{"id":940,"date":"2021-09-09T10:43:47","date_gmt":"2021-09-09T10:43:47","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=940"},"modified":"2021-10-05T22:47:59","modified_gmt":"2021-10-05T22:47:59","slug":"3-7","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch3\/3-7\/","title":{"rendered":"3-7"},"content":{"rendered":"
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3.7<\/b> Karma\u015f\u0131k Say\u0131lar Kullan\u0131larak Konum Analizi<\/h1>\n

Karma\u015f\u0131k say\u0131lar\u0131n devre kapal\u0131l\u0131k denklemlerinde kullan\u0131l\u0131\u015fl\u0131 oldu\u011funu, genellikle polar g\u00f6sterim ile devreyi olu\u015fturan her bir vekt\u00f6r\u00fc karma\u015f\u0131k say\u0131 kullanarak kolayca tan\u0131mlay\u0131p g\u00f6sterebilece\u011fimizi g\u00f6rm\u00fc\u015ft\u00fck. Karma\u015f\u0131k say\u0131larla matematiksel i\u015flemler normal say\u0131larla yap\u0131lan cebir i\u015flemleri ile ayn\u0131 oldu\u011fundan (tek bir fark i2<\/sup>\u00a0=\u00a0 – 1 olacakt\u0131r), karma\u015f\u0131k say\u0131lar ayn\u0131 kolayl\u0131kla mekanizmalarda konum parametrelerinin \u00e7\u00f6z\u00fcm\u00fc i\u00e7in kullan\u0131labilir.<\/p>\n

\"\"<\/p>\n

\u0130lk \u00f6rnek olarak d\u00f6rt-\u00e7ubuk mekanizmas\u0131n\u0131 ele alal\u0131m. Vekt\u00f6r devre denklemi<\/p>\n

A0<\/sub>A<\/strong> + <\/b>AB<\/strong> = <\/b>A0<\/sub>B0<\/sub><\/strong> + <\/b>B0<\/sub>B<\/strong><\/p>\n

olacakt\u0131r. Her bir uzva ba\u011fl\u0131 konum vekt\u00f6r\u00fcn\u00fcn uzunlu\u011fu aj<\/sub>, pozitif x-ekseni ile yapt\u0131\u011f\u0131 a\u00e7\u0131 \u03b81j<\/sub> dersek, karma\u015f\u0131k say\u0131 olarak bu konum vekt\u00f6r\u00fc aj<\/sub>ei\u03b8ij<\/sub><\/sup>\u00a0olarak g\u00f6sterilebilir. Bu durumda devre kapal\u0131l\u0131k denklemi karma\u015f\u0131k say\u0131lar ile:<\/p>\n\n\n\n
a2<\/sub>ei\u03b812<\/sub><\/sup> + a3<\/sub>ei\u03b813<\/sub><\/sup> = a1<\/sub>\u00a0+ a4<\/sub>ei\u03b814<\/sub><\/sup><\/td>\n(1)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Bu denkleminin reel ve sanal k\u0131s\u0131mlar\u0131 ayr\u0131 ayr\u0131 e\u015fitlenerek \u00fc\u00e7 konum parametresini (\u03b812<\/sub>, \u03b813<\/sub> ve \u03b814<\/sub>) i\u00e7eren iki skaler denklem elde edilebilir. Konum parametrelerinden birisi ba\u011f\u0131ms\u0131z parametre olaca\u011f\u0131ndan iki bilinmeyenli iki denklem vard\u0131r. Bu denklemler:<\/p>\n

a2<\/sub>cos\u03b812<\/sub> + a3<\/sub>cos\u03b813<\/sub> = a1<\/sub> + a4<\/sub>cos\u03b814<\/sub><\/p>\n

a2<\/sub>sin\u03b812<\/sub> + a3<\/sub>sin\u03b813<\/sub> = a4<\/sub>sin\u03b814<\/sub><\/p>\n

Bu iki denklemden verilen bir \u03b812<\/sub> a\u00e7\u0131s\u0131 de\u011ferine g\u00f6re \u03b813<\/sub> ve \u03b814<\/sub> de\u011ferlerini bulabilmek deneme-yan\u0131lma gerektirecektir. Bu denklemleri \u03b813<\/sub> ve \u03b814<\/sub>\u00a0de\u011ferlerinin kolayca elde edilebilir hale getirmek gerekmektedir. \u0130sterseniz bu skaler denklemleri kullanarak bu \u00e7\u00f6z\u00fcm\u00fc yapabilirsiniz.<\/p>\n

