[Yt-dev] Radius Calculation

[Matthew Turk]

Hm, this looks wrong to me.  Clearly something I have not considered
-- over here myi mages look like the attached.  What's the value for
the domain width?  This is confusing to me, as I really thought it
would work!

Can you try with a projection?  It should work, right through the
domain, and it should tell us what we're looking for.  Or, jsut
straight-up slices of NewRadius.  I'll try some stuff too.

On Fri, Apr 17, 2009 at 8:48 PM, Britton Smith <brittonsmith at gmail.com> wrote:
> I just tried your script with some data I have here.  I've attached the
> images.  I can't really tell if this is right or not.  I think another way
> to test this would be to do a projection with a periodic sphere as the
> source object.  If you throw a sphere that overlaps the domain boundary to a
> projection, it should be quite clear if it's doing what it's supposed to.
> Does that sound right?
>
> Britton
>
> On Fri, Apr 17, 2009 at 8:17 PM, Matthew Turk <matthewturk at gmail.com> wrote:
>>
>> Oops, just as a note, there's an error in the paste -- the second plot
>> of "NewRadius" should be of "Radius".
>>
>> Thanks guys...
>>
>> -Matt
>>
>> On Fri, Apr 17, 2009 at 7:16 PM, Matthew Turk <matthewturk at gmail.com>
>> wrote:
>> > Hi guys,
>> >
>> > Can somebody review this?  I chatted with Jeff, and I believe it is
>> > correct for periodic boxes, but I need somebody to look at it before I
>> > commit it, as it is a rather crucial aspect of yt.  This changes the
>> > way radius is calculated and should now work with periodic boundary
>> > conditions.
>> >
>> > Here's the new field definition:
>> >
>> > def _NewRadius(field, data):
>> >    center = data.get_field_parameter("center")
>> >    DW = data.pf["DomainRightEdge"] - data.pf["DomainLeftEdge"]
>> >    radius = na.zeros(data["x"].shape, dtype='float64')
>> >    for i, ax in enumerate('xyz'):
>> >        r = na.abs(data[ax] - center[i])
>> >        radius += na.minimum(r, na.abs(DW[i]-r))**2.0
>> >    na.sqrt(radius, radius)
>> >    return radius
>> >
>> > the final argument in na.sqrt is the 'out' argument, to avoid temporary
>> > arrays.
>> >
>> > I've committed a sample script here:
>> >
>> > http://paste.enzotools.org/show/101/
>> >
>> > which loads up a data dump, calculates the radius from [0.1, 0.1, 0.1]
>> > at all points in the root grid, then plots the output.  It gives
>> > correct min & max values for the radius.
>> >
>> > If I can get somebody else to sign off on this, I will commit the
>> > necessary changes to _Radius and _ParticleRadius, as well as grid &
>> > point selection for spheres.  We'll then have a fully-periodic
>> > profiler, which will close #165.
>> >
>> > -Matt
>> >
>> > On Thu, Apr 16, 2009 at 11:39 AM, Matthew Turk <matthewturk at gmail.com>
>> > wrote:
>> >> Hi guys,
>> >>
>> >> I've been exploring the idea of changing radius to be correct for
>> >> periodic boxes.  Right now it is incorrect; each component (x,y,z)
>> >> should not have a distance greater than 0.5 * domain_size.  The
>> >> easiest way to do this would be:
>> >>
>> >> rx = min( abs( x - center_x ), abs( x - center_x - domain_x) )
>> >>
>> >> I think the best way to do this is to set up a NumPy ufunc
>> >>
>> >>
>> >> http://docs.scipy.org/doc/numpy/user/c-info.beyond-basics.html#creating-a-new-universal-function
>> >>
>> >> that accepts three arrays, along with the center and the domain width,
>> >> and then returns the radius.  What do you all think?  Alternatively,
>> >> I'm thinking maybe just a regular function that gets the arrays and
>> >> returns one would be better; the ufunc machinery is a bit complicated
>> >> and I might get confused.  Once I come up with it, will somebody be
>> >> able to look over my work?
>> >>
>> >> -Matt
>> >>
>> >
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