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Log area == twice log d+log 1,8950909, the nat. number of which will

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ameter of greater circumference, and d, that of the lesser circumference.

67. Area of a Sector of a Circle. (See Fig. 8.) Arc E G F is the arc

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but arc E G F

divided by three

8 times the arc E G, less the chord E F, the difference arc E G F (i. e.,)

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69. Area

(8 E G — E F). EG, the chord of the arc,

47. For we have O E

E H,

to the hypothenuse,

E H, . ·. √ (0 E2 — E H2) — O H, and O E — + H G2)

E G.

degrees of the <E OFX diameter by the constant number, or factor 0,008727, i. e., area = = d a × 0,008727 where a < E O F in degrees and decimals of a degree.

70. Segment of a Ring. NK M F G E, the area of this segment may be found by adding the arcs N K M and E G F of the sector O N K M and multiplying their sum by E N, the height of the segment of the arc NK Marc E G F

ring, i. e., area =

71.

2

XG K.

Segment of a Circle. Let E G F be the given segment whose area

8

is required. By 2 67 find the area of the sector O E F, from which take the area of the AO E F, the difference will be the required area.

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the chord by the height, add the cube of the height divided by twice the chord of the segment, the sum will be the required area.

73. Or, divide the height G H by the diameter G L of the circle to three places of decimals. Find the quotient in the column Tabular Heights of Table VII., take out the corresponding area segment; which, when multiplied by the square of the diameter, will give the required

area.

74. When G H, divided by the diameter G L, is greater than,5, take the quotient from 0,7854, and multiply the difference by the square of the diameter as above, when G H divided by G L does not terminate in three places of decimals, take out the quotient to five places of decimals, take out the areas less and greater than the required, multiply their difference by the last two decimals of the quotient, reject two places of decimals, add the remainder of the product to the lesser area, the sum will be the required tabular area.

Example. Let G H 4, and the chord E H

9

E F. By

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diameter.

Euclid III. 35, HGH L EH • HF = E H2 81;

GL

—H L; consequently, by addition, 20,25 +4=24,25 = G L
And 4 divided by 24,25 0,16494 tabular number.
Area corresponding to 0,164 =,084059

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75. Let E F V S (Fig. 8,) be a circular zone, in which E F is parallel to S V, and the perpendicular distance E t is given; consequently E S tv may be found by Euclid I. 47, s t Į (S v — E F)=d, and S v - d

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And by Euclid III. 3, by bisecting the line, O Z is at right angles to F V; and by Euclid III. 31, the UV F is a right angle; and by Euclid VI. 2 and 4, U V

2 o x.

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=

Now having ox and o y — radius, we can find the height of the segment x y; ... having the height of the segment x y, and diameter W F of the segment F Y V, we can find its area as follows:

The area of the trapezium E F VS = (EF + V S) × Et, to which add twice the segment F Y V, the sum will be the required area of the zone E F V S.

In fig. 8, let E F = a, S V = b, E t = p, S t = d — § (S v — E F),

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And by substituting the values of E S, V t and 2 E t, we have

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d, and t v = e, to find the diameter W F and height x y. Here d = 5

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WF

36,07

diameter; and having the diameter W F and height x y, the area of the segment, subtended by the chords F v and E S, can be found by Table VII., and the trapesium E F vt by section 24.

OF A CIRCULAR LUNE.

76. Let A C B D, fig. 10, represent a lune. Find the difference between the segment A C B and A D B, which will be the required area.

77. Hydraulic mean depth of a segment of a circle is found by dividthe area of the segment by the length of the arc of that segment. Of all segments of a circle, the semi-circular sewer or drain, when filled, has the greatest hydraulic mean depth.

78. The greatest isoceles A that can circumscribe a circle will be that whose height or perpendicular C F is equal to 3 times the radius O E.

79. Areas of circles are to one another as the squares of their diameters; i. e., in fig. 8, circle AK BI is to the area of the circle C G V L as the square of A B is to the square of C D.

80. In any circle (fig. 9), if two lines intersect one another, the rectangle contained by the segments of one is to the rectangle contained by the segments of the other; i. e., O M X MC = FMX M H,

or O AXA C = F А Х А Н.

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81. In fig. 8, a TbT-ITX KT

square of the tangent T M.

