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PROBLEM 32.

209. To describe an ellipse with a trammel {fig- 4, pi- IV).

Draw the straight line AB, and make AB equal to the length of the given ellipse. Bisect AB in E, by the perpendicular CD. Make ED and EC each equal to half the lesser axis. Let the trammel-rod be fgh. Suppose the nut J' to be fixed, and the nuts g and h to be moveable. Move the nut g so that the distance fg may be equal to EC or ED; and move the nut h, so that the distance fh may be equal to EA or EB : then set g and h in the grooves of the trammel, and, by moving the end of the rod /, so that the pin g, may slide along the line AB, and the pin h along the line CD; the point or pencil will trace out the curve of the ellipse as required.

Problem M.

210. A rectangle being given, to describe an ellipse, so that the two axes may have the same proportion as the sides of the rectangle {fig- 3, pl. IV).

Let ABCD be the given rectangle; it is required to describe an ellipse that shall pass through the points A, B, C, D, and of which the axes shall have a ratio equal to the sides AB, BC, or to CD, DA, of the rectangle.

Draw the diagonals AC, BD, cutting each other in I; and, through the point I, draw EG, parallel to AB or DC, cutting one of the sides BC in P ; and HF, parallel to AD or BC, cutting one of the sides AB in K. From I, with the radius IK, describe an arc, KL, cutting IG in L. Bisect the arc KL in M, and draw mN, parallel to EG, cutting the diagonal DB in N. Join NP and NK, and draw BF, cutting HF in F, and draw BG, cutting EG in G. Make IE equal to IG, IH equal to IF; then EG and FH are the two axes of the ellipse, which will be described as in the following problem.

Problem 34.

211. To describe an hyperbola, or a figure that may have any curvature at the summit, that we please (fig. 18, pl. II).

Let AC be the base, or what is called a double ordinate: make ED equal to the height, in the middle of AC ; then, upon AC, as a side, describe a rectangle, AFGC, so that the opposite side may pass through D; produce ED to the point B; take the point B, farther or nearer from D, according as the curvature at D is required to be flatter or quicker; and observe that, the quicker the curve is at D the flatter it will be towards A and C. The point B being thus fixed, divide AE into any number of parts, as here into four; also, divide AF into the same number of parts, viz. four; through 1, 2,3, &c, the points of section in AE, draw lines to B ; and through the points of section, 1,2, 3, &c in AF, draw lines to D, cutting the former lines drawn to B, in the points h, i, k, &c; and, through the points A, h, i, k, &c, draw a curve, which will be the half of an hyperbola, or an hyperbolic curve.

Problem M.

212. To describe a, parabola upon a given ordinate, AE, and a given abscissa, ED* [fig. 19, pl. II).

Make EC equal to EA, and complete the rectangle AFGC; so that the opposite side may pass through D. Proceed as in the two former problems, 33 and 34; excepting that, instead of drawing the lines to B, through the points 1, 2, 3, &c, in AE, to draw them parallel to ED.

Problem M.

213. To describe the figure of the Sines f [fig. 20, pl. II).

Describe the quadrant FHG, equal to the height of the figure, and divide the arc HG into any number of equal parts; the more of these the more perfect the operation will be; and extend the chords to double the number of parts upon the line AC, which is a continuation of FH, and mark the points of division. Draw the lines lie, 21,3m, &c perpendicular to AC; and, from the points 1, 2,3, &c of division in the quadrant, draw lines Ik, 21, 3m, &c parallel to AC, and through the points A, k, l, m, &c, draw a curve, which will be the figure of the sines, as required.

* The Parabola is a figure arising from the section of a cone, when cut by a plane parallel to one of its sides. This, with other terms in Conies, is fully described, under Conic Sections, hereafter.

t The Sine, or right sine, of an arch, is a right line drawn from one end of that arch, perpendicular to a radius, drawn to the other end of the same.—See Trigonometry, hereafter.

GEOMETRY OF SOLIDS.

DEFINITIONS OF SOLIDS.

214. A Right Cylinder is that which is formed by the revolution of a rectangle about one of its sides; the line round which the rectangle revolves is called the axis (plural axes); and the circles generated by the two opposite sides of the rectangle, perpendicular to the axes, are termed the ends or bases. The surface of the cylinder, generated by the line parallel to the axis, is termed the curved surface, which is either straight or convex, according as a straight edge is applied, parallel to the axis, or in any other direction.

215. A Right Cone is that which is formed by supposing a right-angled triangle to revolve about one of its legs or perpendicular sides; the fixed leg, or line, is called the axis; the surface generated by the other leg is called the base; and the surface formed by the hypothenuse, or side opposite the right angle, is denominated the curved surface, which is either straight or convex, according as a straight edge is applied upon the surface from the vertex, or in

any other direction.

