{"id":891,"date":"2021-09-09T09:57:36","date_gmt":"2021-09-09T09:57:36","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=891"},"modified":"2022-03-21T06:32:40","modified_gmt":"2022-03-21T06:32:40","slug":"3-6","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch3\/3-6\/","title":{"rendered":"3-6"},"content":{"rendered":"
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3.6<\/b> Devre Kapal\u0131l\u0131k Denklemlerinin Konum De\u011fi\u015fkenleri \u0130\u00e7in \u00c7\u00f6z\u00fcm\u00fc<\/h1>\n

\u00d6nceki k\u0131s\u0131mda g\u00f6rd\u00fc\u011f\u00fcm\u00fcz gibi, geometrik olarak devre kapal\u0131l\u0131k denklemlerinin \u00e7\u00f6z\u00fcm\u00fc yeterli parametresi verilmi\u015f olan bir \u00fc\u00e7genin di\u011fer parametrelerinin belirlenmesinden ibarettir. Analitik geometri derslerinden bildi\u011fimiz gibi, \u00fc\u00e7gen ba\u011flant\u0131lar\u0131 ayn\u0131 zamanda analitik olarak ifade edilerek analitik \u00e7\u00f6z\u00fcmde yap\u0131labilir. Bu yakla\u015f\u0131mda, t\u0131pk\u0131 geometrik y\u00f6ntemde oldu\u011fu, gibi etap etap konum parametrelerinin belirlenmesine \u00e7al\u0131\u015f\u0131lacak ve bunun i\u00e7in seri olarak \u00e7\u00f6z\u00fclebilecek bir dizi denklem elde edilecektir. Her bir konum parametresinin sadece ba\u011f\u0131ms\u0131z parametreye g\u00f6re belirlenmesine \u00e7al\u0131\u015f\u0131lmayacakt\u0131r. Yani, bu \u00e7al\u0131\u015fma sonras\u0131 konum parametrelerini ba\u011f\u0131ms\u0131z parametre de\u011ferine g\u00f6re bulabilece\u011fi bir “algoritma<\/b>” elde edilecektir. Bu algoritmay\u0131 ister bir hesap makinas\u0131 ile ister \u00f6zel bir program yazarak veya ister isek \u00e7e\u015fitli paket programlarda kolayl\u0131kla \u00e7\u00f6zmemiz m\u00fcmk\u00fcn olacakt\u0131r.<\/p>\n

\"\"<\/p>\n

\u0130lk \u00f6rnek olarak yukar\u0131da g\u00f6r\u00fclen bir krank-biyel mekanizmas\u0131n\u0131 ele alal\u0131m. Bilinen uzuv boyutlar\u0131n\u0131 a1<\/sub>, a2<\/sub>, a3<\/sub> olarak g\u00f6sterelim. Amac\u0131m\u0131z 3 ve 4 uzuvlar\u0131n\u0131n her hangi bir krank a\u00e7\u0131s\u0131, \u03b812<\/sub>, de\u011feri i\u00e7in bulmakt\u0131r.<\/p>\n

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

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

Bu vekt\u00f6rler tan\u0131mlanm\u0131\u015f olan sabit ve de\u011fi\u015fken parametrelerle yaz\u0131ld\u0131\u011f\u0131nda:<\/p>\n

A0<\/sub>A<\/strong>\u00a0= a2<\/sub>(cos\u03b812<\/sub>i<\/strong> + sin\u03b812<\/sub>j<\/strong>)<\/p>\n

A0<\/sub>B<\/strong>\u00a0= xi<\/strong>\u00a0+ a1<\/sub>j<\/strong><\/p>\n

BA<\/strong>\u00a0= a3<\/sub>(cos\u03b813<\/sub>i<\/strong> + sin\u03b813<\/sub>j<\/strong>)<\/p>\n

