{"id":1873,"date":"2021-09-30T09:20:54","date_gmt":"2021-09-30T09:20:54","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=1873"},"modified":"2021-10-05T22:10:42","modified_gmt":"2021-10-05T22:10:42","slug":"ek2","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ek2\/","title":{"rendered":"Ek 2"},"content":{"rendered":"
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Ek-2<\/span><\/b><\/p>\n

Basit Mekanizmalar \u0130\u00e7in Bilgisayar Programlar\u0131<\/h1>\n

Uygulamalarda basit d\u00f6rt uzuvlu mekanizmalar veya bu basit d\u00f6rt uzuvlu mekanizmalardan seri ba\u011flanarak elde edilen mekanizmalar \u00e7ok s\u0131k rastlanmaktad\u0131r. Bu mekanizmalar i\u00e7in haz\u0131rlanacak ufak fonksiyon programlar\u0131 gerek Excel paket program\u0131 i\u00e7inde Macro programlar olarak ve gerek VisualBasic lisan\u0131 kullan\u0131larak yaz\u0131lan genel veya \u00f6zel ama\u00e7l\u0131 mekanizma programlar\u0131 i\u00e7in \u00e7ok \u00f6nemli yap\u0131 ta\u015flar\u0131n\u0131 olu\u015fturmaktad\u0131r.<\/p>\n

2.1. Kartezyen koordinatlar\u0131 verilen bir vekt\u00f6r\u00fcn uzunlu\u011fu ve y\u00f6n a\u00e7\u0131s\u0131n\u0131 bulmak:<\/span><\/strong><\/p>\n

x ve y de\u011ferleri verildi\u011finde OP uzunlu\u011fu (r = Boy(x,y)) ve OP do\u011frusunun pozitif x ekseni ile yapt\u0131\u011f\u0131 a\u00e7\u0131y\u0131 radyan olarak (\u03b8 = A\u00e7\u0131(x,y)) verecek iki basit fonksiyon program\u0131d\u0131r. Dikkat edilir ise, a\u00e7\u0131(x,y) program\u0131 basic lisan\u0131nda var olan Atn(x) fonksiyonunu kullanmaktad\u0131r. Ancak a\u00e7\u0131n\u0131n hangi b\u00f6lgede oldu\u011funu x ve y bile\u015fenlerinin i\u015faretlerinden bularak Atn(x) fonksiyonundan farkl\u0131 olarak verilen x ve y de\u011ferine g\u00f6re var olan tek do\u011fru \u00e7\u00f6z\u00fcm\u00fc kullan\u0131c\u0131ya vermektedir. Fonksiyonlar\u0131n d\u0131\u015f\u0131nda PI ve Eps sabit de\u011ferleri tan\u0131mlanm\u0131\u015ft\u0131r. x <\u00a0<\/span>Eps oldu\u011funda x = 0 kabul edilerek y nin i\u015faretine g\u00f6re a\u00e7\u0131 900<\/sup>\u00a0veya -900<\/sup>\u00a0kabul edilmi\u015ftir (bu t\u00fcr bir \u00f6nlem al\u0131nmaz ise k\u00fc\u00e7\u00fck x de\u011ferlerinde s\u0131f\u0131ra b\u00f6lme hatas\u0131 ile kar\u015f\u0131la\u015f\u0131l\u0131r).<\/p>\n

Global Const PI = 3.1415926
\nGlobal Const Eps = 0.0000001\"\"
\nFunction Boy(X, Y)
\nBoy = Sqr(X ^ 2 + Y ^ 2)
\nEnd Function<\/p>\n

Function A\u00e7\u0131(X, Y)
\nDim AA
\nIf Abs(X) > Eps Then
\nAA = Atn(Y \/ X)
\nIf X < 0 Then
\nAA = AA + PI
\nElse:
\nIf Y < 0 Then AA = AA + 2*PI
\nEnd If
\nElse:
\nIf Y > 0 Then AA = PI \/ 2 Else AA = -PI \/ 2
\nEnd If
\nA\u00e7\u0131 = AA
\nEnd Function<\/p>\n

