There are other oval arches, more properly called elliptical; but a description of these does not fall within the limits of the present work.

In Architecture, the doctrine of Tangents is of almost incessant use and application. The Greeks and Romans scarcely used any other form of arch, or any other curve, than the circular, in their buildings; and however beautifully varied their architectural ornaments may appear, the outlines of these are nothing more than simple combinations of circular arches and their tangents, or of right lines merely.

which is not the case with either Euclid's method or the preceding one; PROB. VII. being employed in the former, and PROB. V. in the latter.

The construction of a basket-handle arch mentioned in the above Article will serve to evince the practical utility of this problem. An Architect proposes to himself the following object:

On a given base to construct a basket-handle arch of a given height.

How is he to accomplish this undertaking?

Let AB represent the given base, and ST the given height. Divide AB equally at c, by PROB. V., and draw CL perpendicular to AB, by PROB. VII., and take CL equal to ST. Through L draw LX parallel to

AB, by PROB. IX. Now L must be the highest point of the arch; wherefore, take any portion of it from L, as LV, and measure off AE equal to Lv. Round E as centre describe the circular arch AQ passing through A. Join Ev; divide it equally at o, and draw the right line, or, perpendicular to Ev, by PROB. VII., and meeting LX in P. From P, by PROB. XII., draw a tangent, PG, to the circular arch, aQ. Finally, joining GE, and producing it through E, to meet Lc produced in к, describe round x as a centre a circular arch passing through &; this will also

For example, the opposite diagram represents a column of the Doric order, with its pedestal, shaft, and entablature, and their respective mouldings. The uppermost band,1,is called a Fillet, and is merely a rectangle or oblong. The second band, 2, is called a Cimatium, with its end of a curvilinear shape.

To form this: (see fig. 2) by PROP.VII. we draw be perpendicular to ba, the upper side of the moulding, and equal to half the depth which the moulding is to be. We then draw ca parallel to ba, by PROB. IX.; and round c as a centre describe the quarter-circle be, meeting ca in e. Finally, we take ed equal to ec, and round d as a centre we describe the quarter-circle en, passing through e. We have thus formed the curved outline, ben, of the end of the cimatium.

In this, ab is a tangent at

b to the curvilinear end, by

ART. 55, as bc is perpendicular to ab. Also nm, or the

pass through L, and the quarter-oval AGL be thus formed of the given height, CL.

For: in the triangles, POV, POE, since ov is equal to OE, the angle pov equal to PoE, and the side or common,-the side PV is equal to PE, by ART. 1. Wherefore, in the triangles PLV, FGE, since PV is equal to PE, and likewise the angle VLP to the angle EGP, (both being right,) the squares of PL and of LV together are equal to the squares of PG and of GE together. Consequently, as Lv is equal to GE, the square of PL is equal to that of PG; and therefore PL is equal to PG. Join PK. In the triangles PLK, PGK, the angles PLK, PGK, are both right; consequently, the squares of PL and of LK together are equal to the

lower side of the cimatium, is a tangent at n to the curvilinear end, by the same ART.

Number 3 is another fillet.

Number 4 is a reverse cimatium; its end being curved the opposite way to that of the first cimatium.

To form this: take bc equal to half the depth that the moulding is to be, and round c as a centre describe a quarter-circle be,

passing through b. Produce ce, until ed be equal to ec; and round d as a centre describe a quarter-circle en, passing through e. We have thus formed the curved outline, ben, of the end of the reverse cimatium. The end-line of number 3 is sometimes a tangent at b to be; and the end-line of number 5 a tangent at n to en. Number 6 is denominated an Ovolo, and number 8 a Cavetto; their curved ends are, each, quartercircles, be, described with a radius, bc, equal

8

to the depth of the moulding. The only difference is, that in the former the centre of the curved end is within the moulding; in the latter, without; or, in other words, the first is prominent, the second hollow.

Sometimes a moulding, as the uppermost in the opposite diagram, has an end whose outline, abc, is a half-circle, which is formed by taking its radius, oc, half the depth of the moulding. To this curvilinear end both sides, ad, ce, of the moulding are tangents.

Again: the shaft is often joined to the moulding upon it by means of a quarter-circle,

a

bo

d

mn, whose radius is rn, equal to the projection of the

squares of PG, and of GK, together. But PL has been proved equal to PG, and therefore the square of LK is equal to the square of GK, that is, LK is equal to GK. Hence, the circular arch described round K, with the radius KG, will pass through ; and the quarter-oval, AGI., be therefore of the height required.

In the same manner is the other part, LB, of the semi-oval described, taking FB equal to LV.

moulding beyond the outline, ans, of the shaft.-To this curve, the said outline, ns, and the lowermost side, ms, of the moulding, are tangents at the points n and m respectively.

ART. 57. "If a right line be a tangent to a circle, the perpendicular to it at the point of contact will, if produced sufficiently, pass through the centre."

On this principle we may find the centre of a circle by help of the common Square alone.

Lay the inner edge of one blade at any point of the circumference, B, so that this edge, DC, may just touch the circumference at that point; and so that the outer edge of the other blade may pass through the point of contact. Trace this edge, A, o, B, within

H

the circle.-Now remove the square to any other point, F, of the circumference, and proceeding in like manner, trace EOF, the outer edge of the blade EF. That point, o, where these edge-traces intersect will be the centre of the circle.

For: by ART. 57, as each edge-trace is perpendicular to the tangent lines DC, GH, they both pass through the centre, which must therefore lie in their point of intersection.

Or, we may find the centre without any other assistance than that of a ruler, or straight edge,-provided one of its ends be (as both usually are) exactly square. For we have only by its means to trace two such lines as DC, HG, touching the circumference at в and F respectively; then to set the instrument upright on these lines, with either of its edges running through the said points; and these edges being traced (in the same manner as those of the blades AB, EF), their point of intersection, o, will be the centre of the circle.

LESSON VII.

ART. 58. "The diameter of a circle is the greatest chord which can be drawn in it."

The utility of this consideration in practice is obvious. If the trunk of a tree measure no more than a certain number of inches through the heart or centre, it will not afford a beam of more than that number of inches thickness, shape it how you will. The dimension, or capacity, of a globe is always to be estimated by the length of its diameter, for it is broader from end to end of this, than from end to end of a straight line joining any other two points on its surface. An augre, or boring-piece, will make a hole of no greater width than the diameter of its circular blade, where it is broadest. The calibre of a pistol, gun, or any description of ordnance, is measured by the inner diameter of the barrel; and no ball will fit an engine of this kind if the former be greater in diameter than the latter,-unless first obliged to change its shape by the ramrod. Numberless other examples might be adduced; but these are sufficient to evince the practical tendency of the theorem abovestated.

DEF. XXIII. "A rectilineal figure is said to be inscribed in a circle when the vertices of all its angles are situated in the circumference of that circle."

On the other hand, a circle is said to be circumscribed round a rectilineal figure when it passes through the vertices of all the angles of that figure.

TEACHER. An Upholsterer wishes to insert the end of a round beam into the square aperture of another beam,-the diameter of the former being exactly equal to the diagonal of the square: how should you direct him to proceed?

LEARNER. Suppose acd the cir

cular end of the round beam, whose diameter, ac, is equal to xy, the diagonal of the square aperture in the other beam, AB. Draw the diameter ac, and another diameter bd, perpendicular to the former. Join ab, bc, cd, da, and the figure abcd will be a square exactly equal to the