5 ms·
Oh! This is applying FFT to the images! I thought it was going to be applying to functions f_n : [0,2pi] -> [0,1]^2 where f_n is the n-th level version of the
by drdeca 4y ago
Oh! This is applying FFT to the images! I thought it was going to be applying to functions
f_n : [0,2pi] -> [0,1]^2
where f_n is the n-th level version of the curve.
I guess in that view, higher order curves would have higher frequencies have higher amplitudes, on account of changing direction more often.
I guess that view of things wouldn’t have much visual appeal, as it would just be associating to each integer frequency, an individual vector. (I guess a 2d complex-valued vector?)
Hm, still, I think there would probably be uh, something to see in the directions of these vectors?
And to handle the complex-valued stuff I would think that if you moved from the e^{i t n} basis to the cosines and sines basis, that one could maybe thereby do-away with that? Or...
Well, yeah, that should be true, expressing a real valued periodic function as a weighted sum of sine and cosine functions doesn’t require any complex coefficients, but would doing so actually make the vectors which are the pairs of the coefficients for the two axiis, have a particularly clear meaning?
I would imagine the direction of these vectors should correspond to something like the different directions the curves move in, or differences between these directions.
Oh! One thing that might be nice visually,
If you got the Fourier coefficients for the n-th level of the curve, and then took only the first k coefficients and constructed the curve associated with that, then that might be cool looking.
I wonder, if you held k constant, but increased n, would the first k coefficients converge? If so, what would the curve from those first k coefficients look like?
- yarg 4y agoI've messed about with that sort of thing before: https://www.youtube.com/watch?v=pe4UAmb2wJw https://www.youtube.com/watch?v=pe4UAmb2wJw. I never got that far - I was primarily considering it as a point location definition for an image format, and I wanted a symmetrical curve. You can actually do that, if you relax the constraint that no grid point is shared (it's not really that big a problem, you just need to base the transform on the location between pixels, and not the centres of the pixels).
- EntrePrescott 4y agointeresting. Here's another video where one sees how the precision of this Moore curve drawn with epicycles varies with the frequency range: https://old.reddit.com/r/woahdude/comments/6udj3e/moore_curve_drawn_with_epicycles/ https://old.reddit.com/r/woahdude/comments/6udj3e/moore_curv...
- yarg 4y agoHe's using a different sample set than I did, I only used the centre pixel points, but (I believe) he has also included points in between. Hence his version retains the blockiness of the Moore curve, whereas mine does not.