{"id":1038,"date":"2021-09-09T14:43:59","date_gmt":"2021-09-09T14:43:59","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=1038"},"modified":"2022-08-02T08:43:21","modified_gmt":"2022-08-02T08:43:21","slug":"altbagacisi","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch7\/7-1\/altbagacisi\/","title":{"rendered":"altBagAcisi"},"content":{"rendered":"<div id=\"pl-gb1038-69d746bea5ade\"  class=\"panel-layout\" ><div id=\"pg-gb1038-69d746bea5ade-0\"  class=\"panel-grid panel-no-style\" ><div id=\"pgc-gb1038-69d746bea5ade-0-0\"  class=\"panel-grid-cell\" ><div id=\"panel-gb1038-69d746bea5ade-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>Alt Ba\u011flama A\u00e7\u0131s\u0131 Problemi<\/h1>\n<p>&#8220;Verilen bir sal\u0131n\u0131m a\u00e7\u0131s\u0131n\u0131 (<span style=\"font-family: Symbol\">y<\/span>) ve buna kar\u015f\u0131 gelen bir krank d\u00f6nme a\u00e7\u0131s\u0131n\u0131 (<span style=\"font-family: Symbol\">f<\/span>) sa\u011flayan, ve ayr\u0131ca ba\u011flama a\u00e7\u0131s\u0131n\u0131n 90<sup>0<\/sup> den sapmas\u0131 en az olan kol-sarka\u00e7 oranlar\u0131nda d\u00f6rt \u00e7ubuk mekanizmas\u0131n\u0131n boyutlar\u0131n\u0131 belirleyin.&#8221;<\/p>\n<p>Problem iki k\u0131s\u0131mdan olu\u015fmaktad\u0131r:<\/p>\n<ol>\n<li>Verilen bir sal\u0131n\u0131m a\u00e7\u0131s\u0131n\u0131 (<span style=\"font-family: Symbol\">y<\/span>) ve buna kar\u015f\u0131 gelen bir krank d\u00f6nme a\u00e7\u0131s\u0131n\u0131 (<span style=\"font-family: Symbol\">f<\/span>) sa\u011flayan kol-sarka\u00e7 mekanizmalar\u0131n\u0131n bulunmas\u0131<\/li>\n<li>Elde edilen kol-sarka\u00e7 mekanizmalar\u0131 aras\u0131nda ba\u011flama a\u00e7\u0131s\u0131 en iyi olan kol-sarka\u00e7 mekanizmas\u0131n\u0131n belirlenmesi.<\/li>\n<\/ol>\n<p>1-Verilen bir sal\u0131n\u0131m a\u00e7\u0131s\u0131n\u0131 (<span style=\"font-family: Symbol\">y<\/span>) ve buna kar\u015f\u0131 gelen bir krank d\u00f6nme a\u00e7\u0131s\u0131n\u0131 (<span style=\"font-family: Symbol\">f<\/span>) sa\u011flayan kol-sarka\u00e7 mekanizmalar\u0131n\u0131n bulunmas\u0131:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1045 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img1-9.gif\" alt=\"\" width=\"500\" height=\"285\" \/><\/p>\n<p>Kol-sarka\u00e7 mekanizmas\u0131n\u0131 iki \u00f6l\u00fc konumda \u00e7izelim. Her iki konumda krank ve biyel uzuvlar\u0131 ayn\u0131 do\u011frultudad\u0131r. A\u00e7\u0131k konumda biyel a\u00e7\u0131s\u0131 krank a\u00e7\u0131s\u0131na e\u015fittir, kapal\u0131 konumda ise krank a\u00e7\u0131s\u0131ndan 180<sup>0<\/sup>\u00a0farkl\u0131d\u0131r. Bu konumlar i\u00e7in devre kapal\u0131l\u0131k denklemlerini yazal\u0131m.<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">a<sub>2<\/sub>e<sup>i\u03b2<\/sup> + a<sub>3<\/sub>e<sup>i\u03b2<\/sup> = a<sub>1<\/sub>\u00a0+ a<sub>4<\/sub>e<sup>i\u03c8<sub>1<\/sub><\/sup><\/td>\n<td style=\"text-align: right\">(7)<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\">a<sub>2<\/sub>e<sup>i(\u03b2 + \u03d5)<\/sup> + a<sub>3<\/sub>e<sup>i(\u03b2 + \u03d5 \u2212 \u03c0)<\/sup>\u00a0= a<sub>1<\/sub>\u00a0+ a<sub>4<\/sub>e<sup>i(\u03c8<sub>1<\/sub> + \u03c8)<\/sup><\/td>\n<td style=\"text-align: right\">(8)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>bu denklemler:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">(a<sub>2<\/sub> + a<sub>3<\/sub>)e<sup>i\u03b2<\/sup> \u2212 a<sub>4<\/sub>e<sup>i\u03c8<sub>1<\/sub><\/sup>\u00a0= a<sub>1<\/sub><\/td>\n<td style=\"text-align: right\">(9)<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\">(a<sub>2<\/sub> \u2212 a<sub>3<\/sub>)e<sup>i\u03b2<\/sup>e<sup>i\u03d5<\/sup> \u2212 a<sub>4<\/sub>e<sup>i\u03c8<sub>1<\/sub><\/sup>e<sup>i\u03c8<\/sup>\u00a0= a<sub>1<\/sub><\/td>\n<td style=\"text-align: right\">(10)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\u015feklinde yazabiliriz. Z<sub>1<\/sub>, Z<sub>2<\/sub> ve \u03bb yeni parametreler olup:<\/p>\n<p style=\"padding-left: 40px;text-align: center\">Z<sub>1<\/sub> = a<sub>2<\/sub>e<sup>i\u03b2<\/sup><\/p>\n<p style=\"padding-left: 40px;text-align: center\">Z<sub>2<\/sub> = a<sub>4<\/sub>e<sup>i\u03c8<sub>1<\/sub><\/sup><\/p>\n<p style=\"padding-left: 40px;text-align: center\">\u03bb = a<sub>3<\/sub>\/a<sub>2<\/sub><\/p>\n<p>olarak tan\u0131mlayal\u0131m Z<sub>1<\/sub>\u00a0ve Z<sub>2<\/sub>\u00a0<b>A<sub>0<\/sub>A<sub>e<\/sub><\/b>\u00a0ve\u00a0<b>B<sub>0<\/sub>B<sub>e<\/sub><\/b> vekt\u00f6rlerini g\u00f6steren kompleks say\u0131lard\u0131r <strong>\u03bb<\/strong> ise biyel uzuv boyutunun krank uzuv boyutuna oran\u0131d\u0131r. \u00d6nceden s\u00f6ylenildi\u011fi gibi uzuv boyutlar\u0131 belirli bir \u00f6l\u00e7ekle