The Tiger Explained

Discussion of shapes with curves and holes in various dimensions.

The Tiger Explained

Postby papernuke » Sun Sep 17, 2006 4:24 am

What is the tiger?

moonlord: Split from "The tiger does not exist"
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Postby batmanmg » Sun Sep 17, 2006 6:17 am

if you don't know the answer you probably shouldn't be asking the question...

go look it up somewhere
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Postby Keiji » Sun Sep 17, 2006 8:44 am

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Postby moonlord » Sun Sep 17, 2006 2:52 pm

I doubt the wiki clears anything up for Icon... :roll:
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Postby papernuke » Sun Sep 17, 2006 3:28 pm

but i never studied this stuff, im only in 6th. and i still don't get it.
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Postby Keiji » Sun Sep 17, 2006 3:30 pm

The tiger is what you get when you spherate the duocylinder.

Just like the torus is what you get when you spherate the cylinder.
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Postby papernuke » Sun Sep 17, 2006 3:31 pm

Whats a duocylinder and a torus?
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Postby Keiji » Sun Sep 17, 2006 3:33 pm

http://tetraspace.alkaline.org/wiki/ind ... itle=Torus
http://tetraspace.alkaline.org/wiki/ind ... uocylinder

You should be able to figure out what a torus is from the wiki.

The duocylinder is the cartesian product of two circles, and it's also what you get when you lathe a cylinder in a particular direction.
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Postby papernuke » Sun Sep 17, 2006 3:33 pm

I only read the 1st sentence.
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Postby Keiji » Sun Sep 17, 2006 3:34 pm

and?
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Postby papernuke » Sun Sep 17, 2006 3:35 pm

and i didnt get it so i stopped.
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Postby wendy » Mon Sep 18, 2006 6:12 am

The "tiger" or tri-circular torus (to use the PG term), exists. One can easily visualise it thus:

Start with a sheet of paper in 4d. This is a 3d block, say, xyz.

You roll it into a cylinder, say wx. The sheet is now the cartesian product of a circle in wx, and a square in yz.

You roll it into a clinder in wy. This makes the sheet of paper in the wxy plane, a torus, and in the z plane, a line.

You now roll it into the wz plane, to get a tri-circular torus.

If you want to start with the bi-cylinder margin, and replace each point with a circle, one of the axies lies radially, and the second axis runs from orthogonal circle to orthogonal circle. This gives a set of planes that cross at only one point.

For example, suppose the circles are in the wx and yz planes. The bicylinder-margin becomes then w²+x² = y²+z² = k. We now draw two circles, w²+x² = 4, y = z = 0, and w = x = 0; y²+z² = 4.

Select any point on both of these circles, and draw the line between them. These intersect the margin at exactly one point (which is unique for every pair of points on the two circles). We then draw a second ray from the centre, and these pair of lines define a hedrix "plane", which has one point on the margin.

In this, one draws a circle, and this circle forms the surface of the "tiger".

On the other hand, one notes that even with the normal torus, the direction of pondering (dimension-loss), changes, it is in fact radial, too.

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Postby PWrong » Mon Sep 18, 2006 3:16 pm

That doesn't help much for me...

What I'd like to do is derive the parametric equations for the tiger, using only the equations for the duocylinder and the definition of the torus product. I'll have a go at this tomorrow night. It'll involve finding a basis for the normal plane to the duocylinder.
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Postby papernuke » Mon Sep 18, 2006 11:54 pm

PWrong wrote:That doesn't help much for me...


Me neither.
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Postby wendy » Tue Sep 19, 2006 7:18 am

The duocylinder margin is sqrt(w²+x²)=sqrt(y²+z²) = k

I suppose you could start with the duocylinder-margin in r1, a1, r2, a2. The corresponding equation for the tri-circular torus is then:

(r1-R)²+(r2-R)² = S², where R is the radius of the bicylinder, and S the thickness of the torus.

This becomes:

(sqrt(w+x)²-R)²+(sqrt(y²+z²)-R)² = S²
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Postby PWrong » Tue Sep 19, 2006 4:05 pm

(sqrt(w+x)²-R)²+(sqrt(y²+z²)-R)² = S²

We know that one. I'm talking about the parametric equations. I know not everyone will understand this, but it's worth doing it anyway.

