{"id":978,"date":"2021-09-09T11:14:12","date_gmt":"2021-09-09T11:14:12","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=978"},"modified":"2021-10-05T23:28:24","modified_gmt":"2021-10-05T23:28:24","slug":"3-9","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch3\/3-9\/","title":{"rendered":"3-9"},"content":{"rendered":"
\n
\n\t

3.9<\/b> Devre Kapal\u0131l\u0131k Denklemlerinin \u0130teratif Y\u00f6ntemle \u00c7\u00f6z\u00fclmesi<\/h1>\n

\u00d6nceki k\u0131s\u0131mlarda devre kapal\u0131l\u0131k denklemleri kullan\u0131larak konum de\u011fi\u015fkenleri aras\u0131nda analitik ili\u015fkiler de\u011fi\u015fik y\u00f6ntemlerle bulunmaya \u00e7al\u0131\u015f\u0131lm\u0131\u015ft\u0131r.\u00a0Genel olarak her bir devre kapal\u0131l\u0131k denkleminde iki yeni bilinmeyen konum de\u011fi\u015fkeni bulunuyor ise, bu iki bilinmeyen i\u00e7in analitik bir \u00e7\u00f6z\u00fcm elde edilebilecektir.<\/span> Bu durumlarda denklemleri bilinmeyen konum parametrelerine g\u00f6re \u00f6nceki k\u0131s\u0131mlarda g\u00f6sterildi\u011fi gibi, do\u011frudan \u00e7\u00f6zmek daha uygundur. E\u011fer devre kapal\u0131l\u0131k denklemlerinin her birinde ikiden fazla bilinmeyen konum de\u011fi\u015fkeni var ise, bu konum de\u011fi\u015fkenlerini ba\u011f\u0131ms\u0131z konum de\u011fi\u015fkenine g\u00f6re belirlemek genel olarak m\u00fcmk\u00fcn de\u011fildir (baz\u0131 \u00f6zel durumlarda m\u00fcmk\u00fcn olabilir). Bu durumlar geometrik y\u00f6ntem uyguland\u0131\u011f\u0131nda deneme-yan\u0131lma gerektirmi\u015ftir. Ancak e\u011fer mekanizma serbestlik derecesine uyuyor ise, elde edilen vekt\u00f6r devre denkleminin iki kat\u0131 kadar bilinmeyen konum de\u011fi\u015fkeni ve serbestlik derecesi kadar ba\u011f\u0131ms\u0131z konum de\u011fi\u015fkeni bulunacakt\u0131r. Analitik \u00e7\u00f6z\u00fcm elde edilemedi\u011fi durumlarda veya her t\u00fcrl\u00fc mekanizman\u0131n analizini yapmak i\u00e7in genel bir algoritma olu\u015fturmak istedi\u011fimizde, devre kapal\u0131l\u0131k denklemlerinin say\u0131sal y\u00f6ntemlerle iteratif (yinelemeli) \u00e7\u00f6z\u00fcm\u00fc d\u00fc\u015f\u00fcn\u00fclmelidir.<\/p>\n

Devre kapal\u0131l\u0131k denklemlerine bakt\u0131\u011f\u0131m\u0131zda, her birinden iki lineer olmayan cebirsel denklem elde edece\u011fimize g\u00f6re, bu denklemler:<\/p>\n\n\n\n\n
Fj<\/sub>(x<\/strong>) = 0<\/td>\nj = 1, 2, 3, …, n<\/td>\n(1)<\/td>\n<\/tr>\n
<\/td>\nx<\/strong> = [x1<\/sub>\u00a0x2<\/sub>\u00a0x3<\/sub> … xn<\/sub>]T<\/sup><\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u015feklinde yaz\u0131labilir. Bu g\u00f6sterimde Fk<\/sub>(x<\/strong>) devre kapal\u0131l\u0131k denkleminden elde edilen karma\u015f\u0131k say\u0131lar veya reel say\u0131larla ifade edilebilen devre say\u0131s\u0131n\u0131n iki kat\u0131 kadar lineer olmayan denklem tak\u0131m\u0131d\u0131r. Vekt\u00f6r x<\/b> ise bilinmeyen konum de\u011fi\u015fkenleri: \u03b81j<\/sub>\u00a0veya sij<\/sub>\u00a0den olu\u015fur. Fj<\/sub>(x<\/strong>) ve t\u00fcrevleri, \u00e7\u00f6z\u00fcm\u00fc oldu\u011fu b\u00f6lgede tan\u0131ml\u0131 ise, Fj<\/sub>(x<\/strong>) = 0 denklemlerinin bilinmeyen de\u011ferlere g\u00f6re \u00e7\u00f6z\u00fcm\u00fc Newton-Raphson metodu<\/strong>\u00a0olarak bilinen iterasyon metodu ile yap\u0131labilir. Bu metod denklemleri k\u00fc\u00e7\u00fck aral\u0131klarda lineer olarak kabul eder ve bu lineer denklem tak\u0131m\u0131 i\u00e7in \u00e7\u00f6z\u00fcm bularak lineer olmayan denklem tak\u0131m\u0131n\u0131n \u00e7\u00f6z\u00fcme do\u011fru yakla\u015f\u0131p yakla\u015fmad\u0131\u011f\u0131n\u0131 kontrol eder (Fj<\/sub>(x<\/strong>) = \u03b5 ise, her basamakta lineer denklem \u00e7\u00f6z\u00fcm\u00fcn\u00fcn \u03b5 de\u011ferini daha k\u00fc\u00e7\u00fclt\u00fcp k\u00fc\u00e7\u00fcltmedi\u011fi) \u03b5’un yeteri kadar k\u00fc\u00e7\u00fck de\u011ferini veren x bilinmeyen de\u011ferleri \u00e7\u00f6z\u00fcmd\u00fcr. Bu t\u00fcr lineer olmayan denklem tak\u0131mlar\u0131n\u0131 \u00e7\u00f6zmek i\u00e7in ba\u015fka n\u00fcmerik y\u00f6ntemler de mevcuttur. Burada gerek basit olmas\u0131 ve gerekse mekanizmalar\u0131n \u00e7\u00f6z\u00fcm\u00fcnde olduk\u00e7a iyi sonu\u00e7 vermesi nedeni ile Newton-Raphson metodu a\u00e7\u0131klanacakt\u0131r. Okuyucu herhangi bir n\u00fcmerik metod kitab\u0131ndan di\u011fer y\u00f6ntemleri \u00f6\u011frenebilir.<\/p>\n

