points, a, b, c, &c., from which draw lines perpendicular to the diameter, AB, cutting it in Q, R, S, &c.: and the line of section, DE, in the points, q, r, s, &c.: from the points q, r, s, &c. draw the lines qi, rk, sl, &c. perpendicular to the line of section, DE. Make the ordinates qi rk, sl, &c. each respectively equal to the ordinates Qa, Rb, Sc, &c.; and through the points D, i, k, l, &c. to E, draw a curve, which will evidently be the section of the cylinder, as required.

The curve is an ellipsis; hence the same may be done in this manner, viz.-Bisect the line of section DE in the point t. Draw tm perpendicular to DE. Make tm equal to the radius of the circle which forms the end of the cylinder; then, with the length DE, and the breadth, tm, by any of the methods in Art. 7, and those following it, describe an ellipsis which will be the section of the cylinder required.

39. Given the position of three points, in the circumference of a cylinder, and the heights of the perpendiculars, let fall from these points to the base, to find the section of the cylinder passing through these three points.

Let ABC be the feet of the perpendiculars from the three points, (fig.7, pl. IV,) in the circumference of the base. Join the two points, A and B, and draw AD, CF, and BE, perpendicular to AB. Make AD equal to the height of the point above the base at A, BE equal to the height at B, and CF equal to the height at C. Produce BA and ED to meet each other in H: draw CG parallel to BH, and FG parallel to EH. Join GH. In GH take ary point, G, and draw GK perpendicular to CG, cutting BH in K: from the point K draw KI, perpendicular to EH, cutting EH in L. From H, with the radius HG, describe an arc, cutting KI at I. Join HI. In the circumference of the base ACB, take any number of points a, b, c, &c., at pleasure, and draw ae, bf, cg, &c. parallel to GH, cutting AB at e. f, g, &c. Through the points e, f, g, &c., draw lines ei, fk, gl, &c., parallel to GK, or AD, or BE, cutting DE at i, k, l, &c.; from the points, i, k, l, &c., draw the lines in, ko, lp, &c., parallel to HI. Make the ordinates in, ko, lp, &c. equal to ea, fb, gc, &c.; then, through the points D, n, o, p, &c. draw the curve Dnop, &c. to E, and it will be the section cut by the plane, as required.

The most useful application of this case is to find the moulds for hand-rails of staircases; this application will be shown in treating of that part of our subject.

40. A wedge-formed solid is one ending in a straight line, in which, if any point be taken, a line from that point may be made to coincide with the surface: the end of the figure may be of any form whatever.

The forms which occur in architecture have a semi-circular, a Gothic arched, or a semielliptical end, parallel to the straight line from which the line is applied. The base is generally a triangle.

To find the section of a wedge-formed solid, with a semi-circular end, the given data being a plan, perpendicular to the vertex, or sharp end, and the line of section.

Let ABC, (fig. 6, pl. IV,) be the plan, perpendicular to the sharp edge, and let DE be the line of section.

This construction is similar to that of finding the section of a cone, excepting that, instead of drawing lines to a point, they are, in this figure, drawn parallel to the line of section DE: the ordinates Qa, Rb, Sc, &c., being transferred respectively to qi, rk, sl, &c.; and the curve D, i, k, l, &c. to E, drawn through the points, D; i, k, l, &c., by hand.

41. It must be obvious that, in any of these cases, if the curve DiklE be given, the end AdB may be found by reversing the process.

D

This example applies to drawing the interior elevation of a window or door when the jambs are splayed at the sides, and level at the crown. The interior face of the wall, AB, is commonly parallel to the exterior one DE; but the figure is drawn with the walls oblique to one another to show the general nature of the construction.

Fig. 1, pl. IV, applies in like manner to the case where a window or door is splayed equally

all round,

The construction is not confined to particular curves in any of these examples; that is, whatever form is given to the original curves, the other will be found by the preceding processes to correspond to them.

42. Given the plan of a sphere, and the line of a section at right angles to that plan, to find the form of the section.

Let ABC (fig. 8, pl. IV,) be the plan, and AB the line of section.

