enizor2

Only if you use regular metal discs. Using diamond discs, you break through in less than a minute in those locks: [1]

Comparing with your CyclyngWeekly link:

- the Litelock X3 goes from 5min to cut only 50% of the shackle to 23s fully cut.

- the Abus Granit Super Extreme 2500 also could only be cut to 50% in 5min, while the diamond disc finished in 29s.

- the Hiplok D1000 couldn't be fully cut in 5min with 2 metal discs, it took 52s with a diamond one.

[1]: https://www.youtube.com/watch?v=pbdj3RRbLPQ.

Absolutely not. He refused to defend his classic title in April 2023, citing a lack of motivation for this format.

The "Jeans" controversy happened during the Rapid championship in December 2024, nearly two years after.

It's universally accepted that his streak Classic Championship ended because of his lack of motivation, not on technicalities.

What do you mean ? A K-d tree handles k dimensions. Generating a useful 2-D representation (=projection) of more dimensions is the hard part.

The BYTE_STREAM_SPLIT is a Z-order (also known as Morton encoding). As it is better a preserving locality, it usually performs really well compared to classical "orthogonal" orders. It also is really simple to implement as it boils down to interleaving bits or bytes.

> If you zoom in sufficiently at x = 0, f(x) = sin(x) looks indistinguishable from f(x) = x, whereas g(x) = cos(x) looks indistinguishable from g(x) = 1.

You can't use the plot as you only know the triangle definition yet. (And "looks indistinguidable" is rather handwavy).

> (also, sin(x) is negative approaching 0 from the left and positive approaching 0 from the right)

That only tells you that sin(0)=0

I quote the second paragraph of the Derivatives section. (which was edited to a better, but not yet enough, sin(h)≈h and cos(h)≈1 when h is close to zero).

I perfectly understand that around 0, sin(x) ~ x and cos(x) = 1 + o(x) but it isn't obvious geometrically, unlike what the article implies.

From my point of view, increasing radius / decreasing curvature only gets you sin(x) -> 0 ; cos(x) -> 1, but that isn't enough to obtain the derivatives.

I found a geometric proof in [1] but that part is the longest and hardest of the page. I was wondering whether the author found a clearer way to express is.

[1] https://www.mathsisfun.com/calculus/derivatives-trig-proof.h...

EDIT: after looking at 3B1B's video, the "small" triangle d(sinΘ) by dΘ figure would be a better way to explain the derivative, rather than an "not hard to see geometrically" approximation that isn't enough to conclude.

Could you expand on getting the differential equation from your definition? I don't really where to start from it.

I do not understand this consideration: > By considering a triangle with hypotenuse 1 and a very small “opposite” side, it’s not hard to see geometrically that sin(x)≈x and cos(h)=x when x is small

I fail to see how you can "see" finer than sin(h) -> 0 & cos(h) -> 1

From the limit definitions you actually need :

* (1-cos(h)) / h -> 0

* sin(h)/h -> 1

(which correspond to the derivatives at 0).