F(x) = 0 gibi tek de\u011fi\u015fkenli bir denklemi ele alal\u0131m. Denklem skaler de\u011fi\u015fken x’e g\u00f6re lineer de\u011fildir. Bu denklemde x = x* de\u011ferinin denklemin \u00e7\u00f6z\u00fcm\u00fc (k\u00f6k\u00fc) oldu\u011funu (yani F(x*) = 0) ve x(k)<\/sup> n\u0131n ise, x* de\u011ferine yak\u0131n bir de\u011fer oldu\u011funu kabul edelim. F(x) denkleminin x = x(k)<\/sup> noktas\u0131 civar\u0131nda Taylor serisi a\u00e7\u0131l\u0131m\u0131n\u0131 yaparsak<\/p>\n\n\n\n
F(x) = F(x(k)<\/sup>) + Fx<\/sub>(x(k)<\/sup>)(x \u2212 x(k)<\/sup>) + (y\u00fcksek mertebe terimler)<\/td>\n(2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

olur. Fx<\/sub>(x), F(x) denkleminin x’e g\u00f6re birinci t\u00fcrevidir: Fx<\/sub>(x) = dF(x)\/dx. E\u011fer x(k)<\/sup>\u00a0= x* olsa idi, F(x(k)<\/sup>) = 0 olacak ve x(k)<\/sup>\u00a0arad\u0131\u011f\u0131m\u0131z \u00e7\u00f6z\u00fcm olacak idi. Hatalardan dolay\u0131 bu olmayacakt\u0131r.<\/p>\n

x = x(k+1)<\/sup> in x* \u00e7\u00f6z\u00fcm de\u011ferine x(k)<\/sup>\u00a0dan daha yak\u0131n bir tahmin oldu\u011funu varsayal\u0131m. F(x(k+1)<\/sup>) \u2248 0 oldu\u011fu varsay\u0131l\u0131r ve F(x(k+1)<\/sup>) i\u00e7in Taylor serisi kullan\u0131larak x(k)<\/sup> ya g\u00f6re a\u00e7\u0131l\u0131m\u0131 yaz\u0131ld\u0131\u011f\u0131nda:<\/p>\n\n\n\n
F(x(k+1)<\/sup>) \u2248\u00a0F(x(k)<\/sup>) + Fx<\/sub>(x(k)<\/sup>)(x(k+1)<\/sup> \u2212 x(k)<\/sup>) = 0<\/td>\n(3)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

olacakt\u0131r. x(k+1)<\/sup>\u00a0\u2212 x(k)<\/sup>\u00a0aral\u0131\u011f\u0131 yeterince k\u00fc\u00e7\u00fck ise veya denklem lineer ise, y\u00fcksek mertebe terimlerin ihmal edilmesi b\u00fcy\u00fck bir yanl\u0131\u015fa neden olmayacakt\u0131r (tabi e\u011fer sonsuz say\u0131da terim kullan\u0131p seriyi s\u0131f\u0131ra e\u015fitleyip \u00e7\u00f6z\u00fcm yapsak x*’\u0131 hemen bulabilirdik. Ne yaz\u0131k ki bu m\u00fcmk\u00fcn de\u011fil, ve sadece ilk terimle yetinece\u011fiz). E\u011fer Fx<\/sub>(x(k)<\/sup>) s\u0131f\u0131ra e\u015fit de\u011fil ise, (3) denkleminden x(k+1)<\/sup>\u00a0\u00e7\u00f6z\u00fclebilir:<\/p>\n\n\n\n
x(k+1)<\/sup> = x(k)<\/sup>\u00a0\u2212 F(x(k)<\/sup>)\/Fx<\/sub>(x(k)<\/sup>)<\/td>\n(4)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(4) denklemi Newton-Raphson metodu ile elde edilen tekrarlama denklemidir. Y\u00fcksek mertebe terimler ihmal edilmi\u015f oldu\u011fundan, x(k+1)<\/sup>\u00a0denklemi s\u0131f\u0131ra e\u015fitleyen \u00e7\u00f6z\u00fcm de\u011fildir. Ancak x(k)<\/sup> ya g\u00f6re x*’a daha yak\u0131n bir noktaya gelinebilecektir. Yeni de\u011fer ile denklemin s\u0131f\u0131ra daha yakla\u015ft\u0131\u011f\u0131 g\u00f6r\u00fcl\u00fcr ise, (4) denklemini tekrar tekrar kullanmam\u0131z m\u00fcmk\u00fcn olacak ve m defa uyguland\u0131ktan sonra |F(x(m)<\/sup>)| < \u03b5 olacakt\u0131r. Burada \u03b5, kendisinden k\u00fc\u00e7\u00fck olan de\u011ferleri s\u0131f\u0131r kabul edebilece\u011fimiz k\u00fc\u00e7\u00fck bir de\u011ferdir. Tabi ki \u03b5 probleme g\u00f6re de\u011fi\u015fecek ve baz\u0131 problemlerde m say\u0131s\u0131 ne kadar b\u00fcy\u00fck olursa olsun |F(x(m)<\/sup>)| < \u03b5 \u015fart\u0131 sa\u011flanamayacakt\u0131r (yak\u0131nsama olmayacakt\u0131r). Bir\u00e7ok uygulama i\u00e7in \u00f6rne\u011fin \u03b5 = 10-6<\/sup> gibi k\u00fc\u00e7\u00fck bir de\u011fer kabul edilebilir. Bu durumda F(x) = 0 denklemini sa\u011flayan x de\u011ferini bulmak i\u00e7in bir algoritma ortaya \u00e7\u0131kacakt\u0131r:<\/p>\n

