surface volume

Higher-dimensional geometry (previously "Polyshapes").

surface volume

Postby chickendude » Mon May 31, 2004 4:58 pm

How are the equations for surface volume of tesseracts derived? Some of the simple ones i can understand like for the tetracube but for things like spherinders and cubinders i can't quite understand why they work.
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Postby Aale de Winkel » Fri Jun 04, 2004 12:18 pm

The Surface Area is the total sum of the areas see: http://mathworld.wolfram.com/SurfaceArea.html for sample, according to http://mathworld.wolfram.com/Tesseract.html there are 24 squares in a tesseract so the area is 24a[sup]2[/sup], I don't know how you define "surface volume" so multiply the above by any thickness you want to give a surface?
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Re: surface volume

Postby Euclid » Sat Jun 05, 2004 12:45 am

chickendude wrote:How are the equations for surface volume of tesseracts derived? Some of the simple ones i can understand like for the tetracube but for things like spherinders and cubinders i can't quite understand why they work.


Isn't surface volume an oxymoron? What you may be looking at is what is called content, or n-dimensional volume.
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Postby chickendude » Sat Jun 05, 2004 1:04 am

surface volume is like surface area of a 3D shape but on a 4D

for a tetracube it is 8a^3 because there are 8 volumes in the tetracube and each is a^3

V. The Glome
Other names: hypersphere
Number Series Notation: 4
Algebraic Labelling: (4,4)
Formula: x2 + y2 + z2 + w2 = 1
Surface Volume: 2π2r3
Bulk: (1/2)π^2r^4

and

II. The Cubinder
Other names: hypercylinder
Number Series Notation: 112
Algebraic Labelling: (2,4)
Formula: x^2 + y^2 = 1, |z| = 1, |w| = 1
Surface Volume: 2πrh(h + 2r)
Bulk: πr^2h^2

taken from http://tetraspace.alkaline.org/shapes/rotachora.htm on this site (I put in carrots where the superscript was removed to denote exponents)

The surface volume confused me because i wasnt sure how it was reached
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Postby Aale de Winkel » Tue Jun 15, 2004 11:28 am

which shows that there is some curious thing on that page, shapes are obviously defined with r = h = 1.
In the tetracube (tesseract) I can't even see 8 'volumes', on my own site I need 3 cubes to describe a magic tesseract in full (2<sup>3</sup> = 8 (??)).
that the bulk (or tetra-volume) is a<sup>4</sup> is rather obvious.

for the glome x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup> + w<sup>2</sup> = r<sup>2</sup>, the bulk is obviously defined as a 4 level integral over the fomula, which I believe amounts to an integral over the spheres volume through the remaining w-coordinate and thus:
intrgrl (3/4 πr<sup>3</sup>) dw; = ??? (must check, change in coordinates, and range of variables) !!!

according to http://mathworld.wolfram.com/Hypersphere.html this amounts to:
V<sub>n</sub> = S<sub>n</sub> r<sup>n</sup> / n
with S<sub>n</sub> = 2 π<sup>n/2</sup> / Γ(n/2)
with S<sub>n</sub> the n-sphere surface area and V<sub>n</sub> the n-sphere n-volume;
(for n = 4 this obviously reduces to V<sub>4</sub> = π<sup>2</sup> r<sup>4</sup> / 2)

but why the would be 3π/2 spheres to make up the surface volume illudes me more then the 2<sup>3</sup> = 8 factor in the tesseract.

(or to say it short I simply don't know)

Edit by iNVERTED - fixed <sub> and <sup>
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Postby wendy » Wed Jan 19, 2005 5:55 am

In the Polygloss ( http://www.geocities.com/os2fan2/index.html ) the style of measure of bulk is to use a name like "hedrage" or "chorage" for the extent of 2D and 3D space, and reserve surface, area, volume for those relative to the dimension of discussion.

