Hi
I’m trying to make a rotationally symmetrical polyhedron.
Starting point is a hexagon base with six pentagons attached, I need to rotate these pentagons upwards so that their edges connect to make a bowl.
I’ve come close by trial and error but its very tedious and not quite perfect.
I’ve tried to calculate the angle between the base and the pentagons, but that was clearly beyond my math skills.

How is this done?

The next step would be to create six new planes defined by the upper edges of the pentagons extending up to the intersection of the blue axis.

I’d like to write more about this but I don’t currently have the time. The following link is one of my go to websites whenever I need to reference information on this sort of thing.

I’ll try to post up my thoughts when I can, but it’s unlikely that I’ll have much more to add above and beyond what others will be able to provide in their comments. This forum has a few experts on this topic.

Hi Jim, thanks for the link.
Unfortunately, that only involves known angles and glosses over the part I’m interested in by saying “import an icosahedron”

I did this once upon a long time ago, in order to make a soccer ball…but I honestly can’t remember exactly how. One thing I would say is that I don’t think you can make a polyhedron using pentagons and hexagons in that way. The shape is actually a truncated icosahedron. I think the next level up (after you have hoisted the six pentagons up) will only allow a non-planar shape. I think you need to alternate the two shapes as seen in the pic.

This is a model explaining how to construct one from scratch. It was posted to the Sketchucation forum many years ago by caddict. build icosa.skp (185.7 KB)

Thanks @DaveR
I found a tutorial using those arcs but they never explained how those were derived, they simplified by going to an opposite corner, which makes it parallell to the line of rotation IF you have a pentagon base, but as I have a hexagon I got a few degrees off.
Now that I understand the arc concept that will work out beautifully.

@AlanF
The shape I’m trying to reproduce is more of a six sided pyramid with a rounded bottom so to say.

I was intrigued by these trumpet mutes and thought of trying to make one for myself

The rest of the Dave’s tutorial will make it so much easier to bevel the pieces.

To be honest, I think I’d approach drawing that mute differently. I’d draw the overall shape and pry out the facets to get the angles. that would be much easier than figuring out the rotation angles. I’ll see if I can make an example.

The basics would be two tapered hexagonal prisms, one rotated 30° relative to the other and intersected. You’ll wind up with something like this.

Yep, would definitely do the mutes top-down. If not, then the angle of the pentagonal mitre would dictate the length of the taper, as it has to be coplanar…not something you could visualise in advance.

And I spent all of last evening making my little bowl.

You’re perfectly right, the upper planes are completely dictated by the pentagons and I was worried that the angles of a regular pentagon was wrong, that was what I was trying to find out. It’s now clear that they are not regular pentagons at all

Je découvre ce sujet avec un peut de retard.
Mais mon modèle joint peut-il être utile pour montrer qu’il est possible de déterminer avec précision le point d’intersection commun à deux pentagones.

Pardonnez moi pour mon utilisation du Français, mais mes connaissance en Anglais sont tellement nulle, que je n’ose me lancer dans une traduction, que je ne pourrais d’ailleurs faire qu’avec l’aide de Google … Polyèdre.skp (1.1 MB)

Mettez le pentagone en bas, pas l’hexagone. Alignez le bord de l’hexagone et le faire tourner par 37.3775 degrés.
Suivez le skp.
Le motif est toujours une fleur … avec le pentagone au centre et les hexagones que les pétales. Regardez à nouveau l’image de la balle de football.

J’ai posté cette réponse seulement pour tenter d’apporter une réponse à Ryden, en essayant de démonter qu’il est possible de définir le point d’intersection avec précision, et sans trop de difficultés.
Bien entendu, pour modéliser un icosaèdre tronqué, le pentagone doit être placé en bas (ou en haut…).
Quand à la valeur de l’angle que vous indiquez, je pense qu’il est indispensable d’être plus précis, en appliquant une valeur de 37,377368140649695 degrés. En effet, comme vous pouvez le constater sur votre skp (sur lequel je me suis permis d’ajouter quelques cotes), les dimensions sont quelque peu inégales. trunc_icos_2.skp (62.5 KB)
Pour modéliser un icosaèdre tronqué, la méthode que j’utilise dans la petite vidéo, visible depuis le lien Google Drive ci-après, me semble plus rapide et précise.
(pardonnez moi, mais ne connaissant pas la langue de Shakespeare, je n’ai pas mis de son). https://drive.google.com/file/d/0BwevoKGYYXB3NHRYYWlWeGhEXzg/view?usp=sharing