    \n
  1. F(x) = 0 denklemini sa\u011flayacak x* de\u011feri i\u00e7in x(0)<\/sup>\u00a0de\u011ferini tahmin edelim.<\/li>\n
  2. F(x(k+1)<\/sup>) de\u011ferini k = 0, 1, 2, …, m, basamak kadar (4) numaral\u0131 denklem kullanarak etap etap belirleyelim ve her etapta |F(x(k+1)<\/sup>)| < \u03b5e<\/sub> olup olmad\u0131\u011f\u0131n\u0131 veya |x(k+1)<\/sup>\u00a0\u2212 x(k)<\/sup>| < \u03b5s<\/sub> olup olmad\u0131\u011f\u0131n\u0131 kontrol edelim. Bu iki e\u015fitsizlikten birisi sa\u011flan\u0131yor ise dural\u0131m. x(k+1)<\/sup> arad\u0131\u011f\u0131m\u0131z \u00e7\u00f6z\u00fcmd\u00fcr. (\u03b5e<\/sub> denklem hata tolerans\u0131, ve \u03b5s<\/sub> ise \u00e7\u00f6z\u00fcm hata tolerans\u0131d\u0131r). E\u011fer Fx<\/sub>(x(k+1)<\/sup>) = 0 ve F(x(k+1)<\/sup>) \u2260 0 ise, 1. basama\u011fa gidip yeni bir x(0)<\/sup>\u00a0tahmini yap\u0131lmas\u0131 gereklidir. Aksi takdirde i\u015flem devam eder.<\/li>\n
  3. k+1 yerine k yerle\u015ftirip 2 basamak tekrarlan\u0131r. E\u011fer i\u015flem m defa tekrarland\u0131\u011f\u0131 halde |F(x(k+1)<\/sup>)| < \u03b5e<\/sub> veya |x(k+1)<\/sup>\u00a0\u2212 x(k)<\/sup>| < \u03b5s<\/sub> sa\u011flanamam\u0131\u015f ise, \u00e7\u00f6z\u00fcm bulunamam\u0131\u015ft\u0131r. Durulmas\u0131 gerekir. \u0130\u015fleme yeni bir x(0)<\/sup>\u00a0de\u011feri ile yeniden ba\u015flanabilir veya farkl\u0131 bir y\u00f6ntem denenebilir.<\/li>\n<\/ol>\n

    Newton-Raphson algoritmas\u0131 \u00e7o\u011fu uygulama probleminde gittik\u00e7e \u00e7\u00f6z\u00fcme yakla\u015fan bir dizi sonu\u00e7 verecektir. Genel olarak |F(x)| < \u03b5e<\/sub> \u015fart\u0131na s\u0131n\u0131rl\u0131 say\u0131da basamakta eri\u015filecektir. Ancak e\u011fer x(0)<\/sup> de\u011feri x*’a yeteri kadar yak\u0131n de\u011filse, \u03b5e<\/sub> ve \u03b5s<\/sub> de\u011ferleri eldeki probleme g\u00f6re \u00e7ok k\u00fc\u00e7\u00fck se\u00e7ildi ise, veya \u00e7\u00f6z\u00fclmesi istenilen problemde F(x) fonksiyonu ve t\u00fcrevleri d\u00fczg\u00fcn ve yumu\u015fak bir davran\u0131\u015f g\u00f6stermiyorlarsa, \u00e7\u00f6z\u00fcm bulunamayabilir. Baz\u0131 durumlarda ise, bulunan \u00e7\u00f6z\u00fcm aran\u0131lan x* \u00e7\u00f6z\u00fcm\u00fc olmayabilir (lineer olmayan denklemlerde F(x) = 0 denklemini sa\u011flayan birden fazla k\u00f6k olabilir).<\/p>\n

    Geometrik olarak Newton-Raphson y\u00f6ntemi a\u015fa\u011f\u0131da \u015eekil (a) da g\u00f6sterilmektedir. Her noktada F(x) e\u011frisi, e\u011fimi Fx<\/sub> olan bir do\u011fru oldu\u011fu kabul edilmekte ve o do\u011frunun k\u00f6k\u00fc (s\u0131f\u0131r olan x de\u011feri) bulunmaktad\u0131r. Normal durumlarda k\u00f6k \u00e7ok az say\u0131da basamakta h\u0131zla bulunabilir. Ancak \u015eekil (b),(c), ve (d) de g\u00f6r\u00fcld\u00fc\u011f\u00fc gibi, y\u00f6ntem baz\u0131 durumlarda istenilen \u00e7\u00f6z\u00fcme y\u00f6nelmeyebilir, veya hi\u00e7bir \u00e7\u00f6z\u00fcm bulunamayabilir.<\/p>\n

