{"id":2159,"date":"2022-03-13T13:03:19","date_gmt":"2022-03-13T13:03:19","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=2159"},"modified":"2022-12-06T00:29:55","modified_gmt":"2022-12-06T00:29:55","slug":"4-2-1","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch4\/4-2-1\/","title":{"rendered":"4-2-1"},"content":{"rendered":"
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4.2<\/b> VELOCITY AND ACCELERATION ANALYSIS OF MECHANISMS-1<\/h1>\n

Velocity and acceleration analysis of mechanisms can be performed vectorially using the relative velocity and acceleration concept. Usually we start with the given values and work through the mechanism by way of series of points A, B, C, etc. Solving vector equations in the form:<\/p>\n

v<\/strong>B<\/sub>\u00a0= v<\/strong>A<\/sub>\u00a0+ v<\/strong>B\/A<\/sub><\/p>\n

v<\/strong>C<\/sub>\u00a0= v<\/strong>B<\/sub>\u00a0+ v<\/strong>C\/B<\/sub><\/p>\n

etc. for velocity and<\/p>\n

a<\/strong>B<\/sub>\u00a0= a<\/strong>A<\/sub>\u00a0+ at<\/sup><\/strong>B\/A<\/sub>\u00a0+ an<\/sup><\/strong>B\/A<\/sub><\/p>\n

a<\/strong>C<\/sub>\u00a0= a<\/strong>B<\/sub>\u00a0+ at<\/sup><\/strong>C\/B<\/sub>\u00a0+ an<\/sup><\/strong>C\/B<\/sub><\/p>\n

veya<\/p>\n

a<\/strong>D4<\/sub>\u00a0= a<\/strong>D3<\/sub>\u00a0+ at<\/sup><\/strong>D4\/D3<\/sub> + ac<\/sup><\/strong>D4\/D3<\/sub><\/p>\n

etc for acceleration. The points that one has to use are usually the revolute joint axes between the links since these are the points where the relative velocity or acceleration between the two coincident points on two different links are zero and they have equal velocity and accelerations. If we are to determine the velocity of a point on a link we must first determine the velocity of the points located at the joint axes.<\/p>\n

\"\" \u00a0 \u00a0 Another important consideration is that the acceleration analysis cannot be performed without performing the velocity analysis<\/span><\/strong>\u00a0since the normal and\u00a0Coriolis acceleration components can only be determined after the velocity analysis<\/em><\/strong>.<\/em><\/span><\/b><\/span><\/p>\n

Loop equations can be used very effectively for velocity and acceleration analysis since the loop equations contain the necessary position variables. When the loop equations are differentiated with respect to time we obtain\u00a0\u201cvelocity loop equations<\/strong>\u201d<\/span>. If the position variables are solved beforehand, these\u00a0velocity\u00a0loop equations will always yield a linear set of equations in terms of velocity variables<\/em><\/strong><\/span>\u00a0which are the time rate of change of the position variables of the mechanism. When these velocity variables are solved for a given input condition, the velocity of any point on any link can be determined.<\/p>\n

\"\"<\/p>\n

As a first example, consider a slider-crank mechanism shown above. We shall assume that \u03b812<\/sub> and its derivatives are known: <\/span> \\displaystyle \\text{\u03c9}_{12}=\\dot{\u03b8}_{12}=\\frac{{\\text{d}{{\\text{\u03b8}}_{{12}}}}}{\\text{dt}} <\/span> , \\displaystyle {{\\text{\u03b1}}_{{12}}}={{{\\ddot{\u03b8}}}_{{12}}}=\\frac{{{{\\text{d}}^{2}}{{\\text{\u03b8}}_{{12}}}}}{{\\text{d}{{\\text{t}}^{2}}}} <\/span>. The loop closure and its complex conjugate is:<\/p>\n\n\n\n
\n

s14<\/sub> + ic1<\/sub> + a3<\/sub>ei\u03b813<\/sub><\/sup> = a2<\/sub>ei\u03b812<\/sub><\/sup><\/p>\n

s14<\/sub>\u00a0\u2212 ic1<\/sub> + a3<\/sub>e\u2212<\/sup>i\u03b813<\/sub><\/sup> = a2<\/sub>e\u2212<\/sup>i\u03b812<\/sub><\/sup><\/p>\n<\/td>\n

