{"id":2362,"date":"2022-03-18T21:37:07","date_gmt":"2022-03-18T21:37:07","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=2362"},"modified":"2022-09-02T21:16:24","modified_gmt":"2022-09-02T21:16:24","slug":"translating_roller","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch8\/8-5\/translating_roller\/","title":{"rendered":"translating_roller"},"content":{"rendered":"<div id=\"pl-gb2362-6a1b98d128ea6\"  class=\"panel-layout\" ><div id=\"pg-gb2362-6a1b98d128ea6-0\"  class=\"panel-grid panel-no-style\" ><div id=\"pgc-gb2362-6a1b98d128ea6-0-0\"  class=\"panel-grid-cell\" ><div id=\"panel-gb2362-6a1b98d128ea6-0-0-0\" 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\">Cam Profile for a Radial Cam With Translating Roller Follower<\/strong><\/h1>\n<p>Now let us apply the envelope theory to a basic cam-follower system: a radial cam with a translating roller follower. We shall assume that the follower sliding axis is offset by an amount c from the centre of rotation of the cam (when c = 0, it is an inline follower).<\/p>\n<p>As in the case of geometrical construction, let us use kinematic inversion and fix the cam and release the fixed link. For the same relative motion, if the cam rotates by a counter clockwise angle q, the fixed link will rotate by a clockwise angle\u00a0<span style=\"font-family: Symbol\">q<\/span>\u00a0and the follower\u00a0 will be displaced by an amount s(<span style=\"font-family: Symbol\">q<\/span>) with respect to the fixed link as shown. The roller takes different positions for each\u00a0<span style=\"font-family: Symbol\">q<\/span>\u00a0angle, hence it is the roller profile that is the generating curve and its equation for any cam angle\u00a0<span style=\"font-family: Symbol\">q<\/span>\u00a0is:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\" width=\"90%\">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<\/td>\n<td style=\"justify-content: center;text-align: right\" width=\"10%\">(1)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>where r<sub>r<\/sub>\u00a0is the radius of the roller. P(x<sub>p<\/sub>, y<sub>p<\/sub>) are the coordinates of a point on the pitch curve (the centre of the roller), which is a function of the cam angle\u00a0<span style=\"font-family: Symbol\">q<\/span>. x<sub>p<\/sub>, y<sub>p<\/sub>\u00a0is given by:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td width=\"90%\">\n<p style=\"text-align: center\">x<sub>p<\/sub> = c cos\u03b8 + (k + s) sin\u03b8<\/p>\n<p style=\"text-align: center\">y<sub>p<\/sub> = \u2212c sin\u03b8 + (k + s) cos\u03b8<\/p>\n<\/td>\n<td style=\"vertical-align: middle;text-align: right\" width=\"10%\">(2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>where\u00a0k = <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>.\u00a0(One can also write the coordinates of point P in complex numbers as: z = x<sub>p<\/sub> + iy<sub>p<\/sub> = ce<sup>\u2212i\u03b8<\/sup> + i(k + s)e<sup>\u2212i\u03b8<\/sup>. The real and imaginary parts of this complex number will yield x<sub>p<\/sub>\u00a0and y<sub>p<\/sub>\u00a0coordinates as given in equation 2).<\/p>\n<p align=\"left\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2366\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/03\/Top_clip_image010.gif\" alt=\"\" width=\"900\" height=\"694\" \/><\/p>\n<p align=\"left\">Note that the equation for the roller profile is a generating curve which depends on constant parameters r<sub>b<\/sub>, r<sub>r<\/sub>, and c and on the follower displacement s and cam rotation angle\u00a0<span style=\"font-family: Symbol\">q<\/span>. Since s is also a function of\u00a0<span style=\"font-family: Symbol\">q\u00a0<\/span>defined by the motion curve, the generating curve (roller profile) is in the form f(x, y, <span style=\"font-family: Symbol\">q<\/span>). Where\u00a0<span style=\"font-family: Symbol\">q<\/span>\u00a0is a parameter (for each value of\u00a0<span style=\"font-family: Symbol\">q<\/span>\u00a0we obtain a member within the family of curves). Hence the generating curve is:<\/p>\n<p style=\"text-align: center\" align=\"left\"><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:550px\" 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_6a1b98d12af1f\"><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/09\/kammotione_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\/2022\/09\/kammotione_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\/2022\/09\/kammotione_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\/2022\/09\/kammotione_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\/2022\/09\/kammotione_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\/2022\/09\/kammotione_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\/2022\/09\/kammotione_4.