{"id":504,"date":"2021-09-04T17:41:51","date_gmt":"2021-09-04T17:41:51","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=504"},"modified":"2021-10-05T21:37:25","modified_gmt":"2021-10-05T21:37:25","slug":"top","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch8\/8-5\/top\/","title":{"rendered":"top"},"content":{"rendered":"<div id=\"pl-gb504-69d840ff4a7e2\"  class=\"panel-layout\" ><div id=\"pg-gb504-69d840ff4a7e2-0\"  class=\"panel-grid panel-no-style\" ><div id=\"pgc-gb504-69d840ff4a7e2-0-0\"  class=\"panel-grid-cell\" ><div id=\"panel-gb504-69d840ff4a7e2-0-0-0\" 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><strong data-rich-text-format-boundary=\"true\">Toparlakl\u0131, \u00f6teleme yapan izleyicili radyal kam profilinin elde edilmesi:<\/strong><\/h1>\n<p>Hareket e\u011frisi s(\u03b8) ve t\u00fcrevleri kam\u0131n tam bir devri i\u00e7in bilinmektedir. \u015eekilde g\u00f6sterildi\u011fi gibi, izleyici \u00f6teleme ekseni kam merkezinden, c kadar ka\u00e7\u0131kt\u0131r (c = 0 oldugunda santrik \u00f6teleme yapan izleyici olacakt\u0131r).<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-506\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img9-1.gif\" alt=\"\" width=\"812\" height=\"630\" \/><\/p>\n<p>Her hangi bir \u03b8 a\u00e7\u0131s\u0131 ve buna kar\u015f\u0131 gelen s y\u00fckselme uzunlu\u011funda toparla\u011f\u0131n kama g\u00f6re ba\u011f\u0131l konumu yukar\u0131da g\u00f6r\u00fclmektedir.<\/p>\n<p style=\"text-align: left\" align=\"center\"><span style=\"color: #000000\"><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:700px\" 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_69d840ff4cf56\"><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/kammotion.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\/kammotion.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\/kammotion_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\/kammotion_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\/kammotion_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\/kammotion_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\/kammotion_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\/kammotion_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\/kammotion_4.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"471\" height=\"408\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/kammotion_4.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><\/div><script id=\"su_image_carousel_69d840ff4cf56_script\">if(window.SUImageCarousel){setTimeout(function() {window.SUImageCarousel.initGallery(document.getElementById(\"su_image_carousel_69d840ff4cf56\"))}, 0);}var su_image_carousel_69d840ff4cf56_script=document.getElementById(\"su_image_carousel_69d840ff4cf56_script\");if(su_image_carousel_69d840ff4cf56_script){su_image_carousel_69d840ff4cf56_script.parentNode.removeChild(su_image_carousel_69d840ff4cf56_script);}<\/script><\/span><\/p>\n<p>Yukar\u0131da kam profilinin geometrik olarak elde edilmesi (eksantrikli\u011fin olmad\u0131\u011f\u0131 durum i\u00e7in) a\u00e7\u0131klanm\u0131\u015ft\u0131r. \u015eimdi zarf teorisini \u00f6teleme yapan toparlakl\u0131 radyal kam i\u00e7in uygulayal\u0131m ve bu y\u00f6ntemle kam profilini daha hassas bir \u015fekilde analitik olarak elde edelim.