{"id":826,"date":"2021-09-08T21:49:20","date_gmt":"2021-09-08T21:49:20","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=826"},"modified":"2021-09-30T09:03:26","modified_gmt":"2021-09-30T09:03:26","slug":"3-4","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch3\/3-4\/","title":{"rendered":"3-4"},"content":{"rendered":"<div id=\"pl-gb826-69d7c03a052f3\"  class=\"panel-layout\" ><div id=\"pg-gb826-69d7c03a052f3-0\"  class=\"panel-grid panel-no-style\" ><div id=\"pgc-gb826-69d7c03a052f3-0-0\"  class=\"panel-grid-cell\" ><div id=\"panel-gb826-69d7c03a052f3-0-0-0\" 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><b>3.4<\/b> <b>Mekanizmalarda Vekt\u00f6r Devreleri<\/b><\/h1>\n<p>Mekanizmalarda bulunan uzuvlar ile herhangi bir d\u00fczlemsel hareket yapan cisimler aras\u0131nda en \u00f6nemli fark, mekanizma uzuvlar\u0131 hareketlerini s\u0131n\u0131rlayan ve onlar\u0131 di\u011fer uzuvlara ba\u011flayan mafsallardan dolay\u0131, girdi parametreleri de\u011ferlerine g\u00f6re s\u0131n\u0131rland\u0131r\u0131lm\u0131\u015f bir hareket halindedirler. Birbirlerine mafsallar ile ba\u011fl\u0131 uzuvlar kapal\u0131 \u00e7okgenler olu\u015fturacaklard\u0131r. Bu \u00e7okgenlerin her birine\u00a0<strong>devre<\/strong>\u00a0diyece\u011fiz. Hareket analizinde temel yakla\u015f\u0131m\u0131m\u0131z\u0131n ba\u015flang\u0131\u00e7 noktas\u0131 bu devreleri matematiksel olarak ifade etmek olacakt\u0131r.<\/p>\n<p>Kinematik analize ba\u015flarken her bir uzuvla ilgili t\u00fcm boyutlar\u0131 bildi\u011fimizi kabul edece\u011fiz. Kinematik analiz sadece boyutlar\u0131n\u0131 bildi\u011fimiz (veya boyutlar\u0131n\u0131 bir varsay\u0131m ile tahmin etti\u011fimiz) uzuvlardan olu\u015fan bir mekanizma i\u00e7in yap\u0131labilir. Genel olarak d\u00f6ner mafsal eksenleri aras\u0131nda uzakl\u0131k veya bir d\u00f6ner mafsal orta noktas\u0131ndan ayn\u0131 uzvun ba\u011fland\u0131\u011f\u0131 bir kayar mafsal eksenine olan dik uzakl\u0131k uzuv boyutlar\u0131d\u0131r. Baz\u0131 durumlarda ise verilen boyutlar kullan\u0131larak istenilen ba\u015fka boyutlar\u0131n \u00e7\u0131kar\u0131lmas\u0131 gerekebilir. Ancak her durumda, mekanizma uzuvlar\u0131 hakk\u0131nda yeterli boyut bilgisinin oldu\u011fu kabul edilecektir.<\/p>\n<p>\u00d6nceki k\u0131s\u0131mda, bir uzvun konumu i\u00e7in bu uzvun \u00fczerinde iki noktan\u0131n konumunun bilinmesi gerekti\u011fini ve bu iki noktan\u0131n konumu bilinir ise, ba\u015fka her hangi bir noktan\u0131n konumunun bulanabilece\u011fini g\u00f6stermi\u015ftik. Bu durumda konum analizinde hedefimiz her bir uzuv \u00fczerinde iki noktan\u0131n konumunu belirlemek olacakt\u0131r. Nokta say\u0131s\u0131n\u0131 azaltmak, bir uzvun konumunu belirledi\u011fimizde di\u011fer uzvun konumunun belirlenmesinde kolayl\u0131k sa\u011flamak i\u00e7in bu noktalar\u0131 uzuvlar aras\u0131nda daima \u00e7ak\u0131\u015fan noktalar se\u00e7mek tercih edece\u011fimiz bir husus olacakt\u0131r. Bu \u015fekilde uzuv konumlar\u0131n\u0131 tan\u0131mlamak i\u00e7in gereken parametre say\u0131s\u0131 azalt\u0131labilecektir. Bir uzvun \u00fczerinde tan\u0131mlayaca\u011f\u0131m\u0131z vekt\u00f6r\u00fcn ba\u015flang\u0131\u00e7 noktas\u0131 ba\u015fka bir uzvun \u00fczerinde tan\u0131mlad\u0131\u011f\u0131m\u0131z vekt\u00f6r\u00fcn biti\u015f noktas\u0131 olabilecektir.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-829 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image7.gif\" alt=\"\" width=\"701\" height=\"294\" \/><\/p>\n<p 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:700px\" 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_69d7c03a080df\"><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/fourbar1T1.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\/fourbar1T1.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\/fourbar1T2-1.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\/fourbar1T2-1.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\/fourbar1T3.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"552\" height=\"400\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/fourbar1T3.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\/fourbar1T4.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"552\" height=\"400\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/fourbar1T4.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\/fourbar1T5.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"552\" height=\"400\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/fourbar1T5.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\/fourbar1T6.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"552\" height=\"400\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/fourbar1T6.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\/fourbar1T7.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"552\" height=\"400\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/fourbar1T7.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\/fourbar1T9-1.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"504\" height=\"376\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/fourbar1T9-1.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\/fourbar1T10.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\/fourbar1T10.