{"id":1031,"date":"2021-09-09T13:33:35","date_gmt":"2021-09-09T13:33:35","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=1031"},"modified":"2021-10-08T12:44:58","modified_gmt":"2021-10-08T12:44:58","slug":"4-2-1","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch4\/4-2-1\/","title":{"rendered":"4-2-1"},"content":{"rendered":"<div id=\"pl-gb1031-69d88fea667f2\"  class=\"panel-layout\" ><div id=\"pg-gb1031-69d88fea667f2-0\"  class=\"panel-grid panel-no-style\" ><div id=\"pgc-gb1031-69d88fea667f2-0-0\"  class=\"panel-grid-cell\" ><div id=\"panel-gb1031-69d88fea667f2-0-0-0\" 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>4.2<\/b> Mekanizmalarda H\u0131z ve \u0130vme Analizi -1<\/h1>\n<p>Mekanizmalarda h\u0131z ve ivme analizi vekt\u00f6rel olarak ba\u011f\u0131l h\u0131z ve ivme kavram\u0131 ile yap\u0131l\u0131r. Genel olarak verilen de\u011ferler ile ba\u015flan\u0131l\u0131r ve A, B, C, \u2026 gibi genellikle mafsal eksenlerinin ge\u00e7ti\u011fi noktalar s\u0131ras\u0131 ile kullan\u0131larak analiz ger\u00e7ekle\u015ftirilir. Elde edilecek olan h\u0131z denklemleri:<\/p>\n<p style=\"text-align: center\"><strong>v<\/strong><sub>B<\/sub>\u00a0= <strong>v<\/strong><sub>A<\/sub>\u00a0+ <strong>v<\/strong><sub>B\/A<\/sub><\/p>\n<p style=\"text-align: center\"><strong>v<\/strong><sub>C<\/sub>\u00a0= <strong>v<\/strong><sub>B<\/sub>\u00a0+ <strong>v<\/strong><sub>C\/B<\/sub><\/p>\n<p>ve benzeri denklemler olacakt\u0131r. \u0130vme i\u00e7in ise, benzer \u015fekilde:<\/p>\n<p style=\"text-align: center\"><strong>a<\/strong><sub>B<\/sub>\u00a0= <strong>a<\/strong><sub>A<\/sub>\u00a0+ <strong>a<sup>t<\/sup><\/strong><sub>B\/A<\/sub>\u00a0+ <strong>a<sup>n<\/sup><\/strong><sub>B\/A<\/sub><\/p>\n<p style=\"text-align: center\"><strong>a<\/strong><sub>C<\/sub>\u00a0= <strong>a<\/strong><sub>B<\/sub>\u00a0+ <strong>a<sup>t<\/sup><\/strong><sub>C\/B<\/sub>\u00a0+ <strong>a<sup>n<\/sup><\/strong><sub>C\/B<\/sub><\/p>\n<p>veya<\/p>\n<p style=\"text-align: center\"><strong>a<\/strong><sub>D4<\/sub>\u00a0= <strong>a<\/strong><sub>D3<\/sub>\u00a0+ <strong>a<sup>t<\/sup><\/strong><sub>D4\/D3<\/sub> + <strong>a<sup>c<\/sup><\/strong><sub>D4\/D3<\/sub><\/p>\n<p>benzer denklemler yaz\u0131lacakt\u0131r. Se\u00e7ilen A, B, C, D noktalar\u0131n\u0131n m\u00fcmk\u00fcn oldu\u011funca d\u00f6ner mafsal ekseni \u00fczerinde olmas\u0131n\u0131n en \u00f6nemli nedeni mafsal ekseninde olan bu noktan\u0131n mafsalla birle\u015ftirilen iki uzuv \u00fczerinde daima \u00e7ak\u0131\u015fan nokta olmas\u0131d\u0131r. Daima \u00e7ak\u0131\u015fan bu noktalar aras\u0131nda ba\u011f\u0131l h\u0131z ve ivme olmayaca\u011f\u0131ndan (s\u0131f\u0131r olaca\u011f\u0131ndan) dolay\u0131 her iki uzuvda da bu noktalar\u0131n h\u0131z ve ivmesi e\u015fit olacakt\u0131r. Bu nedenle genel kural olarak bir uzvun \u00fczerinde bulunan her hangi bir noktan\u0131n h\u0131z ve ivmesini belirlemeden \u00f6nce bu uzvu di\u011fer uzuvlar ile birle\u015ftiren mafsallar\u0131n merkezlerinde bulunan noktalar\u0131n h\u0131z ve ivmesi bulunmal\u0131d\u0131r.<\/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\" \/> \u00a0 \u00a0 Dikkat edilmesi gereken di\u011fer bir \u00f6nemli nokta ise, konum h\u0131z ve ivme analizi s\u0131ra ile yap\u0131labilir.<\/b><\/span>\u00a0Konum analizi yap\u0131lmadan h\u0131z analizi, konum ve h\u0131z analizi yap\u0131lmadan ivme analizi yap\u0131lamaz. Bunun nedeni, h\u0131z ve ivmelerin belirlenmesi s\u0131ras\u0131nda gerekli olan y\u00f6nler konum analizi ile belirlenecek, ivme analizi i\u00e7in normal ve Coriolis ivmesi terimleri sadece h\u0131z analizinde elde edilen terimlerin fonksiyonu olacakt\u0131r.<\/p>\n<p>Devre kapal\u0131l\u0131k denklemleri h\u0131z ve ivme analizi i\u00e7in kolayl\u0131kla kullan\u0131labilir \u00e7\u00fcnk\u00fc, bu denklemlerde mekanizmada bulunan t\u00fcm de\u011fi\u015fkenler yer almaktad\u0131r. Devre kapal\u0131l\u0131k denklemlerinin zamana g\u00f6re birinci ve ikinci t\u00fcrevleri al\u0131nd\u0131\u011f\u0131nda, ve konum de\u011fi\u015fkenlerinin birinci ve ikinci t\u00fcrevlerine g\u00f6re \u00e7\u00f6z\u00fcm yap\u0131ld\u0131\u011f\u0131nda bu de\u011ferler her hangi bir noktan\u0131n h\u0131z ve ivmesinin belirlenmesi i\u00e7in gerekli t\u00fcm terimleri belirleyecektir. Vekt\u00f6r devre denkleminin zamana g\u00f6re birinci t\u00fcrevi bize <span style=\"color: #cc0000\"><strong>vekt\u00f6rel h\u0131z devre denklemi<\/strong><\/span>ni, ikinci t\u00fcrevi ise <span style=\"color: #cc0000\"><strong>vekt\u00f6rel ivme devre denklemi<\/strong><\/span>ni verecektir. E\u011fer konum de\u011fi\u015fkenleri vekt\u00f6r devre denklemi kullan\u0131larak \u00e7\u00f6z\u00fclm\u00fc\u015f ise, vekt\u00f6r h\u0131z denklemleri konum de\u011fi\u015fkenlerinin birinci t\u00fcrevlerine g\u00f6re (bu de\u011fi\u015fkenlere <strong><span style=\"color: #cc0000\">h\u0131z de\u011fi\u015fkenleri<\/span><\/strong> diyece\u011fiz) ve vekt\u00f6r ivme denklemleri de konum de\u011fi\u015fkenlerinin ikinci t\u00fcrevine g\u00f6re (bu de\u011fi\u015fkenlere <strong><span style=\"color: #cc0000\">ivme de\u011fi\u015fkenleri<\/span><\/strong>\u00a0diyece\u011fiz) lineer bir denklem tak\u0131m\u0131 olu\u015fturacakt\u0131r. H\u0131z ve ivme de\u011fi\u015fkenleri bu denklemlerden \u00e7\u00f6z\u00fcld\u00fckten sonra herhangi bir uzvun \u00fczerinde her hangi bir noktan\u0131n h\u0131z ve ivmesi kolayl\u0131kla bulunacakt\u0131r.