<\/p>\n<p>Toparlak dairesinin farkl\u0131 konumlari e\u011fri demetini olu\u015fturmaktad\u0131r ve bu e\u011fri demetinin denklemi:<\/p>\n<p style=\"padding-left: 40px\">\u00a0(x \u2212 x<sub>p<\/sub>)<sup>2<\/sup> + (y \u2212 y<sub>p<\/sub>)<sup>2<\/sup> \u2212 r<sub>r<\/sub> = 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (1)<\/p>\n<p>d\u0131r. Bu denklemde r<sub>r<\/sub> toparlak yar\u0131\u00e7ap\u0131 ve P(x<sub>p<\/sub>, y<sub>p<\/sub>) toparla\u011f\u0131n merkez koordinatlar\u0131d\u0131r. x<sub>p<\/sub>\u00a0ve y<sub>p<\/sub>\u00a0kam d\u00f6nme a\u00e7\u0131s\u0131n\u0131n bir fonksiyonudur ve:<\/p>\n<p style=\"padding-left: 40px\">x<sub>p<\/sub> = c cos\u03b8 + (k + s) sin\u03b8<\/p>\n<p style=\"padding-left: 40px\">y<sub>p<\/sub> = \u2212c sin\u03b8 + (k + s) cos\u03b8\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(2)<\/p>\n<p>d\u0131r. Burada<\/p>\n<p style=\"padding-left: 40px\">k = <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle \\sqrt{{{{{\\left( {{{\\text{r}}_{\\text{t}}}+{{\\text{r}}_{\\text{r}}}} \\right)}}^{2}}-{{\\text{c}}^{2}}}} <\/span><\/p>\n<p>dir. Kam daire merkezi istenir ise karma\u015f\u0131k say\u0131lar kullan\u0131larak yaz\u0131labilir. Bu durumda<\/p>\n<p style=\"padding-left: 40px\">z = x<sub>p<\/sub> + iy<sub>p<\/sub> = ce<sup>\u2212i\u03b8<\/sup> + i(k + s)e<sup>\u2212i\u03b8<\/sup><\/p>\n<p>olacakt\u0131r ki reel ve sanal k\u0131s\u0131mlar\u0131 toparlak dairesinin x ve y koordinatlar\u0131n\u0131 verir.<\/p>\n<p>Dikkat edilir ise toparlak dairesinin olusturdu\u011fu e\u011fri demeti fonksiyonu sabit r<sub>t<\/sub>, r<sub>r<\/sub>, c parametreleri d\u0131\u015f\u0131nda s ve \u03b8 ya ba\u011fl\u0131d\u0131r. s y\u00fckselme mesafesinin de hareket diyagram\u0131ndan \u03b8 n\u0131n bir fonksiyonu oldu\u011fu d\u00fc\u015f\u00fcn\u00fcl\u00fcr ise, e\u011fri demeti fonksiyonu f(x, y, \u03b8) \u015feklinde olup her\u00a0\u03b8 a\u00e7\u0131s\u0131na g\u00f6re belirli bir merkezi ve r<sub>r<\/sub>\u00a0yar\u0131\u00e7ap\u0131 olan bir daire denklemi elde edilecektir. \u015eimdi:<\/p>\n<p style=\"padding-left: 40px\">f(x, y, \u03b8) = (x \u2212 x<sub>p<\/sub>)<sup>2<\/sup> + (y \u2212 y<sub>p<\/sub>)<sup>2<\/sup> \u2212 r<sub>r<\/sub> = 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (1)<\/p>\n<p>denkleminin \u03b8 parametresine g\u00f6re k\u0131sm\u0131 t\u00fcrevi al\u0131n\u0131r ise:<\/p>\n<p style=\"padding-left: 40px\">f<sub>\u03b8<\/sub>(x, y, \u03b8) = \u22122(x \u2212 x<sub>p<\/sub>)\u2202x<sub>p<\/sub>\/\u2202\u03b8 \u2212 2(y \u2212 y<sub>p<\/sub>)\u2202y<sub>p<\/sub>\/\u2202\u03b8 = 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (3)<\/p>\n<p>olacakt\u0131r. Bu denklemde:<\/p>\n<p style=\"padding-left: 40px\">\u2202x<sub>p<\/sub>\/\u2202\u03b8 = (\u2202s\/\u2202\u03b8 \u2212 c)sin\u03b8 + (k + s)cos\u03b8<\/p>\n<p style=\"padding-left: 40px\">\u2202y<sub>p<\/sub>\/\u2202\u03b8 = (\u2202s\/\u2202\u03b8 \u2212 c)cos\u03b8 \u2212 (k + s)sin\u03b8\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (4)<\/p>\n<p>d\u0131r. (1) ve (3) denklemlerinden \u03b8 parametresinin yok edilmesi genelde s hareket diyagram\u0131ndan dolay\u0131 m\u00fcmk\u00fcn de\u011fildir (d\u0131yagram karma\u015f\u0131k birka\u00e7 denklemden olu\u015fmaktad\u0131r). Bu nedenle e\u011fri demetinin zarf\u0131 olan kam profil e\u011frisi \u00e7ok basit problemler haricinde parametrik olarak elde edilir. 