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\/fourbar1T11.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\/fourbar1T11.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><\/div><script id=\"su_image_carousel_69d7c03a080df_script\">if(window.SUImageCarousel){setTimeout(function() {window.SUImageCarousel.initGallery(document.getElementById(\"su_image_carousel_69d7c03a080df\"))}, 0);}var su_image_carousel_69d7c03a080df_script=document.getElementById(\"su_image_carousel_69d7c03a080df_script\");if(su_image_carousel_69d7c03a080df_script){su_image_carousel_69d7c03a080df_script.parentNode.removeChild(su_image_carousel_69d7c03a080df_script);}<\/script><\/p>\n<p>Bir \u00f6rnek vermek i\u00e7in yukar\u0131da g\u00f6sterilen d\u00f6rt-\u00e7ubuk mekanizmas\u0131n\u0131 ele alal\u0131m. Bu mekanizmada A<sub>0<\/sub>\u00a0,1 ve 2 uzuvlar\u0131 \u00fczerinde; A, 2 ve 3 uzuvlar\u0131 \u00fczerinde; B, 3 ve 4 uzuvlar\u0131 \u00fczerinde ve B<sub>0<\/sub>\u00a0, 4 ve 1 uzuvlar\u0131 \u00fczerinde daima \u00e7ak\u0131\u015fan noktalard\u0131r. Bir an i\u00e7in \u015eekil B de g\u00f6sterildi\u011fi gibi daima \u00e7ak\u0131\u015fmas\u0131 gereken B<sub>3<\/sub>\u00a0ve B<sub>4<\/sub>\u00a0noktalar\u0131n\u0131 birbirinden ay\u0131rd\u0131\u011f\u0131m\u0131z\u0131 d\u00fc\u015f\u00fcnelim. Bu durumda iki a\u00e7\u0131k zincir elde edilecektir (A<sub>0<\/sub>AB ve B<sub>0<\/sub>B). Bu a\u00e7\u0131k zincirlerde uzuv say\u0131s\u0131 mekanizmada bulunan uzuv say\u0131s\u0131 kadard\u0131r. Bu uzuvlar\u0131n a\u00e7\u0131k zincir durumunda konumlar\u0131n\u0131 belirleyelim. Konumlar\u0131n\u0131 belirlemek i\u00e7in ilk olarak bir sabit referans ekseni se\u00e7memiz gerekir. Her hangi bir referans ekseni se\u00e7ebilir isek de, e\u011fer referans eksenimizin ba\u015flang\u0131\u00e7 noktas\u0131 A<sub>0<\/sub>\u00a0veya B<sub>0<\/sub>\u00a0dan ge\u00e7er ise, ek sabit de\u011ferlere ihtiyac\u0131m\u0131z olmayabilir. \u00d6rne\u011fin eksen ba\u015flang\u0131c\u0131n\u0131 A<sub>0<\/sub>\u00a0olarak ve pozitif x eksenini A<sub>0<\/sub>B<sub>0<\/sub>\u00a0y\u00f6n\u00fcnde se\u00e7er isek, elde edece\u011fimiz ili\u015fkilerin daha basit olaca\u011f\u0131 g\u00f6r\u00fclecektir (e\u011fer ba\u015fka her hangi bir sabit referans ekseni se\u00e7ilir ise, bu durumda se\u00e7im noktas\u0131na ba\u011fl\u0131 olarak bir uzvun veya noktan\u0131n konumu belirlenirken 3 yeni sabit parametre denklemlerde yer alacakt\u0131r). \u015eimdi se\u00e7mi\u015f oldu\u011fumuz referans tak\u0131m\u0131na g\u00f6re 2 uzvunun konumunu belirlemek istersek, 2 uzvu sabit uzva g\u00f6re sadece d\u00f6nme yapabilece\u011finden ve A<sub>0<\/sub>\u00a0noktas\u0131 2 ve 1 uzuvlar\u0131 aras\u0131nda daima \u00e7ak\u0131\u015fan nokta oldu\u011fundan, d\u00f6nme miktar\u0131n\u0131 belirleyen\u00a0\u03b8<sub>12<\/sub> parametresi de\u011ferini belirlersek 2 uzvunun konumunu belirlemi\u015f oluruz. E\u011fer 2 uzvu \u00fczerinde A noktas\u0131n\u0131n yerini bulmak istersek a<sub>2<\/sub> = |A<sub>0<\/sub>A|\u00a0biliniyor kabul edildi\u011finden (uzuv boyutu), A noktas\u0131n\u0131n koordinatlar\u0131 A (a<sub>2<\/sub>cos\u03b8<sub>12<\/sub>\u00a0, a<sub>2<\/sub>sin\u03b8<sub>12<\/sub>) olarak kolayca belirlenir (veya kutupsal olarak a<sub>2<\/sub>\u2220\u03b8<sub>12<\/sub>). A noktas\u0131 2 ve 3 noktalar\u0131 \u00fczerinde \u00e7ak\u0131\u015fan noktalar oldu\u011fundan 3 uzvunun konumunun belirlenmesi s\u0131ras\u0131nda A noktas\u0131 zaten\u00a0\u03b8<sub>12<\/sub>\u00a0parametresi ile bilindi\u011finden, \u015fimdi 3 uzvunun konumunu belirlemek i\u00e7in sadece\u00a0\u03b8<sub>13<\/sub>\u00a0parametresini belirlememiz yeterli olacakt\u0131r. Benzer bir \u015fekilde 4 uzvunun konumunu belirlemek i\u00e7in\u00a0\u03b8<sub>14<\/sub>\u00a0parametresinin belirlenmesi yeterlidir. Bu \u015fekilde elde etti\u011fimiz a\u00e7\u0131k kinematik zincirde b\u00fct\u00fcn uzuvlar\u0131n konumlar\u0131n\u0131n belirlenmesi \u00fc\u00e7 parametre gerekmektedir (\u03b8<sub>12<\/sub>\u00a0,\u00a0\u03b8<sub>13<\/sub>\u00a0ve\u00a0\u03b8<sub>14<\/sub>). G\u00f6r\u00fcld\u00fc\u011f\u00fc gibi bu parametreler mutlaka mafsal serbestlik derecelerine aittir.<\/p>\n<p>Her uzuv bir vekt\u00f6r ile g\u00f6sterilebildi\u011fine g\u00f6re, bu vekt\u00f6rlerin ba\u015flang\u0131\u00e7 ve biti\u015f noktalar\u0131n\u0131 mekanizmada daimi \u00e7ak\u0131\u015fan noktalar olarak se\u00e7elim ve 2 uzvu i\u00e7in <strong>A<sub>0<\/sub>A<\/strong>, 3 uzvu i\u00e7in\u00a0<strong>AB<\/strong>, 1 uzvu i\u00e7in\u00a0<strong>A<sub>0<\/sub>B<sub>0<\/sub><\/strong>\u00a0ve 4 uzvu i\u00e7in\u00a0<strong>AB<\/strong> vekt\u00f6rlerini tan\u0131mlayal\u0131m. Mekanizmam\u0131zda sadece d\u00f6ner mafsallar oldu\u011fu i\u00e7in de\u011fi\u015fken parametreler a\u00e7\u0131 parametreleri olup bu vekt\u00f6rlerin boyutlar\u0131 (\u015fiddeti) sabit uzuv boyutlar\u0131d\u0131r (a<sub>2<\/sub> = |A<sub>0<\/sub>A|, a<sub>3<\/sub> = |AB|, a<sub>1<\/sub> = |A<sub>0<\/sub>B<sub>0<\/sub>| ve a<sub>4<\/sub> = |B<sub>0<\/sub>B|) ve bu vekt\u00f6rlerin y\u00f6nleri (<strong>A<sub>0<\/sub>B<sub>0<\/sub><\/strong>\u00a0sabit vekt\u00f6r\u00fc hari\u00e7) de\u011fi\u015fken a\u00e7\u0131 parametreleri (\u03b8<sub>12<\/sub>\u00a0,\u00a0\u03b8<sub>13<\/sub>\u00a0ve\u00a0\u03b8<sub>14<\/sub>\u00a0) ile