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1171 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img231-1.gif\" alt=\"\" width=\"555\" height=\"257\" \/><\/p>\n<p>\u0130lk \u00f6rnek olarak \u015fekilde g\u00f6sterilen krank-biyel mekanizmas\u0131n\u0131 ele alal\u0131m. Giri\u015f kolunun (2 uzvu) yatay ile yapt\u0131\u011f\u0131 a\u00e7\u0131, \u03b8<sub>12<\/sub>, a\u00e7\u0131sal h\u0131z\u0131 <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{\u03c9}}_{{12}}}={{\\dot{\u03b8}}_{{12}}}=\\frac{{\\text{d}{{\\text{\u03b8}}_{{12}}}}}{\\text{dt}} <\/span> ve a\u00e7\u0131sal ivmesi <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{\u03b1}}_{{12}}}={{{\\ddot{\u03b8}}}_{{12}}}=\\frac{{{{\\text{d}}^{2}}{{\\text{\u03b8}}_{{12}}}}}{{\\text{d}{{\\text{t}}^{2}}}} <\/span> verildi\u011fi kabul edilmektedir. Devre kapal\u0131l\u0131k denklemi ve e\u015fleni\u011fi karma\u015f\u0131k say\u0131lar ile yaz\u0131ld\u0131\u011f\u0131nda:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td>\n<p style=\"text-align: center\">s<sub>14<\/sub> + ic<sub>1<\/sub> + a<sub>3<\/sub>e<sup>i\u03b8<sub>13<\/sub><\/sup> = a<sub>2<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup><\/p>\n<p style=\"text-align: center\">s<sub>14<\/sub>\u00a0\u2212 ic<sub>1<\/sub> + a<sub>3<\/sub>e<sup>\u2212<\/sup><sup>i\u03b8<sub>13<\/sub><\/sup> = a<sub>2<\/sub>e<sup>\u2212<\/sup><sup>i\u03b8<sub>12<\/sub><\/sup><\/p>\n<\/td>\n<td align=\"right\">\n<p style=\"text-align: right\">(1a)<\/p>\n<p style=\"text-align: right\">(1b)<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Verilen bir \u03b8<sub>12<\/sub> de\u011ferine g\u00f6re x ve \u03b8<sub>13<\/sub>\u00a0konum de\u011fi\u015fkenlerinin nas\u0131l bulundu\u011fu bir \u00f6nceki k\u0131s\u0131mda incelenmi\u015fti. Burada bu denklemlerin zamana g\u00f6re birinci t\u00fcrevini alarak vekt\u00f6r h\u0131z devre denklemi ve e\u015fleni\u011fini elde edelim.<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\dot{\\text{s}}}<\/span><sub>14<\/sub> + ia<sub>3<\/sub>\u03c9<sub>13<\/sub>e<sup>i\u03b8<sub>13<\/sub><\/sup> = ia<sub>2<\/sub>\u03c9<sub>12<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup><\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\dot{\\text{s}}}<\/span><sub>14<\/sub> \u2212 ia<sub>3<\/sub>\u03c9<sub>13<\/sub>e<sup>\u2212<\/sup><sup>i\u03b8<sub>13<\/sub><\/sup> = \u2212ia<sub>2<\/sub>\u03c9<sub>12<\/sub>e<sup>\u2212<\/sup><sup>i\u03b8<sub>12<\/sub><\/sup><\/p>\n<\/td>\n<td align=\"right\">\n<p style=\"text-align: right\">(2a)<\/p>\n<p style=\"text-align: right\">(2b)<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Bu denklemlerde \u03c9<sub>12<\/sub> = d\u03b8<sub>12<\/sub>\/dt, \u03c9<sub>13<\/sub> = d\u03b8<sub>13<\/sub>\/dt, <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\dot{\\text{s}}}<\/span><sub>14<\/sub> = ds<sub>14<\/sub>\/dt h\u0131z de\u011fi\u015fkenleridir. (2) denklemi dikkatli bir \u015fekilde incelendi\u011fi takdirde bu vekt\u00f6r denkleminin:<\/p>\n<p style=\"text-align: center\"><strong>v<\/strong><sub>B<\/sub>\u00a0+ <strong>v<\/strong><sub>A\/B<\/sub>\u00a0= <strong>v<\/strong><sub>A<\/sub><\/p>\n<p>ba\u011f\u0131l h\u0131z denkleminden farkl\u0131 bir denklem olmad\u0131\u011f\u0131 g\u00f6r\u00fclecektir. Bu denklemin pratik a\u00e7\u0131klamas\u0131, A ve B noktalar\u0131 3 uzvunda bulunan 2 nokta oldu\u011fu gibi mafsal merkezleri olduklar\u0131ndan dolay\u0131 A noktas\u0131 ayn\u0131 zamanda 2 uzvunda ve B noktas\u0131 4 uzvunda bulunan noktalarla daima \u00e7ak\u0131\u015fan noktalard\u0131r. \u00d6yle ise\u00a0<strong>v<\/strong><sub>A2<\/sub>\u00a0= <strong>v<\/strong><sub>A3<\/sub>\u00a0= <strong>v<\/strong><sub>A<\/sub>\u00a0ve\u00a0<strong>v<\/strong><sub>B3<\/sub>\u00a0= <strong>v<\/strong><sub>B4<\/sub>\u00a0= <strong>v<\/strong><sub>B<\/sub>\u00a0dir. 2 uzvunu ele ald\u0131\u011f\u0131m\u0131zda bu uzuv sabit eksen etraf\u0131nda d\u00f6nme yapt\u0131\u011f\u0131ndan A noktas\u0131n\u0131n h\u0131z\u0131 AA<sub>0<\/sub> do\u011frusuna dik y\u00f6nde (\u03c9<sub>12<\/sub>&#8216;ye g\u00f6re) ve \u015fiddeti \u03c9<sub>12<\/sub>|AA<sub>0<\/sub>|= \u03c9<sub>12<\/sub>a<sub>2<\/sub>\u00a0olan bir vekt\u00f6rd\u00fcr. 3 uzvu dikkate al\u0131nd\u0131\u011f\u0131nda, A noktas\u0131n\u0131n h\u0131z\u0131 B noktas\u0131n\u0131n h\u0131z\u0131 ve ba\u011f\u0131l h\u0131z ile ifade edildi\u011finde:\u00a0<strong>v<\/strong><sub>A3<\/sub>\u00a0= <strong>v<\/strong><sub>B<\/sub>\u00a0+ <strong>v<\/strong><sub>A\/B<\/sub>\u00a0= <strong>v<\/strong><sub>A<\/sub>\u00a0olacakt\u0131r. Bu denklemde\u00a0<strong>v<\/strong><sub>A\/B<\/sub>, A noktas\u0131n\u0131n B noktas\u0131na g\u00f6re ba\u011f\u0131l h\u0131z\u0131 olup, AB do\u011frusuna diktir ve \u015fiddeti \u03b8<sub>13<\/sub>a<sub>3<\/sub>\u00a0dir. B noktas\u0131 3 ve 4 uzuvlar\u0131nda daima \u00e7ak\u0131\u015fan iki noktad\u0131r ve 4 uzvu \u00fczerinde bulunan B noktas\u0131n\u0131n y\u00f6r\u00fcngesi 1 ve 4 uzuvlar\u0131 aras\u0131nda bulunan kayar mafsaldan dolay\u0131, kayar mafsal eksenine paralel bir do\u011frudur. Bu durumda\u00a0<strong>v<\/strong><sub>A<\/sub>\u00a0h\u0131z vekt\u00f6r\u00fcn\u00fcn gerek \u015fiddeti ve gerek y\u00f6n\u00fc bilinmekte,\u00a0<b>v<sub>B<\/sub><\/b>\u00a0ve\u00a0<b>v<sub>A\/B<\/sub><\/b>\u00a0vekt\u00f6rlerinin ise y\u00f6nleri bilinmektedir (<strong>v<\/strong><sub>B<\/sub>\u00a0kayar mafsal eksenine paralel ve\u00a0<strong>v<\/strong><sub>A\/B<\/sub>\u00a0AB do\u011frusuna diktir). Bu durumda\u00a0<strong>v<\/strong><sub>B<\/sub>\u00a0ve<strong>\u00a0v<\/strong><sub>A\/B<\/sub> h\u0131z vekt\u00f6rlerinin \u015fiddeti bilinmemektedir. Bu bilinmeyenler h\u0131z devre denkleminde <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\dot{\\text{s}}}<\/span><sub>14<\/sub>\u00a0ve \u03c9<sub>13<\/sub>a<sub>3<\/sub> de\u011ferleridir.