{"id":2160,"date":"2022-03-13T13:03:44","date_gmt":"2022-03-13T13:03:44","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=2160"},"modified":"2022-12-06T00:35:33","modified_gmt":"2022-12-06T00:35:33","slug":"4-2-2","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch4\/4-2-2\/","title":{"rendered":"4-2-2"},"content":{"rendered":"
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4.2<\/b> VELOCITY AND ACCELERATION ANALYSIS OF MECHANISMS-2<\/h1>\n

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

\"\"<\/p>\n

As for the velocity and acceleration analysis of a four-bar mechanism, a similar approach can be used. The loop closure equation and its complex conjugate is:<\/p>\n

a2<\/sub>ei\u03b812<\/sub><\/sup> + a3<\/sub>ei\u03b813<\/sub><\/sup> = a1<\/sub> + a4<\/sub>ei\u03b814<\/sub><\/sup><\/p>\n

a2<\/sub>e\u2212i\u03b812<\/sub><\/sup> + a3<\/sub>e\u2212<\/sup>i\u03b813<\/sub><\/sup> = a1<\/sub> + a4<\/sub>e-i\u03b814<\/sub><\/sup><\/p>\n

The first derivative of the loop closure equation (velocity loop equation) is:<\/p>\n

ia2<\/sub>\u03c912<\/sub>ei\u03b812<\/sub><\/sup> + ia3<\/sub>\u03c913<\/sub>ei\u03b813<\/sub><\/sup> = ia4<\/sub>\u03c914<\/sub>ei\u03b814<\/sub><\/sup><\/p>\n

\u2212ia2<\/sub>\u03c912<\/sub>e\u2212<\/sup>i\u03b812<\/sub><\/sup> \u2212 ia3<\/sub>\u03c913<\/sub>e\u2212<\/sup>i\u03b813<\/sub><\/sup> = \u2212ia4<\/sub>\u03c914<\/sub>e\u2212<\/sup>i\u03b814<\/sub><\/sup><\/p>\n

Note that the velocity loop equation in vector form is:<\/p>\n

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

In complex numbers, the magnitude and direction of each velocity term can easily be identified. The velocity loop equation is a simple way of writing the velocity vector polygon. The velocity loop equation and its complex conjugate can be used to solve for the speed variables \u03c913<\/sub>\u00a0and \u03c914<\/sub> for a given input speed \u03c912<\/sub>, provided that the position variables \u03b813<\/sub> and \u03b814<\/sub>\u00a0are solved for a given input angle \u03b812<\/sub>:<\/p>\n

\\displaystyle {{\\text{\u03c9}}_{{13}}}=\\frac{{\\left| {\\begin{array}{cc} {-{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} & {-{{\\text{a}}_{4}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{14}}}}}}} \\\\ {-{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} & {-{{\\text{a}}_{4}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{14}}}}}}} \\end{array}} \\right|}}{{\\left| {\\begin{array}{cc} {{{\\text{a}}_{3}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} & {-{{\\text{a}}_{4}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{14}}}}}}} \\\\ {{{\\text{a}}_{3}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} & {-{{\\text{a}}_{4}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{14}}}}}}} \\end{array}} \\right|}}=\\frac{{{{\\text{a}}_{2}}{{\\text{a}}_{4}}\\left( {{{\\text{e}}^{{\\text{i}\\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{14}}}} \\right)}}}-{{\\text{e}}^{{-\\text{i}\\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{14}}}} \\right)}}}} \\right)}}{{{{\\text{a}}_{3}}{{\\text{a}}_{4}}\\left( {-{{\\text{e}}^{{-\\text{i}\\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}}+{{\\text{e}}^{{\\text{i}\\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}}} \\right)}}{{\\text{\u03c9}}_{{12}}}=\\frac{{{{\\text{a}}_{2}}}}{{{{\\text{a}}_{3}}}}\\frac{{\\sin \\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{14}}}} \\right)}}{{\\sin \\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}{{\\text{\u03c9}}_{{12}}} <\/span><\/p>\n