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\/2022\/09\/kammotione_4.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\/2022\/09\/kammotione_5.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\/2022\/09\/kammotione_5.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\/2022\/09\/kammotione_6.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\/2022\/09\/kammotione_6.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\/2022\/09\/kammotione_7.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\/2022\/09\/kammotione_7.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><\/div><script id=\"su_image_carousel_6a1b98d12af1f_script\">if(window.SUImageCarousel){setTimeout(function() {window.SUImageCarousel.initGallery(document.getElementById(\"su_image_carousel_6a1b98d12af1f\"))}, 0);}var su_image_carousel_6a1b98d12af1f_script=document.getElementById(\"su_image_carousel_6a1b98d12af1f_script\");if(su_image_carousel_6a1b98d12af1f_script){su_image_carousel_6a1b98d12af1f_script.parentNode.removeChild(su_image_carousel_6a1b98d12af1f_script);}<\/script><\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\" width=\"90%\">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<\/td>\n<td style=\"justify-content: center;text-align: right\" width=\"10%\">(3)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p align=\"left\">If we take the partial derivative of\u00a0 f(x, y, <span style=\"font-family: Symbol\">q<\/span>) with respect to\u00a0<span style=\"font-family: Symbol\">q\u00a0<\/span>:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\" width=\"90%\">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<\/td>\n<td style=\"justify-content: center;text-align: right\" width=\"10%\">(4)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p align=\"left\">In this equation:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td width=\"90%\">\n<p style=\"text-align: center\">\u2202x<sub>p<\/sub>\/\u2202\u03b8 = (\u2202s\/\u2202\u03b8 \u2212 c)sin\u03b8 + (k + s)cos\u03b8<\/p>\n<p style=\"text-align: center\">\u2202y<sub>p<\/sub>\/\u2202\u03b8 = (\u2202s\/\u2202\u03b8 \u2212 c)cos\u03b8 \u2212 (k + s)sin\u03b8<\/p>\n<\/td>\n<td style=\"vertical-align: middle;text-align: right\" width=\"10%\">(5)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p align=\"left\">Unlike the previous examples where we were able to eliminate the parameter\u00a0<span style=\"font-family: Symbol\">q<\/span>\u00a0from the two equations (Equations 3 and 4), due to the motion curve characteristics, we cannot eliminate the parameter\u00a0<span style=\"font-family: Symbol\">q\u00a0<\/span>from the above two equations. Instead, we obtain the envelope equation in parametric form by eliminating y from the two equations (solve for (y \u2212 y<sub>p<\/sub>) from equation 4 in terms of (x \u2212 x<sub>p<\/sub>) and substitute into equation 3) we obtain:<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\" width=\"90%\">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><\/td>\n<td style=\"justify-content: center;text-align: right\" width=\"10%\">(6)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p align=\"left\">and solving for y<\/p>\n<table border=\"0\" width=\"100%\">\n<tbody>\n<tr>\n<td style=\"text-align: center\" width=\"90%\">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><\/td>\n<td style=\"justify-content: center;text-align: right\" width=\"10%\">(7)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p align=\"left\">The envelope is the cam profile. Note that we have not obtained an explicit expression for the cam profile. Instead we have an algorithm with which we can determine the coordinates on the cam profile in terms of the parameter\u00a0<span style=\"font-family: Symbol\">q<\/span>, which is the cam rotation angle. Also note that for every cam rotation angle we obtain to points on the profile which means that the envelope is of two parts. One is the inner envelope and the other is the outer envelope (external or internal cam profile).