Karma\u015f\u0131k d\u00fczlemde kalarak \u00e7\u00f6z\u00fcm yapmak istedi\u011fimizde, mekanizma i\u00e7in ge\u00e7erli olan ikinci bir denkleme ihtiya\u00e7 vard\u0131r. Bu denklemi elde etmek i\u00e7in Mekanizman\u0131n devre denklemini yazd\u0131\u011f\u0131m\u0131z eksen tak\u0131m\u0131na g\u00f6re, bir aynan\u0131n alt y\u00fczeyini x-ekseni ile \u00e7ak\u0131\u015facak \u015fekilde yerle\u015ftirelim. Ayna \u00fczerinde mekanizman\u0131n g\u00f6r\u00fcnt\u00fcs\u00fc g\u00f6r\u00fclecektir. Ger\u00e7ek mekanizma hareket etti\u011fi zaman aynadaki g\u00f6r\u00fcnt\u00fc de hareket edecektir ve mekanizman\u0131n olu\u015fturdu\u011fu her bir kapal\u0131 devreye kar\u015f\u0131l\u0131k ayna g\u00f6r\u00fcnt\u00fcde bir kapal\u0131 devre olacakt\u0131r.<\/p>\n

\"\"<\/p>\n

Ger\u00e7ek devre ile g\u00f6r\u00fcnt\u00fcde olu\u015fan devrede vekt\u00f6rlerin boyutlar\u0131 ayn\u0131 g\u00f6r\u00fclecek ancak ger\u00e7ek mekanizmada saat yelkovan\u0131na ters y\u00f6nde al\u0131nan a\u00e7\u0131lar, ayna g\u00f6r\u00fcnt\u00fcde saat yelkovan\u0131 y\u00f6n\u00fcnde olacaklard\u0131r. Bu g\u00f6r\u00fcnt\u00fcde olu\u015fan devre i\u00e7in, devre kapal\u0131l\u0131k denklemini yazd\u0131\u011f\u0131m\u0131zda:<\/p>\n\n\n\n
\n

a2<\/sub>e\u2212i<\/sup>\u03b812<\/sub><\/sup> + a3<\/sub>e\u2212i<\/sup>\u03b813<\/sub><\/sup> = a1<\/sub>\u00a0+ a4<\/sub>e\u2212i\u03b814<\/sub><\/sup><\/p>\n<\/td>\n

(2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Elde edilen bu ikinci denkleme devre kapal\u0131l\u0131k denkleminin sanal e\u015fleni\u011fi<\/strong> denmektedir. (1) denkleminin ge\u00e7erli oldu\u011fu her durumda bu sanal e\u015flenik de (2 denklemi) ge\u00e7erlidir.<\/p>\n

Karma\u015f\u0131k say\u0131 d\u00fczleminde (1) ve (2) denklemleri iki konum de\u011fi\u015fkenini bulmak i\u00e7in kullan\u0131labilir. (1) ve (2) denklemlerinin reel ve sanal k\u0131s\u0131mlar\u0131n\u0131 ayr\u0131 ayr\u0131 e\u015fitler ve skaler denklemler elde edersek, her iki denklemden elde edilecek olan denklemlerin ayn\u0131 skaler denklemlere d\u00f6n\u00fc\u015ft\u00fc\u011f\u00fc g\u00f6r\u00fclecektir. Yak\u0131nsama y\u00f6ntemi ile n\u00fcmerik \u00e7\u00f6z\u00fcm, ileride a\u00e7\u0131klanacakt\u0131r. Burada bilinmeyen konum parametrelerinden her biri ba\u011f\u0131ms\u0131z konum de\u011fi\u015fkeninin fonksiyonu olarak nas\u0131l \u00e7\u00f6z\u00fclebilir, onu g\u00f6relim.<\/p>\n

D\u00f6rt \u00e7ubuk mekanizmas\u0131nda \u03b814<\/sub> de\u011fi\u015fkenini \u03b812<\/sub> parametresine g\u00f6re \u00e7\u00f6zmeyi hedefleyelim. Bu durumda \u03b813<\/sub> konum de\u011fi\u015fkenini 1 ve 2 numaral\u0131 denklemlerden yok etmemiz gereklidir. Devre kapal\u0131l\u0131k denklemi ve sanal e\u015fleni\u011finden \u03b813<\/sub> \u00fcn yok edilmesi i\u00e7in yok etmek istedi\u011fimiz a\u00e7\u0131y\u0131 i\u00e7eren terimi denklemin sol taraf\u0131na alarak yazal\u0131m:<\/p>\n\n\n\n\n
\n

a3<\/sub>ei<\/sup>\u03b813<\/sub><\/sup> = a1<\/sub>\u00a0+ a4<\/sub>ei\u03b814<\/sub><\/sup>\u00a0\u2212 a2<\/sub>ei<\/sup>\u03b812<\/sub><\/sup><\/p>\n<\/td>\n