olacakt\u0131r.\u00a0x ve y bile\u015fenleri ayr\u0131 ayr\u0131 birbirlerine e\u015fitlendi\u011finde, devre kapal\u0131l\u0131k denklemi iki skaler denkleme d\u00f6n\u00fc\u015fecektir:<\/p>\n\n\n\n\n
a2<\/sub>cos\u03b812<\/sub>\u00a0= s14<\/sub>\u00a0+ a3<\/sub>cos\u03b813<\/sub><\/td>\n(1)<\/td>\n<\/tr>\n
a2<\/sub>sin\u03b812<\/sub>\u00a0= a1<\/sub>\u00a0+ a3<\/sub>sin\u03b813<\/sub><\/td>\n(2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Bu denklemleri bilinmeyen konum parametreleri i\u00e7in yeniden yazd\u0131\u011f\u0131m\u0131zda:<\/p>\n\n\n\n\n
sin\u03b813<\/sub> = (a2<\/sub>sin\u03b812<\/sub> \u2212 a1<\/sub>)\/a3<\/sub><\/td>\n(3)<\/td>\n<\/tr>\n
s14<\/sub> = a2<\/sub>cos\u03b812<\/sub> \u2212 a3<\/sub>cos\u03b813<\/sub><\/td>\n(4)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Verilen giri\u015f kol a\u00e7\u0131s\u0131 \u03b812<\/sub> de\u011ferine g\u00f6re, \u03b813<\/sub>\u00a0de\u011fi\u015fken de\u011feri (3) denkleminden, s14<\/sub>\u00a0ise (4) denkleminden \u00e7\u00f6z\u00fclebilecektir. s14<\/sub> de\u011fi\u015fken de\u011feri \u03b813<\/sub> de\u011feri (3) denkleminden bulunduktan sonra bulunabilir. Her hangi bir C noktas\u0131n\u0131n koordinatlar\u0131n\u0131, C(xc<\/sub>, yc<\/sub>), bulmak istedi\u011fimizde\u00a0A0<\/sub>C<\/strong>\u00a0= xc<\/sub>i<\/strong>\u00a0+ yc<\/sub>j<\/strong>\u00a0konum vekt\u00f6r\u00fc\u00a0A0<\/sub>C<\/strong> = A0<\/sub>B<\/strong> + BC<\/strong>\u00a0olarak yaz\u0131l\u0131p x ve y bile\u015fenleri ayr\u0131 ayr\u0131 e\u015fitlendi\u011finde:<\/p>\n\n\n\n\n
xC<\/sub>\u00a0= s14<\/sub>\u00a0+ b3<\/sub>cos(\u03b813<\/sub> \u2212 \u03b33<\/sub>)<\/td>\n(5)<\/td>\n<\/tr>\n
yC<\/sub>\u00a0= a1<\/sub>\u00a0+ b3<\/sub>sin(\u03b813<\/sub> \u2212 \u03b33<\/sub>)<\/td>\n(6)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(5) ve (6) numaral\u0131 denklemler (3) ve (4) numaral\u0131 denklemler \u00e7\u00f6z\u00fcld\u00fckten sonra ele al\u0131nabilir.<\/p>\n

\u015eu ana kadar g\u00f6z \u00f6n\u00fcne al\u0131nmam\u0131\u015f olan \u00e7ok \u00f6nemli bir konu bulunmaktad\u0131r. Bu, (3) denkleminde \u03b813<\/sub> konum de\u011fi\u015fkeninin verilen her hangi bir \u03b812<\/sub> a\u00e7\u0131s\u0131 i\u00e7in \u00e7\u00f6z\u00fcm\u00fc s\u0131ras\u0131nda denklemi sa\u011flayan iki farkl\u0131 de\u011ferde olabilece\u011fidir. Bunun nedeni kullan\u0131lacak olan ters sin\u00fcs fonksiyonu \u00e7ift de\u011ferlidir. Yani sin\u00fcs a\u00e7\u0131s\u0131 kar\u015f\u0131 kenar\u0131n hipoten\u00fcse oran\u0131 oldu\u011fundan, verilen bir sin(\u03b8) de\u011ferini \u03b8 a\u00e7\u0131s\u0131 sa\u011flad\u0131\u011f\u0131 gibi (180\u00b0 \u2212 \u03b8) a\u00e7\u0131s\u0131 da sa\u011flayacakt\u0131r. Bu a\u00e7\u0131lardan hangisinin mevcut mekanizma i\u00e7in ge\u00e7erli oldu\u011fu, kullan\u0131c\u0131 taraf\u0131ndan belirlenmesi gerekir, \u00e7\u00fcnk\u00fc yukar\u0131da g\u00f6sterilmi\u015f olan mekanizma uzuv boyutlar\u0131 ayn\u0131 olmak \u00fczere, ayn\u0131 \u03b812<\/sub>\u00a0krank a\u00e7\u0131s\u0131nda istenilir ise, a\u015fa\u011f\u0131da g\u00f6sterildi\u011fi gibi farkl\u0131 bir \u015fekilde de monte edilebilir. Matemati\u011fin mekanizmay\u0131 nas\u0131l monte etti\u011fimizi bilmesine imkan yoktur.<\/p>\n

