boolean operation

[Alexis Berny]

Good day Basilisk user,

I'm working on a geometry, define from the resolution of a differential equation (see in my sandbox, http://basilisk.fr/sandbox/aberny/bubble/bubble.c for the definition of the geometry, and http://basilisk.fr/sandbox/aberny/bubble/burstingBubble.c#initialisation to see how I put it in the simulation).

I want to add a small circle to this geometry with a boolean operation (since I just want to do a reunion between 2 spaces).

According to older discussion, the boolean operation in basilisk should behave like in Gerris. The behaviour is describe here (http://gfs.sourceforge.net/wiki/index.php/GfsSurface). It's said that you need to have explicit formulation, which is not my case.

How can I manage to do that in basilisk? In other word, how can I work with boolean operation between an explicit function and a list of points (assuming it is possible)?

Best

-- 

Alexis Berny

[josé López-Herrera]

Alexis, 

Maybe I am saying an stupidity but...Why not to use two vof scalars: say f1[] for the point curve definition and f2[] defined from the equation surface

and to perform the boolean operation directly with the scalars f1[] and f2[]?

Cheers

Jose

I do not know if

[Alexis Berny]

Hi Jose,

I haven't thought about that, I will try it and let you know if it's work.

Thanks a lot

Best

-- 

Alexis Berny

[Stephane Popinet]

Hi Alexis, Jose,

You cannot use boolean operations between volume fraction fields, only
between distance fields. So the solution is simple, just define a first
distance field (from your geometry) using the distance() function. You
can then use the boolean operations (as described here:
http://gfs.sourceforge.net/wiki/index.php/GfsSurface), between this
distance field and other distance fields to define the resulting
distance function.

This distance function can then be used as input to fraction() to
compute the corresponding volume fraction.

cheers,

Stephane

[Alexis Berny]

Hi Stéphane,

I'm not sure if I get it correctly. Do you mean that an explicit circle equation like this one: x^2+y^2-r^2=0 can be consider as a distance field (if we consider a circle of radius r, in (0, 0)) ?

If so, then indeed the solution and the use of the boolean function is simple.

Best

--

Alexis BERNY

[Stephane Popinet]

> I'm not sure if I get it correctly. Do you mean that an explicit circle
> equation like this one: x^2+y^2-r^2=0 can be consider as a distance
> field (if we consider a circle of radius r, in (0, 0)) ?

yes.