As remarked in the introduction , the four - bar linkage has many uses in mechanical engineering.

Let us propose a hypothetical situation: A four - bar linkage used as a robot arm in your widget factory has fallen off. It is a matter of (perhaps minor) importance, to you, the factory owner, to know whether or not that arm can be untangled from the heap it is lying in on the floor and can then be re - attached.

Thankfully for all those linkage - armed robots, a four - bar
linkage can *always* be opened from a tangled position. The
proof is a simple case analysis that is given here.

A four - bar linkage is an open planar non - crossing polygonal
chain with exactly four edges
*e _{1}*,

**Conjecture:** Given a four - bar linkage with any
vertex angles, the edges can be rotated about the joints in such a way
that all the vertex angles can be made to be 180 degrees. In other
words, given any bent four - bar linkage, it can be straightened to
its maximum length.

Given any four - bar linkage L, we can construct a triangle T with
vertices on each of the three joints of L. Two of the sides of T are
*e _{2}* and

Case 1: Both ends are outside the triangle. In this case, since the ends are non - crossing straight lines, one must have its "cap" (the non - vertex point defining the edge) farther than the other end when seen from T. In other words, the two straight edges cannot block each other from moving apart from each other in the direction away from the triangle.

We rotate the end with the farther cap to a straight angle, as it
is free to move in that direction. Then, the middle joint
*v _{2}* is free to open to a straight angle, because no
edge can possibly wrap around to the outside of the edge we want to
open. Then, three edges are in a straight line and so the fourth is
free to move anywhere, in this case to a straight angle.

Case 2: One end is inside the triangle T and the other is outside. In this case, the end outside T is free to rotate up to its limit when rotated away from T, and is free to rotate towards T as far as T's non - linkage edge. In this case, since straightening the end keeps it outside T, the edges can be straightened as in case 1, starting with the end outside T.

Case 3: Both ends are inside T. In this case, there are two
possibilities: One of the ends can move outside of the triangle, or
the edges are both blocked inside the triangle (as in the diagram).
If the first possibility is correct, then moving the edge outside of T
gives case 2. If the second possibility is correct, then the triangle
T can always be opened by opening vertex *v _{2}*.

To see this, note that, if the two ends are to block each other
from moving outside of T, at least one of them must be able to move
towards the linkage edge of T to which it is attached. The ends can
block each other in
that direction if and only if end A would hit the cap of end B moving
outward, while end B would have to become perpendicular to end A
before escaping, requiring it to shorten to some minimum distance.
This situation can occur only if end B is shorter than the distance
between its joint and the intersection of its attached edge
with any other edge (the hypoteneuse of the triangle in which B can move).
As well, joint *v _{2}* is bent at an acute angle.

In this case, end B can be rotated up to its limit, so it is
parallel to the edge it is attached to. Then joint
*v _{2}* is free to open. Eventually, due to Cauchy's Lemma , the non - linkage edge
of triangle T will be wide enough for end A to pass through and it can
be straightened. Then joint

Since all cases are covered, the proof is complete. The linkage in the diagram, by the way, is a case of a linkage that cannot be straightened from the ends (of importance in the next section, Open Problems).