~ Pergamon Mech. Mach. Theory Vol. 31, No. 6, pp. 763-769, 1996 Copyright © 1996 Elsevier Science Ltd 0094-114X(95)00108-5 Printed in Great Britain. All rights reserved 0094-114X/96 $15.00 + 0.00 THE RSSR MECHANISMS WITH PARTIALLY CONSTANT TRANSMISSION ANGLE ZiYA SAKA Selquk University, Makina Miih. B61iimii, 42040 Konya, Turkey (Received 15 August 1994) Abstract--It is required to calculate roots of a quartic equation for finding extremums of the transmission angle of an RSSR mechanism. If three roots of this equation are equal, the transmission angle will remain nearly constant within an interval. By normalizing the mechanism dimensions, the required conditions for these types of mechanisms are discussed. Copyright © 1996 Elsevier Science Ltd 1. INTRODUCTION The transmission index for spatial linkages is defined by Sutherland and Roth [1]. The movability and transmission angle of the spatial RSSR linkages have been extensively treated by several authors. S6ylemez and Freudenstein [2] give the design charts similar to Alt Charts for spatial four bar mechanisms. Gupta and Kazerounian [3] present a technique to design fully rotatable RSSR linkages with transmission angle control. The extremums of the transmission angle and the analysis of movability depending on this angle are studied by Zhang [4] and Kazerounian and Solecki [5]. An RSSR mechanism with constant transmission angle is investigated in Ref. [6]. In this work, depending on the normalized dimensions, the nearly constant transmission angle within an interval is investigated on the RSSR mechanisms. 2. TRANSMISSION ANGLE FORMULATION AND NORMALIZATION OF DIMENSIONS A spatial RSSR mechanism is shown in Fig. 1, where z 2 and z4 are common perpendicular of input and output axes, al, a2, a3 and a 4 are link lengths, s2 and & are crank and follower axial positions, respectively, ~ is the angle between revolute pair axes, and 0 and ~b are input and output angular positions. Output displacement equation is [3, 4]: E ___ N / E 2 q- F 2 - G 2 tan~= F+G (1) where, E = a4(a 2cos <xsin 0 - s2 sin ~t) F = a4(al + a2 cos 0) G = 0.5t + a2(at cos 0 + s4 sin ~ sin 0) __ 2 2 2 2 2 t - al + a2 -- a3 + a24 + s2 + s4 + 2s2& cos o~ Cosine of the transmission angle is [5]: cos2# = 1 E 2 + F: - G 2 2 2 a3a4 (2) 763