{"id":1134,"date":"2021-09-09T18:33:32","date_gmt":"2021-09-09T18:33:32","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=1134"},"modified":"2021-09-30T20:32:55","modified_gmt":"2021-09-30T20:32:55","slug":"7-2","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch7\/7-2\/","title":{"rendered":"7-2"},"content":{"rendered":"<div id=\"pl-gb1134-69d879ec47e57\"  class=\"panel-layout\" ><div id=\"pg-gb1134-69d879ec47e57-0\"  class=\"panel-grid panel-no-style\" ><div id=\"pgc-gb1134-69d879ec47e57-0-0\"  class=\"panel-grid-cell\" ><div id=\"panel-gb1134-69d879ec47e57-0-0-0\" 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><strong data-rich-text-format-boundary=\"true\">7.2 Krank-Biyel Mekanizmas\u0131<\/strong><\/h1>\n<p>Makina tasar\u0131m\u0131nda yo\u011fun bir \u015fekilde kullan\u0131lan bir ba\u015fka mekanizmada krank-biyel mekanizmas\u0131d\u0131r. Genel olarak bir d\u00f6nme hareketini bir \u00f6teleme hareketine \u00e7evirmek i\u00e7in kullan\u0131ld\u0131\u011f\u0131 gibi bir \u00f6teleme hareketini d\u00f6nme hareketine \u00e7evirmek i\u00e7inde kullan\u0131labilir. \u015eekilde krank-biyel mekanizmas\u0131, de\u011fi\u015fken a\u00e7\u0131lar\u0131 ve boyutlar\u0131 tan\u0131mlayan sabit parametreleri g\u00f6r\u00fclmektedir. D\u00f6rt-\u00e7ubuk mekanizmas\u0131nda oldu\u011fu gibi, \u00f6l\u00fc konumlar krank ile biyelin ayn\u0131 do\u011fru \u00fczerinde oldu\u011fu konumlard\u0131r. Krank\u0131n tam bir d\u00f6nmesi i\u00e7in c eksantrikli\u011finin biyel ile krank uzunluklar\u0131n\u0131n fark\u0131ndan az ve krank\u0131n en k\u0131sa uzuv boyutu olmas\u0131 gerekir. (yani c &lt; a<sub>3<\/sub> \u2212 a<sub>2<\/sub>\u00a0ve a<sub>3<\/sub>\u00a0&gt; a<sub>2<\/sub>\u00a0olmal\u0131d\u0131r).<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1136\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img1-11.gif\" alt=\"\" width=\"655\" height=\"305\" \/><\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1137 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img2-11.gif\" alt=\"\" width=\"692\" height=\"337\" \/><\/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:442px\" 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_69d879ec49a90\"><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/krankbiyel2_1.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"442\" height=\"363\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/krankbiyel2_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\/2021\/09\/krankbiyel2_2.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"442\" height=\"363\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/krankbiyel2_2.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><\/div><script id=\"su_image_carousel_69d879ec49a90_script\">if(window.SUImageCarousel){setTimeout(function() {window.SUImageCarousel.initGallery(document.getElementById(\"su_image_carousel_69d879ec49a90\"))}, 0);}var su_image_carousel_69d879ec49a90_script=document.getElementById(\"su_image_carousel_69d879ec49a90_script\");if(su_image_carousel_69d879ec49a90_script){su_image_carousel_69d879ec49a90_script.parentNode.removeChild(su_image_carousel_69d879ec49a90_script);}<\/script><\/p>\n<p>Krank biyel mekanizmas\u0131nda\u00a0<b>strok<\/b>\u00a0(s) pistonun \u00f6l\u00fc konumlar aras\u0131nda yapt\u0131\u011f\u0131 \u00f6teleme mesafesi olup:<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle \\text{s}={{\\text{a}}_{3}}\\sqrt{{{{{\\left( {1+\\text{\u03bb}} \\right)}}^{2}}-{{\\text{\u03b5}}^{2}}}}-{{\\text{a}}_{3}}\\sqrt{{{{{\\left( {1-\\text{\u03bb}} \\right)}}^{2}}-{{\\text{\u03b5}}^{2}}}} <\/span><\/p>\n<p>dur. Burada\u00a0s = s<sub>e<\/sub> \u2212 s<sub>f<\/sub> (strok), \u03bb = a<sub>2<\/sub>\/a<sub>3<\/sub> ve \u03b5 = c\/a<sub>3<\/sub> t\u00fcr.