{"id":2353,"date":"2022-03-17T22:30:47","date_gmt":"2022-03-17T22:30:47","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=2353"},"modified":"2022-09-02T19:00:26","modified_gmt":"2022-09-02T19:00:26","slug":"altbagacisi","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch7\/7-1\/altbagacisi\/","title":{"rendered":"altBagAcisi"},"content":{"rendered":"
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\n\t

The Classical Transmission Angle Problem:<\/strong><\/h1>\n

The classical transmission angle problem can be stated as follows:
\n“Determine the crank-rocker proportions of a four-bar mechanism with a given swing angle (\u03c8<\/span><\/strong><\/em>) and corresponding\u00a0 crank rotation (\u03d5<\/em>), or time ratio, such that the maximum deviation of the transmission angle from 90\u00b0\u00a0is minimum.”<\/em><\/strong><\/span><\/p>\n

The problem can be considered in two parts. The first part is the synthesis problem in which one must determine four-bar mechanisms with crank-rocker proportions that must have a given swing angle and a corresponding crank rotation. There is an infinite set of solutions for this part of the problem. The second part of the problem is concerned with the optimisation. Out of the infinite possible solutions obtained in the first part, one must determine a particular four-bar mechanism whose maximum transmission angle deviation from 90\u00b0\u00a0is a minimum.<\/p>\n

\"\"<\/p>\n

For the first part of the problem consider the two dead center positions of the crank-rocker mechanism (see figure above). We can write the loop closure equations for these two particular positions (note that the coupler and the crank are collinear at the dead-centers).<\/p>\n\n\n\n\n
a2<\/sub>ei\u03b2<\/sup> + a3<\/sub>ei\u03b2<\/sup> = a1<\/sub>\u00a0+ a4<\/sub>ei\u03c81<\/sub><\/sup><\/td>\n(7)<\/td>\n<\/tr>\n
a2<\/sub>ei(\u03b2 + \u03d5)<\/sup> + a3<\/sub>ei(\u03b2 + \u03d5 \u2212 \u03c0)<\/sup>\u00a0= a1<\/sub>\u00a0+ a4<\/sub>ei(\u03c81<\/sub> + \u03c8)<\/sup><\/td>\n(8)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Rearranging:<\/p>\n\n\n\n\n
(a2<\/sub> + a3<\/sub>)ei\u03b2<\/sup> \u2212 a4<\/sub>ei\u03c81<\/sub><\/sup>\u00a0= a1<\/sub><\/td>\n(9)<\/td>\n<\/tr>\n
(a2<\/sub> \u2212 a3<\/sub>)ei\u03b2<\/sup>ei\u03d5<\/sup> \u2212 a4<\/sub>ei\u03c81<\/sub><\/sup>ei\u03c8<\/sup>\u00a0= a1<\/sub><\/td>\n(10)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Let us define Z1<\/sub>, Z2<\/sub>\u00a0and \u03bb as:<\/p>\n