(3)<\/td>\n<\/tr>\n
\n

a3<\/sub>e\u2212i<\/sup>\u03b813<\/sub><\/sup> = a1<\/sub>\u00a0+ a4<\/sub>e\u2212i\u03b814<\/sub><\/sup>\u00a0\u2212 a2<\/sub>e\u2212i<\/sup>\u03b812<\/sub><\/sup><\/p>\n<\/td>\n

(4)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(yok edece\u011fimiz de\u011fi\u015fkeni i\u00e7eren terim denklemin bir taraf\u0131na al\u0131nm\u0131\u015ft\u0131r). (3) ve (4) denklemlerini taraf tarafa \u00e7arpar isek :<\/p>\n\n\n\n
\n

a3<\/sub>2<\/sup>ei(\u03b813<\/sub> \u2212 \u03b813<\/sub>)<\/sup> = (a1<\/sub>\u00a0+ a4<\/sub>ei\u03b814<\/sub><\/sup>\u00a0\u2212 a2<\/sub>ei<\/sup>\u03b812<\/sub><\/sup>)(a1<\/sub>\u00a0+ a4<\/sub>e\u2212i\u03b814<\/sub><\/sup>\u00a0\u2212 a2<\/sub>e\u2212i<\/sup>\u03b812<\/sub><\/sup>)<\/p>\n<\/td>\n

(5)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

ei(\u03b8 \u2212 \u03b8)<\/sup> = ei0<\/sup> =1 oldu\u011funu hat\u0131rlarsak:<\/p>\n\n\n\n
\n

a3<\/sub>2<\/sup> = a1<\/sub>2<\/sup> + a4<\/sub>2<\/sup> + a2<\/sub>2<\/sup> + a1<\/sub>a4<\/sub>[ei\u03b814<\/sub><\/sup> \u2212 e\u2212i\u03b814<\/sub><\/sup>] \u2212 a1<\/sub>a2<\/sub>[ei<\/sup>\u03b812<\/sub><\/sup> + e\u2212i<\/sup>\u03b812<\/sub><\/sup>] \u2212 a2<\/sub>a4<\/sub>[ei(\u03b814 <\/sub>\u2212\u00a0<\/sub>\u03b812<\/sub>)<\/sup>\u00a0\u2212 e\u2212i(\u03b814 <\/sub>\u2212\u00a0<\/sub>\u03b812<\/sub>)<\/sup>]\n<\/td>\n

(6)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Euler denklemine g\u00f6re \u00a0cos\u03b8 = (ei\u03b8<\/sup> + e\u2212i\u03b8<\/sup>)\/2 olaca\u011f\u0131ndan, (6) numaral\u0131 denklem:<\/p>\n

a3<\/sub>2<\/sup> = a1<\/sub>2<\/sup> + a2<\/sub>2<\/sup> + a4<\/sub>2<\/sup> + 2a1<\/sub>a4<\/sub>cos\u03b814<\/sub> \u2212 2a1<\/sub>a2<\/sub>cos\u03b812<\/sub> \u2212 2a2<\/sub>a4<\/sub>cos(\u03b814<\/sub> \u2212 \u03b812<\/sub>)<\/p>\n

olacakt\u0131r. Bu denklem:<\/p>\n\n\n\n
\n

K1<\/sub>cos\u03b814<\/sub> \u2212 K2<\/sub>cos\u03b812<\/sub> + K3<\/sub> = cos(\u03b814<\/sub> \u2212 \u03b812<\/sub>)<\/p>\n<\/td>\n

