Fairly simple 2D to 3D extrusion with rounding needed

Discussion in '3D Design Requests' started by teralux, Aug 15, 2013.

  1. teralux
    teralux New Member
    I'm great with 2D but still barely scratching the surface of 3D. My end goal is to have this design, but extruded and rounded such that the circles are spheres and the straight arms are actually more tubular in nature, connecting with the sides of the spheres in an organic looking way. It seems like an expert-level 3D artist should be able to accomplish this relatively quickly. Would anyone care to offer me a quote or any feedback on the feasibility of this? I can go into more detail if needed, but that's the basic gist.
     

    Attached Files:

  2. stonysmith
    stonysmith Well-Known Member Moderator
    What overall size do you want the item? What material would you want to print it in?

    If you were to download OpenSCAD from www.openscad.org, then this bit of code will generate your design.

    ====
    Code:
    $fn=64;
    stick_radius=.75;
    color([1,0,0])sphere(r=4,center=true);
    module wing(){
    translate([9,0,0]){
    translate([-5,0,0])rotate([0,90,0])cylinder(r=stick_radius,h=8,center=true);
    sphere(r=3,center=true);
    for (i=[-1,1])
    translate([9,i*5,0]){
    translate([-2.725,0,0])rotate([0,90,0])cylinder(r=stick_radius,h=3.1,center=true);
    translate([-5.5,-i*1.5,0])rotate([-i*45,90,0])cylinder(r=stick_radius,h=4.8,center=true);
    sphere(r=2,center=true);
    for (j=[-1,1])
    translate([6,-j*3,0]){
    translate([-1.5,0,0])rotate([0,90,0])cylinder(r=stick_radius,h=3.6,center=true);
    translate([-4,j,0])rotate([j*45,90,0])cylinder(r=stick_radius,h=3.4,center=true);
    sphere(r=1,center=true);
    }}}}
    color([1,0,0])
    for (i=[0:5])rotate([0,0,60*i])wing();
    
     

    Attached Files:

    Last edited: Aug 15, 2013
  3. INNERLEAF
    INNERLEAF Well-Known Member
    Is there a way to get the code to create a smoother interface between the cylinders as they intersect uncleanly in the STL you have uploaded.
     
  4. JACANT
    JACANT Well-Known Member
    Last edited: Aug 18, 2013
  5. teralux
    teralux New Member
    WOW, THANK YOU! You guys are fantastic. I certainly did not expect to have all free help! So the design is based on phi ratios / the fibonacci sequence. I plan on doing small brass prints between .75 and 1.5 inches in diamater, as well as alumide prints as large as 4 inches in diameter. All the angles are 51.854 degrees to emulate the Pyramid at Giza, and the differences from one iteration to the next are fibonacci proportions. So, starting with a "55" in the sequence as the largest iterations of the fractal and working backwards, one would continue to use whatever ratio is derived from dividing the next smallest number in the sequence, to determine the length of the next part. So, let's say the big sphere is diameter N, then the next smallest sphere would be N x 34/55 [or .61818], the third sphere's size would be obtained by multiplying the second sphere's size by 21/34 [or .61765], and the fourth/smallest sphere's size would be the size of the third's times 13/21 [or .61905]. The legs were determined the same way. If you started with my exact file, though, then it looks like I have gotten exactly what I wanted. SO AGAIN, THANK YOU GUYS FOR BEING AWESOME!
     
  6. teralux
    teralux New Member
    Okay, so I still really appreciate the assistance, but I guess I just realized that the proportions have not been maintained, as the widths of the arms have not been preserved. Any further input is welcome! :)
     
  7. JACANT
    JACANT Well-Known Member
    You can do the maths.
    If you use your original image and mark on it all relevant sizes including angles, diameters of spheres, width of tubes, length of tubes between centres of spheres; keeping in mind the Design Guidelines and minimum wall thickness of the materials you want. Repost the image with all these sizes, then I will do it.