2.2. Ters Sin\u00fcs ve Kosin\u00fcs fonksiyonlar\u0131:<\/span><\/strong><\/p>\n

Normal olarak Pascal ve Basic lisanlar\u0131nda Atn(x) – ters tanjant fonksiyonu d\u0131\u015f\u0131nda ters trigonometrik fonksiyon yoktur. Mekanizma analizi s\u0131ras\u0131nda ters sin\u00fcs (Asin(X)) ve ters kosin\u00fcs (Acos(X)) fonksiyonlar\u0131na ihtiya\u00e7 olmaktad\u0131r. Dikkat edilir ise, bu fonksiyonlar ACOS un ve ASIN in de\u011ferlerinden sadece birini vermektedir. Kullan\u0131c\u0131 e\u011fer di\u011fer \u00e7\u00f6z\u00fcm\u00fc istiyor ise, bu a\u00e7\u0131ya 180 derece eklemeli veya \u00e7\u0131karmal\u0131d\u0131r.<\/p>\n

Function Acos(X)
\nAcos = Atn(-X \/ Sqr(-X * X + 1)) + 2 * Atn(1)
\nEnd Function<\/p>\n

Function Asin(X)
\nAsin = Atn(X \/ Sqr(-X * X + 1))
\nEnd Function<\/p>\n

2.3. \u00dc\u00e7gen \u00c7\u00f6z\u00fcm\u00fc:<\/span><\/strong><\/p>\n

Mekanizmalar\u0131n analizi s\u0131ras\u0131nda \u00fc\u00e7 kenar\u0131 verilen bir \u00fc\u00e7genin bir a\u00e7\u0131s\u0131n\u0131 (A\u00e7\u01313 = A\u00e7\u0131Cos(U1,U2,U3)) veya iki kenar\u0131 ve a\u00e7\u0131s\u0131 verildi\u011finde kar\u015f\u0131 kenar uzunlu\u011funu (U3 = BoyCos(U1,U2;A\u00e7\u01313)) bulmak i\u00e7in haz\u0131rlanm\u0131\u015f fonksiyon programlar\u0131d\u0131r.<\/p>\n

Function A\u00e7\u0131Cos(u1, u2, u3)\"\"
\nDim u u = (u1 * u1 + u2 * u2 – u3 * u3) \/ (2 * u1 * u2)
\nAA = Acos(u)
\nA\u00e7\u0131Cos = AA
\nEnd Function<\/p>\n

Function BoyCos(U1, U2, A\u00e7\u01313)
\nBoyCos = (U1 ^ 2 + U2 ^ 2 – 2 * U1 * U2 * Cos(A\u00e7\u01313)) \/ (2 * U1 * U2)
\nEnd Function<\/p>\n

2.4. Basit Mekanizmalar\u0131n Analizi:<\/b><\/span><\/p>\n

2.4-a) D\u00f6rt-\u00c7ubuk Mekanizmas\u0131<\/span><\/p>\n

\"\"<\/p>\n

Bu fonksiyon program\u0131 uzuv boyutlar\u0131, giri\u015f kolu a\u00e7\u0131s\u0131 ve ba\u011flant\u0131 \u015fekli verilen bir d\u00f6rt-\u00e7ubuk mekanizmas\u0131n\u0131n \u00e7\u0131k\u0131\u015f kolu a\u00e7\u0131s\u0131n\u0131 bulmak i\u00e7in yaz\u0131lm\u0131\u015ft\u0131r. Kullan\u0131c\u0131 s say\u0131s\u0131n\u0131 +1 veya -1 tan\u0131mlayarak mekanizman\u0131n a\u00e7\u0131k veya \u00e7apraz ba\u011fland\u0131\u011f\u0131n\u0131 belirleyecektir.<\/p>\n

