{"id":2302,"date":"2022-03-15T19:58:10","date_gmt":"2022-03-15T19:58:10","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=2302"},"modified":"2022-09-02T19:35:42","modified_gmt":"2022-09-02T19:35:42","slug":"7-5","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch7\/7-5\/","title":{"rendered":"7-5"},"content":{"rendered":"
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7.5 FOUR-BAR AND SLIDER-CRANK COGNATES ROBERTS – CHEBYCHEV THEOREM<\/strong><\/h1>\n

When our concern is only the path traced by the coupler point of a four-bar or a slider-crank mechanism, we can determine other four-bar or slider-crank mechanism proportions that generate identically the same coupler point curve. The four-bar mechanisms that generate the identical coupler point curve are known as the\u00a0cognates<\/span><\/em><\/strong>\u00a0of a four-bar. Let us state a theorem and show how the cognates of a four-bar and slider crank mechanism can be found.<\/p>\n

Roberts-Chebychev Theorem<\/span><\/strong><\/p>\n

There are three different<\/span><\/em>\u00a0four-bar mechanism<\/em>\u00a0proportions that will trace identically the same coupler curve.<\/em><\/span><\/strong><\/p>\n

Consider the four-bar mechanism shown (A0<\/sub>B0<\/sub>=a1<\/sub>, A0<\/sub>A=a2<\/sub>, AB=a3<\/sub>, B0<\/sub>B=a4<\/sub>,\u00a0 AP=b3<\/sub>, BP=c3<\/sub>).\u00a0 The coupler point P traces a curve as shown. For this mechanism the loop equation is:<\/p>\n\n\n\n
a2<\/sub>ei\u03b812<\/sub><\/sup> + a3<\/sub>ei\u03b813<\/sub><\/sup> = a1<\/sub> + a4<\/sub>ei\u03b814<\/sub><\/sup><\/td>\n(1)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The location of coupler point, P, for any crank angle\u00a0q<\/span>12<\/sub>\u00a0can be determined from the equations:<\/p>\n\n\n\n
ZP<\/sub>\u00a0= a2<\/sub>ei\u03b812<\/sub><\/sup> + b3<\/sub>ei(\u03b813 <\/sub>+\u00a0<\/sub>\u03b1)<\/sup><\/td>\n(2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

or<\/p>\n\n\n\n
ZP<\/sub> = a1<\/sub> + a4<\/sub>ei\u03b814<\/sub><\/sup> + c3<\/sub>ei(\u03b813 <\/sub>+\u00a0<\/sub>\u03c0\u00a0<\/sub>\u2212\u00a0<\/sub>\u03b2)<\/sup><\/td>\n(3)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Noting that the angles\u00a0q<\/span>13<\/sub>,\u00a0q<\/span>14<\/sub>, are determined for every crank angle\u00a0q<\/span>12<\/sub>\u00a0from the loop equation.<\/p>\n

\"\"<\/p>\n

Figure\u00a0\u00a07<\/span>.33.<\/strong><\/p>\n

The construction for determining the cognates can be given by the following steps (Fig.7<\/span>.34):<\/p>\n

1. Draw A0<\/sub>A1<\/sub>\/\/AP and B0<\/sub>B1<\/sub>\/\/BP<\/p>\n

2. Draw PA1<\/sub>\/\/A0<\/sub>A and\u00a0 PB1<\/sub>\/\/B0<\/sub>B<\/p>\n

Hence A0<\/sub>A1<\/sub>PA and B0<\/sub>B1<\/sub>PB form parallelograms. Therefore:<\/p>\n

A0<\/sub>A1<\/sub>=b3<\/sub>, A1<\/sub>P=a2<\/sub>, B0<\/sub>B1<\/sub>=c3<\/sub>, B1<\/sub>P=a4<\/sub><\/p>\n

Also:<\/p>\n

< xA0<\/sub>A1<\/sub>=q<\/span>12<\/sub>+a<\/span>, \u00a0< xA1<\/sub>P=q<\/span>12<\/sub>,\u00a0 < xB0<\/sub>B1<\/sub>=q<\/span>13<\/sub>–b<\/span>, < xB1<\/sub>P=q<\/span>14<\/sub><\/p>\n