\n

(1a)<\/p>\n

(1b)<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

For a given value of \u03b812<\/sub>\u00a0we have seen how the position variable\u00a0s<\/span>14<\/sub>\u00a0and \u03b813<\/sub>\u00a0can be solved. Differentiating the loop closure equation with respect to time we obtain the velocity loop equations as:<\/p>\n\n\n\n
\n

\\displaystyle {\\dot{\\text{s}}}<\/span>14<\/sub> + ia3<\/sub>\u03c913<\/sub>ei\u03b813<\/sub><\/sup> = ia2<\/sub>\u03c912<\/sub>ei\u03b812<\/sub><\/sup><\/p>\n

\\displaystyle {\\dot{\\text{s}}}<\/span>14<\/sub> \u2212 ia3<\/sub>\u03c913<\/sub>e\u2212<\/sup>i\u03b813<\/sub><\/sup> = \u2212ia2<\/sub>\u03c912<\/sub>e\u2212<\/sup>i\u03b812<\/sub><\/sup><\/p>\n<\/td>\n

\n

(2a)<\/p>\n

(2b)<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where \u03c912<\/sub> = d\u03b812<\/sub>\/dt, \u03c913<\/sub> = d\u03b813<\/sub>\/dt, \\displaystyle {\\dot{\\text{s}}}<\/span>14<\/sub> = ds14<\/sub>\/dt are the velocity variables. Note that the velocity loop equation is nothing but the vector equation:<\/p>\n

v<\/strong>B<\/sub>\u00a0+ v<\/strong>A\/B<\/sub>\u00a0= v<\/strong>A<\/sub><\/p>\n

Physical explanation of the above equation is that point A is a permanently coincident point between links 2 and 3, since they are points on the revolute joint axis between these two links. Therefore v<\/strong>A2<\/sub>\u00a0= v<\/strong>A3<\/sub>\u00a0= v<\/strong>A<\/sub>. If we consider link 2, it is in a fixed axis of rotation and point A on link 2 has a velocity perpendicular to AA0<\/sub>\u00a0in the sense of \u03c912<\/sub>\u00a0and its magnitude is |AA0<\/sub>|\u03c912<\/sub>. If we consider link 3, it is in a general plane motion. Points A and B are on this link. If we\u00a0 write the velocity of point A on link 3 using point B it is: v<\/strong>A3<\/sub> = v<\/strong>B<\/sub> + v<\/strong>A\/B<\/sub> = v<\/strong>A<\/sub>. v<\/strong>A\/B<\/sub>\u00a0is perpendicular to AB. Point B is a permanently coincident point between links 3 and 4. Therefore v<\/strong>B3<\/sub>\u00a0= v<\/strong>B4<\/sub>\u00a0= v<\/strong>B<\/sub>. If we consider link 4, it is in a translation. Therefore the velocity of every point is tangent to the path, which is the slider axis. In the velocity vector equation the unknowns are the magnitudes of\u00a0v<\/strong>B<\/sub>\u00a0and v<\/strong>A\/B<\/sub>\u00a0(which correspond to \\displaystyle {\\dot{\\text{s}}}<\/span>14<\/sub>\u00a0and a3<\/sub>\u03c913<\/sub>\u00a0in the velocity loop equation).<\/p>\n

If the loop equations are to be solved graphically, unlike the analytical method where we group the unknowns on one side of the equality and the known values on the other side, we leave one unknown on both sides of the equation. The loop equation is rewritten as:<\/p>\n

\\displaystyle {\\dot{\\text{s}}}<\/span>14<\/sub>\u00a0= ia2<\/sub>\u03c912<\/sub>ei\u03b812<\/sub><\/sup>\u00a0\u2212 ia3<\/sub>\u03c913<\/sub>ei\u03b813<\/sub><\/sup><\/p>\n

which is the vector equation:<\/p>\n

v<\/strong>B<\/sub>\u00a0= v<\/strong>A<\/sub>\u00a0+ v<\/strong>A\/B<\/sub><\/p>\n