<\/p>\n<p>E\u011fer devre denklemini grafik olarak \u00e7\u00f6zmek istersek, denklemin her bir taraf\u0131nda bir bilinmeyenin olmas\u0131 daha uygun olacakt\u0131r (analitik \u00e7\u00f6z\u00fcmlerde, genellikle bilinen terimler denklemin sa\u011f\u0131na bilinmeyen terimler solunda bulunmas\u0131 istenilir). Bu durumda h\u0131z devre denklemi:<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\dot{\\text{s}}}<\/span><sub>14<\/sub>\u00a0= ia<sub>2<\/sub>\u03c9<sub>12<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup>\u00a0\u2212 ia<sub>3<\/sub>\u03c9<sub>13<\/sub>e<sup>i\u03b8<sub>13<\/sub><\/sup><\/p>\n<p>olarak yaz\u0131lacakt\u0131r. Bu denklem vekt\u00f6rel olarak<\/p>\n<p style=\"text-align: center\"><strong>v<\/strong><sub>B<\/sub>\u00a0= <strong>v<\/strong><sub>A<\/sub>\u00a0+ <strong>v<\/strong><sub>A\/B<\/sub><\/p>\n<p>dir ve dikkat edilir ise:<\/p>\n<p style=\"text-align: center\"><strong>v<\/strong><sub>B\/A<\/sub> = \u2212<strong>v<\/strong><sub>A\/B<\/sub> = \u2212ia<sub>3<\/sub>\u03c9<sub>13<\/sub>e<sup>i\u03b8<sub>13<\/sub><\/sup><\/p>\n<p>t\u00fcr.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1172\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img231-2.gif\" alt=\"\" width=\"585\" height=\"294\" \/><\/p>\n<p style=\"text-align: justify\" align=\"center\">Bilinmeyen de\u011ferlerin grafik \u00e7\u00f6z\u00fcm\u00fc i\u00e7in ilk olarak\u00a0<strong>v<\/strong><sub>A<\/sub>\u00a0h\u0131z vekt\u00f6r\u00fcn\u00fcn \u015fiddeti, |<strong>v<\/strong><sub>A<\/sub>|= \u03c9<sub>12<\/sub>|AA<sub>0<\/sub>|= \u03c9<sub>12<\/sub>a<sub>2<\/sub>\u00a0bulunur. Belirli bir\u00a0k<sub>v<\/sub> \u00f6l\u00e7e\u011fi kullan\u0131larak (birimi mm\/(mm\/s) = s dir) uzunlu\u011fu <strong>v<\/strong><sub>A<\/sub>\u00a0vekt\u00f6r\u00fc \u015fiddetine orant\u0131l\u0131, y\u00f6n\u00fc\u00a0<strong>v<\/strong><sub>A<\/sub>\u00a0vekt\u00f6r\u00fc y\u00f6n\u00fcnde olan (bu y\u00f6n A<sub>0<\/sub>A do\u011frusuna diktir) bir do\u011fru (vekt\u00f6r) \u00e7izilir. Kullan\u0131lan \u00f6l\u00e7ekten dolay\u0131, \u015fiddet \u00f6l\u00e7\u00fcs\u00fc uzunluk birimidir. Bu\u00a0<strong>v<\/strong><sub>A<\/sub>\u00a0vekt\u00f6r\u00fcn\u00fcn ba\u015flang\u0131\u00e7 noktas\u0131ndan\u00a0<b>v<sub>B<\/sub><\/b>\u00a0vekt\u00f6r\u00fc y\u00f6n\u00fcnde (ki bu kayar mafsal ekseni y\u00f6n\u00fcd\u00fcr) bir do\u011fru, biti\u015f noktas\u0131ndan ise\u00a0<strong>v<\/strong><sub>A\/B<\/sub>\u00a0vekt\u00f6r\u00fc y\u00f6n\u00fcnde bir do\u011fru (ki bu y\u00f6n BA do\u011frusuna diktir) \u00e7izilir. Bu iki do\u011frunun kesi\u015ftikleri nokta\u00a0<strong>v<\/strong><sub>B<\/sub>\u00a0ve\u00a0<strong>v<\/strong><sub>A\/B<\/sub>\u00a0vekt\u00f6rlerinin \u015fiddetini ve y\u00f6nlerini bize verecek, elde edilen \u00fc\u00e7gende\u00a0<strong>v<\/strong><sub>B<\/sub>\u00a0= <strong>v<\/strong><sub>A<\/sub>\u00a0+ <strong>v<\/strong><sub>A\/B<\/sub>\u00a0denklemi sa\u011flanm\u0131\u015f olacakt\u0131r. Bu \u015fekle\u00a0<strong><span style=\"color: #cc0000\">h\u0131z vekt\u00f6r \u00e7okgeni<\/span><\/strong>\u00a0denmektedir. Bu vekt\u00f6rlerin uzunluklar\u0131n\u0131 \u00f6l\u00e7er,\u00a0k<sub>v<\/sub> \u00f6l\u00e7e\u011fi ile b\u00f6ler isek, h\u0131z \u015fiddetlerini belirlemi\u015f oluruz. \u00d6rne\u011fin \u03c9<sub>13\u00a0<\/sub>= d\u03b8<sub>13<\/sub>\/dt a\u00e7\u0131sal h\u0131z\u0131 bulmak i\u00e7in ilk olarak okunan\u00a0<strong>v<\/strong><sub>A\/B<\/sub>\u00a0uzunlu\u011fu\u00a0k<sub>v<\/sub>\u00a0ile b\u00f6l\u00fcnerek h\u0131z \u015fiddeti bulunduktan sonra,\u00a0<strong>v<\/strong><sub>A\/B<\/sub> = \u03c9<sub>13<\/sub>a<sub>3<\/sub>\u00a0oldu\u011fundan ve a<sub>3<\/sub>\u00a0= |AB| oldu\u011fundan, a\u00e7\u0131sal h\u0131z \u03c9<sub>13<\/sub>\u00a0= |<strong>v<\/strong><sub>A\/B<\/sub>|\/a<sub>3<\/sub>\u00a0denklemi kullan\u0131larak belirlenebilir. A\u00e7\u0131sal h\u0131z\u0131n y\u00f6n\u00fcn\u00fc bulmak i\u00e7in ise, bulunan\u00a0<strong>v<\/strong><sub>A\/B<\/sub> h\u0131z vekt\u00f6r\u00fc y\u00f6n\u00fc kullan\u0131larak karar verilebilir. \u00d6rne\u011fin \u015fekilde <strong>v<\/strong><sub>A\/B<\/sub>\u00a0h\u0131z vekt\u00f6r\u00fc a\u015fa\u011f\u0131ya do\u011fru oldu\u011fundan ve bu vekt\u00f6rde B noktas\u0131n\u0131n A noktas\u0131na g\u00f6re ba\u011f\u0131l h\u0131z\u0131 incelendi\u011finden, 3 uzvunun saat yelkovan\u0131 y\u00f6n\u00fcnde bir a\u00e7\u0131sal h\u0131z\u0131 oldu\u011fu g\u00f6r\u00fclecektir.<\/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:550px\" data-flickity-options='{\"groupCells\":true,\"cellSelector\":\".su-image-carousel-item\",\"adaptiveHeight\":true,\"cellAlign\":\"left\",\"prevNextButtons\":true,\"pageDots\":false,\"autoPlay\":false,\"imagesLoaded\":true,\"contain\":false,\"selectedAttraction\":1,\"friction\":1}' id=\"su_image_carousel_69d88fea68b1d\"><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/slider_vel2_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\/slider_vel2_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\/slider_vel2_2.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\/slider_vel2_2.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\/slider_vel2_3.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\/slider_vel2_3.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\/slider_vel2_4.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\/slider_vel2_4.