163. To apply the harmonic-curve to mouldings and. other architectural ornaments. - The use of continuous

mathematical curves for the contour of mouldings, as the

oyee and cima, would produce a more pleasing effect than the common practice of drawing them by arcs of circles.

Let the points C and D be given (figs. 132, 133); draw CF, DF, at right-angles to each other, and bisect DF in E. Draw E B parallel to CF, BC parallel to DF, and bisect BE in A; then A is the point of contraryflexure, EAB the axis, D and C the vertices of the curve, which can be drawn in the manner above described (156).

Fig. 133.

The harmonic-curve can also be employed to form an interlacing ornament (fig. 134), AG being the axis, C

and V the vertices, A, B, E, F, G, the points of contraryflexure.

164. To draw a lemniscate by continuous motion.-Let A C and O B (fig. 135) be two rods or slips of card revolving about the fixed points A and O, the length A C being capable of alteration. Let another rod or slip, PBC, of length equal to twice OB, be attached by one end to a pivot C at the end of the rod A C, and also at its middle point to a pivot B at the end of the rod O B, so as to move freely about those points, as the rods O B, A C, revolve about O and A. When O B and A C coincide with the axis O A, the point B will be at D, OD being equal to twice OB. Now let the rods revolve about O and A, and a pencil at P will trace out the curve DY O. When P reaches the axis again at O, the points B and C will be at E and F, in one straight line with O.

The curve from D to O is one-fourth of the whole lemniscate, the other three parts of which are exactly equal and similar to DY O. Also the point O is a point of contrary

flexure, and the curvature increases from O towards D, where it is greatest.

165. To find points on the contour of a lemniscate.-Draw an axis DOA (fig. 135), and from any point O therein as a centre, and any radius OB, describe a circle HBK. Take OA on the axis greater than O K, and from A as a centre, with A O as radius, describe the circle OCF. From

O as a centre, with OD equal to twice OH as radius, describe an arc cutting the last circle in F; draw OF cutting the first circle in E.

Take any point B on the first circle, between E and H, and from B as a centre, with BO as radius, draw an arc cutting the second circle at C; draw CBP, and make BP equal to BC or BO; then P is a point on the required

lemniscate. Proceeding in this way, and taking other points on the first circle, we can find any number of points upon the curve between O and D, through which the lemniscate can be drawn by hand. The other portions of the curve, being only a repetition of the part DYO, can be drawn from it.

166. To find the algebraical equation to the lemniscate.— Since BP (fig. 135) is always equal to BC and BO, the angle POC is always a right angle (55).

Let OA=a, O A=b, O P=r, the radius-vector, and the angle POD=6. Then OC=2b. sin 6, O P2=PC2—OC2, or 2-4 a 2-4b2. sin 26. If we call a=1, then we have for the equation to the curve,

r= 2√1-b2 sin2 §.

Any varieties of the curve can be obtained by changing the length of b, or O A, as compared with that of a, or O B. 167. To draw a lemniscate of given dimensions.—Let the length OD, and the height XY, be given, Y being the point at which the curve is at its greatest distance from the axis OD. Bisect OD in H, and draw HZ at right-angles to OH, and equal thereto. Draw OZ, and from O as a centre, with OZ as radius, describe the arc ZY Q, cutting the axis at Q. Then this arc passes through the point of greatest distance, Y, from the axis in every variety of the curve, so that by taking HT equal to the given width or height of the curve, and drawing TY parallel to the axis, and cutting the arc at Y, we find Y the required point at which the curve has the greatest distance from the axis. The line TY is also the tangent to the curve at Y. OH is the radius of the smaller circle, and the radius OA of the larger circle is found by taking OH to represent 1000 on any

scale, and O A will be the quotient of 1,000,000 divided by the given height XY, measured to the same scale. For example, let X Y be given equal to one-third of OD, or two-thirds of OH, then by the above rule. we find that O A is one and a half times the length of O H.

The length of AO for any given values of X Y and OH can also be found geometrically, since O A is a third proportional (15) to XY and OH. Take HI on the axis equal to X Y, and draw IM making an acute angle with IO; make IJ equal to OH; join HJ, and draw O M parallel thereto; then JM is the length of OA required.

Having determined the radii of the two generating circles, the curve can be drawn as before described (164, 165).

If OX and X Y only are given, we can find the radius of the smaller generating circle in the following manner. Draw OG at right-angles to OX, and bisect the right-angle GOX by the line OZ. From O as a centre, with OY as radius, describe a circle cutting OZ in the point Z. Drop the perpendicular ZH upon the axis OH, and OH will be the required radius of the smaller circle; from which D is found by taking HD equal to HO.

When both the radii, OH and OA, of the generating circles are given, the length of X Y can be obtained by finding a third proportional to OA and OH. Or, take O H as 1000 on any scale, and from O as a centre with a radius 1414 on the same scale, draw a circle; divide 1,000,000 by the length OA (on the same scale), and take HT equal to the quotient; draw TY cutting the last drawn circle at Y, then YX is the greatest distance of the curve from the axis.

168. To draw a tangent to the lemniscate. The tangent at O (fig. 135) where the curve cuts the axis, and has a