We only need to apply it to a circle and rewrite the solver in C. This will give us a visualizer for a single frequency.
Then we run FFT, run the solver for all frequencies, observe that the final solution is a linear combination of the individual solutions and apply the rainbow coloring to corresponding frequencies.
This solver needs to be kinda fast to run at 15 fps, but luckily, different frequencies can be solved in parallel. Most likely, changing the input frequency a bit will change the solution only a little, and so we could pre-solve the frequency range with sufficient density, cache them and rapidly derive actual solutions by interpolation.
Bonus points for using complex numbers. The boundary condition in a singing bowl is u(x0, t) = A sin(Bt) where x0 denotes the circular boundary. Since real numbers are boring, we could expand the problem into the complex plane: u(x0, t) = A exp(iBt). In this case the solution u(x, t) would be in complex plane also, where the absolute value |u| is the amplitude, or pixel opacity on our visualization, and the angular coordinate arg u would be maybe color of that pixel?
This seems certainly possible. But I don't know how would you display all of the frequencies that are present in music at the same time. Usually, they only play one or two frequencies and look at the result. Mash them all together? Maybe can be tried in the future. I was thinking to do something "inspired by" this cymatics instead (when I get around to it). Thanks for all the interesting thoughts!!
dchyrdvh|5 years ago
http://hplgit.github.io/wavebc/doc/pub/._wavebc_cyborg001.ht...
We only need to apply it to a circle and rewrite the solver in C. This will give us a visualizer for a single frequency.
Then we run FFT, run the solver for all frequencies, observe that the final solution is a linear combination of the individual solutions and apply the rainbow coloring to corresponding frequencies.
This solver needs to be kinda fast to run at 15 fps, but luckily, different frequencies can be solved in parallel. Most likely, changing the input frequency a bit will change the solution only a little, and so we could pre-solve the frequency range with sufficient density, cache them and rapidly derive actual solutions by interpolation.
Bonus points for using complex numbers. The boundary condition in a singing bowl is u(x0, t) = A sin(Bt) where x0 denotes the circular boundary. Since real numbers are boring, we could expand the problem into the complex plane: u(x0, t) = A exp(iBt). In this case the solution u(x, t) would be in complex plane also, where the absolute value |u| is the amplitude, or pixel opacity on our visualization, and the angular coordinate arg u would be maybe color of that pixel?
avaku|5 years ago