\"\"<\/p>\n

Krank biyel mekanizmas\u0131n\u0131n ayn\u0131 krank a\u00e7\u0131s\u0131nda farkl\u0131 monte edili\u015fi<\/span><\/p>\n

Devre kapal\u0131l\u0131k denklemi kullanarak elde etti\u011fimiz konum parametreleri aras\u0131ndaki ili\u015fki lineer de\u011fildir. Bu nedenle her mekanizma i\u00e7in ge\u00e7erli bir y\u00f6ntem s\u00f6ylemek m\u00fcmk\u00fcn de\u011fildir. Yakla\u015f\u0131m genelde \u00fc\u00e7genlerin belirlenerek bu \u00fc\u00e7genlerin gerekli parametrelere g\u00f6re \u00e7\u00f6z\u00fcm\u00fcn\u00fc bulmakt\u0131r.<\/p>\n

D\u00d6RT \u00c7UBUK MEKAN\u0130ZMASI<\/span><\/strong><\/p>\n

D\u00f6rt-\u00e7ubuk mekanizmas\u0131 i\u00e7in kullan\u0131lan y\u00f6nteme Raven metodu<\/strong> denmektedir. A\u015fa\u011f\u0131daki \u015fekilde g\u00f6sterilmi\u015f olan mekanizma i\u00e7in uzuv boyutlar\u0131 (a1<\/sub>, a2<\/sub>, a3<\/sub>, a4<\/sub>) verilmi\u015ftir. T\u00fcm uzuvlar\u0131n konumunu her hangi bir \u03b812<\/sub>\u00a0de\u011feri i\u00e7in bulmak istiyoruz.<\/p>\n

\"\"<\/p>\n

Geometrik y\u00f6ntemde kulland\u0131\u011f\u0131m\u0131z y\u00f6nteme benzer bir \u015fekilde ilk olarak A0<\/sub>AB0<\/sub>\u00a0\u00fc\u00e7genini g\u00f6z \u00f6n\u00fcne alal\u0131m. Bu \u00fc\u00e7gende iki kenar (a1<\/sub>, a2<\/sub>) ve aralar\u0131ndaki a\u00e7\u0131 (\u03b812<\/sub>) bilinmektedir. \u00d6yle ise \u00fc\u00e7\u00fcnc\u00fc kenar (AB0<\/sub> = s) ve \u03d5\u2032 a\u00e7\u0131s\u0131 bu \u00fc\u00e7gende kosin\u00fcs teoremi kullan\u0131larak bulunabilir:<\/p>\n\n\n\n\n
s = \\displaystyle \\sqrt{{{{\\text{a}}_{1}}^{2}+{\\text{a}_{2}}^{2}-2{\\text{a}_{1}}{{\\text{a}}_{2}}\\cos{{\\text{\u03b8}}_{{12}}}}} <\/span><\/td>\n(1)<\/td>\n<\/tr>\n
\u03d5\u2032 = cos-1<\/sup> \\displaystyle \\left[ {\\frac{{{{\\text{a}}_{1}}^{2}+{{\\text{s}}^{2}}-{{\\text{a}}_{2}}^{2}}}{{2{{\\text{a}}_{1}}\\text{s}}}} \\right] <\/span>\u00a0 \u00a0 (\u03d5 = \u03c0 \u2212 \u03d5\u2032)<\/td>\n(2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

