Discussion Forum

Posted: 1 decade ago Dec 14, 2009, 11:09 a.m. EST

Hi I assume you men a a paraboloid of revolution of focal length F=5[m], hence a central radius of R=2*F=10[m]. Now in COMSOL the geometry curves are based on Weighted Bezier Splines, the 2nd order one for weight = 1 are in fact a parabola expression (see command.pdf V3.5a Ch4 "Advanced Geometry topics" "Rational Bezier Curves", pp516-524, for more on Bezier cures try wiki or http://www.cs.mtu.edu/~shene/COURSES/cs3621/NOTES/spline/NURBS/RB-conics.html or http://jtauber.com/blog/2008/11/08/from_focus_and_directrix_to_bezier_curve_parameters/ or http://mathworld.wolfram.com/BezierCurve.html or ...) a 3 point (P0,P1,P2) 2nd order Bezier curve with weight factors=1 are in fact the startingpoint P0 and ending point P2 of a parabola with tangeant of P0 along the line P1-P0 and at P3 along the line P2-P1 if you align these points correctly. Typically if the apex of the prabola is P0=0,0, then the next two points could naturally be (1,0) and (2,4) for a paraboly y=x^2 passing by (0,0) (1,1), (2,4) with a tangeant direction at (0,0) of (1,0) and a tangeant at (2,4) passing by (1,0) or slope of 1/4, hence with a parabolic axis along (1,0) and a focal length of "1/2". Now you need to scale this correctly for your case, see the article of "jtauber" above I believe it's the closest to your demand. So for COMSOL I would make a 2D axial-symmetric shape, and the revolve it into 3D. Now, where to put the points? You want a base of 0.5m in radius along r and 0.15m along z (?) and a parabola of focal length 5[m]. This means 2*5[m] hight (z) at r=10[m] or 0.1[m] at r=1[m] or 0.1/4=0.025[m] at r=1/2=0.5[m] hat is the rim height difference of the paraboly is some 25 mm. The slope at r=0.5m is 1:10 (2*0.5/10) this corresponds then to the three points Apex: (0,0) centre control point (0.25,0) and rim: (0.5,0.025), all with weights = 1. and the remaining 0.125[m] of your cylinder extending in -Z direction Pls check carefully I might have missed a scale factor, as I went through it rather quickly Good luck Ivar

Attachments:

• Parabola.mph

Posted: 1 decade ago Dec 15, 2009, 2:16 a.m. EST

Thanks a lot Ivar. It was really that simple. In fact, simpler than drawing a spherical mirror of similar dimensions. I wish I had paid more attention to conic section in my high school-:) -Simon

Posted: 1 decade ago Jan 12, 2010, 3:26 a.m. EST

Hello Ivar, I just created a a parabola, like you suggested before: Bezier 2nd degree, with the points (0/0), (1/0), (2/4), and the weight of 1. Now i have y(x)=x^2. How can i create a parabola with the formula: y(x)=ax^2, e.g. with a=1/44! I could really need some help on this. Thanks a lot! Moritz

Posted: 1 decade ago Jan 12, 2010, 12:25 p.m. EST

Hi, I used the scale function to scale it up/down, else you need to calculate the derivatives and user the correct point values, the three chosen were the first natural number I could thik of Good luck Ivar

Posted: 1 decade ago Jan 18, 2010, 10:28 a.m. EST

Hi Ivar, I think the construction of the parable succeded, but now I have another problem. Have look in the attachment! I want to create a surface, which looks like the one in the attachment. One of the curves is the parabola, the other a curve, which is quasi parallel to the parabola. This surface needs to be extruded. Do you have an idea, how to construct the second curve? In a normal CAD-Programm I would use the "offset" or "wall-thickness"-Function. Thanks for your help! Regards, Moritz

Attachments:

• g0cop0120100118155823.pdf

Posted: 1 decade ago Jan 19, 2010, 2:31 p.m. EST

Hi It depends on the final precision you want, I would suppose, but if you make the two sides of the wall with a line of the desired wall thickness and then link the two points with a 3 point spline (using the two defined) I beleive you can drag a reasonable shape with your mouse. Or you could try to select your parabola, doa Cntrl-C Cntr-V (or Cut&Paste) and then define the offset vector at a given angle (in one or two steps with further "move") And then finally reconnect the sides. Or import it from a CAD tool ;) Good Luck Ivar

Posted: 1 decade ago Jan 20, 2010, 4:53 a.m. EST

Thanks Ivar, I think I try importing a .dxf file. This might be the easiest way! Ragards, Moritz

Posted: 1 decade ago Jan 20, 2010, 10:16 p.m. EST

Hi understand well your choice, but if you are working in 2D note that the COMSOL geometrical primitives are rather powerfull and not that complex to learn to use via matlab, in 3D it becomes more combersome and length to write out, but still possible for simple geometries. Furthermore in 2D you can run parametrised geometries within the GUI Good luck Ivar