So by this, one might talk of the surface area of a {3,3,4}, measured in cubic units, but if one particualrly wants to designate the extent of the 2D elements, one might use 'bulk of the surhedra', and measure the results in 'hedrage' [hedrix (2d-cloth) +age (measure in)]
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Postby wendy » Fri Apr 29, 2005 12:54 pm

Surface contains volume. If you want to talk of the extent of a four-dimensional polychoron, one might use "surface chorrage" or content of surface for the general term.

In the main, a good number of people think that volume and area should be reserved for 3d and 2d. I use these (sparingly) to mean solid and bounding contents (ie in 4d, solid = 4d, bounding = 3d).

Much of the thrust of the PG is because there simply are not useful words that accurately convey a 5-manifold in 7d, for example (petix) since surface is not suffice: it is no more bounding than a peice of string in three dimensions.

And so it goes.

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Postby RMac2k5 » Tue Aug 09, 2005 4:22 am

This past year I finished taking calculus one, and kind of struggled along in the class. Especialy in the geometrical studies that are associated with a theoretical math curriculum. Well that is until one day things clicked while learning volume by slicing that all objects exist in all dimensions. For instance when learning volume by slicing you're taught that stacking squares next to each other face to face creates a 3 dimensional object. Such as in the situation: When bound by X=2, Y=2, and the X and Y axis, find the volume by slicing using squares. This quite obviousy creates a cube with the dimensions 2X2X2, or a volume of 8units^3. You're taught to solve this by integrals. Where the "variable" Dx comes into play. Dx is just a fancy way of saying one over infinity.

Now this is obviously on the most basic of what I learned in the volume by slicing. But what it shows is that a square has depth. If it had no depth how could it create a volume when stacked face to face. So what this tells me is that a 2 dimensional object exists in threed dimensions, It obviously has a height and width of two. In addition though it has a depth of Dx (one over infinity). So with this being brought into the open. This has been proven to work going from 1 dimension to 2 dimensions as well. Prior to learning volume by slicing we learned area between curves which brought a similar situation into practice using integrals to stack lines next to eachother to create a 2 dimenisional object. Being bound by an X or Y equals function.

With all that in the open, isn't it possible to think that the amount of tetraspace that is filled by a 4 dimensional object could simply be finding the integral of an object bound by the X, Y, and Z equals functions, and use cubes to find the amount of tetraspace taken up? With my lack of experience in calculus I'm not able to calculate out the equations. I can set the formulas up which I will try my hardest to do demonstrate in forum. Oh and all this only comes into play if the 4th dimension is actualy spacial.

So the equations I'm going to use to bound this experiment are Z=X^2Y^3 and Z=2 and the orgin.With cubes with a side paralell to the Z-Axis.

Ok the facts and the variables I use are as follows :
1.) All cubes volume equals s^3. (length of one side cubed.)
2.)s=Z
3.)Z=X^2 Y^3 and bound by the Z axis and the origin.
4.)The volume of one cube equals X^6 Y^9. (This I solved for by figuring that V=Z^3, and since Z=X^2 Y^3 then I simply did the algebra from there.)
5.) According to the basic principles of Volume by slicing we are stacking cubes in such a manner, face to face if you could possiblly say that, building it into a 4th dimensional object.
6.) Lastly I come to the point where I can't progress any further. Theoreticaly I'm supposed to place X^6 Y^9 inside of an integral, which I'm not entirely sure how to form at or solve. So I've come as far as I can go. If anyone out there is a little more versed in Calculus, Maybe you can show us the way this problem is supposed to end.

Anyway this is just my thought on the matter.
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Postby wendy » Tue Aug 09, 2005 6:20 am

Integrals are hard for us who can't do them.

The usual way of finding volume is to thake the dot moment of surface (expressed as a vector), ie

volume = integral position .dot. grad(density) d(surface).

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Postby jinydu » Tue Aug 09, 2005 7:23 am

Hi RMac2k5

Your idea of stacking cubes to build a tesseract should work in principle (sorry, I'm feeling a bit lazy at the moment). Basically, yes, you give a cube an infinitessimal thickness, then integrate.
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