volumes, areas e.t.c.

Higher-dimensional geometry (previously "Polyshapes").

volumes, areas e.t.c.

Postby PWrong » Tue Nov 08, 2005 5:00 pm

We've come pretty far with classifying shapes in 4D, but sooner or later we're going to want to do something with them. So I think it's about time we found out some important information about rotatopes.

volumes and surface areas are the simplest example, but we should also work out things like scale factors, the volume of only parts of a shape, e.t.c. Some members might feel a bit left out if they don't know trig and calculus, but it can't hurt to be introduced to this stuff anyway.

I'll start with scale factors, because they're useful for working out the other stuff. They're not too hard to understand, and you can find them out just by visualising the object. Every nD coordinate system has n scale factors. A scale factor for a coordinate is the distance you travel when you change it.

For instance, imagine you're standing at a point on a circle. If you walk outwards, away from the origin, then the distance you travel is just the normal distance, because you're walking in a straight line. That is, if you change the radius r by dr, then the distance traveled is just 1*dr. So the scale factor is simply h<sub>r</sub> = 1

However, if you walk around the circle, you're no longer travelling in a straight line. If you change the angle a by da, the distance you travel is r*da. So the scale factor is h<sub>a</sub> = r

On the sphere, it's a bit harder. We'll use a for latitude and b for longitude. Like the circle, we have h<sub>r</sub> = 1. When we change the latitude (i.e. walk towards the north or south pole), we're walking around a great circle, so h<sub>a</sub> = r.
Now, longitude is different. The tropic of cancer is shorter than the equator, so the scale factor depends on both the radius and the latitude. If you change the longitude b, the distance is r*cos(a)*db. So our scale factor is h<sub>b</sub> = r cos a

So here's all the 2D and 3D scale factors:

2D
Cartesian (Square):
h<sub>x</sub> = h<sub>y</sub> = 1

Polar (circle):
h<sub>r</sub> = 1
h<sub>a</sub> = r

3D
Cartesian (Cube):
h<sub>x</sub> = h<sub>y</sub> = h<sub>z</sub> =1

Cylindrical:
h<sub>r</sub> = 1, h<sub>a</sub> = r, h<sub>z</sub> = 1

Spherical:
h<sub>r</sub> = 1, h<sub>a</sub> = r, h<sub>b</sub> = r cos a
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Postby wendy » Wed Nov 09, 2005 7:03 am

The volumes and surface area of the rototopes are quite easy, where the thing is defined in terms of the products, eg

[(w,x),(y,z)] = [(diam,diam),(diam,diam)]

for a one inch diameter figure, we have

(w,x) = circular inch = pi/4 sq inch
(y,z) = circular inch = pi/4 sq inch

[(w,x),(y,z)] = pi2/16 biquadrate inches

If one goes the other way:

([w,x],[y,z])

[w,x] = square inch = 4/pi circular inch
([w,x],[y,z]) 4/pi * 4/pi inch(^4) [= glomular inch]

Since 1 inch(^4) = pi2/32 inch[^4] biquadrate inches

we have then the volume = 1/2 biquadrate inch

We then find surface area by noting that the volume is a pyramid product of the radius and surface area. For this we use the tegmic scale.

radus = 1/2 inch<^1> volume = 12 inch <^4>

surface area = 12 / (1/2) = 24 inch<^3> = 4 inch[^3] = 4 cu inches.

Simple really.

The longdome etc needs some rather heavy mathematics, which i don't have.

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Postby PWrong » Thu Nov 10, 2005 7:25 am

Surely we want formulas for volumes and areas, not the volumes and areas of shapes with a specific radius. :? Why are you using inches? :?

Simple really.
The longdome etc needs some rather heavy mathematics, which i don't have.

That's the kind of thing I'm going to try to work out. The reason I brought up scale factors is that they make calculating volumes easy, as well as finding a center of mass, moments of inertia, and rearranging the Laplacian.