On AB, as a diameter, describe a semi-circle, which will be half of the section required: since all the sections of a sphere, or globe, are circles.

43. Given the plan of a spheroid,* and the line of section, at right angles to the plan, to find the form of the section through that line. Let ABCD be the plan, DB the breadth, and EF the line of section. Through the centre of the spheroid draw AC parallel to EF. Bisect EF in H. Join CD; and draw EG parallel to CD, and HG parallel to DB; then HG is half the breadth, and EF the length of the section; and with this length and breadth describe an ellipsis, by Art. 7, or any of those methods already described.

When EF, the line of section, is perpendicular to DB, then AC becomes equal to the length of the plan, and EGF (fig. 9, pl. IV,) represents half the section.

44. To find the section of a ring, the plan and line of section being given.

F

YE

D

Let ABED (fig. 10, pl. IV,) be part of the plan of a ring; AB a straight line, which, if produced would pass through F, the centre of the ring; and DE the line of section.

On AB describe a semi-circle, and take any number of points, a, b, c, d, &c. in its circumference; then draw the ordinates ae, bf, gc, &c. perpendicular to AB. From the points

e, f, g, &c. and centre F, describe arcs of circles, cutting the line of section DE in i, k, l, m, &c.; and from each of these points draw lines perpendicular to DE. Make in equal to ea; ko equal to fb, &c.; and through the points D, n, o, p, &c. draw a curve, which is the boundary of the section required.

Many other examples of sections will be found in different parts of the Work.

Developement of Surfaces.

45. We have already explained that the developement of the surface of any body is the same as describing a flat surface that would cover the body, (Art. 32.) In most works which are bounded by curved surfaces, this mode of drawing is extremely useful; as, for example, in covering domed roofs, centres, and the like, in finding the moulds for arches, in forming the moulds for hand-rails, and the soffits of stairs; and in finding the forms for veneers, and moulds for soffits of windows, arches, and the like.

• A spheroid is a figure generated by the revolution of a semi ellipsis round one of its principal diameters.

The developement of a curved surface might be obtained, in many instances, by the surface being rolled on a plane, so that all its parts should be successively in contact with the plane. 46. To find the developement of the curved surface of a right cylinder.

The surface is evidently of the same length as the cylinder, and of the same breadth as the circumference of the cylinder. And as the circumference of a circle 34th times the diameter, the breadth of the developement will be 34th times the diameter of the cylinder, and its length the length of the cylinder.

If only a portion of the circumference of a cylinder is to be developed, as, for example, the portion Do, (fig. 2, pl. III.) Draw the line DF through the centre C of the circle, divide the radius DC into four equal parts; and make EF equal to three of these parts. Draw Dp perpendicular to DF; and from the point F, and through the point o, draw Fp; then Dp is equal to the arc Do, very nearly.

In the same manner D1 is the length of the arc Da; D2 of the arc Db, &c. to DG, which is equal the quadrant DB. And if the length, D4, of any arc, as Dd, be found, and it is required to divide that arc into any number of equal parts, we have only to divide D4 into the proposed number of parts, and from the points of division to draw lines to F, and these lines will divide the arc into the same number of equal parts.

47. When a portion of a cylinder is to be developed, and its diameter is so great as to render the preceding method troublesome, it may be done in this manner. Let ADB (fig. 3, pl. III,) be the portion of the plan of the cylinder. Join AB, and divide it into four equal parts; set off Af equal to one of these parts, and from the third division draw the line 3f. Then the line 3f Otherwise.-Draw D2 perpendicular to the middle of AB; and with the radius AD, and centre A, describe the arc Dc. Divide 2c into three equal parts, and make cd equal to one of these parts; then Ad is equal to half the length of the arc ADB.

is very nearly equal to half the length of the arc AB.

Hence if a Db be made equal to either, twice Ad, or twice 3f, it will be the developement of the arc ADB.

48. If it be required to develope a curve which is not a circle, it may be done by the operation called stepping. A pair of compasses must be set to such an opening, that a portion of the quickest part of the curve, included between their points, may not sensibly differ from a straight line. Then, beginning at one end, A, step with

the points of the compasses along the curve AB, and suppose the last step to be at D, set off an

equal number of steps on a straight line, and make db equal to DB; then ab will be very nearly equal to the length of the curve AB.

at

b

This mode of finding the length of a curve requires much care in practice, to render it accurate enough for use.

