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From: Sameer Regmi <regmisk@gm...>  20090820 06:01:53
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We are working on plotting mesh (in hermes2d: http://hpfem.math.unr.edu/projects/hermes2dnew/) We created a python function to plot mesh but initially without curves. Later we also tried to work on curved elements but we are having some problems. In the hermes2d examples curves are defined as [4, 7, 45] where 4,7 are vertices indices, and 45 is center angle. But in matplotlib we do not see any function to plot curve with these data, we have to convert them into something more useful to those functions available in matplotlib. For ex: 1) matplot.path porvides a way to plot curve with three points http://matplotlib.sourceforge.net/api/path_api.html#modulematplotlib.path p1 > starting Point p2 > control point p3 > end point iI we want to go this way we do have p1 & p3 but have to calclate p2 (control point) 2) matplot.patch.arc provides a way to plot an arc http://matplotlib.sourceforge.net/api/artist_api.html#modulematplotlib.patches but it needs > center of circle > start angle > end angle > radius of circle we have to calculate all these with data available to us We didn't find other ways of plotting curves using matplotlib...and in our view method 1 is the best way for us because currently we have implemented matplotlib.path to draw mesh without curves, and it will be easy to fill color this way. We tried the method 1 but the result was a garbled mesh. I would appreciate if anybody could guide us how we can use the matplotlib functions to draw curved elements if we are given information as described above. Thanks 
From: JaeJoon Lee <lee.joon@gm...>  20090823 04:12:55

On Thu, Aug 20, 2009 at 2:01 AM, Sameer Regmi<regmisk@...> wrote: > We tried the method 1 but the result was a garbled mesh Please describe what you did and why the result is wrong. The method 1 with quadratic bezier curve should be most straightforward and easy thing to do. Calculating the control points is also straight forward. While you may simply use matplotlib.bezier.get_intersection, you'd better come up with some optimized version since you need to calculate this for lots of positions. It is not clear how you're drawing path currently, but Line2D class is not suitable for bezier lines. You may use PathPatch class. Or you can create your own artist class (maybe this is what you meant by "implemented"). A simple version of bezier artist can be found in mpl_toolkits.axes_grid.axeslines.BezierPath (included in mpl 0.99). If you haven't, please take a look at the tutorial below. http://matplotlib.sourceforge.net/users/path_tutorial.html JJ 
From: Chris Barker <Chris.B<arker@no...>  20090824 00:46:02

Sameer Regmi wrote: > We are working on plotting mesh (in > hermes2d: http://hpfem.math.unr.edu/projects/hermes2dnew/) > In the hermes2d examples curves are defined as [4, 7, 45] where 4,7 > are vertices indices, and 45 is center angle. > 1) matplot.path porvides a way to plot curve with three points > http://matplotlib.sourceforge.net/api/path_api.html#modulematplotlib.path This is a Bezier spline  it can not exactly form a piece of a circle (though it can get pretty close). You can probably find the math somewhere for how to approximate a circle, but... > 2) matplot.patch.arc provides a way to plot an arc > http://matplotlib.sourceforge.net/api/artist_api.html#modulematplotlib.patches > > but it needs > > center of circle > > start angle > > end angle > > radius of circle > we have to calculate all these with data available to us Since this actually plots s circle, I think it's a better bet. It should be pretty straightforward coordinate geometry to find those parameters from the ones you have  and you only have to write that once! This page (or others like it) might help you get started: http://www.codecogs.com/reference/maths/analytical_geometry/the_coordinate_geometry_of_a_circle.php HTH, Chris  Christopher Barker, Ph.D. Oceanographer Emergency Response Division NOAA/NOS/OR&R (206) 5266959 voice 7600 Sand Point Way NE (206) 5266329 fax Seattle, WA 98115 (206) 5266317 main reception Chris.Barker@... 
From: John Hunter <jdh2358@gm...>  20090824 00:54:01