    \"\" \u00a0\u00a0 \u00a0<\/span><\/p>\n

    \u015eu ana kadar Newton-Raphson metodu bir de\u011fi\u015fkenli bir denklem i\u00e7in a\u00e7\u0131klanm\u0131\u015ft\u0131r. Ayn\u0131 y\u00f6ntem kolayl\u0131kla n bilinmeyenli n lineer olmayan denklem tak\u0131m\u0131n\u0131n \u00e7\u00f6z\u00fcm\u00fc i\u00e7in kullan\u0131labilir. Bu durumda Newton-Raphson algoritmas\u0131n\u0131 genelle\u015ftirmemiz gerekecektir. n bilinmeyenli n lineer olmayan denklem tak\u0131m\u0131:<\/p>\n\n\n\n
    F1<\/sub>(x<\/strong>) = F1<\/sub>(x1<\/sub>, x2<\/sub>, x3<\/sub>, …., xn<\/sub>) = 0
    \nF2<\/sub>(x<\/strong>) = F2<\/sub>(x1<\/sub>, x2<\/sub>, x3<\/sub>, …, xn<\/sub>) = 0
    \nF3<\/sub>(x<\/strong>) = F3<\/sub>(x1<\/sub>, x2<\/sub>, x3<\/sub>, …, xn<\/sub>) = 0
    \n……
    \nFn<\/sub>(x<\/strong>) = Fn<\/sub>(x1<\/sub>, x2<\/sub>, x3<\/sub>, …, xn<\/sub>) = 0<\/td>\n    		(5)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    veya:<\/p>\n

    F<\/strong>(x<\/strong>) = 0<\/strong><\/p>\n

    olarak yaz\u0131labilir.\u00a0\u00c7\u00f6z\u00fcm\u00a0x*<\/b>\u00a0= [x*1<\/sub>\u00a0x*2<\/sub>\u00a0x*3<\/sub> … x*n<\/sub>]T<\/sup> dir. Denklemleri k\u00f6k oldu\u011funu tahmin etti\u011fimiz de\u011fer olan x<\/b>(k)<\/sup> = [x(k)<\/sup>1<\/sub>\u00a0x(k)<\/sup>2<\/sub>\u00a0x(k)<\/sup>3 <\/sub>… x(k)<\/sup>n<\/sub>]T<\/sup>\u00a0noktas\u0131nda Taylor serisine g\u00f6re a\u00e7al\u0131m ve seride sadece birinci mertebe terimleri tutal\u0131m (di\u011fer terimleri tek denklem i\u00e7in yapt\u0131\u011f\u0131m\u0131z gibi, s\u0131f\u0131ra \u00e7ok yak\u0131n de\u011ferler oldu\u011funu kabul edip ihmal edece\u011fiz). Bu durumda:<\/p>\n\n\n\n
    F<\/strong>(x<\/strong>) \u2248\u00a0F<\/strong>(x<\/strong>(k)<\/sup>) + Fx<\/sub><\/strong>(x<\/strong>(k)<\/sup>)(x<\/strong> \u2212\u00a0x<\/strong>(k)<\/sup>)<\/td>\n(6)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    olacakt\u0131r. Bu denklemde\u00a0Fx<\/sub><\/strong>(x<\/strong>(k)<\/sup>), x<\/strong>(k)<\/sup>\u00a0noktas\u0131nda hesaplanan\u00a0Jacobian matrisi<\/strong>dir#<\/sup>. E\u011fer\u00a0x<\/strong> =\u00a0x<\/strong>(k+1)<\/sup>\u00a0\u00e7\u00f6z\u00fcme daha yak\u0131n bir de\u011fer olmas\u0131 isteniyor ise,\u00a0x<\/strong>(k)<\/sup>\u00a0dan daha iyi bir de\u011fer:<\/p>\n

    F<\/strong>(x<\/strong>(k+1)<\/sup>) \u2248 F<\/strong>(x<\/strong>(k)<\/sup>) + Fx<\/sub><\/strong>(x<\/strong>(k)<\/sup>)(x<\/strong>(k+1)<\/sup>\u00a0\u2212 x<\/strong>(k)<\/sup>) = 0<\/strong><\/p>\n

    \n

    #<\/sup> Jacobian matrisi: Fx<\/sub><\/strong> = \\displaystyle \\left[ {\\begin{array}{cccc} \\text{F}_{1\\text{x}_1}&\\text{F}_{1\\text{x}_2}&\\cdots&\\text{F}_{1\\text{x}_\\text{n}} \\\\ \\text{F}_{2\\text{x}_1}&\\text{F}_{2\\text{x}_2}&\\cdots&\\text{F}_{2\\text{x}_\\text{n}} \\\\ \\vdots&\\vdots&\\ddots&\\vdots \\\\ \\text{F}_{\\text{nx}_1}&\\text{F}_{\\text{nx}_2}&\\cdots&\\text{F}_{\\text{nx}_\\text{n}} \\end{array}} \\right] <\/span><\/p>\n

    burada Fkx<\/sub>m<\/sub>, Fk<\/sub>\u00a0n\u0131n xm<\/sub>\u00a0e g\u00f6re k\u0131smi t\u00fcrevidir.<\/p>\n\n\n\n
    F<\/strong>x<\/sub>(x<\/strong>(k)<\/sup>)\u0394x<\/strong>(k)<\/sup>= \u2212F<\/strong>(x<\/strong>(k)<\/sup>)<\/td>\n(7)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    yaz\u0131l\u0131r. Burada \u0394x<\/strong>(k)<\/sup>= x<\/strong>(k+1)<\/sup>\u00a0\u2212 x<\/strong>(<\/strong>k)<\/sup>\u00a0veya\u00a0x<\/strong>(k+1)<\/sup>=\u00a0x<\/strong>(k)<\/sup> + \u0394x<\/strong>(k)<\/sup>\u00a0d\u0131r.<\/p>\n

