{"id":859,"date":"2021-09-09T09:29:47","date_gmt":"2021-09-09T09:29:47","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=859"},"modified":"2022-09-27T09:16:39","modified_gmt":"2022-09-27T09:16:39","slug":"3-5","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch3\/3-5\/","title":{"rendered":"3-5"},"content":{"rendered":"<div id=\"pl-gb859-69d7c0146a1e0\"  class=\"panel-layout\" ><div id=\"pg-gb859-69d7c0146a1e0-0\"  class=\"panel-grid panel-no-style\" ><div id=\"pgc-gb859-69d7c0146a1e0-0-0\"  class=\"panel-grid-cell\" ><div id=\"panel-gb859-69d7c0146a1e0-0-0-0\" 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>3.5<\/b> Vekt\u00f6r Devre Denklemlerinin Grafik \u00c7\u00f6z\u00fcm\u00fc<\/h1>\n<p>Bir mekanizman\u0131n kinematik analizini yaparken mekanizmay\u0131 m\u00fcmk\u00fcn olan en basit \u015fekilde \u00e7izmemiz yapaca\u011f\u0131m\u0131z i\u015flerini kolayla\u015ft\u0131r\u0131r. Bu nedenle uzuvlar bir \u00e7izgi ile g\u00f6sterilebilir. Bu \u00e7izgilerin u\u00e7lar\u0131 bir sembolle g\u00f6sterilmi\u015f oldu\u011fundan, bu \u00e7izgiler ger\u00e7ek anlamda vekt\u00f6rd\u00fcrler ve mutlaka bir ok \u015feklinde g\u00f6sterilmelerine gerek yoktur. Yani genel olarak bakt\u0131\u011f\u0131m\u0131zda, mafsal merkezlerini (daima \u00e7ak\u0131\u015fan noktalar\u0131) de\u011fi\u015fik semboller (harfler) kullanarak i\u015faretledi\u011fimizde, elde etti\u011fimiz \u015fekil, vekt\u00f6r devre denklemlerinin grafik g\u00f6sterimidir. Bu nedenle devre kapal\u0131l\u0131k denklemleri yaz\u0131labilmi\u015ftir. Grafik \u00e7\u00f6z\u00fcm i\u00e7in yap\u0131lmas\u0131 gereken, bu \u015feklin belirli bir \u00f6l\u00e7ekle ka\u011f\u0131t \u00fczerine \u00e7izilmesidir (veya bir bilgisayar paket program\u0131 kullanarak bilgisayar ekran\u0131nda benzer bir \u015fekilde \u00e7izim yap\u0131labilir. Bu durumda mekanizma ne kadar k\u00fc\u00e7\u00fck veya ne kadar b\u00fcy\u00fck olursa olsun g\u00f6r\u00fcnt\u00fcy\u00fc \u00f6l\u00e7eklendirebildi\u011fimizden ger\u00e7ek boyutlar kullan\u0131labilir).<\/p>\n<p>Birisi sabit uzuv olmak \u00fczere size d\u00f6rt \u00e7ubuk verilir ve bunlar\u0131 bir zincir haline getirmeniz istenirse, \u00e7\u00f6z\u00fcm tek de\u011fildir.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-866 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-1.gif\" alt=\"\" width=\"594\" height=\"480\" \/><\/p>\n<p>G\u00f6r\u00fcld\u00fc\u011f\u00fc gibi \u00fc\u00e7 farkl\u0131 mekanizma elde edilmi\u015ftir. Sabit uzva ba\u011fl\u0131 uzuvlar\u0131n sa\u011f-sol yer de\u011fi\u015ftirmesi yeni bir mekanizma de\u011fildir (sadece ayn\u0131 mekanizman\u0131n \u00f6n veya arka g\u00f6r\u00fcn\u00fc\u015f\u00fc olur). \u00d6yle ise bir mekanizman\u0131n tan\u0131mlanm\u0131\u015f olmas\u0131 i\u00e7in sadece uzuv boyutlar\u0131n\u0131n bilinmesi yeterli de\u011fildir. Hangi uzvun hangi uzva hangi mafsalla ba\u011fl\u0131 oldu\u011fu da bilinmelidir.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-867 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-2.gif\" alt=\"\" width=\"527\" height=\"229\" \/><\/p>\n<p>Mekanizman\u0131n monte edilmi\u015f bir konumu biliniyor ise, hareket s\u0131ras\u0131nda uzuvlar\u0131n gelebilece\u011fi konumlar s\u0131n\u0131rland\u0131r\u0131lm\u0131\u015f ve mekanizman\u0131n hareketi belirlenmi\u015f olacakt\u0131r. \u015eimdi, bu mekanizman\u0131n ekseni A<sub>0<\/sub> olan d\u00f6ner mafsal\u0131 (2 uzvunu) \u03b8<sub>12<\/sub> a\u00e7\u0131s\u0131ndan \u03b8<sub>12<\/sub>\u2032 a\u00e7\u0131s\u0131na getirdi\u011fimizde di\u011fer uzuvlar\u0131n konumunu bulal\u0131m.<\/p>\n<p>A noktas\u0131n\u0131n yeni konum vekt\u00f6r\u00fc (<strong>A<sub>0<\/sub>A\u2032<\/strong>) \u03b8<sub>12<\/sub>\u2032 verildi\u011finde \u015fiddet ve y\u00f6n\u00fc bilinmektedir. Vekt\u00f6r devre denklemi:<\/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>A noktas\u0131 A&#8217; (biliniyor) ve B noktas\u0131 B&#8217; (bilinmiyor) ile de\u011fi\u015ftirildi\u011finde, B noktas\u0131 daima \u00e7ak\u0131\u015fan nokta oldu\u011fundan bu yeni konum i\u00e7in:<\/p>\n<p style=\"text-align: center\"><strong>A<sub>0<\/sub>A\u2032<\/strong>+ <strong>AB\u2032<\/strong>\u00a0= <strong>A<sub>0<\/sub>B<sub>0<\/sub><\/strong>\u00a0+ <strong>B<sub>0<\/sub>B\u2032<\/strong><\/p>\n<p>olacakt\u0131r. Bu denklem ayn\u0131 zamanda:<\/p>\n<p style=\"text-align: center\"><strong>A\u2032B\u2032<\/strong>\u00a0= <strong>A<sub>0<\/sub>B<sub>0<\/sub><\/strong> \u2212 <strong>A<sub>0<\/sub>A\u2032<\/strong>\u00a0+ <strong>B<sub>0<\/sub>B\u2032<\/strong><\/p>\n<p>veya:<\/p>\n<p style=\"text-align: center\"><strong>A\u2032B\u2032<\/strong>\u00a0= <strong>A\u2032A<sub>0<\/sub><\/strong>\u00a0+ <strong>A<sub>0<\/sub>B<\/strong><sub><strong>0<\/strong>\u00a0<\/sub>+ <strong>B<sub>0<\/sub>B\u2032<\/strong><\/p>\n<p>\u015feklinde yaz\u0131labilecektir.