<\/p>\n<p>Eksantriklik s\u0131f\u0131r ise( c = 0), krank biyel mekanizmas\u0131\u00a0<b>santriktir<\/b>\u00a0ve bu durumda strok iki krank boyu olur (s = 2a<sub>2<\/sub>).<\/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:442px\" 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_69d879ec4a333\"><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/krankbiyel_1.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"461\" height=\"363\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/krankbiyel_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\/2021\/09\/krankbiyel_2.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"461\" height=\"363\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/krankbiyel_2.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><\/div><script id=\"su_image_carousel_69d879ec4a333_script\">if(window.SUImageCarousel){setTimeout(function() {window.SUImageCarousel.initGallery(document.getElementById(\"su_image_carousel_69d879ec4a333\"))}, 0);}var su_image_carousel_69d879ec4a333_script=document.getElementById(\"su_image_carousel_69d879ec4a333_script\");if(su_image_carousel_69d879ec4a333_script){su_image_carousel_69d879ec4a333_script.parentNode.removeChild(su_image_carousel_69d879ec4a333_script);}<\/script><\/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:442px\" 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_69d879ec4ab96\"><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/krankbiyelta1.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"452\" height=\"363\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/krankbiyelta1.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\/krankbiyelta2.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"442\" height=\"363\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/krankbiyelta2.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><\/div><script id=\"su_image_carousel_69d879ec4ab96_script\">if(window.SUImageCarousel){setTimeout(function() {window.SUImageCarousel.initGallery(document.getElementById(\"su_image_carousel_69d879ec4ab96\"))}, 0);}var su_image_carousel_69d879ec4ab96_script=document.getElementById(\"su_image_carousel_69d879ec4ab96_script\");if(su_image_carousel_69d879ec4ab96_script){su_image_carousel_69d879ec4ab96_script.parentNode.removeChild(su_image_carousel_69d879ec4ab96_script);}<\/script><\/p>\n<p>Ba\u011flama a\u00e7\u0131s\u0131, \u03bc, d\u00f6rt \u00e7ubuk mekanizmas\u0131nda verilmi\u015f olan tan\u0131m ile ayn\u0131 olup \u015fekilde g\u00f6r\u00fcld\u00fc\u011f\u00fc gibidir. Ba\u011flama a\u00e7\u0131s\u0131n\u0131n dik a\u00e7\u0131dan sapmas\u0131 krank-biyel mekanizmalar\u0131nda daha \u00f6nemli olup s\u00fcrt\u00fcnmeden dolay\u0131 kilitlenme (veya kas\u0131lma) ihtimali daha fazlad\u0131r. Ayr\u0131ca ba\u011flama a\u00e7\u0131s\u0131n\u0131n dik a\u00e7\u0131dan sapmas\u0131 kayar mafsala etki eden normal kuvveti art\u0131raca\u011f\u0131ndan s\u00fcrt\u00fcnme kuvveti artaca\u011f\u0131ndan, \u00f6nemli enerji kayb\u0131na neden olur. Herhangi bir krank a\u00e7\u0131s\u0131na g\u00f6re ba\u011flama a\u00e7\u0131s\u0131:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">a<sub>3<\/sub>cos\u03bc = a<sub>2<\/sub>sin\u03b8<sub>12<\/sub> \u2212 c<\/td>\n<td style=\"text-align: right\">(1)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Ba\u011flama a\u00e7\u0131s\u0131n\u0131n dik a\u00e7\u0131dan maksimum sapmas\u0131n\u0131 belirlemek i\u00e7in \u03bc n\u00fcn krank a\u00e7\u0131s\u0131 \u03b8<sub>12<\/sub>\u00a0ye g\u00f6re t\u00fcrevi al\u0131n\u0131r s\u0131f\u0131ra e\u015fitlenir. Denklem (1) den:<\/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{{\\text{d\u03bc}}}{{\\text{d}{{\\text{\u03b8}}_{{12}}}}}=\\frac{{-\\cos {{\u03b8}_{{12}}}}}{{\\sin \\text{\u03bc}}}=0 <\/span><\/td>\n<td style=\"text-align: right\">(2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Ba\u011flama a\u00e7\u0131s\u0131n\u0131n maksimum veya minimum de\u011feri \u03b8<sub>12<\/sub>\u00a0= 90\u00b0\u00a0veya \u03b8<sub>12<\/sub>\u00a0= 270\u00b0\u00a0oldu\u011fu konumlard\u0131r ve bu konumlarda ba\u011flama a\u00e7\u0131s\u0131n\u0131n maksimum veya minimum de\u011feri:<\/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{max} \\\\ \\text{min} \\end{smallmatrix}}}=\\frac{{-\\text{c}\\pm {{\\text{a}}_{2}}}}{{{{\\text{a}}_{3}}}} <\/span><\/td>\n<td style=\"text-align: right\">(3)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>dir. E\u011fer c \u015eekilde g\u00f6r\u00fcld\u00fc\u011f\u00fc gibi, pozitif ise ba\u011flama a\u00e7\u0131s\u0131 \u03b8<sub>12<\/sub>\u00a0= 270\u00b0 iken dik a\u00e7\u0131dan maksimum sapar. c negatif ise, en kritik ba\u011flama a\u00e7\u0131s\u0131 \u03b8<sub>12<\/sub>\u00a0= 90\u00b0\u00a0konumundad\u0131r.<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1141\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img3-8.gif\" alt=\"\" width=\"713\" height=\"771\" \/><\/p>\n<p>Eksantriklik s\u0131f\u0131r ise (c = 0), maksimum ve minimum ba\u011flama a\u00e7\u0131s\u0131 de\u011ferlerinin dik a\u00e7\u0131dan sapmas\u0131 e\u015fit olur:<\/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{max} \\\\ \\text{min} \\end{smallmatrix}}}=\\pm \\frac{{{\\text{a}}_{2}}}{{{\\text{a}}_{3}}} <\/span><\/td>\n<td style=\"text-align: right\">(4)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Bir \u00f6rnek olmas\u0131 a\u00e7\u0131s\u0131ndan, pistonlu pompalarda, krank-biyel oran\u0131n\u0131n genellikle \u00bc den az bir oran olmas\u0131 istenilir. Bu ise ba\u011flama a\u00e7\u0131s\u0131n\u0131n dik a\u00e7\u0131dan sapmas\u0131n\u0131 14.48<sup>0<\/sup>\u00a0den az bir de\u011ferde tutacakt\u0131r. Santrik bir krank biyelde krank boyu istenilen stroka ba\u011fl\u0131 oldu\u011fundan (a<sub>2<\/sub>\u00a0= s\/2), bu oran\u0131n elde edilmesi i\u00e7in biyel boyunun b\u00fcy\u00fct\u00fclmesi, dolay\u0131s\u0131 ile mekanizman\u0131n daha fazla hacim kapsamas\u0131n\u0131 gerektirir.<\/p>\n<p>D\u00f6rt \u00e7ubuk mekanizmas\u0131 i\u00e7in a\u00e7\u0131klanm\u0131\u015f olan ba\u011flama a\u00e7\u0131s\u0131 problemi gibi, krank-biyel mekanizmalar\u0131 i\u00e7inde benzer bir problem ortaya konabilir:<\/p>\n<p style=\"text-align: center\"><strong><i><span style=\"color: #cc0000\">Verilen bir strok, s, ve buna kar\u015f\u0131 gelen krank d\u00f6nme a\u00e7\u0131s\u0131n\u0131 (\u03d5) sa\u011flayan ve ayr\u0131ca ba\u011flama a\u00e7\u0131s\u0131 dik a\u00e7\u0131dan en az sapma g\u00f6steren krank biyel mekanizmas\u0131n\u0131 bulun.<\/span><\/i><\/strong><\/p>\n<p>Problemin yine iki k\u0131sm\u0131 bulunmaktad\u0131r. Birinci k\u0131s\u0131m, verilen strok (s) ve kar\u015f\u0131 gelen kol d\u00f6nme a\u00e7\u0131s\u0131n\u0131 (\u03d5) sa\u011flayan krank biyel mekanizmalar\u0131n\u0131n bulunmas\u0131 ikinci k\u0131s\u0131m ise bu kinematik \u00f6zellikleri sa\u011flayan krank-biyel mekanizmalar\u0131 aras\u0131ndan ba\u011flama a\u00e7\u0131s\u0131 dik a\u00e7\u0131dan en az sapma g\u00f6steren mekanizman\u0131n bulunmas\u0131d\u0131r.<\/p>\n<p>Problemin birinci k\u0131sm\u0131 i\u00e7in strokun uzuv boyutlar\u0131 oran\u0131na ba\u011fl\u0131 oldu\u011funa dikkat etmemiz gerekmektedir. \u00d6rne\u011fin uzuv boyutlar\u0131n\u0131 iki misli b\u00fcy\u00fct\u00fcr isek strok iki misli artacakt\u0131r. Bu d\u00f6rt \u00e7ubuk mekanizmalar\u0131nda kol d\u00f6nme a\u00e7\u0131lar\u0131 i\u00e7in olmayan bir durumdur. Bu nedenle strok bir birim (s = 1) olarak ele al\u0131nacak, sonu\u00e7ta elde edilen krank ve biyel boyutlar\u0131 istenilen strok de\u011feri ile \u00e7arp\u0131larak ger\u00e7ek olmas\u0131 gerekli boyutlar bulunacakt\u0131r.