Program incelenir ise, \u00f6nceden yaz\u0131lm\u0131\u015f olan Boy(), A\u00e7\u0131() ve A\u00e7\u0131Cos() fonksiyonlar\u0131 bu fonksiyonda kullan\u0131lm\u0131\u015ft\u0131r. Bu fonksiyonun kullan\u0131labilmesi i\u00e7in di\u011fer fonksiyonlar\u0131nda programda yaz\u0131l\u0131 olmas\u0131 gerekir. Programa basit ekler yaparak biyel a\u00e7\u0131s\u0131n\u0131 ve ba\u011flama a\u00e7\u0131s\u0131n\u0131 bulmak i\u00e7in fonksiyonlara \u00e7evirmek m\u00fcmk\u00fcnd\u00fcr.<\/p>\n

Function D\u00f6rt\u00c7ubuk(Krank, Biyel, Kol, Sabit, s, th)
\nDim dd, fi, si As Double
\nDim xd, yd As Double
\nxd = -Sabit + Krank * Cos(th)
\nyd = Krank* Sin(th)
\ndd = Boy(xd, yd)
\nfi = A\u00e7\u0131(xd, yd)
\nsi = A\u00e7\u0131Cos(Biyel, dd, Kol)
\nD\u00f6rt\u00c7ubuk = fi – s * si
\n<\/b>End Function<\/p>\n

2.4-b) Krank-Biyel Mekanizmas\u0131:<\/span><\/p>\n

Bu fonksiyon program\u0131 uzuv boyutlar\u0131, giri\u015f kolu a\u00e7\u0131s\u0131 ve ba\u011flant\u0131 \u015fekli verilen bir krank-biyel mekanizmas\u0131n\u0131n \u00e7\u0131k\u0131\u015f kolu \u00f6teleme uzakl\u0131\u011f\u0131n\u0131 bulmak i\u00e7in yaz\u0131lm\u0131\u015ft\u0131r.<\/p>\n

Kullan\u0131c\u0131n\u0131n s say\u0131s\u0131n\u0131 +1 veya \u22121 tan\u0131mlayarak pistonun \u015fekilde g\u00f6r\u00fcld\u00fc\u011f\u00fc gibi krank eksenine g\u00f6re pozitif x ekseni y\u00f6n\u00fcnde (s = +1)veya ters y\u00f6nde (s = \u22121) oldu\u011funu belirlemesi gerekir. Program \u00f6nceden yaz\u0131lm\u0131\u015f olan Asin() fonksiyonunu kullanmaktad\u0131r.\"\"<\/p>\n

Function KrankBiyel(Krank, Biyel, Eksantrik, s, th)
\nDim fi As Double
\nfi = Asin((Krank* Sin(th) – Sabit) \/ Eksantrik)
\nIf s = 1 Then fi = 4 * Atn(1) – fi
\nKrankBiyel = Krank * Cos(th) – Biyel * Cos(fi)
\n<\/b>End Function<\/p>\n

2.4-c) Kol-K\u0131zak Mekanizmas\u0131:<\/span><\/p>\n

\"\"<\/p>\n

Kol-k\u0131zak mekanizmas\u0131n\u0131n santrik veya eksantrik olmas\u0131 a3 de\u011ferinin s\u0131f\u0131r olup olmamas\u0131 ile ilgilidir. Eksantrikliklik oldu\u011funda \u00e7\u0131k\u0131\u015f kolu iki de\u011fi\u015fik \u015fekilde monte edilebilece\u011finden s de\u011feri +1 (\u015fekilde g\u00f6sterilen durum) veya -1 olarak girilmelidir (Santrik i\u00e7in s her hangi bir de\u011fer olabilir).<\/p>\n

