3 ms·
Observable is pushing it. Technically observable, sure, but ~1dpi misplaced is… unlikely to be noticeable. Plus, if that level of accuracy is required simply a
by photonerd 3y ago
Observable is pushing it. Technically observable, sure, but ~1dpi misplaced is… unlikely to be noticeable.
Plus, if that level of accuracy is required simply adding more nodes improves accuracy exponentially.
Not least, the “standard” approximation really should be updated in software, as modern 4-node approximations are down to 0.005%, almost an order of magnitude better.
- ars 3y agoIf you are misplaced by 1dpi, then your "1200 dpi" is actually 600 dpi. If you got a machine that's accurate to 1200 dpi presumably you want 1200 dpi and not 600.
- photonerd 3y agoThat’s not true in this case. It would still be 1200 dpi but the approximation of one curve is off by 1 dpi in some dimension. If all the lines were off, then yes, kinda. Ish.
- itishappy 3y agoOh, I completely agree! "Observable" is pushing the limits of the word past absurdity. 1200dpi would put the resolution of the system at just twice the wavelength of their laser. I feel like that's about the limit of what's possible due to physics [citation needed]. I'm fairly confident their actual performance is less.
- versteegen 3y ago> simply adding more nodes improves accuracy exponentially According to another comment, it actually improves at O(n^6).
- photonerd 3y agoI stand corrected! Polynomial..ly