(7)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u015feklinde yaz\u0131labilir. Burada K1<\/sub> = a1<\/sub>\/a2<\/sub> , K2<\/sub> = a1<\/sub>\/a4<\/sub> , K3<\/sub> = (a1<\/sub>2<\/sup> + a2<\/sub>2<\/sup> \u2212 a3<\/sub>2<\/sup> + a4<\/sub>2<\/sup> )\/2a2<\/sub>a4<\/sub> t\u00fcr. (7) numaral\u0131 denkleme, bunu ilk tan\u0131mlayan ki\u015fiye atfen Freudenstein denklemi<\/strong>\u00a0denmektedir. Bu denklem d\u00f6rt-\u00e7ubuk mekanizmalar\u0131n\u0131n sentezinde \u00f6nemli rol oynar. \u03b814<\/sub> ve \u03b812<\/sub> de\u011fi\u015fkenleri aras\u0131nda ili\u015fki bu denklem ile belirlidir. Ancak verilen bir \u03b812<\/sub> de\u011ferine kar\u015f\u0131 gelen \u03b814<\/sub>\u00a0a\u00e7\u0131s\u0131n\u0131 (7) numaral\u0131 denklemden kolayca belirlemek mevcut durumu ile m\u00fcmk\u00fcn de\u011fildir. Freudenstein denklemini<\/p>\n\n\n\n
\n

K1<\/sub>cos\u03b814<\/sub> \u2212 K2<\/sub>sin\u03b812<\/sub> + K3<\/sub> = cos\u03b812<\/sub>cos\u03b814<\/sub>\u00a0+ sin\u03b812<\/sub>sin\u03b814<\/sub><\/p>\n<\/td>\n

(8)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u015feklinde yazabiliriz. t = tan(\u03b814<\/sub>\/2) ise<\/p>\n

sin\u03b814<\/sub> = 2t\/(1 + t2<\/sup>)\u00a0 \u00a0 \u00a0ve\u00a0 \u00a0 \u00a0cos\u03b814<\/sub> = (1 \u2212 t2<\/sup>)\/(1 + t2<\/sup>)<\/p>\n

olur. Bir a\u00e7\u0131n\u0131n yar\u0131s\u0131n\u0131 kullanarak a\u00e7\u0131n\u0131n t\u00fcm trigonemetrik fonksiyonlar\u0131n\u0131 sadece yar\u0131m a\u00e7\u0131n\u0131n tanjant fonksiyonu ile ifade edilmesi\u00a0yar\u0131m tanjant y\u00f6ntemi<\/strong> olarak bilinir). Bu durumda (8) denklemi:<\/p>\n\n\n\n
At2<\/sup> + Bt + C = 0<\/td>\n(9)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

olarak yaz\u0131labilir. Burada<\/p>\n

A = cos\u03b812<\/sub>(1 \u2212 K2<\/sub>) + K3<\/sub> \u2212 K1<\/sub><\/p>\n

B = \u22122 sin\u03b812<\/sub><\/p>\n

C = cos\u03b812<\/sub>(1 + K2<\/sub>) + K3<\/sub>\u00a0+ K1<\/sub><\/p>\n

dir. Dikkat edilirse, \u03b812<\/sub> ba\u011f\u0131ms\u0131z parametre de\u011feri ve uzuv boyutlar\u0131 biliniyor ise A, B ve C parametre de\u011ferlerini hesaplayabiliriz. (9) numaral\u0131 denklem t’ye g\u00f6re ikinci dereceden bir denklemdir ve \u00e7\u00f6z\u00fcm\u00fc<\/p>\n

\\displaystyle \\tan \\left( {\\frac{{{{\\text{\u03b8}}_{{14}}}}}{2}} \\right)=\\frac{{-\\text{B}\\pm \\sqrt{{{{\\text{B}}^{2}}-4\\text{AC}}}}}{{2\\text{A}}} <\/span><\/p>\n

d\u0131r. Bilinmeyen \u03b814<\/sub>\u00a0a\u00e7\u0131s\u0131 bu durumda<\/p>\n\n\n\n
\\displaystyle {{{\\text{\u03b8}}_{{14}}}=2{{\\tan }^{{-1}}}\\left[ {\\frac{{-\\text{B}\\pm \\sqrt{{{{\\text{B}}^{2}}-4\\text{AC}}}}}{{2\\text{A}}}} \\right]} <\/span><\/td>\n(10)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

olur. (10) denklemi diskriminant\u0131n art\u0131 veya eksi i\u015faret almas\u0131na g\u00f6re iki de\u011fi\u015fik \u03b814<\/sub>\u00a0de\u011feri verecektir. Bu mekanizman\u0131n iki farkl\u0131 \u015fekilde monte edilmesi ile ilgilidir.<\/p>\n

