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


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

Again, the diagram contains several 30˚60˚90˚ triangles so we can find various lengths with relative ease.

We might want to show that B'A', the image of BA under shear N, is perpendicular to line n (so that line p must be chosen to be perpendicular to n). This can be done by considering the quadrilateral BB'A'K, which has right-angles at B and K, and showing that BB' = KA'.

We might then want to consider the distance moved by A' under shear P (ie A'A'') and the distance of A' from line p, to find the shear factor for shear P.

Finally, we look at a classic shearing task: PS.