{"id":2137,"date":"2022-03-12T23:55:29","date_gmt":"2022-03-12T23:55:29","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=2137"},"modified":"2022-12-06T01:27:27","modified_gmt":"2022-12-06T01:27:27","slug":"3-7","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch3\/3-7\/","title":{"rendered":"3-7"},"content":{"rendered":"
\n
\n\t

3.7<\/b> Position Analysis of Mechanisms By Means of Complex Numbers<\/h1>\n

We have seen that complex number utilisation is a simple and powerful technique for expressing the loop closure equations. The analytical or numerical solution of planar mechanisms can be easily performed by simple algebraic manipulations in complex numbers.<\/p>\n

Considering the four-bar mechanism shown below, the loop closure equation in vectorial form is:<\/p>\n

A0<\/sub>A<\/strong> + <\/b>AB<\/strong> = <\/b>A0<\/sub>B0<\/sub><\/strong> + <\/b>B0<\/sub>B<\/strong><\/p>\n

or in complex numbers:<\/p>\n\n\n\n
a2<\/sub>ei\u03b812<\/sub><\/sup> + a3<\/sub>ei\u03b813<\/sub><\/sup> = a1<\/sub>\u00a0+ a4<\/sub>ei\u03b814<\/sub><\/sup><\/td>\n(1)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"\"<\/p>\n

If we equate the real and imaginary parts of this equation separately, we obtain two scalar equations in three position variables (\u03b812<\/sub>, \u03b813<\/sub>\u00a0and \u03b814<\/sub>). If one of the position variables is the input variable whose value is given, then we shall be able to solve for the values of the other two variables.<\/p>\n

In complex plane, when we have an equation in complex numbers, the complex conjugate of the equation is also true (see Appendix I<\/b><\/a><\/span>). The complex conjugate yields vectors which are the mirror image of the original vectors with respect to the real axis (x-axis); e.g. in case of a mechanism, if we place a mirror about the real axis, as we move the original mechanism its image will also move and corresponding to the original closed loop, the loop formed on the mirror image will also be closed at every position. Hence we obtain another loop closure equation in terms of complex numbers as:<\/p>\n\n\n\n
a2<\/sub>e\u2212<\/sup>i<\/sup>\u03b812<\/sub><\/sup> + a<\/span>3<\/sub>e\u2212<\/sup><\/span>i\u03b813<\/sub><\/sup> = a<\/span>1<\/sub>\u00a0+ a<\/span>4<\/sub>e\u2212<\/sup><\/span>i\u03b814<\/sub><\/sup><\/td>\n(2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The original equation (1) and its complex conjugate (2) are the two independent equations in the complex plane (if we equate the real and imaginary parts of these equations they will yield the same two scalar equations in the real plane).<\/p>\n

In general, the loop closure equations yield a non-linear relation between the position variables. The closed form solution of these equations is not straight forward. For simple mechanisms, utilising complex algebra one can solve for the unknown position variables. An iterative numerical solution of these equations will be explained in Section 2.5. In this section, closed form solution of the loop closure equations for simple mechanisms will be explained only.<\/p>\n

Using the equations (1) and (2), if we are to find \u03b814<\/sub> as a function of \u03b812<\/sub>, we have to eliminate \u03b813<\/sub>\u00a0from the above equations. We can write the loop closure equations in the form:<\/p>\n\n\n\n\n
\n

a3<\/sub>ei<\/sup>\u03b813<\/sub><\/sup> = a1<\/sub>\u00a0+ a4<\/sub>ei\u03b814<\/sub><\/sup>\u00a0\u2212 a2<\/sub>ei<\/sup>\u03b812<\/sub><\/sup><\/p>\n<\/td>\n

(3)<\/td>\n<\/tr>\n
\n

a3<\/sub>e\u2212i<\/sup>\u03b813<\/sub><\/sup> = a1<\/sub>\u00a0+ a4<\/sub>e\u2212i\u03b814<\/sub><\/sup>\u00a0\u2212 a2<\/sub>e\u2212i<\/sup>\u03b812<\/sub><\/sup><\/p>\n<\/td>\n