Benzer y\u00f6ntem kullan\u0131larak devre kapal\u0131l\u0131k denkleminden \u03b814<\/sub> a\u00e7\u0131s\u0131 yok edilerek \u03b813<\/sub> a\u00e7\u0131s\u0131 i\u00e7in \u03b812<\/sub> ye g\u00f6re ayn\u0131 \u015fekilde elde edilebilir (veya \u03b812<\/sub> de\u011feri ile birlikte bu de\u011fere g\u00f6re bulunan \u03b814<\/sub> a\u00e7\u0131 de\u011feri kullan\u0131larak \u03b813<\/sub> a\u00e7\u0131s\u0131 de\u011feri bulunabilir). Montaj \u015fekline g\u00f6re uygulamada bir tanesi ge\u00e7erli olacakt\u0131r. A, B ve C sadece uzuv boyutlar\u0131na ve ba\u011f\u0131ms\u0131z konum parametresi olan giri\u015f kolu a\u00e7\u0131s\u0131na ba\u011fl\u0131d\u0131r. B2<\/sup> \u2212 4AC < 0 ise, verilmi\u015f olan kol a\u00e7\u0131s\u0131nda (\u03b812<\/sub>) mekanizma monte
\nedilemez.<\/p>\n

Benzer y\u00f6ntem kullan\u0131larak devre kapal\u0131l\u0131k denkleminden \u03b814<\/sub> a\u00e7\u0131s\u0131 yok edilerek \u03b813<\/sub> a\u00e7\u0131s\u0131 i\u00e7in \u03b812<\/sub> ye g\u00f6re ayn\u0131 \u015fekilde elde edilebilir (veya \u03b812<\/sub> de\u011feri ile birlikte bu de\u011fere g\u00f6re bulunan \u03b814<\/sub> a\u00e7\u0131 de\u011feri kullan\u0131larak \u03b813<\/sub> a\u00e7\u0131s\u0131 de\u011feri bulunabilir).<\/p>\n

Denklem (8)’i \u03b814<\/sub> a\u00e7\u0131s\u0131na g\u00f6re \u00e7\u00f6zmek i\u00e7in kullan\u0131labilecek farkl\u0131 bir y\u00f6ntem i\u00e7in bu denklemi<\/p>\n\n\n\n
(K1<\/sub> \u2212 cos\u03b812<\/sub>)cos\u03b814<\/sub> \u2212 sin\u03b814<\/sub>sin\u03b812\u00a0<\/sub>= K2<\/sub>cos\u03b812<\/sub> \u2212 K3<\/sub><\/td>\n(11)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u015feklinde yazal\u0131m. Ayr\u0131ca:<\/p>\n\n\n\n
\n

D cos\u03d5 = K1<\/sub> \u2212 cos\u03b812<\/sub><\/p>\n

D sin\u03d5 = sin\u03b812<\/sub><\/p>\n<\/td>\n

(12)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

der isek:<\/p>\n\n\n\n
\\displaystyle {\\text{D}=\\sqrt{{{{{\\left( {{{\\text{K}}_{1}}-\\cos {{\\text{\u03b8}}_{{12}}}} \\right)}}^{2}}+{{{\\sin }}^{2}}{{\\text{\u03b8}}_{{12}}}}}=\\sqrt{{1+{{\\text{K}}_{1}}^{2}-2{{\\text{K}}_{1}}\\cos {{\\text{\u03b8}}_{{12}}}}}} <\/span>
\n \\displaystyle {\\text{\u03d5} ={{\\tan }^{{-1}}}\\frac{{\\sin {{\\text{\u03b8}}_{{12}}}}}{{{{\\text{K}}_{1}}-\\cos {{\\text{\u03b8}}_{{12}}}}}} <\/span><\/td>\n		(13)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

olacakt\u0131r. Bu tan\u0131mlarla denklem (11):<\/p>\n

Dcos\u03d5cos\u03b814<\/sub> \u2212 Dsin\u03d5sin\u03b814<\/sub>\u00a0= K2<\/sub>cos\u03b812<\/sub> \u2212 K3<\/sub><\/p>\n

t\u00fcr. cos (\u03b8 + \u03d5) = cos\u03b8cos\u03d5 \u2212 sin\u03b8sin\u03d5 trigonometrik e\u015fitli\u011fi kullan\u0131r isek:<\/p>\n\n\n\n
cos(\u03b814<\/sub> + \u03d5) = (K2<\/sub>cos\u03b812<\/sub> \u2212 K3<\/sub>)\/D<\/td>\n(14)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