Similarly:<\/p>\n

\\displaystyle {{\\text{\u03c9}}_{{14}}}=\\frac{{\\left| {\\begin{array}{cc} {{{\\text{a}}_{3}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} & {-{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} \\\\ {{{\\text{a}}_{3}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} & {-{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} \\end{array}} \\right|}}{{\\left| {\\begin{array}{cc} {{{\\text{a}}_{3}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} & {-{{\\text{a}}_{4}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{14}}}}}}} \\\\ {{{\\text{a}}_{3}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} & {-{{\\text{a}}_{4}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{14}}}}}}} \\end{array}} \\right|}}=\\frac{{{{\\text{a}}_{3}}{{\\text{a}}_{2}}\\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{a}}_{3}}{{\\text{a}}_{4}}\\left( {-{{\\text{e}}^{{-\\text{i}\\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}}+{{\\text{e}}^{{-\\text{i}\\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}}} \\right)}}{{\\text{\u03c9}}_{{12}}}=\\frac{{{{\\text{a}}_{2}}}}{{{{\\text{a}}_{4}}}}\\frac{{\\sin \\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}{{\\sin \\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}{{\\text{\u03c9}}_{{12}}} <\/span><\/p>\n

If the velocity of the coupler point C is to be determined, first wee need its position:<\/p>\n

r<\/strong>C<\/sub>\u00a0= a2<\/sub>ei\u03b812<\/sub><\/sup> + b3<\/sub>ei(\u03b813 <\/sub>+\u00a0<\/sub>\u03b2)<\/sup><\/p>\n

The derivative of the position vector will give us the velocity vector:<\/p>\n

v<\/strong>C<\/sub> = \\displaystyle {\\dot{\\text{x}}} <\/span>C<\/sub> + i \\displaystyle {\\dot{\\text{y}}} <\/span>C<\/sub> = ia2<\/sub>\u03c912<\/sub>ei\u03b812<\/sub><\/sup> + ib3<\/sub>\u03c913<\/sub>ei(\u03b813 <\/sub>+\u00a0<\/sub>\u03b2)<\/sup><\/p>\n

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

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

In terms of Cartesian components:<\/p>\n

\\displaystyle {\\dot{\\text{x}}} <\/span>C<\/sub> = \u2212ia2<\/sub>\u03c912<\/sub>sin\u03b812<\/sub> \u2212 ib3<\/sub>\u03c913<\/sub>sin(\u03b813 <\/sub>+\u00a0<\/sub>\u03b2)<\/p>\n

\\displaystyle {\\dot{\\text{y}}} <\/span>C<\/sub> = a2<\/sub>\u03c912<\/sub>cos\u03b812<\/sub> + b3<\/sub>\u03c913<\/sub>cos(\u03b813 <\/sub>+\u00a0<\/sub>\u03b2)<\/p>\n

If the position and the velocity loop equations are solved, one can easily determine the terms on the right hand side of the equations. Once the x and y components of the velocity is determined one can as well transform the velocity vector into polar form to yield the magnitude and direction by writing the velocity vector in the form:<\/p>\n

v<\/strong>C<\/sub> = vC<\/sub>ei\u03b7<\/sup><\/p>\n

where vC<\/sub> = \\displaystyle \\sqrt{{{{{\\dot{\\text{x}}}}_{\\text{C}}}^{2}+{{{\\dot{\\text{y}}}}_{\\text{C}}}^{2}}} <\/span> = magnitude of velocity, and \u03b7 = tan-1<\/sup>( \\displaystyle {\\dot{\\text{y}}} <\/span>C<\/sub>\/ \\displaystyle {\\dot{\\text{x}}} <\/span>C<\/sub>) = the angle the velocity vector makes with respect to positive x-axis.<\/p>\n

For the acceleration analysis, the second derivative of the loop closure equation (acceleration loop equation) is:<\/p>\n

ia2<\/sub>\u03b112<\/sub>ei\u03b812<\/sub><\/sup> \u2212 a2<\/sub>\u03c912<\/sub>2<\/sup>ei\u03b812<\/sub><\/sup> + ia3<\/sub>\u03b113<\/sub>ei\u03b813<\/sub><\/sup> \u2212 a3<\/sub>\u03c913<\/sub>2<\/sup>ei\u03b813<\/sub><\/sup> = ia4<\/sub>\u03b114<\/sub>ei\u03b814<\/sub><\/sup> \u2212 a4<\/sub>\u03c914<\/sub>2<\/sup>ei\u03b814<\/sub><\/sup><\/p>\n

\u2212ia2<\/sub>\u03b112<\/sub>e\u2212<\/sup>i\u03b812<\/sub><\/sup> \u2212 a2<\/sub>\u03c912<\/sub>2<\/sup>e\u2212<\/sup>i\u03b812<\/sub><\/sup> \u2212 ia3<\/sub>\u03b113<\/sub>e\u2212<\/sup>i\u03b813<\/sub><\/sup> \u2212 a3<\/sub>\u03c913<\/sub>2<\/sup>e\u2212<\/sup>i\u03b813<\/sub><\/sup> = \u2212ia4<\/sub>\u03b114<\/sub>e\u2212<\/sup>i\u03b814<\/sub><\/sup> \u2212 a4<\/sub>\u03c914<\/sub>2<\/sup>e\u2212<\/sup>i\u03b814<\/sub><\/sup><\/p>\n

