1. Grashofs criterion
One extreme configuration. If this configuration is reached, further anticlockwise rotation of crank is feasible without snapping the coupler. 3 4 2 1
Another extreme configuration. If this configuration is reached, further clockwise rotation is feasible without snapping the coupler.

2. Full rotation of crank (2)
One extreme configuration. If this configuration is reached, further anticlockwise rotation of crank is feasible without snapping the coupler. 3 l3 4 l4 2 1
l1 + l2
l1 + l2 < l3 + l4
Sum of lengths of two sides of a triangle is greater than the third.

3. Full rotation of crank (2)
Another extreme configuration. If this configuration is reached, further clockwise rotation is feasible without snapping the coupler. 3 l4 4 l3 2 1
l2 – l1
(l2 – l1 ) + l3 > l4
(l2 – l1 ) + l4 > l3
Sum of lengths of two sides of a triangle is greater than the third.

4. Summary of geometrical conditions for full rotation of link 2.
Full rotation of link(2)
Summary of geometrical conditions for full rotation of link 2
l1 + l2 < l3 + l4
(l2 – l1 ) + l3 > l4
(l2 – l1 ) + l4 > l3
Rearranging terms
l1 + l2 < l3 + l4
l1 + l4 < l3 + l2
l1 + l3 < l4 + l2

5. Full rotation of link(4)
Summary of geometrical conditions for full rotation of link 2
l1 + l2 < l3 + l4
(l2 – l1 ) + l3 > l4
(l2 – l1 ) + l4 > l3
Interchange subscripts 2 and 4
l1 + l4 < l3 + l2
(l4 – l1 ) + l3 > l2
(l4 – l1 ) + l2 > l3
Rearranging terms
l1 + l4 < l2 + l3
l1 + l2 < l3 + l4
l1 + l3 < l4 + l2
Identical with (2)!!!

6. Full rotation of link (3)4 5 l3 6 2 1
Construct a six bar mechanism. The two links are parallel to, and hence equal in length to, 2 and 3.
Hence irrespective of the configuration of 4, the four bar 2356 system remains a parallelogram.
As per construction,
complete rotation of 6
implies
complete rotation of 3.

7. Full rotation of link (3)4 5 l2 6 2 1
Consider the 4 bar system 6541.
Note that the motion of this mechanism is independent of 2 and 3 which with 6 and 5 form a continuously changing parallelogram.
Also note that we are effectively replacing 2 by 3 and 3 by 2 in this mechanism.
Hence a new set of conditions for full rotation of 6, and hence of 3, may be obtained by interchanging the subscripts 2 and 3 in the first set of equations.
The explanation follows.

8. Full rotation of link (3)4 5 l2 6 2 1
First we look at the links in terms of their roles.
In the new mechanism 6 is input in place of 2, 5 is coupler in place of 3, 4 is output in place of 4 (unchanged).
Hence we first replace all l2 by l6, all l3 by l5 , all l4 by l4 in the 3 inequations obtained earlier.
l1 + l6 < l5 + l4
(l6 – l1 ) + l5 > l4
(l6 – l1 ) + l4 > l5

9. Full rotation of link (3)4 5 l2 6 2 1
Now we look at the links in terms of their lengths.
l6 = l3, l5 = l2 , l4 = l4.
But we have already substituted
l2 = l6, l3 = l5 , l4 = l4 in the 3 inequations.
Hence effectively
l2 => l3, l3 => l2 , l4 = l4 in the 3 inequations.
l1 + l3 < l2 + l4
(l3 – l1 ) + l2 > l4
(l3– l1 ) + l4 > l2
l1 + l6 < l5 + l4
(l6 – l1 ) + l5 > l4
(l6 – l1 ) + l4 > l5

10. Full rotation of link(3)
Summary of geometrical conditions for full rotation of link 3
l1 + l3 < l2 + l4
(l3 – l1 ) + l2 > l4
(l3– l1 ) + l4 > l2
Rearranging terms
l1 + l3 < l4 + l2
l1 + l4 < l2 + l3
l1 + l2 > l3 + l4
Identical with (2) again!

11. Full rotation of links (2), (3) & (4)
l4 + l3 > l1 + l2 ………. (1)
l3 + l2 > l1 + l4 ………. (2)
l2 + l4 > l1 + l3 ………. (3)
Add (1) and (2) l3 > l1
Add (2) and (3) l2 > l1
Add (3) and (1) l4 > l1
link1 is the shortest link.
Sum of l1 and any other link is smaller than the sum of any other link. Hence
l + s < p + q

12. Hence link 3 cannot rotate fully about link 2
Rotation of link 3 about 2
Extreme configuration. If this configuration is reached, further anticlockwise rotation of link 3 is just feasible without snapping link 4. 3 4 2 1
l1 + l4 > l2 + l3
But as shown earlier
l3 + l2 > l1 + l4
Hence link 3 cannot rotate fully about link 2

13. Hence link 3 cannot rotate fully about link 4
Rotation of link 3 about 4
Extreme configuration. If this configuration is reached, further anticlockwise rotation of crank is just feasible without snapping the coupler. 3 4 2 1
l1 + l2 > l3 + l4
But as shown earlier
l1 + l2 < l1 + l4
Hence link 3 cannot rotate fully about link 4

14. l + s < p + q where s is the shortest link
Conclusions
l + s < p + q where s is the shortest link
All other links rotate fully about s.
Hence if the shortest link is grounded
Both 2 and 4 rotate fully about the ground
making it a Double Crank
l + s < p + q where s is the shortest link
Link 3 does not rotate fully about either 2 or 4.
Hence if link 3 is grounded
i.e. if the shortest link is made the coupler
Neither 2 nor 4 will rotate fully about the ground making it a Double Rocker
l + s < p + q where s is the shortest link
If link 2 (or 4) is grounded
1 rotates fully about 2 (or 4)
3 rotates partially about 2 (or 4)
making it a Crank Rocker