0.5

1 cm

GEOMETRICAL CONSTRUCTIONS—DRAWING OF ELEMENTARY FORMS.

35

division in E F, and from K draw lines to meet those

others successively. The intersections so found are

points in the curve, which may be traced accordingly.

Problem LYIII.—To describe a Gavetto.

The Cavetto is described like the Roman ovolo : by cir-

(Fig. 160.) (Fig- 161.)

cular arcs, as shown in (Figs. 160 and 161). Sometimes

it is composed of two circular arcs united (Fig. 162); set off

b e, two-thirds of the projec-

tion, draw the vertical b d

equal to b e, and on d de-

scribe the arc b i. Join

e d and produce it to p ;

draw i % perpendicular to

e d, set off n o equal to

n i, and draw the hori-

zontal line o p, meeting

e p ; on p describe the arc

i o to complete the curve.

Problem LIX.—To describe & Cyma recta, or Ogee.

The ogee (Fig. 163) is compounded of a concave and a

conves surface. Join a and b, the extremities of the

curve, and bisect a b at c; on a, c, as centres, with the

(Fig. 163.) (Fig. 164.)

radius a c, describe arcs cutting at d; and on b, c, describe

arcs cutting at e. On d and e, as centres, describe the

arcs a c, c b, composing the moulding.

Problem LX.—To describe a Cyma reversa, or Talon.

The talon (Fig. 164), like the ogee, is a compound

curve, and is distinguished from the other, by having the

convex part uppermost. It is described in the same

manner as the ogee.

N ote—If the curve be required to be made quicker, a shorter

radius than a c must be employed. The projection of the

moulding, n b (Fig. 163), is usually

equal to the height, a n.

Second, the Greek talon. Join

the extreme points a, b (Fig. 165);

bisect a b at c, and on a c, c b,

describe semicircles. Draw perpen-

diculars d o, &c., from a number of

points in a c, c b, meeting the cir-

cumferences ; and from the same

points set off horizontal lines equal to the respective

perpendiculars: o n equal to o d, for example. The curve

line b n a, traced through the ends of the lines, will be

the contour of the moulding.

Problem LXI.—To describe a Scotia.

ls£ Method.—Draw the perpendicular A E from the

upper projection A, draw a perpendicular from C, and set

off C F, equal to one-half A E added to two-thirds C E.

On F describe the semicircle C H G, Join G A, and

produce it to H; and draw F H, cutting A E at K. With

K A as radius describe the arc A H, to complete the curve.

2d Method.—Divide the perpendicular A B (Fig. 167),

into three equal parts; and, with the first, A E, as radius,

and on centre E, describe the arc A FH; on the per-

pendicular C 0, set off C L equal to A E, join E L and

bisect it by the perpendicular O D meeting C 0 at 0.

On centre 0, with radius 0 C, describe the arc C H to

complete the figure.

3 d Method.—Join A C, and describe a semicircle, AFC.

In the semicircle draw a number of perpendiculars to

A C, and from the same points in A C, draw horizontal

lines respectively equal in length to the perpendiculars;

and trace the curve A G C through the ends of these lines.

(Fig. 168.) (Fig. 169.)

4th Method.—Join a b (Fig. 169) and bisect it at c;

draw a horizontal line through c, making c d and c e

each equal to the required recess of the curve. Through

d, draw f g parallel to a- b; divide a c and a f into the

same number of equal parts, and to the points of division

draw lines from the points d and e, successively intersect-

ing as shown. Do similarly for the upper paid of the

figure. The curve traced through the points so found is

that required.

Note.—Of these four methods the fourth is the most

generally useful. When the projection of the lower part

is relatively small, the curve obtained by the third

process is nearly of the form of the semicircle used in

finding it. The first method should be employed only for

medium projections, when the projection is about two-fifths

of the height.

(Fig. 162.)

35

division in E F, and from K draw lines to meet those

others successively. The intersections so found are

points in the curve, which may be traced accordingly.

Problem LYIII.—To describe a Gavetto.

The Cavetto is described like the Roman ovolo : by cir-

(Fig. 160.) (Fig- 161.)

cular arcs, as shown in (Figs. 160 and 161). Sometimes

it is composed of two circular arcs united (Fig. 162); set off

b e, two-thirds of the projec-

tion, draw the vertical b d

equal to b e, and on d de-

scribe the arc b i. Join

e d and produce it to p ;

draw i % perpendicular to

e d, set off n o equal to

n i, and draw the hori-

zontal line o p, meeting

e p ; on p describe the arc

i o to complete the curve.

Problem LIX.—To describe & Cyma recta, or Ogee.

The ogee (Fig. 163) is compounded of a concave and a

conves surface. Join a and b, the extremities of the

curve, and bisect a b at c; on a, c, as centres, with the

(Fig. 163.) (Fig. 164.)

radius a c, describe arcs cutting at d; and on b, c, describe

arcs cutting at e. On d and e, as centres, describe the

arcs a c, c b, composing the moulding.

Problem LX.—To describe a Cyma reversa, or Talon.

The talon (Fig. 164), like the ogee, is a compound

curve, and is distinguished from the other, by having the

convex part uppermost. It is described in the same

manner as the ogee.

N ote—If the curve be required to be made quicker, a shorter

radius than a c must be employed. The projection of the

moulding, n b (Fig. 163), is usually

equal to the height, a n.

Second, the Greek talon. Join

the extreme points a, b (Fig. 165);

bisect a b at c, and on a c, c b,

describe semicircles. Draw perpen-

diculars d o, &c., from a number of

points in a c, c b, meeting the cir-

cumferences ; and from the same

points set off horizontal lines equal to the respective

perpendiculars: o n equal to o d, for example. The curve

line b n a, traced through the ends of the lines, will be

the contour of the moulding.

Problem LXI.—To describe a Scotia.

ls£ Method.—Draw the perpendicular A E from the

upper projection A, draw a perpendicular from C, and set

off C F, equal to one-half A E added to two-thirds C E.

On F describe the semicircle C H G, Join G A, and

produce it to H; and draw F H, cutting A E at K. With

K A as radius describe the arc A H, to complete the curve.

2d Method.—Divide the perpendicular A B (Fig. 167),

into three equal parts; and, with the first, A E, as radius,

and on centre E, describe the arc A FH; on the per-

pendicular C 0, set off C L equal to A E, join E L and

bisect it by the perpendicular O D meeting C 0 at 0.

On centre 0, with radius 0 C, describe the arc C H to

complete the figure.

3 d Method.—Join A C, and describe a semicircle, AFC.

In the semicircle draw a number of perpendiculars to

A C, and from the same points in A C, draw horizontal

lines respectively equal in length to the perpendiculars;

and trace the curve A G C through the ends of these lines.

(Fig. 168.) (Fig. 169.)

4th Method.—Join a b (Fig. 169) and bisect it at c;

draw a horizontal line through c, making c d and c e

each equal to the required recess of the curve. Through

d, draw f g parallel to a- b; divide a c and a f into the

same number of equal parts, and to the points of division

draw lines from the points d and e, successively intersect-

ing as shown. Do similarly for the upper paid of the

figure. The curve traced through the points so found is

that required.

Note.—Of these four methods the fourth is the most

generally useful. When the projection of the lower part

is relatively small, the curve obtained by the third

process is nearly of the form of the semicircle used in

finding it. The first method should be employed only for

medium projections, when the projection is about two-fifths

of the height.

(Fig. 162.)