    Denklem tak\u0131m\u0131 i\u00e7in Newton-Raphson algoritmas\u0131:<\/p>\n

      \n
    1. Ba\u015flang\u0131\u00e7ta\u00a0F<\/strong>(x<\/strong>) = 0<\/strong> i\u00e7in x<\/b>(0)<\/sup>\u00a0de\u011ferlerinin \u00e7\u00f6z\u00fcm oldu\u011funu varsayal\u0131m.<\/li>\n
    2. k = 0, 1, 2, …, m iterasyonlar\u0131nda, \u0394x<\/strong>(k)<\/sup>,\u00a0x<\/strong>(k+1)<\/sup>\u00a0ve\u00a0F<\/strong>(x<\/strong>(k+1)<\/sup>) de\u011ferlerini hesaplay\u0131n. E\u011fer |F<\/strong>(x<\/strong>(k+1)<\/sup>)| <\u00a0\u03b5e<\/sub><\/strong> ve |x<\/strong>(k+1)<\/sup>\u00a0\u2212 x<\/strong>(k)<\/sup>| < \u03b5s<\/sub><\/strong>\u00a0ise durun.\u00a0x<\/strong>(k+1)<\/sup> aran\u0131lan sonu\u00e7tur (\u03b5e<\/sub><\/strong> denklem hata tolerans\u0131 ve \u03b5s<\/sub><\/strong>\u00a0ise \u00e7\u00f6z\u00fcm tolerans vekt\u00f6rleridir). E\u011fer |Fx<\/sub><\/strong><\/b>(x<\/strong>(k+1)<\/sup>)| = 0 ve\u00a0F<\/strong>(x<\/strong>(k+1)<\/sup>) \u2260 0 ise, k\u00f6kler i\u00e7in yeni bir tahmin yap\u0131p birinci basama\u011fa d\u00f6n\u00fcl\u00fcr. Aksi takdirde 3. basamakla devam edilir.<\/li>\n
    3. k + 1 de\u011ferlerini k olarak ele al\u0131p 2. basama\u011fa d\u00f6n\u00fcl\u00fcr. E\u011fer bu i\u015flem m defa yap\u0131ld\u0131\u011f\u0131 halde |F<\/strong>(x<\/strong>(k+1)<\/sup>)| < \u03b5e<\/sub><\/strong> veya |x<\/strong>(k+1)<\/sup>\u00a0\u2212 x<\/strong>(k)<\/sup>| < \u03b5s<\/sub><\/strong> sa\u011flanamad\u0131 ise, \u00e7\u00f6z\u00fcm bulunamam\u0131\u015ft\u0131r. Algoritman\u0131n durdurulmas\u0131 gerekir. \u0130\u015fleme yeni bir x<\/strong>(0)<\/sup>\u00a0tahmin vekt\u00f6r\u00fc ile yeniden ba\u015flanabilir veya farkl\u0131 bir y\u00f6ntem denenebilir.<\/li>\n<\/ol>\n

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

      Newton-Raphson algoritmas\u0131n\u0131 g\u00f6stermek i\u00e7in a\u015fa\u011f\u0131daki \u015fekilde g\u00f6r\u00fclen mekanizmay\u0131 ele alal\u0131m. Bu mekanizman\u0131n bilinmeyen konum de\u011fi\u015fkenlerini belirlemek i\u00e7in deneme yan\u0131lma yapmadan verecek kapal\u0131 denklemler elde etmemiz m\u00fcmk\u00fcn de\u011fildir. Karma\u015f\u0131k say\u0131lar ile devre kapal\u0131l\u0131k denklemleri:<\/p>\n

      a2<\/sub>ei\u03b812<\/sub><\/sup> = b1<\/sub>\u00a0\u2212 ic1<\/sub> + a5<\/sub>ei\u03b815<\/sub><\/sup> + s4<\/sub>ei\u03b814<\/sub><\/sup><\/p>\n

      a5<\/sub>ei\u03b815<\/sub><\/sup> + a4<\/sub>ei\u03b814<\/sub><\/sup> = \u2212b1<\/sub> + ia1<\/sub> + s6<\/sub><\/p>\n

      \u015feklindedir. Bu denklemler:<\/p>\n

      b1<\/sub>\u00a0\u2212 ic1<\/sub> \u2212 a2<\/sub>ei\u03b812<\/sub><\/sup> + a5<\/sub>ei\u03b815<\/sub><\/sup> + s4<\/sub>ei\u03b814<\/sub><\/sup>\u00a0= 0<\/p>\n

      b1<\/sub> \u2212 ia1<\/sub> \u2212 s6<\/sub> + a5<\/sub>ei\u03b815<\/sub><\/sup> + a4<\/sub>ei\u03b814<\/sub><\/sup> = 0<\/p>\n

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

      \"\"<\/span><\/p>\n

      veya reel say\u0131larla:<\/p>\n

      b1 <\/sub>\u2212 a2<\/sub>cos\u03b812\u00a0<\/sub>+ a5<\/sub>cos\u03b815\u00a0<\/sub>+ s4<\/sub>cos\u03b814\u00a0<\/sub>= 0<\/p>\n

      \u2212c1 <\/sub>\u2212 a2<\/sub>sin\u03b812\u00a0<\/sub>+ a5<\/sub>sin\u03b815\u00a0<\/sub>+ s4<\/sub>sin\u03b814\u00a0<\/sub>= 0<\/p>\n

      b1 <\/sub>\u2212 s6<\/sub>\u00a0<\/sub>+ a5<\/sub>cos\u03b815\u00a0<\/sub>+ a4<\/sub>cos\u03b814\u00a0<\/sub>= 0<\/p>\n