\u00a0<strong>A\u2032A<sub>0<\/sub><\/strong>\u00a0+ <strong>A<sub>0<\/sub>B<sub>0<\/sub><\/strong>\u00a0vekt\u00f6r toplam\u0131\u00a0<strong>A\u2032B<sub>0<\/sub><\/strong>\u00a0vekt\u00f6r\u00fcd\u00fcr.\u00a0<strong>A\u2032A<sub>0<\/sub><\/strong>\u00a0ve\u00a0<strong>A<sub>0<\/sub>B<sub>0<\/sub><\/strong>\u00a0vekt\u00f6rleri \u015fiddet ve y\u00f6n olarak bilinmektedir.\u00a0<strong>A\u2032B\u2032<\/strong>\u00a0ve\u00a0<strong>B<sub>0<\/sub>B\u2032<\/strong> vekt\u00f6rlerinin boyutlar\u0131 bilinmektedir. Bu vekt\u00f6rlerin ba\u015flang\u0131\u00e7 noktalar\u0131 olan A\u2032 ve B<sub>0<\/sub> noktalar\u0131n\u0131n konumu bilindi\u011fine g\u00f6re, B noktas\u0131n\u0131n yeni konumu B\u2032 n\u00fc belirlemek i\u00e7in A\u2032 merkezli AB yar\u0131 \u00e7apl\u0131 bir yay ile B<sub>0<\/sub>\u00a0merkezli B<sub>0<\/sub>B yar\u0131 \u00e7apl\u0131 bir yay \u00e7izdi\u011fimizde bu yaylar\u0131n kesi\u015ftikleri nokta B\u2032 konumunu belirleyecektir.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-868 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-3.gif\" alt=\"\" width=\"484\" height=\"593\" \/><\/p>\n<p>Bu iki yay\u0131n iki kesi\u015fme noktas\u0131 (B\u2032 ve B\u2033) olacakt\u0131r. Ancak bu kesi\u015fme noktalar\u0131ndan birine mekanizma ilk verilmi\u015f olan A<sub>0<\/sub>ABB<sub>0<\/sub> konumundan ge\u00e7ebilmesi i\u00e7in mafsallardan birinin s\u00f6k\u00fcl\u00fcp bir uzuv hareket ettirildikten sonra yeniden ba\u011flanmas\u0131 ile m\u00fcmk\u00fcn olacakt\u0131r. E\u011fer bu iki daire kesi\u015fmez ise, mekanizma verilmi\u015f olan yeni \u03b8<sub>12<\/sub>\u2032 a\u00e7\u0131s\u0131 i\u00e7in monte edilmi\u015f olarak bulunamaz.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-869\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-4.gif\" alt=\"\" width=\"768\" height=\"294\" \/><\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1565 size-full\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-5-1.gif\" alt=\"\" width=\"354\" height=\"245\" \/><\/p>\n<p>Ba\u015flang\u0131\u00e7ta, ayn\u0131 uzuvlar yukar\u0131da g\u00f6sterildi\u011fi gibi monte edilmi\u015f olarak verilse ve 2 uzvu \u03b8<sub>12<\/sub> a\u00e7\u0131s\u0131ndan \u03b8<sub>12<\/sub>\u2032 a\u00e7\u0131s\u0131na getirdi\u011fimizde di\u011fer uzuvlar\u0131n konumunu bulmak ister isek, bu sefer \u00e7\u00f6z\u00fcm farkl\u0131 olacakt\u0131r.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-871 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-6.gif\" alt=\"\" width=\"461\" height=\"339\" \/><\/p>\n<p>G\u00f6r\u00fcld\u00fc\u011f\u00fc gibi A\u2032 den AB yar\u0131\u00e7apl\u0131 daire ve B<sub>0<\/sub>\u00a0dan B<sub>0<\/sub>B yar\u0131\u00e7apl\u0131 daireler B\u2032 ve B\u2033 noktalar\u0131nda kesi\u015fir. Mekanizman\u0131n verilen A<sub>0<\/sub>ABB<sub>0<\/sub>\u00a0konumundan A<sub>0<\/sub>A\u2032B\u2032B<sub>0<\/sub> konumuna ge\u00e7ebilmesi i\u00e7in demonte edilmesi gerekecektir. Bu sefer verilen ba\u015flang\u0131\u00e7 konumuna g\u00f6re \u00e7\u00f6z\u00fcm A<sub>0<\/sub>A\u2032B\u2033B<sub>0<\/sub>\u00a0konumu olacakt\u0131r.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-872 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-7.gif\" alt=\"\" width=\"489\" height=\"371\" \/><\/p>\n<p>Bu y\u00f6ntem iki kenar\u0131 ve kenarlar aras\u0131nda kalan a\u00e7\u0131s\u0131 verilen bir \u00fc\u00e7genin \u00fc\u00e7\u00fcnc\u00fc kenar\u0131n\u0131n ve a\u00e7\u0131lar\u0131n\u0131n bulunmas\u0131 (AA<sub>0<\/sub>B<sub>0<\/sub>\u00a0\u00fc\u00e7geni: A<sub>0<\/sub>A ve A<sub>0<\/sub>B<sub>0<\/sub> kenar\u0131 ile \u03b8<sub>12<\/sub> a\u00e7\u0131s\u0131 veriliyor, \u00fc\u00e7genin bilinmeyen B<sub>0<\/sub>A kenar\u0131 ve B<sub>0<\/sub>A n\u0131n A<sub>0<\/sub>B<sub>0<\/sub>\u00a0ile yapt\u0131\u011f\u0131 a\u00e7\u0131 belirleniyor) ve ikinci olarak \u00fc\u00e7 kenar\u0131 bilinen bir \u00fc\u00e7genin (ABB<sub>0<\/sub>\u00a0\u00fc\u00e7geni: AB, BB<sub>0<\/sub>\u00a0uzuv boyutlar\u0131 oldu\u011fundan ve AB<sub>0<\/sub>\u00a0ilk \u00fc\u00e7genin \u00e7\u00f6z\u00fcm\u00fc sonucu biliniyor) \u00e7iziminden farkl\u0131 bir y\u00f6ntem de\u011fildir.<\/p>\n<p><strong>\u00d6rnek:<\/strong><\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-873 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-8.gif\" alt=\"\" width=\"299\" height=\"159\" \/><\/p>\n<p>\u015eekilde g\u00f6sterilmi\u015f olan mekanizmay\u0131 ele alal\u0131m. Bu mekanizmada bulunan uzuvlar\u0131n konumunu 2 uzvunun a\u00e7\u0131sal konumu \u03b8<sub>12<\/sub>\u2032 iken bulal\u0131m. Vekt\u00f6r devre denklemi:<\/p>\n<p style=\"text-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>C<\/strong> + <strong>CA<\/strong><\/p>\n<p>olacakt\u0131r.