b\u00fcy\u00fclt\u00fcl\u00fcp k\u00fc\u00e7\u00fclt\u00fcld\u00fc\u011f\u00fc halde d\u00f6nme a\u00e7\u0131lar\u0131 de\u011fi\u015fmiyece\u011finden, sabit uzuv boyutunu bir birim alal\u0131m (a<sub>1<\/sub>=1 ). \u015eimdi \u00f6l\u00fc konumlar i\u00e7in devre kapal\u0131l\u0131k denklemleri:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">(1 + \u03bb)Z<sub>1<\/sub> \u2212 Z<sub>2<\/sub> = 1<\/td>\n<td style=\"text-align: right\">(11)<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\">e<sup>i\u03d5<\/sup>(1 \u2212 \u03bb)Z<sub>1<\/sub> \u2212 Z<sub>2<\/sub>e<sup>i\u03c8<\/sup> = 1<span style=\"font-size: 13.3333px\">\u00a0<\/span><\/td>\n<td style=\"text-align: right\">(12)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>olacakt\u0131r. Kinematik analiz problemlerinin aksine, \u015fimdi bu denklemlerde <span style=\"font-family: Symbol\">f<\/span>\u00a0ve <span style=\"font-family: Symbol\">y<\/span>\u00a0a\u00e7\u0131lar\u0131 bilinen de\u011ferler olup bu de\u011ferler i\u00e7in mekanizma boyutlar\u0131n\u0131 bulmam\u0131z gerekmektedir. Bu iki kompleks say\u0131larla yaz\u0131lm\u0131\u015f olan denklem Z<sub>1<\/sub>\u00a0ve Z<sub>2<\/sub>\u00a0parametreleri i\u00e7in lineer bir denklem tak\u0131m\u0131n\u0131 olu\u015fturur. \u00c7\u00f6z\u00fcm\u00fc bilinen Kramer kural\u0131 ile yap\u0131ld\u0131\u011f\u0131nda:<\/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{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<td style=\"text-align: right\">(13)<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\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<td style=\"text-align: right\">(14)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><span style=\"font-family: Symbol\">fy <\/span>elde edilir. <span style=\"font-family: Symbol\">l<\/span>\u00a0de\u011feri &#8211;<img loading=\"lazy\" decoding=\"async\" src=\"altBagAcisi\/infinity.gif\" width=\"22\" height=\"19\" align=\"absmiddle\" title=\"\" alt=\"\" \/>\u00a0dan +<img loading=\"lazy\" decoding=\"async\" src=\"altBagAcisi\/infinity.gif\" width=\"22\" height=\"19\" align=\"absmiddle\" title=\"\" alt=\"\" \/>a kadar de\u011fi\u015fik de\u011ferler ald\u0131\u011f\u0131nda, Z<sub>1<\/sub>\u00a0ve Z<sub>2<\/sub>\u00a0vekt\u00f6rlerinin u\u00e7 noktalar\u0131 bir e\u011fri \u00e7izecektir. Ve bu u\u00e7 noktalar\u00a0<b>A<sub>e<\/sub><\/b>\u00a0ve\u00a0<b>B<sub>e<\/sub><\/b> noktalar\u0131n\u0131n geometrik yerleridir. Bu geometrik yerler verilen <span style=\"font-family: Symbol\">f <\/span>ve <span style=\"font-family: Symbol\">y <\/span>de\u011ferlerine g\u00f6re bir dairedir. \u00d6rne\u011fin <span style=\"font-family: Symbol\">f<\/span>=160<sup>0<\/sup> ve <span style=\"font-family: Symbol\">y<\/span>=80<sup>0<\/sup>\u00a0i\u00e7in bu iki daire a\u015fa\u011f\u0131da g\u00f6sterildi\u011fi gibi \u00e7izilmi\u015ftir (bu iki daireyi ayn\u0131 sabit eksen referans\u0131nda \u00e7izmek i\u00e7in A<sub>0<\/sub>\u00a0merkezli ve x ekseni sabit uzuv ile \u00e7ak\u0131\u015fan bir referans eksen al\u0131nm\u0131\u015ft\u0131r. Bu durumda Z<sub>2<\/sub>\u00a0vekt\u00f6r\u00fc yerine 1+Z<sub>2<\/sub>\u00a0vekt\u00f6r\u00fc bu eksen tak\u0131m\u0131nda bize\u00a0<b>B<sub>e<\/sub><\/b> nin geometrik yerini verece\u011finden bu vekt\u00f6r \u00e7izilmi\u015ftir). G\u00f6r\u00fcld\u00fc\u011f\u00fc gibi her bir <span style=\"font-family: Symbol\">l <\/span>oran\u0131 farkl\u0131 bir \u00e7\u00f6z\u00fcm verecektir ve sonsuz say\u0131da \u00e7\u00f6z\u00fcm vard\u0131r. Ancak kol sarka\u00e7 oran\u0131 i\u00e7in krank\u0131n en k\u00fc\u00e7\u00fck uzuv olmas\u0131 gerekti\u011finden <span style=\"font-family: Symbol\">l<\/span>&gt;1 gerekli bir \u015fartt\u0131r ve dairelerin belirli bir k\u0131sm\u0131 i\u00e7in bu ge\u00e7erlidir. Elde edilen konum \u00f6l\u00fc konum oldu\u011fundan,\u00a0<b>A<sub>0<\/sub><\/b>,\u00a0<b>A<sub>e<\/sub><\/b>\u00a0ve\u00a0<b>B<sub>e<\/sub><\/b> bir do\u011fru \u00fczerinde olmas\u0131 gerekir. \u00d6yle ise, <span style=\"font-family: Symbol\">l <\/span>parametresi yerine A<sub>0<\/sub>\u00a0dan A<sub>0<\/sub>B<sub>0<\/sub> do\u011frusuna <span style=\"font-family: Symbol\">b<\/span>\u00a0a\u00e7\u0131s\u0131 yapan bir do\u011fru \u00e7izelim. Bu do\u011fru daireleri\u00a0<b>A<sub>e<\/sub><\/b>\u00a0ve\u00a0<b>B<sub>e<\/sub><\/b>\u00a0noktalar\u0131nda kesecektir ve mekanizma \u00f6l\u00fc konumda elde edilecektir. Bu durumda\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"altBagAcisi\/beta.gif\" width=\"15\" height=\"27\" align=\"absmiddle\" title=\"\" alt=\"\" \/>\u00a0ba\u011f\u0131ms\u0131z parametre olacakt\u0131r ve geometrik olarak uzuv boyutlar\u0131 \u015fekilden: A<sub>0<\/sub>A<sub>e<\/sub>=a<sub>2<\/sub>, A<sub>e<\/sub>B<sub>e<\/sub>=a<sub>3<\/sub>, B<sub>0<\/sub>B<sub>e<\/sub>=a<sub>4<\/sub>, A<sub>0<\/sub>B<sub>0<\/sub>=a<sub>1<\/sub>=1 bulunacakt\u0131r. <span style=\"font-family: Symbol\">b <\/span>a\u00e7\u0131s\u0131 krank\u0131n \u00f6l\u00fc konum a\u00e7\u0131s\u0131d\u0131r.