Note
You can download this example as a Python script: :jupyter-download:script:`configuration` or Jupyter Notebook: :jupyter-download📓`configuration`.
Consider the linkage shown below:
a) Shows four links in a plane A, B, C, and N with respective lengths l_a,l_b,l_c,l_n connected in a closed loop at points P_1,P_2,P_3,P_4. b) Shows the same linkage that has been seperated at point P_4 to make it an open chain of links.
This is a planar four-bar linkage with reference frames N,A,B,C attached to each bar. Four bar linkages are used in a wide variety of mechanisms. One you may be familiar with is this rear suspension on a mountain bicycle:
Four bar linkage shown in blue, red, orange, and green used in the rear suspension mechanism of a mountain bicycle.
A four bar linkage is an example of a closed kinematic loop. The case of :numref:`configuration-four-bar` there are two vector paths to point P_4:
\bar{r}^{P_4/P_1} = & l_n \hat{n}_x \\
\bar{r}^{P_4/P_1} = & \bar{r}^{P_2/P_1} + \bar{r}^{P_3/P_2} + \bar{r}^{P_4/P_3} = l_a\hat{a}_x + l_b\hat{b}_x + l_c\hat{c}_x
For the loop to close, the two vector paths must equate. We can resolve this by disconnecting the loop at some location, P_4 in our case, and forming the open loop vector equations to points that should coincide.
.. jupyter-execute:: import sympy as sm import sympy.physics.mechanics as me me.init_vprinting(use_latex='mathjax')
Setup the variables, reference frames, and points:
.. jupyter-execute::
q1, q2, q3 = me.dynamicsymbols('q1, q2, q3')
la, lb, lc, ln = sm.symbols('l_a, l_b, l_c, l_n')
N = me.ReferenceFrame('N')
A = me.ReferenceFrame('A')
B = me.ReferenceFrame('B')
C = me.ReferenceFrame('C')
A.orient_axis(N, q1, N.z)
B.orient_axis(A, q2, A.z)
C.orient_axis(B, q3, B.z)
P1 = me.Point('P1')
P2 = me.Point('P2')
P3 = me.Point('P3')
P4 = me.Point('P4')
SymPy Mechanics will warn you if you try to establish a closed loop among a set of points and you should not do that. Instead you will establish positions among points on one open leg of the chain:
.. jupyter-execute:: P2.set_pos(P1, la*A.x) P3.set_pos(P2, lb*B.x) P4.set_pos(P3, lc*C.x) P4.pos_from(P1)
Now, declare a vector for the other path to P_4:
.. jupyter-execute:: r_P1_P4 = ln*N.x
Now we can form the left hand side of the following equation:
\bar{r}^{P_4/P_1} - \left( \bar{r}^{P_2/P_1} + \bar{r}^{P_3/P_2} + \bar{r}^{P_4/P_3} \right) = 0
Using :external:py:meth:`~sympy.physics.vector.point.Point.pos_from` for the open loop leg made of points and the additional vector:
.. jupyter-execute:: loop = P4.pos_from(P1) - r_P1_P4 loop
This "loop" vector equation must equate to zero for our linkage to always be a closed loop. We have a planar mechanism, so we can extract two scalar equations associated with a pair of unit vectors in the plane of the mechanism:
.. jupyter-execute:: fhx = sm.trigsimp(loop.dot(N.x)) fhx
.. jupyter-execute:: fhy = sm.trigsimp(loop.dot(N.y)) fhy
For the loop to close, these two expressions must equal zero for all values q_1,q_2,q_3. These are two nonlinear equations in three time varying variables. A solution, sometimes analytically but likely only numerical, can be found if we solve for two of the time varying variables. For example, q_2 and q_3 can be solved for in terms of q_1. We would then say that q_2 and q_3 depend on q_1. These two equations are called holonomic constraints, or configuration constraints because they constrain the kinematic configuration to be a loop. Holonomic constraints take the form:
\bar{f}_h(q_1, \ldots, q_n, t) = 0 \textrm{ where } \bar{f}_h \in \mathbb{R}^M
These constraints are functions of configuration variables: time varying angles and distances. In our case of the four-bar linkage:
\bar{f}_h(q_1, q_2, q_3) = 0 \textrm{ where } \bar{f}_h \in \mathbb{R}^2
In SymPy, we'll typically form this column vector as so:
.. jupyter-execute:: fh = sm.Matrix([fhx, fhy]) fh
If you consider a set of v points, P_1,P_2,\ldots,P_v that can move unconstrained in Euclidean 3D space, then one would need 3v constraint equations to fix the points (fully constrain the motion) in that Euclidean space. For the four points in the four-bar linkage, we would then need 3(4)=12 constraints to lock all the points fully in place. The figure below will be used to illustrate the general idea of constraining the configuration of the four bar linkage.
a) Four points in 3D space, b) four points constrained to 2D space, c) points are fixed to adjacent points by a fixed length, d) the first point is fixed at O in two dimensions, e) the fourth point is fixed in the y coordinate relative to O.