<\/p>\n<p><strong><em>Example \u00a08.7.<\/em><\/strong><\/p>\n<p align=\"left\">Design a radial cam with an offset roller follower which rises to 50 mm in 120\u00b0\u00a0cam rotation using simple harmonic motion and dwells at 50 mm for 60\u00b0 cam rotation. For the next 120\u00b0\u00a0cam rotation in simple harmonic motion. Let the roller radius be 20 mm, base circle radius 50 mm and let the eccentricity c = 20 mm.<\/p>\n<ul>\n<li><em><u><strong>Displacement and velocity diagrams:<\/strong><\/u><\/em><\/li>\n<\/ul>\n<p>Let us obtain the displacement of the follower and its velocity for every 10\u00b0\u00a0of cam shaft rotation:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2367\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/03\/Top_clip_image022.gif\" alt=\"\" width=\"723\" height=\"245\" \/><\/p>\n<p>The displacement and velocity diagrams are shown below (we assume <span style=\"font-family: Symbol\">w <\/span>= 1 rad\/s, therefore v = ds\/d<span style=\"font-family: Symbol\">q<\/span>).<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-2369\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/03\/Top_clip_image028.gif\" alt=\"\" width=\"377\" height=\"366\" \/><\/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<ul>\n<li>The coordinates of the centre of the roller and their derivatives with respect to cam angle (use equations 2 and 5) (m is the dummy index: m = 0, 1, \u2026, 360):<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2370\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/03\/Top_clip_image028-1.gif\" alt=\"\" width=\"377\" height=\"366\" \/><\/p>\n<p>dx<sub>p<\/sub>\u00a0and dy<sub>p<\/sub>\u00a0are the partial derivatives of x<sub>p<\/sub>\u00a0and y<sub>p<\/sub>\u00a0with respect to\u00a0<span style=\"font-family: Symbol\">q<\/span>.<\/p>\n<ul>\n<li>Cam profile (use equation 6)<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2371\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/03\/Top_clip_image030.gif\" alt=\"\" width=\"379\" height=\"217\" \/><\/p>\n<p>Note that x<sub>1<\/sub>, y<sub>1<\/sub>\u00a0and x<sub>2<\/sub>, y<sub>2<\/sub> gives us two pairs of cam profile coordinates depending on the + or \u2212 sign in the equation for x.\u00a0 x<sub>1<\/sub>, y<sub>1<\/sub>\u00a0and x<sub>2<\/sub>,y<sub>2<\/sub>\u00a0 will be on the inner envelope or on the outer envelope for certain portions of the curve (one cannot state that x<sub>1<\/sub>, y<sub>1<\/sub>\u00a0is on the inner or outer envelope. x<sub>1<\/sub>, y<sub>1<\/sub>\u00a0will lie on the outer envelope for a certain range of\u00a0<span style=\"font-family: Symbol\">q<\/span>\u00a0and will lie on the inner envelope for the other range of\u00a0<span style=\"font-family: Symbol\">q<\/span>).\u00a0 The result is shown below.<\/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 style=\"text-align: center\" align=\"center\"><span style=\"color: #ff0000\">red: x<sub>1<\/sub>, y<sub>1<\/sub><\/span>:\u00a0<span style=\"color: #1f1fcd\">blue: x<sub>2<\/sub>, y<sub>2<\/sub><\/span><\/p>\n<p align=\"left\">If we want to draw the cam profile in external contact with the roller follower, we can convert the cam profile coordinates to polar form and select the coordinate closer to the origin:<\/p>\n<p style=\"padding-left: 40px\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2372\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/03\/Top_clip_image034.gif\" alt=\"\" width=\"498\" height=\"213\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2373\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/03\/Top_clip_image036.gif\" alt=\"\" width=\"237\" height=\"104\" \/><\/p>\n<p align=\"left\">The plot of the cam profile in polar coordinates (r, x) is shown:<\/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=\"681\" height=\"632\" \/><\/p>\n<p align=\"left\">One can determine the pressure angle for every position as shown.