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><\/div><script id=\"su_image_carousel_69d88fea68b1d_script\">if(window.SUImageCarousel){setTimeout(function() {window.SUImageCarousel.initGallery(document.getElementById(\"su_image_carousel_69d88fea68b1d\"))}, 0);}var su_image_carousel_69d88fea68b1d_script=document.getElementById(\"su_image_carousel_69d88fea68b1d_script\");if(su_image_carousel_69d88fea68b1d_script){su_image_carousel_69d88fea68b1d_script.parentNode.removeChild(su_image_carousel_69d88fea68b1d_script);}<\/script><\/p>\n<p>E\u011fer bilinmeyen ba\u011f\u0131ml\u0131 konum de\u011fi\u015fkenleri verilen bir ba\u011f\u0131ms\u0131z konum de\u011fi\u015fkeni de\u011ferine g\u00f6re \u00e7\u00f6z\u00fclm\u00fc\u015f ise, (x ve \u03b8<sub>13<\/sub> de\u011ferleri verilen bir \u03b8<sub>12<\/sub> de\u011ferine g\u00f6re biliniyor ise) h\u0131z devre denklemi daima h\u0131z de\u011fi\u015fkenleri aras\u0131nda (\u03c9<sub>12<\/sub>, \u03c9<sub>13<\/sub> ve <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\dot{\\text{s}}}<\/span><sub>14<\/sub> de\u011fi\u015fkenleri) bir lineer ili\u015fkiyi belirleyecektir. Giri\u015f a\u00e7\u0131s\u0131n\u0131n zamana g\u00f6re de\u011fi\u015fimi, \u03c9<sub>12<\/sub>, biliniyor ise, h\u0131z devre denkleminden bilinmeyen ba\u011f\u0131ml\u0131 iki h\u0131z de\u011fi\u015fkeni \u03c9<sub>13<\/sub> ve <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\dot{\\text{s}}}<\/span><sub>14<\/sub> lineer cebir kurallar\u0131 kullan\u0131larak (2a) ve (2b) denklemlerinden kolayca \u00e7\u00f6z\u00fclebilir. Genellikle iki bilinmeyenli iki denklem oldu\u011fundan Cramer kural\u0131 kullan\u0131ld\u0131\u011f\u0131nda:<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{{\\dot{\\text{s}}}}_{{14}}}=\\frac{{\\left| {\\begin{array}{cc} {\\text{i}{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} &amp; {\\text{i}{{\\text{a}}_{3}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} \\\\ {-\\text{i}{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} &amp; {-\\text{i}{{\\text{a}}_{3}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} \\end{array}} \\right|}}{{\\left| {\\begin{array}{cc} 1 &amp; {\\text{i}{{\\text{a}}_{3}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} \\\\ 1 &amp; {-\\text{i}{{\\text{a}}_{3}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} \\end{array}} \\right|}}=\\frac{{{{\\text{a}}_{2}}{{\\text{a}}_{3}}\\left( {{{\\text{e}}^{{\\text{i}\\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}}-{{\\text{e}}^{{-\\text{i}\\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}}} \\right)}}{{-\\text{i}{{\\text{a}}_{3}}\\left( {{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{13}}}}}}+{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} \\right)}}{{\\text{\u03c9}}_{{12}}} <\/span><\/p>\n<p>veya<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{{\\dot{\\text{s}}}}_{{14}}}=-{{\\text{a}}_{2}}\\frac{{\\sin \\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}{{\\cos {{\\text{\u03b8}}_{{13}}}}}{{\\text{\u03c9}}_{{12}}} <\/span><\/td>\n<td style=\"text-align: right\" align=\"right\">(4)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>ve<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{\u03c9}}_{{13}}}=\\frac{{\\left| {\\begin{array}{cc} 1 &amp; {\\text{i}{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} \\\\ 1 &amp; {-\\text{i}{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} \\end{array}} \\right|}}{{\\left| {\\begin{array}{cc} 1 &amp; {\\text{i}{{\\text{a}}_{3}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} \\\\ 1 &amp; {-\\text{i}{{\\text{a}}_{3}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} \\end{array}} \\right|}}=\\frac{{-\\text{i}{{\\text{a}}_{2}}\\left( {{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{12}}}}}}+{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} \\right)}}{{-\\text{i}{{\\text{a}}_{3}}\\left( {{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{13}}}}}}+{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} \\right)}}{{\\text{\u03c9}}_{{12}}}=\\frac{{{{\\text{a}}_{2}}}}{{{{\\text{a}}_{3}}}}\\frac{{\\cos {{\\text{\u03b8}}_{{12}}}}}{{\\cos {{\\text{\u03b8}}_{{13}}}}}{{\\text{\u03c9}}_{{12}}} <\/span><\/td>\n<td style=\"text-align: right\" align=\"right\">(5)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Dikkat edilir ise, karma\u015f\u0131k say\u0131larla yaz\u0131lm\u0131\u015f h\u0131z denklemi ve e\u015fleni\u011fi kullan\u0131lmak yerine, istenildi\u011finde vekt\u00f6r h\u0131z denkleminin x ve y bile\u015fenleri (karma\u015f\u0131k say\u0131larla yaz\u0131lm\u0131\u015f h\u0131z devre denkleminin reel ve sanal k\u0131s\u0131mlar\u0131) ayr\u0131 ayr\u0131 e\u015fitlenerek iki skaler denklem elde edilerek de bilinmeyen h\u0131z de\u011fi\u015fkenleri i\u00e7in \u00e7\u00f6z\u00fcm yap\u0131labilecektir. \u00d6rne\u011fin krank biyel mekanizmas\u0131 i\u00e7in bu skaler denklemler:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\dot{\\text{s}}}<\/span><sub>14<\/sub> \u2212 a<sub>3<\/sub>\u03c9<sub>13<\/sub>sin\u03b8<sub>13<\/sub> = \u2212a<sub>2<\/sub>\u03c9<sub>12<\/sub>sin\u03b8<sub>12<\/sub><\/p>\n<\/td>\n<td align=\"right\">(6)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>ve<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td>\n<p style=\"text-align: center\">a<sub>3<\/sub>\u03c9<sub>13<\/sub>cos\u03b8<sub>13<\/sub> = a<sub>2<\/sub>\u03c9<sub>12<\/sub>cos\u03b8<sub>12<\/sub><\/p>\n<\/td>\n<td align=\"right\">(7)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>dir. Bu iki denklem kullan\u0131larak da \u03c9<sub>13<\/sub> ve <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\dot{\\text{s}}}<\/span><sub>14<\/sub> i\u00e7in ayn\u0131 denklemler (4 ve 5 denklemleri) elde edilebilir. Hangi y\u00f6ntem kullan\u0131l\u0131rsa kullan\u0131ls\u0131n, sonu\u00e7ta h\u0131z de\u011fi\u015fkenlerini veren denklemler reel denklemler olmalar\u0131 gerekir (\u03c9<sub>13<\/sub> a\u00e7\u0131sal h\u0131z de\u011feri karma\u015f\u0131k bir de\u011fer olamaz).