3. If you rotated around a BISECTING plane, replace X with 11. If you rotated around a NON-INTERSECTING PLANE, replace X with (11)
Note that the 3-torus has a major radius, a middle radius and a minor radius, while the tiger has two minor radii and a major radius. If the objects are the same, you need a transformation that takes the three radii of the tiger, and returns the three radii of the equivalent 3-torus. If you can visualise the objects as easily as you claim, it shouldn't be hard to find this transformation.
But the torus product is not a matter of spheration. It merges two spaces (and figures inscribed in each), such that the two spaces cross in a line, and one of the figures is centred on the surface of the other, and distorted in the direction of the line.
Suppose you take a polytope P. You can replace the p surtopes of P with a little prism pQ, and fit the whole thing together as a torus.
The 'tetratorus' and the 'tiger' are both formed from a bar whose cross section is a 3d torus.
The main difference is that when we bend this bar into a circle, we impose an axis of compression (pondering) on the torus, so that the circles on the inside are shorter than those on the outside.
Not in any unique way. You can change the angle of pQ however you like, and the result will still be a torus product of your two shapes. Your definition of the torus product is multi-valued. That is, it would be if you had a definition.
Are you talking about the torinder (torus x line)? Why don't you just call it that?
Are you talking about the torinder (torus x line)? Why don't you just call it that?
x²+y² = r², abs(r-n)+abs(z) < E/sqrt(2).
x²+y² = r², max (r-n, z) < E/2
Like the prism product, the tegum product, the pyramid product, and the spheration product, the torus product has additional parameters.
Why not just call it the tri-circular torus. That's what it is.
helix × triangle is not possible, because the product is a surface product. You can only do general shapes if the product is solid.
PWrong wrote:For instance, circle x circle = duocylinder, circle # circle = torus, but the cartesian product of two circles depends on where the circles are.
Btw, you should distinguish between x and #. They are both special cases of the true cartesian product. But the cartesian product depends on the position and orientation of the two objects.
#_ The product is extensive
*_ The product draws (adds a dimension)
_# The surface is mulitopied (product ponders = loses a dimension)
_* The content is multiplied
circle ## circle = torus (3d)
circle #* circle = bi-circular tegum (4d)
circle *# circle = bi-circular prism (4d)
circle ** circle = bi-circular pyramid (5d)
apeirogon ## apeirogon = square lattice (2t = 3d)
apeirogon ## square lat = cubic lattice (3t = 4d)
Why should a product be defined in terms of a finite number of parameters? I don't understand.
These are not even cartesian products! In the Euclidean space, they overlap with the cartesian product, but the current symbols for these products are ## and #* (for prism and comb products). One notes that the cartesian product does not encompass all products, because ** pyramid products and *# tegum product are not cartesian under any stretch of the imagination. Nor is the general () spheration.
. on the other hand, you write of your own understanding:I don't understand any of your products
I'm not sure what the product is I'm thinking of.
is a precise point.'It seems like we're not talking about the same torus product...'
? are you serious ? this thread is over 40 posts long, many of it dealing with that. also, there is POLYGLOSS, which also gives relevant information on this issue. did you look up the products there as well? do you try to understand or to defend old forms of knowledge ? if the latter is true, change it. you write:you refuse to define any of them,...
I don't think I'm doing anything wrong by keeping our old definitions.
When you try to calculate 2*2, you don't expect infinitely many answers, do you? If a product needs a parameter, it's not a product of two shapes, it's a function of two shapes and a parameter.
I don't understand any of your products
. on the other hand, you write of your own understanding:I'm not sure what the product is I'm thinking of.
that's two hands full of uncertainty.
Quote:
you refuse to define any of them,...
? are you serious ? this thread is over 40 posts long, many of it dealing with that. also, there is POLYGLOSS, which also gives relevant information on this issue. did you look up the products there as well? do you try to understand or to defend old forms of knowledge ? if the latter is true, change it. you write:
The products are defined in terms of radiant functions, and in terms of the surtope consist.
The products never were allocated symbols. The thing was always described in text. In practice, the symbol ought be a list operator, eg rss(a,b), rather than a×b.
One notes that a product gives f(AB) = f(A)f(B).
One notes that to spherate a figure is not a product, since you are not actually multiplying anything! It's a function, as s(A). It does not create holes. What it does do is to give substance to thin things: a perfectly flat circle becomes a solid disk that you can fumble with.
The torus product is different, because it's shift-independent. Circle#circle will always be a torus. The x product that we often use is also different, because it always puts the two objects in entirely different dimensions e.g. one circle in the xy plane, the other circle in the zw plane.
I'm talking about a precise mathematical definition, like my torus product which is defined with vectors. I can't find anything like that in the polygloss. I don't even know if Wendy ever uses vectors.
On f(AB) = f(A)f(B): You've got two products and a function there.
It's not a product in the sense of being a binary operation.
The more general product is useless for classifying shapes, which is what we're doing.
This is more a fait of the circle's symmetry than the torus product. The torus product does shift, but the circle makes it less noticable.
Then don't call it a product. If it's not freely open, then it is not a product, really, is it?
First you tell me it does not exist, and then you are kind of splitting hairs as to whether a square/circle torus has this feature or that, based on whether it has flat faces or not
It's a restricted product. It's a special case of a more general product, with the property that for any rotopes A and B, A#B will also be a rotope. It has other nice properties as well. I don't know the definition for the more general product, because I think it uses tensors.
Flat faces have nothing to do with it. There is only one circular torus. There are infinitely many square torii.
But there is no product that takes "circle" and multiplies it by "square".
For example, if one supposes a {12}{5} torus, one starts with a dodecagon. At a corner, one places a pentagon. Through every point of this pentagon, there is a scaled copy of the dodecagon, parallel to the plane of the dodecagon, and having a centre common to the perpendicular of the dodecagon. The figure is bounded by 60 squares.
There are infinitely many tri-circular torus. The tetratorus and the tiger are examples. The flat-face argument is no different to the preserving of the topological torus of the bi-circular prism in the tiger. Ok?
Suppose you have shape A. This has a surface, and some notional centre. Take shape B, and place it so that exactly one point of the surface of B intersects with the surface of A
the intersection of the spaces holding A and B is a line passing through the centre of A.
One then substitutes for the centre of A, an orthogonal through A that spans AB (that is, one can from the set of vectors in A, and a set of vectors in the orthogonal, create by sum, any vector that can be created by A and B together).
One then passes a ray from the orthogonal, to the surface of B, a copy of the surface of A, scaled relative to the lengths of this vector and the original vector of intersection.
One further notes, that there is a hole (ie non-vanishing sphere), formed by the interior of B, that lies wholy inside AB, and that there is a hole (non-vanishing sphere), that lies completely outside AB, caused by the interior of A.
Does this mean you admit the tetratorus and tiger are different objects?
I think I see what you're doing here. You're trying to position B without using the normal space of A. This isn't a good idea, because it makes your product shift-dependent. That is, if you change your origin, you get ugly non-circular shapes.
Ok, do you mean an orthogonal as in, a line perpendicular to something? If so, perpendicular to what? How can a single line span a manifold? Spanning is what a set of vectors does to a vector space. Is AB the product of A and B, A#B?
This isn't the kind of process you can explain in one sentence.
All I can say is, good luck proving that without algebra or calculus.
No, they're the same object with different params.
There's enough bi-pass in mathematics to avoid both of algebra and calculus.
Which is pretty much the same equation
both describe a circle-circle torus, bent into the shape of the torus.
Users browsing this forum: No registered users and 52 guests