\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1048 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img2-9.gif\" alt=\"\" width=\"444\" height=\"352\" \/><\/p>\n<p>Yukar\u0131daki e\u011friyi geometrik olarak \u00e7izme y\u00f6ntemini \u00f6\u011frenmek i\u00e7in alttaki animasyonu inceleyiniz.<\/p>\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_69d746bea7a99\"><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/ALTCONSTRUCTION1.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"550\" height=\"400\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/ALTCONSTRUCTION1.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\/2021\/09\/ALTCONSTRUCTION2.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"550\" height=\"400\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/ALTCONSTRUCTION2.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\/2021\/09\/ALTCONSTRUCTION3.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"550\" height=\"400\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/ALTCONSTRUCTION3.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\/2021\/09\/ALTCONSTRUCTION4.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"550\" height=\"400\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/ALTCONSTRUCTION4.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\/2021\/09\/ALTCONSTRUCTION5.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"550\" height=\"400\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/ALTCONSTRUCTION5.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\/2021\/09\/ALTCONSTRUCTION6.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"550\" height=\"400\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/ALTCONSTRUCTION6.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\/2021\/09\/ALTCONSTRUCTION7.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"550\" height=\"400\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/ALTCONSTRUCTION7.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\/2021\/09\/ALTCONSTRUCTION8.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"324\" height=\"300\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/ALTCONSTRUCTION8.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><\/div><script id=\"su_image_carousel_69d746bea7a99_script\">if(window.SUImageCarousel){setTimeout(function() {window.SUImageCarousel.initGallery(document.getElementById(\"su_image_carousel_69d746bea7a99\"))}, 0);}var su_image_carousel_69d746bea7a99_script=document.getElementById(\"su_image_carousel_69d746bea7a99_script\");if(su_image_carousel_69d746bea7a99_script){su_image_carousel_69d746bea7a99_script.parentNode.removeChild(su_image_carousel_69d746bea7a99_script);}<\/script><\/p>\n<p>Uzuv boyutlar\u0131n\u0131 <span style=\"font-family: Symbol\">b<\/span> parametresine g\u00f6re analitik olarak elde etmek istedi\u011fimizde:<\/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}}_{2}=-\\sin\\left( \\frac{\\text{\u03c8}}{2}\\right)\\frac{{\\cos\\left( {\\frac{\\text{\u03d5}}{2}+\\text{\u03b2}} \\right)}}{{\\sin\\left( {\\frac{{\\text{\u03d5}-\\text{\u03c8}}}{2}} \\right)}}} <\/span><\/td>\n<td style=\"text-align: right\">(15)<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{3}}=+\\sin\\left( \\frac{\\text{\u03c8}}{2}\\right)\\frac{{\\sin\\left( {\\frac{\\text{\u03d5}}{2}+\\text{\u03b2}} \\right)}}{{\\cos\\left( {\\frac{{\\text{\u03d5}-\\text{\u03c8}}}{2}} \\right)}} <\/span><\/td>\n<td style=\"text-align: right\">(16)<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\">a<sub>4<\/sub><sup>2<\/sup> = (a<sub>2<\/sub> + a<sub>3<\/sub>)<sup>2<\/sup> + 1 \u2212 2(a<sub>2<\/sub> + a<sub>3<\/sub>)cos\u03b2<\/td>\n<td style=\"text-align: right\">(17)<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\">a<sub>1<\/sub> = 1<\/td>\n<td style=\"text-align: right\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>denklemleri kullan\u0131labilir. <span style=\"font-family: Symbol\">l <\/span>parametresine g\u00f6re ise uzuv boyutlar\u0131 denklem 13 ve 14 kullan\u0131larak:<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{2}}^{2}={{\\text{Z}}_{1}}\\overline{{{{\\text{Z}}_{1}}}}<\/span><\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{4}}^{2}={{\\text{Z}}_{2}}\\overline{{{{\\text{Z}}_{2}}}}<\/span><\/p>\n<p style=\"text-align: center\">a<sub>3<\/sub>\u00a0= \u03bba<sub>2<\/sub><\/p>\n<p