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, is therefore infinite, and r diminishes as increases, but the curve does not reach the pole for any finite value 1 of 0. Corresponding to the value = 1, r=; i.e. the radius 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. the unit being the length L. In the figure a = 6 1 a Draw 04 perpendicular to OA and with O as centre, and as radius describe a circle 4, 8, 12,... and divide it up into any number of equal parts, as at 1, 2, 3... Draw the line OB making 57.2957... degrees with OA and cutting the circle in B; B will be a point on the curve. Determine the length of radius vector corresponding to any convenient division of the circle-say the radius making 45° with the initial line-i.e. determine Draw the line 02, and produce it to C making OC=7·63 units. C will be a point on the curve. As the angle doubles the radius diminishes one half; so that if OC is bisected in d, the length Od will be the length of radius vector making a right angle with the initial line, i.e. D on the line 04, OD being equal to Od, is another point on the curve, = Bisect OD in e and make OE on 08 Oe ; E will be a point on the curve. OE is also of course = = 100. Similarly OF the radius corresponding to 0 = 2π is 10E or G the point on the curve corresponding to 0 = 3 distance of OC from O. OH the radius corresponding to 0 = 3 is of course OG or of OC. OK the radius corresponding to π 5 = • 2 OM the radius corresponding to is OC, and ON the π radius corresponding to 0 = 5, is 10M or 10C. 4 and so on, and similarly any additional number of points can be obtained. In the figure OV bisects the angle AOG and therefore OV=2.OG, OW bisects the angle AOC and OW = 2.OC. To draw the tangent at any point p. Draw the radius Oq of the circle described with centre O and 1 radius perpendicular to Op. pq will be the tangent at p. a To determine the centre and radius of curvature at any point p. Draw the normal pm perpendicular to the tangent pq and meeting 90 in m. On Pq make pr = Oq = 1 = and pn mq. Then ns drawn through n parallel to rm, meeting pm in 8, determines s the required centre. THE LITUUS. In this curve the radius is inversely proportional to the squareroot of the angle through which it has revolved. Its equation is therefore The radius therefore diminishes as the angle increases and is of infinite length when 0 = 0: it never vanishes however large ✪ may be, so that the spiral never reaches the pole, but makes an infinite series of convolutions round it. PROBLEM 148. To draw the Lituus, the pole, the initial line and the unit and constant of the curve being given (Fig. 151). 1 39 Let O be the pole, and OA the initial line. In the figure a = the unit being the length L. Draw OC perpendicular to OA, π and determine the value of r corresponding to @= i.e. to 0 being the circular measure of a right angle. 2 , Make Oc on OA equal to this length, and make OB on AO produced equal to unity on the scale adopted. A mean proportional between OB and Oc will evidently be the required length OC, i.e. a semi-circle on Bc will cut OC in C, a point on the curve. Draw radii OG, OH bisecting the quadrants COD, DOE. Trisect Oc in e and g, and take two parts measured from 0 as Og. A mean proportional between OB and Og will be equal to the length OG at which the curve cuts the bisector OG of the right angle COB. Bisect Oc in d. A mean proportional between OB and Od will give the length of the radius vector OD corresponding to 0 = =T. Divide Oc into five equal parts, and take two of them from O as Oh. A mean proportional between OB and Oh will give the length OH of the radius vector corresponding to |