216. A Sphere or Globe is that which is formed by supposing a semi-circle to revolve upon its diameter; the diameter upon which the semi-circle revolves is called the axis, and the surface formed by the arc of the semi-circle is called the curved surface, which is convex, in whatever way it may be tried

by a straight edge.

217. An Ellipsoid is formed or generated by supposing a semi-ellipse to revolve upon one its axes; the axis thus fixed is called the axis of the ellipsoid, and the surface generated by the curve is termed the curved surface.

PROBLEM 37.

218. To describe a conic section, from the cone, through a line given in position, in the section passing through the axis.

Let ABC, {figures 1, 2, 3, pl. VI,) be the section of a right cone, and let DE be the line of section. Through the apex or top of the cone, C, draw CF, parallel to the base AB of the section, and produce ED to meet AB in D, as in figures 2 and 3, or AB produced in G, as in fig. 1, as also to meet CF in F. On AB describe a semi-circle, which will be equal to half the base of the cone. In the semi-circle take any number of points, a, b, c, &c Draw Dd, in figures 2 and 3, and Gd in fig. 1, perpendicular to AB and Gd', in fig. 1, per, pendicular to GF ; as, also, D«?', figures 2 and 3, perpendicular to DF. From the points a, b, c, &c, draw lines, ae, bf, eg, &c, cutting Gd {figure 1) and Dd [figures 2 and 3) in the points e,f,g, &c In figure 1, make in Ge', Gf, Gg', &c equal to Ge, Gf, Gg, &c; and in T)d', Figures 2 and 3,) make De', W", T)g', &c. equal to De, D/*, Bg, &c Through the points e', f, g', &c, draw lines to F. Through the points a, b,c,&c. draw lines perpendicular toAB. From the points of section, in AB, draw lines to the vertex C of the cone, cutting ,the sectional line, DE, in /, m, n, &c Through the points of section, l, m, n, &c, draw lh, mi, nk, &c perpendicular to DE. Through the points D, h, i, h, &c in fig. 1, or d', h, i, k, &c, figures 2 and 3, draw a curve, which will be the conic section required.

OBSERVATIONS.

219. In the first of these figures, the line of section cuts both sides of the section of the cone; in this case, the curve D/iik and eE is an Ellipse. In fig. 2, the line of section DE is parallel to the side AC of the section of the cone; in this case, the curve d'hi, &c E, is a Parabola. In fig. 3, the line of section, DE, is not parallel to any side of the cone; it must, therefore, when produced with the sides of the section through the axis, meet each of these two sides in different points: in this case, the section d', h, i, &c, E, is either an Ellipse or Hyperbola; but the case is determined to be an hyperbola by the line of section meeting the opposite side BC at AC, where it cuts above the vertex at the point B'.

Here we may observe, that the line of section, DE, is the same as that which has before been called the abscissa, the part EB produced, contained between the two sides of the section, is called the axis major; and the line Bd, perpendicular to DE, an ordinate.

Hence the same section may be found by the method already shown in the problem; viz. by drawing any straight line, deb, fig. 4: make de equal to DE, fig. 3, and eb equal to EB, fig. 3. Through d draw the straight line DD at right angles to db': make dD equal to Dd', fig. 3; then, with the axis major, b'e, the abscissa ed, and the ordinate dD, on each side of the abscissa describe the curve of the hyperbola, which will be of the same species as that shown in fig. 3.

Problem 38.

220. To describe a cylindric section, through a line given in position, upon the section passing through its axis {fig. 4, pl. VI).

This is no more than a particular case of the last problem. For a cylinder may be considered as a cone, having its apex at an infinite distance from its base; or, practically, at a vast distance from its base. In this case all the lines, for a short distance, would differ insensibly from parallel lines; and this is the construction shown at fig. 5, which is therefore evident. But as the section of a cylinder so frequently occurs, I shall here give a more practical description of it. Thus—

Let ABHI be a section of a right cylinder, passing through its axis, AB being the side which passes through the base, and let DE be the line of section. On AB describe a semi-circle; and, in the arc, take any number of points, a, b, c, &c, from which draw lines perpendicular to the diameter, AB, cutting it in Q, R, S, &c: perpendicular to AB, or parallel to AI or BH, draw the lines Qq, Rr, Ss, &c, cutting the line of section, DE, in the points, q, r, s, &c: from the points of section, q, r, s, &c draw the lines qi, rk, si, &c per

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