      \u2212a1<\/sub>\u00a0<\/sub>+ a5<\/sub>sin\u03b815\u00a0<\/sub>+ a4<\/sub>sin\u03b814\u00a0<\/sub>= 0<\/p>\n

      denklemleri elde edilir (karma\u015f\u0131k say\u0131lar kullanarak denklemlerin karma\u015f\u0131k e\u015flenikleri kullanarak da d\u00f6rt denklem elde edilebilir. Bu \u00f6rnekte reel say\u0131lar kullan\u0131lacakt\u0131r). Elde edilen denklem tak\u0131m\u0131:<\/p>\n

      F1<\/sub>(\u03b814<\/sub>, \u03b815<\/sub>, s4<\/sub>) = 0<\/p>\n

      F2<\/sub>(\u03b814<\/sub>, \u03b815<\/sub>, s4<\/sub>) = 0<\/p>\n

      F3<\/sub>(\u03b814<\/sub>, \u03b815<\/sub>, s6<\/sub>) = 0<\/p>\n

      F4<\/sub>(\u03b814<\/sub>, \u03b815<\/sub>) = 0<\/p>\n

      \u015feklindedir. Verilen \u03b812<\/sub> de\u011ferine g\u00f6re bilinmeyen konum de\u011fi\u015fkenleri: \u03b814<\/sub>, \u03b815<\/sub>, s4<\/sub>\u00a0ve s6<\/sub> de\u011ferlerini bulmak istiyoruz. Bu konum de\u011fi\u015fkenlerinin de\u011feri olarak ba\u015flang\u0131\u00e7ta yapabilece\u011fimiz ilk tahminler: \u03b814<\/sub>0<\/sup>, \u03b815<\/sub>0<\/sup>, s4<\/sub>0<\/sup>, s6<\/sub>0<\/sup> olsun. Bu de\u011ferleri devre kapal\u0131l\u0131k denklemlerine yerle\u015ftirir isek denklemler s\u0131f\u0131ra e\u015fit olmayacak, her biri \u03b5i<\/sub> gibi bir farkl\u0131 de\u011ferlere e\u015fit olacakt\u0131r. Bu de\u011ferin k\u00fc\u00e7\u00fck olmas\u0131 bizim \u00e7\u00f6z\u00fcme ne kadar yak\u0131n oldu\u011fumuzun g\u00f6stergesidir. \u03b814<\/sub>1<\/sup> , \u03b815<\/sub>1<\/sup>, s4<\/sub>1<\/sup>, s6<\/sub>1<\/sup> de\u011ferleri ise aran\u0131lan \u00e7\u00f6z\u00fcm i\u00e7in daha iyi bir tahmin ise, ve \u03b4s4<\/sub> = s4<\/sub>1<\/sup> \u2212 s4<\/sub>0<\/sup>, \u03b4s6<\/sub>\u00a0= s6<\/sub>1<\/sup> \u2212 s6<\/sub>0<\/sup>, \u03b4\u03b814<\/sub>= \u03b814<\/sub>1<\/sup> \u2212 \u03b814<\/sub>0<\/sup>, \u03b4\u03b815<\/sub> = \u03b815<\/sub>1<\/sup> \u2212 \u03b815<\/sub>0<\/sup> de\u011ferleri bu iki tahmin aras\u0131nda fark oldu\u011fu kabul edilir ise bu fark Newton-Raphson algoritmas\u0131na g\u00f6re denklem (7) kullan\u0131larak:<\/p>\n

      \\displaystyle \\left[ {\\begin{array}{cccc} {\\text{i}{{\\text{s}}_{4}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{14}}}^{\\text{k}}}}}} & {\\text{i}{{\\text{a}}_{5}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{15}}}^{\\text{k}}}}}} & {{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{14}}}^{\\text{k}}}}}} & 0 \\\\ {\\text{i}{{\\text{a}}_{4}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{14}}}^{\\text{k}}}}}} & {\\text{i}{{\\text{a}}_{5}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{15}}}^{\\text{k}}}}}} & 0 & {-1} \\\\ {-\\text{i}{{\\text{s}}_{4}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{14}}}^{\\text{k}}}}}} & {-\\text{i}{{\\text{a}}_{5}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{15}}}^{\\text{k}}}}}} & {{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{14}}}^{\\text{k}}}}}} & 0 \\\\ {-\\text{i}{{\\text{a}}_{4}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{14}}}^{\\text{k}}}}}} & {-\\text{i}{{\\text{a}}_{5}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{15}}}^{\\text{k}}}}}} & 0 & {-1} \\end{array}} \\right]\\left[ {\\begin{array}{c} {\\text{\u03b4} {{\\text{\u03b8}}_{{14}}}} \\\\ {\\text{\u03b4} {{\\text{\u03b8}}_{{15}}}} \\\\ {\\text{\u03b4} {{\\text{s}}_{4}}} \\\\ {\\text{\u03b4} {{\\text{s}}_{6}}} \\end{array}} \\right]=-\\left[ {\\begin{array}{c} {{{\\text{b}}_{1}}-\\text{i}{{\\text{c}}_{1}}-{{\\text{a}}_{2}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{12}}}^{\\text{k}}}}}+{{\\text{a}}_{5}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{15}}}^{\\text{k}}}}}+{{\\text{s}}_{4}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{14}}}^{\\text{k}}}}}} \\\\ {{{\\text{b}}_{1}}-\\text{i}{{\\text{a}}_{1}}-{{\\text{s}}_{6}}^{\\text{k}}+{{\\text{a}}_{5}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{15}}}^{\\text{k}}}}}+{{\\text{a}}_{4}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{14}}}^{\\text{k}}}}}} \\\\ {{{\\text{b}}_{1}}+\\text{i}{{\\text{c}}_{1}}-{{\\text{a}}_{2}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{12}}}^{\\text{k}}}}}+{{\\text{a}}_{5}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{15}}}^{\\text{k}}}}}+{{\\text{s}}_{4}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{14}}}^{\\text{k}}}}}} \\\\ {{{\\text{b}}_{1}}+\\text{i}{{\\text{a}}_{1}}-{{\\text{s}}_{6}}^{\\text{k}}+{{\\text{a}}_{5}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{15}}}^{\\text{k}}}}}+{{\\text{a}}_{4}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{14}}}^{\\text{k}}}}}} \\end{array}} \\right] <\/span><\/p>\n