\u00a0<strong>A<sub>0<\/sub>B<sub>0<\/sub><\/strong>\u00a0vekt\u00f6r\u00fcn\u00fcn uzunlu\u011fu ve a\u00e7\u0131s\u0131 sabittir.\u00a0<strong>A<sub>0<\/sub>A<\/strong> vekt\u00f6r\u00fcn\u00fcn ise boyutu sabit olup a\u00e7\u0131sal konumu ba\u011f\u0131ms\u0131z de\u011fi\u015fken \u03b8<sub>12<\/sub>\u00a0nin de\u011feri ile belirlenir.\u00a0<strong>B<sub>0<\/sub>C<\/strong>\u00a0vekt\u00f6r\u00fcn\u00fcn sabit bir boyu vard\u0131r. Buna kar\u015f\u0131n\u00a0<strong>CA<\/strong>\u00a0vekt\u00f6r\u00fcn\u00fcn hem boyutu hemde y\u00f6n\u00fc de\u011fi\u015fecektir. Ancak\u00a0<strong>B<sub>0<\/sub>C<\/strong>\u00a0ve\u00a0<strong>CA<\/strong>\u00a0vekt\u00f6rleri birbirlerine diktir ve daima b\u00f6yle kalacaklard\u0131r (4 uzvu \u00fczerinde A noktas\u0131 anl\u0131k \u00e7ak\u0131\u015fan nokta oldu\u011fundan de\u011fi\u015fecektir, ancak\u00a0<strong>CA<\/strong>\u00a0vekt\u00f6r y\u00f6n\u00fc 4 uzvu \u00fczerinde kayar \u00e7ift ekseni oldu\u011fundan daima\u00a0<strong>B<sub>0<\/sub>C<\/strong>\u00a0vekt\u00f6r\u00fcne diktir). ACB<sub>0<\/sub>\u00a0noktalar\u0131 daima bir dik a\u00e7\u0131l\u0131 \u00fc\u00e7geni olu\u015fturacakt\u0131r. D\u00f6rt \u00e7ubuk mekanizmas\u0131nda yap\u0131ld\u0131\u011f\u0131 gibi A<sub>0<\/sub>\u00a0ve B<sub>0<\/sub> noktalar\u0131n\u0131 yerle\u015ftirdikten sonra ilk olarak verilen \u03b8<sub>12<\/sub>\u2032 a\u00e7\u0131s\u0131 ile A\u2032 noktas\u0131n\u0131 yerle\u015ftirilmi\u015f olan A<sub>0<\/sub>\u00a0ve B<sub>0<\/sub>\u00a0noktalar\u0131na g\u00f6re yerle\u015ftirelim. B<sub>0<\/sub>\u00a0merkezli, B<sub>0<\/sub>C yar\u0131\u00e7apl\u0131 daireyi \u00e7izelim. A\u2032 noktas\u0131ndan bu daireye te\u011fet \u00e7izildi\u011finde kesi\u015fen nokta C noktas\u0131 olacak ve \u2220ACB<sub>0<\/sub> a\u00e7\u0131s\u0131 dik a\u00e7\u0131 olacakt\u0131r. Bu \u015fekilde 3 ve 4 uzuvlar\u0131n\u0131n yeni konumlar\u0131 belirlenmi\u015ftir.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-874 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-9.gif\" alt=\"\" width=\"599\" height=\"213\" \/><\/p>\n<p>Mekanizman\u0131n tam bir devir i\u00e7in hareket analizi yap\u0131lmas\u0131 istenildi\u011finde, belirli derece aral\u0131klarda yukar\u0131da anlat\u0131lm\u0131\u015f olan i\u015flemler tekrarlan\u0131r. B\u00f6yle bir analiz ile mekanizman\u0131n animasyonu ger\u00e7ekle\u015ftirilebilir ve var ise s\u0131n\u0131r konumlar\u0131 (\u00f6rne\u011fin \u03b8<sub>12<\/sub> de\u011ferinden sonra saat yelkovan\u0131 y\u00f6n\u00fcnde veya ters y\u00f6n\u00fcnde mekanizman\u0131n kilitlendi\u011fi konumlar\u0131) veya 4 uzvunun d\u00f6nme y\u00f6n\u00fcn\u00fc de\u011fi\u015ftirdi\u011fi \u03b8<sub>12<\/sub> de\u011ferleri belirlenebilir.<\/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:300px\" 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_69d7c0146bef6\"><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-10.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"337\" height=\"213\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-10.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\/invsl.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"546\" height=\"359\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/invsl.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><\/div><script id=\"su_image_carousel_69d7c0146bef6_script\">if(window.SUImageCarousel){setTimeout(function() {window.SUImageCarousel.initGallery(document.getElementById(\"su_image_carousel_69d7c0146bef6\"))}, 0);}var su_image_carousel_69d7c0146bef6_script=document.getElementById(\"su_image_carousel_69d7c0146bef6_script\");if(su_image_carousel_69d7c0146bef6_script){su_image_carousel_69d7c0146bef6_script.parentNode.removeChild(su_image_carousel_69d7c0146bef6_script);}<\/script><\/p>\n<p style=\"text-align: left\" align=\"center\"><strong>\u00d6rnek:<\/strong><\/p>\n<p>\u015eekilde g\u00f6sterilmekte olan krank-biyel mekanizmas\u0131n\u0131 ele alal\u0131m. Ba\u011f\u0131ms\u0131z de\u011fi\u015fken 2 uzvunun 1 sabit uzvuna g\u00f6re a\u00e7\u0131sal konumudur. Vekt\u00f6r devre denklemi:<\/p>\n<p style=\"text-align: center\"><strong>A<sub>0<\/sub>A<\/strong> + <strong>AB<\/strong> = <strong>A<sub>0<\/sub>Q<\/strong> + <strong>QB<\/strong><\/p>\n<p>Karma\u015f\u0131k say\u0131lar ile (a<sub>2<\/sub> = |A<sub>0<\/sub>A|, a<sub>3<\/sub> = |AB|)<\/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> = ic + s<sub>14<\/sub><\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-876 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-11.gif\" alt=\"\" width=\"354\" height=\"193\" \/><\/p>\n<p>Dikkat edilir ise, s<sub>14<\/sub>\u00a0konum de\u011fi\u015fkeni ile A<sub>0<\/sub>Q sabit uzakl\u0131\u011f\u0131 (eksantriklik; 1 uzvu \u00fczerinde A<sub>0<\/sub> d\u00f6ner mafsal\u0131n\u0131n kayar mafsal eksenine dik uzakl\u0131\u011f\u0131 = c) ay\u0131rmak i\u00e7in <strong>A<sub>0<\/sub>B<\/strong>\u00a0vekt\u00f6r\u00fc dik bile\u015fenleri ile yaz\u0131lm\u0131\u015ft\u0131r. Bu vekt\u00f6r devre denkleminde,\u00a0<strong>AB<\/strong>\u00a0vekt\u00f6r\u00fcn\u00fcn y\u00f6n\u00fc ve\u00a0<strong>QB<\/strong>\u00a0vekt\u00f6r\u00fcn\u00fcn boyutu bilinmeyen konum de\u011fi\u015fkenleridir.