style=\"text-align: center\">a<sub>1<\/sub> = 1<\/p>\n<p>elde edilir. Verilen bir sal\u0131n\u0131m a\u00e7\u0131s\u0131 ve kar\u015f\u0131 gelen bir krank d\u00f6nme a\u00e7\u0131s\u0131na g\u00f6re <span style=\"font-family: Symbol\">b<\/span>\u00a0veya <span style=\"font-family: Symbol\">l <\/span>parametresi kullan\u0131larak istenilen a\u00e7\u0131lar\u0131 sa\u011flayan sonsuz say\u0131da mekanizma elde edilebilmektedir. Kol-sarka\u00e7 oranlar\u0131n\u0131n sa\u011flanabilmesi i\u00e7in ayr\u0131ca sarka\u00e7 sal\u0131n\u0131m a\u00e7\u0131s\u0131:<\/p>\n<p style=\"padding-left: 40px;text-align: center\">0\u00b0 &lt; \u03c8 &lt; 180\u00b0<\/p>\n<p style=\"padding-left: 40px;text-align: center\">90\u00b0 + \u03c8\/2 &lt; \u03d5 &lt; 270\u00b0 + \u03c8\/2<\/p>\n<p>denklemlerini sa\u011flamal\u0131d\u0131r. <span style=\"font-family: Symbol\">f<\/span> ve <span style=\"font-family: Symbol\">y<\/span>\u00a0parametrelerinden elde edilebilecek t,u ve v parametrelerini:<\/p>\n<p style=\"padding-left: 40px;text-align: center\">t = tan(\u03d5\/2)\u00a0 ,\u00a0 u = tan[(\u03d5 \u2212 \u03c8)\/2]\u00a0 ,\u00a0 v = tan(\u03c8\/2)<\/p>\n<p>tan\u0131mlayal\u0131m. Bu yeni parametreler kullan\u0131larak uzuv boyutlar\u0131n\u0131 \u015fu \u015fekilde g\u00f6sterebiliriz:<\/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}}_{1}}^{2}=\\frac{{{{\\text{u}}^{2}}+{{\\text{\u03bb}}^{2}}}}{{1+{{\\text{u}}^{2}}}} <\/span><\/td>\n<td style=\"text-align: right\">(18)<\/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{{{{\\text{v}}^{2}}}}{{1+{{\\text{v}}^{2}}}} <\/span><\/td>\n<td style=\"text-align: right\">(19)<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{3}}^{2}={\\text{\u03bb}}^{2}{{\\text{a}}_{2}}^{2}=\\frac{{\\text{\u03bb}}^{2}{\\text{v}}^{2}}{{1+{{\\text{v}}^{2}}}} <\/span><\/td>\n<td style=\"text-align: right\">(20)<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{4}}^{2}=\\frac{{{{\\text{t}}^{2}}+{{\\text{\u03bb}}^{2}}}}{{1+{{\\text{t}}^{2}}}} <\/span><\/td>\n<td style=\"text-align: right\">(21)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong><em>\u00d6rnek<\/em><\/strong><\/p>\n<p>Sabit uzuv boyu 120 mm olan, sal\u0131n\u0131m a\u00e7\u0131s\u0131 <span style=\"font-family: Symbol\">y<\/span>=40<sup>0<\/sup> ve buna kar\u015f\u0131 gelen krank d\u00f6nme a\u00e7\u0131s\u0131 <span style=\"font-family: Symbol\">f<\/span>=160<sup>0<\/sup>\u00a0olan bir kol-sarka\u00e7 mekanizmas\u0131n\u0131 ve bulunan mekanizman\u0131n kritik ba\u011flama a\u00e7\u0131s\u0131n\u0131 bulun.<\/p>\n<p><b>a)<\/b> Krank \u00f6l\u00fc konum a\u00e7\u0131s\u0131n\u0131 <span style=\"font-family: Symbol\">b <\/span>= 60<sup>0<\/sup>.olarak se\u00e7elim. (15)-(17)denklemlerini kullanarak:<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{2}}=-\\sin \\left( {20{}^\\circ } \\right)\\frac{{\\cos \\left( {80{}^\\circ +60{}^\\circ } \\right)}}{{\\sin \\left( {80{}^\\circ -20{}^\\circ } \\right)}}=0.30254 <\/span><\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{3}}=\\sin \\left( {20{}^\\circ } \\right)\\frac{{\\sin \\left( {80{}^\\circ +60{}^\\circ } \\right)}}{{\\cos \\left( {80{}^\\circ -20{}^\\circ } \\right)}}=0.43969 <\/span><\/p>\n<p>ve<\/p>\n<p style=\"padding-left: 40px;text-align: center\">a<sub>4<\/sub><sup>2<\/sup> = (0.30254 + 0.43969)<sup>2<\/sup> + 1 \u2212 2(0.30254 + 0.43969)cos60\u00b0 = 0.80867\u00a0 \u00a0 \u21d2\u00a0 \u00a0 \u00a0a<sub>4<\/sub> = 0.89926<\/p>\n<p>a<sub>1<\/sub>=120 mm oldu\u011funda , a<sub>2<\/sub>=36.30 mm, a<sub>3<\/sub>=52.76 mm ve a<sub>4<\/sub>=107.91 mm olacakt\u0131r.<\/p>\n<p>Ba\u011flama a\u00e7\u0131s\u0131n\u0131n 90<sup>0<\/sup>\u00a0den en fazla sapmas\u0131 ise denklem (4) kullan\u0131larak:<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\cos \\text{\u03bc}_{\\begin{smallmatrix} \\text{min} \\\\ \\text{max} \\end{smallmatrix}}=\\frac{{107.91}^{2}+{52.76}^{2}-{120}^{2}-{36.30}^{2}}{2 \\cdot 107.91 \\cdot 52.76} \\pm \\frac{120 \\cdot 36.30}{107.91 \\cdot 52.76}=-0.113247 \\pm 0.765106}<\/span><\/p>\n<p>Buradan : <span style=\"font-family: Symbol\">m<\/span><sub>max<\/sub>=151.44<sup>0<\/sup> (<span style=\"font-family: Symbol\">D<\/span><sub>1<\/sub>=61.44<sup>0<\/sup>) ve <span style=\"font-family: Symbol\">m<\/span><sub>min<\/sub>=49.32<sup>0<\/sup> (<span style=\"font-family: Symbol\">D<\/span><sub>2<\/sub>=40.68<sup>0<\/sup>) elde edilecektir. <span style=\"font-family: Symbol\">m<\/span><sub>max<\/sub>\u00a0\u0131n 90<sup>0<\/sup>\u00a0den sapmas\u0131 61.44<sup>0<\/sup> oldu\u011fundan <span style=\"font-family: Symbol\">m<\/span><sub>max<\/sub>\u00a0g\u00f6z \u00f6n\u00fcne al\u0131nmas\u0131 gereken kritik ba\u011flama a\u00e7\u0131s\u0131d\u0131r.