Z1<\/sub> = a2<\/sub>ei\u03b2<\/sup><\/p>\n

Z2<\/sub> = a4<\/sub>ei\u03c81<\/sub><\/sup><\/p>\n

\u03bb = a3<\/sub>\/a2<\/sub><\/p>\n

Z1<\/sub>\u00a0and Z2<\/sub>\u00a0\u00a0are complex numbers which represent the vectors\u00a0A0<\/sub>Ae<\/sub><\/strong>\u00a0and\u00a0B0<\/sub>Be<\/sub><\/strong>\u00a0and \u03bb\u00a0is the ratio of the coupler link to the crank length. Without loss of generality we can let a1<\/sub> = 1 unit (this correspond to the scaling of the mechanism). Now the loop closure equations (9) and (10) for the dead-center positions can be written in \u201cnormalised form\u201d as:<\/p>\n\n\n\n\n
(1 + \u03bb)Z1<\/sub> \u2212 Z2<\/sub> = 1<\/td>\n(11)<\/td>\n<\/tr>\n
ei\u03d5<\/sup>(1 \u2212 \u03bb)Z1<\/sub> \u2212 Z2<\/sub>ei\u03c8<\/sup> = 1\u00a0<\/span><\/td>\n(12)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Unlike the use of the loop equations for kinematic analysis, now \u03d5\u00a0and \u03c8\u00a0are the given swing angle and corresponding crank rotation for which we have to determine the mechanism proportions. The two complex equations are linear in terms of the unknowns Z1<\/sub>\u00a0and Z2<\/sub>\u00a0the solution of which is:<\/p>\n\n\n\n\n
\\displaystyle {{\\text{Z}}_{1}}=\\frac{{1-{\\text{e}^{{\\text{i\u03c8}}}}}}{{{\\text{e}^{{\\text{i\u03d5}}}}-{\\text{e}^{{\\text{i\u03c8}}}}-\\text{\u03bb}\\left( {{\\text{e}^{{\\text{i\u03d5}}}}+{\\text{e}^{{\\text{i\u03c8}}}}} \\right)}}={{\\text{a}}_{2}}{\\text{e}^{{\\text{i\u03b2}}}} <\/span><\/td>\n(13)<\/td>\n<\/tr>\n
\\displaystyle {{\\text{Z}}_{2}}=\\frac{{1-{\\text{e}^{{\\text{i\u03c8}}}}+\\text{\u03bb}\\left( {1+{\\text{e}^{{\\text{i\u03c8}}}}} \\right)}}{{{\\text{e}^{{\\text{i\u03d5}}}}-{\\text{e}^{{\\text{i\u03c8}}}}-\\text{\u03bb}\\left( {{\\text{e}^{{\\text{i\u03d5}}}}+{\\text{e}^{{\\text{i\u03c8}}}}} \\right)}}={{\\text{a}}_{4}}{\\text{e}^{{{{\\text{i\u03c8}}_{1}}}}} <\/span><\/td>\n(14)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

As \u03bb changes from minus infinity to plus infinity, Z1<\/sub>\u00a0and Z2<\/sub>\u00a0describe a curve which is the loci of the tip of the vectors\u00a0A0<\/sub>Ae<\/sub><\/strong>\u00a0and\u00a0B0<\/sub>Be<\/sub><\/strong>. According to equations (13) and (14), these loci are two circles for any given value of \u03d5 and \u03c8. For \u03d5\u00a0<\/span>= 160\u00b0\u00a0and \u03c8 = 80\u00b0 these two circles are as shown below\u00a0<\/span>(in order to draw the two circles with the same reference frame, A0<\/sub>\u00a0as the origin and the real axis is along A0<\/sub>B0<\/sub>; instead of Z2<\/sub>, the vector 1 + Z2<\/sub> is drawn). There is an infinite set of solutions and these can be obtained by drawing a line from A0<\/sub>\u00a0making\u00a0 any angle \u03b2\u00a0with respect to A0<\/sub>B0<\/sub>. The intersection of this line and the circles are Ae<\/sub>\u00a0and Be<\/sub> respectively. Geometrically, the link lengths can be found from the figure as |A0<\/sub>Ae<\/sub>| = a2<\/sub>, |Ae<\/sub>Be<\/sub>| = a3<\/sub>, |B0<\/sub>Be<\/sub>| = a4<\/sub>, |A0<\/sub>B0<\/sub>| = a1<\/sub>\u00a0= 1.<\/p>\n

\"\"<\/p>\n

Analytically, we can determine a2<\/sub>, a4<\/sub>, and a3<\/sub>\u00a0(= \u03bba2<\/sub>) for any value of \u03bb, \u03c8 and \u03d5 using equations (13) and (14) as:<\/p>\n

\\displaystyle {{\\text{a}}_{2}}^{2}={{\\text{Z}}_{1}}\\overline{{{{\\text{Z}}_{1}}}}<\/span><\/p>\n

\\displaystyle {{\\text{a}}_{4}}^{2}={{\\text{Z}}_{2}}\\overline{{{{\\text{Z}}_{2}}}}<\/span><\/p>\n

a3<\/sub>\u00a0= \u03bba2<\/sub><\/p>\n

a1<\/sub> = 1<\/p>\n