Note that the acceleration loop equation is nothing but the acceleration vector equation in the\u00a0 form:<\/p>\n

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

Rearranging the terms so that only the unknown acceleration variables are on the left hand side of the equation:<\/p>\n

\u00a0 ia3<\/sub>\u03b113<\/sub>ei\u03b813<\/sub><\/sup> \u2212 ia4<\/sub>\u03b114<\/sub>ei\u03b814<\/sub><\/sup>\u00a0= \u2212ia2<\/sub>\u03b112<\/sub>ei\u03b812<\/sub><\/sup> + a2<\/sub>\u03c912<\/sub>2<\/sup>ei\u03b812<\/sub><\/sup> + a3<\/sub>\u03c913<\/sub>2<\/sup>ei\u03b813<\/sub><\/sup>\u00a0\u2212 a4<\/sub>\u03c914<\/sub>2<\/sup>ei\u03b814<\/sub><\/sup><\/p>\n

\u2212ia3<\/sub>\u03b113<\/sub>e\u2212<\/sup>i\u03b813<\/sub><\/sup> + ia4<\/sub>\u03b114<\/sub>e\u2212<\/sup>i\u03b814<\/sub><\/sup> = ia2<\/sub>\u03b112<\/sub>e\u2212<\/sup>i\u03b812<\/sub><\/sup> + a2<\/sub>\u03c912<\/sub>2<\/sup>e\u2212<\/sup>i\u03b812<\/sub><\/sup> + a3<\/sub>\u03c913<\/sub>2<\/sup>e\u2212<\/sup>i\u03b813<\/sub><\/sup>\u00a0\u2212 a4<\/sub>\u03c914<\/sub>2<\/sup>e\u2212<\/sup>i\u03b814<\/sub><\/sup><\/p>\n

Solving the two linear equations for the acceleration variables:<\/p>\n

\u03b113<\/sub> = \\displaystyle \\frac{{{{\\text{a}}_{2}}{{\\text{\u03b1}}_{{12}}}\\sin \\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{14}}}} \\right)+{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}^{2}\\cos \\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{14}}}} \\right)+{{\\text{a}}_{3}}{{\\text{\u03c9}}_{{13}}}^{2}\\cos \\left( {{{\\text{\u03b8}}_{{13}}}-{{\\text{\u03b8}}_{{14}}}} \\right)-{{\\text{a}}_{4}}{{\\text{\u03c9}}_{{12}}}^{2}}}{{{{\\text{a}}_{3}}\\sin \\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}} <\/span><\/p>\n

and<\/p>\n

\u03b114<\/sub> = \\displaystyle \\frac{{{{\\text{a}}_{2}}{{\\text{\u03b1}}_{{12}}}\\sin \\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{13}}}} \\right)+{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}^{2}\\cos \\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{13}}}} \\right)+{{\\text{a}}_{3}}{{\\text{\u03c9}}_{{13}}}^{2}-{{\\text{a}}_{4}}{{\\text{\u03c9}}_{{14}}}^{2}\\cos \\left( {{{\\text{\u03b8}}_{{13}}}-{{\\text{\u03b8}}_{{14}}}} \\right)}}{{{{\\text{a}}_{4}}\\sin \\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}} <\/span><\/p>\n

The acceleration of the coupler point C can be obtained from the second derivative of the position vector:<\/p>\n

a<\/strong>C<\/sub> = ia2<\/sub>\u03b112<\/sub>ei\u03b812<\/sub><\/sup> \u2212 a2<\/sub>\u03c912<\/sub>2<\/sup>ei\u03b812<\/sub><\/sup>\u00a0+ ib3<\/sub>\u03b113<\/sub>ei(\u03b813 <\/sub>+\u00a0<\/sub>\u03b2)<\/sup>\u00a0\u2212 b3<\/sub>\u03c913<\/sub>2<\/sup>ei(\u03b813 <\/sub>+\u00a0<\/sub>\u03b2)<\/sup><\/p>\n

Hence, after the solution of the position, velocity and acceleration loop equations, one can easily determine the position, velocity and acceleration of any point in any one of the links of the mechanism. The graphical solution of the velocity and acceleration vector equations are as shown below.<\/span><\/p>\n

\"\"<\/p>\n

The velocity and acceleration loop equations are always linear in terms of the speed and acceleration variables. It is the solution of the loop equations that require more effort since they are nonlinear. The above procedure can be\u00a0<\/span>explained step by step as follows:<\/p>\n