veya<\/p>\n\n\n\n
\u03b814<\/sub> = cos-1<\/sup>[(K2<\/sub>cos\u03b812<\/sub> \u2212 K3<\/sub>)\/D] \u2212 \u03d5<\/td>\n(15)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(13) numaral\u0131 denklemin \u03d5 a\u00e7\u0131s\u0131na g\u00f6re \u00e7\u00f6z\u00fcm\u00fc s\u0131ras\u0131nda a\u00e7\u0131n\u0131n do\u011fru de\u011ferini bulmak i\u00e7in R-P tu\u015fu veya ATAN2(x.y) fonksiyonunun kullan\u0131m\u0131 ihmal edilmemelidir. (15) numaral\u0131 denklemde ters kosin\u00fcs fonksiyonunun iki de\u011fi\u015fik de\u011feri mekanizman\u0131n iki farkl\u0131 montaj \u015feklini verecektir.<\/p>\n

Devre kapal\u0131l\u0131k denklemlerinin konum de\u011fi\u015fkenleri i\u00e7in \u00e7\u00f6z\u00fcm\u00fcne bir ba\u015fka \u00f6rnek olarak kol-k\u0131zak mekanizmas\u0131n\u0131 ele alal\u0131m. Bu mekanizmada \u03b812<\/sub> ba\u011f\u0131ms\u0131z konum de\u011fi\u015fkeni, \u03b814<\/sub>\u00a0ve s43<\/sub> ise ba\u011f\u0131ml\u0131 konum de\u011fi\u015fkenleridir. Amac\u0131m\u0131z her \u03b812<\/sub> a\u00e7\u0131s\u0131na kar\u015f\u0131 gelen \u03b814<\/sub>\u00a0de\u011ferini elde edebilece\u011fimiz denklemi elde etmektir.<\/p>\n

\"\"<\/p>\n

Vekt\u00f6r devre denklemi:<\/p>\n

A0<\/sub>A<\/strong> = A0<\/sub>B0<\/sub><\/strong>\u00a0+ B0<\/sub>B<\/strong> + BA<\/strong><\/p>\n

olacakt\u0131r ve karma\u015f\u0131k say\u0131lar ile yaz\u0131ld\u0131\u011f\u0131nda:<\/p>\n

a2<\/sub>ei<\/sup>\u03b812<\/sub><\/sup>\u00a0= a1<\/sub>\u00a0+ a4<\/sub>ei\u03b814<\/sub><\/sup> + s43<\/sub>ei(\u03b814<\/sub> + \u03c0\/2)<\/sup><\/p>\n

veya<\/p>\n\n\n\n
a2<\/sub>ei<\/sup>\u03b812<\/sub><\/sup>\u00a0= a1<\/sub>\u00a0+ a4<\/sub>ei\u03b814<\/sub><\/sup> + is43<\/sub>ei\u03b814<\/sub><\/sup><\/td>\n(1)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Bu denklemin karma\u015f\u0131k e\u015fleni\u011fi (mekanizman\u0131n ayna g\u00f6r\u00fcnt\u00fcs\u00fcnde devre denklemi):<\/p>\n\n\n\n
a2<\/sub>e\u2212i<\/sup>\u03b812<\/sub><\/sup> = a1<\/sub> + a4<\/sub>e\u2212i<\/sup>\u03b814<\/sub><\/sup> \u2212 is43<\/sub>e\u2212i<\/sup>\u03b814<\/sub><\/sup><\/td>\n(2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

olacakt\u0131r. s43<\/sub>\u00a0parametresini bu iki denklemden yok etmek i\u00e7in bunu ihtiva eden terimi tek ba\u015f\u0131na b\u0131rakal\u0131m.<\/p>\n\n\n\n\n
\n

is43<\/sub>ei\u03b814<\/sub><\/sup> = a2<\/sub>ei<\/sup>\u03b812<\/sub><\/sup> \u2212 a1<\/sub> \u2212 a4<\/sub>ei\u03b814<\/sub><\/sup><\/p>\n<\/td>\n

(3)<\/td>\n<\/tr>\n
\n

is43<\/sub>e\u2212i<\/sup>\u03b814<\/sub><\/sup>\u00a0= a1<\/sub>\u00a0+ a4<\/sub>e\u2212i<\/sup>\u03b814<\/sub><\/sup> \u2212 a2<\/sub>e\u2212i<\/sup>\u03b812<\/sub><\/sup><\/p>\n<\/td>\n