<\/p>\n<p>\u0130vme analizi i\u00e7in en kolay y\u00f6ntem h\u0131z devre denkleminin zamana g\u00f6re t\u00fcrevi al\u0131narak, h\u0131z de\u011fi\u015fkenlerinin zamana g\u00f6re t\u00fcrevi olan\u00a0<strong><span style=\"color: #cc0000\">ivme de\u011fi\u015fkenleri<\/span><\/strong>ni i\u00e7eren <span style=\"color: #cc0000\"><strong>ivme devre denklemi<\/strong><\/span>nin elde edilmesidir. \u00d6rne\u011fin yukar\u0131da verilmi\u015f olan krank-biyel mekanizmas\u0131 i\u00e7in devre kapal\u0131l\u0131k denkleminin t\u00fcrevi olan 2a ve 2b h\u0131z devre denklemlerinin zamana g\u00f6re t\u00fcrevleri al\u0131nd\u0131\u011f\u0131nda:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\ddot{\\text{s}}}<\/span><sub>14<\/sub> + ia<sub>3<\/sub>\u03b1<sub>13<\/sub>e<sup>i\u03b8<sub>13<\/sub><\/sup> \u2212 a<sub>3<\/sub>\u03c9<sub>13<\/sub><sup>2<\/sup>e<sup>i\u03b8<sub>13<\/sub><\/sup> = ia<sub>2<\/sub>\u03b1<sub>12<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup>\u00a0\u2212 a<sub>2<\/sub>\u03c9<sub>12<\/sub><sup>2<\/sup>e<sup>i\u03b8<sub>12<\/sub><\/sup><\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\ddot{\\text{s}}}<\/span><sub>14<\/sub> \u2212 ia<sub>3<\/sub>\u03b1<sub>13<\/sub>e<sup>-i\u03b8<sub>13<\/sub><\/sup> \u2212 a<sub>3<\/sub>\u03c9<sub>13<\/sub><sup>2<\/sup>e<sup>-i\u03b8<sub>13<\/sub><\/sup> = \u2212ia<sub>2<\/sub>\u03b1<sub>12<\/sub>e<sup>-i\u03b8<sub>12<\/sub><\/sup>\u00a0\u2212 a<sub>2<\/sub>\u03c9<sub>12<\/sub><sup>2<\/sup>e<sup>-i\u03b8<sub>12<\/sub><\/sup><\/p>\n<\/td>\n<td align=\"right\">\n<p style=\"text-align: right\">(8a)<\/p>\n<p style=\"text-align: right\">(8b)<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>denklemleri elde edilir bu denklemlerde ivme de\u011fi\u015fkenleri: \u03b1<sub>12<\/sub> = d<sup>2<\/sup>\u03b8<sub>12<\/sub>\/dt<sup>2<\/sup>, \u03b1<sub>13<\/sub> = d<sup>2<\/sup>\u03b8<sub>13<\/sub>\/dt<sup>2<\/sup>, <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\ddot{\\text{\\text{s}}}}<\/span><sub>14<\/sub>\u00a0= d<sup>2<\/sup>s<sub>14<\/sub>\/dt<sup>2 <\/sup>dir. Dikkat edilir ise karma\u015f\u0131k say\u0131 ile yaz\u0131lm\u0131\u015f olan bu denklem vekt\u00f6rel olarak:<\/p>\n<p style=\"text-align: center\"><strong>a<\/strong><sub>B<\/sub> + <strong>a<sup>t<\/sup><\/strong><sub>A\/B<\/sub>\u00a0+\u00a0<strong>a<sup>n<\/sup><\/strong><sub>A\/B<\/sub> = <strong>a<sup>t<\/sup><\/strong><sub>A<\/sub>\u00a0+\u00a0<strong>a<sup>n<\/sup><\/strong><sub>A<\/sub><\/p>\n<p>d\u0131r. \u0130vme devre denklemleri ivme de\u011fi\u015fkenlerine (\u03b1<sub>12<\/sub>, \u03b1<sub>13<\/sub> ve <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\ddot{\\text{s}}}<\/span><sub>14<\/sub>) g\u00f6re daima lineer bir denklem tak\u0131m\u0131 olu\u015fturur. E\u011fer giri\u015f kolunun a\u00e7\u0131sal ivmesi \u03b1<sub>12<\/sub> = d<sup>2<\/sup>\u03b8<sub>12<\/sub>\/dt<sup>2<\/sup> biliniyor ise, (8a) ve (8b) denklemleri kolayl\u0131kla bilinmeyen ivme de\u011fi\u015fkenleri \u03b1<sub>13<\/sub> ve <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\ddot{\\text{s}}}<\/span><sub>14<\/sub>\u00a0i\u00e7in \u00e7\u00f6z\u00fclebilir. \u00d6rne\u011fin Cramer kural\u0131 kullan\u0131larak \u03b1<sub>13<\/sub>\u00a0:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{\u03b1}}_{{13}}}=\\frac{{{{\\text{a}}_{2}}{{\\text{\u03b1}}_{{12}}}\\cos {{\\text{\u03b8}}_{{12}}}-{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}^{2}\\sin {{\\text{\u03b8}}_{{12}}}+{{\\text{a}}_{3}}{{\\text{\u03c9}}_{{13}}}^{2}\\sin {{\\text{\u03b8}}_{{13}}}}}{{{{\\text{a}}_{3}}\\cos {{\\text{\u03b8}}_{{13}}}}} <\/span><\/td>\n<td style=\"text-align: right\" align=\"right\">(9)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\ddot{s}}<\/span><sub>14<\/sub> ise denklem (8a) n\u0131n reel k\u0131sm\u0131ndan:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\ddot{\\text{s}}}<\/span><sub>14<\/sub> = \u2212a<sub>2<\/sub>\u03b1<sub>12<\/sub>sin\u03b8<sub>12<\/sub>\u00a0\u2212 a<sub>2<\/sub>\u03c9<sub>12<\/sub><sup>2<\/sup>cos\u03b8<sub>12<\/sub>\u00a0+ ia<sub>3<\/sub>\u03b1<sub>13<\/sub>sin\u03b8<sub>13<\/sub>\u00a0+ a<sub>3<\/sub>\u03c9<sub>13<\/sub><sup>2<\/sup>cos\u03b8<sub>13<\/sub><\/p>\n<\/td>\n<td style=\"text-align: right\" align=\"right\">(10)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>olarak elde edilir. Bir ba\u015fka yakla\u015f\u0131m ise h\u0131z de\u011fi\u015fkenleri denklemlerinin zamana g\u00f6re t\u00fcrevlerinin al\u0131nmas\u0131d\u0131r. \u00d6rnek olarak denklem (4) ve (5) denklemlerinde verilmekte olan h\u0131z denklemlerinin zamana g\u00f6re t\u00fcrevleri al\u0131nd\u0131\u011f\u0131nda, ivme de\u011fi\u015fkenleri elde edilecektir. (5) denkleminin t\u00fcrevi al\u0131nd\u0131\u011f\u0131nda do\u011frudan elde edilen 3 uzvunun a\u00e7\u0131sal ivmesi<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{\u03b1}}_{{13}}}=\\frac{{{{\\text{a}}_{2}}}}{{{{\\text{a}}_{3}}}}\\frac{1}{{{{{\\cos }}^{2}}{{\\text{\u03b8}}_{{13}}}}}\\left[ {\\cos {{\\text{\u03b8}}_{{12}}}\\cos {{\\text{\u03b8}}_{{13}}}{{\\text{\u03b1}}_{{12}}}-\\sin {{\\text{\u03b8}}_{{12}}}\\cos {{\\text{\u03b8}}_{{13}}}{{\\text{\u03c9}}_{{12}}}^{2}+\\cos {{\\text{\u03b8}}_{{12}}}\\sin {{\\text{\u03b8}}_{{13}}}{{\\text{\u03c9}}_{{12}}}{{\\text{\u03c9}}_{{13}}}} \\right] <\/span><\/p>\n<p>olacakt\u0131r. Benzer bir \u015fekilde (4) denkleminin t\u00fcrevi al\u0131nd\u0131\u011f\u0131nda 4 uzvunun lineer \u00f6teleme ivmesi:<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{{\\ddot{\\text{s}}}}_{{14}}}=-{{\\text{a}}_{2}}\\frac{{\\sin \\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}{{\\cos {{\\text{\u03b8}}_{{13}}}}}{{\\text{\u03b1}}_{{12}}}-{{\\text{a}}_{2}}\\frac{{\\cos \\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}{{\\cos {{\\text{\u03b8}}_{{13}}}}}{{\\text{\u03c9}}_{{12}}}^{2}+{{\\text{a}}_{2}}\\frac{{\\cos {{\\text{\u03b8}}_{{12}}}}}{{{{{\\cos }}^{2}}{{\\text{\u03b8}}_{{13}}}}}{{\\text{\u03c9}}_{{12}}}{{\\text{\u03c9}}_{{13}}} <\/span><\/p>\n<p>olarak elde edilir. Elde edilen denklemde terimler (9) ve (10) denklemine g\u00f6re farkl\u0131d\u0131r. \u0130\u015flemler do\u011fru olarak yap\u0131ld\u0131\u011f\u0131 takdirde, gerekli e\u015fitlikler kullan\u0131l\u0131rsa, \u03b1<sub>13<\/sub> ve <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\ddot{\\text{s}}}<\/span><sub>14<\/sub>\u00a0\u00fc veren denklemlerin birbirine e\u015fit oldu\u011fu g\u00f6sterilebilir. Denklemler kullan\u0131ld\u0131\u011f\u0131nda ayn\u0131 ba\u011f\u0131ms\u0131z parametre de\u011ferlerine g\u00f6re mutlaka ayn\u0131 say\u0131sal sonu\u00e7 al\u0131nacakt\u0131r.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1173 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img231-3.gif\" alt=\"\" width=\"607\" height=\"359\" \/><\/p>\n<p>Grafik olarak \u00e7\u00f6z\u00fcm i\u00e7in ivme devre denklemi vekt\u00f6rel g\u00f6sterim kullan\u0131larak ve baz\u0131 terimlerin yerleri de\u011fi\u015ftirilerek yaz\u0131ld\u0131\u011f\u0131nda:<\/p>\n<p style=\"text-align: center\"><strong>a<\/strong><sub>B<\/sub>\u00a0= <strong>a<sup>t<\/sup><\/strong><sub>A<\/sub>\u00a0+\u00a0<strong>a<sup>n<\/sup><\/strong><sub>A<\/sub>\u00a0+ <strong>a<sup>t<\/sup><\/strong><sub>B\/A<\/sub>\u00a0+\u00a0<strong>a<sup>n<\/sup><\/strong><sub>B\/A<\/sub><\/p>\n<p>Bu denklemde\u00a0<strong>a<\/strong><sub>B\/A<\/sub> = \u2212<strong>a<\/strong><sub>A\/B<\/sub>\u00a0dir. Denklemin bu \u015fekilde yaz\u0131lmas\u0131n\u0131n tek nedeni grafik \u00e7\u00f6z\u00fcm i\u00e7in denklemin her iki taraf\u0131nda tek bilinmeyen b\u0131rak\u0131lmas\u0131 i\u00e7indir.\u00a0<strong>a<\/strong><sub>B<\/sub>\u00a0ivmesinin \u015fiddeti bir bilinmeyen ve\u00a0<strong>a<sup>t<\/sup><\/strong><sub>B\/A<\/sub>\u00a0te\u011fetsel ivmesi \u015fiddeti ise ikinci bilinmeyendir. Giri\u015f uzvu a\u00e7\u0131sal ivmesi bilindi\u011finden\u00a0<strong>a<sup>t<\/sup><\/strong><sub>A<\/sub>\u00a0te\u011fetsel ivmesi; h\u0131z analizi yap\u0131ld\u0131 ise, h\u0131zlar bilindi\u011finden\u00a0<strong>a<sup>n<\/sup><\/strong><sub>A<\/sub>\u00a0ve\u00a0<strong>a<sup>n<\/sup><\/strong><sub>B\/A<\/sub>\u00a0normal ivmeleri hem \u015fiddet ve hem de y\u00f6n olarak bilinmektedir. \u0130vme \u015fiddetleri i\u00e7in belirli bir\u00a0k<sub>a<\/sub>\u00a0\u00f6l\u00e7e\u011fi kullan\u0131lacakt\u0131r (birimi mm\/(mm\/s<sup>2<\/sup>) = s<sup>2<\/sup>), ve bu \u00f6l\u00e7ek ile ivme \u015fiddetleri belirli bir uzunluk boyutuna getirilecektir. <strong>a<sup>t<\/sup><\/strong><sub>A<\/sub>\u00a0veya\u00a0<strong>a<sup>n<\/sup><\/strong><sub>A<\/sub>\u00a0vekt\u00f6rlerinden ba\u015flayarak her iki vekt\u00f6r u\u00e7 uca \u00e7izildi\u011finde A noktas\u0131n\u0131n ivmesi g\u00f6sterilmi\u015ftir. Bu vekt\u00f6rlerin ucundan bilinen\u00a0<strong>a<sup>n<\/sup><\/strong><sub>B\/A<\/sub>\u00a0vekt\u00f6r\u00fc ayn\u0131 \u00f6l\u00e7ek kullan\u0131larak \u00e7izilir. Bundan sonra bu u\u00e7tan y\u00f6n\u00fc BA do\u011frusuna dik bir do\u011fru \u00e7izilir ise, \u015fiddeti bilinmeyen\u00a0<strong>a<sup>t<\/sup><\/strong><sub>B\/A<\/sub> vekt\u00f6r\u00fc y\u00f6n\u00fcnde bir do\u011fru \u00e7izilmi\u015f olacakt\u0131r (\u00fcstteki \u015fekil). B noktas\u0131n\u0131n ivmesi ise kayar mafsal eksenine paralel olmal\u0131 ve yukar\u0131da \u00e7izilmi\u015f olan vekt\u00f6rlerin vekt\u00f6rel toplam\u0131na e\u015fit olmal\u0131d\u0131r. \u00d6yle ise, ba\u015flang\u0131\u00e7 noktas\u0131ndan kayar \u00e7ift eksenine paralel bir do\u011fru \u00e7izdi\u011fimizde, AB do\u011frusuna dik \u00e7izilen do\u011fru ile kesi\u015fti\u011fi nokta h\u0131z devre denkleminin kapal\u0131l\u0131k noktas\u0131d\u0131r. Elde edilen \u015fekil <strong>ivme poligonu<\/strong>\u00a0olarak adland\u0131r\u0131l\u0131r. Bilinmeyen\u00a0<strong>a<sup>t<\/sup><\/strong><sub>B\/A<\/sub>\u00a0ve\u00a0<strong>a<\/strong><sub>B<\/sub>\u00a0ivme \u015fiddetleri \u015fekildeki bu ivmeleri g\u00f6steren uzunluklar\u0131n kullan\u0131lm\u0131\u015f olan\u00a0k<sub>a<\/sub>\u00a0\u00f6l\u00e7e\u011fine b\u00f6l\u00fcnmesi ile elde edilir.<\/p>\n<p style=\"text-align: center\"><div class=\"su-image-carousel  su-image-carousel-has-spacing su-image-carousel-has-lightbox su-image-carousel-has-outline su-image-carousel-adaptive su-image-carousel-slides-style-default su-image-carousel-controls-style-dark su-image-carousel-align-center\" style=\"max-width:550px\" data-flickity-options='{\"groupCells\":true,\"cellSelector\":\".su-image-carousel-item\",\"adaptiveHeight\":true,\"cellAlign\":\"left\",\"prevNextButtons\":true,\"pageDots\":false,\"autoPlay\":false,\"imagesLoaded\":true,\"contain\":false,\"selectedAttraction\":1,\"friction\":1}' id=\"su_image_carousel_69d88fea696c4\"><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/slider_acc1.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\/slider_acc1.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\/slider_acc2.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\/slider_acc2.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\/slider_acc3.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\/slider_acc3.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\/slider_acc4.