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

Multiplying equations (3) and (4):<\/p>\n\n\n\n
\n

a3<\/sub>2<\/sup>ei(\u03b813<\/sub> \u2212 \u03b813<\/sub>)<\/sup> = (a1<\/sub>\u00a0+ a4<\/sub>ei\u03b814<\/sub><\/sup>\u00a0\u2212 a2<\/sub>ei<\/sup>\u03b812<\/sub><\/sup>)(a1<\/sub>\u00a0+ a4<\/sub>e\u2212i\u03b814<\/sub><\/sup>\u00a0\u2212 a2<\/sub>e\u2212i<\/sup>\u03b812<\/sub><\/sup>)<\/p>\n<\/td>\n

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

Noting ei(\u03b8 \u2212 \u03b8)<\/sup> = ei0<\/sup> =1:<\/p>\n\n\n\n
\n

a3<\/sub>2<\/sup> = a1<\/sub>2<\/sup> + a4<\/sub>2<\/sup> + a2<\/sub>2<\/sup> + a1<\/sub>a4<\/sub>[ei\u03b814<\/sub><\/sup> \u2212 e\u2212i\u03b814<\/sub><\/sup>] \u2212 a1<\/sub>a2<\/sub>[ei<\/sup>\u03b812<\/sub><\/sup> + e\u2212i<\/sup>\u03b812<\/sub><\/sup>] \u2212 a2<\/sub>a4<\/sub>[ei(\u03b814 <\/sub>\u2212\u00a0<\/sub>\u03b812<\/sub>)<\/sup>\u00a0\u2212 e\u2212i(\u03b814 <\/sub>\u2212\u00a0<\/sub>\u03b812<\/sub>)<\/sup>]\n<\/td>\n

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

Since cos\u03b8 = (ei\u03b8<\/sup> + e\u2212i\u03b8<\/sup>)\/2 equation (6) reduces to the form:<\/p>\n

a3<\/sub>2<\/sup> = a1<\/sub>2<\/sup> + a2<\/sub>2<\/sup> + a4<\/sub>2<\/sup> + 2a1<\/sub>a4<\/sub>cos\u03b814<\/sub> \u2212 2a1<\/sub>a2<\/sub>cos\u03b812<\/sub> \u2212 2a2<\/sub>a4<\/sub>cos(\u03b814<\/sub> \u2212 \u03b812<\/sub>)<\/p>\n

or, dividing every term by 2a2a4:<\/p>\n\n\n\n
\n

K1<\/sub>cos\u03b814<\/sub> \u2212 K2<\/sub>cos\u03b812<\/sub> + K3<\/sub> = cos(\u03b814<\/sub> \u2212 \u03b812<\/sub>)<\/p>\n<\/td>\n

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

where K1<\/sub> = a1<\/sub>\/a2<\/sub> , K2<\/sub> = a1<\/sub>\/a4<\/sub> , K3<\/sub> = (a1<\/sub>2<\/sup> + a2<\/sub>2<\/sup> \u2212 a3<\/sub>2<\/sup> + a4<\/sub>2<\/sup> )\/2a2<\/sub>a4<\/sub><\/p>\n

Equation (7) is called\u00a0“Freudenstein’s Equation<\/em><\/strong>” which can be used for the synthesis of four-bar mechanisms. It gives an implicit relation between the position variables \u03b814<\/sub> and \u03b812<\/sub>. In order to obtain an explicit expression for \u03b814<\/sub>, Freudenstein’s equation can be written in the form:<\/p>\n\n\n\n
\n

K1<\/sub>cos\u03b814<\/sub> \u2212 K2<\/sub>sin\u03b812<\/sub> + K3<\/sub> = cos\u03b812<\/sub>cos\u03b814<\/sub>\u00a0+ sin\u03b812<\/sub>sin\u03b814<\/sub><\/p>\n<\/td>\n

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

Substituting the trigonometric identities:<\/p>\n

sin\u03b814<\/sub> = 2t\/(1 + t2<\/sup>)\u00a0 \u00a0 \u00a0and\u00a0 \u00a0 \u00a0cos\u03b814<\/sub> = (1 \u2212 t2<\/sup>)\/(1 + t2<\/sup>)<\/p>\n

where t = tan(\u03b814<\/sub>\/2)\u00a0(This is commonly known as half-tangent form of representation of the sine and cosine function).<\/p>\n

