Given that this method uses interpolation to define the surface, can it still have high precision in defining surfaces?
Lockbrains
Not quite sure about what the "surface" means here. Is f(x) = 0 a suface that goes into the screen vertically?
Max
@simply_complicated It certainly can, but higher precision requires a finer grid. That might take a lot of storage if the grid is (for example) a volume that scales with N^3, but on the other hand the cost could be offset by using a multi-resolution grid like a quadtree.
There are also a few outside-the-box approaches to this problem like using multiple channels to simulate precise lines in font rendering.
@Lockbrains the example is 2D; the grid is what you would get if you evaluated f at each of the cells.
Given that this method uses interpolation to define the surface, can it still have high precision in defining surfaces?
Not quite sure about what the "surface" means here. Is f(x) = 0 a suface that goes into the screen vertically?
@simply_complicated It certainly can, but higher precision requires a finer grid. That might take a lot of storage if the grid is (for example) a volume that scales with N^3, but on the other hand the cost could be offset by using a multi-resolution grid like a quadtree. There are also a few outside-the-box approaches to this problem like using multiple channels to simulate precise lines in font rendering.
@Lockbrains the example is 2D; the grid is what you would get if you evaluated f at each of the cells.