tan\u0131mlan\u0131r. A\u00e7\u0131k zincir olarak ele al\u0131nd\u0131\u011f\u0131nda, bu \u00fc\u00e7 parametreye verilen de\u011ferlere ba\u011fl\u0131 olarak B<sub>3<\/sub>\u00a0ve B<sub>4<\/sub>\u00a0noktalar\u0131 farkl\u0131 konumlarda olacakt\u0131r (ve bu sistemin serbestlik derecesi 3 d\u00fcr). B noktas\u0131n\u0131n konumu:<\/p>\n<p style=\"padding-left: 40px;text-align: center\"><strong>A<sub>0<\/sub>A<\/strong> + <strong>AB <\/strong>= <strong>A<sub>0<\/sub>B<sub>3<\/sub><\/strong> (1-2-3 a\u00e7\u0131k zinciri i\u00e7in)<br \/>\n<strong>A<sub>0<\/sub>B<sub>0<\/sub><\/strong> + <strong>B<sub>0<\/sub>B<\/strong> = <strong>A<sub>0<\/sub>B<sub>4<\/sub><\/strong> (1-4 a\u00e7\u0131k zinciri i\u00e7in)<\/p>\n<p>Ancak biliyoruz ki mekanizmada B noktas\u0131 3 ve 4 uzuvlar\u0131 \u00fczerinde daima \u00e7ak\u0131\u015fan iki noktad\u0131r. \u00d6yle ise her iki zincirden ayr\u0131 ayr\u0131 elde edilen\u00a0<strong>A<sub>0<\/sub>B<sub>3<\/sub><\/strong>\u00a0ve\u00a0<strong>A<sub>0<\/sub>B<sub>4<\/sub><\/strong>\u00a0vekt\u00f6rleri daima \u00e7ak\u0131\u015fan iki noktay\u0131 g\u00f6sterdiklerinden birbirlerine e\u015fit olmalar\u0131 gerekir (<strong>A<sub>0<\/sub>B<sub>3<\/sub>\u00a0<\/strong>=\u00a0<strong>A<sub>0<\/sub>B<sub>4<\/sub><\/strong>). Bu de\u011ferlendirme ile:<\/p>\n<p style=\"text-align: center\"><strong>A<sub>0<\/sub>A<\/strong> + <strong>AB<\/strong> = <strong>A<sub>0<\/sub>B<sub>0<\/sub><\/strong> +\u00a0<strong>B<sub>0<\/sub>B<\/strong><\/p>\n<p style=\"text-align: left\">vekt\u00f6r denklemi elde edilir. B noktas\u0131 daima \u00e7ak\u0131\u015fan nokta oldu\u011fundan bu denklem mekanizman\u0131n alabilece\u011fi her konum i\u00e7in ge\u00e7erli olmas\u0131 \u015fartt\u0131r. Aksi durumda mekanizmay\u0131 olu\u015fturan zincir kopmu\u015ftur (mekanizma yoktur).<\/p>\n<p>\u015eekilden g\u00f6r\u00fclece\u011fi gibi, d\u00f6rt-\u00e7ubuk mekanizmas\u0131nda bir kapal\u0131 devre vard\u0131r ve bu kapal\u0131 devreyi g\u00f6steren bir vekt\u00f6r denklemi elde edilmi\u015ftir. Bu denklem bize olu\u015fan zincirin kapal\u0131 bir zincir oldu\u011funu g\u00f6sterir (daima \u00e7ak\u0131\u015fan noktalar). Bu denkleme\u00a0<strong>devre kapal\u0131l\u0131k denklemi<\/strong>\u00a0veya\u00a0<strong>vekt\u00f6r devre denklemi<\/strong>\u00a0diyece\u011fiz. Bu denklem bir vekt\u00f6r denklemi olup denklemdeki de\u011fi\u015fkenler mutlaka mekanizmadaki mafsallar\u0131n serbestlik dereceleri ile ili\u015fkilidir. Bir vekt\u00f6r denkleminden iki skaler denklem elde edilece\u011finden, iki parametre de\u011feri bu denklemler kullan\u0131larak \u00e7\u00f6z\u00fclebilir. Devre kapal\u0131l\u0131k denklemimizde \u00fc\u00e7 parametre (\u03b8<sub>12<\/sub>,\u00a0\u03b8<sub>13<\/sub>\u00a0ve\u00a0\u03b8<sub>14<\/sub>) oldu\u011funa g\u00f6re, e\u011fer bu parametrelerden birisi tan\u0131mlanm\u0131\u015f ise (\u03b8<sub>12<\/sub>\u00a0olsun) di\u011fer iki parametre (\u03b8<sub>13<\/sub>\u00a0ve\u00a0\u03b8<sub>14<\/sub>) bu vekt\u00f6r devre denkleminden \u00e7\u00f6z\u00fclebilmelidir. Tan\u0131mlanmas\u0131 gereken parametre say\u0131s\u0131 mekanizman\u0131n serbestlik derecesine e\u015fittir. Dikkat edilecek husus, bu parametreler aras\u0131nda ili\u015fki basit lineer bir ili\u015fki olmay\u0131p trigonometrik fonksiyonlar\u0131 i\u00e7eren karma\u015f\u0131k bir ili\u015fkidir. Bu parametrelere <strong>konum de\u011fi\u015fkenleri<\/strong>\u00a0diyece\u011fiz.<\/p>\n<p>Devre kapal\u0131l\u0131k denkleminde bulunan vekt\u00f6rleri g\u00f6stermenin bir kolay yolu ise her bir vekt\u00f6r\u00fc karma\u015f\u0131k say\u0131 kullanarak tan\u0131mlamakt\u0131r. \u00d6rne\u011fin A<sub>0<\/sub>A vekt\u00f6r\u00fcn\u00fcn uzunlu\u011fu a<sub>2<\/sub>\u00a0ve yatay eksenle yapt\u0131\u011f\u0131 a\u00e7\u0131\u00a0\u03b8<sub>12<\/sub>\u00a0oldu\u011funa g\u00f6re:<\/p>\n<p style=\"text-align: center\"><strong>A<sub>0<\/sub>A<\/strong> = a<sub>2<\/sub>cos\u03b8<sub>12<\/sub>\u00a0+ ia<sub>2<\/sub>sin\u03b8<sub>12<\/sub><\/p>\n<p>veya Euler form\u00fcl\u00fc kullan\u0131larak \u00fcstel fonksiyon olarak:<\/p>\n<p style=\"text-align: center\"><strong>A<sub>0<\/sub>A<\/strong>\u00a0= a<sub>2<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup><\/p>\n<p>yaz\u0131labilir. Benzer bir \u015fekilde, uzuv boyutlar\u0131 a<sub>i<\/sub> olarak ( a<sub>1<\/sub>\u00a0= |A<sub>0<\/sub>B<sub>0<\/sub>| , a<sub>2<\/sub> = |A<sub>0<\/sub>A|, ve benzeri.) g\u00f6sterilir ise, vekt\u00f6r devre denklemi karma\u015f\u0131k say\u0131lar ile:<\/p>\n<p style=\"text-align: center\">a<sub>2<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup> + a<sub>3<\/sub>e<sup>i\u03b8<sub>13<\/sub><\/sup>\u00a0= a<sub>1<\/sub>\u00a0+ a<sub>4<\/sub>e<sup>i\u03b8<sub>14<\/sub><\/sup><\/p>\n<p>olarak yaz\u0131labilir. \u0130stendi\u011finde bu vekt\u00f6rler x ve y bile\u015fenleri kullan\u0131larak da yaz\u0131labilir. Bu durumda d\u00f6rt-\u00e7ubuk mekanizmas\u0131 i\u00e7in vekt\u00f6r devre denklemi:<\/p>\n<p style=\"text-align: center\">a<sub>2<\/sub>cos\u03b8<sub>12<\/sub><strong>i<\/strong> + a<sub>2<\/sub>sin\u03b8<sub>12<\/sub><strong>j<\/strong> + a<sub>3<\/sub>cos\u03b8<sub>13<\/sub><strong>i<\/strong> + a<sub>3<\/sub>sin\u03b8<sub>13<\/sub><strong>j<\/strong> = a<sub>1<\/sub><strong>i <\/strong>+ a<sub>4<\/sub>cos\u03b8<sub>14<\/sub><strong>i<\/strong> + a<sub>4<\/sub>sin\u03b8<sub>14<\/sub><strong>j<\/strong><\/p>\n<p>olacakt\u0131r. x ve y bile\u015fenleri ayr\u0131 ayr\u0131 e\u015fitlenir ise:<\/p>\n<p style=\"text-align: center\">a<sub>2<\/sub>cos\u03b8<sub>12<\/sub>\u00a0+ a<sub>3<\/sub>cos\u03b8<sub>13<\/sub>\u00a0= a<sub>1<\/sub>\u00a0+ a<sub>4<\/sub>cos\u03b8<sub>14<\/sub><br \/>\na<sub>2<\/sub>sin\u03b8<sub>12<\/sub>\u00a0+ a<sub>3<\/sub>sin\u03b8<sub>13<\/sub>\u00a0= a<sub>4<\/sub>sin\u03b8<sub>14<\/sub><\/p>\n<p>D\u00f6rt-\u00e7ubuk mekanizmas\u0131 i\u00e7in vekt\u00f6rlerin boyutlar\u0131 sabit olup a\u00e7\u0131sal y\u00f6nleri de\u011fi\u015fkendir. Serbestlik derecesi tan\u0131m\u0131na tekrar g\u00f6z atarsak, tan\u0131mlanmas\u0131 gereken parametre say\u0131s\u0131 mekanizman\u0131n tahrik edildi\u011fi mafsala ait olup serbestlik derecesi kadar parametre tan\u0131mlanmal\u0131d\u0131r. Bu parametreler d\u00f6rt-\u00e7ubuk mekanizmas\u0131nda oldu\u011fu gibi daima a\u00e7\u0131sal parametre olmas\u0131 gerekmez.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-830 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image8.gif\" alt=\"\" width=\"628\" height=\"225\" \/><\/p>\n<p>Farkl\u0131 bir basit \u00f6rnek olarak \u015eekil A da g\u00f6r\u00fclen krank-biyel mekanizmas\u0131n\u0131 ele alal\u0131m. B noktas\u0131nda 3 ve 4 uzuvlar\u0131n\u0131 birbirine ba\u011flayan d\u00f6ner mafsal\u0131 s\u00f6kt\u00fc\u011f\u00fcm\u00fcz\u00fc d\u00fc\u015f\u00fcnelim (\u015eekil B). 4 uzvu i\u00e7in yapm\u0131\u015f oldu\u011fumuz incelemeye benzer bir inceleme yapt\u0131\u011f\u0131m\u0131zda A<sub>0<\/sub>AB zincirinde bulunan 2 ve 3 uzuvlar\u0131n\u0131n konumlar\u0131 se\u00e7mi\u015f oldu\u011fumuz referans eksenlerine g\u00f6re\u00a0\u03b8<sub>12<\/sub>\u00a0ve\u00a0\u03b8<sub>13<\/sub>\u00a0a\u00e7\u0131sal konum parametrelerinin tan\u0131mlanmas\u0131 ile belirlenebilir. 4 uzvunun konumu ise, 1 uzvuna g\u00f6re kayar \u00e7ift ekseni y\u00f6n\u00fcnde \u00f6teleme yapaca\u011f\u0131ndan, kayar \u00e7ift ekseni y\u00f6n\u00fcnde tan\u0131mlanan s<sub>14<\/sub> konum de\u011fi\u015fkeni ile belirlenir. Bu konum de\u011fi\u015fkeni \u00f6teleme de\u011fi\u015fkenidir. Bu durumda elde edilen devre kapal\u0131l\u0131k denklemi<\/p>\n<p style=\"text-align: center\"><strong>A<sub>o<\/sub>A<\/strong>\u00a0+\u00a0<strong>AB<\/strong>\u00a0=\u00a0<strong>A<sub>o<\/sub>B<\/strong><\/p>\n<p>olur. Bu denklemde 3 de\u011fi\u015fken (\u03b8<sub>12<\/sub>,\u00a0\u03b8<sub>13<\/sub>\u00a0ve s<sub>14<\/sub>) bulunmakta, bunlardan birinin tan\u0131mlanmas\u0131 ile di\u011fer ikisinin elde edilebilmesi i\u00e7in bir vekt\u00f6r devre denklemi bulunmaktad\u0131r. Tan\u0131mlanmas\u0131 gereken ba\u011f\u0131ms\u0131z parametre, uygulamaya g\u00f6re, \u03b8<sub>12<\/sub>\u00a0(pompa gibi) veya s<sub>14<\/sub> (i\u00e7ten yanmal\u0131 motor gibi) olabilir. Bu \u00f6rnekte <strong>A<sub>o<\/sub>A<\/strong>\u00a0ve\u00a0<strong>AB<\/strong>\u00a0vekt\u00f6rlerinde boyut sabit y\u00f6n de\u011fi\u015fken iken,\u00a0<strong>A<sub>o<\/sub>B<\/strong> vekt\u00f6r\u00fcnde y bile\u015feni (c uzunlu\u011fu) sabit ve x bile\u015feni (s<sub>14<\/sub>) de\u011fi\u015fkendir. Karma\u015f\u0131k say\u0131lar ile bu vekt\u00f6rler yaz\u0131ld\u0131\u011f\u0131nda vekt\u00f6r devre denklemi<\/p>\n<p style=\"text-align: center\">a<sub>2<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup> + a<sub>3<\/sub>e<sup>i\u03b8<sub>13<\/sub><\/sup> = s<sub>14<\/sub> + ic<\/p>\n<p>olur.\u00a0Vekt\u00f6r devre denkleminde kullan\u0131lan de\u011fi\u015fkenler, tan\u0131mlanan boyutlar her zaman ayn\u0131 olmayabilir, problemi tan\u0131mlayan ki\u015fiye g\u00f6re farklar g\u00f6sterebilir. \u00d6rne\u011fin, krank biyel mekanizmas\u0131nda B mafsal\u0131n\u0131 s\u00f6kmektense A mafsal\u0131 s\u00f6k\u00fclebilir (alttaki \u015fekil). Bu durumda 3 uzvunun konumunun belirlenmesi i\u00e7in fakl\u0131 bir a\u00e7\u0131\u00a0\u03b8<sub>13<\/sub>\u2032 = \u2220xBA tan\u0131mlanmas\u0131 gerekebilir. Burada dikkat edilecek husus B mafsal\u0131n\u0131n s\u00f6k\u00fcld\u00fc\u011f\u00fcnde tan\u0131mlanan \u03b8<sub>13<\/sub>\u00a0parametresi ile\u00a0\u03b8<sub>13<\/sub>\u2032 aras\u0131nda sabit bir a\u00e7\u0131 fark\u0131 vard\u0131r (bu durumda 180\u00b0). Elde edilen devre kapal\u0131l\u0131k denklemi ise<\/p>\n<p style=\"text-align: center\"><strong>A<sub>o<\/sub>A<\/strong> = <strong>A<sub>o<\/sub>B<\/strong> +\u00a0<strong>BA<\/strong><\/p>\n<p>d\u0131r. Karma\u015f\u0131k say\u0131lar kullan\u0131larak yaz\u0131ld\u0131\u011f\u0131nda:<\/p>\n<p style=\"text-align: center\">a<sub>2<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup> = s<sub>14<\/sub> + ic + a<sub>3<\/sub>e<sup>i\u03b8<sub>13<\/sub>\u2032<\/sup><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-831\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image9.gif\" alt=\"\" width=\"366\" height=\"195\" \/><\/p>\n<p>Mekanizmalarda \u00e7ok say\u0131da devre bulunabilir ve her bir devre i\u00e7in bir devre kapal\u0131l\u0131k denklemi (vekt\u00f6r devre denklemi) yaz\u0131labilir.