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Through C and D draw JH, and

make AD and AE each equal to rt, fig. 2.

through C and E draw IK; then JH and IK are the assymtotes.

Draw any line, ai, cutting the assymtote IK at h, and the assymtote JH at g. Make hi equal to ag, and i will be a point in the curve. In the same manner we may find as many more points as we please.

Let the axis be cf, fig. 2, cutting the sides of the section AV, BV, at c and f. Draw cd, ef, parallel to AB, cutting AV at e and BV at d.

In fig. 4, draw AB equal to cf, fig. 2. each equal to the half of df, or the half of ce; and focii F, f, describe the ellipse ADBE.

Bisect AB in C. Make CF, Cƒ,

then, with the transverse axis AB,

Again, in fig. 2, let the axis be mn, and let mn be parallel to the side AV of the vertical section, cutting the base AB at m, and the side BV at n. On AB describe the semi-circle AOB, and draw mO perpendicular to AB.

In the straight line AA', fig. 5, take any point, D, and make DA, DA' each equal to mO, fig. 2. Draw DC, perpendicular to AA', and make DC equal to mn, fig. 2. Then, with the abscissa AB, and ordinates DA, DA', describe the curve ACA', which will be the parabola.

105

CHAPTER II.

CARPENTRY.

CARPENTRY is the art of applying timber in the construction of buildings.

The CUTTING OF THE TIMBERS, and adapting them to their various situations, so that one of the sides of every timber may be arranged according to some given surface, as indicated in the designs of the architect, requires profound skill in geometrical construction.

For this purpose it is necessary, not only to be expert in the common problems, generally given in a course of practical geometry, but to have a thorough knowledge of the sections of solids and their coverings. Of these subjects, the first has already been explained in the series of Problems given in the geometrical part of this Work, and we are now about to treat on the other; that is, the METHOD of COVERING them.

As no line can be formed on the edge of a single piece of timber, so as to arrange with a given surface, nor in the intersection of two surfaces, (by workmen called a groin,) without a complete understanding of both, the reader is required not to pass them until the operations are perfectly familiar to his mind. For the more effectually rivetting the principles upon the mind of the student, it is requested that he should model them as he proceeds, and apply the sections and coverings found on the paper to the real sections and surfaces, by bending them around the solid.

The SURFACES, which timbers are required to form, are those of cylinders, cylindroids, cones, cuneoids, spheres, ellipsoids, &c., either entire, or as terminated by cylinders, cylindroids, cones, and cuneoids.

The FORMATION of ARCHES, GROINS, NICHES, ANGLE-BRACKEts, Lunettes, ROOFS, &c. depend entirely upon their Sections, or upon their Covering, or upon both.

This branch of carpentry, from its being subjected to geometrical rules, and described in schemes or diagrams upon a floor, sufficiently large for all the parts of the operation, has been called DESCRIPTIVE CARPENTRY.

In order to prepare the reader's mind for this subject, it will be necessary to point out the figures of the sections, as taken in certain positions.

ALL THE SECTIONS OF A CYLINDER, parallel to its base, are circles. All the sections of a cylinder, parallel to its axis, are parallelograms. And, if the axis of the cylinder be perpendicular to its base, all these parallelograms will be rectangles. If a cylinder be entirely cut through the curved surface, and if the section is not a circle, it is an ellipse.

ALL THE SECTIONS OF A CONE, parallel to its base, are circles: all the sections of a cone, passing through its vertex, are triangles: all the sections of a cone, which pass entirely through the curved surface, and which are not circles, are ellipses: all the sections of a cone, which are parallel to one of its sides, are denominated parabolas; and all the sections of a cone, which are parallel to any line within the solid, passing through the vertex, aré denominated hyperbolas.

ALL THE SECTIONs of a Sphere or GLOBE, made plane, are circles.

The solid formed by a SEMI-ELLIPSE, revolving upon one of its axes, is termed an ellipsoid.