Freudenstein’s equation results:<\/p>\n\n\n\n
At2<\/sup> + Bt + C = 0<\/td>\n(9)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where:<\/p>\n

A = cos\u03b812<\/sub>(1 \u2212 K2<\/sub>) + K3<\/sub> \u2212 K1<\/sub><\/p>\n

B = \u22122 sin\u03b812<\/sub><\/p>\n

C = cos\u03b812<\/sub>(1 + K2<\/sub>) + K3<\/sub>\u00a0+ K1<\/sub><\/p>\n

Equation (9) is a quadratic in terms of<\/p>\n

\\displaystyle \\tan \\left( {\\frac{{{{\\text{\u03b8}}_{{14}}}}}{2}} \\right)=\\frac{{-\\text{B}\\pm \\sqrt{{{{\\text{B}}^{2}}-4\\text{AC}}}}}{{2\\text{A}}} <\/span><\/p>\n

therefore:<\/p>\n\n\n\n
\\displaystyle {{{\\text{\u03b8}}_{{14}}}=2{{\\tan }^{{-1}}}\\left[ {\\frac{{-\\text{B}\\pm \\sqrt{{{{\\text{B}}^{2}}-4\\text{AC}}}}}{{2\\text{A}}}} \\right]} <\/span><\/td>\n(10)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Where the plus or minus sign refers to two different configurations of the four-bar mechanism. Also note that the coefficients of the quadratic (A, B and C) are functions of the link lengths and the input crank angle only. Therefore, if the input crank angle is given, \u03b814<\/sub>\u00a0can be obtained from equation (10) directly.<\/p>\n

One can perform a similar procedure to determine the coupler-link angle, \u03b813<\/sub>, in terms of the input angle, \u03b812<\/sub>. This is left as an exercise.<\/p>\n

Another form of solving equation (8) is to rewrite it in the form:<\/p>\n\n\n\n
(K1<\/sub> \u2212 cos\u03b812<\/sub>)cos\u03b814<\/sub> \u2212 sin\u03b814<\/sub>sin\u03b812\u00a0<\/sub>= K2<\/sub>cos\u03b812<\/sub> \u2212 K3<\/sub><\/td>\n(11)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

If we let:<\/p>\n\n\n\n
\n

D cos\u03d5 = K1<\/sub> \u2212 cos\u03b812<\/sub><\/p>\n

D sin\u03d5 = sin\u03b812<\/sub><\/p>\n<\/td>\n

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

where<\/p>\n\n\n\n
\\displaystyle {\\text{D}=\\sqrt{{{{{\\left( {{{\\text{K}}_{1}}-\\cos {{\\text{\u03b8}}_{{12}}}} \\right)}}^{2}}+{{{\\sin }}^{2}}{{\\text{\u03b8}}_{{12}}}}}=\\sqrt{{1+{{\\text{K}}_{1}}^{2}-2{{\\text{K}}_{1}}\\cos {{\\text{\u03b8}}_{{12}}}}}} <\/span>
\n \\displaystyle {\\text{\u03d5} ={{\\tan }^{{-1}}}\\frac{{\\sin {{\\text{\u03b8}}_{{12}}}}}{{{{\\text{K}}_{1}}-\\cos {{\\text{\u03b8}}_{{12}}}}}} <\/span><\/td>\n		(13)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Equation (11) reduces to:<\/p>\n

D cos\u03d5 cos\u03b814<\/sub> \u2212 D sin\u03d5 sin\u03b814<\/sub>\u00a0= K2<\/sub>cos\u03b812<\/sub> \u2212 K3<\/sub><\/p>\n

Then using the trigonometric identity: cos (\u03b8 + \u03d5) = cos\u03b8cos\u03d5 \u2212 sin\u03b8sin\u03d5 :<\/p>\n\n\n\n
cos(\u03b814<\/sub> + \u03d5) = (K2<\/sub>\u00a0cos\u03b812<\/sub> \u2212 K3<\/sub>)\/D<\/td>\n(14)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

or<\/p>\n\n\n\n
\u03b814<\/sub> = cos-1<\/sup>[(K2<\/sub>cos\u03b812<\/sub> \u2212 K3<\/sub>)\/D] \u2212 \u03d5<\/td>\n(15)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