1 ve 3 denklemlerinden kam profili koordinatlari (x, y) her \u03b8 a\u00e7\u0131s\u0131 i\u00e7in:<\/p>\n<p style=\"padding-left: 40px\">x = x<sub>p<\/sub> \u00b1 <span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle \\frac{{{{\\text{r}}_{\\text{r}}}}}{{\\sqrt{{1+{{{\\left( {\\frac{{\\partial {{\\text{x}}_{\\text{p}}}\\text{\/}\\partial \\text{\u03b8}}}{{\\partial {{\\text{y}}_{\\text{p}}}\\text{\/}\\partial \\text{\u03b8}}}} \\right)}}^{2}}}}}} <\/span><\/p>\n<p style=\"padding-left: 40px\">y = y<sub>p<\/sub> \u2212 (x \u2212 x<sub>p<\/sub>)<span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {\\frac{{\\partial {{\\text{x}}_{\\text{p}}}\\text{\/}\\partial \\text{q}}}{{\\partial {{\\text{y}}_{\\text{p}}}\\text{\/}\\partial \\text{q}}}} <\/span>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (5)<\/p>\n<p>x<sub>p<\/sub>, y<sub>p<\/sub>\u00a0ve \u03b8 ya g\u00f6re t\u00fcrevleri denklem 2 ve 4 de verildi\u011fi gibidir. Dikkat edilir ise, her \u03b8 de\u011feri i\u00e7in iki nokta elde edilece\u011finden kam e\u011fri demetinin zarf\u0131 olan kam profili iki k\u0131s\u0131mdan olu\u015facakt\u0131r. Bunlardan birisi d\u0131\u015ftan temas eden kam profilini digeri ise i\u00e7ten temas eden bir kam profilini verir.<\/p>\n<p><strong>\u00d6rnek 7:<\/strong><\/p>\n<p>120\u00b0 kam d\u00f6nme a\u00e7\u0131s\u0131nda 50 mm basit harmonik hareket ile y\u00fckselen, 60\u00b0 bekleme yapan, 120\u00b0\u00a0kam d\u00f6nme hareketi ile geri d\u00f6nen, toparlak yar\u0131\u00e7ap\u0131 20, temel daire yar\u0131\u00e7ap\u0131 50 mm ve 20 mm eksantrikli\u011fi olan kam sisteminde kam profilini bulal\u0131m.<\/p>\n<p>Bu \u00f6rnek MathCad paket program\u0131 kullan\u0131larak \u00e7\u00f6z\u00fclm\u00fc\u015ft\u00fcr.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignleft wp-image-514\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/eqn33.gif\" alt=\"\" width=\"726\" height=\"155\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignleft wp-image-515\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/eqn34.gif\" alt=\"\" width=\"765\" height=\"167\" \/><\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-527\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img10-1.gif\" alt=\"\" width=\"743\" height=\"267\" \/><\/p>\n<p style=\"text-align: center\">Hareket diyagram\u0131 (dikkat edilir ise, d\u00f6rt ayr\u0131 e\u011friden olu\u015fmaktad\u0131r)<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-528\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img11-1.gif\" alt=\"\" width=\"776\" height=\"280\" \/><\/p>\n<p style=\"text-align: center\">H\u0131z diyagram\u0131 (\u03c9 = 1 rad\/s oldu\u011fundan v = ds\/d\u03b8 d\u0131r)<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-516 alignnone\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/eqn35.gif\" alt=\"\" width=\"767\" height=\"224\" \/><\/p>\n<p style=\"padding-left: 40px\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-517\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/eqn36.gif\" alt=\"\" width=\"705\" height=\"131\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-518\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/eqn37.gif\" alt=\"\" width=\"463\" height=\"291\" \/><\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-529\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img12-1.gif\" alt=\"\" width=\"750\" height=\"776\" \/><\/p>\n<p>Dikkat