This watch was produced by the watchmaker Sinn for the German Special Forces (KSK): http://manufaktuhr.info/wp-content/uploads/2016/10/IMG_7455.jpg

In the picture there are two watches. The one with the green second hand is the one issued to the KSK. The other one is a limited production version for civilians (orange/red second hand). I find it surprising that this $3000 watch would be produced for the KSK as simply a novelty (or, maybe a joke). You can see the official KSK version has additional numbers 1-4 along the innermost part of the circle from 1-5 o’clock, roughly. Is it possible there is some way to correct for the variations to make this method more accurate, and hence the additional numbers printed on the official KSK version? Or, are there 70 KSK members lost in the woods?

]]>The tolerance of diameter has nothing to do with sphere-ness of the ball, but the diameter. Even if the tolerance was 10%, it would just be the difference of a ball that could be 2.03 to 2.47 inches in diameter, but could still be a perfect sphere with perfect smoothness.

I thought of a test that’s simple. The math is already done here as well. Let’s take the earth and convert it’s diameter into mm. Then the height of Everest, then divided by the scale. We’ll even go so far as to say the ocean is perfect smoothness and not count below the surface.

Earth diameter, we only need one: 12742km = 12,742,000m = 12,742,000,000mm

Everest height: 8,848m = 8,848,000mm

To get the Earth and Everest to scale, we need to divide Earth by 57.15mm ( 2.25 inches ) to get the scale and then devide Everest to that scale.

Scale divisor ( Earth diameter / Cueball diameter ):

12,742,000,000mm / 57.15mm = 222,957,130.359mm

Everest scale:

8,848,000mm / 222,957,130.359mm = 0.038541402438mm

So let’s do a test we’ve all done already:

Rub your finger over a roll of scotch tape to find the end. Scotch Tape is 1 mil or 0.0254mm, which is less than 1/10th of the roughness that Everest would add to the Earth Cueball.

I imagine it would be useful to do the same thing here that I did with the allRGB experiment at the larger target resolution: switch to C++ to speed things up.

]]>Q = (2n^2 + 8n – 17 + (-1)^n) / 8

So Q is the minimum number of evaluations of P that absolutely need to be made. One might notice that the algorithmic complexity mentioned in my previous comment is actually LESS than this. This is not an error. That is simply due to another slight optimization I made.

Basically, for all P(2, K), the 2nd line of the recurrence (the part that looks like “(1 – (k/m))[…..]”) cancels out.

So in my program I replaced the rule “P(1,K) = 1 – K” with “P(2, K) = (K / 2) * ((K / 2) * (2 – K) + (1 – (K / 2)) * (1 – K))” and simply disallowed all evaluations of the form P(1, K).

So now I can’t calculate the odds of the game failing when there is only one participant anymore. But that’s not interesting since that probability is always 100%. But I save up on a little bit more redundant calculations. The difference is negligible in practice, but I wanted to opt for the absolute minimum number of calculations required. And when calculating with rationals, instead of floating point numbers, the difference is actually measurable for very large values of N (though still negligible, since the numbers involved are very small anyway)

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