Box - Ball intersection in 2D

[aj-fleming]

The documentation for intersection reads

Compute the intersection of two geometries or domains g₁ and g₂ and apply function f to it. Default function is identity.

One might then expect to be able to run

orb = Meshes.Ball((0., 0.), 1.0)
square = Meshes.box((0.0, 0.0), (1.0, 1.0))
intersection(measure, orb, square)

which results in a StackOverflowError.

Perhaps "intersection" is not intended for this purpose?

[juliohm]

We really need a FAQ for this. There are multiple issues closed about the same stack overflow.

The pair Ball + Box is not implemented currently. What are you trying to achieve?

[aj-fleming]

If it's a known issue, would it be appropriate to add a warning in the documentation?

I would like to generate a Cartesian grid of data points, and clip it using a ball. Ideally, the segments would be cut by the ball and turn the cartesian mesh into a "mostly cartesian" mesh, with some Quadrangles converted into other NGons by the intersection with the ball.

Perhaps I'm approaching the problem completely wrong, I've only spent an afternoon with the package.

[juliohm]

If it's a known issue, would it be appropriate to add a warning in the documentation?

That would be great, though I feel that people will continue opening issues on GitHub. The problem lies in the fact that we exploit the symmetry of the intersection and write a fallback intersection(f, a, b) = intersection(f, b, a). When the fallback doesn't exist, Julia goes in an infinity loop flipping the arguments.

We tried to address this issue in the past, but the design became much more convoluted. We aborted the attempt. If you have a better suggestion that avoid code repetition, please feel free to submit a PR.

I would like to generate a Cartesian grid of data points, and clip it using a ball. Ideally, the segments would be cut by the ball and turn the cartesian mesh into a "mostly cartesian" mesh

I think you can already achieve this if you convert the Ball into a polygon:

using Meshes

b = Ball((0, 0), 1)

# sample points on boundary
s = sample(boundary(b), HomogeneousSampling(10))

# approximate ball with polygon
p = PolyArea(collect(s))

The intersection between the Quadrangles of the CartesianGrid and the PolyArea above is available.

Related discussion: #402