New here, have multi dimensional formula for bashing :)

[meckano]

Based on Pi*r^2 and 4/3*Pi*r^3.
I found they are dimensionally related:

1D: (2^1*Pi*r)/(2*1)
2D: (2^2*Pi*r^2)/(2*2)
3D: (2^3*Pi*r^3)/(2*3)

4D: (2^4*Pi*r^4)/(2*4)
- is what I'm looking at now.

I prefer my short verision for circles of:
dia^dimension*Pi/(2*dimension)

I came to those indirectly via:
2D, Area:
A circle with a diameter of 4 has an area of... π * r^2 = 12.5664
I like 1, so
I found how many circles of diameter 1 were needed to equal the area
of that 4d circle.
The answer is 16: 12.5664 / 0.7854
a) 16 is also the square of the diameter.
b) 0.7854 = area of circle with a diameter of 1.
c) 0.7854 happens to be π / 4 or

3D, Volume:
I did the same check with 3D: (4/3) * π * r^3 and
I found that it takes d^3 circles with a diameter of 1.
so again
4^3 = 64,
Volume = 64 * 0.5256(Volume for dia. of 1) = 33.5103

1D, Vector:
If you travel straight for 4 feet, how many circular feet have you traveled?
4^1 * (π / 2 * 1) = 4 * (π / 2) = 2π
What does this mean?
well we know that the circumference is π * d,
and we don't need to go in a complete circle,
we only need half the circle, hence C = 4π, 4π / 2 = 2π.
- it essentially scribes the shell of the sphere in any direction.

Here is where I would like some comments:
If the 4D is calculated on a circle with a d of 4, and a circle with a d of 1,
it would take 256 of the circle with a d of 1 to fill??? the circle with a d of 4.
What is going on that it takes 256?

[wendy]

The actual formulae can be derived by adding 0d and 1d to the equation.


0d S = 1
1d S = 2
2d S = 2pi C = pi
3d S = 4pi C = 4pi/3
4d S = 2pi2 C = pi2/2
5d S = 8pi/3 C = 8pi2/15
6d S = pi3 C = pi3/6

The general formula is

n D S = 2pi * C(n-2), C = S / (n)

Wendy

[meckano]

Do you have a link about that formula?

in mine i get div.by zero for 0D
thanks

[wendy]

no link here. just a known result.

W

[jinydu]

no link here. just a known result.

W

Allow me:

http://mathworld.wolfram.com/Hypersphere.html

However, wendy, I'm not sure about your formulas for 0D and 1D...

A 1D circle should be the set of all points on a line that are some fixed distance from a fixed point on that line (the center). If I'm not mistaken, there are only two points that meet that condition. So are you saying that all 1D circles have a "surface area" of 2?

As for 0D, I'm afraid you've lost me there. I thought that 0D was just a single point. Since there are no other points, at any distance from that single point, how can a 0D "circle" be formed at all? Plugging in n = 0 into formula on the Mathworld page leads to an error, since (-1)! is undefined...

[meckano]

sweeeeet, ty
and I must say thanks for validating my near-solo effort.
I got somewhere close to what I show here, a few years ago.
now you tell me it's valid and give me a formula that incorporates my mish-mash.
Today is a good day to be a 'mad man'. lol
cheers

A mad man knows he always puts his hat on more than once each time.
It ain't deja-vu, it's just another planar ol day.

Addit:
1D to me is a vector, with circles it is arc'ed. Fitting the diameter in question.
The answer I get is equivalent to 1/2 C (C=Circumference?).
It is the vector length which can be used to scribe the shell of the sphere.

1D: vector (arc, from my circle formula)
2D: area (now you can have a circle)
3D: sphere ( now you can climb in )
4D: tba ( now you can climb in and do a disappearing act?

[wendy]

The standard incidence table for polytopes has a top and bottom closure. The top closure is the content of the figure (ie the bulk), and the bottom closure is a nulloid or as Norman Johnson calls it 'nullitope'.

If one includes one of each in the dimensional equation, one gets the Euler characteristic to balance exactly, without alternating 2 and zero.

The dimension of the nulloid is -1, and therefore a point is "bounded" by this single nulloid.

The lowest observed dimension that has been observed is -2d.