s ve \u03d5\u2032 de\u011ferlerini belirlemek i\u00e7in bir ba\u015fka y\u00f6ntem ise (ki bu y\u00f6ntemde ters kosin\u00fcs fonksiyonu kullan\u0131ld\u0131\u011f\u0131nda ortaya \u00e7\u0131kacak olan \u00e7ifte de\u011ferlilik olmayacakt\u0131r) B0<\/sub>A<\/strong>\u00a0vekt\u00f6r\u00fcn\u00fc\u00a0B0<\/sub>A<\/strong> = B0<\/sub>A0<\/sub><\/strong>\u00a0+ A0<\/sub>B<\/strong>\u00a0olarak yazd\u0131ktan sonra :<\/p>\n\n\n\n\n
s\u00a0<\/sub>cos\u03d5 = a2<\/sub>cos\u03b812<\/sub>\u00a0\u2212 a1<\/sub>\u00a0(B0<\/sub>A<\/strong>\u00a0vekt\u00f6r\u00fcn\u00fcn yatay bile\u015feni )<\/td>\n(1\u2032)<\/td>\n<\/tr>\n
s\u00a0<\/sub>sin\u03d5 = a2<\/sub>sin\u03b812<\/sub>\u00a0(B0<\/sub>A<\/strong>\u00a0vekt\u00f6r\u00fcn\u00fcn dikey bile\u015feni)<\/td>\n(2\u2032)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Her iki denklemin sa\u011f taraf\u0131 bir \u03b812<\/sub>\u00a0a\u00e7\u0131s\u0131 de\u011ferine g\u00f6re\u00a0B0<\/sub>A<\/strong> vekt\u00f6r\u00fcn\u00fcn yatay ve dikey bile\u015fenlerini verecektir. Vekt\u00f6r\u00fc kutupsal koordinatlarda g\u00f6steren s ve \u03d5 de\u011ferlerini bulmak ise dik koordinat sisteminden kutupsal koordinat sistemine d\u00f6n\u00fc\u015f\u00fcmd\u00fcr. Hesap makinalar\u0131nda bu i\u015flem (R-P) ile g\u00f6sterilen tu\u015f ile (dik koordinat sisteminden kutupsal koordinat sistemine d\u00f6n\u00fc\u015f\u00fcm) yap\u0131l\u0131r ise s ve \u03d5 de\u011ferleri elde edilir. E\u011fer Excel paket program\u0131 kullan\u0131l\u0131r ise \u03d5 a\u00e7\u0131s\u0131n\u0131n bulunmas\u0131 i\u00e7in ATAN2(X, Y) fonksiyonu kullan\u0131labilir. (s i\u00e7in her iki terimin karelerinin toplam\u0131n\u0131n kare k\u00f6k\u00fc al\u0131n\u0131r). Mekanizma i\u00e7in s ve ge\u00e7erli \u03d5 de\u011feri bulunduktan sonra, ABB0<\/sub> \u00fc\u00e7geninin her \u00fc\u00e7 kenar uzunlu\u011fu bilindi\u011finden (s = |AB0<\/sub>| (\u03b812<\/sub> nin fonksiyonu olarak), |AB| = a3<\/sub>, |BB0<\/sub>| = a4<\/sub>), kosin\u00fcs teoremi ile bu \u00fc\u00e7genin a\u00e7\u0131 de\u011ferleri bulunabilir:<\/p>\n\n\n\n\n
\u03bc = cos-1<\/sup> \\displaystyle \\left[ {\\frac{{{{\\text{a}}_{3}}^{2}+{{\\text{a}}_{4}}^{2}-{{\\text{s}}^{2}}}}{{2{{\\text{a}}_{3}}{{\\text{a}}_{4}}}}} \\right] <\/span><\/td>\n(3)<\/td>\n<\/tr>\n
\u03c8 = cos-1<\/sup> \\displaystyle \\left[ {\\frac{{{{\\text{a}}_{4}}^{2}+{{\\text{s}}^{2}}-{{\\text{a}}_{3}}^{2}}}{{2{{\\text{a}}_{4}}\\text{s}}}} \\right] <\/span><\/td>\n(4)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u03bc ve \u03c8 a\u00e7\u0131lar\u0131 ters kosin\u00fcs fonksiyonu ile elde edildiklerinden 0 ile 180\u00b0\u00a0aras\u0131nda de\u011fer al\u0131rlar. Bilinmeyen \u03b813<\/sub> ve \u03b814<\/sub>\u00a0konum parametreleri e\u011fer mekanizma A0<\/sub>ABB0<\/sub>\u00a0\u015feklinde (a\u00e7\u0131k durum) monte edilmi\u015f ise:<\/p>\n\n\n\n\n
\u03b814<\/sub>\u00a0= \u03d5 \u2212 \u03c8<\/td>\n(5)<\/td>\n<\/tr>\n
\u03b813<\/sub> = \u03b814<\/sub> \u2212 \u03bc<\/td>\n(6)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