Note that<\/p>\n

v<\/strong>B\/A<\/sub> = \u2212v<\/strong>A\/B<\/sub> = \u2212ia3<\/sub>\u03c913<\/sub>ei\u03b813<\/sub><\/sup><\/p>\n

In order to determine the unknowns graphically, we first determine the magnitude of |v<\/strong>A<\/sub>|= |AA0<\/sub>|\u03c912<\/sub>. Then using a certain scale factor,\u00a0kv<\/sub>, a directed line (vector) whose magnitude is proportional to\u00a0v<\/strong>A<\/sub>\u00a0and direction of v<\/strong>A<\/sub> is drawn (see figure below).\u00a0From the starting point of this line, a line whose direction is that of\u00a0v<\/strong>B<\/sub>, and from the tip a line whose direction is that of\u00a0v<\/strong>B\/A<\/sub>is drawn (v<\/strong>B<\/sub>\u00a0is along the slider axis and v<\/strong>B\/A<\/sub>\u00a0is perpendicular to AB). The point of intersection gives us the tips of the vectors representing the velocities v<\/strong>B<\/sub>\u00a0and v<\/strong>B\/A<\/sub>. If we measure these lengths and then divide by the scale factor we have used for\u00a0v<\/strong>A<\/sub>, the unknown magnitudes will be solved. The diagram thus obtained is known as the\u00a0velocity polygon<\/strong>. For example if \u03c913<\/sub>\u00a0= d\u03b813<\/sub>\/dt\u00a0 is to be determined, first the length of the vector\u00a0v<\/strong>B\/A<\/sub>\u00a0\u00a0measured on the velocity polygon is divided by\u00a0kv<\/sub>\u00a0to determine the actual value of\u00a0v<\/strong>B\/A<\/sub>. Since vB\/A<\/sub>\u00a0= a3<\/sub>\u03c913<\/sub>\u00a0and a3<\/sub>\u00a0= |AB|, then \u03c913<\/sub> = vB\/A<\/sub>\/a3<\/sub>. The direction of the angular velocity of link 3 is determined by the direction of\u00a0v<\/strong>B\/A<\/sub>. In the figure, since the velocity of point B relative to A (v<\/strong>B\/A<\/sub>) is downwards, in order to have this velocity at this position link 3 must have a clockwise angular velocity.<\/p>\n

\"\"<\/p>\n

I<\/span>f the dependant position variables (s14<\/sub>\u00a0and \u03b813<\/sub>) are to be solved analytically for a corresponding input position \u03b812<\/sub>, the velocity loop equation will always give a linear relation between the velocity variables (\u03c912<\/sub>, \u03c913<\/sub> and \\displaystyle {\\dot{\\text{s}}}<\/span>14<\/sub>). If the rate of change of the input variable (\u03c912<\/sub>) is given, one can solve for the other two velocity variables \u03c913<\/sub>) and \\displaystyle {\\dot{\\text{s}}}<\/span>14<\/sub> from the velocity equations by the methods of linear algebra. Since we have two equations (equations 2a and 2b) in two unknowns, we can apply Cramer\u2019s rule:<\/p>\n

\\displaystyle {{{\\dot{\\text{s}}}}_{{14}}}=\\frac{{\\left| {\\begin{array}{cc} {\\text{i}{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} & {\\text{i}{{\\text{a}}_{3}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} \\\\ {-\\text{i}{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} & {-\\text{i}{{\\text{a}}_{3}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} \\end{array}} \\right|}}{{\\left| {\\begin{array}{cc} 1 & {\\text{i}{{\\text{a}}_{3}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} \\\\ 1 & {-\\text{i}{{\\text{a}}_{3}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} \\end{array}} \\right|}}=\\frac{{{{\\text{a}}_{2}}{{\\text{a}}_{3}}\\left( {{{\\text{e}}^{{\\text{i}\\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}}-{{\\text{e}}^{{-\\text{i}\\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}}} \\right)}}{{-\\text{i}{{\\text{a}}_{3}}\\left( {{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{13}}}}}}+{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} \\right)}}{{\\text{\u03c9}}_{{12}}} <\/span><\/p>\n