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-877 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-12.gif\" alt=\"\" width=\"410\" height=\"236\" \/><\/p>\n<p>Bu mekanizman\u0131n verilen ba\u011f\u0131ms\u0131z parametre de\u011ferine g\u00f6re her hangi bir konumunu bulmak i\u00e7in ilk olarak A noktas\u0131n\u0131n konumu belirlenir (A\u2032). B noktas\u0131n\u0131n y\u00f6r\u00fcngesi sabit uzva g\u00f6re kayar \u00e7ift ekseni olaca\u011f\u0131ndan ve A noktas\u0131na uzakl\u0131\u011f\u0131 (AB) sabit oldu\u011fundan, belirlenen A noktas\u0131 merkezli AB yar\u0131\u00e7apl\u0131 yay \u00e7izildi\u011finde kayar \u00e7ift eksenini B noktas\u0131nda kesecektir (B\u2032).<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-878 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-13.gif\" alt=\"\" width=\"452\" height=\"536\" \/><\/p>\n<p>Problemi sistematik olarak \u00e7\u00f6zmek i\u00e7in, A<sub>0<\/sub>\u00a0merkezli A<sub>0<\/sub>A yar\u0131\u00e7apl\u0131 daire, A noktas\u0131n\u0131n de\u011fi\u015fik krank a\u00e7\u0131lar\u0131 i\u00e7in geometrik yerini verecektir. A\u00e7\u0131y\u0131 belirli say\u0131da aral\u0131klara b\u00f6lelim (genellikle bu b\u00f6lme, istenilen hassasiyete g\u00f6re 5\u00b0\u00a0veya daha ufak aral\u0131klarda yap\u0131l\u0131r. Ancak y\u00f6ntemi g\u00f6stermesi ve \u015feklin dahada karma\u015f\u0131k olmamas\u0131 i\u00e7in \u015fekilde 30\u00b0\u00a0aral\u0131klar se\u00e7ilmi\u015ftir). Belirlenen bu A noktalar\u0131n\u0131n de\u011fi\u015fik konumlar\u0131na A<sub>i<\/sub> diyelim (i = 1, &#8230;, 12). A<sub>i<\/sub>\u00a0merkezli ve AB yar\u0131\u00e7apl\u0131 daire yaylar\u0131 \u00e7izdi\u011fimizde her bir yay kayar mafsal ekseni ile B<sub>i<\/sub>\u00a0noktas\u0131nda kesi\u015fecektir. Bundan sonra pistonun (4 uzvu) krank d\u00f6nmesine g\u00f6re hareket diyagram\u0131 isteniyor ise, her konum i\u00e7in s<sub>14i<\/sub> uzakl\u0131\u011f\u0131 \u00f6l\u00e7\u00fclerek belirli bir \u00f6l\u00e7ekte \u00e7izilebilir.<\/p>\n<p>Herhangi bir konum de\u011fi\u015fkeninin ba\u011f\u0131ms\u0131z konum de\u011fi\u015fkenine g\u00f6re de\u011fi\u015fimini g\u00f6steren diyagrama <strong>hareket diyagram\u0131<\/strong>\u00a0denir. E\u011fer ba\u011f\u0131ms\u0131z de\u011fi\u015fkenin zamana g\u00f6re de\u011fi\u015fimi sabit ise (sabit h\u0131zda tahrik) bu durumda hareket diyagram\u0131nda her hangi bir noktada e\u011fim o konum de\u011fi\u015fkeninin o an i\u00e7in zamana g\u00f6re de\u011fi\u015fimini (t\u00fcrevini, a\u00e7\u0131sal h\u0131z\u0131n\u0131) verecektir. Bu diyagramlar kullan\u0131larak mekanizmalar\u0131n s\u0131n\u0131r konumlar\u0131n\u0131 veya kritik durumlar\u0131 belirlemekte m\u00fcmk\u00fcnd\u00fcr.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-879 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-14.gif\" alt=\"\" width=\"570\" height=\"414\" \/><\/p>\n<p>Yukar\u0131da bir d\u00f6rt \u00e7ubuk mekanizmas\u0131nda 2 uzvunun a\u00e7\u0131sal yer de\u011fi\u015fimine g\u00f6re 4 uzvunun a\u00e7\u0131sal yer de\u011fi\u015fimi g\u00f6sterilmektedir. Ba\u015fka uygulamalarda sabit uzva ba\u011fl\u0131 olmayan uzvun (biyel uzvu) her hangi bir noktas\u0131n\u0131n y\u00f6r\u00fcngesi \u00f6nemli olabilir.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-880 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-15.gif\" alt=\"\" width=\"490\" height=\"408\" \/><\/p>\n<p style=\"text-align: center\" align=\"center\">(|A<sub>0<\/sub>A| = 1, |A<sub>0<\/sub>B<sub>0<\/sub>| = 2, |AB| = |B<sub>0<\/sub>B| = |BC| = 2.5 birim)<\/p>\n<p>Yukar\u0131da hareketinin belirli bir k\u0131sm\u0131nda biyel noktas\u0131 y\u00f6r\u00fcngesi yakla\u015f\u0131k bir do\u011fru olacak \u015fekilde tasarlanm\u0131\u015f bir d\u00f6rt-\u00e7ubuk mekanizmas\u0131nda bu biyel noktas\u0131n\u0131n y\u00f6r\u00fcngesi g\u00f6r\u00fclmektedir. Bu e\u011friler\u00a0<strong>biyel e\u011frileri<\/strong>\u00a0olarak adland\u0131r\u0131l\u0131r.<\/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:300px\" 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_69d7c0146c78d\"><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-16.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"208\" height=\"205\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-16.