edilir ise, denklemde bulunan + ve &#8211; de\u011ferlere g\u00f6re her \u03b8\u00a0a\u00e7\u0131s\u0131na g\u00f6re iki kam e\u011frisi koordinat\u0131 elde edilmektedir. x<sub>1<\/sub>, y<sub>1<\/sub>\u00a0ve x<sub>2<\/sub>, y<sub>2<\/sub>\u00a0koordinatlar\u0131 belirli bir aral\u0131kta d\u0131\u015f e\u011friyi, belirli bir aral\u0131kta ise i\u00e7 e\u011friyi belirler.<\/p>\n<p>Sadece d\u0131\u015ftan temas eden kam\u0131n \u00e7izilmesi i\u00e7in ise, bu koordinatlardan merkeze daha yak\u0131n olan koordinat se\u00e7ilmelidir. Bunun i\u00e7in her \u03b8\u00a0a\u00e7\u0131s\u0131na kar\u015f\u0131 gelen iki koordinat\u0131n merkezden uzakl\u0131\u011f\u0131 bulunarak birbiri ile kar\u015f\u0131la\u015ft\u0131r\u0131l\u0131r, merkeze daha yak\u0131n koordinat arad\u0131\u011f\u0131m\u0131z kam e\u011frisi koordinat\u0131d\u0131r:<\/p>\n<p style=\"padding-left: 40px\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-519\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/eqn38.gif\" alt=\"\" width=\"695\" height=\"210\" \/><\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-530\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img13-1.gif\" alt=\"\" width=\"680\" height=\"631\" \/><\/p>\n<p>Dikkat edilir ise kam profili kartezyen koordinatlara g\u00f6re (x, y) polar koordinatlara g\u00f6re ise (\u03c1, \u03be) olarak elde edilmi\u015ftir. Polar koordinatlar kullan\u0131larak \u00e7izim \u015fekilde g\u00f6r\u00fclmektedir:<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-531\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img14-1.gif\" alt=\"\" width=\"628\" height=\"631\" \/><\/p>\n<p>Bu kam mekanizmas\u0131nda ba\u011flama a\u00e7\u0131s\u0131n\u0131n bir devirde de\u011fi\u015fimini belirlemek i\u00e7in yukar\u0131daki \u015fekilden yararlanarak her bir konumda ba\u011flama a\u00e7\u0131s\u0131n\u0131 bulabiliriz.<\/p>\n<p>Mathcad kullan\u0131larak bu a\u00e7\u0131 hesapland\u0131\u011f\u0131nda:<\/p>\n<p style=\"padding-left: 40px\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-521\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/eqn40.gif\" alt=\"\" width=\"327\" height=\"102\" \/><\/p>\n<p>Ba\u011flama a\u00e7\u0131s\u0131n\u0131n 0\u00b0 &lt; \u03bc &lt; 90\u00b0 olmas\u0131 i\u00e7in:<\/p>\n<p style=\"padding-left: 40px\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-522\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/eqn41.gif\" alt=\"\" width=\"273\" height=\"113\" \/><\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-532\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img15-1.gif\" alt=\"\" width=\"704\" height=\"222\" \/><\/p>\n<p>Ba\u011flama a\u00e7\u0131s\u0131n\u0131n bir devirde de\u011fi\u015fimi \u015fekilde g\u00f6sterilmi\u015ftir. G\u00f6r\u00fcld\u00fc\u011f\u00fc gibi, y\u00fckseli\u015f s\u0131ras\u0131nda \u03bc<sub>min<\/sub> = 75\u00b0, geri d\u00f6n\u00fc\u015f s\u0131ras\u0131nda ise \u03bc<sub>min<\/sub> = 58.23\u00b0 olmaktad\u0131r.