For instance, the volume of a sphere is simply the integral of h<sub>r</sub> h<sub>a</sub> h<sub>b</sub> dr da db.
You have to integrate over the right range though. Incidentally, I think one of my scale factors is wrong. Latitude is the angle measure from the north pole, so I should have sin rather than cos. In fact, my whole coordinate system is probably wrong.

To tell the truth, I'm not sure I understand the dome shapes yet. Is the dome a bit like the intersection between two cylinders?

I realised yesterday that there are several different coordinate systems that describe the glome. I was thinking about the duocylindrical coordinate system, and how it's basically two separate polar coordinates.
x = r_1 cos a
y = r_1 sin a
z = r_2 cos a
w = r_2 sin a

Then I realised you could treat r_1 and r_2 as a third set of 2D coordinates, and convert them to polar. So we have
x = R cos c cos a
y = R cos c sin a
z = R sin c cos a
w = R sin c sin a

I think this looks a bit nicer than the current glomar coordinates (Some of these cosines should be sines, e.t.c):
x = R cos c cos b cos a
y = R cos c cos b sin a
z = R cos c sin b
w = R sin c
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Postby wendy » Thu Nov 10, 2005 8:39 am

The thing is that [(w,x),(y,z)] is a specific formula for content of a general bi-elliptical prism, and ([w,x],[y,z]) is a formula for a bi-square-spheroid.

I use feet and inches in atomic constants as well. it's no big deal.

In any case, an ellipse is a stretched circle, and gives a circle with suitable coordinate changes.

I really don't have any heavy mathematics, except that one can make inch^4 reoresent the fourth prismic power, the fourth tegmic power, the fourth spheric power, etc of an inch. The nature of multiplication is that there are prehaps 6,220,800,000 different forms, each to itself coherent.

As for the calculus, and all that sort of stuff, i don't know. Even moments of inertia will change depending on the mode of operation: there are also things like clifford-modes of rotation, and hence clifford modes of rotation. Ye might as well have moment of content (which i have used too), but the content of these things are not zero, so the moment is as useless as a choclate teapot.

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Postby PWrong » Fri Nov 18, 2005 12:51 pm

The thing is that [(w,x),(y,z)] is a specific formula for content of a general bi-elliptical prism, and ([w,x],[y,z]) is a formula for a bi-square-spheroid.


I wouldn't really call it a formula. It's a useful classification, but it's not really general enough. You couldn't use it to describe, say, a quartic surface or a 4D harmonic function.

I use feet and inches in atomic constants as well. it's no big deal.

I don't have a problem with what units you use. My point is we don't want the volume of "a one inch diameter figure", we want the volume of a figure with radius R.

Even moments of inertia will change depending on the mode of operation

Still, it's interesting to work these things out. I showed in another thread (that noone replied to), that a duocylinder rolls faster than a glome.

The longdome etc needs some rather heavy mathematics, which i don't have.

How much maths do you have? Basic calc and trig? If we have a strict enough definition of the longdome, I'm sure it's possible to work out the content.
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Postby Batman3 » Fri Nov 18, 2005 2:15 pm

Is it possible to find the relative dimensions(i.e. units such as feet, pounds, kilograms, moles, degrees kelvin etc.) for a 4d planet by using the 4d tensile strength of 4d-carbon bone,etc. and then assuming the existence of a 4d-man or 4d-animal or 4d-tree? And then find the ratio between the mass of a 4d-man and the mass of a 4d planet? And the ratio between his height and the diameter of the planet? The w/out loss of generality, we could assign units such as 4d-feet or 4d-kilograms.

The mass of a 3d-man is 100 kg and the mass of Earth is 6*10^24 kg.

btw, if man ever gets to Mars, he will have to translate the first verse of the Bible as:"In the beginning God created the heavens and Mars" and the Lord's prayer might read,"Thy will be done on MARS as it is done in heaven". Or an other denomination might translate it as, "Thy will be done on Earth as it is done on Mars.", thus providing a basis for interplanetary relations.
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Postby wendy » Sat Nov 19, 2005 7:19 am

First you have to make assumptions about the tensile strength of 4d carbon, etc.
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Postby wendy » Sat Nov 19, 2005 7:31 am

[(w,x),(y,z)] is a formula for a specific class of polytopes, such as formed by the nested spheric and prism products. You can add even tegum products to it. Not everything is one of these products.