82. In a circle (fig 9), the angle at the centre is double the angle at the circumference; i. e., < C A B — 2 < C O B.

2 <CO B.

Euclid III. 20.

83. By Euclid III. 21, equal angles stand upon equal circumferences;

i. e., < C O B = < C L B.

84. By Euclid III. 26, the <BCL=<B LC
< BLC = < C O B.

85. By Euclid III. 32, the angle contained by a tangent to a circle, and a chord drawn from the point of contact, is equal to the angle in the alternate segment of the circle; i. e., in fig. 9, the < T B C = <BOC = } < C A B. This theorem is much used in railway engineering.

86. The angle T B C is termed by railroad engineers the tangential angle, or angle of half deflection.

87. To draw a tangent to a circle from the point T without the circle. (See fig. 9.) Join the centre A and the point T, on the line A T describe a semi-circle, where A cuts the circle, in B. Join T and B, the line T B will be the required tangent or the square root of any line Q TH=TB; i. e., √ (Q T H) = T M.

Then from the point T with the distance T B, describe a circle, cutting the circle in the point B, the line T B is the required tangent.

In Section 81, we have T a • TB = T M2, T M2, ... √(T a • √(Ta • TB) T M, and a circle describe with T as centre and T M as radius will determine the point M.

OF THE ELLIPSE.

88. An ellipse is the section of a cone, made by a plane cutting the cone obliquely from one side to the other.

DE

the transverse axis, and

Let fig. 39 represent an ellipse, where A B the conjugate axis. F and G the foci, and C the centre. Construction.-An ellipse may be described as follows: Bisect the transverse axis in C, erect the perpendicular C D equal to the semi-conjugate, from the point D, as centre with AC as distance describe arcs cutting the transverse axis in the foci F and G. Take a fine cord, so that when knotted and doubled, will be equal to the distance A G or FB. At

the points or foci F and G put small nails or pins, over which put the line, and with a fine-pointed pencil describe the curve by keeping the line tight on the nails and pencil at every point in the curve.

89. Ordinates are lines at right angles to the axis, as O I is an ordinate to the transverse axis A B.

90. Double ordinates are those which meet the curve on both sides of the axis, as HV is a double ordinate to the transverse axis.

91. Abscissa is that part of the axis between the ordinate and vertex, as A O and O B are the abscissas to the ordinate O I; and A G and G B are abscissas to the ordinate G H.

92. Parameter or Latus rectum is that ordinate passing through the focus, and meeting the curve at both sides, as H. V.

93. Diameter is any line passing through the centre and terminated by the curve, as Q X or R I.

94. Ordinate to a diameter is a line parallel to the tangent at the vertex of that diameter, as Z T is the ordinate being parallel to the tangent X Y drawn to the vertex X of the diameter X Q.

95. Conjugate to a diameter is that line drawn through the centre, terminated by the curve, and parallel to the tangent at the vertex of that diameter, as C b is the semi-conjugate to the diameter Q X.

96. Tangent to any point H' in the curve, join H F and G H, bisect the angle L H G by the line H K, then H K will be the required tangent.

97. Tangent from a point without, let P be the given point, (see fig. 40) join PF; on P F and A B describe circles cutting one another in X, join PX and produce it to meet the ellipse in T, then P T will be the required tangent, and H K tangent to the point h.

98. Focal tangents, are the tangents drawn through the points where the latus rectum meets the curve, K H is the focal tangent to the point H.

99. Normal is that line drawn from the point of contact of the tangent with the curve, and at right angles to the tangent, H N is normal to K H.

100. Subnormal is the intercepted, distance between the point where the normal meets the axis, and that point where an ordinate from the point of tangents contact with the curve meets the axis, as N O' is the subnormal to the point H.

101. Eccentricity is the distance from the focus to the centre, as C G. 102. All diameters bisect one another in the centre C; that is, CX= CQ and CI=CR.

103. To find the centre of an ellipse. Draw any two cords parallel to onc another, bisect them, join the points of bisection and produce the line both ways to the curve, bisect this last line drawn, and the point of bisection will be the centre of the ellipse.

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AB=FD+GB=FI+GI=FH+GH, etc.; that is, the sum of any two lines drawn from the foci to any point in the curve, is equal to the transverse axis.

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