Figure with one face, one edge and two vertices.

[tar]

Accidently made a figure in clay, attempting to cut the surface off a sphere in one large piece, that has one face, one edge and two vertices. Sort of neat. Like a solid Moibus strip. Looks a litte like a cone and then another cone, with the two tips the ends of the same edge. The edge being s shaped with the center of the edge and the two vertices making sort of a triangle, causing the shape to look a little tetrahedralish.

 

Anyone know off hand what this figure is called? Could not find it with a quick couple searches.

 

Seems it most probably has been made before and has a name, but I do not remember seeing it before.

 

Regards, TAR

Here are three pictures of it. 1 standing on the semicircular one end of the face. 2 rolled halfway along the face. 3 standing on the other semicircular end of the face.

[Commander]

tar :

 

I am struggling to visualize what happens when we cut off the 12 Squares drawn on a Sphere.

 

What kind of polyhedron will be left ?

 

Also using ping pong balls or identical spheres I find :

 

Making 3 balls touch each other and connecting their centers with straight lines results in an equilateral Triangle.

 

Making 4 balls touch each other and joining their centers with straight lines results in a tetrahedron.

 

No more than 4 balls can be made to touch each other in real space.

 

If a tetrahedron is boxed out with plane surfaces on all the four edges it creates a Cube inside. Or in other words if a triangular roof with three 90 Deg Vertices is mounted on each of the 4 faces of the tetrahedron the cube is created, [Cube A]

 

The size of the above cube may be the same as adding 4 balls to the all touching 4 balls where each of the added ball touches 3 other balls and then connecting all the centers with straight lines. These 8 balls are perhaps the most compactly constructed as possible.

 

Another arrangement of just arranging 4 balls on a plane each touching 2 others and their center connected resulting in a square. An exact set of another 4 such balls kept on top of these 4 by which each of the 8 balls touch 3 others and if their centers are connected it results in a cube [say Cube B] which must not be same as the above mentioned cube A.

 

The size of Cube A must be less than that of Cube B [if they are of different sizes] and that is why I say that Cube A is coupled tighter than Cube B.

 

I am trying out some free modeling tools to study these intricacies but have not found one with which I can create any solid shape at will and cut it the way i want and visualize the fall outs !

 

Regards

 

Thomas

 

PS : I think the size of Cube B is 8 R³ [R being the radius of the ping pong/ sphere] and the volume of Cube A is 2 x 2^1/2 R³.

 

It all started [these sleepless visualizations] when I was trying to box out Cube A with planes on its edges just like how the tetrahedron was boxed out to produce Cube A.

 

As a four faced roof on a square face can not have 90 Deg angle [has to be less] the boxed out shape should have some diamond shapes. 6 x 4 / 2 = 12 in all

 

Perhaps this will be what we get cutting off the 12 squares drawn on the surface of the sphere.

Edited January 15, 2015 by Commander

[tar]

Commander,

 

I had the same thought. Matter of fact, I still have the clay piece I cut off, from around the figure, with the spherical rhombic dodecahedral markings on it. I will fix it ( piece got cut and fell off) and take a picture of it, tonight.

Since it is one piece, and the figure has one face, there is probably a one to one correspondence between the demarkations on the outside of the sphere, and the points on the 2d surface of the singular face of the figure offered in this thread. A lot of possibilities for the projection of spherical surface characteristics onto the two dimensional space that I rolled the face along, above. Interesting, because once I see where I made my cuts and such I can draw out the twelve sections onto the above 3d figure's face, as well as onto that s shaped 2d figure with the circular ends. All sorts of applications might come from that. I will retain rights to having come up with the system though, if it turns out to be as useful as its appearing to me at the moment. Your discovery however was/is coincidental so I suppose you are a co-discoverer, should this be a new discovery.

 

As I would like to call the 1/12 solid section of the sphere, the Janus, should it not yet have a name, I would like to call the figure depicted in this thread, the Sedge, after the S shaped edge, (and a co-worker I talk about this stuff with,) should it not already have a name.

 

Regards, TAR

 

It is, after all however a geometrical shape, which belongs to everybody, so only its applications to 3D graphics, photography and printing, 3 dimensional coordinate systems, and projections and such would be things carrying any ownership rights. I will retain any such rights due to me, while giving due credit and ownership to those on this thread and the 12 section thread that have assisted in the development the associated ideas, models and diagrams.

By the way, it appears that you should be able to start your cut at any of twelve identical spots on the spherical rhombic dodecahedron, and come up with the same figure.
Being that the figure is mirror image of itself, that might mean you could get 6 different cuts off the surface of the sphere, depending on where you started (where you end, meaning the same thing as if you started there.)

