cutting the axis at A.

into a straight line of FG, each equal to A B.

from O as a centre, with OD as radius, describe a circle Take A B equal to the development the quadrant AD; make BE, EF, Then A G represents the development of the whole circle. Divide the quadrant AD into any number of equal parts at K, L, &c.; and also A B into the same number of equal parts at H, I, &c. Draw verticals at H, I, &c., cutting horizontals at K, L, &c., in the points P, Q, &c., then these will be upon the required curve; and for every point on the quadrant of the circle we can find a corresponding point on the curve. The horizontal DC will be a tangent to the curve at C, and the part of the curve CE is an exact repetition of CA, so also is EV and V G; so that, having drawn the part A C, we can obtain any number of undulations.

The curvature is greatest at the vertex C, and decreases towards A, at which point the radius of curvature becomes of infinite length, A being a point of contrary-flexure; so also are the points E, G, and any other point where the curve cuts the axis.

157. To find points on the harmonic-curve by a scale of parts.-Take FG (fig. 127) as an axis, and draw FV at right angles thereto. Measure FV equal to 1000 on any scale, and FG equal to 1571 on the same scale. Divide FG into twelve equal parts, and draw the ordinates; measuring their lengths by the same scale, according to the numbers marked on the figure; then eleven points will be found upon the curve between the axis and the vertex.

Any variety of the harmonic-curve can be produced by measuring the abscissæ on one scale, and the ordinates on some other scale. Let A B (fig. 128) represent the length of a quadrant of the generating circle, draw BC at right

angles to AB, and equal to its radius. Then AC is the curve, having its abscissæ and ordinates measured on the same scale as in fig. 127; the curve A C has the scale of the ordinates half that of the abscissæ; the curve A C2 has the Fig. 128.

scales in the ratio of 3 to 2; the curve A C has the scale of the ordinates double that of the abscissæ.

158. To draw the tangent at any point of the harmoniccurve. The tangent at A (fig. 127), the point of contraryflexure, is drawn by taking A N equal to BC, drawing NT vertical cutting the horizontal through C in the point T; then TA is the tangent at A.

If we draw the horizontals through C1, C2, C3 (fig. 128), meeting NT produced in T1, T2, T3; then T, A, T, A, T, A, are the tangents to the curves A C1, A C2, A C,, at the point A. The tangent at the vertex C is always parallel to the axis A B.

The tangent at any other point of the curve can be drawn

by first drawing the normal (159) at that point, and then drawing a perpendicular to the normal at the given point.

159. To draw the normal at any given point of the harmonic-curve.-Let P (fig. 127) be the given point on the curve, HP the ordinate perpendicular to AB. On BA take BJ equal to A H, and draw the ordinate JR, meeting the curve at R. Produce BC to U, making CU equal to JR; draw CPW cutting the axis in the point W, and join W U. Produce HP to meet W U at S, and take Hn on the axis equal to PS; then Pn is the normal at the point P; Hn, called the subnormal, being a fourth proportional (16) to BC, JR, and HP.

The normal can be drawn in the same way in any variety of the curve (fig. 128), the subnormal being the fourth proportional in every case to BC, JR, and HP; BC being taken the same length in all varieties of the curve.

160. To set out the harmonic-curve as a railway curve. -Let it be required to join the points E and G (fig. 127) by a curve, the tangents at those points meeting at e. Join EG, and bisect EG in F, draw the perpendicular Fe. Let FG represent 1571 on any scale, and take Ga, representing 1000, on the same scale. Draw a b perpendicular to Fa, meeting the tangent Ge in b. Draw bV parallel to FG. Divide FV into 1000 equal parts, and set up the twelve equidistant ordinates, as figured along FG, to the same scale as that on which FV represents 1000; then twelve points will be determined on the harmonic curve. By repeating the process, the curve from V to E can be set out in like manner.

Where an S curve is required from A to G, the tangents at those points being parallel; divide A G into four equal parts, A B, BE, EF, FG. Draw Fe at right-angles to F G, and meeting the tangent at G in the point e; e E will be

the tangent at E. Then proceed to set out the curve G V,

as above described, and the remaining parts, VE, EC, CA, will be similar to V G, and can be set out in the same manner.

By using this curve for railways instead of circular arcs, the sudden change in the direction of the train, which occurs in passing from a straight line to a circle, is avoided, and greater safety insured.

161. To apply the harmonic-curve to form the entasis of a column.-Let CB, AE (fig. 129), be the radii at bottom and top of the shaft, BA the axis of the column, BC and AE at right-angles to A B. Take BD equal to A E, then CD is the difference between the radii at bottom and top. Draw DE parallel to BA, and divide it into any number of equal parts at G, H, &c. From B as a centre, with BC as radius, describe a circle cutting DE in the point F. Divide the arc CF into as many equal parts as DE is divided into, by the points a, b, &c. Through a, b, &c., draw lines parallel to DE, intersecting lines parallel to

Fig. 129.

BC through G, H, &c., in the points P, Q, &c.; these will be points upon the required curve, through which it can be drawn by bending a lath.

I

162. To apply the harmonic-curve to form the contour of an ogival arch or pinnacle.-Let CF (fig. 130) be the given Fig. 130.

span of the arch.

angles to CF.

at the centre.

Fig. 131.

Bisect CF in E, and draw ED at rightLet ED be the given height of the arch Bisect CE in B and draw BA at rightangles to CE, and equal to half ED. Draw CK parallel to A B, and take AH to AB in the ratio of 1000 to 1571. Draw H K parallel to CE and cutting CK in K; then KA is the tangent at the point of contrary-flexure A. Taking CB to represent 1000 on any scale, we can find twelve points on CA, as in VG (fig. 127), by measuring ordinates to the same scale. curve from A to D is similar to A C, and the curve DF is a repetition of DAC. The joints of the youssoirs of the arch must be in the direction of the normals to the curve (158).

The

The outline of an ogival pinnacle (fig. 131) can be formed in the same way.