49. To find the developement of the curved surface of a cone.

With the

Let AFB (fig. 1, pl. V,) be the plan of the cone, and AEB a plan of half its base. radius AF, and F as a centre, describe the arc Aeb. Divide the arc AE of the plan of the base into any number of equal parts, and set off the same number of these parts from A to e; and make be equal to Ae. Join Fb, and FAeb is the developement of half the surface of

the cone.

Otherwise.-Multiply half the diameter of the base in inches by 360; and divide the product by the length AF in inches, and the quotient will express the degrees, and parts of a degree,

contained in the angle AFb; therefore, if AFb be made equal to that angle, it will be the boundary of half the developement. Thus, if half AB be 12 inches, and AF be 42 inches, then 360 X 12-102 degrees 51 minutes, for the angle AFb.

42

When only part of a conic surface is to be developed, and ABCD represents the plan of the part, then proceed as before; and the covering for the whole cone being found, with a radius FD, and centre F, describe the arc Dc, and AD cb will be the developement of the part ABCD of the cone.

If ABCD be the plan of the walls of a semi-circular headed window, which is splayed equally all round the head, then AD cb is the lining of the soffit, which is lightly tinted to show it more distinctly.

50. To find the lining of a soffit formed by a circular aperture in a circular wall, when splayed equally all round.

Produce the lines AD and BC to With the radius AE, and centre F,

Let ABCD be the plan of the aperture, (fig. 2, pl. V.) meet in F; also, join AB, and describe the arch AEB. describe the arc Ab; and make its length equal to that of the arch AEB, as in the preceding example, marking the points of division on both arcs, as at 1, 2, 3, &c. From the points

1, 2, 3, &c. in Ab, draw lines to F; and from the points of division in AE, draw lines perpendicular to AB; also, from the points in which these perpendiculars cut the line AB, draw lines to F; and from the points where the lines to F cut the plan of the wall, draw lines parallel to AB to meet the line AD, then from the points of meeting, and centre F, describe arcs of circles, each circle to meet its corresponding line 1F, 2F, 3F, &c. which will determine the form of the edge of the lining of the soffit. The whole lining is shown by the lightly tinted part, AD cb.

The problem stated in general terms, is to find the developement of the interior surface of the aperture, formed by piercing a conical hole through a hollow cylinder.

51. To develope the soffit of a circular arch, which cuts obliquely through a straight wall. Let ABCD be the plan of the wall. (fig. 3, pl. V.) From the point C, draw CGc perpendicular to CB, and produce AD to G. On CG describe a semicircle CFG, which is the curvature of the arch. Divide the semicircle GFC into any number of equal parts, so small in practice that the distance between two points on the arc may be considered a straight line, and extend the same number of parts along from G to c. From each point, both in the arch and in the line Gc, draw a line parallel to GA; and from each point, where the parallels from the divisions in the arc cut the line AB of the wall, draw a line parallel to Cc, which will meet the corresponding parallel of the developement in the edge of the soffit; and the line Aeb being drawn through the points thus found, will be the form of the edge corresponding to the side AB. The breadth of the soffit measured on the parallels will be every where the same, and equal to AD, or BC; therefore, setting off this breadth on each parallel will give the other edge, and completes the soffit ADcbe.

If AGCH had been the plan of the opening, then AGch would have been the lining of the soffit, as it would in that case have been half a cylinder.

52. To develope the soffit of a circular arch, which cuts obliquely through a circular wall. Let ABCD, (fig. 4, pl. V,) be the plan of the wall. Draw Dd perpendicular to DA, and produce BC to meet Dd in G. On DG, as a base, describe the arch DFG, and divide it into equal parts; and extend those parts on the line Gd, so that Gd may be equal to the length of the curve DFG. From each point of division draw a line parallel to GB; and from the points where the parallels, from the divisions in the arch, cut the lines of the wall in the plan,