<\/p>\n<p>\u015eekilde g\u00f6sterilmi\u015f olan \u00f6l\u00fc konumlar i\u00e7in vekt\u00f6r devre denklemi:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\"><strong>A<sub>0<\/sub>B<sub>e<\/sub><\/strong>\u00a0+ <strong>B<sub>e<\/sub>A<sub>e<\/sub><\/strong>\u00a0+ <strong>A<sub>e<\/sub>A<sub>0<\/sub><\/strong>\u00a0= <strong>0<\/strong><\/td>\n<td style=\"text-align: right\">(5)<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\"><strong>A<sub>0<\/sub>B<sub>f<\/sub><\/strong>\u00a0+ <strong>B<sub>f<\/sub>A<sub>f<\/sub><\/strong>\u00a0+ <strong>A<sub>f<\/sub>A<sub>0<\/sub><\/strong>\u00a0= <strong>0<\/strong><\/td>\n<td style=\"text-align: right\">(6)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>veya karma\u015f\u0131k say\u0131lar kullan\u0131larak<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">ic + s<sub>e<\/sub> + (a<sub>3<\/sub> + a<sub>2<\/sub>)e<sup>i\u03d5<sub>1<\/sub><\/sup> = 0<\/td>\n<td style=\"text-align: right\">(7)<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\">ic + s<sub>e<\/sub> + (a<sub>3<\/sub> \u2212 a<sub>2<\/sub>)e<sup>i(\u03d5<sub>1 <\/sub>+<sub>\u00a0<\/sub>\u03d5<sub>\u00a0<\/sub>\u2212<sub>\u00a0<\/sub>\u03c0)<\/sup> = 0<\/td>\n<td style=\"text-align: right\">(8)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>d\u0131r.\u00a0Denklem (8) i denklem (7) den \u00e7\u0131kar\u0131p s<sub>e<\/sub> \u2212 s<sub>f<\/sub>\u00a0= s = 1 olarak al\u0131r isek:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">1 + (a<sub>3<\/sub> + a<sub>2<\/sub>)e<sup>i\u03d5<sub>1<\/sub><\/sup> + (a<sub>3<\/sub> \u2212 a<sub>2<\/sub>)e<sup>i(\u03d5<sub>1 <\/sub>+<sub>\u00a0<\/sub>\u03d5)<\/sup> = 0<\/td>\n<td style=\"text-align: right\">(9)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>olacakt\u0131r. \u03bb = a<sub>3<\/sub>\/a<sub>2<\/sub> ve Z = a<sub>3<\/sub>e<sup>i\u03d5<sub>1<\/sub><\/sup> olarak tan\u0131mlar isek, denklem (9)<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">(1 + \u03bb)Z + (1 \u2212 \u03bb)e<sup>i\u03d5<\/sup>Z\u00a0+ 1= 0<\/td>\n<td style=\"text-align: right\">(10)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>olarak yaz\u0131labilir. Krank\u0131n tam bir d\u00f6nme yapabilmesi i\u00e7in gerek \u015fart (yeter \u015fart de\u011fil) |\u03bb|&lt; 1 dir. Denklem (10) Z i\u00e7in \u00e7\u00f6z\u00fcld\u00fc\u011f\u00fcnde:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">Z = <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle \\frac{{-1}}{{1+{{\\text{e}}^{{\\text{i\u03d5}}}}+\\text{\u03bb}\\left( {1-{{\\text{e}}^{{\\text{i\u03d5}}}}} \\right)}} <\/span><\/td>\n<td style=\"text-align: right\">(10)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>elde edilir. E\u011fer \u03bb ba\u011f\u0131ms\u0131z parametre olarak kabul edilir ise, \u03bb n\u0131n de\u011fi\u015fik de\u011ferleri ile Z vekt\u00f6r\u00fcn\u00fcn u\u00e7 noktas\u0131 bir daire \u00e7izecektir (k<sub>a<\/sub>\u00a0dairesi). Bu, mekanizman\u0131n \u00f6l\u00fc konumunda A<sub>e<\/sub>\u00a0noktas\u0131n\u0131n B<sub>e<\/sub> noktas\u0131na g\u00f6re geometrik yeridir. Sabit d\u00f6ner mafsal ekseni ise (1 + \u03bb)Z\u00a0vekt\u00f6r\u00fc ile tan\u0131mlanan farkl\u0131 bir daire (k<sub>o<\/sub>\u00a0dairesi) olacakt\u0131r. Her iki vekt\u00f6rde merkezi B<sub>e<\/sub> den ge\u00e7en ve reel ekseni piston \u00f6teleme ekseni ile \u00e7ak\u0131\u015fan sabit koordinat eksenine g\u00f6re \u00e7izilebilir. \u015eekilde bu daireler \u03d5 = 160<sup>0<\/sup> durumu i\u00e7in \u00e7izilmi\u015ftir. B<sub>e<\/sub>\u00a0den \u00e7izilen herhangi bir do\u011fru bu daireleri A<sub>e<\/sub>\u00a0ve A<sub>o<\/sub>\u00a0noktalar\u0131nda kesecektir.