105. The square of half the transverse, is to the square of half the conjugate, as the rectangle of any two abscissas is to the square of the ordinate to these abscissas; i. e., 106. Rectangles of the abscissas are to one another as the squares of their ordinates; i. e., AO. OB: A G · G B :: 0 12 : G H⁄2 107. The square of any diameter is to the square of its conjugate, as the rectangle of the abscissas to that, diameter, is to the square of the ordinate to these abscissas; i. e., Q X2: H/ b2 :: Q T • TX: T Z2; i. e., C X2 : C b2 :: QT • TX: T Z2. 108. To find where the tangent to the point I will meet the transverse axis produced: :: : CO: A C A C C K'. Substituting x for C O', and a for A C Having the semi-transverse axis = a, the semi-conjugate — b 00 Hany ordinate, x= = CO co-ordinate of y. Let A0=8= greater abscissa, and O B = 8 = lesser abscissa. We will from the above deduce formulas for finding either a, b, S, s, O or x. 115. Area of an ellipse = A BXDEX,7854—4 a b • 78543,1416 Xab. 116. Area of an elliptical segment.-Let h height of the segment. Divide the height h, by the diameter of which it is a part; find the tabular area corresponding to the quotient taken from tab. VII; this area multiplied by the two axes will give the required area, i. e., 4 a b, when the base is parallel to the conjugate axis; 118. Application.-Let the transverse = 35, and conjugate 35 X 25 X,7854 875 X,7854 = 687,225. Area 25. 352 +252. Circumference = -) • 3·1416 — 22.09 × 3·1416 69,3979. 2 Let A 0=28=greater abscissa, then 7 — the lesser abscissa, to find the ordinate O H. 122. A parabola is the section of a cone made by a plane cutting it parallel to one of its sides (see fig. 41). 123. To describe a parabola.—Let D C = directrix and F = focus bisect AF in V; then V: vertex; apply one side of a square to the directrix C D; attach a fine line or cord to the side H I; make it fast to the end I and focus F; slide one side of the square, along the edge of a ruler laid on the derectrix; keep the line by a fine pencil or blunt needle close to the side of the square, and trace the curve on one side of the axis. Otherwise, Assume in the axis the points F B B' B'' B/// B//// etc., at equal distances from F; from these points erect perpendicular ordinates to the axis, as F Q, BP, B' O, B'' N, B/// M; from the focus F, with the distances A F, A B, A B', A B', describe arcs cutting the above ordinates in the points Q, P, (), N, M, etc., which points will be in the curve of the required parabola; by marking the distances F B = B B′ = B' B'', etc., each distance equal about two inches, the curve can be drawn near enough; but where strict accuracy is required, that method given in sec. 122 is the best. 124. Definitions.-C D is the directrix, F = focus, V = vertex, A B axis. The lines at right angles to the axis are called ordinates. The double ordinate Q R through the focus is equal to four times F V, and is called parameter, or latus rectum. Diameter to a parabola is a line drawn from any point in the curve parallel to the axis, as S Y. Ordinate to a diameter is the line terminated by the curve and bisected by the diameter. Abscissa is the distance from the vertex of any diameter to the intersection of an ordinate to that diameter, as V B is the abscissa to the ordinate P. B. 124a. Every ordinate to the axis is amean proportional between its abscissa and the latus rectum; that is 4 VFX B V B/ N2, consequently having the abscissa and ordinate given, we find the latus rectum B// N2 B// V 4 VF: B// N2 ; also the distance of the focus F from the vertex = F V = 4 B/ N 125. Squares of the ordinates are to one another as their abscissas; i. e., B P2: B' 02 :: V B: V B'. 127. The ordinate B S2 VB.4 VF; hence, the equation to the curve is y2 = p x, where y = ordinate B S, and x abscissa V B, and p=parameter or latus rectum. = 128. To draw a tangent to any point S in the curve, join S F; draw Y S L parallel to the axis A B; bisect the angle F S L by the line X S, which will be the required tangent. Otherwise, Draw the line from the focus to the derectrix, as F L; bisect F L in m; draw m X at right angles to F L; then m X S will be the tangent required, because S L S F. Otherwise, Let S be the point from which it is required to draw a tangent to the curve; draw the ordinate S B, produce B/ V to G, making VG = V B; then the line G S will be the required tangent. 129. Area of a parabola is found by multiplying the height by the base, and taking two-thirds of the product for the area; i. e., the area of the parabola N V U (B// VN W). 130. To find the length of the curve N V B of a parabola : Rule. To the square of the ordinate N B// add four thirds of the square of the abscissa V B; the square root of the product multiplied by 2 will be the required length. Or, by putting a = abscissa V B', and d = a d2 + 4 a2 ordinate N B’; length of the curve N V U = √(a Length of the curve N V U = √(3 d2 + 4 a2) × 1,155. .). 2, i. e., Rule 11.-The following is more accurate than the above rule, but is more difficult. 131. By sec. 57, of all triangles the equilateral contains the greatest area enclosed by the same perimeter; therefore, in sewerage, the sewer having its double ordinate, at the spring of the arch, equal to d; then its depth or abscissa will be,866 d; i. e., multiply the width of the sewer at the spring of the arch by the decimal,866. The product will be the depth of that sewer, approximately for parabolic sewer. 132. The great object in sewerage is to obtain the form of a sewer, such that it will have the greatest hydraulic mean depth with the least possible surface in contact. OF THE PARABOLIC SEWER. 183. Given the area of the parabolic sewer, NV U = a to find its abscissa V B and ordinate B// N such that the hydraulic mean depth of the sewer will be the greatest possible. ↑ rejecting the denominator and bringing to the same common denominator. Now we have the abscissa x = 1,612, and ordinate — 1,863. By Sec. 130, we find the length of the curve N V U = 5,26; and by dividing the perimeter, 5,26, into the area of the sewer, we will have the 4 5,16 hydraulic mean depth 0,76 feet. 134. The circular sewer, when running half full, has a greater hydraulic mean depth than any other segment; but as the water falls in the sewer, the difference between the circular and parabolic hydraulic mean depths, decreases until in the lower segments, where the debris is more concentrated in the parabolic, than in the circular, the parabolic sewer with the same sectional area will give the greatest hydraulic mean depth. This will appear from the following calculations: Where the segment of a circle is assumed equal to a segment of a parabola, which parabola is equal to one-half of the given circle. The method of finding the length of the curve, area and hydraulic mean depth, will also appear. |