When solving for \u03d5<\/span>\u00a0in equation (13), one must remember that for a correct quadrant of the angle either the sign of the x and y components must be checked or double argument inverse tangent function must be used. In equation (15) the two different solutions for the inverse cosine function (= \u00b1\u03d5) will yield open or crossed configuration of the four-bar mechanism.<\/p>\n

\"\"<\/p>\n

As another example for the solution of loop closure equations in complex numbers consider an inverted slider-crank mechanism where \u03b814<\/sub>\u00a0is the input variable and \u03b814<\/sub>\u00a0is the output variable. We would like to determine \u03b814<\/sub>\u00a0as a function of \u03b812<\/sub>.<\/p>\n

Vector loop closure equation is:\u00a0<\/strong><\/p>\n

A0<\/sub>A<\/strong> = A0<\/sub>B0<\/sub><\/strong>\u00a0+ B0<\/sub>B<\/strong> + BA<\/strong><\/p>\n

\u00a0 and in complex numbers:<\/p>\n

a2<\/sub>ei<\/sup>\u03b812<\/sub><\/sup>\u00a0= a1<\/sub>\u00a0+ a4<\/sub>ei\u03b814<\/sub><\/sup> + s43<\/sub>ei(\u03b814<\/sub> + \u03c0\/2)<\/sup><\/p>\n

or<\/p>\n\n\n\n
a2<\/sub>ei<\/sup>\u03b812<\/sub><\/sup>\u00a0= a1<\/sub>\u00a0+ a4<\/sub>ei\u03b814<\/sub><\/sup> + is43<\/sub>ei\u03b814<\/sub><\/sup><\/td>\n(1)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

and its complex conjugate is:<\/p>\n\n\n\n
a2<\/sub>e\u2212i<\/sup>\u03b812<\/sub><\/sup> = a1<\/sub> + a4<\/sub>e\u2212i<\/sup>\u03b814<\/sub><\/sup> \u2212 is43<\/sub>e\u2212i<\/sup>\u03b814<\/sub><\/sup><\/td>\n(2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

We would like to eliminate s43<\/sub>\u00a0from the equations. Therefore:<\/p>\n\n\n\n\n
\n

is43<\/sub>ei\u03b814<\/sub><\/sup> = a2<\/sub>ei<\/sup>\u03b812<\/sub><\/sup> \u2212 a1<\/sub> \u2212 a4<\/sub>ei\u03b814<\/sub><\/sup><\/p>\n<\/td>\n

(3)<\/td>\n<\/tr>\n
\n

is43<\/sub>e\u2212i<\/sup>\u03b814<\/sub><\/sup>\u00a0= a1<\/sub>\u00a0+ a4<\/sub>e\u2212i<\/sup>\u03b814<\/sub><\/sup> \u2212 a2<\/sub>e\u2212i<\/sup>\u03b812<\/sub><\/sup><\/p>\n<\/td>\n

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

Taking the ratio of the two sides and cross-multiplying:<\/p>\n

a2<\/sub>ei(\u03b812 <\/sub>\u2212\u00a0<\/sub>\u03b814<\/sub>)<\/sup>\u00a0\u2212 a1<\/sub>e\u2212i<\/sup>\u03b814<\/sub><\/sup> \u2212 a4<\/sub> = a1<\/sub>ei<\/sup>\u03b814<\/sub><\/sup> + a4<\/sub>\u00a0\u2212 a2<\/sub>e\u2212i(\u03b812 <\/sub>\u2212\u00a0<\/sub>\u03b814<\/sub>)<\/sup><\/p>\n

or<\/p>\n\n\n\n
\n

a2<\/sub>[ei(\u03b812 <\/sub>\u2212\u00a0<\/sub>\u03b814<\/sub>)<\/sup> + e\u2212i(\u03b812<\/sub> \u2212 \u03b814<\/sub>)<\/sup>]\u2212 a1<\/sub>[ei<\/sup>\u03b814<\/sub><\/sup> + e\u2212i<\/sup>\u03b814<\/sub><\/sup>] \u2212 2a4<\/sub> = 0<\/p>\n<\/td>\n

