To solve this example by the sliding rule:-Find the diameter of the equal circle, then place 1 on C over 14.28 on D, (the gauge point for acres in a circle,) and over the diameter of the equal circle found on D, will be found the area on C. 3. What is the area of an ellipse whose axes are 70 and 50 rods? Ans. 17 acres, 29 rods, nearly. 4. What is the circumference of an ellipse whose axes are 40 and 60? By the first rule we find the circumference to be 157.555, and by the second, 160,22, the mean between which is 158.8875, which is the perimeter of the ellipse, nearly. When the ellipse does not differ much from a circle, the second rule will be found the most accurate; but when the eccentricity is great, the first rule should only be used. 5. What is the circumference of an ellipse whose axes are 40 and 10? Ans. 85.8. 6. The axes of an ellipse are 8 and 6; what is the length of the curve? Ans. 22.2, nearly. 7. What is the eccentricity of an ellipse whose axes are 30 and 50? Ans. 20. 8. The major and minor axes of an ellipse are 60 and 20, and one absciss is 12; it is required to find the ordinate. Ans. 8. 9. The axes are 45 and 15, and one absciss is 9; what is the ordinate? Ans. 6. 10. The axes are 45 and 15, and the ordinate 6; what are the abscisses? Ans. 36 and 9. 11. The axes are 30 and 50; what is the parameter, or double ordinate passing through the focus? 50 30 30 is to the parameter. 12. The minor axis is 15, an ordinate 6, and sciss 9; what is the major axis? Ans. 18. the less ab Ans. 45. 13. The minor axis is 50, an ordinate 20, and the less absciss 14; the major axis is required. Ans. 70. 14. When the major axis is 15, an ordinate 2, and an absciss 3, what is the minor axis? Ans. 5. 15. If the major axis is 70, and an ordinate 20, and one of the abscisses 14, what is the minor axis? Ans. 50. The ellipse is the curve in which the planets perform their revolutions round the sun, and its properties enter into almost every investigation connected with physical astronomy. It is a singular and curious property of the ellipse, that if a moving or generating circle roll along the concave circumference of another circle in the same plane, the radius of the generating circle being half that of the other, any given point in the circumference or plane of the moving circle will describe an ellipse. An ellipse is most conveniently drawn by an instrument called the trammel, usually possessed by master builders. ELLIPTICAL SEGMENTS. An elliptical segment bears the same proportion to the corresponding circular segment, that the whole ellipse does to the whole circle described on the axis of which the height of the segment is a part. Hence, if either axis of an ellipse be made the diameter of a circle, and a line perpendicular to this axis cuts off a segment from the ellipse and from the circle; THE DIAMETER OF THE CIRCLE WILL BE TO THE OTHER AXIS OF THE ELLIPSE, AS THE CIRCULAR SEGMENT IS TO THE ELLIPTIC SEGMENT. EXAMPLES. 1. What is the area of an elliptic segment, cut off by an ordinate or chord perpendicular to the major axis, the major axis being 415 feet, the minor 332 feet, and the height, or absciss of the segment, 96 feet? Ans. 18876.8 square feet. In this example, the radius of the corresponding circular segment will be 415, or 207.5, and its versed sine 96 feet; and consequently, the apothegm, or height of the triangle formed by the radii of the circular sector and the segment's base, will be 111.5; and twice the square root of the difference of the height of the triangle and the radius will be the base of the circular segment, viz. 350, and the square root of the sum of the squares of the versed sine and half the base will be the chord of half the arc, viz. 199.6, and the arc line will = 415.6, and the area of the circular segment will = 23596. Then, as 415 332 23596 to the area of the elliptic segment. Or, to find the segment of an ellipse:— Divide the height of the segment by the axis of the ellipse, which is perpendicular to the base of the segment; find the tabular area corresponding to this quotient, in the table of circular segments, and multiply it by the product (or rectangle) of the two axes of the ellipse, and the result will be the area. 2. What is the area of an elliptic segment, whose base is parallel to the minor axis, the height of the segment being 10 feet, and the axes of the ellipse 35 and 25? Ans. 162.021. = Solution.- .2857, and the tabular area còrresponding to this number is .185167, and .185167 × 35 × 25=162.02. 3. Find the area of an elliptic segment, whose base is parallel to the major axis, its height being 2, and the diameters 14 and 10. Ans. 15.6553. 29. PARABOLA. A parabola is a figure formed by cutting a section from a cone by a plane parallel to one of its sides. The nature of the curve is such that every point in it is equally distant from a certain point F, and a given straight line GH. The given point F is called the focus of the parabola, and the given right line GH, is called the directrix. The point D is called the vertex of the parabola; and the line DC, which is perpendicular to the directrix, is called the axis, or principal diameter. An ordinate is a perpendicular from any point in the curve on the axis, as SN; and when the ordinate is produced to meet the curve on the other side of the axis, it is called a double ordinate; and the portion of the axis intercepted between the ordinate and the vertex of the curve, is called the absciss, as DS or DC. The double ordinate PQ, passing through the focus, is called the parameter, or latus rectum; this line is always equal to four times the absciss DF, (or to four times its distance from the vertex,) and its distance from the vertex is always half its distance from the directrix: consoquently, DF and DI are equal. To construct a parabola : Draw the base AB of any required length, and on the extremities of the base, or double ordinate; erect perpendiculars, and make BL and AK equal to the absciss of the parabola, and complete the rectangle by joining KL. Then divide the lines AB and KL into an even number of equal parts, and through each of the divisions, or points, draw lines parallel to the sides LB or AK; then divide the other two sides of the rectangle into half as many equal parts as the former, and from each of the divisions in the sides LB and AK, draw lines meeting at the centre of the line KL, and the vertex of the parabola will be at this point, and its curve will pass through the points .3 where the parallel lines are cut by the converging: consequently, commence at the vertex, and draw lines through these points, and the figure will be completed. The greater the number of equal parts into which the sides of the rectangle are divided, the greater will be the number of points found in the curve, and the more accurately it may be drawn. When an ordinate of a parabola and its absciss are given, to find the parameter :— Divide the square of the ordinate by the absciss, and the quotient will be the parameter. Any two abscisses are to each other as the squares of their respective ordinates; and the absciss of any given ordinate, is to that of any required ordinate, as the square of the given ordinate to the square of the required ordinate; and the square of any given ordinate is to its absciss, as the square of any other given ordinate is to its required absciss. The parabola is exactly two-thirds of its least circumscribing rectangle. Therefore, to find its area : Multiply its base, or double ordinate, by two-thirds of its altitude or absciss. Or, Multiply any double ordinate by two-thirds of its absciss, and the product will be the area of that part of the parabola between the ordinate and vertex. To find the area of a zone of a parabola : From the area of the whole parabola subtract the area of the segment or part above the less ordinate. Or, To one of the parallel sides add the quotient obtained by dividing the square of the other by the sum of the two parallel sides; and multiply this sum by of the altitude of the zone. To find the length of the parabolic curve: To the square of the ordinate, or half the base, add fourthirds of the square of the absciss, and the square root of the sum, multiplied by two, will be the length of the curve, nearly. |