After seeing this in LEGO House:
https://photos.app.goo.gl/bAAXaEebvrTMJfZZ6... I feeled urged to create a LDraw model!
I was too lazy to create a flexed axle template for LDCad so I used 12L rigid tubing instead, but it does the job!
(2019-06-04, 18:03)Philippe Hurbain Wrote: [ -> ]After seeing this in LEGO House: https://photos.app.goo.gl/bAAXaEebvrTMJfZZ6... I feeled urged to create a LDraw model!
I was too lazy to create a flexed axle template for LDCad so I used 12L rigid tubing instead, but it does the job!
Very cool, how did you get the positions of those axle joiners?
awesome.
I also wonder like Roland how you found the positions of the connectors in advance
(2019-06-04, 22:29)Steffen Wrote: [ -> ]awesome.
I also wonder like Roland how you found the positions of the connectors in advance
Problem is not so much to get connector positions (dodecahedron vertices coordinates are known:
https://en.wikipedia.org/wiki/Regular_do...oordinates), but to have them in the right orientation.
Here is the method I used:
- Get the distance between connectors. I used a visual method for that, placing two connectors with roughly the right orientation, then changing the distance until I got the desired length (96mm = 12L) for the flex part I put between them. Turns out the distance is 252 LDU.
- Calculate the radius r of the circumscribed sphere[url=https://en.wikipedia.org/wiki/Circumscribed_sphere][/url] (
https://en.wikipedia.org/wiki/Regular_do...Dimensions): 353.117LDU.
- Because of symmetry, a connector must be tangent to this sphere. So I placed a connector at y=-353.117, above the middle of top face.
- Then calculate the angle I must turn that connector around origin to place it on one top pentagon vertex. We get alpha = arcsin(r1/r)=37.377° (r1 is radius of circumscribed circle of pentagon with side a = a/2/sin(36°) )
[
attachment=3719]
After that, it's only a matter of duplication, rotation by 72° around world Y axis and rotation by 120° around of Y local axis of already placed connectors...
(2019-06-05, 7:01)Philippe Hurbain Wrote: [ -> ]Problem is not so much to get connector positions (dodecahedron vertices coordinates are known: https://en.wikipedia.org/wiki/Regular_do...oordinates), but to have them in the right orientation.
Here is the method I used:
- Get the distance between connectors. I used a visual method for that, placing two connectors with roughly the right orientation, then changing the distance until I got the desired length (96mm = 12L) for the flex part I put between them. Turns out the distance is 252 LDU.
- Calculate the radius r of the circumscribed sphere[url=https://en.wikipedia.org/wiki/Circumscribed_sphere][/url] (https://en.wikipedia.org/wiki/Regular_do...Dimensions): 353.117LDU.
- Because of symmetry, a connector must be tangent to this sphere. So I placed a connector at y=-353.117, above the middle of top face.
- Then calculate the angle I must turn that connector around origin to place it on one top pentagon vertex. We get alpha = arcsin(r1/r)=37.377° (r1 is radius of circumscribed circle of pentagon with side a = a/2/sin(36°) )
After that, it's only a matter of duplication, rotation by 72° around world Y axis and rotation by 120° around of Y local axis of already placed connectors...
Don't we all just LOVE maths?
Using a similar method (a bit complicated because all edges are not equivalent) I also made a truncated icosahedron (soccer balls, geodesic domes, fullerenes, etc...). Enjoy!
[
attachment=3724]
Witchcraft. I love it.
If I remember correct, Isn't there a way to inscribe all the Platonic solids into one another?
There is a cube inside the dodecahedron, and there's a octahedron inside the cube, and so on...
(2019-06-06, 21:24)Magnus Forsberg Wrote: [ -> ]Witchcraft. I love it.
If I remember correct, Isn't there a way to inscribe all the Platonic solids into one another?
There is a cube inside the dodecahedron, and there's a octahedron inside the cube, and so on...
Yah. Stellate the corner point of the faces out to a center point and then connect those points.
Pyramid -> Pyramid
Cube -> Octohedron
Dodecahedron -> Icosahedron.
Oh, yes.
Beautiful visualisation. I have to make something like that one day.
Thank you for finding it.
yes.
beautiful.
love it also.