Comments by "WhatAboutTheBee" (@WhatAboutTheBee) on "Machine Thinking" channel.

  1. From the peanut gallery: In Martin's companion video, we learn that each tooth corresponds to ~0.010 seconds (ten milliseconds). From observation then, each engraved number represents 100 milliseconds. If I've got this wrong, it is entirely my fault. I do not think the numbers have absolute meaning though. For each Beat Per Minute, there should then be a corresponding setting on this clutch. So it is simply a reference, like 90 BPM=6 and 140BPM=3, or something like that. Hopefully, the experts will correct me if I am wrong. This is just my understanding.
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  2.  @opendstudio7141  Yes on the correction for each numbered increment. 100 ms per numbered increment. Yet I am unsure of the maximum correction. Suppose we turn the clutch one full turn. That provides zero correction, but does advance the music by one turn. There is a subtle relationship here which is hard to wrap my head around. I think the maximum correction is 1/2 turn in either direction, for a +/- delta seconds from nominal. So +/- 0.300 seconds. At least, that is my understanding so far. [BTW the abbreviation for milliseconds is ms, not m/s. The m is a prefix for the unit, while the s is the unit. The "/" is an abbreviation for "per". Therefore, milli per second (m/s) is not a valid unit. This is meant in a friendly way, not in a grammar nazi sense. I hope you take it that way.]
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  3. At 17:17, you ask for the number of combinations the gauge blocks can form. That is defined by N factorial, denoted "N!". N is the number of blocks. For example, if N=5 blocks, then (N)! = 5×4×3×2×1= 120 combinations. This will over count duplicate lengths but every combination of blocks is accounted for. For example, we can first have .250 and then .125 as one arrangement, except first have .125 then .250 as a second arrangement. There are probably other combinations of blocks which reach .375 as well. Are those to be counted as a unique arrangement, or are you only looking for unique lengths. Let's suppose there are ~50 blocks in that set. 50! = 3 × 10 to the 64th power or 3 followed by 64 zeros. That's just a pinch more than 100,000. Ha! You'd have to write a bit of S/W to find block combinations that have unique lengths. Ditto eliminating different orders of the same block. No matter. The end result will be far and away much greater than 100K.
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