Centroid of a Linkage (Part I)

Point I in Figure 1 is an instantaneous centroid of the four-bar linkage. The centroid of a linkage is important in mechanical design because when the mechanism is subjected to a combined force that passes through the centroid, it will act like a rigid structure. That is, the system will not deform.

Figure 1

In a design related course, the class was asked to find both the fixed (space) centroid and the body (moving) centroid of a mechanical structure. The fixed centroid is the locus of the instantaneous centroid of the structure at different geometry configurations. The body centroid, on the other hand, is the position of the fixed centroid when viewed from the joint of the stricture (in the case of the four-bar linkage shown in Figure 1, the body centroid is viewed from points (e, f) or (g, h)).

Figure 2

The meaning of this definition will be apparent as we explore more on the subject in Part II. First, we will get familiar of the subject by an example of what the centroid locus look like when drawn in the same picture. Figure 2 is an example of the centroid calculated for the particular linkage. The body centroid (red) will move with the linkage (blue). At each given instant, one of the points of the body centroid will coincide with a point on the fixed centroid. Again, how the calculation is done will be covered in Part II.

Newton-Raphson Method: Application

This is an application of the Newton-Raphson method for finding the joint angles of a mechanism. This was adapted from a note I made in a class called Mechanics of Machines in my sophomore year. It has proven useful to me for numerous times since then. The Newton-Raphson method can be used on any closed-loop mechanisms, including non-inverted and inverted slider-crank mechanisms. It also converges relatively fast in many common cases which makes it such a convenient tool. Throughout the years the Newton-Raphson method one of my favorite tools as a ME student. 

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Consider the double-loop mechanism:

Link 1 is the input, i.e. the rotation angle of link 1 is specified. Since this mechanism have one degree of freedom the positions of all other links can be found. Realizing that the two loops shown in the figure are in fact two four-bar linkages, we can solve them independently.

First of all, to find the joint angles of the 4-bar linkage made by links 1, 2, 3, and the ground link,notice that the vectors R₁, R₂, R₃, and R₆' have the following relationships:

$R_1+R_2-R_3-R_6=0$.

This is the loop equation for Loop 1. One must be aware of the direction of the vectors. Rewriting this equation in terms of link lengths r₁, r₂, r₃, r₆', and joint angles $\theta_1,\ \theta_2,\ \theta_3,$ and $\phi_6'$, it becomes:

$r_1\cos\theta_1+r_2\cos\theta_2-r_3\cos\theta_3-r_6\cos\phi_6=0$

$r_1\sin\theta_1+r_2\sin\theta_2-r_3\sin\theta_3-r_6\sin\phi_6=0$.

There are two unknowns, $\theta2$ and $\theta3$. The goal here is to find solutions to these two equations:

$F_1(\theta_2,\ \theta_3)=r_1\cos\theta_1+r_2\cos\theta_2-r_3\cos\theta_3-r_6\cos\phi_6$

$F_2(\theta_2,\ \theta_3)=r_1\sin\theta_1+r_2\sin\theta_2-r_3\sin\theta_3-r_6\sin\phi_6$

Here we will apply the Newton-Raphson method.

1. Find estimated values $\theta_2$ and $\theta_3$. Let's call them $\theta_2'$ and $\theta_3'$.
2. The partial differivative of F1 and F2 with respect to $\theta1$ and $\theta2$ are
   
   $\frac{\partial F_1}{\partial \theta_2}=-r_2\sin\theta_2$,    $\frac{\partial F_1}{\partial \theta_3}=r_3\sin\theta_3$
   
    $\frac{\partial F_2}{\partial \theta_2}=r_2\cos\theta_2$,   $\frac{\partial F_2}{\partial \theta_2}=-r_3\cos\theta_3$.
3. Since the values are estimated, it is highly unlikely that they will be the exact solution to F1 and F2. That is, F1 and F2 will not be zero when substituting these values into the equations. We write:
   
   $F_1(\theta_2',\ \theta_3')=\epsilon_1'$
   $F_2(\theta_2',\ \theta_3')=\epsilon_2'$.
4. Then Newton-Raphson method yields that
   
   $-r_2\sin\theta2'\times(\Delta\theta_2)+r_3\sin\theta_3\times(\Delta\theta_3)=\epsilon_1-\epsilon_1'$
   $r_2\cos\theta2'\times(\Delta\theta_2)-r_3\cos\theta_3\times(\Delta\theta_3)=\epsilon_2-\epsilon_2'$

   where $\epsilon_1=\epsilon_2=0$ in this case (the cost of F1 and F2).
5. Solving the equations in 4. gives the change for $\theta_2$ and $\theta_3$.
   .
6. The new estimation for the two values are $\theta_2''=\theta_2'+\Delta\theta_2$ and $\theta_3''=\theta_3'+\Delta\theta_3$, respectively. Using these new values we can start the process from step 3 and find new values again. The values will converge to the exact value of  $\theta_2$ and $\theta_3$. Once the values are withing tolerance, the procedure is terminated and we had our answer.

The procedure for Loop 2 is the same as that described above. Just use the Newton-Raphason method again to find  $\theta_4$ and $\theta_5$.