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\/slider_acc4.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\/slider_acc5.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\/slider_acc5.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><\/div><script id=\"su_image_carousel_69d88fea696c4_script\">if(window.SUImageCarousel){setTimeout(function() {window.SUImageCarousel.initGallery(document.getElementById(\"su_image_carousel_69d88fea696c4\"))}, 0);}var su_image_carousel_69d88fea696c4_script=document.getElementById(\"su_image_carousel_69d88fea696c4_script\");if(su_image_carousel_69d88fea696c4_script){su_image_carousel_69d88fea696c4_script.parentNode.removeChild(su_image_carousel_69d88fea696c4_script);}<\/script><\/p>\n<p>H\u0131z ve ivme de\u011fi\u015fkenlerinin de\u011ferleri o konum i\u00e7in elde edildikten sonra, her hangi bir C noktas\u0131n\u0131n h\u0131z ve ivmesi, o noktan\u0131n konum vekt\u00f6r\u00fcn\u00fcn yaz\u0131lmas\u0131 ve zamana g\u00f6re bu konum vekt\u00f6r\u00fcn\u00fcn t\u00fcrevlerinin al\u0131narak h\u0131z ve ivmesinin bulunmas\u0131ndan ibarettir. \u015eekilde g\u00f6sterilmi\u015f olan mekanizmada biyel uzvunu geni\u015fleterek bir uzuv \u00fczerinde bir C noktas\u0131n\u0131 ele alal\u0131m. \u015eekilde <strong>r<\/strong><sub>C<\/sub>\u00a0konum vekt\u00f6r\u00fc:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td>\n<p style=\"text-align: center\"><strong>r<\/strong><sub>C<\/sub>\u00a0= s<sub>14<\/sub> + ic<sub>1<\/sub> + b<sub>3<\/sub>e<sup>i(\u03b8<sub>13 <\/sub>\u2212<sub>\u00a0<\/sub>\u03b3<sub>3<\/sub>)<\/sup><\/p>\n<\/td>\n<td style=\"text-align: right\" align=\"right\">(11)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>C noktas\u0131n\u0131n h\u0131z ve ivmesi:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td>\n<p style=\"text-align: center\"><strong>v<\/strong><sub>C<\/sub> = <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\dot{\\text{s}}}<\/span><sub>14<\/sub> + ib<sub>3<\/sub>\u03c9<sub>13<\/sub>e<sup>i(\u03b8<sub>13 <\/sub>\u2212<sub>\u00a0<\/sub>\u03b3<sub>3<\/sub>)<\/sup><\/p>\n<p style=\"text-align: center\"><strong>a<\/strong><sub>C<\/sub> = <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\ddot{\\text{s}}}<\/span><sub>14<\/sub> + ib<sub>3<\/sub>\u03b1<sub>13<\/sub>e<sup>i(\u03b8<sub>13 <\/sub>\u2212<sub>\u00a0<\/sub>\u03b3<sub>3<\/sub>)<\/sup> \u2212 b<sub>3<\/sub>\u03c9<sub>13<\/sub><sup>2<\/sup>e<sup>i(\u03b8<sub>13 <\/sub>\u2212<sub>\u00a0<\/sub>\u03b3<sub>3<\/sub>)<\/sup><\/p>\n<\/td>\n<td align=\"right\">\n<p style=\"text-align: right\">(12)<\/p>\n<p style=\"text-align: right\">(13)<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1174 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img231-4.gif\" alt=\"\" width=\"417\" height=\"212\" data-wp-editing=\"1\" \/><\/p>\n<p>Yukar\u0131da karma\u015f\u0131k say\u0131lar ile yaz\u0131lm\u0131\u015f olan denklemler vekt\u00f6rel olarak:<\/p>\n<p style=\"text-align: center\"><strong>v<\/strong><sub>C<\/sub> = <strong>v<\/strong><sub>B<\/sub> + <strong>v<\/strong><sub>C\/B<\/sub><br \/>\nve<br \/>\n<strong>a<\/strong><sub>C<\/sub>\u00a0= <strong>a<\/strong><sub>B<\/sub>\u00a0+ <strong>a<sup>t<\/sup><\/strong><sub>C\/B<\/sub>\u00a0+ <strong>a<sup>n<\/sup><\/strong><sub>C\/B<\/sub><\/p>\n<p>vekt\u00f6rel h\u0131z ve ivme denklemlerinden farkl\u0131 denklemler de\u011fillerdir. E\u011fer verilen giri\u015f de\u011ferleri ile (\u03b8<sub>12<\/sub>, \u03c9<sub>12<\/sub> ve \u03b1<sub>12<\/sub>), konum de\u011fi\u015fkenleri (\u03b8<sub>13<\/sub>\u00a0ve s<sub>14<\/sub>), h\u0131z de\u011fi\u015fkenleri (\u03c9<sub>13<\/sub> ve <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\dot{\\text{s}}}<\/span><sub>14<\/sub>) ile ivme de\u011fi\u015fkenleri (\u03b1<sub>13<\/sub> ve <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\ddot{\\text{s}}}<\/span><sub>14<\/sub>) i\u00e7in \u00e7\u00f6z\u00fcm yap\u0131ld\u0131 ise, denklemin sa\u011f\u0131nda bulunan t\u00fcm terimler bilinmektedir. Bu nedenle C noktas\u0131n\u0131n konumu, h\u0131z\u0131 ve ivmesi kolayca vekt\u00f6rel toplam yap\u0131larak bulunabilir. C noktas\u0131n\u0131n konumu, h\u0131z\u0131 ve ivmesi devre kapal\u0131l\u0131k denklemi, h\u0131z devre denklemi ve ivme devre denklemi \u00e7\u00f6z\u00fclmeden belirlenemez.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1175 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img231-5.gif\" alt=\"\" width=\"474\" height=\"654\" \/><\/p>\n<p>Grafik \u00e7\u00f6z\u00fcm i\u00e7in ise, C noktas\u0131n\u0131n h\u0131z\u0131 ve ivmesi i\u00e7in verilmi\u015f olan denklemdeki vekt\u00f6rler belirli bir \u00f6l\u00e7ek kullan\u0131larak \u00f6nceden \u00e7izilmi\u015f olan h\u0131z ve ivme poligonu \u00fczerine \u00e7izilerek bulunabilir. \u00c7\u00fcnk\u00fc, C noktas\u0131n\u0131n h\u0131z ve ivmesini veren terimler aras\u0131nda h\u0131z ve ivme devre denklemlerini \u00e7\u00f6zd\u00fc\u011f\u00fcm\u00fczde elde etmi\u015f oldu\u011fumuz terimler bulunmaktad\u0131r. Bu nedenle h\u0131z devre denkleminde elde edilmi\u015f olan\u00a0<strong>v<\/strong><sub>B<\/sub> h\u0131z vekt\u00f6r\u00fcn\u00fcn u\u00e7 noktas\u0131ndan <strong>v<\/strong><sub>C\/B<\/sub>\u00a0h\u0131z vekt\u00f6r\u00fcn\u00fc \u00e7izdi\u011fimiz takdirde C noktas\u0131n\u0131n h\u0131z\u0131n\u0131 bulabiliriz. Benzer bir \u015fekilde ivme devre denklemi i\u00e7in \u00e7izmi\u015f oldu\u011fumuz ivme poligonu \u00fczerinde bulunan\u00a0<strong>a<\/strong><sub>B<\/sub>\u00a0ivme vekt\u00f6r\u00fcn\u00fcn u\u00e7 noktas\u0131ndan\u00a0<strong>a<sup>n<\/sup><\/strong><sub>C\/B<\/sub>\u00a0ve\u00a0<strong>a<sup>t<\/sup><\/strong><sub>C\/B<\/sub>\u00a0ivme