Disagreement with the Duocylinder

Discussion of shapes with curves and holes in various dimensions.

Postby pat » Wed Mar 01, 2006 10:11 pm

moonlord wrote:Now I remember that during a boring history class, I wrote the cartesian for the duocylinder, as I think it is (x**2+y**2<=1, z**2+w**2<=1) and, by projecting it on the four mutually perpendicular hyperplanes Oxyz, Oxyw, Oxzw and Oyzw, I got only cylinders, which seemed a little odd to me. Is it correct?


That's correct.... for the same reason that projecting a normal cylinder: x**2 + y**2 <= 1, |z| <=1 down to Oxz or Oyz gives only squares.

If you projected to some hyperplane that doesn't go through the origin, you'll have slightly more interesting results.... and if your hyperplane isn't axis-aligned, it'll be even better.
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Postby Neues Kinder » Thu Mar 02, 2006 1:40 am

I think what you got were the cylindrical cross-sections of the figure. If you string them together you should get the duocylinder. If not, I don't know what's wrong, because that type of stuff is beyond me.
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Postby Marek14 » Thu Mar 02, 2006 7:54 am

No, that's correct. Duocylinder has symbol 22, and each coordinate hyperplane cut is got by reducing one of the numbers in symbol by 1 - in all four cases you get 21 which is a cylinder.
When you look at cuts with six coordinate planes, you find that two of them are circles 2 and four are squares 11
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Postby moonlord » Thu Mar 02, 2006 7:07 pm

Interesting. I'll try to see what do i get if I project it on a hyperplane at 45 degrees in a future class, but I've got a test next time I have history :).
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Postby Marek14 » Fri Mar 03, 2006 7:30 am

Let's see...

Duocylinder's equation is max(x^2+y^2, z^2+w^2) = 1. We can rotate it in six coordinate planes, but obviously rotating in xy or zw does nothing. Let's try xz:

x' = x cos alpha - z sin alpha
z' = x sin alpha + z cos alpha

For alpha = 45 degrees, we have
x' = sqrt(2)/2 * (x - z)
z' = sqrt(2)/2 * (x + z)

So the new equations are:

max((x-z)^2/2 + y^2,(x+z)^2/2 + w^2) = 1

max(x^2/2 + y^2 + z^2/2 - xz,x^2/2 + z^2/2 + w^2 + xz) = 1

Let's put x=0 here:

max(y^2 + z^2/2,z^2/2 + w^2) = 1

simplifying to

max(y^2,w^2) = 1-z^2/2

This 3D object is a sort of dome (one of my extended rotatopes) which is made of square slices with square of edge 2 in yw plane and reducing to point in z=sqrt(2) and z=-sqrt(2). This is what you get slicing through hyperplane yzw. The slices of this object are: square max(y^2,w^2) = 1, and two ellipses, y^2+z^2/2 = 1 and z^2/2 + w^2 = 1.

Let's now try to eliminate z from the original equation:

max(x^2/2 + y^2 + z^2/2 - xz,x^2/2 + z^2/2 + w^2 + xz) = 1
z=0

max(x^2/2+y^2,x^2/2+w^2)=1
max(y^2,w^2)=1-x^2/2

giving exact analogue, which was to be expected.

I didn't try slicing through y or w (the formulas become more complex there), but I would expect that it will still be cylinder, just rotated (after all, we rotated in other directions).

Would the same method work for other rotatopes rotated through 45 degrees? Since equations of the majority of them only contain squares, we have transformations

x1'^2 = (x1 - x2)^2/2
x2'^2 = (x1 + x2)^2/2

If we put x2=0 here, we get equations
x1'^2 = x1^2/2
x2'^2 = x1^2/2

and with this replacement in original equation we should get the equation of x1-slive after rotating 45 degrees in x1x2 plane.

Let's try it...

Square: max(x^2,y^2) = 1. Rotating through xy.
max(x^2/2,x^2/2) = 1 => x^2=2 => x=sqrt(2), x=-sqrt(2)
We get two points of distance 2 sqrt(2), which is correct.