<\/p>\n<p><strong>\u00d6rnek:<\/strong> T\u0131klay\u0131n\u0131z: <span style=\"text-decoration: underline\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch3\/3-4\/3-4_ornek\/\"><span style=\"color: #0000ff\"><b>Alt\u0131 uzuvlu bir mekanizmadaki devreler<\/b><\/span><\/a><\/span><\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-832 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image10.gif\" alt=\"\" width=\"298\" height=\"195\" \/><\/p>\n<p>D\u00f6rt-\u00e7ubuk mekanizmas\u0131na bakt\u0131\u011f\u0131m\u0131zda, geometrik olarak<\/p>\n<p style=\"text-align: center\"><strong>A<sub>o<\/sub>A<\/strong> +\u00a0<strong>AB\u00a0<\/strong>=\u00a0<strong>A<sub>o<\/sub>B<\/strong><\/p>\n<p>\u015feklinde bir vekt\u00f6r denklemi yaz\u0131labilir. Ancak bu denklemde <b>A<sub>o<\/sub>B<\/b> vekt\u00f6r\u00fcn\u00fc inceledi\u011fimizde bu vekt\u00f6r\u00fcn hem y\u00f6n\u00fc ve hem de boyutu (x ve y bile\u015fkeleri) de\u011fi\u015fken olup bu de\u011fi\u015fkenler hi\u00e7 bir \u015fekilde bir mafsal serbestli\u011fi ile ili\u015fkili de\u011fildir. Bu denklem sadece di\u011fer iki vekt\u00f6r\u00fcn \u015fiddeti ve y\u00f6n\u00fc biliniyor ise, B noktas\u0131n\u0131n konumunu veren <strong>A<sub>o<\/sub>B<\/strong>\u00a0vekt\u00f6r\u00fcn\u00fc bulmaya yarar. Bu denklem ge\u00e7erli bir vekt\u00f6r denklemi ise de vekt\u00f6r devre denklemi de\u011fildir. Bu t\u00fcr denklemlerle devre kapal\u0131l\u0131k denklemleri aras\u0131nda en \u00f6nemli fark, de\u011fi\u015fken parametrelerin mafsal serbestlik derecesine ba\u011fl\u0131 olup olmad\u0131\u011f\u0131 ve bu denklemlerin bir mafsal\u0131n s\u00f6k\u00fclmesi ile yok olup olmad\u0131\u011f\u0131d\u0131r.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-833 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image11.gif\" alt=\"\" width=\"269\" height=\"220\" \/><\/p>\n<p>Benzer bir de\u011ferlendirme C biyel noktas\u0131 (3 uzvu) kullan\u0131larak yaz\u0131lan vekt\u00f6r denklemleri<\/p>\n<p style=\"text-align: center\"><strong>A<sub>o<\/sub>A<\/strong>\u00a0+ <strong>AC<\/strong> = <strong>A<sub>o<\/sub>C<\/strong>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0(i)<\/p>\n<p style=\"text-align: center\"><strong>AB<\/strong>\u00a0+\u00a0<strong>BC<\/strong>\u00a0=\u00a0<strong>AC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/strong>(ii)<\/p>\n<p style=\"text-align: left\">i\u00e7in de yap\u0131labilir. (i) ve (ii) vekt\u00f6r denklemleri , vekt\u00f6r devre denklemleri de\u011fildirler ve konum parametrelerinin \u00e7\u00f6z\u00fcm\u00fcnde kullan\u0131lmalar\u0131 m\u00fcmk\u00fcn de\u011fildir. (ii) denkleminde \u00fc\u00e7 vekt\u00f6r de ayn\u0131 uzuv \u00fczerindedir ve birbirlerine g\u00f6re sabit a\u00e7\u0131lar olu\u015ftururlar. \u0130kinci denklemde \u00fc\u00e7 vekt\u00f6r de ayn\u0131 uzuv \u00fczerindedir ve birbirlerine g\u00f6re sabit a\u00e7\u0131lar olu\u015ftururlar. Bu vekt\u00f6rlerden herhangi birinin konumu ve yatay ile yapt\u0131\u011f\u0131 a\u00e7\u0131 bilinir ise, di\u011fer vekt\u00f6rlerin ba\u015flang\u0131\u00e7 konumu ve yatay ile yapt\u0131klar\u0131 a\u00e7\u0131 bulunabilir.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-834 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image12.gif\" alt=\"\" width=\"485\" height=\"230\" \/><\/p>\n<p>Ba\u015fka baz\u0131 mekanizmalar\u0131n analizi i\u00e7in iki uzuv \u00fczerinde bulunan iki noktan\u0131n anl\u0131k \u00e7ak\u0131\u015fma konumu bir vekt\u00f6r\u00fcn ba\u015flang\u0131\u00e7 veya biti\u015f noktas\u0131 olarak kullan\u0131labilir. Kol-k\u0131zak mekanizmas\u0131nda g\u00f6r\u00fcld\u00fc\u011f\u00fc gibi, bu durumda kullan\u0131lmas\u0131 gereken parametre bir uzvun di\u011fer uzva g\u00f6re ba\u011f\u0131l konumunu belirleyen de\u011fi\u015fkendir.\u00a0<strong>B<sub>o<\/sub>A<\/strong> vekt\u00f6r\u00fc <strong>B<sub>o<\/sub>C<\/strong>\u00a0ve\u00a0<strong>CA<\/strong> olmak \u00fczere iki bile\u015fene ayr\u0131labilir. Bu bile\u015fenlere ay\u0131rma nedeni kullan\u0131lacak olan de\u011fi\u015fken parametrelerin mafsal serbestlikleri ile uyumlu hale getirilmesidir. \u00d6rne\u011fimizde <strong>B<sub>o<\/sub>C<\/strong> sabit uzunluk de\u011fi\u015fken a\u00e7\u0131 (B<sub>o<\/sub> da bulunan d\u00f6ner mafsal serbestisi) ve <strong>CA<\/strong> vekt\u00f6r\u00fc ise de\u011fi\u015fken uzunluk (kayar mafsal serbestisi) ve <strong>B<sub>o<\/sub>C<\/strong> ye g\u00f6re sabit bir a\u00e7\u0131d\u0131r. \u015eekilde g\u00f6r\u00fclen iki a\u00e7\u0131k zincir kullan\u0131larak (ve sonu\u00e7ta her iki a\u00e7\u0131k zincirden elde edilen A noktas\u0131n\u0131n daima \u00e7ak\u0131\u015fan nokta olmas\u0131 gerekti\u011fi g\u00f6z \u00f6n\u00fcne al\u0131narak) vekt\u00f6r devre denklemi<\/p>\n<p style=\"text-align: center\"><strong>A<sub>o<\/sub>A<\/strong>\u00a0=\u00a0<strong>A<sub>o<\/sub>B<sub>o<\/sub><\/strong>\u00a0+\u00a0<strong>B<sub>o<\/sub>C<\/strong>\u00a0+\u00a0<strong>CA<\/strong><\/p>\n<p>dir. Karma\u015f\u0131k say\u0131lar ile yaz\u0131ld\u0131\u011f\u0131nda:<\/p>\n<p style=\"text-align: center\">a<sub>2<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup> = a<sub>1<\/sub> + a<sub>4<\/sub>e<sup>i\u03b8<sub>14<\/sub><\/sup> + s<sub>43<\/sub>e<sup>i(\u03b8<sub>14 <\/sub>+<sub>\u00a0<\/sub>\u03b1<sub>4<\/sub>)<\/sup><\/p>\n<p>A<sub>2<\/sub>\u00a0ve A<sub>3<\/sub>\u00a0noktalar\u0131 daima \u00e7ak\u0131\u015fan noktalard\u0131r ve \u015eekil B de s\u00f6k\u00fclerek elde edilen iki a\u00e7\u0131k zincir kullan\u0131larak devre kapal\u0131l\u0131k denklemi yaz\u0131lm\u0131\u015ft\u0131r. A<sub>4<\/sub>\u00a0noktas\u0131 ise A<sub>2<\/sub>\u00a0ve A<sub>3<\/sub>\u00a0noktalar\u0131 ile sadece anl\u0131k \u00e7ak\u0131\u015fmakta,\u00a0\u0394t s\u00fcre sonra mekanizmada uzuvlar\u0131n yer de\u011fi\u015fimi ile 3 uzvu 4 uzvuna g\u00f6re kayar \u00e7ift ekseni y\u00f6n\u00fcnde\u00a0\u0394s kadar ilerlemesi sonucunda A<sub>2<\/sub>\u00a0ve A<sub>3<\/sub> noktalar\u0131 ile 4 uzvu \u00fczerinde farkl\u0131 bir nokta (A\u2032) anl\u0131k \u00e7ak\u0131\u015fm\u0131\u015f olacakt\u0131r (\u015eekil C).