<\/p>\n<p><strong>b)<\/strong> E\u011fer biyel uzunlu\u011funun krank uzunlu\u011funa oran\u0131 <span style=\"font-family: Symbol\">l<\/span>= 1.4 olarak se\u00e7ilir ise, (18)-(21) denklemlerini kullanarak:<\/p>\n<p style=\"padding-left: 40px;text-align: center\">t = tan(80\u00b0) = 5.671282\u00a0 ,\u00a0 u = tan(60\u00b0) = 1.732051\u00a0 ,\u00a0 v = tan(20\u00b0) = 0.36397<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{1}}^{2}=\\frac{{{{{1.732051}}^{2}}+{{{1.4}}^{2}}}}{{1+{{{1.732051}}^{2}}}}=1.24\\text{ }\\Rightarrow \\text{ }{{\\text{a}}_{1}}=1.113553 <\/span><\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{2}}^{2}=\\frac{{{{{0.363970}}^{2}}}}{{1+{{{0.363970}}^{2}}}}=0.116978\\text{ }\\Rightarrow \\text{ }{{\\text{a}}_{2}}=0.342020 <\/span><\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{3}}=1.4\\cdot 0.342020=0.478828 <\/span><\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{4}}^{2}=\\frac{{{{{5.671282}}^{2}}+{{{1.4}}^{2}}}}{{1+{{{5.671282}}^{2}}}}=1.028948\\text{ }\\Rightarrow \\text{ }{{\\text{a}}_{4}}=1.0114371 <\/span><\/p>\n<p>elde edilir. a<sub>1<\/sub>=120 mm oldu\u011funda: a<sub>2<\/sub>=(0.342020\/1.113553)\u00d7120=36.86 mm, a<sub>3<\/sub>=51.60 mm, a<sub>4<\/sub>=109.31 mm olacakt\u0131r. Bu mekanizma i\u00e7in kritik ba\u011flama a\u00e7\u0131s\u0131 denklem (4) den:<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\cos \\text{\u03bc}_{\\begin{smallmatrix} \\text{min} \\\\ \\text{max} \\end{smallmatrix}}=\\frac{{{{{109.31}}^{2}}+{{{51.60}}^{2}}-{{{120}}^{2}}-{{{36.86}}^{2}}}}{{2\\cdot 109.31\\cdot 51.60}}\\pm \\frac{{120\\cdot 36.86}}{{109.31\\cdot 51.60}}=-0.101715 \\pm 0.784200}<\/span><\/p>\n<p>Buradan : <span style=\"font-family: Symbol\">m<\/span><sub>max<\/sub>=152.36<sup>0<\/sup> (<span style=\"font-family: Symbol\">D<\/span><sub>1<\/sub>=62.36<sup>0<\/sup>) ve <span style=\"font-family: Symbol\">m<\/span><sub>min<\/sub>=49.96<sup>0<\/sup> (<span style=\"font-family: Symbol\">D<\/span><sub>2<\/sub>=43.04<sup>0<\/sup>). <span style=\"font-family: Symbol\">m<\/span><sub>max<\/sub>\u00a0\u0131n dik a\u00e7\u0131dan sapmas\u0131 62.36<sup>0<\/sup> oldu\u011fundan <span style=\"font-family: Symbol\">m<\/span><sub>max<\/sub>\u00a0kritik ba\u011flama a\u00e7\u0131s\u0131d\u0131r.<\/p>\n<p>G\u00f6r\u00fcld\u00fc\u011f\u00fc gibi, istenilen sarka\u00e7 sal\u0131n\u0131m a\u00e7\u0131s\u0131 ve kar\u015f\u0131 gelen kol d\u00f6nme a\u00e7\u0131s\u0131n\u0131 sa\u011flayan sonsuz say\u0131da kol-sarka\u00e7 mekanizmas\u0131 elde edilebilecektir. Elde edilen her mekanizman\u0131n ba\u011flama a\u00e7\u0131s\u0131 farkl\u0131 olaca\u011f\u0131ndan 90<sup>0<\/sup>\u00a0den sapmalar\u0131da farkl\u0131 olacakt\u0131r. Basit bir \u00e7al\u0131\u015fma olarak l veya b parametrelerinden birisi kullan\u0131larak bu parametre serbest oldu\u011fundan de\u011fi\u015fik de\u011ferler verelim. Her bir de\u011fer i\u00e7in ba\u011flama a\u00e7\u0131s\u0131n\u0131n 90<sup>0<\/sup>\u00a0den maksimum sapmas\u0131 farkl\u0131 olacakt\u0131r. Bu sapmalar\u0131n de\u011fi\u015fimini kulland\u0131\u011f\u0131m\u0131z parametreye g\u00f6re \u00e7izelim. \u015eekilde g\u00f6r\u00fcld\u00fc\u011f\u00fc gibi bir e\u011fri elde edilecektir. Dik a\u00e7\u0131dan sapman\u0131n en az oldu\u011fu mekanizma ba\u011flama a\u00e7\u0131s\u0131 bak\u0131m\u0131ndan en iyi mekanizmad\u0131r ve onu se\u00e7ti\u011fimizde uygulamada en az sorun ile kar\u015f\u0131la\u015f\u0131laca\u011f\u0131 b\u00fcy\u00fck bir ihtimaldir.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1061 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img3-7.gif\" alt=\"\" width=\"316\" height=\"294\" \/><\/p>\n<p>Bu en iyi ba\u011flama a\u00e7\u0131s\u0131na sahip mekanizmay\u0131 bulmak i\u00e7in her durumda yukar\u0131da a\u00e7\u0131klanan y\u00f6ntemi izlemektense, uzuv boyutlar\u0131 tek bir parametre ile (<span style=\"font-family: Symbol\">l<\/span>) ifade edilebildi\u011fine g\u00f6re ba\u011flama a\u00e7\u0131s\u0131 da sadece bu parametrenin fonksiyonu olacakt\u0131r. (4) numaral\u0131 denklemde:<\/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 {\\cos \\text{\u03bc}_{\\begin{smallmatrix} \\text{min} \\\\ \\text{max} \\end{smallmatrix}}=\\frac{{{{\\text{a}}_{3}}^{2}+{{\\text{a}}_{4}}^{2}-{{\\text{a}}_{1}}^{2}-{{\\text{a}}_{2}}^{2}}}{{2{{\\text{a}}_{3}}{{\\text{a}}_{4}}}} \\pm \\frac{{{{\\text{a}}_{1}}{{\\text{a}}_{2}}}}{{{{\\text{a}}_{3}}{{\\text{a}}_{4}}}}}<\/span><\/td>\n<td style=\"text-align: right\">(4)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>uzuv boyutlar\u0131n\u0131 <span style=\"font-family: Symbol\">l<\/span> ile