NOTE: this is a still image, not a movie!
MN-analysis
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.
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