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\/frbrstraightline.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"225\" height=\"233\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/frbrstraightline.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><\/div><script id=\"su_image_carousel_69d7c0146c78d_script\">if(window.SUImageCarousel){setTimeout(function() {window.SUImageCarousel.initGallery(document.getElementById(\"su_image_carousel_69d7c0146c78d\"))}, 0);}var su_image_carousel_69d7c0146c78d_script=document.getElementById(\"su_image_carousel_69d7c0146c78d_script\");if(su_image_carousel_69d7c0146c78d_script){su_image_carousel_69d7c0146c78d_script.parentNode.removeChild(su_image_carousel_69d7c0146c78d_script);}<\/script><\/p>\n<p>Grafik analiz pergel, cetvel gibi \u00e7izim aletleri ile yap\u0131labildi\u011fi gibi, AutoCad<sup>\u00ae<\/sup>, SolidWorks<sup>\u00ae<\/sup>, Creo<sup>\u00ae<\/sup>, Catia<sup>\u00ae<\/sup>, NX<sup>\u00ae<\/sup> gibi \u00e7e\u015fitli paket programlar kullan\u0131larak bilgisayar ortam\u0131nda da yap\u0131labilir ve mekanizman\u0131n \u015fekli veya montaj resmi \u00e7izildikten sonra sim\u00fclasyonu da yap\u0131labilir. Grafik analiz bilhassa konuya yeni ba\u015flayanlara kavramlar\u0131 g\u00f6rsel olarak verdi\u011finden \u00e7ok yararl\u0131 olmaktad\u0131r.<\/p>\n<p>Farkl\u0131 uygulamalar internette \u00e7e\u015fitli sitelerde bulunmaktad\u0131r. Geogebra ve Excel kullanarak hareket, h\u0131z ve ivme analizi konular\u0131nda \u00f6rnekler haz\u0131rlanm\u0131\u015f ve: <a href=\"https:\/\/www.youtube.com\/c\/MuhendislerIcinExcelveGeogebraEresSoylemez\">https:\/\/www.youtube.com\/c\/MuhendislerIcinExcelveGeogebraEresSoylemez<\/a> Youtube kanal\u0131nda yay\u0131nlanm\u0131\u015ft\u0131r. Okuyucular\u0131n bu kanaldaki videolar\u0131 izlemesi tavsiye olunur.<\/p>\n<p><strong>\u00d6rnek:<\/strong><\/p>\n<p><strong>Geogebra<\/strong> yaz\u0131l\u0131m\u0131 kullan\u0131c\u0131n\u0131n bir\u00e7ok siteden yasal olarak \u00fccretsiz bir programd\u0131r. Bu program son y\u0131llarda bir\u00e7ok \u00fclkede lise seviyesinde bile matematik ve geometriyi \u00f6\u011fretmek i\u00e7in kullan\u0131lmaktad\u0131r. Program istendi\u011finde ki\u015fisel bilgisayar\u0131n\u0131za indirilebildi\u011fi gibi, internet ortam\u0131nda da kullan\u0131labilmektedir. Burada basit bir \u00f6rnekle Geogebran\u0131n mekanizmalar\u0131n analizinde ve sim\u00fclasyonunda kullan\u0131m\u0131 anlat\u0131lacakt\u0131r.<\/p>\n<p>\u015eekilde g\u00f6sterilen mekanizman\u0131n her \u03b8<sub>12<\/sub> a\u00e7\u0131s\u0131nda t\u00fcm uzuvlar\u0131n\u0131n konumunu bulmak istiyoruz. Mekanizma uzuv boyutlar\u0131: |A<sub>0<\/sub>A| = 400 , |AB| = 300 , |AC| = 300 , |BC| = 150 , |B<sub>0<\/sub>B| = 350 , |A<sub>0<\/sub>B<sub>0x<\/sub>| = 200 , |A<sub>0<\/sub>B<sub>0y<\/sub>| = 100 , |CD| = 1000 , a<sub>1<\/sub> = 170.<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1545\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-g1.gif\" alt=\"\" width=\"535\" height=\"235\" \/><\/p>\n<\/div>\n<\/div><\/div><\/div><\/div><div id=\"pg-gb859-69d7c0146a1e0-1\"  class=\"panel-grid panel-no-style\" ><div id=\"pgc-gb859-69d7c0146a1e0-1-0\"  class=\"panel-grid-cell panel-grid-cell-empty\" ><\/div><div id=\"pgc-gb859-69d7c0146a1e0-1-1\"  class=\"panel-grid-cell panel-grid-cell-mobile-last\" >        <div id=\"panel-gb859-69d7c0146a1e0-1-1-0\" class=\"so-panel widget widget_wylwidget panel-first-child panel-last-child widgetopts-SO\" data-index=\"1\" >        <h3 class=\"widget-title\">Geogebra&#8217;da mekanizma analizinin video anlat\u0131m\u0131<\/h3>        <div class=\"lyte-wrapper lidget\" style=\"width:711px; height:400px; min-width:200px; max-width:100%;\"><div class=\"lyMe\" id=\"YLW_6x0MqWHvgz0\"><div id=\"lyte_6x0MqWHvgz0\" data-src=\"https:\/\/img.youtube.com\/vi\/6x0MqWHvgz0\/hqdefault.jpg\" class=\"pL\"><div class=\"play\"><\/div><div class=\"ctrl\"><div class=\"Lctrl\"><\/div><\/div><\/div><\/div><noscript><a href=\"https:\/\/youtu.be\/6x0MqWHvgz0\"><img decoding=\"async\" src=\"https:\/\/img.youtube.com\/vi\/6x0MqWHvgz0\/hqdefault.jpg\" alt=\"\" \/><\/a><\/noscript><\/div>\n        <div><\/div>\n        <\/div>        <\/div><div id=\"pgc-gb859-69d7c0146a1e0-1-2\"  class=\"panel-grid-cell panel-grid-cell-empty\" ><\/div><\/div><div id=\"pg-gb859-69d7c0146a1e0-2\"  class=\"panel-grid panel-no-style\" ><div id=\"pgc-gb859-69d7c0146a1e0-2-0\"  class=\"panel-grid-cell\" ><div id=\"panel-gb859-69d7c0146a1e0-2-0-0\" class=\"so-panel widget widget_sow-editor panel-first-child panel-last-child widgetopts-SO\" data-index=\"2\" ><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<p>Geogebra ekran\u0131 a\u00e7\u0131ld\u0131\u011f\u0131nda