ifade eden (18)-(21) numaral\u0131 denklemler kullan\u0131l\u0131r ise, ba\u011flama a\u00e7\u0131s\u0131 bu parametre ile ifade edilmi\u015f olacakt\u0131r. Ba\u011flama a\u00e7\u0131s\u0131n\u0131n dik a\u00e7\u0131dan sapmas\u0131n\u0131 minimum veya maksimum edecek <span style=\"font-family: Symbol\">l<\/span> de\u011ferine (<span style=\"font-family: Symbol\">l<\/span><sub>opt<\/sub>) dersek, denklem (4) \u00fcn <span style=\"font-family: Symbol\">l<\/span> ya g\u00f6re t\u00fcrevini al\u0131p s\u0131f\u0131ra e\u015fitledi\u011fimizde elde etti\u011fimiz de\u011fer bu en iyi durumu i\u00e7erecektir. Bu i\u015flem yap\u0131ld\u0131\u011f\u0131nda <span style=\"font-family: Symbol\">l<\/span><sub>opt<\/sub>\u00a0de\u011feri:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">Q<sup>3<\/sup> +2Q<sup>2<\/sup> \u2212 t<sup>2<\/sup>Q \u2212 t<sup>2<\/sup>(1 + t<sup>2<\/sup>)\/u<sup>2<\/sup> = 0<\/td>\n<td style=\"text-align: right\">(22)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\u00dc\u00e7\u00fcnc\u00fc derece denklemin k\u00f6klerinden birisi olarak elde edilir. Bu denklemde Q = t<sup>2<\/sup>\/\u03bb<sub>opt<\/sub><sup>2<\/sup> dur ve denklemin \u00fc\u00e7 k\u00f6k\u00fcnden 1\/u<sup>2<\/sup> &lt; Q &lt; t<sup>2<\/sup> olan\u0131 kol-sarka\u00e7 oran\u0131n\u0131 sa\u011flayaca\u011f\u0131 gibi, bu (<span style=\"font-family: Symbol\">l<\/span><sub>opt<\/sub>) de\u011feri kullan\u0131larak elde edilen kol-sarka\u00e7 mekanizmas\u0131n\u0131n ba\u011flama a\u00e7\u0131s\u0131 di\u011fer <span style=\"font-family: Symbol\">l <\/span>de\u011ferleri kullan\u0131larak elde edilecek kol-sarka\u00e7 mekanizmalar\u0131na g\u00f6re dik a\u00e7\u0131dan en az sapma g\u00f6sterecektir.<\/p>\n<p>(22) numaral\u0131 denklem basit Newton-Raphson iteratif y\u00f6ntemi ile \u00e7\u00f6z\u00fclebilir. K\u00f6k aral\u0131\u011f\u0131 1\/u<sup>2<\/sup> &lt; Q &lt; t<sup>2 <\/sup>orta noktas\u0131 ilk tahmin al\u0131narak (Q<sub>0<\/sub> = (1\/u<sup>2<\/sup> + t<sup>2<\/sup>)\/2) ve bu k\u00f6k\u00fcn bulunmas\u0131 i\u00e7in tekrarlama denklemi:<\/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 {{Q}_{{i+1}}}=\\frac{{2{{Q}_{i}}\\left( {{{Q}_{i}}+1} \\right)+{{\\text{t}}^{2}}\\left( {1+{{\\text{t}}^{2}}} \\right)\/{{\\text{u}}^{2}}}}{{{{Q}_{i}}\\left( {3{{Q}_{i}}+4} \\right)-{{\\text{t}}^{2}}}} <\/span><\/td>\n<td style=\"text-align: right\">(23)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>kullan\u0131larak yeni k\u00f6ke daha yak\u0131n de\u011ferler elde edilebilir. \u0130terasyona (Q<sub>i+1 <\/sub>&#8211; Q<sub>i<\/sub>)\/Q<sub>i<\/sub>\u00a0yeteri kadar k\u00fc\u00e7\u00fck oldu\u011funda (\u00f6rne\u011fin 10<sup>-6<\/sup>dan k\u00fc\u00e7\u00fck) son verilir.<\/p>\n<p>Bir sal\u0131n\u0131m a\u00e7\u0131s\u0131 ve kar\u015f\u0131 gelen kol a\u00e7\u0131s\u0131na g\u00f6re en iyi ba\u011flama a\u00e7\u0131s\u0131 de\u011ferlerini veren kol \u00f6l\u00fc konum a\u00e7\u0131s\u0131 ve en iyi ba\u011flama a\u00e7\u0131s\u0131 de\u011ferler grafik olarak Abak 1 de g\u00f6r\u00fclmektedir (bu abak ilk olarak Alt taraf\u0131ndan haz\u0131rlanm\u0131\u015f, soradan Volmer taraf\u0131ndan d\u00fczenlenmi\u015ftir. Bu nedenle Alt diyagram\u0131 denmektedir). Bu abak kullan\u0131larak kol \u00f6l\u00fc konum a\u00e7\u0131s\u0131 <span style=\"font-family: Symbol\">b<\/span> ve en kritik ba\u011flama a\u00e7\u0131s\u0131n\u0131n en iyi de\u011feri max<span style=\"font-family: Symbol\">m<\/span>min verilmektedir. Uzuv boyutlar\u0131 bu abaktan elde edilen <span style=\"font-family: Symbol\">b<\/span>\u00a0de\u011ferinin (15)-(17) numaral\u0131 denklemde kullan\u0131lmas\u0131 ile elde edilir.<\/p>\n<p><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/AltTas.xls\">You can download the Excel file for the design of crank rocker mechanisms when the swing angle and the corresponding crank rotations are given.<\/a><\/p>\n<p><strong><em>\u00d6rnek<\/em><\/strong><\/p>\n<p>Sabit uzuv boyutu 120 mm, sal\u0131n\u0131m a\u00e7\u0131s\u0131 <span style=\"font-family: Symbol\">y<\/span>=40<sup>0<\/sup> ve kar\u015f\u0131 gelen krank d\u00f6nme a\u00e7\u0131s\u0131 <span style=\"font-family: Symbol\">f<\/span>=160<sup>0<\/sup>\u00a0olan ayr\u0131ca en iyi ba\u011flama a\u00e7\u0131s\u0131na sahip kol-sarka\u00e7 mekanizmas\u0131n\u0131 bulun.