APPLIED MECHANICS. 



[RCPERPIOIAI. MENSURATION. 



Example 4. To exprem 23" 27' 68" 14 English in 

 French. The quantity, decimally expressed, u 



23-4fi613 English. 

 Add one-ninth . 2 -60735 



26-0735 French. 



Tlie circnmference of every circle is 3-1416 times its 

 diameter, or 6 2832 times its radius or half diameter. 



Hence to find the circumference of a circle whose 

 diameter or whose radius is given, 



-Multiply the diameter by 3'1416, or the radius 



by 6-2832. 



Exam^e 1. To find the circumference of a pulley 

 whose diameter is 4 feet 6 inches. 



4 feet 6 ins., or 54 ins. X 3 1416=169 -6464 ins., or 



14 feet 1* inches nearly. 



Example 2. To find the circumference of the pitch 

 circle of a wheel, the pitch radius being 11 inches. 



11 ins. X 6-2832=69-1152 ins., or nearly 69i ing. 

 Since the decimal '1416 is nearly equal to ftfi (which 

 is accurately -142587), we have the approximate rule for 

 finding the circumference of a circle. 



Rvle. Multiply the diameter by 3i, or (since 3> = V) 

 multiply the diameter by 22 and divide by 7, or multiply 

 the radius by 44, and divide by 7- 



Example 3. To find the circumference of a pulley 

 4 feet 6 inches diameter. 



4 feet 6 ins. X 3 = 13 feet 6 ins. 



4 feet 6 ins. x (or -i- 7) = 7| ins. nearly. 



Circumference 14 feet 1| ins. nearly. 



Example 4. To find the circuinfereuce of a wheel 

 whose radius is 11 inches. 



_69 ins. nearly. 



The circumference of an ellipse is that of a circle whose 

 diameter is a mean between the two axes of the ellipse. 



Example 5. To find the circumference of an ellipse 

 having a longer axis 18 ins., and a shorter axis 12 ins. 



MeM) 1 l^=15 ing. x 3-1416 = 47 '124 ins. or 3 feet 



J 



lliins. 



The converse operation of finding the diameter of a 

 circle having a given circumference is, 



Rule. Divide the circumference by 31416, or mul- 

 tiply it by -31831, the reciprocal of 3-1416. 



Example 6. To find the diameter of a pulley whose 

 circumference is 14 feet 1 in. or 1C9"625 ins. 



^ -54 ins., or 4 feet 6 ins. very nearly. 



o ' 14 1 b 



or 169 625 X '31831=54 ins. nearly. 

 The following approximation is generally sufficiently 

 seen for practical purposes, 



Rule Multiply the circnmference by 7, and divide by 

 22, the quotient is the diameter. 



Example 7 To find the diameter of a wheel having 

 23 teeth, each of 3 ins. pitch. 



23 X 3 ins. 69 ins. the circumference of the pitch 

 69x7 



circle, and 



22 



"22 ins. nearly. 



The pitch radius is | of 22 ins. = 11 ins. 



MENSURATION OF SUPERFICIES. 



The object of Mensuration of Superficies is to diwover 

 the number of square units in a figure, the form and 

 dimensions of whose boundary are known. The simplest 

 kind of figure U that bounded by three straight lines, the 

 triangle. 



To find the area of a triangle. 



Jtule Multiply the length of any one of the sides by 

 that of the perpendicular let fall upon it from the oppo- 

 site angle (or, briefly, multiply the base by the height), 

 and halve the rvmilt. 



Rxamptt \. To find the area of a triangle having base 

 2 feat 3 ins. , and height 1 foot 8 ins. 



2 feet 3 ins. - 27 ins., and 1 foot 8 ins. 20 ins. 



Area, *f--270sq. in, 



Example 2. To find the area of a triangular field, hav- 

 ing base 23 chains 8 links, and height 14 chains, 73 links. 



23-08 ch. X.14-73ch. 



I 



- 169 9842 sq. chains ; or 16-99842 



acres, very nearly 17 acres. 



The parallelogram is a fijrnre bounded by four straight 

 lines, the opposite sides being equal and parallel to one 

 another. When the angles of the parallelogram :ir> 

 right angles, or, as it is said in ordinary language, when 

 thu aides are square to one another, the figure is called 

 a rectangle or oblong. 



To find the area of a parallelogram. 



Ride. Multiply the base by the perpendicular height. 

 When the figure is an oblong, we say, multiply the 

 length by the breadth. 



'tmple 3. To find the area of a parallelogram, base 

 2 feet 3 ins. , height I foot 8 ins. 



27 ins. X 20 ins. = 640 sq. ins. 



Example 4. To find the surface of a cylinder, diameter 

 4 feet, length 14 feet. 



The surface of a cylinder, if developed or unrolled, 

 would form a rectangle whose length is the length of the 

 cylinder, and breadth the circumference. Its area is 

 therefore found by multiplying the length by the dia- 

 meter, and by 3f. 



14 feet x 4 feet X 3f = 176 sq. feet 



The trapezoid is a figure bounded by four straight 

 lines, two of which are parallel. To find the area of a 

 trapezoid. 



Rule. Multiply half the sum of the two parallel sides by 

 their perpendicular distance apart. 



Example 5. Required the area of a trapezoid having 

 parallel sides, respectively 5 feet and 7 feet, 3 feet apart. 



ft 



Any figure bounded by straight lines may bo divided 

 into triangles ; and the sum of the areas of all the tri- 

 angles into which it is divided, is the area of the whole 

 figure. 



Of surfaces bounded by curved lines, the most regular 

 and frequent is the circle. 



To find the area of a circle. 



Rule. Multiply the square of the diameter by "7854. 

 Or, multiply the circumference by the radius, and halve 

 the product. 



Example 6. Required the area of a circle 4ft. diara. 

 4 X 4 X "7854 = 12-5664 sq. feet. 



Otherwise, the radius is 2 feet, and the circumference 

 is 4 X 3-1416 = 12-5G64, and the area is 



12-5604 circumf. X rad. 



= 12 -5664 sq.ft. 



The following method furnishes a near approximation 

 to the area. 



Rule. Multiply the diameter by itself. 

 Take half the product, to which add its half, and the 

 seventh part of that half.* 



Example 7. Required the area of a circle 4 feet 

 diameter. 



4 X 4 = 16 and i of 16 = 8 

 * of 8 = 4 

 f Of 4 = 57 



Area 125" sq. ft. nearly. 



Example 8. Required the area of a circle 37| ins. 

 diameter. 



Tlii approximate ml* i that derived : 



Since one-hair, decimally, b -5000 

 One-half of -6000 = -2.MO 



One-wrenthof -2500 =-0367 



Their nm = -7857 



which 1< newly equal to -78M, the proper multiplier of the tqnaro of tht 

 diameter. 



