NOTE: this is a still image, not a movie!

**MN-analysis**

This diagram shows the result of shear *M* then *N* on square ABCD.

The diagram contains several 30˚60˚90˚ triangles so we can find the lengths of various line segments with relative ease.

For example, if we let the square have sides of length 1 unit, then, in the 30˚60˚90˚ triangle A'GD, A'D=2 while in the 30˚60˚90˚ triangle C''CD, C''D = 0.5. So the final image of ABCD is a 2 by 0.5 rectangle.

Also, in the 30˚60˚90˚ triangle A'GD, DG = √3 = AA', which ties in with the fact that under shear *M* (invariant line *m*, shear factor √3), the distance moved by point A is AD x √3.

Similarly, we can use the 30˚60˚90˚ triangle CC''D to find the shear factor for shear *N* by considering the distance moved by C (ie the distance CC'') and the distance of C from the invariant line *n*.

Next, we take a static view of shear *N* then *P*: NP-analysis.