Cube: max(x^2,y^2,z^2) = 1. Rotating through xy.
max(x^2/2,x^2/2,z^2) = 1 => max(x^2/2,z^2) = 1 => rectangle with boundaries z=1,z=-1,x=sqrt(2),x=-sqrt(2)

Cylinder: max(x^2+y^2,z^2) = 1. Rotating through xz.
max(x^2/2+y^2,x^2/2) = 1 => x^2/2 + y^2 = 1. Ellipse.

Tesseract: max(x^2,y^2,z^2,w^2) = 1. Rotating through xy.
max(x^2/2,x^2/2,z^2,w^2) = 1 => max(x^2/2,z^2,w^2) = 1. Cuboid of edges 2, 2, and 2 sqrt(2)

Cubinder: max(x^2+y^2,z^2,w^2) = 1. Rotating through xz.
max(x^2/2+y^2,x^2/2,w^2) = 1 => max(x^2/2+y^2,w^2) = 1. Elliptical cylinder with height 2 and base ellipse 2 x 2 sqrt(2)

Cubinder: max(x^2+y^2,z^2,w^2) = 1. Rotating through zw.
max(x^2+y^2,z^2/2,z^2/2) = 1 => max(x^2+y^2,z^2/2) = 1. Cylinder of height 2 sqrt(2).

This is sufficient to make a conjecture, which provides the following way to construct the slice:

1. Represent the rotatope in graph form (for normal rotatopes, the graph will be comprised of several complete graphs Kn). Write number 1 to each node (it represents half the diameter of the rotatope along that particular coordinate axis).
2. Pick any two nodes.
3. Identify both nodes. New node will have all the edges the previous nodes had.
4. Mark the new node with sqrt(2) if the original nodes weren't connected or with 1 if they were.

So, shortening sqrt(2) to just "2" for square (1) (1) we get (2).
For circle (1)-(1) we get (1) - the same thing as if we just omitted one node.

For cube (1) (1) (1) we get rectangle (2) (1)
For cylinder (1)-(1) (1) we get either square (1) (1) or ellipse (2)-(1)
For sphere (1)-(1)-(1)- we get circle (1)-(1). In summary, rotating in circular plane has no effect on slice in 3D (it's not the case in higher dimensions!)

(For dome (1)-(1)-(1) we get either circle (1)-(1) (if we rotate it in one of its circular planes, the cut will still be circle), or ellipse (2)-(1))

In 4 dimensions:
For tesseract (1) (1) (1) (1) we get cuboid (2) (1) (1)
For cubinder (1)-(1) (1) (1) we get either:
- cube (1) (1) (1)
- elliptic cylinder (2)-(1) (1)
- or cylinder (1)-(1) (2)
For duocylinder (1)-(1) (1)-(1) we get either:
- cylinder (1)-(1) (1)
- or dome (1)-(2)-(1)

So far it matches. Let's verify it for spherinder:
For spherinder (1)-(1)-(1)- (1) we get either:
- cylinder (1)-(1) (1)
- or ellipsoid (2)-(1)-(1)-

Let's look at the equations:

Spherinder: max(x^2+y^2+z^2,w^2) = 1. Rotating through xw.
max(x^2/2+y^2+z^2,x^2/2) = 1 => max(x^2/2+y^2+z^2) = 1. Yep, it's ellipsoid.

For glome, we always get sphere.

How about other 4D extended rotatopes (a.k.a. graphotopes)?