vekt\u00f6rlerini arka arkaya \u00e7izdi\u011fimizde (s\u0131ra \u00f6nemli de\u011fil)\u00a0<strong>a<\/strong><sub>B<\/sub>\u00a0vekt\u00f6r\u00fcn\u00fcn ba\u015flang\u0131\u00e7 noktas\u0131n\u0131n elde edilen son u\u00e7 noktaya birle\u015ftiren do\u011fru bize C noktas\u0131n\u0131n ivmesini verecektir. Alternatif olarak C noktas\u0131n\u0131n konumu, h\u0131z\u0131 ve ivmesi 3 uzvu \u00fczerinde bulunan B noktas\u0131 yerine ayn\u0131 uzuv \u00fczerinde bulunan A noktas\u0131 kullan\u0131larak yaz\u0131labilir. Bu durumda C noktas\u0131n\u0131n konumu, h\u0131z\u0131 ve ivmesi:<\/p>\n<p style=\"text-align: center\"><strong>r<\/strong><sub>C<\/sub> = = a<sub>2<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup> + c<sub>3<\/sub>e<sup>i(\u03b8<sub>13<\/sub> + \u03b2<sub>3<\/sub> \u2212 \u03c0)<\/sup> = a<sub>2<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup> \u2212 c<sub>3<\/sub>e<sup>i(\u03b8<sub>13 <\/sub>+<sub>\u00a0<\/sub>\u03b2<sub>3<\/sub>)<\/sup><\/p>\n<p style=\"text-align: center\"><strong>v<\/strong><sub>C<\/sub> = ia<sub>2<\/sub>\u03c9<sub>12<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup> \u2212 ic<sub>3<\/sub>\u03c9<sub>13<\/sub>e<sup>i(\u03b8<sub>13 <\/sub>+<sub>\u00a0<\/sub>\u03b2<sub>3<\/sub>)<\/sup><\/p>\n<p style=\"text-align: center\"><strong>a<\/strong><sub>C<\/sub> = ia<sub>2<\/sub>\u03b1<sub>12<\/sub>e<sup>i\u03b8<sub>12<\/sub><\/sup> \u2212 a<sub>2<\/sub>\u03c9<sub>12<\/sub><sup>2<\/sup>e<sup>i\u03b8<sub>12<\/sub><\/sup> \u2212 ic<sub>3<\/sub>\u03b1<sub>13<\/sub>e<sup>i(\u03b8<sub>13 <\/sub>+<sub>\u00a0<\/sub>\u03b2<sub>3<\/sub>)<\/sup>\u00a0+ c<sub>3<\/sub>\u03c9<sub>13<\/sub><sup>2<\/sup>e<sup>i(\u03b8<sub>13 <\/sub>+<sub>\u00a0<\/sub>\u03b2<sub>3<\/sub>)<\/sup><\/p>\n<p>Dikkat edilir ise bu h\u0131z ve ivme denklemleri ayn\u0131 zamanda:<\/p>\n<p style=\"text-align: center\"><strong>v<\/strong><sub>C<\/sub> = <strong>v<\/strong><sub>A<\/sub> + <strong>v<\/strong><sub>C\/A<\/sub><br \/>\nve<br \/>\n<strong>a<\/strong><sub>C<\/sub>\u00a0= <strong>a<sup>n<\/sup><\/strong><sub>A<\/sub>\u00a0+ <strong>a<sup>t<\/sup><\/strong><sub>A<\/sub>\u00a0+ <strong>a<sup>t<\/sup><\/strong><sub>C\/A<\/sub>\u00a0+ <strong>a<sup>n<\/sup><\/strong><sub>C\/A<\/sub><\/p>\n<p>vekt\u00f6r denklemleri ile ayn\u0131d\u0131r.<\/p>\n<p>H\u0131z ve ivme poligonlar\u0131nda Vekt\u00f6rlerin son u\u00e7 noktalar\u0131n\u0131, h\u0131z\u0131n\u0131 veya ivmesini g\u00f6sterdikleri noktalar\u0131n harfleri ile i\u015faretleyelim (ivme poligonunda A veya C noktas\u0131n\u0131n ivmesi bir ka\u00e7 vekt\u00f6r\u00fcn toplam\u0131ndan olu\u015faca\u011f\u0131ndan son \u00e7izilen vekt\u00f6r\u00fcn u\u00e7 noktas\u0131 a veya c olarak i\u015faretlenecektir). <strong>v<\/strong><sub>A<\/sub>, <strong>v<\/strong><sub>B<\/sub>,\u00a0<strong>v<\/strong><sub>C<\/sub>\u00a0h\u0131z vekt\u00f6rlerinin u\u00e7 noktalar\u0131n\u0131 ve\u00a0<strong>a<\/strong><sub>A<\/sub>,\u00a0<strong>a<\/strong><sub>B<\/sub>,\u00a0<strong>a<\/strong><sub>C<\/sub>\u00a0ivme vekt\u00f6rlerinin u\u00e7lar\u0131n\u0131 birle\u015ftirdi\u011fimizde her iki poligonda abc gibi iki farkl\u0131 \u00fc\u00e7gen olu\u015facakt\u0131r. Bu \u00fc\u00e7genle ilgili olarak \u00f6nemli bir teorem bulunmaktad\u0131r:<\/p>\n<p><i><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\" \/><\/b> \u00a0 \u00a0 \u00a0<strong>H\u0131z ve ivme poligonunda olu\u015fan abc \u00fc\u00e7geni mekanizma uzvunda bulunan A, B ve C noktalar\u0131n\u0131n olu\u015fturdu\u011fu \u00fc\u00e7gen ile benzerdir. ABC \u00fc\u00e7geni ile abc \u00fc\u00e7geninin y\u00f6n\u00fc ayn\u0131d\u0131r (A dan B ye sonra C ye giderken yap\u0131lan d\u00f6nme saat yelkovan\u0131na ters ise, h\u0131z veya ivme poligonunda a dan b ye sonra c ye giderken d\u00f6nme saat yelkovan\u0131 y\u00f6n\u00fcne ters olacakt\u0131r).<\/strong><\/span><\/i><\/p>\n<p>Yukar\u0131da belirtilmi\u015f olan teorem <strong>Mehmke Teoremi<\/strong> veya\u00a0<strong>h\u0131z ve ivme g\u00f6r\u00fcnt\u00fc prensibi<\/strong>\u00a0olarak adland\u0131r\u0131l\u0131r. Bu teoremin kullan\u0131m yeri bir cisim \u00fczerinde iki noktan\u0131n h\u0131z ve ivmesini biliniyor ise, \u00fc\u00e7\u00fcnc\u00fc bir noktan\u0131n h\u0131z ve ivmesinin bu \u00fc\u00e7gen benzerli\u011finden belirlenmesidir.<\/p>\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n<p> <a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch4\/4-1-2\/\" 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\/ch4\/\" 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\/ch4\/4-2-2\/\" 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>4.2 Mekanizmalarda H\u0131z ve \u0130vme Analizi -1 Mekanizmalarda h\u0131z ve ivme analizi vekt\u00f6rel olarak ba\u011f\u0131l h\u0131z ve ivme kavram\u0131 ile yap\u0131l\u0131r. Genel olarak verilen de\u011ferler ile ba\u015flan\u0131l\u0131r ve A, B, C, \u2026 gibi genellikle mafsal eksenlerinin ge\u00e7ti\u011fi noktalar s\u0131ras\u0131 ile &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch4\/4-2-1\/\"> <span class=\"screen-reader-text\">4-2-1<\/span> Devam\u0131n\u0131 Oku &raquo;<\/a><\/p>\n","protected":false},"author":7747,"featured_media":0,"parent":1027,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-1031","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1031","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=1031"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1031\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1027"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/media?parent=1031"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}