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-835 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image13.gif\" alt=\"\" width=\"520\" height=\"297\" \/><\/p>\n<p><strong>CA<\/strong>\u00a0vekt\u00f6r\u00fcn\u00fcn hem uzunlu\u011fu ve hemde a\u00e7\u0131sal konumu de\u011fi\u015fecektir. Ancak\u00a0<strong>CA<\/strong>\u00a0vekt\u00f6r\u00fcn\u00fcn\u00a0<strong>B<sub>o<\/sub>C<\/strong>\u00a0vekt\u00f6r\u00fcne g\u00f6re ba\u011f\u0131l a\u00e7\u0131sal konumu, her iki vekt\u00f6r ayn\u0131 uzuvda bulunan noktalar aras\u0131nda oldu\u011fundan daima sabittir. E\u011fer\u00a0<strong>B<sub>o<\/sub>C<\/strong>\u00a0vekt\u00f6r\u00fcn\u00fcn sabit x ekseni ile yapt\u0131\u011f\u0131 a\u00e7\u0131\u00a0\u03b8<sub>14<\/sub>\u00a0ise,\u00a0<strong>CA<\/strong>\u00a0vekt\u00f6r\u00fcn\u00fcn sabit x ekseni ile yapt\u0131\u011f\u0131 a\u00e7\u0131\u00a0\u03b8<sub>14<\/sub> + \u03b1<sub>4<\/sub>\u00a0olacakt\u0131r ve\u00a0\u03b1<sub>4<\/sub>\u00a0a\u00e7\u0131s\u0131 4 uzvu \u00fczerinde bulunan B<sub>o<\/sub>C ve CA do\u011frular\u0131n\u0131n aras\u0131nda kalan sabit bir uzuv parametresidir. Bu durumda devre kapal\u0131l\u0131k denkleminde konum de\u011fi\u015fkenleri \u03b8<sub>12<\/sub>,\u00a0\u03b8<sub>14<\/sub>\u00a0ve s<sub>43<\/sub> t\u00fcr.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-836 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image14.gif\" alt=\"\" width=\"478\" height=\"250\" \/><\/p>\n<p>Yukar\u0131da g\u00f6sterilmi\u015f olan mekanizmay\u0131 inceler isek, 2 ve 3 uzuvlar\u0131 \u00fczerinde daima \u00e7ak\u0131\u015fan A noktas\u0131n\u0131 s\u00f6k\u00fcp elde edilen a\u00e7\u0131k zincirleri kullanarak devre kapal\u0131l\u0131k denklemlerini yazd\u0131\u011f\u0131m\u0131zda, Bir \u00f6nceki \u00f6rnekte (kol-k\u0131zak) g\u00f6sterilmi\u015f olan mekanizma i\u00e7in elde edilen denklemlerin aynen elde edildi\u011fi g\u00f6r\u00fclecektir.Bu nedenle, mekanizma g\u00f6r\u00fcn\u00fcmleri farkl\u0131 olsa da kinematik a\u00e7\u0131dan ayn\u0131d\u0131r. Tasar\u0131m olarak 3 ve 4 uzuvlar\u0131 farkl\u0131 \u015fekillerde \u00fcretilmi\u015fseler de, her iki mekanizmada e\u011fer sabit uzuv parametreleri (a<sub>1<\/sub>, a<sub>2<\/sub>, a<sub>4<\/sub>,\u00a0\u03b1<sub>4<\/sub>) ayn\u0131 ise, ayn\u0131 harekete sahiptirler.<\/p>\n<p>Devre kapal\u0131l\u0131k denklemleri yaz\u0131l\u0131rken denklemin sadece o konumda ge\u00e7erli olmay\u0131p her konum i\u00e7in kullan\u0131laca\u011f\u0131 unutulmamal\u0131d\u0131r. Verilmi\u015f olan mekanizma \u00f6zel bir konumda olabilir veya \u00f6zel boyutlara sahip olabilir. \u00d6rne\u011fin yukar\u0131daki \u015fekilde mekanizman\u0131n \u00e7izildi\u011fi konumda 2 ve 1 uzuvlar\u0131 ayn\u0131 do\u011frultuda bulunmaktad\u0131r. 2 uzvu k\u0131sa bir s\u00fcre sonra bir miktar d\u00f6nebilir ve farkl\u0131 bir a\u00e7\u0131da olabilir. Bu durumlarda mekanizmay\u0131 bu \u00f6zel konumdan biraz kayd\u0131rarak \u00e7izmemizde yarar vard\u0131r . Di\u011fer baz\u0131 durumlarda ise devre kapal\u0131l\u0131k denklemleri m\u00fcmk\u00fcn olabilecek en basit \u015fekilde yaz\u0131labilmesi i\u00e7in mekanizman\u0131n \u00f6zel boyutlar\u0131 denklemde kullan\u0131larak yaz\u0131labilir. \u00d6rne\u011fin kol k\u0131zak mekanizmas\u0131nda g\u00f6sterilen mekanizma i\u00e7in e\u011fer \u03b1<sub>4<\/sub> a\u00e7\u0131s\u0131n\u0131 \u00f6nceden bir dik a\u00e7\u0131 olaca\u011f\u0131 biliniyor ise, bunun devre kapal\u0131l\u0131k denklemi kullan\u0131l\u0131rken kullan\u0131lmas\u0131nda yarar vard\u0131r. Bu durumda bu mekanizma i\u00e7in karma\u015f\u0131k say\u0131larla devre kapal\u0131l\u0131k denklemi:<\/p>\n<p style=\"text-align: center\">a<sub>2<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup> = a<sub>1<\/sub> + a<sub>4<\/sub>e<sup>i\u03b8<sub>14<\/sub><\/sup> + is<sub>43<\/sub>e<sup>i\u03b8<sub>14<\/sub><\/sup><\/p>\n<p>Verilen her ba\u011f\u0131ms\u0131z parametre i\u00e7in devre kapal\u0131l\u0131k denklemlerinden di\u011fer konum de\u011fi\u015fkenleri i\u00e7in \u00e7\u00f6z\u00fcm elde edilmesi m\u00fcmk\u00fcn olmayabilir. Bu durumlar mekanizman\u0131n o ba\u011f\u0131ms\u0131z parametre de\u011ferinde ba\u011flanamayaca\u011f\u0131n\u0131 veya mekanizman\u0131n o ba\u011f\u0131ms\u0131z parametre de\u011feri konumunda olamayaca\u011f\u0131n\u0131 g\u00f6sterir.