282 SPIRAL OF ARCHIMEDES. In this equation ✪ is the circular measure of the position angle, and therefore ra when is unity, i.e. when the number of degrees in the position angle is 57-2957... i.e. corresponding to this angle measured from the initial line, the tracing point is at a distance of a units (inch or any other that may be chosen) from the pole; when r0, 0-0, or the initial line is the position of the revolving line when the travelling point is at the pole. PROBLEM 145. To describe the spiral of Archimedes, the pole, two points on the curve, and the unit of the curve being given (Fig. 148). Let O be the pole, P and Q the two points on the curve which we will suppose to be on the same convolution; and let OQ be greater than OP; let ◊ be the angle between OP and the initial line, and the length L the given unit. OP=a9 OQ=a (0 + QOP); therefore or OQ – OP = a × circ. meas. of QOP OQ-OP a circ. meas. of QOP' OQ - OP can be measured by scale, the number of degrees in the angle QOP can be measured by a protractor and its circular measure can be obtained from a table of the circular measures of OP angles, and the numerical value of a thus calculated: then 0 a the length OP being of course measured on the same scale as that used for determining OQ – OP, which gives the circular measure of the angle between OP and the initial line, and the corresponding number of degrees can be obtained from the table. To take a numerical example: Let the unit of length be an inch. Suppose OQ = 2, OP = 1.5, and the angle QOP = 60o, the circular measure of which is the number of degrees corresponding to which may be taken 180o. The initial line will therefore be the line OA. If the tracing point after one complete revolution of the generating line cuts OP again in P' we have and therefore OP = að = a (0 + 2π), OP' - OP = 2πα. Successive points on the curve may at once be found thus: Construct an angle QOR= angle QOP; with centre O and radius OP describe an arc cutting OQ in p; on OQ produced make 284 TANGENT AND RADIUS OF CURVATURE. Qr=Qp and with centre O and radius Or describe an arc cutting OR in R a point of the curve. Similarly if ROS = PÔQ, and Qs on OQ produced = 2Qp, an arc described with centre O and radius Os will cut OS in S a point of the curve. (In the figure S coincides with A on the initial line.) In like manner points can be found nearer the pole than P by constructing angles on the side of OP remote from Q equal respectively to POQ, 2POQ, 3POQ, &c., and diminishing the radii vectores by the constant difference pQ. To draw the tangent at any point of the curve. A known expression for the angle which the tangent at any point makes with the radius vector is tan i.e. the tangent ༡༩ 1 , a of the angle is the radius vector divided by the given constant of the curve. Therefore to draw the normal at any point Q, on the radius OG at right angles to OQ measure a length OG = a, the constant of the curve, and QG will be the normal at Q, for evidently Hence if a circle be drawn with centre O, and radius a, nor mals at all the points on the curve can at once be drawn by merely joining them to the corresponding points in which such circle cuts the perpendicular radii. The initial line is a tangent at the pole. If is the radius of curvature at any point P p: √ a2 + r2 :: a2 + r2 : 2a2 + r2, so that p can be calculated without much difficulty. PROBLEM 146. To describe the spiral of Archimedes, the pole, the initial line and the constant of the curve being given (Fig. 149). Here a is given in the equation r=a0. Let O be the pole, and ÒA the initial line. In the figure, the unit being the length L, a = '239. Determine some convenient length of radius corresponding to a multiple (n) of 4 right angles; say the greatest distance to which it is proposed to draw the curve. In the figure e.g. A is taken at angular distances of 8 right angles from the initial line (i.e. n = 2), so that OA = ·239 × 4π Draw OD at right angles to OA and divide up the quadrants formed at O into any number (m) of equal parts (in the figure m = 3) and draw the radii OB, OC, &c. through the points of division. Divide OA into 4.m.n equal parts. In the figure therefore OA is divided into 24 equal parts. Then arcs drawn through the successive points on OA with centre O will intersect the corresponding radii in points on the curve. The point P in the figure of course bisects OA, and after one complete convolution has been found the curve can be completed by measuring from B, C, &c. on the successive radii a constant distance BQ, CR, &c. = AP. THE RECIPROCAL OR HYPERBOLIC SPIRAL. DEF. In this curve the length of the radius vector is inversely proportional to its position angle. where r is the length of any radius vector, ◊ the circular measure of the angle it makes with the initial line, and a a numerical constant. When 0=0, is therefore infinite, and r diminishes as increases, but the curve does not reach the pole for any finite value of 0. Corresponding to the value 0 = 1, r = ; i.e. the radius 1 a vector making 57.2957... degrees with the initial line is long. A line parallel to the initial line and an asymptote to the curve. 1 units distant from it, is a PROBLEM 147. To draw the reciprocal spiral, the pole, the initial line and the unit and constant of the curve being given (Fig. 150). Let O be the pole and OA the initial line. 1 In the figure a = 6 |