      denklem tak\u0131m\u0131ndan (karma\u015f\u0131k say\u0131lar ile) veya:<\/p>\n

      \\displaystyle \\left[ {\\begin{array}{cccc} {-{{\\text{s}}_{4}}\\sin {{\\text{\u03b8}}_{{14}}}^{k}}&{-{{\\text{a}}_{5}}\\sin {{\\text{\u03b8}}_{{15}}}^{\\text{k}}}&{\\cos {{\\text{\u03b8}}_{{14}}}^{\\text{k}}}&0 \\\\ {{{\\text{s}}_{4}}\\cos {{\\text{\u03b8}}_{{14}}}^{\\text{k}}}&{{{\\text{a}}_{5}}\\cos {{\\text{\u03b8}}_{{15}}}^{\\text{k}}}&{\\sin {{\\text{\u03b8}}_{{14}}}^{\\text{k}}0}&0 \\\\ {-{{\\text{a}}_{4}}\\sin {{\\text{\u03b8}}_{{14}}}^{\\text{k}}}&{{{\\text{a}}_{5}}\\sin {{\\text{\u03b8}}_{{15}}}^{\\text{k}}}&0&1 \\\\ {{{\\text{a}}_{4}}\\cos {{\\text{\u03b8}}_{{14}}}^{\\text{k}}}&{{{\\text{a}}_{5}}\\cos {{\\text{\u03b8}}_{{15}}}^{\\text{k}}}&0&0 \\end{array}} \\right] \\left[ {\\begin{array}{c} {\\text{\u03b4} {{\\text{\u03b8}}_{{14}}}} \\\\ {\\text{\u03b4} {{\\text{\u03b8}}_{{15}}}} \\\\ {\\text{\u03b4} {{\\text{s}}_{4}}} \\\\ {\\text{\u03b4} {{\\text{s}}_{6}}} \\end{array}} \\right]=-\\left[ {\\begin{array}{c} {{{\\text{b}}_{1}}-{{\\text{a}}_{2}}\\cos {{\\text{\u03b8}}_{{12}}}^{\\text{k}}+{{\\text{a}}_{5}}\\cos {{\\text{\u03b8}}_{{14}}}^{\\text{k}}+{{\\text{s}}_{4}}\\cos {{\\text{\u03b8}}_{{14}}}^{\\text{k}}} \\\\ {-{{\\text{c}}_{1}}-{{\\text{a}}_{2}}\\sin {{\\text{\u03b8}}_{{12}}}^{\\text{k}}+{{\\text{a}}_{5}}\\sin {{\\text{\u03b8}}_{{14}}}^{\\text{k}}+{{\\text{s}}_{4}}\\sin {{\\text{\u03b8}}_{{14}}}^{\\text{k}}} \\\\ {{{\\text{b}}_{1}}-{{\\text{s}}_{6}}^{\\text{k}}+{{\\text{a}}_{5}}\\cos {{\\text{\u03b8}}_{{15}}}^{\\text{k}}+{{\\text{a}}_{4}}\\cos {{\\text{\u03b8}}_{{14}}}^{\\text{k}}} \\\\ {-{{\\text{a}}_{1}}+{{\\text{a}}_{5}}\\sin {{\\text{\u03b8}}_{{15}}}^{\\text{k}}+{{\\text{a}}_{4}}\\sin {{\\text{\u03b8}}_{{14}}}^{\\text{k}}} \\end{array}} \\right] <\/span><\/p>\n

      denklem tak\u0131m\u0131ndan (reel say\u0131lar ile) elde edilebilir. Dikkat edilir ise, denklem tak\u0131m\u0131 bilinmeyen farklara g\u00f6re lineer olup:<\/p>\n

      A\u03b4<\/strong> = E<\/strong><\/p>\n

      \u015feklinde yaz\u0131labilir. A<\/strong> “katsay\u0131 matrisi<\/strong>” (Jacobian<\/strong>) reel veya karma\u015f\u0131k \u015fekilde olabilir. \u03b5i<\/sub> elemanlar\u0131ndan olu\u015fan E<\/strong> = [\u03b51<\/sub>\u00a0\u03b52<\/sub>\u00a0\u03b53<\/sub> \u2026 \u03b5n<\/sub>]T<\/sup> kolon vekt\u00f6r\u00fc ise, konum de\u011fi\u015fkenlerinin ilk tahmin de\u011ferleri ile devre kapal\u0131l\u0131k denklemlerini hesaplad\u0131\u011f\u0131m\u0131zda ortaya \u00e7\u0131kan denklem hatas\u0131d\u0131r (yani her bir denklem s\u0131f\u0131r olmay\u0131p farkl\u0131 bir \u03b5i<\/sub>\u00a0de\u011feri alacakt\u0131r ve bu da devrenin kapal\u0131 olmad\u0131\u011f\u0131n\u0131 g\u00f6steren hata de\u011ferleridir). Bu denklem tak\u0131m\u0131ndan daha iyi olaca\u011f\u0131n\u0131 bekledi\u011fimiz tahmin ile o an yapt\u0131\u011f\u0131m\u0131z tahmin aras\u0131nda fark ise:<\/p>\n