(4)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

s43<\/sub> parametresini yok etmek i\u00e7in 3 denklemini e\u2212i<\/sup>\u03b814<\/sub><\/sup> ve 4 denklemini ei<\/sup>\u03b814<\/sub><\/sup> ile \u00e7arp\u0131p birbirlerine e\u015fitleyelim:<\/p>\n

a2<\/sub>ei(\u03b812 <\/sub>\u2212\u00a0<\/sub>\u03b814<\/sub>)<\/sup>\u00a0\u2212 a1<\/sub>e\u2212i<\/sup>\u03b814<\/sub><\/sup> \u2212 a4<\/sub> = a1<\/sub>ei<\/sup>\u03b814<\/sub><\/sup> + a4<\/sub>\u00a0\u2212 a2<\/sub>e\u2212i(\u03b812 <\/sub>\u2212\u00a0<\/sub>\u03b814<\/sub>)<\/sup><\/p>\n

veya<\/p>\n\n\n\n
\n

a2<\/sub>[ei(\u03b812 <\/sub>\u2212\u00a0<\/sub>\u03b814<\/sub>)<\/sup> + e\u2212i(\u03b812<\/sub> \u2212 \u03b814<\/sub>)<\/sup>]\u2212 a1<\/sub>[ei<\/sup>\u03b814<\/sub><\/sup> + e\u2212i<\/sup>\u03b814<\/sub><\/sup>] \u2212 2a4<\/sub> = 0<\/p>\n<\/td>\n

(5)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

ei<\/sup>\u03b8<\/sup>\u00a0+ e\u2212i<\/sup>\u03b8<\/sup> = 2cos\u03b8 oldu\u011fu hat\u0131rlan\u0131r ise:<\/p>\n\n\n\n
a2<\/sub>cos(\u03b814<\/sub> \u2212 \u03b812<\/sub>) \u2212 a1<\/sub>cos\u03b814<\/sub> \u2212 a4<\/sub>\u00a0= 0<\/td>\n(6)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Denklem 6 y\u0131 kullanarak verilen bir \u03b812<\/sub> de\u011ferine g\u00f6re \u03b814<\/sub> de\u011ferini \u00e7\u00f6zmemiz i\u00e7in cos(\u03b814<\/sub> \u2212 \u03b812<\/sub>) yi trigonometrik e\u015fitlik kullanarak a\u00e7al\u0131m:<\/p>\n

a2<\/sub>cos\u03b814<\/sub>cos\u03b812<\/sub>\u00a0+ a2<\/sub>sin\u03b814<\/sub>sin\u03b812<\/sub> \u2212 a1<\/sub>cos\u03b814<\/sub> \u2212 a4<\/sub>\u00a0= 0<\/p>\n

veya:<\/p>\n

a2<\/sub>(cos\u03b812<\/sub> \u2212 a1<\/sub>)cos\u03b814<\/sub>\u00a0+ a2<\/sub>sin\u03b814<\/sub>sin\u03b812<\/sub> \u2212 a4<\/sub>\u00a0= 0<\/p>\n

olacakt\u0131r. D\u00f6rt-\u00e7ubuk mekanizmas\u0131 i\u00e7in elde edilmi\u015f olan denklemde yap\u0131lm\u0131\u015f oldu\u011fu gibi, cos\u03b814<\/sub> ve sin\u03b814\u00a0<\/sub>fonksiyonlar\u0131 t = tan(\u03b814<\/sub>\/2) ile g\u00f6sterilerek gerekli basitlemeler yap\u0131ld\u0131\u011f\u0131nda:<\/p>\n\n\n\n
At2<\/sup> + Bt + C = 0<\/td>\n(7)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

bu denklemde:<\/p>\n

A = a1<\/sub> \u2212 a4<\/sub> \u2212 a2<\/sub>cos\u03b812<\/sub><\/p>\n

B = 2sin\u03b812<\/sub><\/p>\n

C = a2<\/sub>cos\u03b812 <\/sub>\u2212 a1<\/sub> \u2212 a4<\/sub><\/p>\n

Bu ikinci derece denklemin \u03b814<\/sub> a\u00e7\u0131s\u0131 i\u00e7in \u00e7\u00f6z\u00fcm\u00fc kolayca yap\u0131labilir. Genel olarak yar\u0131m tanjant y\u00f6ntemi kapal\u0131 \u00e7\u00f6z\u00fcm elde edilmesi i\u00e7in kuvvetli bir y\u00f6ntem olarak g\u00f6r\u00fclmektedir (e\u011fer istenilir ise d\u00f6rt-\u00e7ubuk mekanizmas\u0131 i\u00e7in verilen ve (11)-(15) numaral\u0131 denklemlerle a\u00e7\u0131klanm\u0131\u015f olan farkl\u0131 y\u00f6ntem de kullan\u0131labilir).<\/p>\n

