Locks that work

This required a Maths answer, and the man for that is actually Tommy(TheBrick), but since he hasn't seen it, I've had to try to answer it myself.

If the circular bar has a diameter of 13mm, we can Imagine this:

But if the bar is square shaped then the 13mm width (as claimed by Abus) would give a through-diameter of 18.38mm (approx). The width of the square finds the diameter by multiplying the 13mm side by the squareroot of 2 (1.41421356...).

So even though the sidea are all 13mm, cutting at any angle at all, encounters 18mm of hardened steel.

I disagree.

The hardened steel encountered when cutting through the lock shackle is a volume of material which has to be removed, effectively the dimensions we care about are the cross-sectional area of the shackle and the width of the cutting tool. Assuming both are cut with the same cutting tool then we can unitise the width of the material removed and consider the cross-sectional area for comparison.

In this case, the cross-sectional area's are Pi * r^2 for the circle (3.14159 * 9^2) = 254.5mm2 and 13 ^ 2 for the square section shackle 169mm2.

So, for the same steel then for locks with the dimensions stated above, the round cross-section shackle should be tougher to cut through.

If you find this difficult to believe then consider cutting through both with a hacksaw (from a corner on the square i.e. cutting through the 18.4mm diagonal). Cutting the round section shackle you quickly encounter a greater section of material and your cutting rate will slow faster than cutting through the square section. Superimposing scale drawings of the two cross-sections should help elucidate.

So, at a cross sectional area of 132.7mm the Kryptonite Evo Mini has 80% of the hardened steel of the very tough sounding Abus Granit-X Plus 54. Seems to me that the lock manufacturers have managed to find a way to charge people more and use less material. Smart.