E\u011fer mekanizma A0<\/sub>AB’B0<\/sub>\u00a0\u015feklinde (\u00e7apraz durum) monte edilmi\u015f ise:<\/p>\n\n\n\n\n
\u03b814<\/sub>\u2032 = \u03d5 + \u03c8<\/td>\n(5\u2032)<\/td>\n<\/tr>\n
\u03b813<\/sub>\u2032 = \u03b814<\/sub> + \u03bc<\/td>\n(6\u2032)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

olacakt\u0131r. Kol-sarka\u00e7 boyutlar\u0131nda mekanizmalar i\u00e7in her \u03b812<\/sub>\u00a0a\u00e7\u0131s\u0131na kar\u015f\u0131 iki de\u011fi\u015fik \u03b813<\/sub> ve \u03b814<\/sub> de\u011feri elde edilecektir. Ancak, e\u011fer denklem (3) ve (4) te 1’den b\u00fcy\u00fck bir de\u011ferin ters kosin\u00fcs fonksiyonunun de\u011feri bulunmak istenir ise, \u00e7\u00f6z\u00fcm elde edilemez. Bu, mekanizman\u0131n verilmi\u015f olan \u03b812<\/sub> a\u00e7\u0131s\u0131nda kapal\u0131 bir devre olu\u015fturamayaca\u011f\u0131n\u0131 g\u00f6sterir.<\/p>\n

Bu \u015fekilde, konum de\u011fi\u015fkenleri de\u011ferlerini hesaplayabilece\u011fimiz bir denklem dizisi elde etmi\u015f bulunuyoruz.<\/p>\n

KOL-KIZAK MEKAN\u0130ZMASI<\/span><\/strong><\/p>\n

\"\"<\/p>\n

Yukar\u0131da kol-k\u0131zak mekanizmas\u0131 g\u00f6sterilmektedir. Konum analizi i\u00e7in (konum de\u011fi\u015fkenlerini bulmak i\u00e7in) gereken denklemler:<\/p>\n\n\n\n\n
p\u00a0<\/sub>cos\u03d5 = a2<\/sub>cos\u03b812<\/sub> \u2212 a1<\/sub><\/td>\n(1)<\/td>\n<\/tr>\n
p\u00a0<\/sub>sin\u03d5 = a2<\/sub>sin\u03b812<\/sub><\/td>\n(2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Excel paket program\u0131nda \u03d5 nin \u00e7\u00f6z\u00fcm\u00fc i\u00e7in ATAN2(X,Y) fonksiyonu kullan\u0131l\u0131r ise her durumda do\u011fru sonu\u00e7 elde edilebilecektir.<\/p>\n\n\n\n\n\n
\n

\u03c8 = cos-1<\/sup>(a4<\/sub>\/p)<\/p>\n

s = \\displaystyle \\sqrt{{{{\\text{p}}^{2}}-{{\\text{a}}_{4}}^{2}}} <\/span><\/p>\n

\u03b814<\/sub> = \u03d5 \u00b1 \u03c8<\/p>\n<\/td>\n

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

Bu denklemlerin \u00e7\u0131kar\u0131lmas\u0131 okuyucuya b\u0131rak\u0131lm\u0131\u015ft\u0131r (s\u0131ra ile A0<\/sub>AB ve AB0<\/sub>B \u00fc\u00e7genlerinin \u00e7\u00f6z\u00fcm\u00fcn\u00fc d\u00fc\u015f\u00fcnmek gerekir).<\/p>\n

\u00c7OK UZUVLU MEKAN\u0130ZMALAR<\/span><\/strong><\/p>\n

Genellikle \u00e7ok uzuvlu mekanizmalar dikkatlice incelenir ise, b\u00fcy\u00fck bir k\u0131sm\u0131n\u0131n d\u00f6rt uzuvlu devrelerden olu\u015ftu\u011fu g\u00f6r\u00fclecektir. Her bir devre yukar\u0131da \u00fc\u00e7 temel mekanizma i\u00e7in vermi\u015f oldu\u011fumuz y\u00f6ntem ile \u00e7\u00f6z\u00fclebilir. Bu durumda giri\u015f uzvunu i\u00e7eren devreden ba\u015flayarak devreler s\u0131rayla \u00e7\u00f6z\u00fclebilecektir.<\/p>\n