On Sun, Aug 23, 2009 at 7:45 PM, Chris Barker<Chris.Barker@...> wrote: > This is a Bezier spline  it can not exactly form a piece of a circle > (though it can get pretty close). You can probably find the math > somewhere for how to approximate a circle, but... somewhere like ... matplotlib.path.unit_circle (thanks to Michael D) @classmethod def unit_circle(cls): """ (staticmethod) Returns a :class:`Path` of the unit circle. The circle is approximated using cubic Bezier curves. This uses 8 splines around the circle using the approach presented here: Lancaster, Don. `Approximating a Circle or an Ellipse Using Four Bezier Cubic Splines <http://www.tinaja.com/glib/ellipse4.pdf>`_. """ if cls._unit_circle is None: MAGIC = 0.2652031 SQRTHALF = np.sqrt(0.5) MAGIC45 = np.sqrt((MAGIC*MAGIC) / 2.0) vertices = np.array( [[0.0, 1.0], [MAGIC, 1.0], [SQRTHALFMAGIC45, SQRTHALFMAGIC45], [SQRTHALF, SQRTHALF], [SQRTHALF+MAGIC45, SQRTHALF+MAGIC45], [1.0, MAGIC], [1.0, 0.0], [1.0, MAGIC], [SQRTHALF+MAGIC45, SQRTHALFMAGIC45], [SQRTHALF, SQRTHALF], [SQRTHALFMAGIC45, SQRTHALF+MAGIC45], [MAGIC, 1.0], [0.0, 1.0], [MAGIC, 1.0], [SQRTHALF+MAGIC45, SQRTHALF+MAGIC45], [SQRTHALF, SQRTHALF], [SQRTHALFMAGIC45, SQRTHALFMAGIC45], [1.0, MAGIC], [1.0, 0.0], [1.0, MAGIC], [SQRTHALFMAGIC45, SQRTHALF+MAGIC45], [SQRTHALF, SQRTHALF], [SQRTHALF+MAGIC45, SQRTHALFMAGIC45], [MAGIC, 1.0], [0.0, 1.0], [0.0, 1.0]], np.float_) codes = cls.CURVE4 * np.ones(26) codes[0] = cls.MOVETO codes[1] = cls.CLOSEPOLY cls._unit_circle = cls(vertices, codes) return cls._unit_circle 
From: Sameer Regmi <regmisk@gm...>  20090824 16:05:00
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Thank you Lee. Thank you Chris and John, the problem is solved. Chris, we did the method 1) as this was easier for us. Sameer On Sun, Aug 23, 2009 at 5:53 PM, John Hunter <jdh2358@...> wrote: > On Sun, Aug 23, 2009 at 7:45 PM, Chris Barker<Chris.Barker@...> > wrote: > > > This is a Bezier spline  it can not exactly form a piece of a circle > > (though it can get pretty close). You can probably find the math > > somewhere for how to approximate a circle, but... > > somewhere like ... matplotlib.path.unit_circle (thanks to Michael D) > > @classmethod > def unit_circle(cls): > """ > (staticmethod) Returns a :class:`Path` of the unit circle. > The circle is approximated using cubic Bezier curves. This > uses 8 splines around the circle using the approach presented > here: > > Lancaster, Don. `Approximating a Circle or an Ellipse Using Four > Bezier Cubic Splines <http://www.tinaja.com/glib/ellipse4.pdf>`_. > """ > if cls._unit_circle is None: > MAGIC = 0.2652031 > SQRTHALF = np.sqrt(0.5) > MAGIC45 = np.sqrt((MAGIC*MAGIC) / 2.0) > > vertices = np.array( > [[0.0, 1.0], > > [MAGIC, 1.0], > [SQRTHALFMAGIC45, SQRTHALFMAGIC45], > [SQRTHALF, SQRTHALF], > > [SQRTHALF+MAGIC45, SQRTHALF+MAGIC45], > [1.0, MAGIC], > [1.0, 0.0], > > [1.0, MAGIC], > [SQRTHALF+MAGIC45, SQRTHALFMAGIC45], > [SQRTHALF, SQRTHALF], > > [SQRTHALFMAGIC45, SQRTHALF+MAGIC45], > [MAGIC, 1.0], > [0.0, 1.0], > > [MAGIC, 1.0], > [SQRTHALF+MAGIC45, SQRTHALF+MAGIC45], > [SQRTHALF, SQRTHALF], > > [SQRTHALFMAGIC45, SQRTHALFMAGIC45], > [1.0, MAGIC], > [1.0, 0.0], > > [1.0, MAGIC], > [SQRTHALFMAGIC45, SQRTHALF+MAGIC45], > [SQRTHALF, SQRTHALF], > > [SQRTHALF+MAGIC45, SQRTHALFMAGIC45], > [MAGIC, 1.0], > [0.0, 1.0], > > [0.0, 1.0]], > np.float_) > > codes = cls.CURVE4 * np.ones(26) > codes[0] = cls.MOVETO > codes[1] = cls.CLOSEPOLY > > cls._unit_circle = cls(vertices, codes) > return cls._unit_circle > 