Here's the equations for the duocylinder, with parameters u and v. The tiger will have parameters u, v and t.

x = a cos u
y = a sin u
z = b cos v
w = b sin v

now the tangent vectors are:
r<sub>u</sub> = {-a sin u, a cos u, 0, 0}
r<sub>v</sub> = {0, 0, -b sin v, b cos v}

Now we want a pair of vectors perpendicular to these two. We'll call them m and n. We want to have:

r<sub>u</sub>.m = r<sub>u</sub>.n = r<sub>v</sub>.m = r<sub>v</sub>.n = m.n = 0

This should be ok:
m= {a cos u, a sin u, 0, 0}
n = {0, 0, b cos u, b sin u}
It's easy to show that the four vectors are all perpendicular.
Now we have a new plane with a basis, on which we can put a circle. Note that it doesn't matter what basis we choose, because the circle is symmetric.

We can represent any vector in this new plane by r' = x' m + y' n
These dashes aren't derivatives, they just mean x' is a different variable from x.

Now the equations for our circle are:
x' = R cos t
y' = R sin t

In vector form:
r' = x' m + y' n
r' = (R cos t) m + (R sin t) n
r' = {R a cos(t) cos(u), R a cos(t) sin(u), R b sin(t) cos(v), R b sin(t) sin(v)}

Now, by the definition of the torus product, the final parametric equations for the tiger are given by r + r'

So we have:
x = a cos u + R a cos(t) cos(u)
y = a sin u + R a cos(t) sin(u)
z = b cos v + R b sin(t) cos(v)
w = b sin v + R b sin(t) sin(v)

And this is exactly what we expected :D.
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Postby papernuke » Tue Sep 19, 2006 9:40 pm

i still dont get it.
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Postby batmanmg » Tue Sep 19, 2006 11:40 pm

ok since i might be as clueless about tigers as icon... but i know geomety, and toruses. maybe i can explain it to him and myself.

Please correct me if im completely out of my wits.

as far as i know. a torus is donut shaped. to make one in your mind you can take a circle and a line outside that circle. now rotate the circle around that line 3 dimensionaly. that gets you a donut shaped torus.

Image

a tiger. Im guessing. is a 4d like torus. well all you need to do is bump up the dimensions and repeat.

take a sphere. and a plain outside that spere. now rotate the sphere around that plain 4 dimensionaly. theres your tiger.

my probleme would be imagining this 4d rotation. can anyone draw a picture of what the spere would look like at different points througout the rotation, or explain what this would look like?


[Edit] All the red stuff is wrong but at least I got the torus stuff right
Last edited by batmanmg on Wed Sep 20, 2006 8:00 pm, edited 2 times in total.
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Postby moonlord » Wed Sep 20, 2006 6:43 am

Actually, you're not correct. Sorry. A torus and a tiger are unrelated. There are several types of torii in 4D, but the tiger isn't one of them.
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Postby wendy » Wed Sep 20, 2006 8:08 am

The 'tiger' is a tri-circular torus. It is the torus-product of three circles: its surface corresponds to the identification of the six opposite faces of a cube, in the same manner that the regular 3d torus does a square. Now how more torus-y can you get?

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Postby PWrong » Wed Sep 20, 2006 11:29 am

Actually, you're not correct. Sorry. A torus and a tiger are unrelated. There are several types of torii in 4D, but the tiger isn't one of them.
There's a few different definitions for "torus" as a category of objects. I think the best one is "an object who's RNS notation has brackets around it". This includes tiger, but not torinder. The tiger is also a "beast", but thanks to RNS notation, beasts aren't as scary as they used to be.

The 'tiger' is a tri-circular torus. It is the torus-product of three circles: its surface corresponds to the identification of the six opposite faces of a cube, in the same manner that the regular 3d torus does a square.

No it's not. You're thinking of the 3-torus (which Rob calls tetratorus on the wiki). The 3-torus is the torus product of three circles: (2#2)#2 = ((21)1), and it does correspond to a rolled up cube.

The tiger is the torus product of a duocylinder and a circle: (2x2)#2. It might be topologically equivalent, I'm not sure. But it's not the same object.

Icon wrote:i still dont get it.

I wrote:I know not everyone will understand this, but it's worth doing it anyway.

My post wasn't meant to help you understand, it was mostly for my own benefit. Here's my advice:

Try to understand all the other 4D shapes first. The tiger is the most difficult 4D rotope to visualise. Start with the tetracube, then spherinder, cubinder, glome and duocylinder in that order. You can find descriptions of these on the wiki or on alkaline's main site. Those are the rotatopes. Then have a go at the other rotopes: torinder, 3-torus, toracubinder, toraspherinder, again in that order. Only then will you be ready to battle the tiger!