You can invent formulae for radii dependence. It depends on how many radii you put into a diameter. However, the tradition for circular inch, spheric inch, glomic inch, etc, is that one uses the diameter of the unit, and we have circ inch = inch^2, spheric inch = in^3, glomic inch = in^4. when the spheric product is used to define volume. Since an inch refers to a diameter of a line (rather than the radius), this propagates upwards.

I don't know how much maths is needed to handle a longdome, or a spherotegmate figure. Sometimes they are dirt-easy, sometimes, they are hard. But in these, the volume can be written as the tegmate product of surface and radius, and so finding one finds the other.
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Postby Batman3 » Sat Nov 19, 2005 5:28 pm

I was thinking we might get the tensile strength od 4d-carbon from the physics and chemistry of 4d-carbon. You would use QM but I might require a periodic table. I don't know. Any ideas?
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Postby wendy » Sun Nov 20, 2005 7:12 am

Ye would prolly be better trying to glark the required stuff from dimensional modelling then fudging QM. In any case, QM isn't going to pull a number from nothing.

Most of the time, we assume that at the level we live at, the numbers are pretty much what dimensional modeling sets. Anything else is then calculated to fit.

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Postby Batman3 » Sun Nov 20, 2005 1:38 pm

What do ye mean, "dimensional modelling"?

I am thinking that if we set tensile strength at ts(units of force), and if a 4d-man stands 2 meters and if he weighs say, 100 kg, and m*g=m*G*mp/r^3 then ts=100kg*G*mass of planet/ radius^3 . Then the mass of the planet is a simple function of the radius and G.

The tensile strength force of 4d-carbon is presumably related to the tensile strength force of 3d-carbon. I.e. by replacing '3' with '4' like when we replaced 1/r^2 with 1/r^3. If the radius(4d-carbon)=radius(3d-carbon)=1*10^-10 meters...??

weight(4d-man)=g*m(4d-man). If g==9.8 m/s^2 then 9.8 m/s^2=G Mp /r^3. Can we pin G down by considering the requirements of a earth-supporting 4d-sun?
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Postby PWrong » Sun Nov 20, 2005 7:17 pm

What's the point of putting numbers in? If every object in the universe double in size, along with all the constants, who would notice? The only thing that counts is the way they relate to each other. That's why we use algebra.

I was thinking we might get the tensile strength od 4d-carbon from the physics and chemistry of 4d-carbon. You would use QM but I might require a periodic table. I don't know. Any ideas?


What kind of carbon? Graphite, soot, diamond, buckyballs or nanotubes? I think there's more than that, and there's probably hundreds of arrangements in 4D. :shock:

I don't know how much maths is needed to handle a longdome, or a spherotegmate figure. Sometimes they are dirt-easy, sometimes, they are hard. But in these, the volume can be written as the tegmate product of surface and radius, and so finding one finds the other.

Ok, we'll start with something simpler. Is the original dome the same as the intersection of two cylinders? If it is, I can find the volume and the SA.
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Postby Marek14 » Sun Nov 20, 2005 10:33 pm

Dome is not an intersection of two cylinders, but its volume can be found pretty easily:

When I inscribe the dome in a cube, the ratio of dome:cube volume is the same as ratio of sphere:cylinder volume for sphere inscribed in cylinder.