<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1147 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img4-8.gif\" alt=\"\" width=\"437\" height=\"382\" \/><\/p>\n<p>c eksantriklik de\u011feri ise:<\/p>\n<p style=\"padding-left: 40px;text-align: center\"><strong>B<sub>e<\/sub>A<sub>0<\/sub><\/strong> = <strong>B<sub>e<\/sub>A<sub>e<\/sub><\/strong>\u00a0+ <strong>A<sub>e<\/sub>A<sub>0<\/sub><\/strong><\/p>\n<p>vekt\u00f6r\u00fcn\u00fcn sanal bile\u015feni olarak elde edilecektir. Bir kompleks say\u0131dan kompleks e\u015fleni\u011fi \u00e7\u0131kar\u0131l\u0131r ise, sanal k\u0131sm\u0131n iki kat\u0131 olaca\u011f\u0131ndan bu vekt\u00f6r denkleminden:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">2ic = (a<sub>3<\/sub> + a<sub>2<\/sub>)e<sup>i\u03d5<sub>1<\/sub><\/sup> \u2212 (a<sub>3<\/sub> + a<sub>2<\/sub>)e<sup>\u2212i\u03d5<sub>1<\/sub><\/sup> = 0<\/td>\n<td style=\"text-align: right\">(12)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>elde edilir. Bu denklemde \u00f6nceden tan\u0131mlanm\u0131\u015f olan Z ve \u03bb parametreleri kullan\u0131ld\u0131\u011f\u0131nda:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">2ic = Z(1 + \u03bb) \u2212 <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\bar{\\text{Z}}}} <\/span>(1 + \u03bb)<\/td>\n<td style=\"text-align: right\">(13)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>olacakt\u0131r. Denklem (11) den elde edilen Z de\u011feri kullan\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{c}=\\frac{1}{2}\\frac{{\\left( {1-{{\\text{\u03bb}}^{2}}} \\right)\\sin \\text{\u03d5}}}{{1+{{\\text{\u03bb}}^{2}}+\\left( {1-{{\\text{\u03bb}}^{2}}} \\right)\\cos \\text{\u03d5}}} <\/span><\/td>\n<td style=\"text-align: right\">(14)<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{3}}^{2}=\\frac{1}{2}\\frac{1}{{1+{{\\text{\u03bb}}^{2}}+\\left( {1-{{\\text{\u03bb}}^{2}}} \\right)\\cos \\text{\u03d5}}}=\\frac{1}{4}\\frac{1}{{{{{\\cos }}^{2}}\\left( {\\text{\u03d5}\/2} \\right)+{{\\text{\u03bb}}^{2}}{{{\\sin }}^{2}}\\left( {\\text{\u03d5}\/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}}_{2}}^{2}={{\\text{\u03bb}}^{2}}{{\\text{a}}_{3}}^{2}=\\frac{1}{2}\\frac{{{{\\text{\u03bb}}^{2}}}}{{1+{{\\text{\u03bb}}^{2}}+\\left( {1-{{\\text{\u03bb}}^{2}}} \\right)\\cos \\text{\u03d5}}}=\\frac{1}{4}\\frac{{{{\\text{\u03bb}}^{2}}}}{{{{{\\cos }}^{2}}\\left( {\\text{\u03d5}\/2} \\right)+{{\\text{\u03bb}}^{2}}{{{\\sin }}^{2}}\\left( {\\text{\u03d5}\/2} \\right)}} <\/span><\/td>\n<td style=\"text-align: right\">(16)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>olacakt\u0131r. (14-16) denklemleri kullan\u0131larak verilen krank d\u00f6nme a\u00e7\u0131s\u0131 ve stroku bir birim olan krank biyel mekan\u0131zmalar\u0131 \u03bb ba\u011f\u0131ms\u0131z parametresine g\u00f6re elde edilmi\u015ftir ve sonsuz say\u0131da \u00e7\u00f6z\u00fcm vard\u0131r. \u0130stenildi\u011finde \u03bb\u00a0parametresi yerine eksantriklik, krank veya biyel boyu ba\u011f\u0131ms\u0131z parametre olarak kullan\u0131labilir.<\/p>\n<p><span style=\"color: #cc0000\">Problemin geometrik olarak \u00e7\u00f6z\u00fcm\u00fc i\u00e7in:<\/span><\/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:450px\" 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_69d879ec4b8b1\"><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/altslider1.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\/altslider1.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\/altslider2.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\/altslider2.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\/altslider3.