolled, 

 into the form of a sector of a circle, and its area found 

 by the rule. Or thus, 



Rule. Multiply the diameter of the base by the 

 length of the side (from the apex to the edge of the base), 

 and by 1-5708. 



Example 14. Required the surface of a cone, having 

 a base 12 inches in diameter, and 18 inches length of 

 side. 



12 x 18 X 1-5708 = 339-2928. 



The surface of a sphere is 4 times that of any 

 of its circular sections passing through the centre. 

 Hence, to compute the surface of a sphere 



Rule. Multiply the square of the diameter by 

 31416. 



Example 15. Required the surface of a sphere 

 18 inches diameter. 



---V 



IS X 18 X 3-1416 = 1017-8784 square inches. 

 MENSDRATION OF SOLIDS. 



irregular curvature, as A L B S, the most convenient 

 method is to draw through the figure a straight line A B, 

 the longest possible, and to divide it into numerous equal 



* This rule, like the former, is thus derived, 

 1 ' = 1-000 



J of 1 = 0-250 

 T\, of 0-25 = 0025 



1-275, which i nearly 1-273. 



By mensuration of solids we discover the number of 

 cubic units in a body of given form and dimensions. A 

 parallelepiped, or prism, has two opposite sides equal 

 and parallel to one another. Either of these is called a 

 base, and their perpendicular distance apart is called tiie 

 height or altitude of the prism. A cylinder has a circular 

 base. 



For finding the solid contents of a prism or cylinder 



Rule. Multiply the area of the base by the height. 



Example 1. Required the capacity of a cistern, whose 