If we put all relevant edge lengths / diameters as 2, we get

cube: V = 8

cylinder: V = 2*pi

Sphere: V = 4/3*pi

Sphere/cylinder: 2/3
Dome/cube: 2/3

Dome: V = 16/3

Surface are is then equal to 16 (volume times dimension). I'm not sure what are the exact condition for this simple surface formula to hold, but according to Wendy it should work for all rotatopes as long as they have all cross-dimensions 2.
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Postby PWrong » Mon Nov 21, 2005 7:59 am

Dome is not an intersection of two cylinders, but its volume can be found pretty easily:

What is it then? :? From your description in the other thread, it sounds exactly like the Steinmetz Solid. It has the same volume and surface area, and satisfies all of these properties from your description:
First, we review its properties: It has to cast two circular shadows (when we remove the end-nodes) and one square shadow (removing the middle-node). Passing it through the plane will result in two possibilities: either a square that will start from a point, grow, then shrink...
...or we start with a line. This will grow into circle (originally I thought it was through ellipse, but in fact the intermediate shapes are circles with their top and bottom cut). Then it shrinks back into line.


If we put all relevant edge lengths / diameters as 2, we get

I still don't understand why we can't be more general. It's only baby algebra.

cube: V = x*y*z, A = 2xy + 2xz +2yz

cylinder: V = pi r^2 z, A = 2 pi r z + 2 pi r^2

Sphere: V = 4/3*pi r^3, A = 4 pi r^2

Sphere/cylinder: 4/3*r/z

Surface are is then equal to 16 (volume times dimension). I'm not sure what are the exact condition for this simple surface formula to hold, but according to Wendy it should work for all rotatopes as long as they have all cross-dimensions 2.


The surface area of an n-sphere is the derivative of the volume with respect to radius. So for the 3D sphere, SA = d/dr (4/3 pi r^3) = 4 pi r^2.

It turns out the ratio of volumes between two nD shapes is the Jacobian. Even though this is taught in more advanced calculus (above my level anyway), it's not hard to understand if you have basic calculus. I'll find the jacobians of some rotatopes and toratopes, and I'll see if I can find a pattern.
Last edited by PWrong on Mon Nov 21, 2005 9:18 am, edited 1 time in total.
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Postby Marek14 » Mon Nov 21, 2005 8:52 am

Eh, my bad. It seems that it's indeed the Steinmetz solid, and so it is the intersection of two cylinders. It's just that I never thought of it as such.
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Postby PWrong » Mon Nov 21, 2005 4:00 pm

I wonder if this means all the graphotopes are the intersections of rotatopes?

I finally found the simple way to find scale factors. Mathworld skips around it for some reason. You just take the parametric equations as a vector (x, y, z), and differentiate with respect to whatever you want the scale factor for. Then the scale factor is just the magnitude.
h<sub>u</sub> = |d/du (x,y,z)|

So for the sphere:
x = r sin b cos a
y = r sin b sin a
z = r cos b

the position vector is (x,y,z) = (r sin b cos a, r sin b sin a, r cos b), so we have:
h<sub>r</sub> = |d/dr (x, y, z)| = |(sin b cos a, sin b sin a, cos b)|
h<sub>r</sub> = 1
h<sub>a</sub> = |(-r sin b sin a, r sin b cos a, 0)| = r sin b
h<sub>b</sub> = |(r cos b cos a, r cos b sin a, -r sin b)| = r

I've almost finished a table of the scale factors and jacobians for all the 4D shapes.
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Postby Batman3 » Thu Nov 24, 2005 8:37 pm

Pwrong asked,"what kind of Carbon?". Well I don't know what bone is made of but I imagine it is carbon. There are 3 carbon elemnts in 4d and these would(?) be used for bone. Maybe we could get tensile strength of bone from that sort of carbon's properties.

Anyway, I am not sure that is rellevant to my question becuse...
The density of 4d-man is (100 kg)/(2 m)/(.2 m)/.1m)/(.1 m)=.25*10^5 kg/m^4=25 g/cm^4. This gives a measure of the density of Carbon. Perhaps someone knows how to look that number up in 3d.

Thus, 9.8 m/sec^2=G mp /r^3 = G(10^5 kg/m^4 *r^4)/r^3=G(10^5 kg/m^4)r, thus giving an inverse realtion between r & G :rp = 4*10^-4/G/(kg/m^4), based merely on an earthlike man and some constant G.
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Postby PWrong » Sat Dec 10, 2005 2:36 pm

The density of 4d-man is (100 kg)/(2 m)/(.2 m)/.1m)/(.1 m)=.25*10^5 kg/m^4=25 g/cm^4. This gives a measure of the density of Carbon.