Btw, if you get confused on the way, just ask us and we'll try to help. You have an advantage in that you're seeing this stuff at an early age. Older people are too used to living in the 3rd dimension, so they don't like to think outside of it.
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Postby moonlord » Wed Sep 20, 2006 6:29 pm

PWrong wrote:
Actually, you're not correct. Sorry. A torus and a tiger are unrelated. There are several types of torii in 4D, but the tiger isn't one of them.
There's a few different definitions for "torus" as a category of objects. I think the best one is "an object who's RNS notation has brackets around it". This includes tiger, but not torinder. The tiger is also a "beast", but thanks to RNS notation, beasts aren't as scary as they used to be.


Hey, which are the other beasts? 2#3 and 3#2? Some other?
BTW, by your definition, the sphere is a torus also - (111) - so... Or you only meant ones who ACTUALLY have brackets around them? Besides, I think the torinder should be a torus. I see it as a lathed cylinder.

PWrong wrote:The tiger is the torus product of a duocylinder and a circle: (2x2)#2.


Maybe you can insist on the torus product, I'm not familiar with it yet.

Pwrong wrote:Try to understand all the other 4D shapes first. The tiger is the most difficult 4D rotope to visualise. Start with the tetracube, then spherinder, cubinder, glome and duocylinder in that order. You can find descriptions of these on the wiki or on alkaline's main site. Those are the rotatopes. Then have a go at the other rotopes: torinder, 3-torus, toracubinder, toraspherinder, again in that order. Only then will you be ready to battle the tiger!


That's my advice too ;) -- I'm currently at the 3-torus...
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Postby batmanmg » Wed Sep 20, 2006 8:08 pm

just a quick question. where did you guys look up information about these things before you put the info on a wiki?

did you guys make all this up? not saying its fake just that you came up with it all. the math for it too? and the language? soo all of these shape concepts aren't readily accepted by the scientific community? or are they?
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Postby papernuke » Thu Sep 21, 2006 2:23 am

now i know what a torus is, but i still don't understand the tiger.
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Postby batmanmg » Thu Sep 21, 2006 3:13 am

ok.. so is the 3d projection of a duocylinder a torus? but with the whole filled in yet still a wall there? its rotation looks like this. This
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Postby PWrong » Thu Sep 21, 2006 3:23 am

did you guys make all this up? not saying its fake just that you came up with it all. the math for it too? and the language? soo all of these shape concepts aren't readily accepted by the scientific community? or are they?

Most of them are made up by people here at the forum. They've probably been mentioned in mathematical papers, but they've never been named or classified anywhere.

Hopefully one day they will be accepted. I might even write a paper on them when I'm a mathematician :D.
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Postby wendy » Thu Sep 21, 2006 7:53 am

The figure described by spherating the margin of the bi-circular prism, and the figure derived by closing three ends of the cube in torus-manner do indeed lead to the same figure. Not only are the surface the same, but the topological genus of the solid the same: one can be readily be deformed into the other.

At the moment, the genus appears to be 3ch+3hc. At least, this is the list of closures that i have managed to not find answers for.

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Postby PWrong » Thu Sep 21, 2006 8:16 am

The tiger and the 3-torus are not the same. The cartesian and parametric equations are different. I will believe that they're topologically equivalent when someone proves it. But it's obvious from the equations that they're not the same object.
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Postby papernuke » Fri Sep 22, 2006 1:20 am

hey, can someone give me a picture of the tiger? if its a shape?
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Postby wendy » Fri Sep 22, 2006 7:53 am

PWrong writes:

The tiger and the 3-torus are not the same. The cartesian and parametric equations are different. I will believe that they're topologically equivalent when someone proves it. But it's obvious from the equations that they're not the same object.


We note that the margin of the duocylinder is inscrible in a glome, and that the tiger is the spherate of this.

Imagine now that you punch a hole in this glome, and flatten it onto a chorix "realm". As it does, the margin becomes the surface of a torus, and the tiger becomes the spherate of a torus-surface: ie a 3-torus.

Another consideration, is that if you take a 3-torus-prism, that is, a bar in the cross-section of a 3d-torus, and bend it into a circle, then it depends on whether you get a tiger or 3-torus. The axis of bending makes a hedrix (or 2fabric or 'plane'). This intersects the cross-section in a line. If this line runs through the torus as a puncture might, then the result is a 3-torus.

If it runs through the torus as an axle of a wheel does, then the result is a tiger. One notes that the circles that form the tread of the tire are not distorted as the bend occurs, and one is left with a shape that is the product of two undistorted circles: ie a duocylinder-margin.
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