Problem 14. To find the Length of the Transverse and Conjugate Axis of an Hyperbola, having the Bafe and Height given, with the Height also above the Place where the Hyperbola measures across exactly equal half the Breadth of that Bafe. (Ift.) To find the Transverse Axis. Rule. From the Square of the whole Height, take 4 Times the Square of the leffer Height'; divide the Remainder by 4 Times the lesser Height, lessened by the whole Height, and the Quotient will give the Length of the Transverse Axis required. Crample. Suppose an Hyperbola, whose Base A B is 96 Inches, and Height VH 40; and suppose the Height VG, above the Breadth of 48 Inches, equal half the Base, to be 12.111 Inches; what is the Length of the Transverse Axis TV, above the Vertex V of the Hyperbola? The Square of the whole Height 1600, lessened by 4 Times the Square of the lefler Height 586.7052, &c. leaves for a Remainder 1013.2147, &c. which divided by 4 Times the lefler Height 48.444, lessened by the whole Height 40, which is 8.444, gives in the Quotient 120, the Length of the Transverse Axis fought. (zdly.) To find the Conjugate Axis. Rule. Multiply the Height of the Hyperbola by the Sum of the Tranfverfe and Height: Make the Square Root of that Product a Divisor to the Product of the Transverse and half the Breadth of the Base, so will the Quotient arifing be the Conjugate Axis required. Example. Suppose, as in the last Example, the Transverse Axis TV to be 120 Inches; the Height of the Hyperbola V H 40 Inches; and the Base A B 96, consequently its half, 48 Inches; what is the Length of the Conjugate Axis KF? Operation. The Height 40, multiplied by the Sum of the Tranfverse and Height, is = 1600, the Square Root of which is = 80: By which Root divide the Product of the Tranfverse and half the Base multiplied together, = 5760, and the Quotient = 72 gives the Length of the Conjugate Axis fought. Problem 15. To delineate or describe an Hyperbola by having the Tranfverfe Diameter TV, and the Conjugate Diameter KL given. Construction. Draw the Line TV for the Transverse Diameter of Axis, and continue it upwards and downwards at Pleasure. Take the Conjugate Diameter or Axis KL, and place the Middle of it on the End of the Transverse at V, so that those Lines may stand at right Angles to each other. Divide TV into 2 equal Parts at S; and from S, with the Compasses opened to Kor L, describe the Circle KfLF, cutting TV continued in F and f, which are the Focii of the Hyperbola. In TV continued downwards take any Number of Points, as a, a, a, &c. and from F and f, as Centers, with the Distances Ta and V a in the Compaffes, describe Arches cutting each other in e, e, e, &c. Then, through the several Points e, e, e, draw the Curve e Ve, and it will be the Hyperbola required. Note. If two right Lines be drawn from the Point S, by the Ends K and L of the Conjugate Diameter, they will be the Afymtotes of the Hyperbola, whose Property it is to approach continually nearer the Curve, yet never to Problem 16. meet it. To find the Solidity of a Circular, Elliptical, Parabolical, or Hyperbolical Spindle. General Rule. To the Square of the Diameter in the Middle of the Spindle, add the Square of double the Diameter at, or one fourth of the Length, (i. e. exactly between the Middle and one of its Ends) multiply the Sum by the Length, and the Product again by .1309, and it will give the Solidity very nearly. Example. What is the Solidity of a Spindle (of any of the above Forms) whose Length AB is 20 Inches, the greatest Diameter CD 6, and the Diameter EF at Length 4.74 Inches ? of the The Diameter CD = 36, added to double the Dia = 89.8704, makes 125.8704; which x by meter EF 2517.408; this × again by .1309, gives 329.5287072 Inches, for the Solid Content Problem 17. required. To find the Solidity of a Frustum or Segment of an Elliptical, Parabolical, or Hyperbolical Spindle. General Rule. Add together the Squares of the greatest and least Diameters, and the Square of double the Diameter in the Middle between the two; multiply the Sum by the Length, and the Produst again by .1309 for the Solidity. Example. What is the Solidity of the Middle Frustum of any Kind of Spindle, whose Length AB is 20 Inches; the middle, or greater Diameter CD 16 Inches; the Diameter at each End, E F and GH, 12 Inches; and the Diameter at of the Length, (i. e.) between the Middle and the End, IK, 14.5 Inches ? The of CD = 256 + EF = 144 + twice I K = 841 is = 1241. This x by 20, the Length, A Bis = 24.820, which multiplied again by 1309, gives in the Product 3248.9380 Inches, for the Solid Content. |