ilk olarak A<sub>0<\/sub> ve B<sub>0<\/sub> d\u00f6ner mafsal eksenlerini yerle\u015ftirmek i\u00e7in A_0=(0,0) ve B_0=(200,100) olarak yazal\u0131m. Kayar mafsal ekseni i\u00e7in Q=(0,100) diyelim ve bu noktay\u0131 belirledikten sonra do\u011fru komutunda bulunan dik \u00e7izgi komutu ile y eksenine dik Q dan ge\u00e7en do\u011fruyu \u00e7izelim (veya, Q notas\u0131ndan ge\u00e7en x eksenine paralel do\u011fru olarak da bu do\u011fruyu \u00e7izebiliriz). Ekran a\u015fa\u011f\u0131daki \u015fekilde de g\u00f6r\u00fcld\u00fc\u011f\u00fc gibi olacakt\u0131r. Cebir k\u0131sm\u0131nda girdi\u011finiz bilgiler ve Grafik k\u0131sm\u0131nda ise \u015fu anda belirledi\u011finiz noktalar ve \u00e7izgi g\u00f6r\u00fclecektir.<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1548 size-full aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-g2.gif\" alt=\"\" width=\"853\" height=\"628\" \/><\/p>\n<p>\u015eimdi S\u00fcrg\u00fc komutunu se\u00e7elim. \u015eekil 2.28 de g\u00f6r\u00fclen k\u00fc\u00e7\u00fck ekranda de\u011fi\u015fken olarak a\u00e7\u0131y\u0131 se\u00e7elim. Bu a\u00e7\u0131n\u0131n ba\u015flang\u0131\u00e7 ve biti\u015f de\u011ferlerini ve de\u011fi\u015ftirme birimini girebilirsiniz. Ayr\u0131ca, genellikle program\u0131n kendisinin verdi\u011fi bir sembol yerine siz isterseniz kendi sembol\u00fcn\u00fcz\u00fc girebilirsiniz. Bir sembol girmek i\u00e7in Alt tu\u015fu ile birlikte \u03b8 i\u00e7in t, \u03d5 i\u00e7in f, v.b. harflerini se\u00e7ersiniz. \u0130ndis yazmak i\u00e7in \u03b8<sub>12<\/sub> i\u00e7in Alt+t den sonra _1_2 \u015feklinde yazman\u0131z gerekir.<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1549\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-g3.gif\" alt=\"\" width=\"582\" height=\"312\" \/><\/p>\n<p>\u015eimdi girdi olarak A=(400; Alt+t_1_2) yazal\u0131m. Noktal\u0131 virg\u00fcl Geogebra&#8217;ya boyut ve a\u00e7\u0131 girdi\u011fimizi belirtir (kutupsal koordinatlar). Bu \u015fekilde A<sub>0<\/sub> dan 400 birim uzakta ve x ekseni ile \u03b8<sub>12<\/sub> a\u00e7\u0131s\u0131 yapan bir nokta belirlemi\u015f oluruz. A ve A<sub>0<\/sub> noktalar\u0131 aras\u0131nda \u00e7izgi \u00e7izdi\u011fimizde krank\u0131m\u0131z\u0131 g\u00f6stermi\u015f olaca\u011f\u0131z. \u015eimdi merkezi ve yar\u0131 \u00e7ap\u0131 verilen bir daire komutunu se\u00e7elim ve A dan AB uzunlu\u011funda (300) ve B<sub>0<\/sub>\u00a0dan B<sub>0<\/sub>B uzunlu\u011funda (350) daireleri \u00e7izelim. Bu iki daire iki noktada kesi\u015fecektir. \u015eekilde verilen mekanizma g\u00f6r\u00fcnt\u00fcs\u00fcne g\u00f6re uygun kesim noktas\u0131n\u0131 se\u00e7elim. Geogebra farkl\u0131 isim verebilir. Yeniden adland\u0131rarak bu noktay\u0131 B olarak belirleyelim. Daireleri \u201cShow object\u201d komutu ile gizleyebiliriz. C noktas\u0131n\u0131 belirlemek i\u00e7in 3 uzvu \u00fczerinde iki nokta tan\u0131mlanm\u0131\u015f oldu\u011fundan (A ve B noktalar\u0131) A dan AC yar\u0131\u00e7apl\u0131, B den BC yar\u0131\u00e7apl\u0131 dairelerin kesim noktas\u0131 bize C noktas\u0131n\u0131 verecektir. Yine daireler iki noktada kesi\u015fti\u011finden uygun olan kesi\u015fme noktas\u0131n\u0131 se\u00e7memiz gerekir. \u015eimdi \u00c7okgen komutunu kullanarak ABC \u00fc\u00e7genini ve \u00e7izgi iki nokta aras\u0131 do\u011fru komutu ile B<sub>0<\/sub>B do\u011frusunu \u00e7izersek, mekanizmay\u0131 olu\u015fturan birinci devreyi \u00e7izelim. \u015eimdi, C noktas\u0131ndan CD yar\u0131\u00e7apl\u0131 (1000) daireyi \u00e7izelim ve Kayar mafsal ekseni i\u00e7in \u00e7izdi\u011fimiz \u00e7izgi ile kesim noktas\u0131n\u0131 Belirleyip bu noktaya D diyelim. D noktas\u0131ndan k\u0131zak eksenine dik bir do\u011fru \u00e7izdikten sonra P_1 = D + (100,50) diyerek bir P<sub>1<\/sub> noktas\u0131n\u0131 belirleyelim (P<sub>1<\/sub>, \u00e7izece\u011fimiz dikd\u00f6rtgenin bir kenar\u0131 olacakt\u0131r, D den x ve y uzakl\u0131\u011f\u0131n\u0131 istedi\u011finiz de\u011ferler se\u00e7ebilirsiniz). P<sub>1<\/sub> noktas\u0131n\u0131 belirledikten sonra bu noktan\u0131n yatay ve dikey eksenlerde g\u00f6r\u00fcnt\u00fclerini al\u0131rsak d\u00f6rtgenin d\u00f6rt k\u00f6\u015fesini elde edip bir \u00e7okgen \u00e7izebiliriz. Sonu\u00e7 a\u015fa\u011f\u0131daki \u015fekilde g\u00f6r\u00fclmektedir.