Do you mean to say subtract the profile from the pentagon??Using the solid editing, subtract the pentagon from revolved profile. Fillet the corners of the face only of the pentagon
Btw, that looks awesome!


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I thought I share this to you guys; It’s my first tutorial contribution to this forum.
I did this using 2000i. If you find it hard to follow; my apology.
The following figures are icosahedron and truncated icosahedron.
You can find the coordinates of these from this site.
http://en.wikipedia.org/wiki/icosahedron
http://en.wikipedia.org/wiki/Truncated_icosahedron
I used the regular icosahedron coordinates and truncate them but you can also used the coordinates of truncated icosahedron as well.
As you can see none of the centroid of the pentagons and hexagons are in any axes (x,y,z axis) as shown from top view.
I rotated the truncated icosahedron so that one of the pentagons is on the z-axis.
See top view with pentagon in red and hexagon in green
You have to manipulate the UCS so that you can use the regular rotate command. (You can use the 3d rotate as well but I prefer the 2d rotate)
Once you have that, you have to isolate the pentagon and hexagon to make a solid. You have to make a multiple copies first before you can isolate them. And you have to enlarge the image larger than the ball.
Isolate first the pentagon on the z-axis and erase the rest. Retrace the pentagon to create a polyline for extrusion. In this case I used the centroid of the pentagon, draw a circle and redraw using the polygon command. I also created a line (red) in z-axis from the centroid of pentagon to center of the ball at 0,0,0. This line also represents the radius of the ball (Ball size is from 27” to 28” in circumference). I also created two lines (in red) from the centroid to the vertex of the pentagon and from the vertex to the 0,0,0. This line represents the taper angle of the extrusion.
Draw another profile as region or boundary to represents the size of the ball or to shape the top face of the pentagon. (in magenta).
Extrude the pentagon with the tapered length and revolved the region.
Using the solid editing, subtract the pentagon from revolved profile. Fillet the corners of the face only of the pentagon. This fillet represents the stitching of the ball.
Next is the Hexagon, isolate the one on the YZ plane (see top view of icosahedron) then rotate it to the Z-axis. Do the same procedure as above for the pentagon. Once you have the solid, you have to rotate it back to their original location.
Array the hexagon.
You can manipulate the others by using copy, rotate and array.
This is the half of the soccer ball. The other half is the mirror image.
Hope you guys enjoyed the games in last world cup.
Larry
Do you mean to say subtract the profile from the pentagon??Using the solid editing, subtract the pentagon from revolved profile. Fillet the corners of the face only of the pentagon
Btw, that looks awesome!
very nice work! i'm going to try it out when i get some time
umm....for sleepy people like me, how exactly do we construct the truncated icosahedron again? i looked it up on wikipedia like you said, and then my mind just went
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its just one of those things that as soon as you see the solution, you're like 'd'oh! i knew that...
'
thanks![]()
Hyperactivity + Insomnia = me doing stupid things
Sorry
X, I believe you take a look at the ratios (golden ration=1.61then you can easily construct the three rectangles on each plane to make up the vertices of the icosahedron. Once you've done that simply slice the corners properly and then you'll notice you've made the proper amount of hexagons and octagons.
"This polyhedron can be constructed from an icosahedron with the 12 vertices truncated (cut off) such that one third of each edge is cut off at each of both ends."
oh, alright. i didnt know if there was a command or something that you cold just plug some stuff in and it would make it for you.