It's probably closer to the density of water. Any animal is mostly water.

Thus, 9.8 m/sec^2=G mp /r^3 = G(10^5 kg/m^4 *r^4)/r^3=G(10^5 kg/m^4)r, thus giving an inverse realtion between r & G :rp = 4*10^-4/G/(kg/m^4), based merely on an earthlike man and some constant G.

Again, why are you putting numbers in everywhere? It's an interesting problem though. Maybe we should split the thread.
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Postby Nick » Sun May 07, 2006 2:26 pm

"Perimeter" of line = l
Perimeter of square = 2l + 2w
Surface area of cube = (2l + 2w) * 6
"Volume" of Hypercube (following the pattern) = ({2l+2w}*6)*8

Perimeter of circle = circumference
Surface area of sphere = 4 * {pi * (r ^ 2)}
"Volume" of Glome = ???????

Volume of cylinder = h * {pi * (r ^ 2)}
"Volume" of Cubinder = trength * {h * (pi * {r ^ 2})}

That's some... could someone fill in the Glome, please?

This would be excellent to put on the wiki, if we could figure it out :wink:.
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Postby moonlord » Sun May 07, 2006 3:53 pm

They appear in alkaline's introduction. So that's no problem.

You forgot the bulk: for a cube, it's a^4 (easy to find out why), for a 13 it's 4/3 pi r^3 h, for a 112 it's pi r^2 h^2 I assume, but it's more difficult for the 22 and the glome. Basically to find the bulk of a glome you must calculate Integral[-r, r; -r, r; -r, r]{sqrt( r^2 - x^2 - y^2 - z^2) dx dy dz}. You also have to add some more conditions. Unfortunatelly I don't know double or triple integrals.

There is another approach. You can transform a circle until it is a triangle of base 2 pi r and height r. It's area is 2 pi r * r /2 = pi r^2, which is the same as the circle's. You can transform a sphere until it is a cone of base area 4 pi r^2 (area of sphere) and height r. It's volume is 4 pi r^2 * r /3 = 4/3 pi r^3, which is the same as the sphere's.

Now, you have a glome and transform it into a hypercone of base volume equal with the volume of a glome and height r. It's bulk (and the glome's) will be V r /4. The problem is, I'm not yet sure what's the volume of a glome.

EDIT: Just saw a simpler way. You have to integrate the volume of a sphere over [-r, r]. Right now I'm checking how does the volume depend on the distance from the glome's center. As soon as I get the equation, I'll post it.
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Postby jinydu » Sun May 07, 2006 4:22 pm

All of these questions about higher-dimensional versions of the sphere have already been investigated and thoroughly answered:

http://mathworld.wolfram.com/Hypersphere.html
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Postby Nick » Sun May 07, 2006 4:56 pm

moonlord wrote:They appear in alkaline's introduction. So that's no problem.


Really? I don't see them...

Anyway, here's the "volume", or "surface area", or whatever you want to call it, of the first three polychora.

Pentatope: {(1/2 base * height) * 4} * 4, which simplifies to:
{2 * (base * height)} * 4, which simplifies to:
8 * (base * height)

Hypercube: {(2l + 2w) * 6} * 8, since w and l are the same:
(4l * 6) * 8, which simplifies to:
4l * 48

16-cell: {(1/2 base * height) * 4} * 16, which simplifies to:
{1/2 base * height) * 64, which simplifies to:
32 * (base * height)

I don't really understand the rest... their too confusing :?. [/quote]
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Postby moonlord » Mon May 08, 2006 2:22 pm

"God does not play dice." -- Albert Einstein, early 1900's.
"Not only does God play dice, but... he sometimes throws them where we cannot see them." -- Stephen Hawking, late 1900's.
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