[tar]

I will have to get back with you, on this. I can not figure out how I did it. My markings were obscured, so I am not sure what lines I followed or why. It was sort of accidental. I will have to continue to try and reconstruct the situation.

By the way, just put six spheres around a center ball and three on top and then three on the bottom diametricallly opposing the three on the top. Nice close packed situation with 4 intersecting hexagonal planes and three intersecting square planes, is seeded and can extend out indefinitely as Janus rendered in the other thread.

 

Balls in such an arrangement are at the centers of the diamonds, spacing wise, as the diamonds are drawn by placing a line exactly between each ball and the closest other four.

[Commander]

tar :

 

I am struggling to visualize what happens when we cut off the 12 Squares drawn on a Sphere.

 

What kind of polyhedron will be left ?

 

Also using ping pong balls or identical spheres I find :

 

Making 3 balls touch each other and connecting their centers with straight lines results in an equilateral Triangle.

 

Making 4 balls touch each other and joining their centers with straight lines results in a tetrahedron.

 

No more than 4 balls can be made to touch each other in real space.

 

If a tetrahedron is boxed out with plane surfaces on all the four edges it creates a Cube inside. Or in other words if a triangular roof with three 90 Deg Vertices is mounted on each of the 4 faces of the tetrahedron the cube is created, [Cube A]

 

The size of the above cube may be the same as adding 4 balls to the all touching 4 balls where each of the added ball touches 3 other balls and then connecting all the centers with straight lines. These 8 balls are perhaps the most compactly constructed as possible.

 

Another arrangement of just arranging 4 balls on a plane each touching 2 others and their center connected resulting in a square. An exact set of another 4 such balls kept on top of these 4 by which each of the 8 balls touch 3 others and if their centers are connected it results in a cube [say Cube B] which must not be same as the above mentioned cube A.

 

The size of Cube A must be less than that of Cube B [if they are of different sizes] and that is why I say that Cube A is coupled tighter than Cube B.

 

I am trying out some free modeling tools to study these intricacies but have not found one with which I can create any solid shape at will and cut it the way i want and visualize the fall outs !

 

Regards

 

Thomas

 

PS : I think the size of Cube B is 8 R³ [R being the radius of the ping pong/ sphere] and the volume of Cube A is 2 x 2^1/2 R³.

 

It all started [these sleepless visualizations] when I was trying to box out Cube A with planes on its edges just like how the tetrahedron was boxed out to produce Cube A.

 

As a four faced roof on a square face can not have 90 Deg angle [has to be less] the boxed out shape should have some diamond shapes. 6 x 4 / 2 = 12 in all

 

Perhaps this will be what we get cutting off the 12 squares drawn on the surface of the sphere.

 

tar :

 

Thank you for your response which I will study and reply to.

 

In the mean time :

 

What I said :

 

"The size of the above cube may be the same as adding 4 balls to the all touching 4 balls where each of the added ball touches 3 other balls and then connecting all the centers with straight lines. These 8 balls are perhaps the most compactly constructed as possible."

 

I want to admit that after adding these 4 balls and connecting their centers it may not result in a cube but just a figure where four tetrahedrons are mounted on the 4 faces of a central tetrahedron.

 

Otherwise Cube A and B as mentioned above and other points made remain same.

 

Error is regretted.

I will have to get back with you, on this. I can not figure out how I did it. My markings were obscured, so I am not sure what lines I followed or why. It was sort of accidental. I will have to continue to try and reconstruct the situation.

By the way, just put six spheres around a center ball and three on top and then three on the bottom diametricallly opposing the three on the top. Nice close packed situation with 4 intersecting hexagonal planes and three intersecting square planes, is seeded and can extend out indefinitely as Janus rendered in the other thread.

 

Balls in such an arrangement are at the centers of the diamonds, spacing wise, as the diamonds are drawn by placing a line exactly between each ball and the closest other four.

 

tar :

 

Yes it is an interesting shape you have proposed and I am still to visualize the two semi circular faces , two conic surfaces with opposite conical apex so that while rolling it out the bottom of the cone covers area on the plane while the apex point is stationary and at the end of the first conical section's roll the apex point shifts to the other side and the area is covered on the opposite side.

 

Perhaps you want to end with a shape as below while covering the entire surface area of the proposed solid.

 

If what you have found is possible to carve out the following shapes can occur depending on the starting position of the semi circular face.

 

 

At this point frankly I can not see or visualize that shape !