Function KolK\u0131zak(Krank, Sabit, Eksantrik, s,th)
\nDim Xd,Yd as double
\nXd = Krank * Cos(th) – Sabit
\nYd = Krank * Sin(th)
\nSi= A\u00e7\u0131(Xd,Yd)
\nIf Eksantrik=0 then
\nFi=Si
\nElse
\nDd=Boy(Xd,Yd)
\nFi=si-s*Asin(Eksantrik\/Dd)
\nEnd If
\nKolK\u0131zak = Fi
\n<\/b>End Function<\/p>\n

Yukar\u0131da yaz\u0131lm\u0131\u015f olan fonksiyon programlar\u0131 Excel paket program\u0131 i\u00e7inde, kullan\u0131c\u0131 taraf\u0131ndan tan\u0131mlanm\u0131\u015f fonksiyon olarak t\u0131pki di\u011fer Excel fonksiyonlar\u0131 gibi kullan\u0131labilir. Ayr\u0131ca, yukar\u0131da yaz\u0131lm\u0131\u015f olan programlar\u0131n Excelde birle\u015ftirilerek de kullan\u0131lmas\u0131 m\u00fcmk\u00fcnd\u00fcr. Ayr\u0131ca bu programlar\u0131n VisualBasic i\u00e7inde de kullan\u0131lmas\u0131 m\u00fcmk\u00fcnd\u00fcr. VisualBasic platformunun grafik \u00f6zellikleri ile birlikte kullan\u0131c\u0131 animasyon yapabilecek programlarda haz\u0131rlayabilir.<\/p>\n

\u00d6rnek 1:<\/strong><\/p>\n

Yukar\u0131da haz\u0131rlanm\u0131\u015f olan programlar ile \u00e7ok say\u0131da de\u011fi\u015fik mekanizman\u0131n analizi kolayca yap\u0131labilecektir. A\u015fa\u011f\u0131da g\u00f6sterilen mekanizman\u0131n bir d\u00f6rt-\u00e7ubuk ve bir krank-biyel mekanizmas\u0131n\u0131n seri olarak ba\u011flanmas\u0131 ile elde edildi\u011fi g\u00f6r\u00fcl\u00fcr ise, ilk olarak d\u00f6rt-\u00e7ubuk mekanizmas\u0131 \u00e7\u0131k\u0131\u015f a\u00e7\u0131s\u0131 (th4) D\u00f6rt\u00c7ubuk() fonksiyonu kullanarak belirlenir. Th4-25\u00b0\u00a0giri\u015f a\u00e7\u0131s\u0131 olarak KrankBiyel() fonksiyonu kullan\u0131larak strok \u00e7\u00f6z\u00fclebilir.<\/p>\n

\"\"<\/p>\n

\u0130lk olarak sabit uzuv boyutlar\u0131n\u0131 (sabit a\u00e7\u0131lar da dahil) Excel \u00fczerinde yerle\u015ftirelim.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n
<\/td>\nA<\/td>\nB<\/td>\nC<\/td>\nD<\/td>\nE<\/td>\n<\/tr>\n
1<\/td>\nD\u00f6rt-\u00c7ubuk<\/td>\n<\/td>\nKrank-Biyel<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
2<\/td>\nKrank<\/td>\n320<\/td>\nKrank 2<\/td>\n1050<\/td>\n<\/td>\n<\/tr>\n
3<\/td>\nBiyel<\/td>\n1180<\/td>\nBiyel2<\/td>\n1140<\/td>\n<\/td>\n<\/tr>\n
4<\/td>\n\u00c7\u0131k\u0131\u015fKolu<\/td>\n528<\/td>\nEksantrik<\/td>\n400<\/td>\n<\/td>\n<\/tr>\n
5<\/td>\nSabitU_y<\/td>\n1250<\/td>\nA\u00c7I 4<\/td>\n0.43633231<\/td>\n25<\/td>\n<\/tr>\n
6<\/td>\nSabitU_d<\/td>\n546<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
7<\/td>\nSabitUzuv<\/td>\n1364.044<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
8<\/td>\nSabitU A\u00e7\u0131<\/td>\n2.72977<\/td>\n156.40429<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
9<\/td>\nFaz<\/td>\n0.411823<\/td>\n23.595705<\/td>\n<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