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

noting ei<\/sup>\u03b8<\/sup>\u00a0+ e\u2212i<\/sup>\u03b8<\/sup> = 2cos\u03b8:<\/p>\n\n\n\n
a2<\/sub>cos(\u03b814<\/sub> \u2212 \u03b812<\/sub>) \u2212 a1<\/sub>cos\u03b814<\/sub> \u2212 a4<\/sub>\u00a0= 0<\/td>\n(6)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Using half-tangent form for cosq<\/span>\u00a014<\/sub>\u00a0and sinq<\/span>\u00a014<\/sub> equation (6) reduces to a quadratic equation in t = tan(\u03b814<\/sub>\/2)\u00a0 as<\/p>\n\n\n\n
At2<\/sup> + Bt + C = 0<\/td>\n(7)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where:<\/p>\n

A = a1<\/sub> \u2212 a4<\/sub> \u2212 a2<\/sub>cos\u03b812<\/sub><\/p>\n

B = 2sin\u03b812<\/sub><\/p>\n

C = a2<\/sub>cos\u03b812 <\/sub>\u2212 a1<\/sub> \u2212 a4<\/sub><\/p>\n

The solution for Equation (7) can be obtained as in the four-bar example.<\/p>\n

If we are to determine s34<\/sub>\u00a0as a function of\u00a0q<\/span>12<\/sub>, the loop closure equation and its complex conjugate can be written in the form:<\/p>\n

(a4<\/sub>\u00a0+ is43<\/sub>)ei\u03b814<\/sub><\/sup> = a2<\/sub>ei<\/sup>\u03b812<\/sub><\/sup> \u2212 a1<\/sub><\/p>\n

(a4<\/sub> \u2212 is43<\/sub>)e\u2212i\u03b814<\/sub><\/sup> = a2<\/sub>e\u2212i<\/sup>\u03b812<\/sub><\/sup> \u2212 a1<\/sub><\/p>\n

Multiplying the two equations yields:<\/p>\n

s43<\/sub>2<\/sup> = a1<\/sub>2<\/sup> + a2<\/sub>2<\/sup> \u2212 a4<\/sub>2<\/sup> \u2212 2a1<\/sub>a2<\/sub>cos\u03b812<\/sub><\/p>\n

In general, for closed form solutions, utilisation of half-tangent form for the trigonometric functions is preferred. The main reason is that the inverse sine and cosine functions are double valued and usually the inverse functions in calculators or computers yield only one of the values as the output (for y1<\/sub> = cos-1<\/sup>(x), y2<\/sub> = sin-1<\/sup>(x), y3<\/sub> = tan-1<\/sup>(x), the range of function values are 0< yl<\/sub>\u00a0<\u00a0p<\/span>, \u2212p<\/span>\/2< y2\u00a0<\/sub><\u00a0p<\/span> \/2 and \u2212p<\/span>\/2 < y3<\/sub><\u00a0p<\/span>\u00a0\/2 for any value of x). If one of these inverse functions are used, one must first determine the quadrant in which the variable angle lies for each position and then change the value obtained by adding or subtracting p. Because of this calculation difficulty, either half tangent form of the trigonometric functions or double argument inverse tangent (e.g.\u00a0q<\/span>\u00a0= tan-1<\/sup>(x, y)) or polar-to-rectangular conversion is used.<\/p>\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n

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

3.7 Position Analysis of Mechanisms By Means of Complex Numbers We have seen that complex number utilisation is a simple and powerful technique for expressing the loop closure equations. The analytical or numerical solution of planar mechanisms can be easily …<\/p>\n

3-7<\/span> Devam\u0131n\u0131 Oku »<\/a><\/p>\n","protected":false},"author":7747,"featured_media":0,"parent":1950,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"full-width-page.php","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-2137","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2137","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=2137"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2137\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1950"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/media?parent=2137"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}