<\/p>\n<p>Verilen a\u00e7\u0131lar kullan\u0131larak t=5.671282 ve u=1.732051 bulunur, \u00f6yle ise aranan k\u00f6k: 0.333333&lt;Q&lt;32.163437 aral\u0131\u011f\u0131ndad\u0131r. Q<sub>0<\/sub>=16.248 alal\u0131m ve titerasyon denklemi kullanarak:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">i<\/td>\n<td style=\"text-align: center\">0<\/td>\n<td style=\"text-align: center\">1<\/td>\n<td style=\"text-align: center\">2<\/td>\n<td style=\"text-align: center\">3<\/td>\n<td style=\"text-align: center\">4<\/td>\n<td style=\"text-align: center\">5<\/td>\n<td style=\"text-align: center\">6<\/td>\n<td style=\"text-align: center\">7<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\">\u00a0Q<\/td>\n<td style=\"text-align: center\">\u00a016.248<\/td>\n<td style=\"text-align: center\">\u00a011.47207<\/td>\n<td style=\"text-align: center\">\u00a08.095663<\/td>\n<td style=\"text-align: center\">\u00a07.982059<\/td>\n<td style=\"text-align: center\">\u00a07.857866<\/td>\n<td style=\"text-align: center\">\u00a07.855707<\/td>\n<td style=\"text-align: center\">\u00a07.855706<\/td>\n<td style=\"text-align: center\">\u00a07.855706<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\u00d6yle ise aranan k\u00f6k (6 basamak do\u011frulukta): Q=7.855706 dir. Buna g\u00f6re <span style=\"font-family: Symbol\">l<\/span><sub>opt<\/sub>=2.023432 bulunur. Bu <span style=\"font-family: Symbol\">l<\/span><sub>opt<\/sub>\u00a0de\u011feri uzuv boyutlar\u0131n\u0131 bulmak i\u00e7in kullan\u0131ld\u0131\u011f\u0131nda:<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{1}}^{2}=\\frac{{{{{1.732051}}^{2}}+{{{2.023432}}^{2}}}}{{1+{{{1.732051}}^{2}}}}=1.773569\\text{ }\\Rightarrow \\text{ }{{\\text{a}}_{1}}=1.331754 <\/span><\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{2}}^{2}=\\frac{{{{{0.363970}}^{2}}}}{{1+{{{0.363970}}^{2}}}}=0.116978\\text{ }\\Rightarrow \\text{ }{{\\text{a}}_{2}}=0.342020 <\/span><\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{3}}=2.023432\\cdot 0.342020=0.692054 <\/span><\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{4}}^{2}=\\frac{{{{{5.671282}}^{2}}+{{{2.023432}}^{2}}}}{{1+{{{5.671282}}^{2}}}}=1.093304\\text{ }\\Rightarrow \\text{ }{{\\text{a}}_{4}}=1.045612 <\/span><\/p>\n<p>a<sub>1<\/sub>=120 mm oldu\u011funda : a<sub>2<\/sub>=(0.342020\/1.331754)\u00d7120=30.82 mm a<sub>3<\/sub>=62.36 mm, a<sub>4<\/sub>=94.22 mm olarak bulunur. Bu mekanizma i\u00e7in ba\u011flama a\u00e7\u0131s\u0131n\u0131n dik a\u00e7\u0131dan en fazla sapmas\u0131 ise:<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\cos \\text{\u03bc}_{\\begin{smallmatrix} \\text{min} \\\\ \\text{max} \\end{smallmatrix}}=\\frac{{{{{94.22}}^{2}}+{{{62.36}}^{2}}-{{{120}}^{2}}-{{{30.82}}^{2}}}}{{2\\cdot 94.22\\cdot 62.36}}\\pm \\frac{{120\\cdot 30.82}}{{94.22\\cdot 62.36}}=0.219938\\pm 0.629455}<\/span><\/p>\n<p>Bu denklemden : <span style=\"font-family: Symbol\">m<\/span><sub>max<\/sub>=114.170 (<span style=\"font-family: Symbol\">D<\/span><sub>1<\/sub>=24.170) ve <span style=\"font-family: Symbol\">m<\/span><sub>min<\/sub>=31.850 (<span style=\"font-family: Symbol\">D<\/span>=58.150). <span style=\"font-family: Symbol\">m<\/span><sub>min<\/sub> daha fazla sapma g\u00f6sterdi\u011finden <span style=\"font-family: Symbol\">m<\/span><sub>min<\/sub> kritik ba\u011flama a\u00e7\u0131s\u0131d\u0131r. Verilen sal\u0131n\u0131m a\u00e7\u0131s\u0131 ve kar\u015f\u0131 gelen kol d\u00f6nme a\u00e7\u0131s\u0131n\u0131 sa\u011flayan ve ba\u011flama a\u00e7\u0131s\u0131 dik a\u00e7\u0131dan en az sapan mekanizma boyutlar\u0131 bu elde edilmi\u015f olan boyutlard\u0131r. Farkl\u0131 bir <span style=\"font-family: Symbol\">l<\/span> veya <span style=\"font-family: Symbol\">b<\/span>\u00a0parametre de\u011feri ile bulunan kol-sarka\u00e7 mekanizmas\u0131n\u0131n ba\u011flama a\u00e7\u0131s\u0131n\u0131n dik a\u00e7\u0131dan sapmas\u0131 bu bulunan mekanizmadaki sapmadan mutlaka daha fazla olacakt\u0131r.<\/p>\n<p>Abak 1&#8217;e bakt\u0131\u011f\u0131m\u0131zda ise, verilen <span style=\"font-family: Symbol\">y<\/span>=40<sup>0<\/sup> ve <span style=\"font-family: Symbol\">f<\/span>=160<sup>0<\/sup> de\u011ferleri i\u00e7in abakdan: max(<span style=\"font-family: Symbol\">m<\/span><sub>min<\/sub>)\u224832<sup>0<\/sup> ve <span style=\"font-family: Symbol\">b\u2248<\/span>50.5<sup>0<\/sup>\u00a0okunabilir. Tabii ki abaktan s\u0131n\u0131rl\u0131 bir hassasiyetle okumak m\u00fcmk\u00fcnd\u00fcr. Uzuv boyutlar\u0131n\u0131 bu b de\u011feri ile buldu\u011fumuzda: a<sub>1<\/sub>=1, a<sub>2<\/sub>=0.2565, a<sub>3<\/sub>=0.5201 and a<sub>4<\/sub>=0.784 de\u011ferlerini elde ederiz. a<sub>1<\/sub>=120 mm oldu\u011funda; a<sub>2<\/sub>=30.78 mm, a<sub>3<\/sub>=62.41 mm and a<sub>4<\/sub>=94.08 mm bulunur. Sonu\u00e7 mekanizma \u015fekilde g\u00f6sterilmi\u015ftir.