E\u011fer s34<\/sub> konum de\u011fi\u015fkenini \u03b812<\/sub> ye g\u00f6re bulmam\u0131z gerekiyor ise bu sefer \u03b814<\/sub> parametresini devre denkleminden yok etmemiz gerekecektir. Bu durumda devre kapal\u0131l\u0131k denklemi ve onun karma\u015f\u0131k e\u015fleni\u011fi:<\/p>\n

(a4<\/sub>\u00a0+ is43<\/sub>)ei\u03b814<\/sub><\/sup> = a2<\/sub>ei<\/sup>\u03b812<\/sub><\/sup> \u2212 a1<\/sub><\/p>\n

(a4<\/sub> \u2212 is43<\/sub>)e\u2212i\u03b814<\/sub><\/sup> = a2<\/sub>e\u2212i<\/sup>\u03b812<\/sub><\/sup> \u2212 a1<\/sub><\/p>\n

\u015feklinde yaz\u0131labilir. Taraf tarafa \u00e7arp\u0131m yap\u0131ld\u0131\u011f\u0131nda \u03b814<\/sub>\u00a0a\u00e7\u0131s\u0131 yok edilecek ve s43<\/sub>\u00a0i\u00e7in \u00e7\u00f6z\u00fcm:<\/p>\n

s43<\/sub>2<\/sup> = a1<\/sub>2<\/sup> + a2<\/sub>2<\/sup> \u2212 a4<\/sub>2<\/sup> \u2212 2a1<\/sub>a2<\/sub>cos\u03b812<\/sub><\/p>\n

olacakt\u0131r. Bu \u00e7\u00f6z\u00fcmleri elde ederken sin-1<\/sup>, cos-1<\/sup>, tan-1<\/sup> gibi ters trigonometrik fonksiyonlar kullan\u0131l\u0131r iken bu fonksiyonlar\u0131n iki de\u011ferli oldu\u011fu ve hesap makinesi, veya ba\u015fka herhangi bilgisayar program paketinin bu de\u011ferlerden birini verece\u011fini, mekanizman\u0131n ba\u011flan\u0131\u015f \u015fekline g\u00f6re hangi de\u011ferin ge\u00e7erli oldu\u011funa kullan\u0131c\u0131 olarak bizim karar vermemiz gerekti\u011fi unutulmamal\u0131d\u0131r.<\/p>\n

Geometrik \u00e7\u00f6z\u00fcmde g\u00f6rd\u00fc\u011f\u00fcm\u00fcz gibi, uygulamada bir\u00e7ok \u00e7ok uzuvlu mekanizma basit mekanizmalar\u0131n ba\u011flanmas\u0131 ile elde edilmi\u015ftir. Basit mekanizmalar i\u00e7in elde edilen bu \u00e7\u00f6z\u00fcmler daha karma\u015f\u0131k mekanizmalar\u0131n \u00e7\u00f6z\u00fcm\u00fcnde kolayca kullan\u0131labilir.<\/p>\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n

\"\"<\/a>\"\"<\/a>\"\"<\/a>\"\"<\/a>\"\" <\/p>\n","protected":false},"excerpt":{"rendered":"

3.7 Karma\u015f\u0131k Say\u0131lar Kullan\u0131larak Konum Analizi Karma\u015f\u0131k say\u0131lar\u0131n devre kapal\u0131l\u0131k denklemlerinde kullan\u0131l\u0131\u015fl\u0131 oldu\u011funu, genellikle polar g\u00f6sterim ile devreyi olu\u015fturan her bir vekt\u00f6r\u00fc karma\u015f\u0131k say\u0131 kullanarak kolayca tan\u0131mlay\u0131p g\u00f6sterebilece\u011fimizi g\u00f6rm\u00fc\u015ft\u00fck. Karma\u015f\u0131k say\u0131larla matematiksel i\u015flemler normal say\u0131larla yap\u0131lan cebir i\u015flemleri ile ayn\u0131 …<\/p>\n

3-7<\/span> Devam\u0131n\u0131 Oku »<\/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-940","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/940","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=940"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/940\/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=940"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}