PLATE 59.

FIG. 1.-To describe the circumference of a circle of a given radius GH through a given point D, to touch a given circle ABC, whose centre is F; provided that the sum of the difference between the two given radii and the radius of the required circle shall exceed a straight line, drawn from the given point to the centre of the circle given in position.

Join FD; from GH cut off the part GI, equal to the radius of the circle ABC, and with the difference IH, on the centre F, describe an arc at E; on D, as a centre with a radius GH, describe an arc cutting the former at E; lastly, on E, as a centre with the distance ED, describe a circle, it will touch a circle ABC, as was required.

FIG. 2.-To describe a circle, to touch a given circle ABC, whose centre is K, and to touch also a given straight line DE, in the point F.

From F draw FI perpendicular to DE; from the point F, and from the line FI, cut off the part FG, equal to the radius of the circle ABC; join KG, and bisect it by a perpendicular, cutting FI at I; through the points I and K draw the straight line IKA; lastly, with the radius 1A or IF describe the circle required.

FIG. 3.-To describe a circle, to touch a given circle ABC, whose centre shall be in a given straight line FE, and pass through a given point F in that line.

Through D, the centre of the given circle ABC, draw DA parallel to FE, cutting the circumference of the circle ABC at A; through the points A and F draw the straight line AFB, cutting the circumference of the circle ABC at B; draw the straight line BED, cutting FE at E; then on E, as a centre with the distance EB or EF, describe the circle required.

PLATE 60.

FIGS. 1, 2, 3.-To describe a parabola by means of tangents, having a double ordinate AB, and a diameter ED to that double ordinate.

Produce the diameter DE towards c; make EC equal to ED; join AC and BC, divide each of the lines AC and Bc into a like

number of equal parts, and join the corresponding points of the divisions, and it is done.

Although any mathematical reader may easily prove that a figure described in this manner is a parabola, yet such is the ignorance of many architects, and of workmen in general, that they pretend they can by this method describe a figure almost of any form they please.

If the parabola be very flat, as in Fig. 3, it may be used in practice, without any sensible error, for the segment of a circle; or it may be described as in Fig. 4, by describing each half of the figure as before; it has also been applied in describing an ellipsis, as is shown at Fig. 5; but the difference of a figure compounded of parabolas from the true ellipsis is very visible to every discerning eye, as is shown in Fig. 5, where the inside curve is an ellipsis taken from Fig. 6; the parabolic curve seems more particularly adapted to the description of gothic arches, as applied in some of the following examples.

FIG. 7 is another ridiculous application of parabolic curves, in order to produce the form of an egg, or oval; but to an eye that has been accustomed to curves generated by a continued motion of an instrument, this curve is very ungraceful. FIG. 8 shows a method, by means of sincs, of inscribing an ellipsis within a trapezoid.

PLATE 61.

FIG. 1.-To find the joints of an arch ABD, that is the segment of a circle.

Divide the arch into as many equal parts as you intend to have joints to make a joint at any point d, that is not at one of the extremes of the arch, take any two equal distant points a and b, on each side of the point d; and with any radius greater than bd, or da, on the points 6 and a, as centres, describe two arcs, cutting each other at c; draw cd, and it will be one of the joints at the point d; in the same manner find all the other joints except the two extremes, which may be found as follows. Suppose it were required to find the extreme joint at B, take any distance de, on any one of the joints that is found, and with that radius on в describe an arc at e; and with the distance BC, on the point d describe an arc cutting the former, and draw EB, and it will be the joint required.

FIG. 3.-To find the joints of an elliptic arch ABD.

Find the foci E and F, by Problem 1, page 15; then to draw a joint through any point G, in the arch, draw the straight lines FG and EG; and bisect the angle aGb by the line co, and it will be a joint at the point G in the same manner all the other joints may be found.

FIG. 2 is an arch made of the sectors of circles, representing an elliptical arch, the joints in any of the sectors will be found, by drawing them to their corresponding centres.