or<\/p>\n\n\n\n
\\displaystyle {{{\\dot{\\text{s}}}}_{{14}}}=-{{\\text{a}}_{2}}\\frac{{\\sin \\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}{{\\cos {{\\text{\u03b8}}_{{13}}}}}{{\\text{\u03c9}}_{{12}}} <\/span><\/td>\n(4)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

and<\/p>\n\n\n\n
\\displaystyle {{\\text{\u03c9}}_{{13}}}=\\frac{{\\left| {\\begin{array}{cc} 1 & {\\text{i}{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} \\\\ 1 & {-\\text{i}{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} \\end{array}} \\right|}}{{\\left| {\\begin{array}{cc} 1 & {\\text{i}{{\\text{a}}_{3}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} \\\\ 1 & {-\\text{i}{{\\text{a}}_{3}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} \\end{array}} \\right|}}=\\frac{{-\\text{i}{{\\text{a}}_{2}}\\left( {{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{12}}}}}}+{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} \\right)}}{{-\\text{i}{{\\text{a}}_{3}}\\left( {{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{13}}}}}}+{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} \\right)}}{{\\text{\u03c9}}_{{12}}}=\\frac{{{{\\text{a}}_{2}}}}{{{{\\text{a}}_{3}}}}\\frac{{\\cos {{\\text{\u03b8}}_{{12}}}}}{{\\cos {{\\text{\u03b8}}_{{13}}}}}{{\\text{\u03c9}}_{{12}}} <\/span><\/td>\n(5)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Note that one can as well obtain two scalar equations by equating the real and imaginary parts of the velocity loop equation and solve for the velocity variables as well. For the slider-crank mechanism these two scalar equations will be:<\/p>\n\n\n\n
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\\displaystyle {\\dot{\\text{s}}}<\/span>14<\/sub> \u2212 a3<\/sub>\u03c913<\/sub>sin\u03b813<\/sub> = \u2212a2<\/sub>\u03c912<\/sub>sin\u03b812<\/sub><\/p>\n<\/td>\n

(6)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

and<\/p>\n\n\n\n
\n

a3<\/sub>\u03c913<\/sub>cos\u03b813<\/sub> = a2<\/sub>\u03c912<\/sub>cos\u03b812<\/sub><\/p>\n<\/td>\n

(7)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

One will obtain exactly the same result from the solution of these two equations for the velocity variables \u03c913<\/sub> and \\displaystyle {\\dot{\\text{s}}}<\/span>14<\/sub>.<\/p>\n

For the acceleration analysis the velocity loop equations can be differentiated with respect to time to yield\u00a0acceleration loop equations<\/strong>\u00a0in terms of\u00a0acceleration variables<\/em>\u00a0which are the second rate of change of the position variables. For the slider crank mechanism given, differentiating equations 2a and 2b with respect to time:<\/p>\n\n\n\n
\n

\\displaystyle {\\ddot{\\text{s}}}<\/span>14<\/sub> + ia3<\/sub>\u03b113<\/sub>ei\u03b813<\/sub><\/sup> \u2212 a3<\/sub>\u03c913<\/sub>2<\/sup>ei\u03b813<\/sub><\/sup> = ia2<\/sub>\u03b112<\/sub>ei\u03b812<\/sub><\/sup>\u00a0\u2212 a2<\/sub>\u03c912<\/sub>2<\/sup>ei\u03b812<\/sub><\/sup><\/p>\n

\\displaystyle {\\ddot{\\text{s}}}<\/span>14<\/sub> \u2212 ia3<\/sub>\u03b113<\/sub>e-i\u03b813<\/sub><\/sup> \u2212 a3<\/sub>\u03c913<\/sub>2<\/sup>e-i\u03b813<\/sub><\/sup> = \u2212ia2<\/sub>\u03b112<\/sub>e-i\u03b812<\/sub><\/sup>\u00a0\u2212 a2<\/sub>\u03c912<\/sub>2<\/sup>e-i\u03b812<\/sub><\/sup><\/p>\n<\/td>\n