<\/p>\n<p>T\u0131klay\u0131n\u0131z: <a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch3\/3-4\/3-4_bds\/\">&#8211; Ba\u011f\u0131ms\u0131z devre say\u0131s\u0131 ve dolay\u0131s\u0131 ile yaz\u0131lmas\u0131 gerekli ba\u011f\u0131ms\u0131z devre denklemleri say\u0131s\u0131 \u00f6nceden belirlenebilir &#8211;<\/a><\/p>\n<p>D\u00fczlemsel mekanizmalar i\u00e7in mekanizmada bulunan devre say\u0131s\u0131 (L) kadar vekt\u00f6r devre denklemi yaz\u0131ld\u0131\u011f\u0131nda, devre say\u0131s\u0131n\u0131n iki kat\u0131 kadar skaler denklem elde edilmi\u015ftir. Denklemlerde kullan\u0131lan konum de\u011fi\u015fkenlerinden F serbestlik derecesine kar\u015f\u0131 gelen parametreler \u00f6nceden belirlenmelidir (bunlar ba\u011f\u0131ms\u0131z konum de\u011fi\u015fkenleridir). Bu denklemlerde ba\u011f\u0131ms\u0131z konum de\u011fi\u015fkenleri d\u0131\u015f\u0131nda e\u011fer 2L kadar konum de\u011fi\u015fkeni bulunuyor ise, teorik olarak mekanizmada hareket belirlidir ve devre kapal\u0131l\u0131k denklemleri kullan\u0131larak di\u011fer konum de\u011fi\u015fkenleri de\u011ferleri o konum i\u00e7in bulunabilir. Ba\u011f\u0131ms\u0131z konum de\u011fi\u015fkeni verilen s\u0131n\u0131rlar i\u00e7inde belirli aral\u0131klarla de\u011fi\u015ftirilir ve di\u011fer konum de\u011fi\u015fkenleri de\u011ferleri de\u011fi\u015fik ba\u011f\u0131ms\u0131z de\u011fi\u015fken de\u011ferleri i\u00e7in bulunur ise, incelemi\u015f oldu\u011fumuz mekanizman\u0131n t\u00fcm \u00e7al\u0131\u015fma aral\u0131\u011f\u0131nda kinematik analizi yap\u0131lm\u0131\u015f olur. E\u011fer ba\u011f\u0131ms\u0131z parametre bir giri\u015f kolu ise (krank) kol a\u00e7\u0131s\u0131 0 ile 360\u00ba aral\u0131\u011f\u0131nda de\u011fi\u015ftirilir. Ba\u011f\u0131ms\u0131z parametre bir hidrolik veya pn\u00f6matik pistonun hareketi ise, hareket s\u0131n\u0131rlar\u0131 pistonun kapal\u0131 boyutundan ba\u015flayarak pistonun \u00f6nceden bilinen stroku kadar de\u011fi\u015febilir.<\/p>\n<p>Devre kapal\u0131l\u0131k denklemlerinden konum de\u011fi\u015fkenleri \u00e7\u00f6z\u00fcld\u00fckten sonra her hangi bir uzvun \u00fczerinde bulunan her hangi bir noktan\u0131n konumu bu konum parametreleri kullan\u0131larak ba\u011f\u0131ms\u0131z de\u011fi\u015fken de\u011ferine g\u00f6re kolayl\u0131kla bulunabilir.<\/p>\n<p><span style=\"color: #cc0000\"><b><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-19\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/04\/important.gif\" alt=\"\" width=\"28\" height=\"27\" \/> Devre kapal\u0131l\u0131k denklemlerinin elde edilmesi i\u00e7in her zaman mekanizmada mafsallar\u0131n s\u00f6k\u00fclm\u00fc\u015f halde g\u00f6sterilmesine gerek yoktur. Devaml\u0131 \u00e7ak\u0131\u015fan iki noktay\u0131 birbirinden ay\u0131rma (s\u00f6kme) ve sonra birle\u015ftirme kavramsal olarak yap\u0131lmas\u0131 gereken bir i\u015flemdir. Ba\u015flang\u0131\u00e7 \u00f6rneklerinde konunun g\u00f6rsel olarak anlat\u0131labilmesi i\u00e7in, mafsallar s\u00f6k\u00fcl\u00fc olarak g\u00f6sterilmi\u015ftir.<\/b><\/span><\/p>\n<p>A\u015fa\u011f\u0131da de\u011fi\u015fik mekanizmalar i\u00e7in yaz\u0131lm\u0131\u015f olan devre kapal\u0131l\u0131k denklemleri g\u00f6r\u00fclmektedir. Sizlerin bu denklemlerin yaz\u0131lmas\u0131 s\u0131ras\u0131nda hangi mafsal\u0131n s\u00f6k\u00fcl\u00fcp tak\u0131ld\u0131\u011f\u0131n\u0131 belirlemeniz gerekmektedir.<\/p>\n<p>Bilhassa son y\u0131llarda geli\u015ftirilmi\u015f olan paket programlar\u0131n kullan\u0131lmas\u0131 s\u0131ras\u0131nda en \u00f6nemli husus bu devre denklemlerinin do\u011fru de\u011fi\u015fkenler kullan\u0131larak do\u011fru yaz\u0131lmas\u0131d\u0131r. G\u00f6receksiniz ki e\u011fer devre denklemini yanl\u0131\u015f yazar isek yapaca\u011f\u0131m\u0131z di\u011fer t\u00fcm hesaplar yanl\u0131\u015f olacakt\u0131r. Bu y\u00f6ntem ile, somut bir mekanizman\u0131n matematiksel denklemleri yaz\u0131ld\u0131\u011f\u0131nda soyut matematiksel bir probleme indirgenebilmi\u015ftir. Bundan sonra \u00e7e\u015fitli y\u00f6ntemlerle bu denklemleri \u00e7\u00f6zebiliriz.<\/p>\n<p><strong>\u00d6rnek I:<\/strong><\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-838 size-full aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image16.gif\" alt=\"\" width=\"476\" height=\"319\" \/><\/p>\n<p style=\"text-align: center\" align=\"center\">(a<sub>2<\/sub> = |A<sub>0<\/sub>A|, a<sub>3<\/sub> = |AB|, a<sub>4<\/sub> = |BC|)<\/p>\n<p style=\"text-align: center\" align=\"center\"><strong>A<sub>0<\/sub>A<\/strong>\u00a0+ <strong>AB<\/strong> +\u00a0<strong>BC<\/strong>\u00a0= <strong>A<sub>0<\/sub>C<\/strong>\u00a0 \u21d2\u00a0 a<sub>2<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup> + a<sub>3<\/sub>e<sup>i(\u03b8<sub>13<\/sub> \u2212 \u03b1<sub>3<\/sub>)<\/sup>\u00a0+ a<sub>4<\/sub>e<sup>i\u03b8<sub>14<\/sub><\/sup> = s<sub>15<\/sub> + ia<sub>1<\/sub><\/p>\n<p style=\"text-align: center\" align=\"center\"><strong>A<sub>0<\/sub>A<\/strong>\u00a0+\u00a0<strong>AD <\/strong>= <strong>A<sub>0<\/sub>D<sub>0<\/sub><\/strong>\u00a0+ <strong>D<sub>0<\/sub>D<\/strong> \u21d2\u00a0 a<sub>2<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup> + s<sub>36<\/sub>e<sup>i\u03b8<sub>13<\/sub><\/sup> = c<sub>1<\/sub> + ib<sub>1<\/sub> \u2212 ia<sub>6<\/sub>e<sup>i\u03b8<sub>13<\/sub><\/sup><\/p>\n<p><strong>\u00d6rnek II:<\/strong><\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1483 