<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2376\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/03\/Top_clip_image040.gif\" alt=\"\" width=\"802\" height=\"836\" \/><\/p>\n<p>Using MathCad:<\/p>\n<p style=\"padding-left: 40px\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2375\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/03\/Top_clip_image042-1.gif\" alt=\"\" width=\"285\" height=\"198\" \/><\/p>\n<p align=\"left\">The variation of the pressure angle within one cycle is shown below. Note that with c=20 mm the pressure angle during the rise is within \u221216\u00b0 &lt;\u00a0<span style=\"font-family: Symbol\">a<\/span> &lt; 11\u00b0\u00a0and during the return motion\u00a0<span style=\"font-family: Symbol\">a<\/span><sub>max<\/sub> = 34\u00b0. The pressure angle curve when the eccentricity is zero is also shown.\u00a0The maximum pressure angle is 22.3\u00b0 for both rise and return periods. By employing eccentricity, the maximum pressure angle during the rise portion is decreased at the expense of the maximum pressure angle during the return motion (which is less critical).<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2377\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/03\/Top_clip_image046.gif\" alt=\"\" width=\"792\" height=\"252\" \/><\/p>\n<p style=\"text-align: center\" align=\"center\">with eccentricity c = 20 mm<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2378\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/03\/Top_clip_image043.gif\" alt=\"\" width=\"774\" height=\"238\" \/><\/p>\n<p style=\"text-align: center\" align=\"center\">no eccentricity (c = 0)<\/p>\n<p align=\"left\">If you are to cut this cam profile using a cutter of radius of r<sub>c<\/sub>\u00a0(usually r<sub>c<\/sub> &gt; r<sub>r<\/sub>), then we have to know the centre of the cutter. In such a case the cutter must be tangent to the cam profile. Since the roller has the same tangent at the point of contact due to the envelope theory, the centre of the cutter must be at a distance r<sub>c<\/sub> from the point of contact (x, y) and it must be along the line defined by the contact point (x, y) and the centre of the roller (x<sub>p<\/sub>, y<sub>p<\/sub>)<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2379\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2022\/03\/Top_clip_image048.gif\" alt=\"\" width=\"570\" height=\"493\" \/><\/p>\n<p style=\"padding-left: 40px\">x<sub>c<\/sub> = x + r<sub>c<\/sub> cos\u03c8<\/p>\n<p style=\"padding-left: 40px\">y<sub>c<\/sub> = y + r<sub>c<\/sub> sin\u03c8<\/p>\n<p>where:<\/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>In our example, if we are to cut the cam profile with a cutter diameter of 60 mm, the coordinates of the centre of the cutter will be given by:<\/p>\n<p style=\"padding-left: 40px\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-525 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/eqn45.gif\" alt=\"\" width=\"477\" height=\"108\" \/><\/p>\n<p style=\"padding-left: 40px\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-526 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/eqn46.gif\" alt=\"\" width=\"533\" height=\"74\" \/><\/p>\n<p>The path traced by the center of the cutter (dashed line) and the cam profile is as shown.<\/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<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n<p><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/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\/mechanisms\/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\/mechanisms\/\" 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>Cam Profile for a Radial Cam With Translating Roller Follower Now let us apply the envelope theory to a basic cam-follower system: a radial cam with a translating roller follower. We shall assume that the follower sliding axis is offset &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch8\/8-5\/translating_roller\/\"> <span class=\"screen-reader-text\">translating_roller<\/span> Devam\u0131n\u0131 Oku &raquo;<\/a><\/p>\n","protected":false},"author":7747,"featured_media":0,"parent":2313,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"full-width-page.php","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-2362","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2362","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=2362"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2362\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2313"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/media?parent=2362"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}