\n

(8a)<\/p>\n

(8b)<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where \u03b112<\/sub> = d2<\/sup>\u03b812<\/sub>\/dt2<\/sup>, \u03b113<\/sub> = d2<\/sup>\u03b813<\/sub>\/dt2<\/sup>, \\displaystyle {\\ddot{\\text{\\text{s}}}}<\/span>14<\/sub>\u00a0= d2<\/sup>s14<\/sub>\/dt2 <\/sup>are the acceleration variables.. Note that this equation can be written as a vector equation in the form:<\/p>\n

a<\/strong>B<\/sub> + at<\/sup><\/strong>A\/B<\/sub>\u00a0+\u00a0an<\/sup><\/strong>A\/B<\/sub> = at<\/sup><\/strong>A<\/sub>\u00a0+\u00a0an<\/sup><\/strong>A<\/sub><\/p>\n

The acceleration loop equations are linear in terms of the acceleration variables (\u03b112<\/sub>, \u03b113<\/sub> and \\displaystyle {\\ddot{\\text{s}}}<\/span>14<\/sub>). If the input angular acceleration, \u03b112<\/sub>\u00a0is known, these two equations can be solved for the unknowns \u03b113<\/sub> and \\displaystyle {\\ddot{\\text{s}}}<\/span>14<\/sub>\u00a0using Cramer\u2019s rule :<\/p>\n\n\n\n
\\displaystyle {{\\text{\u03b1}}_{{13}}}=\\frac{{{{\\text{a}}_{2}}{{\\text{\u03b1}}_{{12}}}\\cos {{\\text{\u03b8}}_{{12}}}-{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}^{2}\\sin {{\\text{\u03b8}}_{{12}}}+{{\\text{a}}_{3}}{{\\text{\u03c9}}_{{13}}}^{2}\\sin {{\\text{\u03b8}}_{{13}}}}}{{{{\\text{a}}_{3}}\\cos {{\\text{\u03b8}}_{{13}}}}} <\/span><\/td>\n(9)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n\n
\n

\\displaystyle {\\ddot{\\text{s}}}<\/span>14<\/sub> = \u2212a2<\/sub>\u03b112<\/sub>sin\u03b812<\/sub>\u00a0\u2212 a2<\/sub>\u03c912<\/sub>2<\/sup>cos\u03b812<\/sub>\u00a0+ ia3<\/sub>\u03b113<\/sub>sin\u03b813<\/sub>\u00a0+ a3<\/sub>\u03c913<\/sub>2<\/sup>cos\u03b813<\/sub><\/p>\n<\/td>\n

(10)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Another approach is to differentiate the equations 4 and 5 directly to obtain acceleration variables. For example, differentiating equation 5, angular acceleration of link 3 will be obtained in the form:<\/p>\n

\\displaystyle {{\\text{\u03b1}}_{{13}}}=\\frac{{{{\\text{a}}_{2}}}}{{{{\\text{a}}_{3}}}}\\frac{1}{{{{{\\cos }}^{2}}{{\\text{\u03b8}}_{{13}}}}}\\left[ {\\cos {{\\text{\u03b8}}_{{12}}}\\cos {{\\text{\u03b8}}_{{13}}}{{\\text{\u03b1}}_{{12}}}-\\sin {{\\text{\u03b8}}_{{12}}}\\cos {{\\text{\u03b8}}_{{13}}}{{\\text{\u03c9}}_{{12}}}^{2}+\\cos {{\\text{\u03b8}}_{{12}}}\\sin {{\\text{\u03b8}}_{{13}}}{{\\text{\u03c9}}_{{12}}}{{\\text{\u03c9}}_{{13}}}} \\right] <\/span><\/p>\n

Although the terms may look different one will obtain the same result in either case.<\/p>\n

Graphically, the acceleration loop equation can be solved by rewriting the acceleration vector loop equation as:<\/p>\n

a<\/strong>B<\/sub>\u00a0= at<\/sup><\/strong>A<\/sub>\u00a0+\u00a0an<\/sup><\/strong>A<\/sub>\u00a0+ at<\/sup><\/strong>B\/A<\/sub>\u00a0+\u00a0an<\/sup><\/strong>B\/A<\/sub><\/p>\n