{"id":2146,"date":"2022-03-13T12:43:49","date_gmt":"2022-03-13T12:43:49","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=2146"},"modified":"2022-12-06T02:18:10","modified_gmt":"2022-12-06T02:18:10","slug":"3-9","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch3\/3-9\/","title":{"rendered":"3-9"},"content":{"rendered":"
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3.9<\/b> Iterative Solution of the Loop Closure Equations<\/h1>\n

In the previous sections we were able to solve the loop closure stepwise or in closed form. Usually such a solution can be found if in each loop closure equation there are two unknowns. In cases where the loop closure equations are coupled i.e. there are more than two unknowns in each loop closure equation, simultaneous solution of the loop closure equations will be required, in which case a search for a stepwise or closed form solution is not practical and iterative solutions are used.<\/p>\n

If we look at the loop closure equations, there is a set of nonlinear algebraic equations in the form:<\/p>\n\n\n\n\n
Fj<\/sub>(x<\/strong>) = 0<\/td>\nj = 1, 2, 3, …, n<\/td>\n(1)<\/td>\n<\/tr>\n
<\/td>\nx<\/strong> = [x1<\/sub>\u00a0x2<\/sub>\u00a0x3<\/sub> … xn<\/sub>]T<\/sup><\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Where in mechanism analysis Fk<\/sub>(x<\/strong>) are the loop equations written in complex numbers or in terms of their real and imaginary parts and\u00a0x <\/strong>is the vector formed by the position variables such as \u03b81j<\/sub>\u00a0or sij<\/sub>. When the set of equations and their derivatives are continuous, a simple numerical iterative method known as\u00a0Newton-Raphson method<\/em><\/strong>\u00a0may be used to find a solution of such an equation set. There are other more elegant numerical methods which can be used but this is beyond the topic (any textbook on numerical methods will contain a chapter on the solution of nonlinear equations).<\/p>\n

Let us consider a single equation F(x) = 0, that is nonlinear in the scalar variable x. Let x = x* be the solution of this equation and let x(k)<\/sup> be an approximation to x*. The Taylor series expansion of F(x) about x = x(k)<\/sup>\u00a0is:<\/p>\n\n\n\n
F(x) = F(x(k)<\/sup>) + Fx<\/sub>(x(k)<\/sup>)(x \u2212 x(k)<\/sup>) + (higjer order terms)<\/td>\n(2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where Fx<\/sub> \u00a0is the derivative of the function with respect to x, Fx<\/sub>(x) = dF(x)\/dx.<\/p>\n

Let x = x(k+1)<\/sup>\u00a0be an improved approximation that is to be determined from the condition:<\/p>\n\n\n\n
F(x(k+1)<\/sup>) \u2248\u00a0F(x(k)<\/sup>) + Fx<\/sub>(x(k)<\/sup>)(x(k+1)<\/sup> \u2212 x(k)<\/sup>) = 0<\/td>\n(3)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

If x(k+1)<\/sup> \u2212 x(k)<\/sup> is small, higher order terms can be neglected. If Fx<\/sub>(x(k)<\/sup>) is not equal to zero, equation (3) can be solved for x(k+1)<\/sup>\u00a0as:<\/p>\n\n\n\n
x(k+1)<\/sup> = x(k)<\/sup>\u00a0\u2212 F(x(k)<\/sup>)\/Fx<\/sub>(x(k)<\/sup>)<\/td>\n(4)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Equation 4 is the basis for Newton-Raphson method (this is also known as the recursive formula). Since higher order terms in Taylor series has been neglected, x(k+1)<\/sup>\u00a0will not be the exact solution. If the problem is well behaving , x(k+1)<\/sup>\u00a0will be a better approximation, and when this equation is applied repetitively, at an mth step,\u00a0 F(x(m)<\/sup>) < \u03b5, where \u03b5\u00a0is a very small quantity which may be assumed to be equal to zero for the engineering application. Hence Newton-Raphson algorithm can be summarised as:<\/p>\n