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\/altslider3.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\/altslider4.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\/altslider4.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\/altslider5.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\/altslider5.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\/altslider6.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\/altslider6.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\/altslider7.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"450\" height=\"221\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/altslider7.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><\/div><script id=\"su_image_carousel_69d879ec4b8b1_script\">if(window.SUImageCarousel){setTimeout(function() {window.SUImageCarousel.initGallery(document.getElementById(\"su_image_carousel_69d879ec4b8b1\"))}, 0);}var su_image_carousel_69d879ec4b8b1_script=document.getElementById(\"su_image_carousel_69d879ec4b8b1_script\");if(su_image_carousel_69d879ec4b8b1_script){su_image_carousel_69d879ec4b8b1_script.parentNode.removeChild(su_image_carousel_69d879ec4b8b1_script);}<\/script><\/p>\n<p><strong>\u00d6rnek:<\/strong><\/p>\n<p>\u00d6rnekde strok ve krank d\u00f6nme a\u00e7\u0131s\u0131 olan ancak krank uzunlu\u011funun biyel uzunlu\u011funa g\u00f6re oran\u0131 tan\u0131mlanm\u0131yarak, eksantrikli\u011fi c = 20 mm olmas\u0131 istenilen krank biyel mekanizmas\u0131 boyutlar\u0131n\u0131 bulal\u0131m.<\/p>\n<p>Birim stok i\u00e7in c = 20\/120 = 0.16667 dir. Denklem (10) u \u03bb i\u00e7in \u00e7\u00f6zd\u00fc\u011f\u00fcm\u00fczde:<\/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{{1-2\\text{c}\\cot \\left( {\\text{\u03d5}\/2} \\right)}}{{1+2\\text{c}\\tan \\left( {\\text{\u03d5}\/2} \\right)}} <\/span><\/td>\n<td style=\"text-align: right\">(17)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>olacakt\u0131r. c = 0.16667 ve \u03d5 =160<sup>0<\/sup> i\u00e7in \u03bb<sup>2<\/sup> = 0.325635 (\u03bb = 0.5706) bulunur. (11) ve (12) numaral\u0131 denklemler kullan\u0131larak a<sub>2<\/sub>\u00a0=\u00a0 0.48508 ve a<sub>3<\/sub>\u00a0= 0.85006 bulunur. s = 120 mm i\u00e7in c = 20mm , a<sub>2<\/sub>\u00a0= 58.21 mm ve a<sub>3<\/sub> = 102.01 mm olacakt\u0131r. Bu krank biyel mekanizmas\u0131 i\u00e7in minimum ba\u011flama a\u00e7\u0131s\u0131 de\u011feri: \u03bc<sub>min<\/sub>\u00a0= 39.94<sup>0<\/sup>. dikkat edilir ise, biyel uzunlu\u011fu veya krank uzunlu\u011fu \u00f6nceden verilir ise, benzer bir y\u00f6ntem kullan\u0131labilir.<\/p>\n<p>\u03b8 = \u03c0\/2 iken pozitif c de\u011ferleri i\u00e7in minimum ba\u011flama a\u00e7\u0131s\u0131 olacakt\u0131r ve de\u011feri:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">\u03bc<sub>min<\/sub> = cos<sup>-1<\/sup><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle \\left( {\\frac{{\\text{c}+\\text{a}}}{\\text{b}}} \\right)<\/span><\/td>\n<td style=\"text-align: right\">(18)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>dir. Ayr\u0131ca, krank\u0131n tam d\u00f6nme yapabilmesi i\u00e7in c + a &lt; b veya c &lt; b \u2212 a olmal\u0131d\u0131r. S\u0131n\u0131r \u015fart olarak (c = b \u2212 a) al\u0131nd\u0131\u011f\u0131nda \u03bc<sub>min<\/sub> = 0 olur. Denklem (10), (11) ve (12) kullan\u0131larak bu \u015fartlar \u00f6l\u00fc konumlar ras\u0131nda krank d\u00f6nme a\u00e7\u0131s\u0131 \u03d5\u00a0i\u00e7in s\u0131n\u0131rlar getirecektir:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">\u03c0\/2 \u2264 \u03d5 \u2264 tan<sup>-1<\/sup>(\u22121\/c)<\/td>\n<td style=\"text-align: right\">(19)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>ve