Dominder (y)-(x)-(z) (w): Rotating through xy or xz leads to ordinary cylinder. Rotating through xw leads to dome (1)-(2)-(1), rotating through yz to elliptic cylinder (2)-(1) (1), rotating through yw or zw to dome (2)-(1)-(1)

Tridome (y)-(x*)-(z) *-(w): Rotating through xy, xz, or xw leads to ordinary dome. Rotating through yz, yw, or zw leads to dome (2)-(1)-(1)

Longdome (z)-(x)-(y)-(w): Rotating through xy leads to dome (1)-(1)-(1)! This is significant because ordinary slice through either x or y without rotation leads to just a cylinder. Rotating through xz or yw leads to ordinary dome. Rotating through xw or yz leads to dome (1)-(2)-(1), and rotating through zw leads to ellipsoid (2)-(1)-(1)-*

Spheridome (w)-(x*)-(y)-(z*)-: Rotating through xy or xz leads to ordinary dome. Rotating through xw leads to ordinary sphere. Rotating through yz leads to ordinary dome. Rotating through yw or zw leads to ellipsoid (2)-(1)-(1)-.

Cyclodome (x)-(y)-(w)-(z)-: Rotating through xy,xz,yw, or zw, leads to sphere. Rotating through xw or yz leads to dome (1)-(2)-(1)

Semiglome: three possible results are sphere, dome, or ellipsoid.
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Postby moonlord » Fri Mar 03, 2006 4:32 pm

What is a dome? Is it an ellipse rotated around the longer axis? Or is it also sectioned? Maybe you can explain further... I totally lost the thread when you started talking about graphotopes...

As far as I understand, a dome is something like:

{(x,y,z) | (x/a)^2+(y/b)^2+(z/c)^2<=1, a!=b=c} intersected with {(x,y,z) | x>=-a/k, k fixed (0<k<1)}

Is it correct?
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Postby Marek14 » Fri Mar 03, 2006 7:38 pm

Dome can be imagined as taking a square in xy plane, and then dragging it in z direction, shrinking it along the way, in such a way that edge midpoints trace circles. Dome cut along xy plane yields square, cut along xz or yz yields circle. It's equation is max(x^2,y^2)+z^2 = 1.

Graphotopes are described to greater extent in the threads. To say it simply, they are shapes whose intersections with coordinate planes are certain collection of squares and circles. Therefore, they can be drawn like graphs with each coordinate axis represented by node, and edge connecting nodes iff the cut along the coordinate plane determined by these two coordinates gives circle. This graph approach has some advantages (for example, it allows to quickly determine shape of any slice parallel to a coordinate hyperplane).
Last edited by Marek14 on Fri Mar 03, 2006 7:45 pm, edited 1 time in total.
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Postby moonlord » Fri Mar 03, 2006 7:41 pm

So, let me see if I got it... Is a crind (intersection of two perpendicular cylinders) the same with a dome?
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Postby Marek14 » Fri Mar 03, 2006 7:48 pm

Yes, it is. I came upon the shape independently, and gave it different name, that's all. For the purpose of graphotope naming, "dome" is better name, since otherwise I'd have things like "crindinder" which doesn't look good. :)
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Postby moonlord » Fri Mar 03, 2006 7:53 pm

I see... I still try to understand how can you get a dome from a duocylinder... Perhaps the reason lies in it's name...
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Postby Marek14 » Fri Mar 03, 2006 8:24 pm

moonlord wrote:I see... I still try to understand how can you get a dome from a duocylinder... Perhaps the reason lies in it's name...


I was surprised too, but when I transformed the coordinates, that is what I got.
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Postby moonlord » Sat Mar 04, 2006 3:31 pm

Really interesting... I suddenly realised my picture of the duocylinder was incorrect...
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Postby Marek14 » Sun Mar 05, 2006 7:39 am

How so?
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Postby moonlord » Sun Mar 05, 2006 2:42 pm

I was sure it looks somewhat like a crind (dome), but now I see is more appropiate to consider it as a circle of circles... somewhat like a sliced and opened torus...
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Postby Neues Kinder » Mon Mar 06, 2006 1:31 am

No, that's correct. Each of the wrapped-around surfaces of the duocylinder is a circle of circles. Just like the wrapped-around surface of a cylinder is a circle of lines.
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Postby moonlord » Mon Mar 06, 2006 2:47 pm

Yes, I understood now...
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