Starting with a), there are the four points in 3D Euclidean space that are free to move. Moving to b), each of the four points can be then constrained to be in a plane with:
\bar{r}^{P_1/O}\cdot\hat{n}_z = 0 \\
\bar{r}^{P_2/O}\cdot\hat{n}_z = 0 \\
\bar{r}^{P_3/O}\cdot\hat{n}_z = 0 \\
\bar{r}^{P_4/O}\cdot\hat{n}_z = 0
where O is a point fixed in N. This applies four constraints leaving 8 coordinates for the planar location of the points. Now at c) we constrain the points with:
|\bar{r}^{P_2/P_1}| = l_a \\
|\bar{r}^{P_3/P_2}| = l_b \\
|\bar{r}^{P_4/P_3}| = l_c \\
|\bar{r}^{P_4/P_1}| = l_n
These four constraint equations keep the points within the specified distances from each other leaving 4 coordinates free. In d) point P_1 is fixed relative to O with 2 scalar constraints:
\bar{r}^{P_1/O}\cdot\hat{n}_x = 0 \\
\bar{r}^{P_1/O}\cdot\hat{n}_y = 0
Finally in e), P_4 is constrained with the single scalar:
\bar{r}^{P_4/P_1} \cdot \hat{n}_y = 0
Notice that we did not need \bar{r}^{P_4/P_1} \cdot \hat{n}_x = 0, because :numref:math:`length-constraint` ensures the x coordinate of P_4 is in the correct location.
These 11 constraints leave a single free coordinate to describe the orientation of A, B, and C in N. When we originally sketched :numref:`configuration-four-bar` most of these constraints were implied, i.e. we drew a planar mechanism with points P_1 and P_4 fixed in N, but formally there are 12 coordinates needed to locate the four points and 11 constraints that constrain them to have the configuration of a four-bar linkage.
A general holonomic constraint for a set of v points with Cartesian coordinates is then ([Kane1985]_ pg. 35):
f_h(x_1, y_1, z_1, \ldots, x_v, y_v, z_v, t) = 0
We include t as it may also be possible that the constraint is an explicit function of time (instead of only implicit, as seen above).
If a set of v points are constrained with M holonomic constraints then only n of the Cartesian coordinates are independent of each other. The number of independent coordinates is then defined as ([Kane1985]_ pg. 37):
n := 3v - M
These n independent Cartesian coordinates can also be expressed as n functions of time q_1(t),q_2(t),\ldots,q_n(t) in such a way that the constraint equations are always satisfied. These functions q_1(t),q_2(t),\ldots,q_n(t) are called generalized coordinates and it is possible to find n independent coordinates that minimize the number of explicit constraint equations needed to describe the system's configuration at all times t. These generalized coordinates are typically determined by inspection of the system and there is a bit of an art to choosing the best set. But you can always fall back to the formal process of constraining each relevant point.
Take this simple pendulum with points O and P as an example:
If the pendulum length l is constant and the orientation between A and N can change, then the location of P relative to O can be described with the Cartesian coordinates x and y. It should be clear that x and y depend on each other for this system. The constraint relationship between those two coordinates is:
x^2 + y^2 = l^2
This implies that only one coordinate is independent, i.e. n=1. More formally, the two points give 3v=3(2)=6 and there are 2 constraints for the planar motion of each point, 2 constraints fixing O in N, and 1 constraint fixing the distance from O to P, making M=5 and thus confirming our intuition n=6-5=1.
But there may be functions of time that relieve us from having to consider Eq. :math:numref:`pendulum-length-constraint`. For example, these two coordinates can also be written as as functions of the angle q:
x = l\cos q \\ y = l\sin q
and if we describe the configuration with only q, the constraint is implicitly satisfied. q is then a generalized coordinate because it satisfies n=1 and we do not have to explicitly define a constraint equation.
Now, let's return to the four-bar linkage example in :numref:`configuration-four-bar` and think about what the generalized coordinates of this system are. We know, at least intuitively, that n=1 for the four bar linkage. The four-bar linkage in :numref:`configuration-four-bar` is described in a way that assumes a number of constraints are fulfilled, such as Eqs. :math:numref:`planar-constraints` and :math:numref:`p1-constraint`, so we do not have to formally consider them.
Exercise
Are q_1,q_2,q_3 generalized coordinates of the four-bar linkage? If not, why?
Solution
Any one of the q_1,q_2,q_3 can be a generalized coordinate, but only one. The other two are depdendent due to the two constraints. We started with three coordinates q_1,q_2,q_3 describing the open chain P_1 to P_2 to P_3 to P_4. Then we have two scalar constraint equations, leaving n=1. Thus we can choose q_1, q_2, or q_3 to be the indepdendent generalized coordinate.
If we take the formal approach, starting with four unconstrained points, we need 11 constraints to describe the system, but if we select generalized coordinates to describe the system we only need 2 constraint equations (Eq. :math:numref:`four-bar-constraints`)! This simplifies the mathematical problem description and, as we will later see, is essential for obtaining the simplest forms of the equations of motion of a multibody system.