<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1550 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-g4.gif\" alt=\"\" width=\"1074\" height=\"899\" \/><\/p>\n<p>6 uzvunun krank d\u00f6nme a\u00e7\u0131s\u0131na g\u00f6re hareket e\u011frisini g\u00f6rmek istedi\u011fimizde, ilk olarak A<sub>0<\/sub>A do\u011frusunun x eksenine g\u00f6re a\u00e7\u0131s\u0131n\u0131 ve D noktas\u0131n\u0131n Q noktas\u0131ndan uzakl\u0131\u011f\u0131n\u0131 belirleriz. \u0130kinci bir grafik alan\u0131 a\u00e7al\u0131m. Bu yeni grafik alanda sadece P=(a*180\/p, DistanceQD) olarak bir nokta \u00e7izelim. Mekanizman\u0131n animasyonu i\u00e7in s\u00fcrg\u00fcy\u00fc oynama durumuna getirdi\u011fimizde P noktas\u0131n\u0131n konumu ikinci grafikte de\u011fi\u015fecektir. Bu noktay\u0131 \u201c\u0130zle\u201d meye ald\u0131\u011f\u0131m\u0131zda a\u015fa\u011f\u0131daki \u015fekilde g\u00f6r\u00fcld\u00fc\u011f\u00fc gibi, hareket e\u011frisi elde edilecektir. Bu noktan\u0131n koordinatlar\u0131n\u0131 ba\u015fka bir yerde kullanmak istedi\u011finizde listeleme (SpreadSheet) sayfas\u0131 a\u00e7\u0131rak P noktas\u0131n\u0131n koordinatlar\u0131 bir dosyaya yaz\u0131labilir, ba\u015fka bir program taraf\u0131ndan kullan\u0131labilir).<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1551 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-g5.gif\" alt=\"\" width=\"648\" height=\"532\" \/><\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-882 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-17.gif\" alt=\"\" width=\"500\" height=\"349\" \/><\/p>\n<p>Mekanizmada birden \u00e7ok devre var ise, bu devreler \u00e7o\u011funlukla teker teker \u00e7\u00f6z\u00fclebilir. Yukar\u0131da g\u00f6sterilen mekanizma bir kar\u0131\u015ft\u0131r\u0131c\u0131y\u0131 yukar\u0131-a\u015fa\u011f\u0131 hareket ettirmek i\u00e7indir. A<sub>0<\/sub>A uzvuna bir kol ba\u011fl\u0131 olup hareket elle bu uzva verilir.<\/p>\n<p>A<sub>0<\/sub>ABB<sub>0<\/sub> (1, 2, 3 ve 4 uzuvlar\u0131) birinci devreyi olu\u015fturmaktad\u0131r. Vekt\u00f6r devre denklemi (B mafsal\u0131n\u0131 s\u00f6k\u00fcp takarak):<\/p>\n<p style=\"text-align: center\"><strong>A<sub>0<\/sub>A<\/strong> + <strong>AB<\/strong> = <strong>A<sub>0<\/sub>B<\/strong><sub><strong>0<\/strong>\u00a0<\/sub>+ <strong>B<sub>0<\/sub>B<\/strong><\/p>\n<p>Bu vekt\u00f6r devre denklemi A<sub>0<\/sub>A uzvunun yeni bir konumu i\u00e7in bir d\u00f6rt-\u00e7ubuk mekanizmas\u0131nda g\u00f6sterildi\u011fi gibi \u00e7\u00f6z\u00fclebilir ve B noktas\u0131n\u0131n konumu bulunur. C noktas\u0131 3 uzvu ve D noktas\u0131 4 uzvu \u00fczerinde oldu\u011fundan yeni AB konumlar\u0131 belirlendikten sonra sabit uzuv boyutlar\u0131 kullan\u0131larak C ve D noktalar\u0131n\u0131n yeni konumlar\u0131 bulunur (3 uzvu \u00fczerinde B ve A noktalar\u0131n\u0131n konumu bilindi\u011fine ve C noktas\u0131 BA do\u011frusu \u00fczerinde olup B den 90 mm uzakta oldu\u011funa g\u00f6re C noktas\u0131n\u0131n konumu ve D noktas\u0131 4 uzvu \u00fczerinde oldu\u011funa, bu uzuv \u00fczerinde B ve B<sub>0<\/sub>\u00a0noktalar\u0131n\u0131n konumu bilindi\u011fine, D noktas\u0131 B<sub>0<\/sub>B do\u011frusu \u00fczerinde ve B<sub>0<\/sub> dan 189.5 mm uzakta oldu\u011funa g\u00f6re D noktas\u0131n\u0131n konumu belirlenir).<\/p>\n<p>\u0130kinci olarak BCEDB (3, 6, 5 ve 4 uzuvlar\u0131) devresini g\u00f6z \u00f6n\u00fcne alal\u0131m. Bu devre i\u00e7in vekt\u00f6r devre denklemi:<\/p>\n<p style=\"text-align: center\"><strong>BC<\/strong> + <strong>CE<\/strong> = <strong>BD<\/strong> + <strong>DE<\/strong><\/p>\n<p>yaz\u0131labilir. Bu vekt\u00f6r devre denkleminde\u00a0<strong>BC<\/strong>\u00a0ve\u00a0<strong>BD<\/strong>\u00a0gerek uzunlu\u011fu ve gerek y\u00f6n\u00fc bilinen iki vekt\u00f6rd\u00fcr.\u00a0<strong>CE<\/strong>\u00a0ve\u00a0<strong>DE<\/strong>\u00a0vekt\u00f6rlerinin ise boyutlar\u0131 uzuv boyutu oldu\u011fundan bilinmektedir. C ve D noktalar\u0131 birinci devre denkleminin \u00e7\u00f6z\u00fclmesi ile yukar\u0131da a\u00e7\u0131kland\u0131\u011f\u0131 gibi belirlendikten sonra, \u00fc\u00e7 kenar\u0131 bilinen CED \u00fc\u00e7genini \u00e7izmemiz gerekmektedir. \u00d6yle ise C merkezli CE yar\u0131 \u00e7apl\u0131 bir daire yay\u0131 ile D merkezli DE yar\u0131 \u00e7apl\u0131 daire yay\u0131n\u0131n kesi\u015fme noktas\u0131 E noktas\u0131n\u0131n ve dolay\u0131s\u0131 ile t\u00fcm uzuvlar\u0131n konumunu belirler (her uzuv \u00fczerinde iki noktan\u0131n konumu belirlenmi\u015f olur).<\/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:300px\" 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_69d7c0146d97a\"><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-mixer.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"230\" height=\"225\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-mixer.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\/mixer.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"256\" height=\"256\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/mixer.