\"\"<\/p>\n

Yukar\u0131da g\u00f6sterilmi\u015f olan var-gel mekanizmas\u0131n\u0131 ele alal\u0131m. 1, 2, 3 ve 4 uzuvlar\u0131ndan olu\u015fan A0<\/sub>AB0<\/sub> devresi \u00f6nceden g\u00f6sterilmi\u015f olan kol-k\u0131zak mekanizmas\u0131n\u0131n a4<\/sub>\u00a0= 0 durumundan ba\u015fka bir \u015fey de\u011fildir. Bu devrenin \u00e7\u00f6z\u00fcm\u00fcnden \u03b814<\/sub>\u00a0a\u00e7\u0131s\u0131 ve s43<\/sub> elde edilebilecektir. 1, 4, 5 ve 6 uzuvlar\u0131n\u0131n olu\u015fturdu\u011fu B0<\/sub>BCPB0<\/sub> devresi ise, bir krank-biyel mekanizmas\u0131d\u0131r. Bu devrenin herhangi bir \u03b814<\/sub>\u00a0a\u00e7\u0131s\u0131na g\u00f6re \u00e7\u00f6z\u00fcm\u00fc ise s16<\/sub> ve \u03b815<\/sub>\u00a0konum parametrelerini verir. Se\u00e7ilen referans eksenlerine g\u00f6re konum parametrelerinin tan\u0131m\u0131na dikkat edilmesi gerekir.<\/p>\n

\u00d6yle ise, bilinen \u03b812<\/sub>\u00a0a\u00e7\u0131s\u0131na g\u00f6re di\u011fer konum de\u011fi\u015fkenleri:<\/p>\n

s43<\/sub>cos\u03b814<\/sub> = a2<\/sub>cos\u03b812<\/sub> \u2212 a1<\/sub><\/p>\n

s43<\/sub>sin\u03b814<\/sub> = a2<\/sub>sin\u03b812<\/sub><\/p>\n

\u03b815<\/sub> = sin-1<\/sup> \\displaystyle \\left[ {\\frac{{{{\\text{a}}_{4}}\\cos{{\\text{\u03b8}}_{{14}}}+{{\\text{b}}_{1}}}}{{{{\\text{a}}_{5}}}}} \\right] <\/span><\/p>\n

s16<\/sub> = a4<\/sub>sin\u03b814<\/sub> + a5<\/sub>cos\u03b815<\/sub><\/p>\n

denklemleri kullan\u0131larak elde edilebilir.<\/p>\n

\u0130lk iki denklem kol k\u0131zak mekanizmas\u0131 i\u00e7in A0<\/sub>AB0<\/sub>\u00a0\u00fc\u00e7geninin \u00e7\u00f6z\u00fcm\u00fcn\u00fc, son iki denklem ise krank biyel mekanizmas\u0131n\u0131n (B0<\/sub>BC)\u00e7\u00f6z\u00fcm\u00fcn\u00fc vermektedir. \u03b812<\/sub> ve \u03b814<\/sub> a\u00e7\u0131lar\u0131n\u0131n istendi\u011finde dikey bir do\u011fruya g\u00f6re \u00f6l\u00e7\u00fclmeyip kolayl\u0131kla yatay bir do\u011fruya g\u00f6re \u00f6l\u00e7\u00fclebilece\u011fi a\u00e7\u0131kt\u0131r. Ayr\u0131ca \u03b815<\/sub> de\u011feri belirlenirken ters sin\u00fcs kullan\u0131ld\u0131\u011f\u0131ndan \u015fekil ile uyumlu olabilmesi i\u00e7in \u03b815<\/sub>\u00a0a\u00e7\u0131 de\u011feri 90\u00b0\u00a0den k\u00fc\u00e7\u00fck olmal\u0131d\u0131r.<\/p>\n

Bu k\u0131s\u0131mda a\u00e7\u0131klanm\u0131\u015f olan y\u00f6ntem pratikte bir\u00e7ok mekanizman\u0131n analizinde kullan\u0131lan \u00e7ok basit bir y\u00f6ntemdir. M\u00fchim olan mekanizmay\u0131 olu\u015fturan basit devreleri belirlemek ve bu basit devrelerin her birini ayr\u0131 ayr\u0131 \u00e7\u00f6zebilmektir.<\/p>\n

\u00d6rnek:<\/strong><\/p>\n