cot<sup>2<\/sup>(\u03d5\/2) &lt; \u03bb &lt; 1 olur. Denklem (18) i \u03bb ve \u03d5 parametreleri ile yazd\u0131\u011f\u0131m\u0131zda<\/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}}_{{\\text{min}}}}=\\text{\u03bb}+\\frac{1}{2}\\frac{{\\left( {1-{{\\text{\u03bb}}^{2}}} \\right)\\sin \\text{\u03d5}}}{{\\sqrt{{1+{{\\text{\u03bb}}^{2}}+\\left( {1-{{\\text{\u03bb}}^{2}}} \\right)\\cos \\text{\u03d5}}}}}\u00a0<\/span><\/td>\n<td style=\"text-align: right\">(20)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>olacakt\u0131r. \u03bb ba\u011f\u0131ms\u0131z parametre oldu\u011fundan, en k\u00fc\u00e7\u00fck ba\u011flama a\u00e7\u0131s\u0131 de\u011ferinin maksimum olmas\u0131 i\u00e7in gerekli \u015fart d\u03bc<sub>min<\/sub>\/d\u03bb = 0 d\u0131r. Bu t\u00fcrevi s\u0131f\u0131r yapan \u03bb de\u011ferine \u03bb<sub>opt<\/sub> dersek, (20) denkleminin t\u00fcrevini al\u0131p d\u03bc<sub>min<\/sub>\/d\u03bb = 0 olur ise<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\">t<sup>2<\/sup>Q<sup>3<\/sup> + (1 \u2212 t<sup>2<\/sup>)Q<sup>2<\/sup> \u2212 (1\u00a0+ t<sup>2<\/sup> + t<sup>4<\/sup>)Q + 1 + t<sup>2<\/sup> = 0<\/td>\n<td style=\"text-align: right\">(21)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Bu denklemde Q = \u03bb<sub>opt<\/sub><sup>2<\/sup>t<sup>2<\/sup>\u00a0 ve\u00a0 t = tan(\u03d5\/2) dir. Denklem (21) in \u00fc\u00e7 k\u00f6k\u00fc:<\/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{Q}}_{1}}=-\\frac{1}{2}+\\frac{1}{2}\\sqrt{{5+4{{\\text{t}}^{2}}}} <\/span>\u00a0 \u00a0 \u00a0;\u00a0 \u00a0 \u00a0<span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{Q}}_{2}}=-\\frac{1}{2}-\\frac{1}{2}\\sqrt{{5+4{{\\text{t}}^{2}}}} <\/span>\u00a0 \u00a0 \u00a0;\u00a0 \u00a0 \u00a0<span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{Q}}_{3}}=-\\frac{1}{{{{\\text{t}}^{2}}}} <\/span><\/td>\n<td style=\"text-align: right\">(22)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Q pozitif olmas\u0131 gerekti\u011finden (\u03bb<sub>opt<\/sub><sup>2<\/sup>t<sup>2<\/sup>\u00a0daima pozitiftir), Q<sub>2<\/sub>\u00a0\u00e7\u00f6z\u00fcm de\u011fildir. Q<sub>3<\/sub>, de\u011ferinden \u03bb = 1\/t<sup>2<\/sup> elde edilir. Bu de\u011fere g\u00f6re elde edilen mekanizmada ba\u011flama a\u00e7\u0131s\u0131n\u0131n dik a\u00e7\u0131dan sapmas\u0131 maksimumdur (cos\u03bc<sub>min<\/sub> = 1) ve bu de\u011fer de optimum \u00e7\u00f6z\u00fcm de\u011fildir. Q<sub>1<\/sub> k\u00f6k\u00fcnden elde edilen \u03bb<sub>opt<\/sub>\u00a0de\u011feri (1\/t<sup>2<\/sup>, 1) aral\u0131\u011f\u0131ndad\u0131r ve ba\u011flama a\u00e7\u0131s\u0131n\u0131n dik a\u00e7\u0131dan sapmas\u0131n\u0131 en aza indirir. Bu de\u011fer:<\/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}}_{{\\text{opt}}}}^{2}=\\frac{{\\sqrt{{5+4{{\\text{t}}^{2}}}}-1}}{{2{{\\text{t}}^{2}}}} <\/span><\/td>\n<td style=\"text-align: right\">(21)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>dir ve aran\u0131lan \u00e7\u00f6z\u00fcm tektir.<\/p>\n<p><strong>\u00d6rnek:<\/strong><\/p>\n<p>Stroku s = 120 mm ve krank d\u00f6nme a\u00e7\u0131s\u0131 \u03d5 = 160<sup>0<\/sup>\u00a0istenilmektedir. Bu kinematik \u00f6zellikleri sa\u011flayan ve en iyi ba\u011flama a\u00e7\u0131s\u0131 \u00f6zelliklerine sahip krank biyel mekanizmas\u0131n\u0131 bulun.<\/p>\n<p>Denklem (23) den, \u03bb<sub>opt<\/sub> = 0.405185. (14), (15) ve (16) numaral\u0131 denklemler kullan\u0131larak birim strok i\u00e7in uzuv boyutlar\u0131: a<sub>2<\/sub> = 0.465542, a<sub>3<\/sub>\u00a0= 1.14896, c = 0.377378 dir. s = 120 mm strok oldu\u011funda ise<\/p>\n<p style=\"padding-left: 40px;text-align: center\">a<sub>2<\/sub> = 55.87 mm, a<sub>3<\/sub> = 137.88 mm, c = 42.81 mm<\/p>\n<p>Sonu\u00e7 mekanizma \u015fekilde g\u00f6r\u00fclmektedir. Mekanizman\u0131n minimum ba\u011flama a\u00e7\u0131s\u0131 \u03bc<sub>min<\/sub>\u00a0= 42.8\u00b0 dir.