822 



THE IRKIGATION AGE. 



THE PRIMER OF HYDRAULICS* 



By FREDERICK A. SMITH, C. E. 



Fig. 32. 



8. The circle and its properties. 



The circle is an uniformly curved line, each point of 

 which has the same distance from a point within called the 

 center. Thus in Fig. 31 let O 

 be the center of the circle, draw 

 any line OA to the circumference 

 of the circle, then OA is called a 

 radius ; draw another radius OB, 

 then AO = BO; the part of the 

 circle line lying between A and B 

 is called an arc, and the angle AOB 

 is called a central angle, spanning 

 arc AB; a straight line AB is called 

 a chord. Equal chords span equal 

 arcs and ct^ual central angles. Thus, 

 F '8- si- if AB and JJP are equal, then the 



arcs AB and CD are equal, and the central angles A OB 

 and COD are also equal. The area inclosed between two 

 radii and an arc is called a sector, and the area lying between 

 a chord and arc is called a segment. Thus sectors and 

 segments of equal central angles or having equal arcs or 

 chords in the same circle are equal. If a line ST is drawn 

 so as to touch the circle line in 

 just one point A as indicated in 

 Fig. 32, it is called a tangent, and 

 if A is connected to O then AO 

 stands perpendicular to ST 

 in point A. A line like CD which 

 cuts the circle line in two points 

 E and F is called a secant. A 

 chord GH passing through the 

 center of the circle is called a 

 diameter. An angle like AHG 

 having its vertex in the circum- 

 ference of the c ; rcle is called a 

 periphery angle and is equal to 54 of the central angle AOG, 

 both standing over the same arc AC; thus central angles are 

 measured by the full arcs and periphery angles by l / 2 of the 

 arcs upon wh : ch they stand. Any periphery angle standing 

 over the diameter is consequently equal to 90. 



Thus in Fig. 33 if AB is a di- 

 ameter then angle ACB = 90 ; no 

 matter which point of the circumfer- 

 ence is connected with the ends of a 

 diameter rt always gives a right 

 angled triangle as ADB. 



The ratio of the circumference 

 of a circle to its diameter is the deci- 

 mal fraction 3.1416; this number, 

 which is also called v (pi), is very 

 important and enters in all circular 

 problems, so that it should be known 

 by heart; for ordinary work the fraction 3.14 or 31-7 may be 

 used. Thus if the diameter of a circle is given the circum- 

 ference is formed by multiplying the diameter by ; inversely 

 the diameter of a circle is found by dividing the circumference 

 by f. Thus let the diameter of a circle be 3 ft., then the cir- 

 cumference equals 3 X 3.14 == 9.42 ft. To find the diameter 

 of a circle having a circumference of 10 ft., divide 10 by 3.14 

 = 3.184 ft. 



The area of a circle is found by multiplying the square of 

 the radius by T . To illustrate their relations by formulae let 

 TT = 3.1416, the circumference /> the radius r and the area a, 

 then the following formulae obtain : 

 /> = 2,r r. (1.) 

 a=vr\ (2.) 

 r-P/Z v (3.) 



= V 0/1T. (4.) 



Applied problems. 



1. What is the circumference of a circle having a radius 

 of 5 inches? 



Solution. Substitute in formula 1 the given quantities, then : 



p = 2 X 3.1416 X 5 = 31.416 inches. 

 2. What is the area of a circle having a radius of 5 inches? 



Copyright by D. H. Anderson. 



Fig. 33. 



Solution. Substitute in formula 2 the given quantities : 



a =: 3.1416 X 5 s = 3.1416 X 35 = 78.54 square inches. 



:!. What is the radius of a circle, having a circumference 

 rn.416 inches? 



Solution. Substitute the given quantities in formula 3 : 



r = 31.416 -r- 2 X 3.1416 = 5 inches. 



4. What is the radius of a circle having an area of 78.54 

 square inches? 



Solution. Substitute the given quantities in formula 4 : 



r = -\/ 78.54 -f- 3.1416. 



r = y' 25 = 5 inches. 



~>. Find the diameter of a cast iron pipe having a cross 

 sectional area of 32 square inches. 



Solution. Here the area is given and the radius wanted, 

 so use formula 4 and substitute the given quantities : 



r = \/ 32 -f- 3.1416. 



Fig. 34. 



ft. 



to 

 let 



r = v 10.186 = 3.191 inches. 



This makes the diameter 2X3.191=6.382 in., or 6f inches. 



.">. Find the area of a Segment in a circle, Fig. 34, having 

 a radius of 3 ft. and in which the 

 central angle ACB = 90. 



Solution. The area of the whole 

 circle is equal to ^ r"; hence the area 

 of Sector ACB = v r*/i; the area 

 of triangle ACB = r/2, hence the 

 area of the segment equals 77 r 2 /4 - 

 r/2 ; substitute given values : 



1. Principles and Definitions. 

 3.1416 X3X3-=-4 3X3 

 -4- 2. 



7.0686 4.5 = 2.5686 square 



Article IV. Trigonometry. 



Trigonometry teaches how 

 compute the sides and angles of triangles. In Fig. 35. 



ABC be any angle a and draw 

 from any point A a line . / /) 

 perpendicular to EC, then ABD 

 is a right angled triangle ; drop 

 another perpendicular E F from 

 any point E, then E B F is also 

 a right angled triangle, and the 

 "ft 1"* D C, two triangles are similar ; hence : 



E F : B E = A D : A B ; 

 also B F : B E - B D -. A B; 

 also' F : B F = A D : B D and 

 B F : E F = B D : A D. 



No matter where a perpendicular might be drawn paral- 

 lel to A D, the above ratios remain the same so long as the 

 angle a remains the same ; it is also seen as soon as the 

 angle a changes then the ratios of the sides change also; 

 thus the ratio EF/BE changes for every degree of angular 

 measure of a, and if a table be computed giving the value 

 of this ratio for all the degrees of angle from to 90 

 we can from such table draw conclusion as to the value of 

 the angle if the sides EF and BE are known, or we can 

 find the sides when the angles and one side are given. 



These ratios of the sides of a right angle triangle are 

 called trigonometric functions and are shown in a table 

 embodied in this work. 

 2. The Functions. 



In Fig. 36 let the angle ACB 90, and let the sides 

 of the triangle be a, b and c re- 

 spectively, then the following ratios 

 can be formed between the 3 sides : 

 a/c = sinus a, abbreviated sin a 

 b/c = cosinus a, abbreviated cos a 

 a/b = tangent a, abbreviated tg a 

 b/a = cotangent a, abbreviated cotg a 

 c/a = secant a, abbreviated sec a 

 c/b = cosecant a, abbreviated cosec a 

 The secant is the reciprocal of 

 the cosine and the cosecant is the reciprocal of the sine, 

 and both are but little used. 



If the side c is = 1 then the side n is the sine and the 

 side b is the cosine of a ; in the relation to angle the side a 

 K the cosine and the side b the sine of (3; since a and $ are 

 complementary angles it is seen that: 

 sin a = cos |3 

 cos a = sin 

 tg o. = cotg 

 cotg a = tg 

 Thus it follows that for instant, sin 1 = cos 89, or tg 