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><\/div><script id=\"su_image_carousel_69d7c0146d97a_script\">if(window.SUImageCarousel){setTimeout(function() {window.SUImageCarousel.initGallery(document.getElementById(\"su_image_carousel_69d7c0146d97a\"))}, 0);}var su_image_carousel_69d7c0146d97a_script=document.getElementById(\"su_image_carousel_69d7c0146d97a_script\");if(su_image_carousel_69d7c0146d97a_script){su_image_carousel_69d7c0146d97a_script.parentNode.removeChild(su_image_carousel_69d7c0146d97a_script);}<\/script><\/p>\n<p>&nbsp;<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-883 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/image215-18.gif\" alt=\"\" width=\"311\" height=\"428\" \/><\/p>\n<p>Baz\u0131 karma\u015f\u0131k mekanizmalarda konum analizi her bir devrenin ayr\u0131 ayr\u0131 ve s\u0131ra ile \u00e7\u00f6z\u00fclmesi m\u00fcmk\u00fcn olmay\u0131p, \u00f6rne\u011fin iki devrenin birlikte \u00e7\u00f6z\u00fcm\u00fc gerekebilir. Bu durumlarda devre denkleminin grafik \u00e7\u00f6z\u00fcm\u00fc deneme-yan\u0131lma y\u00f6ntemini gerektirecekti. \u00d6rne\u011fin yukar\u0131da g\u00f6r\u00fclen mekanizma i\u00e7in devre denklemleri:<\/p>\n<p style=\"text-align: center\"><strong>PA<sub>0<\/sub><\/strong>\u00a0+ <strong>A<sub>0<\/sub>A<\/strong> = <strong>PB<\/strong><sub><strong>0<\/strong>\u00a0<\/sub>+ <strong>B<sub>0<\/sub>B<\/strong> + <strong>BA<\/strong><\/p>\n<p style=\"text-align: center\"><strong>PQ<\/strong> + <strong>QC<\/strong> = <strong>PB<\/strong><sub><strong>0<\/strong>\u00a0<\/sub>+ <strong>B<sub>0<\/sub>B<\/strong> + <strong>BC<\/strong><\/p>\n<p>\u015feklinde yaz\u0131labilir. Her iki denklemde de \u00fc\u00e7 bilinmeyen konum parametresi bulunmakta, toplamda ise d\u00f6rt bilinmeyen konum parametresi bulunmaktad\u0131r. (A<sub>0<\/sub>A kol a\u00e7\u0131s\u0131 verildi\u011fi kabul edilmektedir. Birinci denklemde:\u00a0<strong>B<sub>0<\/sub>B<\/strong>\u00a0nin a\u00e7\u0131sal konumu,\u00a0<strong>BA<\/strong>\u00a0n\u0131n hem a\u00e7\u0131sal konumu ve hemde boyutu; \u0130kinci denklemde ise,\u00a0<strong>QC<\/strong>\u00a0boyutu,\u00a0<strong>B<sub>0<\/sub>B<\/strong>\u00a0nin a\u00e7\u0131sal konumu,\u00a0<strong>BC<\/strong>\u00a0nin a\u00e7\u0131sal konumu bilinmeyen konum parametreleridir.\u00a0<strong>BC<\/strong>\u00a0ve\u00a0<strong>BA<\/strong>\u00a0vekt\u00f6rlerinin a\u00e7\u0131sal konumu ayn\u0131 oldu\u011fundan, toplam 4 bilinmeyen konum parametresi vard\u0131r.<\/p>\n<p>B noktas\u0131n\u0131n geometrik yeri B<sub>0<\/sub>\u00a0merkezli B<sub>0<\/sub>B yar\u0131\u00e7apl\u0131 bir daire yay\u0131d\u0131r. C noktas\u0131n\u0131n geometrik yeri ise kayar mafsal ekseni olan do\u011frudur. Giri\u015f kolunun a\u00e7\u0131sal konumu belirlendi\u011finde konumu belli olan A noktas\u0131 ise BC do\u011frusu \u00fczerinde olmal\u0131d\u0131r. Deneme yan\u0131lma ile BC uzunlu\u011fu sabit do\u011fruyu B ve C y\u00f6r\u00fcngelerini sa\u011flayacak \u015fe-kilde hareket ettirerek A noktas\u0131n\u0131n Bu do\u011fru \u00fczerinde bulundu\u011fu konumu bulmam\u0131z gerekmektedir. Sa\u011flanmas\u0131 gerekenler:<\/p>\n<p style=\"padding-left: 40px\">a) B noktas\u0131 B<sub>0<\/sub>B yar\u0131\u00e7apl\u0131 ve B<sub>0<\/sub> merkezli bir yay \u00fczerinde olmal\u0131d\u0131r.<\/p>\n<p style=\"padding-left: 40px\">b) C noktas\u0131 kayar mafsal ekseni ile belirtilen do\u011fru \u00fczerinde olmal\u0131d\u0131r.<\/p>\n<p style=\"padding-left: 40px\">c) 4 uzvunu belirleyen BC do\u011frusunun A<sub>0<\/sub>A boyutu ve giri\u015f kolu a\u00e7\u0131s\u0131 ile belirlenmi\u015f olan A noktas\u0131ndan ge\u00e7mesi gerekmektedir.<\/p>\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n<p>  <a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch3\/3-4\/\" 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\/3-6\/\" 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\n\n\n<p><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch3\/\" data-type=\"page\" data-id=\"52\"><\/a><\/p>\n\n\n\n<p>    <\/p>\n","protected":false},"excerpt":{"rendered":"<p>3.5 Vekt\u00f6r Devre Denklemlerinin Grafik \u00c7\u00f6z\u00fcm\u00fc Bir mekanizman\u0131n kinematik analizini yaparken mekanizmay\u0131 m\u00fcmk\u00fcn olan en basit \u015fekilde \u00e7izmemiz yapaca\u011f\u0131m\u0131z i\u015flerini kolayla\u015ft\u0131r\u0131r. Bu nedenle uzuvlar bir \u00e7izgi ile g\u00f6sterilebilir. Bu \u00e7izgilerin u\u00e7lar\u0131 bir sembolle g\u00f6sterilmi\u015f oldu\u011fundan, bu \u00e7izgiler ger\u00e7ek anlamda vekt\u00f6rd\u00fcrler &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch3\/3-5\/\"> <span class=\"screen-reader-text\">3-5<\/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-859","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/859","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=859"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/859\/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=859"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}