913 



APPLIED MECHANICS. 



[DUODECIMALS. 



6. The result of the two multiplications are added, 

 carried by twelves. 



ii-ing, the same process is repeated, remembering 

 that in the cubed result, or the result of multiplying 

 throe quantities of feet and inches, there are four 

 places viz., cubic feet ; twelfths, each 144 cubic inches; 



4 ft 8 iu. 

 3 10 



twelfths of twelfths, each 12 cubic inches ; and cubic 

 inches. 



Example. Required the cubic contents of a cistern, 

 the base of which is 4 ft. 8 ins. long by 3 ft. 10 ins. wide, 

 and thu height 5 ft. 7 ins. 



We first hud the area of the base. 



3r. 106. 

 14r. Od. 



So. 



17 ft 10 pts. 8 ins. 



80 



12 " 6 P* 8 - 8 ins - 



Operation. 8 ins. X 10 ins. = 80 sq. ins. and 

 carry 6 and write 8, a. 

 4 ft. X 10 ins. = 40, add 6 - 46, and = 3 ft. 10 pts. b, c. 



8 iiis. X 3 ft. - 24, and 



24 



'., =. 2 ft. pta. 



carry 2, and write 0, d. 

 4 ft X 3 ft. = 12 ft, add 2, 14 ft, . 

 Sum the two lines. 



We now multiply the area of base by the height 

 ft. pu. ins. 



Operation. 8 X 7 = 56, and 



Area 17 10 8 

 6 7 



10 6 2 8 



89 5 4 



99 10 6 8 



56 



= 4 ft. 8 ins , carry 4. 



10 X 7 + 4 

 17 X 7 + 6 



74 



74, and 1 , t = 6 f t 2 ins., carry 6. 



125, and "^ = 10 ft. 5 ins. 



8 X 6 = 40, and 



10 X 5 + 3 

 17 X 5 + 4 



40 



j2 = 3 ft. 4 ins. carry 3. 



to 



53, and ^ = 4 ft. 5 ins., carry 4. 



89. Add up the lines. 

 The cubic contents are therefore 

 99 cub. ft 10 twelfths, 6 hundred and forty fourths, 8 cub. ins., or 99 ft. 1520 cub. ins., because 10 twelfths, 



each of 144 cubic inches = 1440 cubic inches. 



hundred and forty-fourths, each 12 cub. ins. = 72 



8 cubic inches 



Water is generally taken as the standard of specific 

 gravity, and it fortunately happens that 1 cubic foot oi 

 water weighs very nearly 1000 ounces. By reference to 

 any table of specific gravities, the weight of a cubic foot 

 of each material is given in ounces ; and from this may 

 be readily derived 



1. The weight of 1 cubic foot in lbs. , vjz. , ^th of the 

 specific gravity or weight in oz. , because 1 oz. =^- of 1 Ib. 



2. The weight of 1 cubic inch in ounces viz., the 

 specific gravity divided by 1728, because there are 1728 

 cubic inches in 1 foot and the weight of 1 cubic inch in 

 lbs, viz., the specific gravity divided by 1728, aii'l tint 

 quotient by 16, or the specific gravity divided by 27648. 



3. The number of cubic inches in 1 Ib., viz., 10 times 

 1728, or 27648 divided by the specific gravity. 



4. The number of cubic feet in 1 ton, viz., 2240 lbs. X 

 16 oz., or 35840 oz. divided by the specific gravity. 



Example 1. Required the weight in lbs. of 7 cubic 

 feet of English oak. 



Spec. erav. 900 



~16 x 7=393$ lbs., or 3cwt.2qr. IJlbs. 



Esample 2. Required the weight of 178 cubic inches 

 of cunt-iron. 



Spe* grav. 7200 



01 



Example 3. Required the number of cubic inches In 

 2 cwt. (or 224 lbs.) of gun metal. 

 27C48 



X 



Spec. grv. 8800 



Example 4. Required the number of cubic feet in 20 

 tons of granite. 



Spec. 



X 2 - 



Cubi feefc 



., 



1520 



In estimating the weight of wrought-iron cylindrical 

 bars, the following method is a very near approxima- 

 tion : 



Rule. Square the diameter reduced to eighths of an 

 inch, multiply by the length in feet and by 4y, and puiut 

 off two decimal places for the weight in Ibs. 



Example 1. Required the weight of a cylindrical bar 

 of wrought-iron 5J ins. diameter, and 7 ft. 7 ins. long. 

 5t ins. =47 eighths, and 7 ft 7 ins. =7^j ft 

 47x47x7jVx4}.<=6U4<.)l. 



The weight is therefore 694 lbs., or 6 cwt qr. 22 lbs. 



For wrought-irou bars of rectangular section : The 

 width of these advances by quarters, and the thickness by 

 eighths of an inch. 



Rule. Multiply the breadth in quarters of an inch, 

 by the thickness in eighths, by the length in feet, and by 

 10)i, and point off two decimal places for the weight 

 in Ibs. 



Example 2. Required the weight of a bar of wrought- 

 iron 2i inches wide, J in. thick, aud 3 feet long. 



Width 9 qrs. X thickness 3 eighths X length 3 ft. x 

 101=850 ; therefore the weight is 8 '5 lbs. 



For wrought-iron plates. 



Rule. Multiply the area in square feet by the thick- 

 ness in eighths of an inch, and by 5 for the weight in Ibs. 



J-'xample 3. Required the weight of 123 square feet of 

 boiler-plate J and jL in. thick. 



Thickness } and -,', or 2 eighths of an inch. 

 123 X 2J X 5 = 1538 Ibs., or 13 cwt. 2 .jr. .'0 lbs. 



For cast-iron balls. 



Rule. To the cube of half the diameter (in inches) add 

 .ts eleventh part for the weight in Ibs. 



Example 4. Required the weight of a cast-iron ball 

 ) ins. diameter. Half the diameter is 3 ins. 

 And 3x3x3 . . = 27 

 Add one- eleventh of 27 * 2 '5 



Weight 



20'51bi. 