PLATE 62.

FIG. 1.—To draw a gothic arch of any height CD, and width AB, and to touch two given lines DG and DH, making equal angles with CD.

Draw AG and BH perpendicular to AB, cutting DG and DH, at G and H; join GH, cutting CD at E; then apply the distance ED, from c to F, in the same straight line; join FB and FA; divide each of the lines FB and FA, AG and BH, into a like number of equal parts, draw lines through the points 1, 2, 3, 4, 5, in AG and BH to the vertex D; also draw lines through the points 1, 2, 3, 4, 5, in the lines AF and BF to 1, cutting the former, as is shown in the figure, and these will be the points in the curve.

If the point E be distant from c more than one half of CD, the two sides of the gothic arch will be two parts of an ellipsis; but if the point E falls in the middle of CD, then the two sides of the arch are parabolical; lastly, if E fall under one half of CD, then each side of the arch is hyperbolical.

FIG. 2.-The same things being given, to describe the representation of such an arch with compasses.

Make AI equal to AG, and describe on 1, with the radius 14, a part of a circle AOL; then describe a circle to touch the straight line DG, at the point D, and the circle AOL, and it will complete one side of the arch, the other side may be completed in the same manner, as is plain by the figure.

FIGS. 1, 2, 3.—To describe parabolical gothic arches, whether right or rampant, to any two lines meeting at the vertex, and also to touch any two lines from the ends of the arch. Proceed with each half of the arch, as in Plate 60, page 166; that is, describe a parabola on each side, and it is done.

PLATE 64.

FIG. 1. To describe a gothic arch by finding points in the curves, the tangent DE and DF at the vertex of the arch being given.

Join DA and DB, and bisect them at the points G and H; join also EG and HF; bisect them at the points K and 1; then will the lines KG and HI be diameters; DA and DB will be a double ordinate corresponding to each; (then by Problem 1, page 23,) describe each parabola, and they will form the gothic arch required.

FIG. 2.-To find the joints of a gothic arch described as above.

Suppose it were required to draw a joint to a given point a; through a draw ac, parallel to AD, cutting the diameter EG at c; from the point d, where EG cuts the arch, make de equal to de; join ea, and it will be a tangent at the point a; a line drawn through a, perpendicular to ae, will be the joint sought; in the same manner all the other joints are to be found.

The joints are to be found in Fig. 3 in the same manner.

MOULDINGS.

DEFINITIONS.

1. MOULDINGS are profiles composed of various curves and straight lines.

If the mouldings are only composed of parts of a circle and straight lines they are called Roman, because the Romans, in their buildings, seldom or never employed any other curve for mouldings than that of a circle; but if a moulding be made of part of an ellipsis, or a parabola, or an hyperbola, the mouldings are then in the Grecian taste.

Corollary. Hence it appears, that mouldings in the Greek taste are of a much greater variety than those of the Roman, where only parts of circles are concerned.

Mouldings have various names, according to the manner in which they are curved.

2. The straight-lined part under or above a moulding in general is called a fillet.

3. If the contour of the moulding be convex, and a part of a circle equal to or less than a quadrant, then the moulding is called a Roman ovolo, or an echinus, such as Fig. 2, Plate 66.

4. If the contour of the moulding be concave, and equal to or less than a quadrant, it is called a cavetto, or hollow, such as Fig, 3, Plate 66.

Corollary. Hence a cavetto is just the reverse of an ovolo. 5. A bead is a moulding, whose contour is simply a convex semicircle.

6. If the contour be convex, and a complete semicircle, or a semiellipsis, having a fillet above or below it, the moulding is called a torus, as Fig. 1, Plate 66.

Corollary. Hence a torus is a bead with a fillet; and is more particularly distinguished in an assemblage of mouldings from a bead, by its convex part being much greater.

7. If the contour of a moulding be a concave semiellipsis, it is called a scotia, as Fig. 6, Plate 68.