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image17.png\" alt=\"\" width=\"204\" height=\"356\" srcset=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image17.png 204w, https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image17-172x300.png 172w, https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image17-100x175.png 100w, https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image17-150x262.png 150w, https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image17-200x349.png 200w\" sizes=\"auto, (max-width: 204px) 100vw, 204px\" \/><\/p>\n<p style=\"text-align: center\" align=\"center\">(c<sub>1<\/sub> = |PB<sub>0<\/sub>|, a<sub>2<\/sub> = |A<sub>0<\/sub>A|, a<sub>4<\/sub> = |BC|, a<sub>5<\/sub> = |B<sub>0<\/sub>B|)<\/p>\n<p style=\"text-align: center\" align=\"center\"><strong>A<sub>0<\/sub>A<\/strong> = <strong>A<sub>0<\/sub>B<sub>0<\/sub><\/strong>\u00a0+<strong>B<sub>0<\/sub>B <\/strong>+ <strong>BA<\/strong>\u00a0 \u21d2\u00a0 a<sub>2<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup> = c<sub>1<\/sub> \u2212 ia<sub>1<\/sub> + a<sub>5<\/sub>e<sup>i\u03b8<sub>15<\/sub><\/sup> + s<sub>43<\/sub>e<sup>i\u03b8<sub>14<\/sub><\/sup><\/p>\n<p style=\"text-align: center\" align=\"center\"><strong>A<sub>0<\/sub>C <\/strong>=\u00a0<strong>A<sub>0<\/sub>B<sub>0<\/sub><\/strong> +\u00a0<strong>B<sub>0<\/sub>B<\/strong> +\u00a0<strong>BC<\/strong>\u00a0 \u21d2\u00a0 s<sub>16<\/sub> + ib<sub>1<\/sub>\u00a0= c<sub>1<\/sub> \u2212 ia<sub>1<\/sub> + a<sub>5<\/sub>e<sup>i\u03b8<sub>15<\/sub><\/sup> + a<sub>4<\/sub>e<sup>i\u03b8<sub>14<\/sub><\/sup><\/p>\n<p style=\"text-align: left\" align=\"center\"><strong>\u00d6rnek III:<\/strong><\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-840 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image18.gif\" alt=\"\" width=\"441\" height=\"351\" \/><\/p>\n<p style=\"text-align: center\" align=\"center\"><strong>B<sub>0<\/sub>B <\/strong>+ <strong>BC <\/strong>+ <strong>CD <\/strong>= <strong>B<sub>0<\/sub>A<sub>0<\/sub><\/strong> +\u00a0<strong>A<sub>0<\/sub>A <\/strong>+ <strong>AD<\/strong>\u00a0 \u21d2\u00a0 a<sub>3<\/sub>e<sup>i\u03b8<sub>13<\/sub><\/sup> + a<sub>4<\/sub>e<sup>i\u03b8<sub>14<\/sub><\/sup> + a<sub>6<\/sub>e<sup>i\u03b8<sub>16<\/sub><\/sup> = a<sub>1<\/sub> + ib<sub>1<\/sub> + a<sub>2<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup> + a<sub>7<\/sub>e<sup>i\u03b8<sub>17<\/sub><\/sup><\/p>\n<p style=\"text-align: center\" align=\"center\"><strong>B<sub>0<\/sub>B <\/strong>+ <strong>BC <\/strong>= <strong>B<sub>0<\/sub>C<sub>0<\/sub><\/strong> +\u00a0<strong>C<sub>0<\/sub>C<\/strong>\u00a0 \u21d2\u00a0 a<sub>3<\/sub>e<sup>i\u03b8<sub>13<\/sub><\/sup> + a<sub>4<\/sub>e<sup>i\u03b8<sub>14<\/sub><\/sup> = \u2212d<sub>1<\/sub> + ic<sub>1<\/sub> + a<sub>5<\/sub>e<sup>i\u03b8<sub>15<\/sub><\/sup><\/p>\n<p>Ayr\u0131ca di\u015fli \u00e7iftten dolay\u0131 : r<sub>3<\/sub>\u03b8<sub>13<\/sub> = \u2212r<sub>2<\/sub>(\u03b8<sub>12<\/sub> \u2212 \u03b1<sub>2<\/sub>) (\u03b8<sub>13<\/sub>\u00a0= 0 iken \u03b8<sub>12<\/sub>\u00a0= \u03b1<sub>2<\/sub>)<\/p>\n<p>Mekanizmalarda konum analizi, g\u00f6r\u00fcld\u00fc\u011f\u00fc gibi, devre kapal\u0131l\u0131k denklemlerinin \u00e7\u00f6z\u00fcm\u00fcne indirgenmi\u015ftir. Konum analizinin basit fakat \u00e7ok etkili bir y\u00f6ntemi, incelenmesi istenilen bir mekanizman\u0131n, kartondan, tahtadan veya herhangi kolay bulunabilir uygun bir maddeden basit \u00f6l\u00e7ekli modelini yapmakt\u0131r (son y\u0131llarda h\u0131zl\u0131 prototipleme y\u00f6ntemleri de bu model yap\u0131m\u0131nda kullan\u0131lmaktad\u0131r). Bu tip bir modelin yap\u0131lmas\u0131 konumuz d\u0131\u015f\u0131d\u0131r. Burada devre kapal\u0131l\u0131k denklemlerinin \u00e7\u00f6z\u00fcm y\u00f6ntemleri a\u00e7\u0131klanacakt\u0131r.<\/p>\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n<p>  <a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch3\/3-3\/\" 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\/ch3\/\" data-type=\"page\" data-id=\"52\"><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><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch3\/3-5\/\" data-type=\"page\" data-id=\"92\"><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>3.4 Mekanizmalarda Vekt\u00f6r Devreleri Mekanizmalarda bulunan uzuvlar ile herhangi bir d\u00fczlemsel hareket yapan cisimler aras\u0131nda en \u00f6nemli fark, mekanizma uzuvlar\u0131 hareketlerini s\u0131n\u0131rlayan ve onlar\u0131 di\u011fer uzuvlara ba\u011flayan mafsallardan dolay\u0131, girdi parametreleri de\u011ferlerine g\u00f6re s\u0131n\u0131rland\u0131r\u0131lm\u0131\u015f bir hareket halindedirler. Birbirlerine mafsallar ile &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch3\/3-4\/\"> <span class=\"screen-reader-text\">3-4<\/span> Devam\u0131n\u0131 Oku &raquo;<\/a><\/p>\n","protected":false},"author":7747,"featured_media":0,"parent":370,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-826","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/826","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=826"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/826\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/370"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/media?parent=826"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}