<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-533\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img16-1.gif\" alt=\"\" width=\"402\" height=\"387\" \/><\/p>\n<p>Bu kam y\u00fczeyini r<sub>k<\/sub> yar\u0131\u00e7apl\u0131 bir kesici ile numerik kontrollu bir freze tezgah\u0131nda i\u015fleyece\u011fimizi d\u00fc\u015f\u00fcnelim. G\u00f6sterilen konumda kesicinin kam profilinde (x, y) koordinatlar\u0131n\u0131 sa\u011flamas\u0131 i\u00e7in freze mili O\u2032 merkez koordinatlar\u0131 (x<sub>k<\/sub>, y<sub>k<\/sub>), bilinen toparlak merkezi ve kam profil e\u011frisine g\u00f6re:<\/p>\n<p style=\"padding-left: 40px\">x<sub>k<\/sub>\u00a0= x + r<sub>k<\/sub> cos\u03c8<\/p>\n<p style=\"padding-left: 40px\">y<sub>k<\/sub> =\u00a0y + r<sub>k<\/sub> sin\u03c8<\/p>\n<p>olacakt\u0131r. Bu denklemde:<\/p>\n<p style=\"padding-left: 40px\">\u03c8 = tan<sup>-1<\/sup>[(y<sub>p<\/sub> \u2212 y)\/(x<sub>p<\/sub> \u2212 x)]\n<p>d\u0131r. \u00d6rne\u011fimizde, kam profilini 60 mm \u00e7apl\u0131 bir kesici ile i\u015fleyece\u011fimiz var say\u0131l\u0131r ise, MathCad kullan\u0131larak kesici eksen koordinatlar\u0131:<\/p>\n<p style=\"padding-left: 40px\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-525\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/eqn45.gif\" alt=\"\" width=\"477\" height=\"108\" \/><\/p>\n<p>Polar koordinatlara g\u00f6re:<\/p>\n<p style=\"padding-left: 40px\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-526\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/eqn46.gif\" alt=\"\" width=\"533\" height=\"74\" \/><\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-534\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img17.gif\" alt=\"\" width=\"722\" height=\"717\" \/><\/p>\n<p>Elde edilen kesici eksen y\u00f6r\u00fcngesi, kam e\u011frisi ile birlikte (kesik \u00e7izgi) \u015fekilde g\u00f6sterilmi\u015ftir. Dikkat edilir ise, bu y\u00f6r\u00fcnge kam e\u011frisine paralel bir e\u011fri de\u011fildir (bekleme konumlar\u0131 hari\u00e7).<\/p>\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n<p><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch8\/8-5\/\" 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\/ch8\/\" data-type=\"page\"><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><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>Toparlakl\u0131, \u00f6teleme yapan izleyicili radyal kam profilinin elde edilmesi: Hareket e\u011frisi s(\u03b8) ve t\u00fcrevleri kam\u0131n tam bir devri i\u00e7in bilinmektedir. \u015eekilde g\u00f6sterildi\u011fi gibi, izleyici \u00f6teleme ekseni kam merkezinden, c kadar ka\u00e7\u0131kt\u0131r (c = 0 oldugunda santrik \u00f6teleme yapan izleyici olacakt\u0131r). &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch8\/8-5\/top\/\"> <span class=\"screen-reader-text\">top<\/span> Devam\u0131n\u0131 Oku &raquo;<\/a><\/p>\n","protected":false},"author":7747,"featured_media":0,"parent":479,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-504","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/504","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=504"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/504\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/479"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/media?parent=504"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}