      \u03b4<\/strong> = A<\/strong>-1<\/sup>E<\/strong><\/p>\n

      olarak \u00e7\u00f6zebiliriz. Bu fark de\u011ferleri, kullanmakta oldu\u011fumuz \u00f6rnekte \u03b4\u03b814<\/sub>, \u03b4\u03b815<\/sub>, \u03b4s4<\/sub>\u00a0ve \u03b4s5<\/sub>\u00a0olsun. Bu durumda konum de\u011fi\u015fkenleri i\u00e7in bir sonraki etapta yapaca\u011f\u0131m\u0131z daha iyi tahmin:<\/p>\n

      s4<\/sub>(k+1)<\/sup> = s4<\/sub>k<\/sup> + \u03b4s4<\/sub><\/p>\n

      s5<\/sub>(k+1)<\/sup> = s5<\/sub>k<\/sup> + \u03b4s5<\/sub><\/p>\n

      \u03b814<\/sub>(k+1)<\/sup> = \u03b414<\/sub>k<\/sup> + \u03b4\u03b814<\/sub><\/p>\n

      \u03b815<\/sub>(k+1)<\/sup> = \u03b815<\/sub>k<\/sup> + \u03b4\u03b815<\/sub><\/p>\n

      olacakt\u0131r. Yukar\u0131da a\u00e7\u0131klanm\u0131\u015f olan i\u015flem, istenildi\u011fi kadar tekrarlanabilir. Durma kriteri olarak |F<\/strong>(x<\/strong>(k+1)<\/sup>)| < \u03b5e<\/sub><\/strong> veya |x<\/strong>(k+1)<\/sup> \u2212 x<\/strong>(k)<\/sup>| < \u03b5s<\/sub><\/strong>\u00a0kullan\u0131labilir. Mekanizmalarda devre denklemlerinin \u00e7\u00f6z\u00fcm\u00fc s\u0131ras\u0131nda genellikle \u00e7ok k\u00fc\u00e7\u00fck hata de\u011ferleri kullan\u0131lsa bile yakla\u015f\u0131k 10 etapta \u00e7\u00f6z\u00fcm bulunabilmektedir. Bunun m\u00fcmk\u00fcn olmad\u0131\u011f\u0131 durumlar mekanizman\u0131n kilitlendi\u011fi veya kilitlenmeye yak\u0131n oldu\u011fu konumlar veya ilk yapm\u0131\u015f oldu\u011fumuz tahminin ger\u00e7ek konum de\u011fi\u015fkenlerine g\u00f6re \u00e7ok uzak oldu\u011fu durumlard\u0131r. Bir mekanizman\u0131n \u00f6l\u00e7ekli \u00e7izimi \u00fczerinde g\u00f6zle bile yapabilece\u011fimiz ilk tahminler genel olarak uygun bir ilk tahmin olabilir.<\/p>\n

      E\u011fer mekanizma t\u00fcm bir \u00e7evrim i\u00e7in incelenecek ise, bu i\u015flem her bir krank a\u00e7\u0131s\u0131 i\u00e7in yap\u0131lmal\u0131 ve her bir krank a\u00e7\u0131s\u0131nda konum de\u011fi\u015fkenleri de\u011ferleri i\u00e7in tahminde bulunulmal\u0131d\u0131r. Ancak uygulamada mekanizman\u0131n birinci konumu i\u00e7in tahmin yap\u0131lmas\u0131, ikinci konum, e\u011fer mekanizma ba\u011f\u0131ms\u0131z parametresinin ufak bir de\u011fi\u015fimi sonucu olu\u015fuyor ise (\u00f6rne\u011fin krank d\u00f6nmesi iki konum aras\u0131nda 10\u00b0\u00a0az oldu\u011funda), bir \u00f6nceki konumda elde edilen konum de\u011ferlerini bir sonraki konum de\u011fi\u015fkenleri i\u00e7in tahmin edilen de\u011ferler olarak kullan\u0131lmas\u0131n\u0131n m\u00fcmk\u00fcn oldu\u011fu g\u00f6r\u00fclm\u00fc\u015ft\u00fcr. Bu farklar\u0131n b\u00fcy\u00fck oldu\u011fu durumlarda \u00e7\u00f6z\u00fcm yak\u0131nsamayabilir veya mekanizman\u0131n farkl\u0131 bir montaj \u015fekline (ilk konumdan bu \u015fekle mekanizma s\u00f6k\u00fcl\u00fcp tak\u0131lmadan getirilemez) yak\u0131nsama olabilir. Benzer bir \u015fekilde Jacobian matrisinin s\u0131f\u0131r veya s\u0131f\u0131ra yak\u0131n oldu\u011fu durumlarda baz\u0131 sorunlar ortaya \u00e7\u0131kabilir. Genellikle lineer olmayan denklemlerin \u00e7\u00f6z\u00fcm\u00fc s\u0131ras\u0131nda kar\u015f\u0131la\u015f\u0131lan problemler aynen mekanizmalar\u0131n \u00e7\u00f6z\u00fcm\u00fc i\u00e7inde ge\u00e7erlidir.<\/p>\n

      Bu y\u00f6ntem \u00e7ok genel olup d\u00fczlemsel veya uzaysal, kayar, d\u00f6ner veya kamal\u0131 silindir mafsall\u0131 her t\u00fcrl\u00fc mekanizma i\u00e7in kolayca uygulanabilir. Uzaysal mekanizmalar i\u00e7in elde edilecek olan denklem say\u0131s\u0131 tabi ki daha fazla olacakt\u0131r. Ayn\u0131 y\u00f6ntem g\u00fcn\u00fcm\u00fczde kullan\u0131lmakta olan \u00e7e\u015fitli mekanizma analiz programlar\u0131nda da kullan\u0131lmaktad\u0131r.<\/p>\n

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