[tar]

Commander.

Perhaps the semicicular remark was misleading. It might not be a semicircle, might be a third of a circle, for instance, But perhaps, since the shape works, it has to be a semicircle.

 

The first of your three drawiings is the closest to the situation though.

 

Your reference to the two semicircular faces, was incorrect. The two "semicircles" are on the opposite ends of the same face, in such as to cause the bottom of the cone to roll out while the apex is stationary for half the trip along the face, and cause then the apex to be stationary on the near side, while the edge rolls out on the other side.

 

Not well described, but I have to get ready for work.

 

Later.

 

Regards, TAR

remember that the "semicircles: are the edge and there is only one edge

Commander,

 

 

OK, couple more hints. When I started I cut shallow great circle lines at the North Pole four point and the South pole to allow the surface to move and open up as I made my normal to the center type cuts to remove the surface diamonds. My depth of cut was such as to just get under a complete pair of diamonds. This is an interesting statement, but in the other thread, somebody pointed out that the four points of the diamond are not all on the same plane and whatever my rule was, to pass as close beneath the diamonds as possible,without breaking the surface, somehow picked a path of cut that respected all four sides of a diamond at once.

 

Another hint, as to the path I may have taken is that one can get around a sphere laid out in the spherical rhombic dodecahedron pattern in four diameter length moves along the surface. Start at a four point, travel one radii along the sides of touching diamonds, take a 120 degree left, travel 2 radii, take a 120 degree right, travel 2 radii, take a 120 degree left, travel 2 radii, take a right and travel 1 radii back to your starting point.

 

I have not done this yet, cutting below the diamonds, but I traced it out on the sphere, and it does account for all 12 diamonds and may be the path I took to cut the surface off and be left with the sedge.

 

Regards, TAR

[tar]

Ok,

 

Forget the left right left right stuff.

 

It looks like its related to the three diamond sections, but although I have not nailed it down completely, I believe you start cutting pretty agressively under two diamonds, possibly starting at a four point, and work your way around a three point to your right until there is not a surface diamond left to your right, at which point you turn left and take the diamonds left on the surface.

 

Or it might be the same idea but working your way around a four point. In any case, I think I have the basic idea down, and I just have to reproduce the thing, to tell you exactly how its done.

 

All I know is its possible, because I have the figure right here, and it absolutely has only one face and one edge that terminates at each end.

although I just made something like it, without marking any diamonds on the sphere at all

 

seems its just what you get if you start cutting the peel off around a point and then switch directions around a point on your left to complete the peel

 

The only help the diamonds might give is to provide the width of the peel that will make the thing just exactly symetrical.

Actually the figure does not have that much to do with even the sphere, except for possibility of projecting a sphere unto that two dimensional shape it rolls out onto.

 

I just made something with the same properties by starting with a tetrahedron, and marking three of the edges (in like an S) for edge duty and the other three for destruction (rounding off,) Then its just a matter of pushing the clay around until its in the same shape as pictured above.

[tar]

So,

 

Bottom line, this figure is not so very odd, unique or surprising as I initially thought. But its still interesting, as each point on the face has its unique position on the surface of the sphere, thus allowing one to consider projections from two to three or three to two. And it probably therefore has something to do with a sphere, especially since it shares the property of having one face, and since it has the circle built in.

 

Regards, TAR

But,

 

As my wife pointed out when I showed her the version I crudely made from the tetrahedron. "It's not as pretty as the original one".

 

The first (the one pictured,) was neat to hold in your hand and turn and feel and look at. Very balanced, the lines where/are very satisfying. Sort of a 3D yin yang.

 

I'll still call it the Sedge.

 

Regards, TAR

Well, Got it figured,

 

It is definitely neat. And does go along with the twelve sections of the Sphere, Cube, Tetraherdron, Octahedron. Same internal angles, same 12 diamonds.

 

I give you the 12 divisions of the Sedge.

 

 

 

 

Regards, TAR

 

 

 

 

 

 

Wow!

 

Just found that a sphere made from 1/4lb of clay and marked out with the diamonds, rolls out (what appears like might be) exactly on the Sedge imprint, made with a Sedge, made from a quarter pound of clay.

 

Might be even more interesting a figure than I originally sensed.

 

Let me clean up the middle of that last diagram and try things out a bit. Wierd, but this might even be a basis for laying out small distortion 3D images on 2D surfaces. (360 degree images, at that.)

 

Regards, TAR

 

By the way, I claim copyright on the images of the Sedge, and the shape it rolls out on, included in this thread. (Not Commander's of course, that is his.)