B8 h\u00fccresinde g\u00f6r\u00fclen a\u00e7\u0131 radyan olarak B0<\/sub>A0<\/sub> vekt\u00f6r\u00fcn\u00fcn yatay ile yapt\u0131\u011f\u0131 a\u00e7\u0131d\u0131r (bu de\u011feri ATAN2(-B5;B6) \u015feklinde ATAN2() Excel fonksiyonu kullanarak elde edebiliriz. Faz a\u00e7\u0131s\u0131 ise bu a\u00e7\u0131y\u0131 tamlar a\u00e7\u0131 olup B9 h\u00fccresine =pi()-B8 yazarak elde edilebilir. D5 h\u00fccresinde ise 4 uzvunda \u2220BB0<\/sub>C a\u00e7\u0131s\u0131 radyan olarak bulunmaktad\u0131r. B ve D kolonunda g\u00f6r\u00fclen h\u00fccrelere A kolonunda g\u00f6r\u00fclmekte olan isimler etiket ad\u0131 olarak verilmi\u015ftir (\u00f6rne\u011fin B2 h\u00fccresi ismi Krank, D2 h\u00fccresi ismi ise Krank2 dir).<\/p>\n

Sabit de\u011ferler girildikten sonra, A16-A40 h\u00fccrelerine 15\u00ba\u00a0aral\u0131klarla th krank a\u00e7\u0131s\u0131 0\u00ba ila 360\u00ba aras\u0131nda girilmi\u015ftir.<\/p>\n

\"\"<\/p>\n

A kolonunda verilmekte olan a\u00e7\u0131lar yatay ile krank\u0131n yapt\u0131\u011f\u0131 a\u00e7\u0131lard\u0131r. B kolonunda ise bu krank a\u00e7\u0131s\u0131 radyana \u00e7evrildikten sonra sabit uzva g\u00f6re a\u00e7\u0131 olmas\u0131 i\u00e7in B9 h\u00fccresinde bulunan Faz a\u00e7\u0131s\u0131 de\u011feri \u00e7\u0131kar\u0131lm\u0131\u015ft\u0131r (isim vermedi iseniz B9 h\u00fccresini sabitlemek i\u00e7in $B$9 olarak girin). \u015eimdi C16 h\u00fccresine gelinir ve
\n“=D\u00f6rt\u00c7ubuk(Krank;Biyel;\u00c7\u0131k\u0131\u015fKolu;SabitUzuv;1;B16)<\/b>“
\nyaz\u0131l\u0131r ise, th4 \u00e7\u0131k\u0131\u015f kolu a\u00e7\u0131s\u0131 hesap edilir. Form\u00fcl\u00fcn C16-C40 h\u00fccrelerine kopya edilmesi ile t\u00fcm th a\u00e7\u0131lar\u0131na g\u00f6re Th4 a\u00e7\u0131s\u0131 elde edilmi\u015f olur. D kolonunda bu a\u00e7\u0131 dereceye \u00e7evrilmi\u015ftir. \u015eimdi krank-biyel mekanizmas\u0131 i\u00e7in \u00e7\u00f6z\u00fcm yap\u0131labilir. Bunun i\u00e7in E16 h\u00fccresine
\n“=KrankBiyel(Krank2;Biyel2;Eksantrik;1;(C15-A\u00e7\u01314)) + SabitU_y<\/b>“
\nkomutunu yazal\u0131m. Bununla piston \u00f6telemesini veren s parametresi de\u011feri bulunacakt\u0131r. Denklemi E16-E40 h\u00fccrelerine kopyalar isek her 15\u00b0\u00a0krank a\u00e7\u0131s\u0131 i\u00e7in \u00f6teleme uzunlu\u011funu belirlemi\u015f oluruz. Sonucu grafik olarak Excel Chart komutu kullanarak g\u00f6stermemiz m\u00fcmk\u00fcnd\u00fcr.<\/p>\n

