{"id":2359,"date":"2022-03-17T23:03:07","date_gmt":"2022-03-17T23:03:07","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=2359"},"modified":"2022-09-02T19:02:35","modified_gmt":"2022-09-02T19:02:35","slug":"eniyibagacisi","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch7\/7-1\/eniyibagacisi\/","title":{"rendered":"enIyiBagAcisi"},"content":{"rendered":"<div id=\"pl-gb2359-69d73736e854e\"  class=\"panel-layout\" ><div id=\"pg-gb2359-69d73736e854e-0\"  class=\"panel-grid panel-no-style\" ><div id=\"pgc-gb2359-69d73736e854e-0-0\"  class=\"panel-grid-cell\" ><div id=\"panel-gb2359-69d73736e854e-0-0-0\" class=\"so-panel widget widget_sow-editor panel-first-child panel-last-child widgetopts-SO\" data-index=\"0\" ><div\n\t\t\t\n\t\t\tclass=\"so-widget-sow-editor so-widget-sow-editor-base\"\n\t\t\t\n\t\t>\n<div class=\"siteorigin-widget-tinymce textwidget\">\n\t<h1>Design of Drag-link Mechanisms with optimum transmission angle<\/h1>\n<p>Drag-link mechanisms are used to convert a uniform continuous rotation into a nonuniform continuous rotary motion. They can be used in series with crank-rocker mechanism to obtain a different motion characteristics. In\u00a0<span style=\"font-family: Arial, Helvetica, sans-serif\">figure below\u00a0<\/span>the four-bar mechanism with drag-link proportions is shown in positions where the deviation of the transmission angle from 90<sup>0<\/sup>\u00a0is maximum (the input crank and the fixed link are collinear). It develops that the transmission angle is optimum when the two maximum deviations are equal (<span style=\"font-family: Symbol\">m<\/span><sub>max<\/sub>-90<sup>0<\/sup>=90<sup>0<\/sup>&#8211;<span style=\"font-family: Symbol\">m<\/span><sub>min<\/sub>).<\/p>\n<p>Corresponding to 180<sup>0<\/sup>\u00a0input crank rotation, the output link will be required to rotate by\u00a0<span style=\"font-family: Symbol\">y<\/span>\u00a0 (Note that\u00a0<span style=\"font-family: Symbol\">y\u00a0<\/span>&lt;180<sup>0<\/sup>, since the output link will rotate 360<sup>0<\/sup>&#8211;<span style=\"font-family: Symbol\">y<\/span>\u00a0for the next 180<sup>0<\/sup>\u00a0rotation of the crank). Let us also assume a minimum value of the transmission angle\u00a0<span style=\"font-family: Symbol\">m<\/span><sub>min<\/sub>.\u00a0 The two extreme positions will be as shown in Fig.\u00a0<span style=\"font-family: Arial, Helvetica, sans-serif\">7<\/span>.16. For the best transmission angle characteristics, at these two extreme positions if we equate the transmission angle deviations, we obtain the loop equations as:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">\u2212a<sub>1<\/sub> \u2212 a<sub>2<\/sub> + a<sub>3<\/sub>e<sup>i(\u03c8<sub>1<\/sub> \u2212 \u03c0 + \u03bc<sub>min<\/sub>)<\/sup>\u00a0 = a<sub>4<\/sub>e<sup>i\u03c8<sub>1<\/sub><\/sup><\/td>\n<td style=\"text-align: right\">(1)<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\">\u2212a<sub>1<\/sub> + a<sub>2<\/sub> + a<sub>3<\/sub>e<sup>i(\u03c8<sub>1<\/sub> + \u03c8 \u2212 \u03bc<sub>min<\/sub>)<\/sup> = a<sub>4<\/sub>e<sup>i(\u03c8<sub>1<\/sub> + \u03c8)<\/sup><\/td>\n<td style=\"text-align: right\">(2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>In addition, in order to obtain equal transmission angle deviations at the two extreme positions:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">a<sub>1<\/sub><sup>2<\/sup> + a<sub>2<\/sub><sup>2<\/sup> = a<sub>3<\/sub><sup>2<\/sup> + a<sub>4<\/sub><sup>2<\/sup><\/td>\n<td style=\"text-align: right\">(3)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>using\u00a0 a<sub>1\u00a0<\/sub>=1 (scaling) , \u03bb = a<sub>4<\/sub>\/a<sub>3<\/sub> and letting Z\u00a0= a<sub>3<\/sub>e<sup>i\u03c8<sub>1<\/sub><\/sup>, equations (1) and (2) can be written as:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">\u22121\u00a0\u2212 a<sub>2<\/sub> \u2212 Ze<sup>i\u03bc<sub>min<\/sub><\/sup>\u00a0 = \u03bbZ<\/td>\n<td style=\"text-align: right\">(4)<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\">\u22121\u00a0+ a<sub>2<\/sub> + Ze<sup>i(\u03c8 \u2212 \u03bc<sub>min<\/sub>)<\/sup> = \u03bbZe<sup>i\u03c8<\/sup><\/td>\n<td style=\"text-align: right\">(5)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center\"><div class=\"su-image-carousel  su-image-carousel-has-spacing su-image-carousel-has-lightbox su-image-carousel-has-outline su-image-carousel-adaptive su-image-carousel-slides-style-default su-image-carousel-controls-style-dark su-image-carousel-align-center\" style=\"max-width:550px\" data-flickity-options='{\"groupCells\":true,\"cellSelector\":\".su-image-carousel-item\",\"adaptiveHeight\":true,\"cellAlign\":\"left\",\"prevNextButtons\":true,\"pageDots\":false,\"autoPlay\":false,\"imagesLoaded\":true,\"contain\":false,\"selectedAttraction\":1,\"friction\":1}' id=\"su_image_carousel_69d73736e9c96\"><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/09\/ciftkole_1.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"605\" height=\"440\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/09\/ciftkole_1.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/09\/ciftkole_2.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"605\" height=\"440\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/09\/ciftkole_2.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><\/div><script id=\"su_image_carousel_69d73736e9c96_script\">if(window.SUImageCarousel){setTimeout(function() {window.SUImageCarousel.initGallery(document.getElementById(\"su_image_carousel_69d73736e9c96\"))}, 0);}var su_image_carousel_69d73736e9c96_script=document.getElementById(\"su_image_carousel_69d73736e9c96_script\");if(su_image_carousel_69d73736e9c96_script){su_image_carousel_69d73736e9c96_script.parentNode.removeChild(su_image_carousel_69d73736e9c96_script);}<\/script><\/p>\n<p>Eliminating Z from equations (4) and (5):<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle \\frac{{1+{{\\text{a}}_{2}}}}{{1-{{\\text{a}}_{2}}}}=\\frac{{\\text{\u03bb}+{{\\text{e}}^{{\\text{i}{{\\text{\u03bc}}_{{\\text{min}}}}}}}}}{{{{\\text{e}}^{{\\text{i\u03c8}}}}\\left( {\\text{\u03bb}-{{\\text{e}}^{{\\text{i}{{\\text{\u03bc}}_{{\\text{min}}}}}}}} \\right)}} <\/span><\/td>\n<td style=\"text-align: right\">(6)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Which can be put into the form:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">a<sub>2<\/sub>[\u03bbcos(\u03c8\/2) \u2212 i sin(\u03c8\/2 \u2212 \u03bc<sub>min<\/sub>)] + i \u03bbsin(\u03c8\/2) \u2212 cos(\u03c8\/2 \u2212 \u03bc<sub>min<\/sub>) = 0<\/td>\n<td style=\"text-align: right\">(7)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1123\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img1-10.gif\" alt=\"\" width=\"637\" height=\"562\" \/><\/p>\n<p>Equating the real and imaginary parts to zero and solving for <span style=\"font-family: Symbol\">l<\/span>\u00a0and a<sub>2<\/sub>\u00a0yields:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{\u03bb}}^{2}}=\\frac{{\\sin \\left( {\\text{\u03c8}-2{{\\text{\u03bc}}_{{\\text{min}}}}} \\right)}}{{\\sin \\left( \\text{\u03c8} \\right)}} <\/span><\/td>\n<td style=\"text-align: right\">(8)<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{2}}^{2}=\\frac{{\\tan \\left( {\\text{\u03c8}\/2} \\right)}}{{\\tan \\left( {\\text{\u03c8}\/2-{{\\text{\u03bc}}_{{\\text{min}}}}} \\right)}} <\/span><\/td>\n<td style=\"text-align: right\">(9)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>since 0&lt;\u00a0<span style=\"font-family: Symbol\">y\u00a0<\/span>&lt;180<sup>0<\/sup>,\u00a0 sin<span style=\"font-family: Symbol\">y<\/span>&gt;0 for<span style=\"font-family: Symbol\">\u00a0l<\/span><sup>2<\/sup>=0 ,\u00a0<span style=\"font-family: Symbol\">y<\/span>-2<span style=\"font-family: Symbol\">m<\/span><sub>min<\/sub>\u00a0&gt;0 or\u00a0<span style=\"font-family: Symbol\">m<\/span><sub>min<\/sub>&lt;\u00a0<span style=\"font-family: Symbol\">y<\/span>\u00a0\/2. In other words, the minimum transmission angle can not be better than\u00a0<span style=\"font-family: Symbol\">y<\/span>\u00a0\/2 for drag link proportions. Hence, as we decrease\u00a0<span style=\"font-family: Symbol\">y<\/span>, the maximum deviation of the transmission angle from 90<sup>0<\/sup>\u00a0\u00a0will tend to increase (when\u00a0<span style=\"font-family: Symbol\">l<\/span>=1, the mechanism will form a folded position i.e. the sum of the lengths of the longest and shortest link lengths will be equal to the sum of the two intermediate link lengths)<\/p>\n<p>Since <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\text{a}_{3}}^{2}=\\text{Z}\\bar{\\text{Z}} <\/span>, we have:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{3}}^{2}=\\frac{{{{{\\left( {{{\\text{a}}_{2}}+1} \\right)}}^{2}}}}{{\\left( {\\text{\u03bb}+{{\\text{e}}^{{\\text{i}{{\\text{\u03bc}}_{{\\text{min}}}}}}}} \\right)\\left( {\\text{\u03bb}+{{\\text{e}}^{{\\text{-i}{{\\text{\u03bc}}_{{\\text{min}}}}}}}} \\right)}} <\/span><\/td>\n<td style=\"text-align: right\">(10)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>or<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{3}}^{2}=\\frac{{\\sin \\left( \\text{\u03c8} \\right)}}{{\\sin \\left( {2{{\\text{\u03bc}}_{{\\text{min}}}}} \\right)}}\\left[ {\\frac{{\\tan \\left( {\\text{\u03c8}\/2} \\right)}}{{\\tan \\left( {\\text{\u03c8}\/2-{{\\text{\u03bc}}_{{\\text{min}}}}} \\right)}}-1} \\right]=\\frac{{\\sin \\left( \\text{\u03c8} \\right)}}{{\\sin \\left( {2{{\\text{\u03bc}}_{{\\text{min}}}}} \\right)}}\\left( {{{\\text{a}}_{2}}^{2}-1} \\right) <\/span><\/td>\n<td style=\"text-align: right\">(11)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>and a<sub>4\u00a0<\/sub>=<span style=\"font-family: Symbol\">l<\/span>a<sub>3<\/sub>. Note that the resulting mechanism must be checked for movability.<\/p>\n<p><strong><em>Example 4.5.<\/em><\/strong><\/p>\n<p>Determine the four-bar mechanism with drag-link proportions for which for half a revolution of the input link the output link rotates by 150<sup>0<\/sup>. The minimum transmission angle must be greater than 45<sup>0<\/sup>. Let the distance between the cranks be equal to 100 mm<\/p>\n<p>Taking y=1500 and mmin=45<sup>0<\/sup>\u00a0and a<sub>1<\/sub>=100 mm, from equations (8), (9) and (11):<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{\u03bb}}^{2}}=\\frac{{\\sin \\left( {\\text{150}{}^\\circ -90{}^\\circ } \\right)}}{{\\sin \\left( {150{}^\\circ } \\right)}}=1.73205\\text{\u00a0 \u00a0}\\Rightarrow \\text{\u00a0 \u00a0\u03bb}=1.31607 <\/span><\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{2}}^{2}=\\frac{{\\tan \\left( {75{}^\\circ } \\right)}}{{\\tan \\left( {75{}^\\circ -45{}^\\circ } \\right)}}=6.464101\\text{\u00a0 \u00a0}\\Rightarrow \\text{\u00a0 \u00a0}{{\\text{a}}_{2}}=2.54246 <\/span><\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{3}}^{2}=\\frac{{\\sin \\left( {150{}^\\circ } \\right)}}{{\\sin \\left( {90{}^\\circ } \\right)}}\\left( {\\frac{{\\tan \\left( {75{}^\\circ } \\right)}}{{\\tan \\left( {75{}^\\circ -45{}^\\circ } \\right)}}-1} \\right)=2.73205\\text{\u00a0 \u00a0}\\Rightarrow \\text{\u00a0 \u00a0}{{\\text{a}}_{3}}=1.65289 <\/span><\/p>\n<p style=\"text-align: center\">a<sub>4<\/sub> = \u03bba<sub>3<\/sub> = 2.17533<\/p>\n<p>a<sub>4\u00a0<\/sub>=<span style=\"font-family: Symbol\">l<\/span>a<sub>3\u00a0<\/sub>=2.17533. When a<sub>1\u00a0<\/sub>=100 mm a<sub>2\u00a0<\/sub>=254 mm a<sub>3\u00a0<\/sub>=165 mm and a<sub>4<\/sub>=218 mm. According to Grashof&#8217;s rule, the mechanism is of drag-link proportions.<\/p>\n<p><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/ciftkoltasarim.xls\">The procedure described above can be implemented on an Excel sheet. Click here to download this file.<\/a><\/p>\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n<p><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch7\/7-1\/\" data-type=\"page\"><img loading=\"lazy\" decoding=\"async\" width=\"38\" height=\"38\" class=\"wp-image-16\" style=\"width: 38px\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/04\/back_button.gif\" alt=\"\" \/><\/a><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch7\/\" data-type=\"page\"><img loading=\"lazy\" decoding=\"async\" width=\"38\" height=\"38\" class=\"wp-image-17\" style=\"width: 38px\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/04\/contents_button.gif\" alt=\"\" \/><\/a><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/\" data-type=\"page\" data-id=\"47\"><img loading=\"lazy\" decoding=\"async\" width=\"38\" height=\"38\" class=\"wp-image-18\" style=\"width: 38px\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/04\/home_button.gif\" alt=\"\" \/><\/a><img loading=\"lazy\" decoding=\"async\" width=\"119\" height=\"40\" class=\"wp-image-15\" style=\"width: 119px\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/04\/ceres.gif\" alt=\"\" \/><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Design of Drag-link Mechanisms with optimum transmission angle Drag-link mechanisms are used to convert a uniform continuous rotation into a nonuniform continuous rotary motion. They can be used in series with crank-rocker mechanism to obtain a different motion characteristics. In\u00a0figure &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch7\/7-1\/eniyibagacisi\/\"> <span class=\"screen-reader-text\">enIyiBagAcisi<\/span> Devam\u0131n\u0131 Oku &raquo;<\/a><\/p>\n","protected":false},"author":7747,"featured_media":0,"parent":2298,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"full-width-page.php","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-2359","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2359","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/users\/7747"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/comments?post=2359"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2359\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2298"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/media?parent=2359"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}