ALL THE SECTIONS OF AN ELLIPSOID are similar figures: those sections, perpendicular to the fixed axis, are circles; and those parallel thereto are similar to the generating figure

OF THE COVERINGS OF SOLIDS.

PROBLEM 1.

To find the covering of the frustum of a right cone.

Let ABCD (fig. 1, pl. VII,) be the generating section of the frustum. On BC describe the semi-circle BEC, and produce the sides BA and CD, of the generating section ABCD, to meet each other in F. From the centre F, with the radius FA, describe the arc AH; and, from the same centre, F, with the radius FB, describe the arc BG; divide the arc, BEC, of the semi-circle, into any number of equal parts; the more, the greater truth will result from the operation; repeat the chord of one of these equal arcs upon the arc BG, as often as the arc BEC contains equal parts; then, through G, the extremity of the last part, draw GF, cutting the arc AH at H; then will ABGH be the covering required.

PROBLEM 2.

To find the covering of the frustum of a right cone, when cut by two concentric cylindric surfaces, perpendicular to the generating section.

Let ABCD (fig. 2, pl. VII,) be the given section, and AD, BC, the line on which the cylindric surface stands. Find the arc BG, as before, in problem 1, and mark the points, 1, 2, 3, &c. of division, both in the arc BG, and in the semi-circumference; from the points 1, 2, 3, &c. draw lines to F; also from the points 1, 2, 3, &c. in the semi-circumference, draw lines perpendicular to BC; so that each line thus drawn may meet or cut it. From the points of division in BC, draw more lines to F, cutting the arc BC in a, b, c, &c. From the points a, b, c, &c. draw lines parallel to BC, to cut the side BA from the centre F, through each point of section in BA, describe an arc, cutting the lines drawn from each of the points 1, 2, 3, &c., in BG, at a, b, c, &c.; then will Be G be the curve, which will cover the line BC on the plan, or BC will be the seat of the line BeG.

In the same manner AH, the original of the line AD, will be found; and, consequently, Be GHA will form the covering over the given seat, ABCD, as

PROBLEM 3.

To find the covering of a right cylinder.

Let ABCD (fig. 3, pl. VII,) be the seat or generating section. Produce the sides DA and CB to H and G, and on BC describe a semi-circle, and make the straight line BG equal to the semi-circumference: draw GH parallel to AB, and AH parallel to BG; then will ABGH be the covering required.

PROBLEM 4.

To find the covering of a right cylinder contained between two parallel planes, perpendicular to the generating section (fig. 4, pl. VII).

Through the point B draw IK, perpendicular to AB, and produce DC to K; on BK describe a semi-circle, and make BI equal to the length of the arc of the semi-circle, by dividing it into equal parts, and extending them on the line BI. Through the points of section, 1, 2, 3, &c., in the line BI, draw lines, 1a, 2b, 3c, &c., parallel to BA, and through the points 1, 2, 3, &c., in the arc of the semi-circle, draw the other lines la, 2b, 3c, &c., parallel to BA, cutting AD in a, b, c, &c. Draw aa, bb, cc, &c., parallel to BK; then, through the points, a, b, c, &c., draw the curve AH, and AH will be the edge of the covering over AD.

In the same manner the other opposite edge BG will be found, and the whole covering will therefore be ABGH.

PROBLEM 5.

ABCD (fig. 5, pl. VII,) being the seat of the covering of a semi-cylindric surface, contained between the surfaces of two other concentric cylinders, of which the axis is perpendicular to the given seat; it is required to find the covering.

Through B draw IK, perpendicular to AB; and produce DC to K. On BK describe a semi-circle, and divide its circumference into equal parts, at the points 1, 2, 3, &c. ; the more of these the truer will be the operation; and repeat the chord on the straight line BI, as often as the arc contains equal parts, and mark the points 1, 2, 3, &c., of division. Through the points

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