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1085\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img4-7.gif\" alt=\"\" width=\"592\" height=\"397\" \/><\/p>\n<p>Yukar\u0131da anlat\u0131lmaya \u00e7al\u0131\u015f\u0131lm\u0131\u015f olan \u00f6l\u00fc konumlara g\u00f6re(verilen herhangi bir <span style=\"font-family: Symbol\">f <\/span>ve <span style=\"font-family: Symbol\">y <\/span>de\u011ferine g\u00f6re) bir kol-sarka\u00e7 mekanizmas\u0131 tasar\u0131m\u0131 i\u00e7in haz\u0131rlanm\u0131\u015f olan\u00a0<a href=\"altBagAcisi\/AltTas.xls\">Excel k\u00fct\u00fc\u011f\u00fcn\u00fc alabilirsiniz<\/a>.<\/p>\n<p>Yukar\u0131da verilmi\u015f olan genel \u00e7\u00f6z\u00fcm\u00fcn d\u0131\u015f\u0131nda iki \u00f6zel durum olu\u015fmaktad\u0131r. Bunlardan birisi santrik kol-sarka\u00e7 mekanizmas\u0131d\u0131r.bu durumda \u00f6l\u00fc konumlar aras\u0131nda kol d\u00f6nme a\u00e7\u0131s\u0131 <span style=\"font-family: Symbol\">f<\/span>=180<sup>0<\/sup>\u00a0dir ve ba\u011flama a\u00e7\u0131s\u0131n\u0131n maksimum ve minimum de\u011ferleri:<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\cos \\text{\u03bc}_{\\begin{smallmatrix} \\text{min} \\\\ \\text{max} \\end{smallmatrix}}= \\pm \\frac{{{{\\text{a}}_{1}}{{\\text{a}}_{2}}}}{{{{\\text{a}}_{3}}{{\\text{a}}_{4}}}}}<\/span><\/p>\n<p>dir. E\u011fer dik a\u00e7\u0131dan sapman\u0131n en az olmas\u0131 istenir ise a<sub>2<\/sub>=0 olmas\u0131 ( a<sub>1<\/sub>=1 al\u0131nm\u0131\u015ft\u0131) veya a<sub>3<\/sub>\u00a0veya a<sub>4<\/sub> uzuv boyutlar\u0131ndan birisinin sonsuz olmas\u0131 gerekir. Pratikte bu durumlar m\u00fcmk\u00fcn olam\u0131yaca\u011f\u0131ndan, ba\u011flama a\u00e7\u0131s\u0131 ve uzuv boyutlar\u0131 aras\u0131nda oran makul de\u011ferlerde olacak \u015fekilde kol-sarka\u00e7 mekanizmas\u0131 boyutlar\u0131n\u0131 arar\u0131z. (\u00f6rne\u011fin ba\u011flama a\u00e7\u0131s\u0131n\u0131n dik a\u00e7\u0131dan sapmas\u0131 <span style=\"font-family: Symbol\">l<\/span>\u00a0de\u011ferinin b\u00fcy\u00fcmesi ile azalacakt\u0131r).<\/p>\n<p>\u0130kinci \u00f6zel durum ise <span style=\"font-family: Symbol\">f<\/span>&#8211;<span style=\"font-family: Symbol\">y<\/span>=180<sup>0<\/sup> olmas\u0131nda rastlan\u0131l\u0131r ( $latex \\displaystyle =\\tan \\left[ {\\frac{1}{2}\\left( {\\phi -\\psi } \\right)} \\right]=\\infty $<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2847\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/08\/eqn34.gif\" alt=\"\" width=\"149\" height=\"48\" srcset=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/08\/eqn34.gif 149w, https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/08\/eqn34-100x32.gif 100w\" sizes=\"auto, (max-width: 149px) 100vw, 149px\" \/>). Ba\u011flama a\u00e7\u0131s\u0131n\u0131n optimum de\u011feri: <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{\u03bb}}_{{\\text{opt}}}}=1+\\frac{1}{{\\sin \\left( {\\text{\u03c8}\/2} \\right)}} <\/span>\u00a0olacakt\u0131r ve uzuv boyutlar\u0131:<\/p>\n<p style=\"text-align: center\">a<sub>1<\/sub>\u00a0= 1<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\text{a}_{2}}=\\sin \\left( {\\text{\u03c8}\/2} \\right) <\/span><\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\text{a}_{3}}=\\sqrt{{{\\text{a}_{2}}\\left( {1+{\\text{a}_{2}}} \\right)}} <\/span><\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\text{a}_{4}}=\\sqrt{{1+{\\text{a}_{2}}}} <\/span><\/p>\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n<p><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/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\/mekanizma-teknigi\/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\/mekanizma-teknigi\/\" 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>Alt Ba\u011flama A\u00e7\u0131s\u0131 Problemi &#8220;Verilen bir sal\u0131n\u0131m a\u00e7\u0131s\u0131n\u0131 (y) ve buna kar\u015f\u0131 gelen bir krank d\u00f6nme a\u00e7\u0131s\u0131n\u0131 (f) sa\u011flayan, ve ayr\u0131ca ba\u011flama a\u00e7\u0131s\u0131n\u0131n 900 den sapmas\u0131 en az olan kol-sarka\u00e7 oranlar\u0131nda d\u00f6rt \u00e7ubuk mekanizmas\u0131n\u0131n boyutlar\u0131n\u0131 belirleyin.&#8221; Problem iki k\u0131s\u0131mdan olu\u015fmaktad\u0131r: &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch7\/7-1\/altbagacisi\/\"> <span class=\"screen-reader-text\">altBagAcisi<\/span> Devam\u0131n\u0131 Oku &raquo;<\/a><\/p>\n","protected":false},"author":7747,"featured_media":0,"parent":863,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-1038","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1038","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=1038"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1038\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/863"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/media?parent=1038"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}