that would make a pretty nice little lisp
Hyperactivity + Insomnia = me doing stupid things
Sorry
X, if you'd like a dwg of the truncated icosahedron you could PM me with your e-mail address. Or if you'd like the satisfaction of doing it yourself that's just as well![]()


Hi guys,
Here is the permutations of coordinates of truncated icosahedron;
And this is the permutation of coordinates of regular icosahedron;Code:(0.00000000 1.00000000 4.85410197) (0.00000000 -1.00000000 -4.85410197) (0.00000000 1.00000000 -4.85410197) (0.00000000 -1.00000000 4.85410197) (1.00000000 4.85410197 0.00000000) (1.00000000 -4.85410197 0.00000000) (-1.00000000 4.85410197 0.00000000) (-1.00000000 -4.85410197 0.00000000) (4.85410197 0.00000000 1.00000000) (4.85410197 0.00000000 -1.00000000) (-4.85410197 0.00000000 1.00000000) (-4.85410197 0.00000000 -1.00000000) (2.00000000 4.23606798 1.61803399) (-2.00000000 4.23606798 1.61803399) (2.00000000 4.23606798 -1.61803399) (-2.00000000 4.23606798 -1.61803399) (2.00000000 -4.23606798 1.61803399) (-2.00000000 -4.23606798 1.61803399) (2.00000000 -4.23606798 -1.61803399) (-2.00000000 -4.23606798 -1.61803399) (4.23606798 1.61803399 2.00000000) (-4.23606798 1.61803399 2.00000000) (4.23606798 1.61803399 -2.00000000) (-4.23606798 1.61803399 -2.00000000) (4.23606798 -1.61803399 2.00000000) (-4.23606798 -1.61803399 2.00000000) (4.23606798 -1.61803399 -2.00000000) (-4.23606798 -1.61803399 -2.00000000) (1.61803399 2.00000000 4.23606798) (-1.61803399 2.00000000 4.23606798) (1.61803399 2.00000000 -4.23606798) (-1.61803399 2.00000000 -4.23606798) (1.61803399 -2.00000000 4.23606798) (-1.61803399 -2.00000000 4.23606798) (1.61803399 -2.00000000 -4.23606798) (-1.61803399 -2.00000000 -4.23606798) (1.00000000 3.61803399 3.23606798) (-1.00000000 3.61803399 3.23606798) (1.00000000 3.61803399 -3.23606798) (-1.00000000 3.61803399 -3.23606798) (1.00000000 -3.61803399 3.23606798) (-1.00000000 -3.61803399 3.23606798) (1.00000000 -3.61803399 -3.23606798) (-1.00000000 -3.61803399 -3.23606798) (3.61803399 3.23606798 1.00000000) (-3.61803399 3.23606798 1.00000000) (3.61803399 3.23606798 -1.00000000) (-3.61803399 3.23606798 -1.00000000) (3.61803399 -3.23606798 1.00000000) (-3.61803399 -3.23606798 1.00000000) (3.61803399 -3.23606798 -1.00000000) (-3.61803399 -3.23606798 -1.00000000) (3.23606798 1.00000000 3.61803399) (-3.23606798 1.00000000 3.61803399) (3.23606798 1.00000000 -3.61803399) (-3.23606798 1.00000000 -3.61803399) (3.23606798 -1.00000000 3.61803399) (-3.23606798 -1.00000000 3.61803399) (3.23606798 -1.00000000 -3.61803399) (-3.23606798 -1.00000000 -3.61803399)
And this is the lisp I found way back in 1991 from Cadence Magazine;Code:(0.0000000000 1.0000000000 1.6180339887) (0.0000000000 -1.0000000000 -1.6180339887) (0.0000000000 1.0000000000 -1.6180339887) (0.0000000000 -1.0000000000 1.6180339887) (1.0000000000 1.6180339887 0.0000000000) (-1.0000000000 -1.6180339887 0.0000000000) (1.0000000000 -1.6180339887 0.0000000000) (-1.0000000000 1.6180339887 0.0000000000) (1.6180339887 0.0000000000 1.0000000000) (-1.6180339887 0.0000000000 -1.0000000000) (-1.6180339887 0.0000000000 1.0000000000) (1.6180339887 0.0000000000 -1.0000000000)
Have fun and show us your team colors when you're done!Code:(defun C:pt_in1 (/ i file fp pt) (setvar "cmdecho" 0) (setq i 1) (while i (setq fp (findfile (getstring "\Input file: "))) (if file (progn (setq fp (open file "r")) (setq i nil) ) (princ "\nInvalid filename.") ) ) (setq pt 1) (while pt (setq pt (read-line fp)) (if pt (progn (setq pt (read pt)) (command "point" pt) ) ) ) (closed fp) (princ) )
I'm making another one that shows the maple leaf.
Larry
so how do we load the file into the lisp? i saved the coordinates in a txt file, but nothing i tried to load it into the lisp worked! what do i need to do?
Hyperactivity + Insomnia = me doing stupid things
Sorry
Yeah I did the same thing and it just tells me that it's an invalid input file. What extension do we need to save it as?
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LOL, you need to go buy a book on AutoLISP, which is the programming language behind AutoCAD. walk before you run.... lolOriginally Posted by X
Tannar Frampton | Facilities Engineering | Revit 2013
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