\"\"<\/p>\n

\"\" \u015eekil a<\/span><\/p>\n

\"\" \u015eekil b<\/span><\/p>\n

Mekanizmay\u0131 olu\u015fturan krank-biyel mekanizmas\u0131n\u0131 yukar\u0131daki \u015fekillerden biri gibi de\u011fi\u015ftirdi\u011fimizde, dikkat edilir ise d\u00f6rt \u00e7ubuk mekanizmas\u0131 ve dolay\u0131s\u0131 ile th4 ayn\u0131 kalacakt\u0131r. Krank-Biyel mekanizmas\u0131nda ise, giri\u015f kolu a\u00e7\u0131s\u0131n\u0131 kayar \u00e7ift eksenine paralel bir do\u011fruya g\u00f6re \u00f6l\u00e7memiz gerekece\u011finden giri\u015f kolu a\u00e7\u0131s\u0131 (th4 – \u2220BB0C + pi\/2) olmal\u0131d\u0131r. Bu durumda \u015eekil a da E16 h\u00fccresine s de\u011ferini bulmak i\u00e7in E16 h\u00fccresine
\n“=KrankBiyel (Krank2 ;Biyel2 ;Eksantrik ;1;(C15-A\u00e7\u01314+PI()\/2))<\/b>“
\nkomutunu tu\u015flay\u0131p ve bu komutu E16-E40 h\u00fccrelerine kopyalamam\u0131z yeterlidir. \u015eekil b de g\u00f6r\u00fclen mekanizma i\u00e7in ise E16 h\u00fccresine
\n“= -KrankBiyel(Krank2 ;Biyel2 ;Eksantrik ;-1;(C15-A\u00e7\u01314+PI()\/2))\u00a0<\/b>“
\nkomutunu tu\u015flamam\u0131z gerekecektir. Dikkat edilir ise her iki durumda pozitif x ekseni dikey a\u015fa\u011f\u0131ya do\u011frudur. Ikinci durumda mekanizman\u0131n \u015feklinden dolay\u0131 ba\u011flant\u0131 durumunu g\u00f6steren s de\u011feri \u22121 girilmi\u015f ve fonksiyonun eksi de\u011feri piston \u00f6telemesi olarak al\u0131nm\u0131\u015ft\u0131r. Sonu\u00e7ta piston \u00f6telemesi \u015eekil a i\u00e7in \u015fekil c de ve \u015fekil b i\u00e7in de \u015fekil d de g\u00f6sterildi\u011fi gibi elde edilmi\u015ftir. (\u015eekil d de g\u00f6r\u00fclen iki h\u00f6rg\u00fc\u00e7l\u00fc hareketin nedenini de\u011ferlendirin).<\/p>\n

\"\" \u00a0\u015eekil c<\/span><\/p>\n

\"\"\u015eekil d<\/span><\/p>\n

Excel K\u00fct\u00fc\u011f\u00fc i\u00e7in t\u0131klay\u0131n\u0131z: sixlink.xls<\/a>.<\/p>\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n

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

Ek-2 Basit Mekanizmalar \u0130\u00e7in Bilgisayar Programlar\u0131 Uygulamalarda basit d\u00f6rt uzuvlu mekanizmalar veya bu basit d\u00f6rt uzuvlu mekanizmalardan seri ba\u011flanarak elde edilen mekanizmalar \u00e7ok s\u0131k rastlanmaktad\u0131r. Bu mekanizmalar i\u00e7in haz\u0131rlanacak ufak fonksiyon programlar\u0131 gerek Excel paket program\u0131 i\u00e7inde Macro programlar olarak …<\/p>\n

Ek 2<\/span> Devam\u0131n\u0131 Oku »<\/a><\/p>\n","protected":false},"author":7747,"featured_media":0,"parent":47,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"full-width-page.php","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-1873","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1873","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=1873"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1873\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/47"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/media?parent=1873"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}