<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1161 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img5-6.gif\" alt=\"\" width=\"501\" height=\"284\" \/><\/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_69d879ec4c175\"><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/krankbiyel5_1.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\/krankbiyel5_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\/2021\/09\/krankbiyel5_2.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"570\" height=\"358\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/krankbiyel5_2.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><\/div><script id=\"su_image_carousel_69d879ec4c175_script\">if(window.SUImageCarousel){setTimeout(function() {window.SUImageCarousel.initGallery(document.getElementById(\"su_image_carousel_69d879ec4c175\"))}, 0);}var su_image_carousel_69d879ec4c175_script=document.getElementById(\"su_image_carousel_69d879ec4c175_script\");if(su_image_carousel_69d879ec4c175_script){su_image_carousel_69d879ec4c175_script.parentNode.removeChild(su_image_carousel_69d879ec4c175_script);}<\/script><\/p>\n<p>Elde edilen sonu\u00e7 \u00f6l\u00fc konumlar aras\u0131 kol d\u00f6nme a\u00e7\u0131s\u0131na g\u00f6re bulunabilir. Bu sonu\u00e7lar kitapta Abak 2 de g\u00f6r\u00fclmektedir. Krank biyel mekanizmas\u0131 uzuv boyutlar\u0131 (a<sub>2<\/sub>, a<sub>3<\/sub>, c) ve \u03bb<sub>opt<\/sub> de\u011feri ile birlikte en k\u00fc\u00e7\u00fck ba\u011flama a\u00e7\u0131s\u0131 de\u011feri mmin krank\u0131n \u00f6l\u00fc konumlar aras\u0131nda d\u00f6nme a\u00e7\u0131s\u0131na (\u03d5) g\u00f6re verilmi\u015ftir. Abak 3 te ise t\u00fcm sonu\u00e7lar eksantrikli\u011fin stroka oran\u0131 (c\/s) ve \u00f6l\u00fc konumlar artas\u0131nda kalan krank d\u00f6nme a\u00e7\u0131s\u0131na g\u00f6re (\u03d5) verilmektedir (yatay eksen c\/s olup lineer bir \u00f6l\u00e7ekte de\u011fildir).<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1162\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/Abak2-3.gif\" alt=\"\" width=\"1549\" height=\"2216\" \/><\/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\"><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><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch7\/7-3\/\"><img loading=\"lazy\" decoding=\"async\" width=\"38\" height=\"38\" class=\"wp-image-20\" style=\"width: 38px\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/04\/next_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>7.2 Krank-Biyel Mekanizmas\u0131 Makina tasar\u0131m\u0131nda yo\u011fun bir \u015fekilde kullan\u0131lan bir ba\u015fka mekanizmada krank-biyel mekanizmas\u0131d\u0131r. Genel olarak bir d\u00f6nme hareketini bir \u00f6teleme hareketine \u00e7evirmek i\u00e7in kullan\u0131ld\u0131\u011f\u0131 gibi bir \u00f6teleme hareketini d\u00f6nme hareketine \u00e7evirmek i\u00e7inde kullan\u0131labilir. \u015eekilde krank-biyel mekanizmas\u0131, de\u011fi\u015fken a\u00e7\u0131lar\u0131 ve &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch7\/7-2\/\"> <span class=\"screen-reader-text\">7-2<\/span> Devam\u0131n\u0131 Oku &raquo;<\/a><\/p>\n","protected":false},"author":7747,"featured_media":0,"parent":853,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-1134","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1134","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=1134"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1134\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/853"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/media?parent=1134"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}